DE SECTORE & RADIO. The description and use of the Sector in three books. The description and use of the cross-staff in other three books. For such as are studious of Mathematical practice. geometric illustration LONDON, Printed by WILLIAM JONES. and are to be sold by JOHN TAP at Saint Magnus' corner. 1623. THE DESCRIPTION AND USE OF THE SECTOR, For such as are studious of Mathematical practice. geometric illustration LONDON, Printed by WILLIAM JONES. 1623. THE FIRST BOOK OF THE SECTOR. CHAP. I. The Description, the making, and the general use of the Sector. A Sector in Geometry, is a figure comprehended of two right lines containing an angle at the centre, and of the circumference assumed by them. This Geometrical instrument having two legs containing all variety of angles, & the distance of the feet, representing the subtenses of the circumference, is therefore called by the same name. It containeth 12 several lines or scales, of which 7 are general, the other 5 more particular. The first is the scale of Lines divided into 100 equal parts, and numbered by 1.2.3.4.5.6.7.8.9.10. The second, the lines of Superficies divided into 100 unequal parts, and numbered by 1.1.2.3.4.5.6.7.8.9.10. The third, the lines of Solids, divided into 1000 unequal parts, & numbered by 1. 1. 1. 2. 3. 4. 5. 6. 7. 8. 9 10. The fourth, the lines of Sines and Chords divided into 90 degrees, and numbered with 10. 20. 30. unto 90. These four lines of Lines, of Superficies, of Solids, and of Sines, are all drawn from the centre of the Sector almost to the end of the legs. They are drawn on both the legs, that every line may have his fellow. All of them are of one length, that they may answer one to the other: And every one hath his parallels, that the eye may the better distinguish the divisions. But of the parallels those only which are inward most contain the true divisions. There are three other general lines, which because they are infinite are placed on the side of the Sector. The first a line of Tangents, numbered with 10.20.30.40.50.60. signifying so many degrees from the beginning of the line, of which 45 are equal to the whole line of Sines, the rest follow as the length of the Sector will bear. The second, a line of Secants, divided by pricks into 60 degrees, whose beginning is the same, with that of the line of Tangents, to which it is joined. The third, is the Meridian line, or line of Rumbs, divided unequally into degrees, of which the first 70 are almost equal to the whole line of Sines, the rest follow unto 84 according to the length of the Sector. Of the particular lines inserted among the general, because there was void space, the first are the lines of Quadrature placed between the lines of Sines, and noted with 10.9.8.7. S. 6.5.90. Q. The second, the lines of Segments placed between the lines of Sines and Superficies, divided into 50 parts, and numbered with 5.6.7.8.9.10. The third, the lines of Inscribed bodies in the same Sphere, placed between the scales of Lines, and noted with D. S. I. C. O. T. The fourth, the lines of Equated bodies, placed between the lines of Lines and Solids, and noted with D. I. C. S. O. T. The fift, are the lines of Metals, inserted with the lines of Equated bodies (there being room sufficient) and noted with these Characters. ☉. ☿. ♄. ☽. ♀. ♂. ♃. There remain the edges of the Sector, and on the one I have set a line of Inches, which are the twelfth parts of a foot English: on the other a lesser line of Tangents, to which the Gnomon is Radius. 2 Of the making of the Sector. LEt a Ruler be first made either of brass or of wood, like unto the former figure, which may open and shut upon his centre. The head of it may be about the twelfth part of the whole length, that it may bear the movable foot, and yet the most part of the divisions may fall without it. Then let a movable Gnomon be set at the end of the movable foot, and there turn upon an Axis, so as it may sometime stand at a right angle with the feet, and sometimes be enclosed within the feet. But this is well known to the workman. For drawing of the lines. Upon the centre of the Sector, and semidiameter somewhat shorter than one of the feet, draw an occult ark of a circle, crossing the closure of the inward edges of the Sector about the letter T. In this ark, at one degree on either side from the edge, draw right lines from the Centre, fitting them with Parallels and divide them into an hundred equal parts, with subdivisions into 2.5. or 10. as the line will bear, but let the numbers set to them, be only 1.2.3.4. etc. unto 10. as in the example. These lines so divided, I call the lines or scales of Lines; and they are the ground of all the rest. In this Ark at 5 degrees on either side, from the edge near T, draw other right lines from the Centre, and fit them with Parallels. These shall serve for the lines of Solids. Then on the other side of the Sector in like manner, upon the Centre & equal Semidiameter, draw another like Ark of a circle: & here again at one near degree on either side from the edge near the letter Q draw right lines from the Centre, and fit them with parallels. These shall serve for the lines of Sines. At 5 Degrees on either side from the edge near Q draw other right lines from the centre, and fit them with parallels: these shall serve for the lines of Superficies. These four principal lines being drawn, and fitted with parallels, we may draw other lines in the middle between the edges and the lines of Lines, which shall serve for the lines of inscribed bodies, and others between the edges and the Sins for the lines of quadrature. And so the rest as in the example. 3 To divide the lines of Superficies. SEeing like Superficies do hold in the proportion of their homologal sides duplicated, by the 29 Pro. 6 lib. Euclid. If you shall find mean proportionals between the whole side, and each hundred part of the like side, by the 13 Pro. 6 lib. Euclid. all of them cutting the same line, that line so cut shall contain the divisions required. wherefore upon the centre A and Semidiamiter equal to the line of Lines, describe a Semicircle ACBD, with AB perpendicular to the diameter CD. And let the Semidiameter AD he divided as the line of Lines into an hundred parts, & A the one half of AC divided also into an hundred parts, so shall the divisions in A be the centres from whence you shall describe the Semicircles C 10. C 20. C 30. &c. dividing the lin AB into an hundred unequal parts: and this line AB so divided shall be the line of Superficies, and must be transferred into the Sector. But let the numbers set to them be only 1.1.2.3. unto 10, as in the example, geometric illustration 4 To divide the lines of Solids. seeing like Solids do hold in the proportion of their homologal sides triplicated, if you shall find two mean proportionals between the whole side & each thousand part of the like side: all of them cutting the same two right lines, the former of those lines so cut, shall contain the divisions required. Wherefore upon the centre A & Semidiameter equal to the line of Lines. describe a circle and divide it into 4 equal parts CEBD, drawing the cross diameters CB, ED. Then divide the semidiameter AC, first into 10 equal parts, and between the whole line AD & OF the tenth part of AC, seek out two mean proportional lines AI and AH. again between AD and AGNOSTUS being two tenth parts of AC, seek out two mean proportionals ALL and AK, and so forward in the rest. So shall the line AB be divided into 10 unequal parts. geometric illustration Secondly, divide each tenth part of the line AC into 10 more, and between the whole line AD, and each of them, seek out two mean proportionals as before: So shall the line AB be divided now into an hundred unequal parts. Thirdly, If the length will bear it, subdivide the line AC once again, each part into ten more: and between the whole line AD and each subdivision, seek two mean proportionals as before. So should the line AB be now divided into 1000 parts. But the ruler being short, it shall suffice, if those 10 which are nearest the centre be expressed, the rest be understood to be so divided, though actually they be divided into no more than 5 or 2. and this line AB so divided shall be the line of Solids, and must be transferred into the Sector: But let the numbers set to them be only 1.1.1.2.3. etc. unto 10. as in the example. 5 To divide the lines of Sines and Tangents on the side of the Sector. Upon the centre A, and semidiameter equal to the line of Lines, describe a semicircle ABCD, with AB, perpendicular to the diameter CD. Then divide the quadrants CB, BD, each of them into 90. and subdivide each degree into 2 parts: For so, if straight lines be drawn parallel to the diameter CD, through these 90, and their subdivisions they shall divide the perpendicular AB unequally into 90. geometric illustration And this line AB so divided shall be the line of Sines; and must be transferred into the Sector. The numbers set to them are to be 10.20.30. etc. unto 90 as in the example. If now in the point D, unto the diameter CD, we shall raise a perpendicular DE, and to it draw straight lines from the centre A, through each degree of the quadrant DB. This perpendicular so divided by them shall be the line of Tangents, & must be transferred unto the side of the Sector. The numbers set to them, are to be 10.20.30. etc. as in the example. If between A and D, another straight line GF, be drawn parallel to DE, it will be divided by those lines from the centre in like sort as DE is divided, and it may serve for a lesser line of Tangents, to be set on the edge of the Sector. These lines of Sines and Tangents, may yet otherwise be transferred into the Sector out of the line of Lines, (or rather out of a diagonal Scale equal to the line of Lines) by tables of Sines and Tangents. In like manner may the lines of Superficies, be transferred by tables of square roots; and the line of Solids, by tables of cubique roots: which I leave to others to extract at leisure. 6 To show the ground of the Sector. LEt AB, AC, represent the legs of the Sector: then severing these two AB, AC, are equal, and their sections AD, A, also equal, they shall be cut proportionally: and if we draw the lines BC, DE, they will be parallel by the second Pro. 6 lib. of Euclid, and so the Triangles ABC, ADE, shallbe equiangle; by reason of the common angle at A, and the equal angles at the base, and therefore shall have the sides proportional about those equal angles, by the 4 Pro. 6 lib. of Euclid. geometric illustration The side AD, shallbe to the side AB, as the basis DE, unto the parallel basis BC, and by conversion AB, shall be unto AD, as BC, unto DE: and by permutation AD, shall be unto DE, as AB, to BC. etc. So that if AD, be the fourth part of the side AB, then DE, shall also be the fourth part of his parallel basis BC, The like reason holdeth in all other sections. 7 To show the general use of the Sector. THere may some conclusions be wrought by the Sector, even then when it is shut, by reason that the lines are all of one length: but generally the use hereof consists in the solution of the Golden rule, where three lines being given of a known denominaton, a fourth proportional is to be found. And this solution is divers in regard both of the lines, and of the entrance into the work. The solution in regard of the lines is sometimes simple, as when the work is begun and ended upon the same lines. Sometimes it is compound, as when it is begun on one kind of lines, and ended on another. It may be begun upon the lines of Lines; & finished upon the lines of Superficies. It may begin on the Sins, and end on the Tangents. The solution in regard of the entrance into the work, may be either with a parallel or else lateral on the side of the Sector, I call it parallel entrance, or entering with a parallel, when the two lines of the first denomination are applied in the parallels, and the third line, and that which is sought for, are on the side of the Sector. I call it lateral entrance, or entering on the side of the Sector, when the two lines of the first denomination are one the side of the Sector, and the third line and that which is to be found out, do stand in the parallels. geometric illustration As for example, let there be given three lines A, B, C, to which I am to find a fourth proportional. let A, measured in the line of lines, be 40, B 50, and C 60, and suppose the question be this. If 40 Months give 50 pounds, what shall 60? Here are lines of two denominations, one of months, another of pounds, and the first with which I am to enter must be that of 40 months. If then I would enter with a parallel, first I take A, the line of 40, and put it over as a parallel in 50, reckoned in the line of lines, on either side of the Sector from the centre, so as it may be the Base of an Isoscheles triangle BAC, whose sides AB, AC are equal to B, the line of the second denomination. Then the Sector being thus opened, I take C the line of 60, between the feet of the compasses, and carrying them parallel to BC, I find them to cross the lines AB, AC, on the side of the Sector in D and E, numbered with 75, wherefore I conclude the line AD, or A, is the fourth proportional and the correspondent number 75 which was required. But if I would enter on the side of the Sector, then would I dispose the lines of the first denomination A and C, in the line of Lines, on both sides of the Sector, in AB, AC, & in AD, A, so as they should all meet in the centre A, and then taking B the line of the second denomination put it over as a parallel in BC, that it may be the Basis of the Isocheles triangle BAC, whose sides AB, AC, are equal to A, the first line of the first denomination, for so the Sector being thus opened, the other parallel from D to E, shall be the fourth proportional which was required, and if it be measured with the other lines, it shall be 75, as before. In both this manner of operations, the two first lines do serve to open the Sector to his due angle, the difference between them is especially this, that in parallel entrance, the two lines of the first denomination, are placed in the parallels B, C, D, E, & in latterall entrance they are placed on both sides of the Sector, in AB, AD and in AC, A geometric illustration Now in simple solution which is begun and ended, upon the same kind of lines, it is all one which of the two latter lines be put in the second or third places. As in our example we may say, as 40 are to 50, so 60 unto 75, or else as 40 are to 60, so 50 unto 75. And hence it comes that we may enter both with a parrallell, & on the sides two manner of ways at either entrance, and so the most part of questions may be wrought 4 several ways, though in the propositions following, I mention only that which is most convenient. Thus much for the general use of the Sector, which being considered and well understood, there is nothing hard in that which followeth. CHAP. II. The use of the Scale of Lines. 1. To set down a Line, resembling any given parts or fraction of parts. numbered line segment THe lines of Lines are divided actually into 100 parts, but we have put only 10 numbers to them. These we would have to signify either themselves alone, or ten times themselves, or an hundred times themselves, or a thousand times themselves, as the matter shall require. As if the numbers given be no more than 10, than we may think the lines only divided into 10 parts according to the numbers set to them. If they be more than 10, and not more than 100, than either line shall contain 100 parts, and the numbers set by them shall be in value 10.20.30. etc. as they are divided actually. If yet they be more than 100, than every part must be thought to be divided into 10, and either line shall be 1000 parts, and the numbers set to them shall be in value 100.200.300, and so forward still increasing themselves by 10. This being presupposed, we may number the parts and fraction of parts given in the line of lives; and taking out the distance with a pair of compasses, set it by, for the line so taken shall resemble the number given. In this manner may we set down a line resembling 75, if either we take 75 out of the hundred parts, into which one of the line of lines is actually divided, and note it in A, or 7 ½. of the first 10 parts, and note it in B, or only ¾. of one of those hundred parts, and note it in C. Or if this be either to great or to small, we may run a Scale at pleasure, by opening the compass to some small distance, and running it ten times over, then opening the compass to these ten, run them over nine times more, & set figures to them as in this example, and out of this we may take what parts we will as before. To this end I have divided the line of inches on the edge of the Sector, so as one inch containeth 8 parts, another 9, another 10, etc. according as they are figured, and as they are distant from the other end of the Sector, that so we might have the better estimate. 2 To increase a line in a given proportion. 3 To diminish a line in a given proportion. TAke the line given with a pair of compasses, and open the Sector, so as the feet of the compasses may stand in the points of the number given, then keeping the Sector at this angle, the parallel distance of the points of the number required, shall give the line required. line segment Let A, be a line given to be increased in the proportion of 3 to 5. First I take the line A, with the compasses, and open the Sector till I may put it over in the points of 3 and 3, so the parallel between the points of 5 & 5, doth give me the line B, which was required. In like manner, if B, be a line given to be diminished in the proportion of 5 to 3, I take the line B & to it open the Sector in the points of 5, so the parallel between the points of 3, doth give me the line A, which was required. If this manner of work doth not suffice, we may multiply or divide the numbers given by 1, or 2, or 3, etc. And so work by their numbers equimultiplices, as for 3 and 5, we may open the Sector in 6 and 10, or else in 9 and 15, or else in 12 and 20, or in 15 and 25, or in 18. and 30. etc. 4 To divide a line into parts given. TAke the line given, and open the Sector according to the length of the said line in the points of the parts, whereunto the line should be divided, then keeping the Sector at this angle, the parallel distance between the points of 1 and 1 shall divide the line given into the parts required. line segment Let AB, be the line given to be divided into five parts, first I take this line AB, and to it open the Sector in the points of 5 and 5, so the parallel between the points of 1 and 1, doth give me the line AC, which doth divide it into the parts required. line segment Or let the like line AB, be to be divided into twenty three parts. First I take out the line and put it upon the Sector in the points of 23, then may I by the former proposition diminish it in AC, CD, in the proportion of 23, to 10, and after that divide the line AC into 10, etc. As before. 5 To find a proportion between two or more right lines given. TAke the greater line given, and according to it open the Sector in the points of 100 and 100, then take the lesser lines severally, & carry them parallel to the greater, till they stay in like points, so the number of points wherein they stay, shall show their proportion unto 100 line segment Let the lines given be AB, CD, first I take the line CD, & to it open the Sector in the points of 100, and 100, then keeping the Sector at this angle, I enter the lesser line AB, parallel to the former, and find it to cross the lines of Lines in the points of 60. Wherefore the proportion of AB to CD, is as 60 to 100 Or if the line CD, be greater than can be put over in the points of 100, than I admit the lesser line AB, to be 100, & cutting off CE equal to AB, I find the proportion of CE, unto ED, to be as 100, almost to 67; wherefore this way the proportion of AB unto CD, is as 100 unto almost 167. This proposition may also not unfitly be wrought by any other number, that admits several divisions, and namely, by the numbers of 60. And so the lesser line will be found to be 36, which is as before in lesser numbers, as 3 unto 5. It may also be wrought without opening the Sector. For if the lines between which we seek a proportion, be applied to the lines of Lines, (or any other Scale of equal parts) there will be such proportion found between them, as between the lines to which they are equal. 6 Two lines being given to find a third incontinuall proportion. FIrst place both the lines given, on both sides of the Sector from the Centre, and mark the terms of their extension, then take out the second line again, and to it open the Sector, in the term of the first line, so keeping the Sector at this angle, the parallel distance between the terms of the second line, shall be the third proportional. geometric illustration Let the two lines given be AB, AC, which I take out and place on both sides of the Sector, so as they all meet in the centre A, let the terms of the first line be B and B, the terms of the second C and C. Then do I take out AC the second line again, and to it open the Sector in the terms BB. So the parallel between C and C doth give me the third line in continual proportion. For as AB is unto AC, so BB, equal to AC, is unto CC. 7 Three lines being given to find the fourth in discontinuall proportion. HEre the first line and the third are to be placed on both sides of the Sector from the centre, then take out the second line, and to it open the Sector in the terms of the first line. For so keeping the Sector at this angle, the parallel distance between the terms of the third line, shallbe the fourth proportional. Let the three lines given be A, B, C. geometric illustration First I take out A and C, and place them on both sides of the Sector, in AB, AC, and AD, A, laying the beginning of both lines at the centre A, then do I take out B the second line, according to it I open the Sector in B and C, the terms of the first line: so the parallel between D and E, doth give me the fourth proportional which was required. As in Arithmetic, it sufficeth if the first and third number given be of one denomination, the second & the fourth which is required be of another. For one and the same denomination is not required necessarily in them all. So in Geometry, it sufficeth if the sides AB, AD, resembling the first and third lines given be measured in one Scale, and the parallels BC, DE be measured in another. Wherefore knowing the proportion of A the first line, and C the third line, by the fift prop. before. Which is here as 8 to 12, & descending in lesser numbers is as 4 to 6, or as 2 to 3, or ascending in greater numbers, as 16 unto 24, or 18 to 27, or 20 to 30, or 30 to 45, or 40 to 60 etc. If the Sector be opened in the points of 8 and 8, to the quantity of B, the second line given, than a parallel between 12 and 12, shall give DE, the fourth line required. So likewise if it be opened in 4 and 4, than a parallel between 6 and 6, or if in 16 and 16, than a parallel between 24 and 24 shall give the same DE. And so in the rest. 8 To divide a line in such sort as another line is before divided. FIrst take out the line given, which is already divided, and laying it on both sides of the Sector from the centre, mark how fare it extendeth. Then take out the second line which is to be divided, and to it open the Sector in the terms of the first line. This done, take out the parts of the first line, and place them also on the same side of the Sector from the centre. For the parallels taken in the terms of these parts, shallbe the correspondent parts in the line which is to be divided. Let AB, be a line divided in D and E, and BC, the line which I am to divide in such sort, as AB is divided. geometric illustration If the line AB, were longer than one of the sides of the Ruler, then should I find what proportion it hath to his parts AD, A, and that known I may work as before in the former proposition. 9 Two numbers being given to find a third in continual proportion. FIrst reckon the two numbers given on both sides of the lines of Lines from the centre, and mark the terms to which either of them extendeth, then take out a line resembling the second number again, and to it open the Sector in the terms of the first number, for so keeping the Sector at this angle, the parallel distance between the terms of the second lateral number, being measured in the same Scale, from whence his parallel was taken, shall give the third number proportional. Let the two numbers given be 18, 24, these being resembled in lines, the work will be in a manner all one, with that in the sixth Prop. and so the third proportional number will be found to to be 32. 10. Three numbers being given to find a fourth in discontinuall proportion. THe solution of this proposition, is in a manner all one with that before in the seventh Prop. only there may be some difficulty in placing of the numsers. To avoid this, we must remember that three number being given, the question is annexed but to one, and this must always be placed in the third place, that which agrees with this third number in denomination, shallbe the first number, and that which remaineth the second number. This being considered, reckon the first, and third numbers, which are of the first denomination on both sides of the lines of Lines from the centre, and mark the terms to which either of them extendeth, then take out a line resembling the second number, and to it open the Sector in the terms of the first number, for so keeping the Sector at this angle, the parallel distance between the terms of the third lateral number, being measured in the same Scale from whence his parallel was taken, shall give the fourth number proportional. As if a question were proposed in this manner, 10 yards cost 8 l, how many yards may we buy for 12 l? here the question is annexed to 12; and therefore it shall be the third number, and because 8 is of the same denomination, it shall be the first number, than 10 remaining, it must be the second number, so will they stand in this order, 8, 10, 12. These being resembled in lines, the work will be in a manner the same, with that in the seventh Prop. and the fourth proportional number will be found to be 15. For as 8 are to 10, so 12 unto 15. And this holdeth in direct proportion, where, as the first number is to the second, so the third to the fourth. So that if the third number be greater than the first, the fourth will be greater than the second, or if the third number be less than the first, the fourth will be less than the second, but it● reciprocal proportion, commonly called the Back rule, where by how much the first number is greater than the third, so much the second will be less than the fourth, or by how much the first number is less than the third, so much the second will be greater than the fourth. The manner of working must be contrary, that is; the Sector is to be opened in the terms of the third number, and the parallel resembling the number required, is to be found between the terms of the first number, the rest may be observed as before, as for example. If twelve men would raise a frame in ten days, in how many days would eight men raise the same frame? Here, because the fewer men would require the longer time, though the numbers be 12, 10, 8, yet the fourth proportional will be found to be 15. So if 60 yards, of three quarters of a yard in breadth, would hang round about a room, and it were required to know how many yards of half a yard in breadth, would serve for the same room. The fourth proportional would be found to be 90. So if to make a foot superficial, 12 inches in breadth do require 12 inches in length, and the breadth being 16 inches, it were required to know the length. Here, because the more breadth, the less length, the fourth proportional will be found to be 9 So if to make a Solid foot, a base of 144 inches, require 12 inches in height, and a base given being 216 inches, it were required to know how many inches it shall have in height. The fourth proportional would be found to be 8. This last proposition of finding a fourth proportional number, may be wrought also by the lines of Superficies, and by the lines of Solids. CHAP, III. The use of the lines of Superficies. 1 To find a proportion between two or more like Superficies. TAke one of the sides of the greater Superficies given, and according to it open the Sector in the points of 100 and 100, in the lines of Superficies, then take the like sides of the lesser Superficies severally, and carry them parallel to the former, till they stay in like points, so the number of points wherein they stay, shall show their proportion unto 100 line segment Let A and B, be the sides of like Superficies, as the sides of two squares, or the diameters of two circles, first I take the side A, and to it open the Sector in the points of 100, then keeping the Sector to this angle, I enter the lesser side B, parallel to the former, and find it to cross the lines of Superficies in the points of 40, wherefore the proportion of the Superficies, whose side is A, to that whose side is B, is as 100 unto 40, which is in lesser numbers, as 5 unto 2. This proposition might have been wrought by 60, or any other number that admits several divisions. It may also be wrought without opening the Sector, for if the sides of the Superficies given, be applied to the lines of Superficies beginning always at the centre of the Sector, there will be such proportion found between them, as between the number of parts whereon they fall. 2 To augment a Superficies in a given Proportion. 3 To diminish a Superficies in a given Proportion. TAke the side of the Superficies, and to it open the Sector in the points of the numbers given; then keeping the Sector at that angle, the parallel distance between the points of the number required, shall give the like side of the Superficies required. line segments Let A be the side of a Square to be augmented in the proportion of 2 to 5. First I take the side A, and put it over in the lines of Superficies, in 2 and 2; so the parallel between 5 and 5, doth give me the side B, on which if I should make a Square, it would have such proportion to the square of A, as 5 unto 2. In like manner if B were the semidiameter of a circle to be diminished in the proportion of 5 unto 2, I would take out B, and put it over in the lines of Superficies, in 5 and 5; so the parallel between 2 and 2, would give me A; on which Semidiameter if I should make a circle, it would be less than the circle made upon the Semidiameter B, in such proportion as 2 less than 5. For variety of work the like caution may be here observed to that which we gave in the third Prop. of Lines. 4 To add one like Superficies to another. 5 To subtract one like Superficies from another. FIrst, the proportion between like sides of the Superficies given, is to be found by the first Prop. of Superficies, then add or subtract the numbers of those proportions, and accordingly augment or diminish by the former Prop. line segments As if A and B were the side of two Squares, and it were required to make a third Square equal to them both. First the proportion between the squares of A and B, would be found to be as 100 unto 40, or in the lesser numbers as 5 to 2; then because 5 and 2 added do make 7, I augment the side A in the proportion of 5 to 7, and produce the side C, on which if I make a square, it will be equal to both the squares of A and B, which was required. In like manner A and B being the sides of two Squares, if it were required to subtract the square of B out of the square of A, and to make a square equal to the remainder, here the proportion being as 5 to 2, because 2 taken out of 5, the remainder is 3, I would diminish the side A in the proportion of 5 to 3, and so I should produce the side D, on which if I make a square, it will be equal to the remainder when the square of B is taken out of the square of A, that is, the two squares made upon B and D, shall be equal to the first square made upon the side A. 4 To find a mean proportional between two lines given. FIrst find what proportion is between the lines given, as they are lines, by the fifth Prop. of Lines, then open the Sector in the lines of Superficies, according to his number, to the quantity of the one, and a parallel taken between the points of the number belonging to the other line shall be the mean proportional. line segments Let the lines given be A and C. The proportion between them as they are lines will be found by the fifth Prop. of Lines to be as 4 to 9 Wherefore I take the line C, and put it over in the lines of Superficies between 9 and 9, and keeping the Sector at this angle, his parallel between 4 and 4 doth give me B for the mean proportional. Then for proof of the operation I may take this line B, and put it over between 9 and 9: so his parallel between 4 and 4, shall give me the first line A. Whereby it is plain that these three lines do hold in continual proportion; and therefore B is a mean proportional between A and C the extremes given. Upon the finding out of this mean proportion depend many Corollaries, as To make a Square equal to a Superficies given. IF the Superficies given be a rectangle parallellogram, a mean proportional between the two unequal sides shall be the side of his equal square. If it shall be a triangle, a mean proportion between the perpendicular and half the base shall be the side of his equal square. If it shall be any other rightlined figure, it may be resolved into triangles, and so a side of a square found equal to every triangle; and these being reduced into one equal square, it shall be equal to the whole rightlined figure given. To find a proportion between Superficies, though they be unlike one to the other. IF to every Superficies we find the side of his equal square, the proportion between these squares, shall be the proportion between the Superficies given. geometric illustration Let the Superficies given, be the oblong A, and the triangle B. First between the unequal sides of A, I find a mean proportional, and note it in C: this is the side of a square equal unto A. Then between the prependicular of B, and half his base, I find a mean proportional, and note it in B: this is the side of a Square equal to B: but the proportion between the squares of C and B, will be found by the first Prop. of Superficies to be as 5 to 4: and therefore this is the proportion between those given Superficies. To make a Superficies like to one Superficies and equal to another. geometric illustration First between the perpendicular and the base of B, I find a mean proportional, and note it in B, as the side of his equal square: then between the perpendicular of the triangle A, and half his base, I find a mean proportional, and note it in A, as the side of his equal square. Wherefore now as the side B is to the side A, so shall the sides of the Rhomboides given be to C and D, the sides of the Romboides required, & his pendicular also to E, the perpendicular required. Having the sides and the perpendicular, I may frame the Rhomboides up, and it will be equal to the triangle A. If the Superficies given had been any other rightlined figures, they might have been resolved into triangles, and then brought into squares as before. Many such Corollaries might have been annexed, but the means of finding a mean proportional being known, they all follow of themselves. 7 To find a mean proportional between two numbers given. FIrst reckon the two numbers given on both sides of the Lines of Superficies, from the centre, and mark the terms whereunto they extend; then take a line out of the Line of Lines, or any other scale of equal parts resembling one of those numbers given, and put it over in the terms of his like number in the lines of Superficies; for so keeping the Sector at this angle, the parallel taken from the terms of the other number and measured in the same scale from which the other parallel was taken, shall here show the mean proportional which was required. Let the numbers given be 4 and 9 If I shall take the line A, in the Diagram of the sixth Prop. resembling 4 in a scale of equal parts, and to it open the Sector in the terms of 4 and 4, in the lines of Superficies, his parallel between 9 and 9 doth give me B for the mean proportional. And this measured in the scale of equal parts doth extend to 6, which is the mean proportional number between 4 and 9 For as 4 to 6, so 6 to 9 In like manner if I take the line C, resembling 9 in a scale of equal parts, and to it open the Sector in the terms of 9 and 9, in the lines of Superficies, his parallel between 4 and 4 doth give me the same line B, which will prove to be 6, as before, if it be measured in the same scale whence D was taken. 8 To find the square root of a number. 9 The root being given to find the square number of that root. IN the extraction of a square root it is usual to set pricks under the first figure, the third, the fifth, the seventh, and so forward, beginning from the right hand toward the left, and as many pricks as fall to be under the square number given, so many figures shall be in the root: so that if the number giuen be less than 100, the root shall be only of one figure; if less than 10000, it shall be but two figures; if less than 1000000, it shall be three figures, etc. Thereupon the lines of Superficies are divided first into an hundred parts, and if the number giuen be greater than 100, the first division (which before did signify only one) must signify 100, and the whole line shall be 10000 parts: if yet the number giuen be greater than 10000, the first division must now signify 10000, and the whole line be esteemed at 1000000 parts: and if this be too little to express the number given, as oft as we have recourse to the beginning, the whole line shall increase itself an hundred times. By this means if the last prick to the left hand shall fall under the last figure, which will be as oft as there be odd figures, the number given shall fall out between the centre of the Sector and the tenth division: but if the last prick shall fall under the last figure but one, which will be as oft as there be even figures, than the number given shall fall out between the tenth division and the end of the Sector. This being considered, when a number is given and the square root is required, take a pair of compasses and setting one foot in the centre, extend the other to the term of the number given in one of the lines of Superficies; for this distance applied to one of the Lines of Lines, shall show what the Square root is, without opening the Sector. Thus 64 doth give a root of 8, and 860 a root of almost 19, and 1296 a root of 36, and 7056 a root of 84, and 62500 a root of 250, and 714000 a root of about 845, and so in the rest. On the contrary, a number given may be squared, if first we extend the compasses to the number given in the lines of Lines, and then apply the distance to the Lines of Superficies, as may appear by the former examples. 10 Three numbers being given to find the fourth in a duplicated proportion. IT is plain by the 19 and 20 Prop. 6. Lib. of Euclid, that like Superficies do hold in a duplicated proportion of their homologal sides, whereupon a question being moved concerning Superficies and their sides. It is usual in Arithmetic that the proportion be first duplicated before the question be resolved, which is not necessary in the use of the Sector, only the numbers which do signify Superficies must be reckoned in the lines of Superficies, and they which signify the sides of Superficies, in the lines of Lines, after this manner. If a question be made concerning a Superficies, the two numbers of the first denomination must be reckoned in the lines of Lines, and the Sector opened in the terms of the first number to the quantity of a line out of the scale of Superficies resembling the second number; so his parallels taken between the terms of the third number, being measured in the same scale of Superficies, shall give the Superficial number which was required. As if a Square, whose side is forty perches in length, shall contain ten acres in the Superficies, and it be required to know how many acres the Square should contain, whose side is sixty perches. Here if I took 10 out of the line of Superficies, and put it over in 40 in the lines of Lines, his parallel between 60 and 60 measured in the line of Superficies, would be 22 ½; and such is the number of acres required. For Squares do hold in a duplicated proportion of their sides; wherefore when the proportion of their sides is as 4 to 6, and 4 multiplied into 4 become 16, and 6 multiplied into 6 become 36, the proportion of their squares shall be as 16 to 36; and such is the proportion of 10 to 22 ½. If a field measured with a statute perch of 16 ½ foot, shall contain 288 acres, and it be required to know how many acres it would contain if it were measured with a woodland perch of 18 foot. Here because the proportion is reciprocal, if I took 288 out of the line of Superficies, and put it over in 18, in the lines of Lines, his parallel between 16 ½ and 16 ½ measured in the line of Superficies, would be 242; and such is the number of acres required. For seeing the proportion of the sides is as 16 to 18, or in lesser numbers as 11 to 12, and that 11 multiplied into 11 become 121, and 12 into 12 become 144, the proportion of these Superficies shall be as a 121 to 144, and so have 288 to 242, in reciprocal proportion. On the contrary, if a question be proposed concerning the side of a Superficies, the two numbers of the first denomination must be reckoned in the lines of Superficies, and the Sector opened in the terms of the first number, to the quantity of a line, out of the line of Lines, or some Scale of equal parts, resembling the second number; so his parallel taken between the terms of the third number being measured in the same scale with the second number, shall give the fourth number required. As if a field contained 288 acres when it was measured with a statute perch of 16 ½, and being measured with another perch, was found to contain 242 acres, it were required to know what was the length of the perch with which it was so measured. Here because the proportion is reciprocal, if I took 16 ½ out of the line of Lines, and put it over in 242 in the lines of Superficies, his parallel between 288 and 288, being measured in the line of Lines, would be 18, and such is the length of the perch in foot whereby the field was last measured. For seeing the proportion of the acres is as 288 unto 242, or in the least numbers as 144 to 121, and that the root of 144 is 12, and the root of 121 is 11, the proportion of roots and consequently of the perches shall be as 12 to 11, and so are 16 ½ to 18, in reciprocal proportion. If 360 men were to be set in form of a long square, whose sides shall have the proportion of 5 to 8; and it were required to know the number of men to be placed in front and file: if the sides were only 5 and 8, there should be but 40 men; but there are 360: therefore working as before, I find that As 40 to the square of 5, so 360 to the square of 15. As 40 to the square of 8, so 360 to the square of 24. and so 15 and 24 are the sides required. If 1000 men were lodged in a square ground, whose side were 60 paces, and it were required to know the side of the square wherein 5000 might be so lodged, here working as before, I should find that As 1000 are to the square of 60: so 5000 to the square of 134. And such very near is the number of paces required▪ CHAP. IU. The use of the lines of Solids. To find a proportion between two or more like Solids. IN the Sphere, in regular, parallel, and other like bodies, whose sides next the equal angles are proportional, the work is in a manner the same, with that in the first Prop. of Superficies, but that it is wrought on other lines. Take one of the sides of the greater Solid, & according to it open the Sector in the points of 1000 and 1000, in the lines of Solids, then take the like sides of the lesser Solids severally, and carry them parallel to the former, till they stay in like points, so the number of points wherein they stay, shall show their proportion to 1000 line segments Let A and B, be the like sides of like Solids, either the diameters, or semidiameters of two spheres, or the sides of two cubes, or other like. First I take the side A, and to it open the Sector in the points of 1000, then keeping the Sector at this angle, I enter the lesser side B, parallel to the former, and find it to cross the line of Solids in the points of 400, and such is the proportion between the Solids required, which in lesser number is as 5 to 2. This proposition might have been wrought by 60, or any other number that admits several divisions. It may also be wrought without opening the Sector, for if the sides of the Solids given, be applied to the lines of Solids, beginning always at the centre of the Sector, there will be such proportion between them, as between the numbers of parts whereon they fall. 2 To augment a Solid in a given proportion. 3 To diminish a Solid in a given proportion. TAke the side of the Solid given, and to it open the Sector, in the points of the number given: then keeping the Sector at that angle, the parallel distance between the points of the number required, shall give the like side of the Solid required. If it be a paralleleopipedon, or some irregular Solid, the other like sides may be found out in the same manner, and with them the Solids required, may be made up with the same angles. line segments Let A be the side of a cube, to be augmented in the proportion of 2 to 3. First I take the side A, and put it over in the lines of Solids in 2 and 2, so the parallel between 3 and 3, doth give me the side B, on which if I make a cube, it will have such proportion to the cube of A, as 3 to 2. In like manner, if B were the diameter of a Sphere, to be diminished in the proportion of 3 to 2. I would take out B, and put it over in the lines of Solids, in 3 and 3, so the parallel between 2 and 2, would give me A: to which diameter if I should make a Sphere, it would be less than the Sphere, whose diameter is B, in such proportion as 2 less than 3. Here also for variety of work, may the like caution be observed to that which we gave in the third Prop. of Lines. 4 To add one like Solid to another. 5 To subtract one like Solid from another. FIrst the proportion between the sides of the like Solids given, is to be found by the first Prop. of Solids: then add or subtract those proportions, and accordingly augment or diminish by the former Prop. line segments As if A and B were the sides of two cubes, and it were required to make a third cube equal to them both: first the proportion between the sides A and B, would be found to be as 100 to 40, or in lesser terms as 5 to 2. Then because 5 and 2 being added do make 7, I augment the side A in the proportion of 5 to 7, and produce the side C, on which if I make a cube, it will be equal to both the cubes of A and B, which was required. In like manner A and B being the fides of two cubes, if it were required to subtract the cube of B out of the cube of A, and to make a cube equal to the remainder. Here the proportion being as 5 to 2, because 2 taken out of 5, the remainder is 3, I should diminish the side A in the proportion of 5 to 3, and so I should have the side D, on which if I make a cube, it will be equal to the remainder when the cube of B is taken out of the cube of A, that is the two cubes made upon B and D, shall be equal to the first cube made upon the side A. 6 To find two mean proportional lines between two extreme lines given. FIrst I find what proportion is between the two extreme lines given as they are lines, by the fifth Prop. of Lines, then open the Sector in the lines of Solids, to the quantity of the former extreme, and a parallel between the points of the number belonging to the other extreme, shall be that mean proportional which is next the former extreme. This done, open the Sector again to this mean proportional in the points of the former extreme, and the parallel distance between the points of the latter extreme, shall be the other mean proportional required. line segments Let the two extreme lines given be A and D, the proportion between them, as they are lines, will be found to be as 27 to 8. Wherefore I take the line A, and put it over in the lines of Solids between 27 and 27, and keeping the Sector at this angle, his parallel between 8 and 8, doth give me B, the mean proportional next unto A. Then put I over this line B, between the aforesaid 27 and 27, and his parallel between 8 and 8 doth give me the line C, the other mean proportional which was required. Again, for proof of the operation I put over this line C in the aforesaid 27 and 27, and his parallel between 8 and 8 doth give me the very line D: whereby it is plain that these four lines do hold in continual proportion; and so B and C are found to be the mean proportionals between A and D the extremes given. 7 To find two mean proportional numbers between two extreme numbers given. FIrst reckon the numbers given on both sides of the lines of Solids, beginning from the centre, and marking the terms whereto they extend: then take a line out of the line of Lines, or any other scale of equal parts resembling the former of those numbers, and put it over in the lines of Solids, between the points of his like number, and a parallel between the points belonging to the other extreme, measured in the scale from whence the other parallel was taken, shall give that mean proportional number which is next the former extreme. This done, open the Sector again to this mean proportional in the points of the former extreme, and the parallel distance between the points of the latter extteame, measured in the same scale as before, shall there show the other mean proportional required. line segments Let the two extreme numbers given be 27 and 8; if I shall take the line A, resembling 27 in a scale of equal parts, and to it open the Sector in 27 and 27, in the line of Solids, his parallel between 8 and 8 doth give me B for his next mean proportional, and this measured in the former scale doth extend to 18. Then put I over this line B between the aforesaid 27 and 27, and his parallel between 8 and 8 doth give me C for the other mean proportional, and this measured in the former scale doth extend to 12. Again, for proof of my work, I put over this line C between 27 and 27, as before, and his parallel between 8 and 8 doth give me D, which measured in the former scale doth extend to 8, which was the latter extreme number given; whereby it is plain that these four numbers do hold in continual proportion: and therefore 18 and 12 are mean proportionals between 27 and 8, which was required. 8 To find the cubique root of a number. 9 The root being given to find the cube number of that root. IN the extraction of a cubique root, it is usual to set prick● under the first figure, the fourth, the seventh, the tenth, and so forward, omitting two, and pricking the third from the right-hand toward the left; and as many pricks as fall to be under the cubique number, so many figures shall be in the root. So that if the number giuen be less than 1000, the root shall be only of one figure; if less than 1000000, it shall be but of two figures; if above these, and less than 1000000000, it shall be but three figures; etc. whereupon the lines of Solids are divided, first into 1000 parts, and if the numbers given be greater than 1000, the first division (which before did signify only one) must signify 1000, and the whole line shall be 1000000: if yet the number giuen be greater than 1000000, the first division must now signify 1000000, and the whole line be esteemed at 1000000000 parts, and if these be to little to express the numbers given, as oft as we have recourse to the beginning, the whole line shall increase itself a thousand times. By these means, if the last prick, to the left hand, shall fall under the last figure, the number given shall be reckoned at the beginning of the lines of Solids, from 1 to 10, and the first figure of the root shall be always either 1, or 2. If the last prick shall fall under the last figure but one, than the number given shall be reckoned in the middle of the line of Solids, between 10 and 100, and the first figure of the root shall be always either 2, or 3, or 4. But if the last prick shall fall under the last figure but two, than the number given, shall be reckoned at the end of the line of Solids, between 100, and 1000 This being considered when a number is given, and the cubique root required: Set one foot of the compasses in the centre of the Sector, extend the other in the line of Solids, to the points of the number given: for this distance applied to one of the line of Lines, shall show what the cubique root is, without opening the Sector. So the nearest root of 8490000, is about 204. The nearest root of 84900000, is about 439. The nearest root of 849000000, is about 947. On the contrary, a number may be cubed, if first we extend the compasses to the number given, in the line of Lines, and then apply the distance to the lines of Solids; as may appear by the former examples. 10 Three numbers being given to find a fourth in a triplicated proportion. AS like Superficies do hold in a duplicated proportion, so like solids in a triplicated proportion of their homologal sides: and therefore the same work is to be observed here on the lines of Solids, as before in the lines of Superficies; as may appear by these two examples. If a cube whose side is 4 inches, shall be 7 pound weight, and it be required to know the weight of a cube whose side is 7 inches; here the proportion would be, As 4 are to a cube of 7: so 7 to a cube of 37 ½. And if I took 7 out of the lines of Solids, and put it over in 4 and 4, in the lines of Lines, his parallel between 7 and 7 measured in the lines of Solids, would be 37 ½; and such is the weight required. If a bullet of 27 pound weight have a diameter of 6 inches, and it be required to know the diameter of the like bullet, whose weight is 125 pounds; here the proportion would be, As the cubique root of 27 is unto 6: so the cubique root of 125 is unto 10. And if I took 6 out of the line of Lines, and put it over in 27 and 27 of the lines of Solids, his parallel between 125 and 125 measured in the line of Lines, would be 10; and such is the length of the diameter required. The end of the first book. THE SECOND BOOK OF THE SECTOR, Containing the use of the Circular Lines. CHAP. I. Of the nature of Sines, Chords, Tangents and Secants, fit to be known before hand in reference to right-line Triangles. IN the Canon of Triangles, a circle is commonly divided into 360 degrees, each degree into 60 minutes, each minute into 60 seconds. geometric illustration A semicircle therefore is an ark of 180 gr. A quadrant is an ark of 90 gr. The measure of an angle is the ark of a circle, described out of the angular point, intercepted between the sides sufficiently produced. So the measure of a right angle is always an ark of 90 gr. and in this example the measure of the angle BAD is the ark BC of 40 gr; the measure of the angle BAG, is the ark BF of 50 gr. The compliment of an ark or of an angle doth commonly signify that ark which the given ark doth want of 90 gr: and so the ark BF is the compliment of the ark BC; & the angle BAF, whose measure is BF, is the compliment of the angle BAC; and on the contrary. The compliment of an ark or angle in regard of a semicircle, is that ark which the given ark wanteth to make up 180 gr: and so the angle EAH is the compliment of the angle EAF, as the ark EH is the compliment of the ark FE, in which the ark CE is the excess above the quadrant. The proportions which these arkes (being the measures of angles) have to the sides of a triangle, cannot be certain, unless that which is crooked be brought to a strait line; and that may be done by the application of Chords, Right Sines, versed Sines, Tangents and Secants, to the semidiameter of a circle. A chord is a right line subtending an ark: so BE is the chord of the ark BE, and BF a chord of the ark BF. A right Sine is half the chord of the double ark, viz. the right line which falleth perpendicularly from the one extreme of the given ark, upon the diameter drawn to the other extreme of the said ark. So if the given ark be BC, or the given angle be BAC, let the diameter be drawn through the centre A unto C; and a perpendicular BD be let down from the extreme B, upon AC; this perpendicular BD shall be the right sine both of the ark BC, and also of the angle BAC: and it is also the half of the chord BE, subtending the ark BE, which is double to the given ark BC. In like manner, the semidiameter FAVORINA, is the right sine of the ark FC, and of the right angle FAC; for it falleth perpendicularly upon AC, and it is the half of the chord FH. This whole Sine of 90 gr. is hereafter called Radius; but the other Sins take their denomination from the degrees and minutes of their arks. Sinus versus, the versed sine is a segment of the diameter, intercepted between the right sine of the same ark, and the circumference of the circle. So DC is the versed sine of the ark CB, and GF the versed sine of the ark BF, and GH the versed sine of the ark BH. A Tangent is a right line perpendicular to the diameter, drawn by the one extreme of the given ark, and terminated by the secant drawn from the centre through the other extreme of the said ark. A Secant is a right line drawn from the centre, through one extreme of the given ark, till it meet with the tangent raised from the diameter at the other extreme of the said ark. So if the given ark be CE, or the given angle be CAESAR, let the diameter be drawn through the centre A to C, and in C to AC, be raised a perpendicular CI. Then let another line be drawn from the centre A through E, till it meet with the perpendicular CI in I; the line CI is a Tangent, and AI is the Secant both of the ark CE, and of the angle CAESAR. CHAP. II. Of the general use of Sines and Tangents. 1 The Radius being known to find the right sine of any ark or angle. IF the Radius of the circle given be equal to the lateral Radius, that is, to the whole line of Sins on the Sector, there needs no farther work, but to take the other sins also out of the side of the Sector. But if it be either greater or lesser, then let it be made a parallel Radius, by applying it over in the lines of Sines, between 90 and 90; so the parallel taken from the like lateral sins, shall be the sine required. As if the given Radius be AC, and it were required to find the sine of 50 Gr. & his compliment agreeable to that radius. geometric illustration Let AB, AB represent the lines of sins on the Sector, and let BB, the distance between 90 and 90, be equal to the given radius AC. Here the lines A 40, A 50, A 90, may be called the lateral sins of 40, 50, & 90; in regard of their place on the sides of the Sector. The lines between 40 and 40, between 50 and 50, between 90 and 90, may be called the parallel sins of 40, 50, and 90; in regard they are parallel one to the other. The whole sine of 90 Gr. here standing for the semidiameter of the circle, may be called the Radius. And therefore if AC be put over in the line of Sins in 90 and 90, and so made a parallel radius, his parallel sine between 50 and 50, shall be BD, the sine of 50 required. And because 50 taken out of 90, the compliment is 40; his parallel sine between 40 and 40, shall be BG, the sine of the compliment which was required. 2 The right sine of any ark being given to find the Radius. Turn the sine given into a parallel sine, and his parallel Radius shall be the Radius required. As if BD were the given sine of 50 Gr. and it were required to find the Radius: let BD be made a parallel sine of 50 Gr. by applying it over in the lines of Sines, between 50 and 50; so his parallel Radius between 90 and 90 shall be AC, the Radius required. 3 The Radius of a circle, or the right Sine of any ark being given, and a straight line resembling a Sine, to find the quantity of that unknown Sine. LEt the Radius or right sine given be turned into his parallel; then take the right line given, and carry it parallel to the former, till it stay in like Sins: so the number of degrees and minutes where it stayeth, shall give the quantity of the Sine required. As if BD were the given sine of 50 Gr. and BG the straight line given: first I make BD a parallel sine of 50 Gr; then keeping the Sector at this angle, I carry the line BG parallel, and find it to stay in no other but 40 and 40; and therefore 40 gr. is his quantity required. 4 The Radius or any right Sine being given, to find the versed sine of any ark. IF the ark, whose versed sine is required, be less than the quadrant, take the sine of the compliment out of the radius, and the remainder shall be the sinus versus, the versed sine of that ark. As if AB being the lateral Radius, it were required to find to find the versed line of 40 gr; here the sine of the compliment is A 50, and therefore B 50 is the versed sine required. Or if I reckon from B, at the end of the Sector, toward the centre, the distance from 90 to 80, is the versed sine of 10 gr; from 90 to 70, the versed sine of 20 gr; from 90 to 60, is the versed sine of 30 gr; and so in the rest. If AD be the given sine of 50 gr. and it be required to find the versed sine of 50 gr; here because AD is unequal to the lateral sine of 50 gr, I make it a parallel. And first I find the radius AC; then the sine of the compliment A 40, which being taken out of AC, leaveth C 40 for the versed sine of 50 gr. which was required. But if the ark, whose versed sine is required, be greater than the quadrant, his versed sine also is greater than the Radius, by the right sine of his excess above 90 gr. As if AC being the Radius given, it were required to find the versed sine of 130 gr: here the excess above 90 gr. is 40 gr; and therefore the versed sine required is equal to the Radius AC and A 40, both being set together. 5 The Diameter or Radius being given to find the Chords of every ark. The sins may be fitted many ways to serve for chords. 1 A sine being the half of the chord of the double ark, if the sine be doubled, it giveth the chord of the double ark, a Sine of 10 gr. doubled giveth a Chord of 20 gr; and a Sine of 15 gr. being doubled giveth a Chord of 30 gr; and so in the rest. As here BD, the sine of BC, an ark of 40 gr. being doubled giveth BE the chord of BE, which is an ark of 80 gr. Wherefore if the Radius of the circle given be equal to the lateral Radius, let the Sector be opened near unto his length, so that both the lines of Sines may make but one direct line: so the distance on the sins between 10 and 10, shall be a chord of 20; the distance between 20 and 20, shall be a chord of 40; and the distance between 30 and 30, shall be a chord of 60; and so in the rest. 2 Because a sine is the half of the chord of the double ark, the proportion holdeth. geometric illustration As the diameter FH unto the radius AH, so the chord BE unto the sine DE, or the chord GL unto the sine ALL: and then if the radius AH, be put for the diameter, which is a chord of 180 gr, the sine DE or ALL shall serve for a chord of 80 gr, and the semiradius which is the sine of 30 gr, shall serve for a chord of 60 gr, and go for the semidiameter of a circle, and so in the rest. So that by these means we shall not need to double the lines of Sins as before, but only to double the numbers. And to this purpose I have subdivided each degree of the sins into two, that so they might show how far the half degrees do reach in the sins, and yet stand for whole degrees when they are used as chords. Wherefore if the Radius of the circle given be equal to the lateral semiradius (the sine of 30 Gr. and chord of 60 Gr.) there needs no farther work then to take the sine of 10 Gr. for a chord of 20 Gr. and a sine of 15 Gr. for a chord of 30 Gr. etc. But if the Radius of the circle given be either greater or lesser than the lateral semiradius, take the diameter of it, and make it a parallel chord of 180 Gr. by applying it over the lines of Sins between 90 and 90: or take the Radius or Semidiameter which is equal to the chord of 60 Gr. and make it a parallel Radius of 60 Gr. by applying it over in the sins of 30 and 30, and keep the Sector at this angle. The parallels taken from the lateral chords shall be the chords required. As if the diameter of a circle given were the line AB, and it were required to find the chord of 80 gr: first, I make AB a parallel chord of 180 Gr. or the half of it a parallel chord of 60 Gr; so his parallel LG doth give me FG the chord of 80 Gr. which was required. 3 Seeing that as the sine of the compliment of the half ark is unto the Radius, so the sine of the same whole ark is unto the chord of it: if we seek but for one single chord, we may find it without either doubling the sins, or doubling the number. For applying over the Radius given in the sine of the compliment of half the ark required, his parallel sine shall be the chord required. As if the semidiameter of the circle given were AC, and it were required to find the chord of 40 Gr: the half of 40 Gr. is 20 Gr. the compliment of 20 Gr. is 70 Gr. Wherefore I make AC a parallel sine of 70 Gr. and his parallel sine GL doth give me FG the chord of 40 Gr. agreeable to the semidiameter AC. geometric illustration 6 The chord of any ark being given to find the diameter and Radius. Turn the chord given unto a parallel chord, and his parallel semiradius shall be the semidiameter, and the parallel radius shall be the diameter. As if FG be the chord of 80 gr. I put this over in G and L, the sine of 40, and chord of 80 gr. and the parallel chord of 180 gr. giveth me AB the diameter required. Or if I turn the chord given into a parallel sine of the same quantity, his parallel sine of the compliment of half the ark, doth give me the semidiameter. As if FG be the given chord of 40 gr. I put it over in G and L, the sins of 40 gr; then because the half of 40 gr. is 20 gr. and the compliment of 20 gr. is 70 gr. I take out the parallel sine of 70 gr. and it giveth me AC for the semidiameter, agreeable to that chord of 40 gr. 7 To open the Sector to the quantity of any angle given. 8 The Sector being opened, to find the quantity of the angle. IT is one thing to open the edges of the Sector to an angle, and another thing to open the lines on the Sector to the same angle. For the lines of lines on the one side, & the lines of sins on the other side, do make an angle of 2 gr. when the Sector is close shut, and the edges do make no angle at all. So likewise the lines of Superficies and the lines of Solids do make an angle of 10 gr, which are to be allowed to the edges. The lines of lines may be opened to a right angle, if the whole line of 100 parts be applied over in 80 and 60. The lines of sins may be opened to a right angle, if the large secant of 45 gr. be applied over in the sins of 90 gr. or if the sine of 90 gr. be applied over in the sins of 45 gr. or if the sine of 45 gr. be applied over in the sins of 30 gr. If it be required to open those lines to any other angle, take out the chord thereof, and apply it over in the semiradius, and those lines shall be opened to that angle. As if it were required to open the Sector in the lines of sins to an angle of 40 gr, take out the chord of 40 gr, and to it open the Sector in the chord of 60 gr; so shall the lines of sins be opened to the angle required. Or if the same chord of 40 Gr. be applied over between 50 and 50, in the lines of lines, they shall also be opened to the same angle. If it be applied over in 25 of the lines of Superficies, or 125 in the lines of Solids, they also shall be opened to the same angle: because the chord of 60 Gr. or sine of 30 Gr. and 50 in the lines of lines, and 25 in the lines of Superficies, and 125 in the Solids, are all of the same length with the semiradius. Or if the Semiradius be applied over between ●he sine of 30 Gr. and the sine of the compliment of the angle required, it will open the lines of Sins to that angle. As if the semiradius be applied over in the sins of 30 Gr. and the sine of 50 Gr. it shall open the lines of Sins to an angle of 40 Gr. On the contrary, if the Sector be opened to an angle, and it be required to know the quantity thereof, open the compasses to the semiradius, and setting one foot in the sine of 30 Gr. turn the other toward the other line of sins, and it shall fall there in the compliment of the angle; if it fall on 50 Gr. the angle is 40 Gr; if on 60 Gr. the angle is 30 Gr. etc. Or take over the parallel chord of 60 Gr. and measure it in the lateral chord, and it shall there show the quantity of the angle. As if the Sector being opened to an angle, I should take over the parallel of 30 Gr. of the sins, and 60 Gr. of the chords, and measure it in the lateral chords, find it to be 40 Gr; the angle comprehended between the lines of Sines is 40 Gr. but the angle between the edges of the Sector is 2 Gr. less, and therefore but 38 Gr. 9 To find the quantity of any angle given. IF out of the angular point, to the quantity of the Semiradius, be described an occult ark that may cut both sides of the angle, the chord of this ark measured in the lateral chord, shall give the quantity of the angle. Let the angle giuen be BAC: first I take the Semiradius with the compasses, and setting one foot in A, I cut the sides of the angle in B and C; then I take the chord BC, and measure it in the lateral chord, and I find it to be 11 Gr. and 15 M. and such is the quantity of the angle given. geometric illustration Or if the ark be described out of the angular point at any other distance, let the semidiameter be turned into a parallel chord of 60 Gr. then take the chord of this ark, and carry it parallel till it cross in like chords: so the place where it stayeth shall give the quantity of the angle. As in the former example, if I make the semidiameter AB a parallel chord of 60 Gr. and then keeping the Sector at that angle, carry the chord BC parallel, till it stay in like chords; I shall find it to stay in no other but 11 Gr. 15 M and such is the angle BAC. 10 Upon a right line and a point given in it, to make an angle equal to any angle given. FIrst out of the point giuen describe an ark, cutting the same line: then by the 5. Prop afore, find the chord of the angle given agreeable to the semidiameter, and inscribe it into this ark: so a right line drawn through the point given, and the end of this chord, shall be the side that makes up the angle. Let the right line given be AB, and the point given in it be A, and let the angle given be 11 gr. 15 m. Here I open the compasses to any semidiameter AB, (but as oft as I may conveniently to the lateral semiradius) and setting one foot in A, I describe an occult ark BC; then I seek out the chord of 11 gr. 15 m. and taking it with the compasses, I set one foot in B, the other crosseth the ark in C, by which I draw the line AC, and it makes up the angle required. 11 To divide the circumference of a circle into any parts required. IF 360 the measure of the whole circumference be divided by the number of parts required, the quotient giveth the chord, which being found will divide the circumference. So a chord of 120 gr. will divide the circumference into 3 equal parts; a chord of 90 gr. into 4 parts; a chord of 72 gr, into 5 parts; a chord of 60 gr. into 6 parts; a chord of 51 gr. 26. into 7 parts; a chord of 45 gr. into 8 parts; a chord of 40 gr. into 9 parts; a chord of 36 gr. into 10 parts; a chord of 32 gr. 44 m. into 11 parts; a chord of 30 gr. into 12 parts. In like manner if it be required to divide the circumference of the circle, whose semidiameter is AB, into 32: first I take the semidiameter AB, and make it a parallel chord of 60 gr; then because 360 gr. being divided by 32, the quotient will be 11 gr. 15 m. I find the parallel chord of 11 gr. 15 m. and this will divide the circumference into 32. But here the parts being many, it were better to divide it first into fewer, and after to come over it again. As first to divide the circumference into 4, and then each 4 parts into 8, or otherwise, as the parts may be divided. 12 To divide a right line by extreme and mean proportion. THe line to be divided by extreme and mean proportion, hath the same proportion to his greater segment, as in figures inscribed in the same circle, the side of an hexagon a figure of six angles, hath to a side of a decagon a figure of ten angles: but the side of a hexagon is a chord of 60 gr. and the side of a decagon is a chord of 36 gr. Let AB be the line to be divided: if I make AB a parallel chord of 60 gr. and to this semidiameter find AC a chord of 36 gr. this AC shall be the greater segment, dividing the whole line in C, by extreme and mean proportion. So that, As AB the whole line is unto AC the greater segment: so AC the greater segment unto CB the lesser segment. Or let AC be the greater segment given: if I make this a parallel chord of 36 gr. the correspondent semidiameter shall be the whole line AC, and the difference CB the lesser segment. line segment Or let CB be the lesser segment given: if I make this a parallel chord of 36 gr. the correspondent semidiameter shall be greater segment AC, which added to CB, giveth the whole line AB. To avoid doubling of lines or numbers, you may put over the whole line in the Sins of 72 gr. and the parallel sine of 36 gr. shall be the greater segment. Or if you put over the whole line in the sins of 54 gr. the parallel sine of 30 gr. shall be the greater segment, and the parallel sine of 18 gr. shall be the lesser segment. CHAP. III. Of the projection of the Sphere in Plano. 1 THe Sphere may be projected in Plano in straight lines, as in the Analemma, if the semidiameter of the circle given be divided in such sort as the line of Sins on the Sector. As if the Radius of the circle given were A, the circle thereon described may represent the plane of the general meridian, which divided into four equal parts in E, P, A, S, and crossed at right angles with EAE and PS, the diameter EAE shall represent the equator, and PS the circle of the hour of 6. And it is also the axis of the world, wherein P stands for the North pole, and S for the South pole. Then may each quarter of the meridian be divided into 90 gr. from the equator towards the poles. In which if we number 23 gr. 30 m. the greatest declination of the Sun from E to 69 Northwards, from A to ♑ Southward, the line drawn from 69 to ♑ shall be the ecliptic, and the lines drawn parallel to the equator through ♋ and ♑ shall be the tropiques. Having these common sections with the plane of the meridian, if we shall divide each diameter of the Ecliptic into 90 gr. in such sort as the Sins are divided on the Sector. The first 30 gr. from A toward 69, shall stand for the sine of ♈. The 30 gr. next following for ♉. The rest for ♊. ♋. ♌. etc. in their order. So that by these means we have the place of the Sun for all times of the year. geometric illustration If again we divide AP, AS, in the like sort, and set to the numbers 10. 20. 30. etc. unto 90 gr. the lines drawn through each of these degrees parallel to the equator, shall show the declination of the Sun, and represent the parallels of latitude. If farther we divide A, AAE, and his parallels in the like sort, and then carefully draw a line through each 15 gr. so as it makes no angles; the lines so drawn shall be elliptical and represent the houre-circles. The meridian PES the hour of 12 at noon; that next unto it drawn through 75 gr. from the centre the hours of 11 and 1, that which is drawn through 60 gr. from the centre the hours of 10 and 2. etc. Then having respect unto the latitude, we may number it from E Northward unto Z, and there place the zenith: by which and the centre the line drawne ZAN shall represent the vertical circle, passing through the zenith and nadir East and West, and the line MAH crossing it at right angles shall represent the horizon. These two being divided in the same sort as the ecliptic and the equator, the line drawn through each degree of the semidiameter AZ, parallel to the horizon, shall be the circles of altitude, and the divisions in the horizon and his parallels shall give the azimuth. Lastly, if through 18 gr. in AN, be drawn a right line IK parallel to the horizon, it shall show the time when the day breaketh, and the end of the twilight. For example of this projection, let the place of the Sun be the last degree of ♉, the parallel passing through this place is LD, and therefore the meridian altitude ML, and the depression below the horizon at midnight HD: the semidiurnal ark LC, the seminocturnal ark CD, the declination AB, the ascentionall difference BC, the amplitude of ascenon AC. The difference between the end of twilight and the day break is very small; for it seems the parallel of the Sun doth hardly cross the line of twilight. If the altitude of the Sun be given, let a line be drawn for it parallel to the horizon; so it shall cross the parallel of the Sun, and there show both the azimuth and the hour of the day. As if the place of the Sun being given as before, the altitude in the morning were found to be 20 gr. the line FG drawn parallel to the horizon through 20 gr. in AZ, would cross the parallel of the Sun in ☉. Wherefore F ☉ showeth the azimuth, & L ☉ the quantity of hours from the meridian. It seems to be about half an hour past 6 in the morning, and yet more than half a point short of the East. The distance of two places may be also showed by this projection, their latitudes being known, and their difference of longitude. For suppose a place in the East of Arabia, having 20 gr. of North latitude, whose difference of longitude from London is found by an eclipse to be 5 ho. ½. Let Z be the zenith of London, the parallel of latitude for that other place must be LD, in which the difference of longitude is L ☉. Wherefore ☉ representing the site of that place, I draw through ☉ a parallel to the horizon MH, crossing the vertical AZ near about 70 gr. from the zenith, which multiplied by 20, showeth the distance of London, and that place to be 1400 leagues. Or multiplied by 60, to be 4200 miles. 2 The Sphere may be projected in plano by circular lines, as in the general astrolabe of Gemma Frisius, by the help of the tangent on the side of the Sector. For let the circle giuen represent the plane of the general meridian as before; let it be divided into four parts, and crossed at right angles with EAE the equator, and PS the circle of the hour of 6, wherein P stands for the North pole, and S for the South pole. Let each quarter of the meridian be divided into 90 gr. and so the whole into 360, beginning from P, and setting to the numbers of 10, 20, 30. etc. 90 at A, 180 at S, 270 at E, 360 at P. The semidiameters AP, AAE, AS, A E, may be divided according to the tangents of half their arkes, that is a tangent of 45 gr. which is always equal to the Radius, shall give the semidiameter of 90 gr; a tangent of 40 gr. shall give 80 gr. in the semidiameter: a tangent of 35 gr. shall give 70. etc. So that the semidiameters may be divided in such sort as the tangent on the side of the Sector, the difference being only in their numbers. Having divided the circumference and the semidiameters, we may easily draw the meridians and the parallels by the help of the Sector. geometric illustration The meridians are to be drawn through both the poles P and S, and the degrees before graduated in the equator. The distance of the centre of each meridian from A the centre of the plane, is equal to the tangent of the same meridian, reckoned from the general meridian PAESE, and the semidiameter equal to the secant of the same degree. As for example, if I should draw the meridian PBS, which is the tenth from PAES, the tangent of 10 gr. giveth me AC, and the secant of 10 gr. giveth me SC, whereof C is the centre of the meridian PBS, and CS his semidiameter: so OF a tangent of 20 gr. showeth F to be the centre of PDS, the twentith meridian from PAES, and AGNOSTUS a tangent of 23 gr. 30 M. sheweth G to be the centre of P 69 S. etc. The parallels are to be drawn through the degrees, in AP, AS, and their correspondent degrees in the general meridian. The distance of the centre of each parallel from A the centre of the plane, is equal to the secant of the same parallel from the pole, and the semidiameter equal to the tangent of the same degree. As if I should draw the parallel of 80 gr. which is the tenth from the pole S, first I open the compasses unto AC the tangent of 10 gr. and this giveth me the semidiameter of this parallel, whose centre is a little from S, in such distance as the secant SC is longer than the radius SA. The meridians and parallels being drawn, if we number 23 gr. 30 m. from E to ♋ Northward, from A to ♑ Southward, the line drawn from ♋ to ♑ shall be the ecliptic: which being divided in such sort as the semidiameter AP, the first 30 gr. from A to ♋ shall stand for the sine of ♈; the 30 gr. next following for ♉; the rest for ♊. ♋. ♌. etc. in their order. If farther we have respect unto the latitude, we may number it from E Northward unto Z, and there place the zenith, by which and the centre, the line drawne ZAN shall represent the vertical circle, and the line MAH crossing it at right angles, shall represent the horizon; and these divided in the same sort as AP, the circles drawn through each degree of the semidiameter AZ, parallel to the horizon, shall be the circles of altitude: and the circles drawn through the horizon and his poles, shall give the azimuths. For example of this projection, let the place of the Sun be in the beginning of ♒, the parallel passing through this place is ♒ ☉ L; and therefore the meridian altitude ML, and the depression below the horizon at midnight H ♒, the semidiurnal ark L ☉, the seminocturnal ark O ♒, the declination ARE, the ascensional difference R ☉, the amplitude of ascension A ☉. Or if A be put to represent the pole of the world, then shall PAESE stand for the equator, and P ♋ S ♑ for the ecliptic, and the rest which before stood for meridians, may now serve for particular horizons, according to their several elevations. Then suppose the place of the Sun given to be 24 gr. of ♉, his longitude shall be PI, his right ascension PH, his declination HI. And if the place giuen be 19 gr. of ♌, his longitude shall be PK, his right ascension PN, his declination NK. Again, the declination brought to the horizon of the place, shall there show the ascentionall difference, amplitude of ascension, and the like conclusions of the globe. But I intent not here to show the use of the Astrolabe, but the use of the Sector in projection. And after this manner may a nocturnal be projected to show the hour of the night, whereof I will set down a type for the use of Seamen. geometric illustration It consists as you see of two parts, the one is a plane, divided equally according to the 24 hours of the day, and each hour into quarters or minutes, as the plane will bear: the line from the centre to XII, stands for the meridian, and XII stands for the hour of 12 at midnight. The other part is a rundle for such stars as are near the North pole, together with the twelve months, and the days of each month fitted to the right ascension of the stars. Those that have occasion to see the South pole, may do the like for the Southern constellations, and put them in a rundle on the back of this plane, and so it may serve for all the world. The use of this nocturnal is easy and ready. For look up to the pole, and see what stars are near the meridian, then place the rundle to the like situation, so the day of the month will show the hour of the night. 3 The Sphere may be projected in plano by circular lines, as in the particular Astrolabe of joh. Stophlerin, by help of the tangent, as before. For let the circle giuen represent the tropic of ♑, let it be divided into four parts, and crossed at right angles with AC the equinoctial colour, and MB the solstitial colour, and general meridian, the centre P representing the pole of the world. Let each quarter be divided into 90 gr. and so the whole into 360, beginning from A towards B. The meridian PM, or PB, may be divided according to the tangent of half his ark. So as the ark from the North pole to the tropic of ♑, being 90 gr. and 23 gr. 30 m. that is 113 gr. 80 m. and the half ark 56 gr. 45 m. the meridian shall be divided into 90 gr. and 23 gr. 30 m. in such sort as the tangent of 56 gr. 45 m. on the side of the Sector is divided into degrees and half degrees; of which PAE the ark of the equator 90 gr. from the pole, shall be given by the tangent of 45 gr. And P 69 the ark of the Summer tropic 66 gr. 30 m. from the pole, shall be given by the tangent of 33 gr. 15 m. And the circles drawn upon the centre P through A and ♋, shall be the equator, and the Summer tropic. Having the equator and both the tropiques, the ecliptic ♈ ♋ ♎ ♑ shall be drawn from the one tropic to the other, through the intersection of the equator and the equinoctial colour. And it may be divided first into the twelve Signs after this manner: PE the ark of the pole of the ecliptic 23 gr. 30 m. from the pole of the world, shall be given by the tangent of 11 gr. 45 m. The centre of the circle of longitude passing through this pole E ♈ and ♎, shall be found at D (somewhat below B) by the tangent of 66 gr. 30 m. Then through D draw an occult line parallel to AC, and divide it on each side from D, in such sort as the tangent is divided on the side of the Sector, allowing 45 gr. to be equal to DE. So the thirtieth degree from D toward the right hand, shall be the centre of the circle of longitude passing through E ♉ and ♏. The sixtith degree, the centre of ♊ E ♐. The thirtieth degree from D toward the left hand, the centre of ♓ E ♍. The sixtith, the centre of ♒ E ♌. And the other intermediate degrees shall be the centres to divide each Sign into 30 gr. If farther we have respect unto the latitude, we may (the meridian being before divided) number it from P Northward unto H, and there place the North intersection of the meridian and horizon: then the compliment of the latitude being numbered from P Southward unto Z, shall there give the zenith; and 90 gr. from Z Southward unto F, shall there give the South intersection of the meridian and horizon. The middle between F and H shall be G the centre of the horizon ♈ H ♎ F, passing through the beginning of ♈ and ♎, unless there be some former error. All parallels to the horizon may be found in like sort by their intersections with the meridian, and the middle between those intersections is always the centre. geometric illustration For example of this projection, let ☉ the place of the Sun giuen be 10 gr. of ♉: a right line drawn from P through this place unto the equator, shall there show his right ascension ♈ K, and his declination K ☉. Then may we on the centre P and semidiameter ☉ P, draw an occult parallel of declination, crossing the horizon in L and M, the meridian in G and N. So the right lines PL and PM produced, shall show the time of the Sun's rising and setting, ♈ Q the difference of ascension, ♎ R the difference of descension, ♈ L the amplitude of his rising, and ♎ M the amplitude of his setting. LGM showeth the length of the day, LN M the length of the night. ZG showeth his distance from the zenith at noon, HN his depression below the horizon at midnight. And then having the altitude of the Sun at any time of the day, the intersection of the parallel of altitude with the parallel of declination, showeth the azimuth, and a right line drawn from P through this intersection, giveth the hour of the day. 4 The Sphere may be projected in plano by circular lines, after the manner of the old concave hemisphere, by the help of the tangent on the side of the Sector. For let the circle giuen represent the plane of the horizon, let it be divided into four parts, and crossed at right angles with SN the meridian, and EV the vertical; so as S may stand for the South, N for the North, E the East, V the West part of the horizon, and the centre Z representeth the zenith. Let each quarter of the horizon be divided into 90 gr. and so the whole into 360 gr. beginning from N, and setting to the numbers of 10. 20. 30. etc. 90 at E, 180 at S, 270 at V, 360 at N. The semidiameters ZN, ZS, may be divided according to the tangent of half their arkes: So as the ark from the zenith to the horizon being 90 gr. and the half ark 45 gr. the semidiameters are to be divided in such sort as the tangent of 45 gr. as was showed before in the second projection. And if from Z we draw circles through each of these divisions, they shall be parallels of altitude. geometric illustration The hour circles may be here drawn as the azimuths in the third projection. For the centre of EPV, the hour of 6 will be found at B (somewhat near unto N) by the tangent of the latitude. And if through B we draw an occult line parallel unto EV, and divide it on each side from B, in such sort as the tangent is divided on the side of the Sector, allowing 45 gr. to be equal to BP, and 15 gr. for every hour: those divisions shall be the centres, and the distance from these divisions unto P, shall be the semidiameters, whereon to describe the rest of the hour circles. The ecliptic may be drawn as the equator. For the centre of that half which hath Southern declination, shall be given by the tangent of the altitude, which the Sun hath in his entrance into ♑. And the centre of the other half, by the tangent of his altitude, at his entrance into ♋. And it may be divided, as in the former projection, or else by tables calculated to that purpose. To these circles thus drawn, if we shall add the months of the year, and the days of each month, as we may well do, at the horizon, on either side between the tropiques; this projection shall be fitted for the most useful conclusions of the globe. For the day of the month being given, the parallel that shooteth on it, doth show what declination the Sun hath at that time of the year. And where this parallel crosseth the ecliptic, there is the place of the Sun. Or the place of the Sun being first given, the parallel which crosseth it shall at the horizon show the day of the month. Either of these then being given, or only the parallel of declination, we may follow it first unto the horizon, there the distance of 〈…〉 of the parallel from E or V, sheweth the amply 〈…〉 same among the hour circles showeth the time 〈…〉 ●●e rises or setteth. Then having the altitude 〈…〉 any time of the day, the intersection of the 〈…〉 on with the parallel of altitude, showeth 〈…〉 ●ay; and a right line drawn from Z through this intersection to the horizon, giveth the azimuth. Thus in either of these projections, that which is otherwise most troublesome, is easily done by the help of the tangent line: and what I have said of this line, the same may be wrought by scale and numbers out of the table of Tangents. CHAP. IU. Of the resolution of right-line Triangles. IN all Triangles there being six parts, viz. three angles and three sides, any three of them being given, the rest may be found by the Sector. As in a Rectangle triangle, 1 To find the base, both sides being given. Let the Sector be opened in the lines of Lines to a right angle, (as before was showed Cap. 2. Prop. 7.) then take out the sides of the triangle, and lay them, one on one line, the other on the other line, so as they meet in the centre, and mark how fare they extend. For the line taken from the terms of their extension, shall be the base required, viz. the side opposite to the right angle. Or add the squares of the two sides (as in Prop. 4. Superf.) and the side of the compound square shall be the base. 2 To find the base by having the angles, and one of the sides given. Take the side given, and turn it into the parallel sine of his opposite angle; so the parallel Radius shall be the base. 3 To find a side by having the base, and the other side given. Let the Sector be opened in the lines of lines to a right angle, and the side given laid on one of those lines from the centre; then take the base with a pair of compasses, and setting one foot in the term of the given side, turn the other to the other line of the Sector, and it shall there show the side required. Or take the square of the side out of the square of the base (as in Prop. 4. Superf.) and the side of the remaining square shall be the side required. 4 To find a side having the base and the angles given. Take the base given, and make it a parallel Radius, so the parallel sins of the angles, shall be the opposite sides required. 5 To find a side by having the other side and the angles given. Take the side given, and turn it into his parallel sine of his opposite angle; so the parallel sine of the compliment shall be the side required. 6 To find the angles by having the base and one of the sides given. First takeout the base given, and laying it on both sides of the Sector, so as they may meet in the centre, and mark how fare it extendeth. Then take out the lateral Radius, and to it open the Sector in the terms of the base. This done, take out the side given, and place it also on the same lines of the Sector from the centre. For the parallel taken in the terms of this side, shall be the sine of his opposite angle. Or take the base given, and make it a parallel Radius; then take the side given, and carry it parallel to the base, till it stay in like sins: so they shall give the quantity of the opposite angle. 7 To find the angles by having both the sides given. Take out the greater side, and lay it on both sides of the Sector, so as they meet in the centre, and mark how fare it extendeth. Then take the other side, and to it open the Sector in the terms of the greater side; so the parallel Radius shall be the tangent of the lesser angle. The third angle is always known by the compliment. 8 The Radius being given, to find the tangent, and secant of any ark. 9 The tangent of any ark being given, to find the tangent thereof, and the Radius. 10 The secant of any ark being given, to find the tangent thereof, and the Radius. The tangent, and the secant, together with the Radius of every ark, do make a right angle triangle; whose sides are the Radius and tangent, and the base always the secant; and the angles always known by reason of the given arkes. Wherefore the solution is the same with those before. In any rightlined triangle whatsoever, 11 To find a side by knowing the other two sides, and the angle contained by them. Let the Sector be opened in the lines of lines to the angle given, then take out the sides of the triangle, & laying them the one on the one line, the other on the other, so as they meet in the centre, mark how far they extend. For the line taken between the terms of their extension, shall be the third side required. 12 To find a side by having the other two sides, and one of the adjacent angles, so it be known which of the other angles is acute or obliqne. Let the Sector be opened in the lines of lines to the angle given, and the adjacent side laid on one of those lines from the centre; then take the other side with a pair of compasses, and setting one foot in the term of the former given side, turn the other to the other line of the Sector which here representeth the side required, and it shall cross it in two places; but with which of them is the term of the side required, must be judged by the angle. As if in the triangle following, the side AC being given, and the side CD and the angle GOD 18 gr. 40 m. it were required to find the side AD. First I open the Sector in the lines of lines to an angle of 18 gr. 40 m. and laying the adjacent side from the centre A, it extendeth to 800 in C. Then I take the other side CD with the compasses, and setting one foot in C, and turning the other to the other line of the Sector, I find that it doth cross it both in B and D; so that it is uncertain whither the side required be AB or AD, only it may be judged by the angle. For if the inward angle where they cross be obtuse, the side required is the lesser; if it be acute, it is the greater. 13 To find a side by having the angles and one of the other sides given. Take the side given, and turn it into the parallel sine of his opposite angle; so the parallel sins of the other angle shall be the opposite sides required. 14 To find the proportion of the sides by having the three angles. Take the lateral sins of the angles, and measure them in the line of lines. For the numbers belonging to those lines do give the proportion of the sides. 15 To find an angle by knowing the three sides. Let the two containing sides be laid, on the lines of the Sector from the centre, one on one line, and the other on the other; and let the third side, which is opposite to the angle required, be fitted over in their terms: so shall the Sector be opened in those lines to the quantity of the angle required. The quantity of this angle is found as in Cap. 2. Prop. 8. 16 To find an angle by having two sides and one adjacent angle. First take out the side opposite to the angle given, and laying it on both sides of the Sector, so as they meet in the centre, mark how fare it extendeth; then take out the lateral sine of the angle, and to it open the Sector in the terms of the first side: this done, take out the other side given, and place it also on the same lines of the Sector from the centre, for the parallels taken in the terms of this side, shall be the sine of the angle opposite to the second side. Or take out the side opposite to the angle given, and make it a parallel sine of that angle; then take the other side given and carry it parallel to the former, till it stay in like sins, so they shall give the quantity of the angle opposite to the second side. 17 To find an angle by having two sides, and the angle contained by them. First find the third side by the 11. Prop. and then the angles may be found by the 15. or 16. Prop. For practise in each of these cases, we may use the examples following, wherein CEA, CEB, CED are rectangle in E; the rest consist of obliqne angles. CAB 18 gr. 40 m. ABC 126 52 ACB 34 28 ACD 108 12 ADC 53 8 BCD 73 44 geometric illustration For observation of angles, the Sector may have sights set on the movable foot; so that by looking through them, the edges of the Sector may be applied to the sides of the angle. For measuring of the sides of lesser triangles, any scale may suffice, either of feet, or inches, or lesser parts. But for greater triangles, especially for plotting of grounds, I hold it fit to use a chain of four perches in length, divided into an hundred links. For so the length being multiplied into the breadth, the five last figures give the content in roods and perches by this Table; the other figures toward the left hand, do show the number of acres directly. Links R P 100000 4 0 90000 3 24 80000 3 8 70000 2 32 60000 2 16 50000 2 0 40000 1 24 30000 1 8 20000 32 10000 16 9375 15 8750 14 8125 13 7500 12 6875 11 6250 10 5625 9 5000 8 4375 7 3750 6 3125 5 2500 4 1875 3 1250 2 625 1 Ass if in the former triangle ACD, the length AD be 9 chains and 50 links, the breadth CE be 2 chains and 56 links; these multiplied give the content for the long square 2.43200, the half whereof for the triangle is 1.21600, that is 1 acre, 21600 parts of 100000, of which last five figures, 20000 give 32 perches, and the remainder 1600 give better than two perches more. CHAP. V Of the resolution of spherical triangles. FOr our practice in spherical triangles, let A be the equinoctial point, AB an ark of the ecliptic representing the longitude of the Sun in the beginning of ♉, BC an ark of the declination from the Sun to the equator, and AC an ark of the equator representing the right ascension. geometric illustration Let BD be an ark of the horizon representing the amplitude of the Sun's rising from the East, and BE an ark of the horizon for his setting from the West: so DC shall be the difference of ascension, and CE the difference of descension; AD the obliqne ascension, and A the obliqne descension of the same place of the Sun in our latitude at Oxford of 51 gr. 45 m. whose compliment 38 gr. 15 m. is the angle at E and D. The triangles ACB, D CB, ECB, are rectangle in C: the other AD B, AEB, consist every way of obliqne angles. geometric illustration Or to fit an example nearer to the latitude of London. Let ZPS represent the zenith pole and Sun, ZP being 38 gr. 30 m. the compliment of the latitude, PS 70 gr. the compliment of the declination, and ZS 40 gr. the compliment of the Sun's altitude. The angle at Z shall show the azimuth, and the angle at P, the hour of the day from the meridian. Then if from Z to PS we let down a perpendicular ZR, we shall reduce the obliqne triangle into two rectangle triangles ZRP, ZRS. Or if from S to ZP we set down a perpendicular SM, we shall reduce the same ZPS into two other triangles, SMZ, SIMP, rectangle at M: whatsoever is said of any of these triangles, the same holdeth for all other triangles in the like cases. For the resolution of each of these, there be several ways. I only choose those which are fittest for the Sector, wherein if that be remembered which before is showed in the general use of the Sector concerning lateral and parallel entrance, it may suffice only to set down the proportion of the three parts given to the fourth required, and so I show first by the sins alone. In a rectangle triangle 1 To find a side by knowing the base, and the angle opposite to the required side. As the Radius is to the sine of the base: So the sine of the opposite angle to the sine of the side required. As in the rectangle ACB, having the base AB, the place of the Sun 30 gr. from the equinoctial point, and the angle BAC of 23 gr. 30 m. the greatest declination, if it were required to find the side BC the declination of the Sun. Take either the lateral sine of 23 gr. 30 m. and make it a parallel Radius; so the parallel sine of 30 gr. taken and measured in the side of the Sector, shall give the side required 11 gr. 30 m. Or take the sine of 30 gr. and make it a parallel Radius; so the parallel sine of 23 gr. 30 m. taken and measured in the lateral sins, shall be 11 gr. 30 m. as before. So in the triangle ZPS having ZP 38 gr. 30 m. and the angle P 31 gr. 34 m. given, we shall find the perpendicular Z R to be 19 gr. 1 m; or having PS 70 gr. and the said angle P 31 gr. 34 m. given, we may find the perpendicular SM to be 29 gr. 28 m. 2 To find a side by knowing the base and the other side. As the sine of the compliment of the side given is to the Radius: So the sine of the compliment of the base to the sine of the compliment of the side required. So in the rectangle ACB, having AB 30 gr. and BC 11 gr. 30 m. given, the side AC will be found 27 gr. 54 m. Or in the rectangle ZRP having ZP 38 gr. 30 m. and Z R 29 gr. 1 m. given, the side RP will be found 34 gr. 7 m. 3 To find a side by knowing the two obliqne angles. As the sine of either angle to the sine of the compliment of the other angle: So is the Radius to the sine of the compliment of the side opposite to the second angle. So in the rectangle ACB, having CAB for the first angle 23 gr. 30 m. and ABC for the second 69 gr. 21 m. the side AC will be found 27 gr. 54 m. Or making ABC the first angle, and CAB the second, the side BC will be found 11 gr. 30 m. 4 To find the base by knowing both the sides. As the Radius to the sine of the compliment of the one side: So the sine of the compliment of the other side, to the sine of the compliment of the base required. So in the rectangle ACB having AC 27 gr. 54 m. and BC 11 gr. 30 m. the base AB will be found 30 gr. 5 To find the base by knowing the one side, and the angle opposite to that side. As the sine of the angle given, to the sine of the side given: So is the Radius to the sine of the base required. So in the rectangle BCD, knowing the latitude and the declination, we may find the amplitude; as having BC the side of the declination 11 gr. 30 m. and BDC the angle of the compliment of the latitude 38 gr. 15 m. the base BD which is the amplitude, will be found to be 18 gr. 47 m. 6 To find an angle by the other obliqne angle, and the side opposite to the inquired angle. As the Radius to the sine of the compliment of the side: So the sine of the angle given, to the sine of the compliment of the angle required. So in the rectangle ACB, having the angle BAC 23 gr. 30 m. and the side AC 27 gr. 54 m. the angle ABC will be found 69 gr. 21 m. 7 To find an angle by the other obliqne angle, and the side opposite to the angle given. As the sine of the compliment of the side to the side of the compliment of the angle given: So is the Radius to the sine of the angle required. So in the rectangle ACB, having BAC 23 gr. 30 m. and BC 11 gr. 30 m. the angle ABC will be found 69 gr. 21 m. 8 To find an angle by the base, and the side opposite to the inquired angle. As the sine of the base is to the Radius: So the sine of the side to the sine of the angle required. So in the rectangle BCD, having BD 18 gr. 47 m. and BC 11 gr. 30 m. the angle BDC will be found 38 gr. 15 m. These eight Propositions have been wrought by the sins alone; those which follow require joint help of the tangent. And forasmuch as the tangent could not well be extended beyond 63 gr. 30 m. I shall set down two ways for the resolution of each Proposition; if the one will not hold, the other may. 9 To find a side by having the other side, and the angle opposite to the inquired sine. 1 As the Radius to the sine of the side given: So the tangent of the angle, to the tangent of the side required. 2 As the sine of the side given, is to the Radius: So the tangent of the compliment of the angle, to the tangent of the compliment of the side required. So in the rectangle ACB, having the right side AC 27 gr. 54 m, and the angle BAC 23 gr. 30 m. the side BC will be sound to be 11 gr. 30 m. 10 To find a side, by having the other side, and the angle adjacent next to the inquired side. 1 As the tangent of the angle, to the tangent of the side given: So is the Radius to the sine of the side required. 2 As the tangent of the compliment of the side, to the tangent of the compliment of the angle: So is the Radius to the sine of the side required. This and the like, where the tangent standeth in the first place, are best wrought by parallel entrance. And so in the rectangle BCD, having BC the side of declination 11 gr. 30 m. and BDC the angle of the compliment of the latitude 38 gr. 15 m. the side DC, which is the ascensional difference, will be found 14 gr. 57 m. By the ascensional difference is given the time of the Sun's rising and setting, and length of the day; allowing an hour for each 15 gr. and 4 minutes of time for each several degree. As in the example the difference between the Sun's ascension in a right sphere, which is always at 6 of the clock, and his ascension in our latitude being 14 gr. 57 m. it showeth that the Sun riseth very near an hour before 6, because of the Northern declination; or after 6, if the Sun be declining to the Southward. 11 To find a side by knowing the base, and the angle adjacent next to the inquired side. 1 As the Radius to the sine of the compliment of the angle: So is the tangent of the base, to the tangent of the side required. 2 As the sine of the compliment of the angle is to the Radius: So the tangent of the compliment of the base, to the tangent of the compliment of the side required. So in the rectangle ACB, knowing the place of the Sun from the next equinoctial point, and the angle of his greatest declination, we may find his right ascension: viz. the base AB 30 gr. and the angle BAC 23 gr. 30 m. being given, the right ascension AC will be found 27 gr. 54 m. 12 To find the base by knowing the obliqne angles. As the tangent of the one angle, to the tangent of the compliment of the other angle: So is the Radius to the sine of the compliment of the base. So in the rectangle ACB, having BAC 23 gr. 30 m. and ABC 69 gr. 21 m. the base AB will be found 30 gr. 13 To find the base, by one of the sides, and the angle adjacent next that side. 1 As the Radius is to the sine of the compliment of the angle: So the tangent of the compliment of the side, to the tangent of the compliment of the base. 2 As the sine of the compliment of the angle is to the Radius: So the tangent of the side given, to the tangent of the base required. So in the rectangle ACB, having AC 27 gr. 54 m. and BAC 23 gr. 30 m. the base AB will be found 30 gr. 0 m. 14 To find an angle, by knowing both the sides. 1 As the Radius is to the sine of the side next the inquired angle: So the tangent of the compliment of the opposite side, to the tangent of the compliment of the angle required. 2 As the sine of the side next the inquired angle, is to the Radius: So the tangent of the opposite side, to the tangent of the angle required. So in the rectangle ACB, having AC 27 gr. 54 m. and BC 11 gr. 30 m. the angle at A will be found 23 gr. 30 m. and the angle at B 69 gr. 21 m. 15 To find an angle, by the base, and the side adjacent to the inquired angle. 1 As the tangent of the compliment of the side, to the tangent of the compliment of the base: So is the Radius to the sine of the compliment of the angle required. 2 As the tangent of the base, to the tangent of the side: So is the Radius, to the sine of the compliment of the angle required. So in the rectangle BCD, having the base BD 18 gr. 47 m. and the side BC 11 gr. 30 m. the angle DBC between them will be found 53 gr. 15 m. 16 To find an angle, by knowing the other obliqne angle, and the base. 1 As the Radius, to the sine of the compliment of the base: So the tangent of the angle given, to the tangent of the compliment of the angle required. 2 As the sine of the compliment of the base, is to the Radius: So the tangent of the compliment of the angle given, to the tangent of the angle required. So in the rectangle ACB, having the angle at A 23 gr. 30 m. and the base AB 30 gr. the angle ABC will be found 69 gr. 21 m. These sixteen cases are all that can fall out in a rectangle triangle: those which follow do hold In any spherical triangle whatsoever 17 To find a side opposite to an angle given, by knowing one side, and two angles, whereof one is opposite to the given side, the other to the side required. As the sine of the angle opposite to the side given, is to the sine of that side given: So the sine of the angle opposite to the side required, to the sine of the side required. So in the triangle ABE, having the place of the Sun, the latitude, and the greatest declination, we may find the amplitude. As having AB 30 gr. BAE 23 gr. 30 m. and AEB 38 gr. 15 m. the side BE which is the amplitude, will be found 18 gr. 47 m. 18 To find an angle opposite to a side given, by having one angle and two sides, the one opposite to the given angle, the other to the angle required. As the sine of the side opposite to the angle given, is to the sine of that angle given: So the sine of the side opposite to the angle required, to the sine of the angle required. So in the triangle ZPS, having the azimuth, and latitude, and declination, we may find the hour of the day. As having PZS 130 gr. 3 m. PS 70 gr. and ZS 40 gr. the angle ZPS, which showeth the hour from the meridian shall be found 31 gr. 34 m. 19 To find an angle by knowing the three sides. This proposition is most useful, but most difficult of all others: as in Arithmetic, so by the Sector, yet may it be performed several ways. 1 According to Regiomontanus and others. As the sine of the lesser side next the angle required, to the difference of the versed sins of the base and difference of the sides▪ So is the Radius to a fourth proportional. Then as the sine of the greater side next the angle required is to that fourth proportional: So is the Radius to the versed sine of the angle required. So in the triangle ZPS, having the side PS, the compliment of the declination 70 gr. 0 m. the side ZP the compliment of the latitude 38 gr. 30 m. and the base ZS the compliment of the altitude 40 gr. the angle of the hour of the day ZPS will be found 31 gr. 34 m. which is 2 h. 6 m. from the meridian. For the base being 40 gr. 0 m. and the difference of the sides 38 gr. 30 m. and 70 gr. 0 m. being 31 gr. 30 m. the difference of their versed sins will be the same with the distance between the right sine of 50 gr. and 58 gr. 30 m. This difference I take out, and make it a parallel sine of the lesser side 38 gr. 30 m. so the parallel Radius will be the fourth proportional. Then coming to the second operation, I make this fourth proportional, a parallel sine of the greater side of 70 gr. 0 m. and take out his parallel Radius. For this measured from 90 gr. toward the centre, will be the versed sine of 31 gr. 34 m. In the like sort in the same triangle Z PS, having the same compliments given, the angle PZ S which is the azimuth from the North part of the meridian, will be found 130 gr. 3 m. For here the base opposite to the angle required being 70 gr. and the difference of the sides 38 gr. 30 m. and 40 gr. being 1 gr. 30 m. the difference of their versed sins will be the same with the distance between the right sins of 20 gr. and 88 gr. 30 m. This difference I take, and make it a parallel sine of the lesser side 38 gr. 30 m. so the parallel Radius will be the fourth proportional. Then coming to the second operation, I make this fourth proportional a parallel sine of the greater side 40 gr. and take out his parallel Radius. For this measured from 90 gr. beyond the centre in the lines of sins stretched forth at their full length, will be the versed sine of 130 gr. 3 m. 2 I may find an angle by knowing three sides, by that which I have elsewhere demonstrated upon Barth. Pitiscus, and that at one operation in this manner. As the sine of the greater side is to the secant of the compliment of the other side: So the difference of sins of the compliment of the base, and the ark compounded of the lesser side with the compliment of the greater, to the versed sine of the angle required. So in the same triangle ZPS, having the same compliments given, the angle at P, which showeth the hour from the meridian, will be found as before 31 gr. 34 m. For the sides being 38 gr. 30 m. and 70 gr. 0 m. I take the secant of the compliment of 38 gr. 30 m. and make it a parallel sine of 70 gr; then keeping the Sector at this angle, I consider that the compliment of 70 gr. being 20 gr. added unto 38 gr. 30 m. the compounded side (which is here the meridian altitude) will be 58 gr. 30 m; and that the base being 40 gr. the difference of sins of the compounded side and the compliment of the base will be (as before) the distance between the sins of 50 gr. and 58 gr. 30 m. Wherefore I take out this difference, and lay it on both the lines of sins from the centre: so the parallel taken in the terms of this difference, and measured from 90 gr. toward the centre, doth give the versed sine of 31 gr. 34 m. The other angel's PZ S, PSZ, may be found in the same sort; but having the sides and one angle, it will be sooner done by that which we shown before in the 18. Prop. 20 To find a side by knowing the three angles. If for the greater angle we take his compliment to 180 gr. the angles shall be turned into sides, and the sides into angles, & the operation shall be the same, as in the former Prop. 21 To find a side, by having the other two sides, and the angle comprehended. This proposition being the converse of the nineteenth, may be wrought accordingly; but the best way both for it and those which follow, is to resolve them into two rectangles, by letting down a perpendicular, as was showed in the first Prop. So in the triangle Z PS, having Z P the compliment of the latitude, and PS the compliment of the declination, with Z PS the angle of the hour from the meridian, we may find Z S the compliment of the altitude of the Sun. For having let down the perpendicular Z R by the first Prop. we have two triangles, Z RP, Z RS, both rectangle at R. Then may we find the side PR, either by the second, or tenth, or eleventh Prop; which taken out of PS, leaveth the side RS: with this RS and Z R we may find the base Z S by the fourth Prop. Or having let down the perpendicular SM, we have two rectangle triangles SMZ, SIMP. Then may we find MP, from which if we take Z P, there remaineth MZ: but with MZ and SM, we may find the base Z S. 22 To find a side, by having the other two sides, and one of the angles next the inquired side. So in the triangle ZPS, having Z P the compliment of the latitude, and PS the compliment of the declination, with PZ S the angle of the azimuth, we may find Z S the compliment of the altitude of the Sun. For having Z P, and the angle at Z, we may to SZ produced, let down a perpendicular PV. Then we have two rectangle triangles, PUZ, PUS, wherein if we find the sides US, US, and take the one out of the other, there will remain the side inquired Z S. 23 To find a side, by having one side, and the two angles next the inquired side. So in the triangle ABDELLA, having AB the place of the sun, and BAD the angle of the greatest declination, and ADB the angle of the equator with the horizon, we may find AD the obliqne ascension. For having let down BC the perpendicular of declination, we have two rectangle triangles, ACB, DCB. Then may we find AC the right ascension, and DC the ascentionall difference; and comparing the one with the other, there remaineth AD. 24 To find a side, by having two angles, and the side enclosed by them. So in the triangle ZPS, having the angles at Z and P, with the side intercepted ZP, we may find the side PS. For having let down the perpendicular PV, we have two rectangles PUZ, PUS. Then may we find the angle VPZ, ei- by the seventh, or fifteenth, or sixteenth Prop. which added to ZPS, maketh the angle VPS: with this VPS and PV, we may find the base PS, according to the 13 Prop. 25 To find an angle by having the other two angles and the side enclosed by them. So in the triangle ZPS, having the angles at Z and P, with the side intercepted ZP, we may find the other angle Z SP. For having let down the perpendicular ZR, we have two rectangles ZRP, ZR S. Then may we find the angle PZR by the sixteenth Prop. and that compared with PZ S, leaveth the angle RZ S: with this RZS and ZR we may find the angle required ZSR, according to the sixth Prop. 26 To find an angle, by having the other two angles, and one of the sides next the inquired angle. So in the triangle ABDELLA, having the angles at A and D, with the side AB, we may find the angle ABDELLA. For having let down the perpendicular BC, we have two rectangles, ACB, DCB. Then may we find the angles ABC, DBC, and take DBC out of ABC; for so there remaineth the angle required ABDELLA. 27 To find an angle, by knowing two sides, and the angle contained by them. So in the triangle ZPS, having the sides ZP, PS, with the angle comprehended ZPS, we may find the angle PZS. For having let down the perpendicular SM, we have two rectangles SMZ, SIMP. Then may we find the side MP, and taking ZP out of MP, there remaineth MZ: with this MZ and the perpendicular MS, we may find the angle MZS, by the fourteenth Prop. This angle MZ S, taken out of 180 gr. there remaineth PZ S. 28 To find an angle by knowing the two sides next it, and one of the other angles. So in the triangle ZPS, having the sides ZP and PS, with the angle PZS, we may find the angle Z PS. For having let down the perpendicular PV, we have two rectangles PUZ, PUS. Then may we find the angles VPZ, VPS; and taking UP Z out of VPS, there remaineth ZPS, which was required. These 28 cases are all that can fall out in any spherical triangle: if any do not presently understand them, let them once more read over the use of the globes, and they shall soon become easy unto them▪ CHAP. VI. Of the use of the Meridian line in Navigation. THe Meridian line is here set on the side of the Sector, stretched forth at full length, on the same plane with the line of lines and Solids, and is divided unequally toward 87 gr. (whereof 70 gr. are about one half) in such sort as the Meridian in the cart of Mercators' projection. The use of it may be 1 To divide a sea-chart according to Mercators' projection. If a degree of the equator on the sea-chart be equal to the hundred part of the line of lines in the Sector, the degrees of the Meridian upon the Sector, shall give the like degrees upon the sea-chart: if otherwise they be unequal, then may the meridians of the sea-chart be divided in such sort as the line of Meridian's is divided on the Sector, by that which we shown before in the 8 Prop. of the line of lines. But to avoid error, I have here set down a Table, whereby the Meridian line may be divided out of the degrees of the equator, supposing each degree to be subdivided into a thousand parts. By which Table, & the usual Table of Sines, Tangents and Secants, the proportions following may be also resolved arithmetically. For the manner of division, let the equator (or one of the parallels if it be a particular chart) be drawn, and divided, and crossed with parallel meridians, as in the common sea-chart: then look into the Table, and let the distance of 40 gr. in the meridian, from the equator, be equal to 43 gr. 711 parts of the equator; let 50 gr. in the meridian from the equator, be equal to 57 gr. 909 parts of the equator; and so in the rest. M Gr Par M Gr Par M Gr Par M Gr Par M Gr Par 0 0 0 3 3 001 6 6 011 9 9 037 12 12 088 100 3 101 6 111 9 138 12 190 200 3 201 6 212 9 239 12 293 300 3 301 6 312 9 341 12 395 400 3 402 6 413 9 442 12 497 500 3 502 6 514 9 543 12 600 600 3 602 6 614 9 645 12 702 700 3 702 6 715 9 746 12 805 800 3 803 6 816 9 848 12 907 900 3 903 6 916 9 949 13 010 1 1 000 4 4 003 7 7 017 10 10 051 13 13 112 1 100 4 103 7 118 10 152 13 215 1 200 4 204 7 219 10 254 13 318 1 300 4 304 7 319 10 355 13 421 1 400 4 404 7 420 10 457 13 523 1 500 4 504 7 521 10 559 13 626 1 600 4 605 7 622 10 661 13 729 1 700 4 705 7 723 10 762 13 832 1 800 4 805 7 824 10 864 13 935 1 900 4 906 7 925 10 966 14 038 2 2 000 5 5 006 8 8 026 11 11 068 14 14 141 2 100 5 106 8 127 11 170 14 244 2 200 5 207 8 228 11 272 14 347 2 300 5 307 8 329 11 374 14 450 2 400 5 408 8 430 11 476 14 553 2 500 5 508 8 531 11 578 14 656 2 601 5 609 8 632 11 680 14 760 2 701 5 709 8 733 11 782 14 863 2 801 5 810 8 834 11 884 14 967 2 901 5 910 8 936 11 986 15 070 3 3 001 6 6 011 9 9 037 12 12 088 15 15 174 M Gr Par M Gr Par M Gr Par M Gr part M Gr part 15 15 174 18 18 303 21 21 486 24 24 734 27 28 058 15 277 18 408 21 593 24 844 28 171 15 381 18 513 21 701 24 953 28 283 15 485 18 619 21 808 25 063 28 396 15 588 18 724 21 915 25 173 28 508 15 692 18 830 21 023 25 282 28 621 15 796 18 935 22 130 25 392 28 734 15 900 19 041 22 238 25 502 28 847 16 004 19 146 22 345 25 613 28 959 16 107 19 251 22 453 25 723 29 072 16 16 211 19 19 356 22 22 561 25 25 833 28 29 186 16 316 19 463 22 669 25 943 29 299 16 420 19 569 22 777 26 054 29 413 16 524 19 675 22 885 26 164 29 526 16 628 19 781 22 993 26 275 29 640 16 732 19 887 23 101 26 386 29 753 16 836 19 993 23 210 26 497 29 867 16 941 20 100 23 318 26 608 29 981 17 045 20 206 23 427 26 719 30 095 17 150 20 312 23 535 26 830 30 300 17 17 255 20 20 419 23 23 643 26 26 941 29 30 324 17 359 20 525 23 752 27 052 30 438 17 464 20 632 23 861 27 164 30 553 17 568 20 738 23 970 27 275 30 667 17 673 20 845 24 079 27 387 30 782 17 778 20 952 24 188 27 499 30 897 17 883 21 059 24 297 27 610 31 012 17 988 21 165 24 406 27 722 31 127 18 093 21 272 24 515 27 834 31 242 18 198 21 379 24 624 27 946 31 357 18 18 303 21 21 486 24 24 734 27 28 058 30 31 473 M Gr Par M Gr Par M Gr Par M Gr Par M Gr Par 30 31 473 33 34 992 36 38 633 39 42 415 42 46 362 31 588 35 111 38 757 42 544 46 496 31 704 35 231 38 880 42 673 46 631 31 820 35 350 39 004 42 802 46 766 31 936 35 470 39 129 42 931 46 902 32 052 35 590 39 253 43 061 47 037 32 168 35 710 39 377 43 191 47 173 32 284 35 830 39 502 43 320 47 309 32 400 35 950 39 627 43 451 47 445 32 517 36 071 39 752 43 581 47 581 31 32 633 34 36 191 37 39 877 40 43 711 43 47 718 32 750 36 312 40 002 43 842 47 855 32 867 36 433 40 128 43 973 47 992 32 984 36 554 40 253 44 104 48 129 33 101 36 675 40 379 44 235 48 267 33 218 36 796 40 505 44 366 48 404 33 336 36 917 40 631 44 498 48 542 33 453 37 039 40 757 44 630 48 681 33 571 37 161 40 884 44 762 48 819 33 688 37 283 41 011 44 894 48 958 32 33 806 35 37 405 38 41 137 41 45 026 44 49 097 33 924 37 527 41 264 45 159 49 236 34 042 37 649 41 392 45 292 49 375 34 161 37 771 41 519 45 425 49 515 34 279 37 894 41 646 45 558 49 655 34 397 38 017 41 774 45 691 49 795 34 516 38 140 41 902 45 825 49 935 34 635 38 263 42 030 45 959 50 076 34 754 38 386 42 158 46 093 50 217 34 873 38 509 42 287 46 227 50 358 33 34 992 36 38 633 39 42 415 42 46 362 45 50 499 M Gr Par M Gr Par M Gr Par M Gr part M Gr part 45 50 499 48 54 860 51 59 481 54 64 412 57 69 711 50 641 55 010 59 640 64 582 69 895 50 783 55 160 59 800 64 753 70 080 50 925 55 310 59 960 64 924 70 263 51 068 55 460 60 120 65 096 70 449 51 210 55 611 60 280 65 268 70 635 51 353 55 762 60 441 65 440 70 821 51 496 55 913 60 602 65 613 71 008 51 639 56 065 60 763 65 786 71 195 51 783 56 217 60 925 65 960 71 383 46 51 927 49 56 369 52 61 088 55 66 134 58 71 572 52 071 56 522 61 250 66 308 71 761 52 215 56 675 61 413 66 483 71 950 52 360 56 828 61 577 66 659 72 140 52 505 56 981 61 740 66 835 72 331 52 650 57 135 61 904 67 011 72 522 52 795 57 289 62 069 67 188 72 714 52 941 57 444 62 234 67 365 72 906 53 087 57 598 62 399 67 543 73 099 53 233 57 754 62 564 67 721 73 292 47 53 380 50 57 909 53 62 730 56 67 900 59 73 486 53 526 58 065 62 897 68 079 73 680 53 673 58 221 63 063 68 258 73 875 53 821 58 377 63 231 68 438 74 071 53 968 58 534 63 398 68 618 74 267 54 116 58 691 63 566 68 799 74 464 54 264 58 848 63 734 68 981 74 661 54 413 59 006 63 903 69 163 74 859 54 562 59 164 64 072 69 345 75 057 54 711 59 322 64 242 69 528 75 256 48 54 860 51 59 481 54 64 412 57 69 711 60 75 456 M Gr Par M Gr Par M Gr Par M Gr Par M Gr. Par 60 75 450 63 81 749 66 88 725 69 96 575 72 105 579 75 656 81 970 88 971 96 854 105 904 75 857 82 191 89 219 97 135 106 230 76 059 82 413 89 467 97 418 106 558 76 201 82 635 89 716 97 701 106 888 76 464 82 860 89 967 97 986 107 220 76 667 83 084 90 218 98 272 107 553 76 871 83 310 90 470 98 560 107 888 77 076 83 536 90 723 98 849 108 226 77 281 83 763 90 978 99 139 108 565 61 77 487 64 83 990 67 91 232 70 99 431 73 108 906 77 694 84 219 91 489 99 724 10 249 77 901 84 448 91 746 100 018 109 594 78 109 84 678 92 005 100 314 109 941 78 317 84 909 92 264 100 612 110 290 78 526 85 141 92 525 100 910 110 641 78 736 85 374 92 787 101 211 110 994 78 947 85 607 93 050 101 513 111 349 79 158 85 842 93 314 101 816 111 707 79 370 86 077 93 579 102 121 112 066 62 79 583 65 86 313 68 93 846 71 102 427 74 112 428 79 796 86 550 94 113 102 735 112 792 80 010 86 788 94 382 103 044 113 158 80 225 87 027 94 652 103 356 113 526 80 441 87 267 94 923 103 668 113 897 80 657 87 508 95 195 103 983 114 270 80 874 87 749 95 468 104 299 114 645 81 091 87 992 95 743 104 616 115 023 81 310 88 235 96 019 104 936 115 403 81 529 88 480 96 296 105 257 115 786 63 81 749 66 88 725 69 96 575 72 105 579 75 116 171 M Gr. Par M Gr. Par M Gr. Par M Gr. Par M Gr. part 75 116 171 78 129 075 81 145 650 84 168 947 87 208 705 116 559 129 558 146 292 169 912 210 649 116 949 130 045 146 942 170 893 212 668 117 342 130 536 147 600 171 891 214 745 117 737 131 031 148 265 172 907 216 909 118 135 131 530 148 937 173 941 219 158 118 536 132 034 149 618 174 994 221 498 118 939 132 542 150 307 176 067 223 938 119 345 133 055 151 003 177 160 226 486 119 755 133 572 151 709 178 275 229 153 76 120 160 79 134 094 82 152 423 85 179 411 88 231 950 120 581 134 620 153 147 180 569 234 891 121 000 135 151 153 878 181 752 237 991 121 420 135 687 154 620 182 960 241 268 121 843 136 228 155 372 184 194 244 744 122 270 136 775 156 132 185 454 248 445 122 700 137 326 156 903 186 743 252 402 123 133 137 883 157 685 188 062 256 652 123 570 138 445 158 478 189 411 261 243 124 009 139 012 159 281 190 793 266 235 77 124 452 80 139 585 83 160 096 86 192 210 89 271 705 124 898 140 164 160 922 193 661 277 753 125 348 140 748 161 761 195 151 284 517 125 801 141 339 162 612 196 680 292 191 126 258 141 936 163 475 198 251 301 058 126 718 142 538 164 352 199 867 311 563 127 182 143 147 165 242 201 529 324 455 127 649 143 763 166 146 203 240 341 166 128 121 144 385 267 065 205 005 365 039 128 596 145 014 167 999 206 825 408 011 78 129 075 81 145 650 84 168 947 87 208 705 90 Infinite geometric illustration geometric illustration If any desire to have his chart to agree with his Sector, he may make each degree of longitude equal to the tenth part of the line of lines, and divide the meridian of his chart out of the Sector: so shall each degree of the chart, be ten times as large as the like degree on the Sector, and the work be easy from the one to the other. 2 To find how many leagues answer to one degree of longitude in every several latitude. In sailing by the compass, the course holds sometime upon a great circle, sometime upon a parallel to the equator; but most commonly upon crooked lines winding towards one of the poles, which lines are well known by the name of Rumbs. If the course hold upon a great circle, it is either North or South, under some meridian, or East or West, under the equator. And in these cases, every degree requires an allowance of twenty leagues, every twenty leagues will make a degrees difference in the sailing: so that here needs no further precept than the rule of proportion in the Chapter of lines. But if the course hold East or West, on any of the parallels to the equator; As the Radius is to twenty leagues, the measure of one degree at the equator: So the sine of the compliment of the latitude to the measure of leagues answering to one degree in that latitude. Wherefore I take 20 leagues out of the line of lines, and make it a parallel Radius, by fitting it over in the sins of 90 and 90: so his parallel sine taken out of the compliment of the latitude, and measured in the line of lines, shall show the number of leagues required. Thus in the latitude of 18 gr. 12 m. we shall find 19 leagues answering to one degree of longitude, and 18 leagues in the latitude of 25 gr. 15 m. and as in this Table. Gr. 1 Lg 0 0 20 18 12 19 25 15 18 31 48 17 36 52 16 41 25 15 45 34 14 49 28 13 53 8 12 56 38 11 60 0 10 63 15 9 66 25 8 69 30 7 72 32 6 75 31 5 78 28 4 81 23 3 84 15 2 87 8 1 This may be done more readily without opening the Sector, by doubling the sine of the compliment of the latitude, as may appear in the same example. It may also be done by the line of meridians, either upon the Sector, or upon the chart. For if we open a pair of compasses to the quantity of one degree of longitude in the equator, and measure it in the meridian line, setting one foot as much above the latitude given, as the other falleth beneath it, so that the latitude may be in the middle between the feet of the compasses, the number of leagues intercepted shall be that which was required. But if the course hold upon any of the rumbs, between a parallel of the equator and the meridian, we are to consider beside the quarter of the world to which we tend, which must be always known. 1 The difference of longitude at least in general. 2 The difference of latitude, and that in particular. 3 The rumb whereon the course holds. 4 The distance upon the rumb, which is the distance, which we are here to consider, and is always somewhat greater than the like distance upon a greater circle. And for these first I show in general this third Prop. 3 To find how many leagues do answer to one degree of latitude in every several Rumb. As the sine of the compliment of the rumb from the meridian, is to 20 leagues the measure of one degree at the meridian: So the Radius to the leagues answering to one degree upon the Rumb. Wherefore I take 20 leagues out of the line of lines, and make it a parallel sine of the compliment of the Rumb from the meridian; so his parallel Radius taken and measured in the line of lines, shall show the number of leagues required. Thus in the first Rumb from the meridian, we shall find 20 lgs 39 parts answering to one degree of latitude, and 21 lgs 65 parts in the second Rumb, etc. as in this Table, where we subdivide each league into a hundred parts, and show beside what inclination the rumb hath to the meridian. This may be done more readily without opening the Sector, by doubling the secant of the latitude, as may appear in the same example. It may also be done upon the chart, if we take the distance upon the Rumb between two parallels, and measure it in the meridian line, as fare above the greater latitude as beneath the lesser. For so the number of leagues intercepted, shall be that which was required. This considered in general, I show more particularly in twelve Prop. following, how of these four any two being given, the other two may be found, both by Mercators' chart, and by this Sector. Rumbs. Inclination to the meridian Number of leagues. Gr. Mi. Lgs Par 2 49 20 02 5 37 20 10 8 26 20 22 1 11 15 20 39 14 4 20 62 16 52 20 90 19 41 21 24 2 22 30 21 65 25 19 22 12 28 7 22 68 30 56 23 32 3 33 45 24 05 36 34 24 90 39 22 25 87 42 11 26 99 4 45 0 28 28 47 49 29 78 50 37 31 52 53 26 33 57 5 56 15 36 00 59 4 38 90 61 52 42 43 64 41 36 78 6 67 30 52 26 70 19 59 37 73 7 68 90 75 56 82 31 7 78 45 102 52 81 34 136 30 84 22 205 24 87 11 407 60 8 90 0 Infinita. 1 By one latitude Rumb and distance to find the difference of latitudes. As the Radius to the sine of the compliment of the Rumb from the meridian: So the distance upon the Rumb, to the difference of latitudes. Let the place given be A in the latitude of 50 gr. C in a greater latitude, but unknown, the distance upon the Rumb being 6 gr. between them, and the Rumb the third from the meridian. First I take 6 gr. for the distance upon the Rumb, out of the line of lines, and make it a parallel Radius, by putting it over in the sins of 90. and 90. Then keeping the Sector at this angle, I take out the parallel sine of 56 gr. 15 m. which is the sine of the compliment of the third Rumb from the meridian, and measuring it in the line of lines, I find it to be 5 gr. and such is the difference of latitude required. Or I may take out the sine of 56 gr. 15 m. for the compliment of the third Rumb from the meridian, make it a parallel Radius; then keeping the Sector at this angle, I take 6 gr. for the distance, either out of the line of lines, or any other scale of equal parts, or else out of the meridian line, and lay it on both sides of the Sector from the centre, either on the line of lines or sins: so the parallel taken from the terms of this distance, and measured in the same scale wherein the distance was measured, shall show the difference of latitude to be 5 gr. as before. But in shorter distances, such as fall within the compass of a days sailing, this work will hold much better. As may appear by comparing the work with the Table following where the numbers in the front do signify the leagues; those in the side, the Rumb; and the rest in the middle, the difference of latitude. Lgs 100 80 60 40 20 19 18 17 16 15 Rum. G. M G. M G. M G. M M M M M M M 5 0 4 0 3 0 2 0 60 57 54 51 48 45 4 59 3 59 2 59 1 59 60 57 54 51 48 45 4 58 3 58 2 59 1 59 60 57 54 51 48 45 4 56 3 57 2 58 1 58 59 56 53 50 47 44 1 4 54 3 55 2 56 1 57 59 56 53 50 47 44 4 51 3 53 2 55 1 56 58 56 52 50 47 43 4 47 3 50 2 52 1 55 57 55 52 49 46 43 4 42 3 46 2 49 1 53 56 54 51 48 45 42 2 4 37 3 42 2 46 1 51 55 53 50 47 44 41 4 31 3 37 2 43 1 48 54 52 49 46 43 40 4 25 3 32 2 39 1 46 53 50 48 45 42 39 4 17 3 26 2 34 1 43 51 49 46 44 41 38 3 4 10 3 20 2 30 1 40 50 47 45 42 40 37 4 1 3 13 2 25 1 36 48 46 43 41 39 36 3 52 3 5 2 19 1 32 46 44 42 39 37 35 3 42 2 58 2 13 1 28 44 42 40 38 36 33 4 3 32 2 50 2 7 1 25 42 40 38 36 34 32 3 22 2 41 2 1 1 21 40 38 36 34 32 30 3 10 2 32 1 54 1 16 38 36 34 32 30 28 2 59 2 23 1 47 1 12 36 34 32 30 29 27 5 2 47 2 14 1 40 1 7 33 32 30 28 27 25 2 34 2 3 1 32 1 2 31 29 28 26 25 23 2 22 1 53 1 25 0 57 28 27 25 24 23 22 2 8 1 43 1 17 0 52 26 24 23 22 21 19 6 1 55 1 32 1 8 0 46 23 22 21 20 18 17 1 41 1 20 1 0 0 40 20 19 18 17 16 15 1 27 1 9 0 52 0 35 17 16 16 15 14 13 1 13 0 58 0 44 0 30 15 14 13 12 12 11 7 0 59 0 47 0 35 0 24 12 11 11 10 9 9 0 44 0 36 0 26 0 18 9 8 8 7 7 7 0 30 0 24 0 18 0 12 6 6 5 5 5 4 0 15 0 12 0 9 0 9 3 3 3 3 2 2 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Lg M M M M M M M M M M M M M M Rum. 40 39 36 33 30 27 24 21 18 15 12 9 6 3 42 39 36 33 30 27 24 21 18 15 12 9 6 3 42 39 36 33 30 27 24 21 18 15 12 9 6 3 42 39 36 33 30 27 24 21 18 15 12 9 6 3 41 38 35 32 29 26 24 21 18 15 12 9 6 3 1 41 38 35 32 29 26 23 20 17 15 12 9 6 3 40 37 34 32 29 26 23 20 17 14 11 9 6 3 40 37 34 31 28 25 23 20 17 14 11 8 6 3 39 36 33 31 28 25 22 19 17 14 11 8 6 3 2 38 35 33 30 27 24 22 19 16 14 11 8 5 3 37 34 32 29 26 24 21 19 16 13 11 8 5 3 36 33 31 28 26 23 21 18 15 13 10 8 5 3 35 32 30 27 25 22 20 17 15 12 10 7 5 2 3 34 30 29 26 24 22 19 17 14 12 10 7 5 2 33 30 28 26 23 21 19 16 14 12 9 7 5 2 31 29 27 24 22 20 18 16 13 11 9 7 4 2 30 28 25 23 21 19 17 15 13 11 8 6 4 2 4 28 26 24 22 20 18 16 14 12 10 8 6 4 2 27 25 23 21 19 17 15 13 11 10 8 6 4 2 25 23 21 20 18 16 14 13 11 9 7 5 4 2 23 22 20 18 17 15 13 12 10 8 7 5 3 2 5 22 20 18 17 15 14 12 11 9 8 6 5 3 2 20 18 17 16 14 13 11 10 8 7 6 4 3 1 18 17 15 14 13 12 10 9 8 6 5 4 3 1 16 15 14 13 11 10 9 8 7 6 5 3 2 1 6 14 13 12 11 10 9 8 7 6 5 4 3 2 1 12 11 10 10 9 8 7 6 5 4 3 3 2 1 10 9 9 8 7 7 6 5 4 4 3 2 1 1 8 8 7 6 6 6 5 4 3 3 2 2 1 1 7 6 6 5 5 4 5 4 3 3 2 2 1 1 1 4 4 4 3 3 3 2 2 2 1 1 1 1 0 2 2 2 2 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 In the Chart let a meridian AB be drawn through A, and in A with AB make an angle of the Rumb BAC. Then open the compasses, according to the latitude of the places, to OF the quantity of 6 gr. in the meridian, transferring them into the Rumb from A to C, and through C draw the parallel BC, crossing the meridian AB in B: so the degrees in the meridian from A to B, shall show the difference of latitude to be 5 gr. 2 By the Rumb and both latitudes to find the distance upon the Rumb. As the sine of the compliment of the Rumb from the meridian, is to the Radius: So the difference of latitudes, to the distance upon the Rumb. As if the places given were A in the latitude of 50 gr. C in the latitude of 55 gr. and the Rumb the third from the meridian. Here I may take 5 gr. for the difference of latitude out of the line of lines, and put it over in the sine of 56 gr. 15 m. for the compliment of the third Rumb from the meridian. Then keeping the Sector at this angle, I take out the parallel Radius, and measuring it in the line of lines, I find it to be 6 gr. and such is the distance upon the Rumb, which was required. Or I may take the lateral Radius, and make it a parallel sine of 56 gr. 15 m. the compliment of the Rumb from the meridian: then keeping the Sector at this angle, I take 5 gr. for the difference of latitude, either out of the line of lines, or out of some other scale of equal parts, and lay it on both sides of the Sector from the centre, either on the line of lines or of sins: so the parallel taken from the terms of this difference, and measured in the same scale with the difference, shall show the distance upon the Rumb to be 6 gr. or 120 leagues. Or keeping the Sector at this angle, I may take the difference between 50 gr. and 55 gr. out of the Meridian line, and measuring it in the equator, I shall find it to be equal to 8 gr. 22 p. of the equator. Wherefore I take the parallel between 822 and 822 out of the line of lines, and measuring it in the line of lines I shall find it to be 989; which shows that according to this projection, the distance upon this third Rumb, answerable to the former distance of latitudes, will be equal to 9 gr. 89 p. of the equator. Or the Sector remaining at this angle, I may take the difference between 50 gr. and 55 gr. out of the Meridian line, and lay it from the centre on both sides of the Sector, either on the line of lines or of sins: so the parallel taken from the terms of this difference, shall be the very line of distance required, the same with AC or OF upon the chart; which may serve for the better pricking down of the distance upon the Rumb, without taking it forth of the Meridian line, as in the former Prop. Or if the Rumb fall nearer to the equator, that the lateral Radius cannot be fitted over in it, this proposition may be wrought by parallel entrance. For if I first take out the sine of 56 gr. 15 m. and make it a parallel Radius, by fitting it over in the sins of 90 and 90, or in the ends of the lines of lines, and then take 5 gr. for the difference of latitudes out of the line of lines, and carry it parallel to the former, I shall find it to cross both lines of lines in the points of 6: and so it gives the same distance as before. Or if the distance be small, it may be found by the former Table. For the Rumb being found in the side of the Table, and the difference of latitude in the same line; the top of the column wherein the difference of latitude was found, shall give the number of leagues in the distance required. Or we may find this distance in the Table of Rumbs in the fift Prop. following. For according to the example look into the Table of the third Rumb for 5 gr. of latitude, and there we shall find 6 gr. 01 parts under the title of distance. So if the difference of latitude upon the same Rumb were 50 gr. the distance would be 60 gr. 13 parts. If the difference of latitude upon the same Rumb were only ½ of a degree, the distance would be only 60 parts, such as 100 do make a degree. In the chart let a Meridian AB be drawn through A, and parallels of latitude through A and C; & then in A with AB make an angle of the Rumb BAC: so the distance taken from A to C, and measured in the Meridian line, according to the latitude of the places, shall be found to be 6 gr. or 120 leagues. And such is the distance required. 3 By the distance and both latitudes to find the Rumb. As the distance upon the Rumb, to the difference of latitudes: So is the Radius to the sine of the compliment of the Rumb from the Meridian, As if the places given were A in the latitude of 50 gr. C in the latitude of 55 gr. the distance between them being 6 gr. upon the Rumb. First I take 6 gr. for the distance upon the Rumb, and lay it on both sides of the Sector from the centre; then out of the same scale I take 5 gr. for the difference of latitude, and to it open the Sector in the terms of the former distance: so the parallel Radius taken and measured in the sins, doth give 56 gr. 15 m. the compliment whereof 33 gr. 45 m. is the angle of the Rumbs inclination to the Meridian, which was required. In the chart let a meridian AB be drawn through A, and parallels of latitude both through A and C; then open the compasses according to the latitude of the places to OF the quantity of 6 gr. in the meridian, and setting one foot in A, turn the other till it cross the parallel BC in C, and draw the right line AC: so the angle BAC shall show the inclination of the Rumb to the Meridian to be 33 gr. 45 m. as before. These three last Prop. depend one on the other, and may be wrought as truly by the common sea-chart as by this of Mercators' projection: and therefore in working them by the Sector, the distance and the difference of latitudes may as well or better be taken out of the line of lines (which here representeth the equator) or any other line of equal parts, as out of the enlarged degrees in the meridian line. But in the propositions following, the difference of longitude must be taken out of the equator; the difference of latitudes and distance upon the Rumb, must always be taken out of the meridian line; which I therefore call the proper difference, and proper distance. 4 By the longitude and latitude of two places to find the Rumb. As if the places given were A in the latitude of 50 gr. C in the latitude of 55 gr. and the difference of longitude between them were 5 gr. 30 m. In the chart let meridians and parallels be drawn through A and C, and a strait line for the Rumb from A to C; then by that we shown Cap. 2. Prop. 9 inquire the quantity of the angle BAC, and it shall be found to be 33 gr. 45 m. which is the third Rumb from the Meridian. Wherefore the proportion holds for the Sector, As AB the proper difference of latitude, is to BC the difference of longitude: So AB as Radius, to BC the tangent of the Rumb from the Meridian. According to this I take the proper difference of latitude from 50 gr. to 55 gr. out of the line of meridians, and lay it on both sides of the Sector from the centre; then I take the difference of longitude 5 gr. ½ out of the line of lines, and to it open the Sector in the terms of the former difference of latitudes: so the parallel Radius taken from between 90 and 90, and measured in the greater tangent on the side of the Sector, doth give 33 gr. 45 m. for the Rumb required. But if the Rumb fall nearer to the equator; As AD the difference of longitudes, is to DC the proper difference of latitudes: So AD as Radius, to DC the tangent of the rumb from the equator. According to this I take the former difference of latitudes from 50 gr. to 55 gr. out of the line of Meridian's, and to it open the Sector in the terms of the difference of longitude reckoned in the line of lines from the centre: so the parallel Radius taken and measured in the tangent, doth give 56 gr. 15 m. for the rumb from the equator; which is the compliment to the former 33 gr. 45 m: and so both ways it is found to be the third rumb from the Meridian. But if this rumb were to be found in the common sea-chart, it should seem to be above 47 gr. which is more than the fourth rumb from the meridian. 5 By the Rumb and both latitudes to find the difference of longitude. As if the places given were A in the latitude of 50 gr. and C in the latitude of 55 gr. and the rumb the third from the meridian. In the chart, let a meridian be drawn through A, and a parallel of latitude through C; then in A with AB make the angle of the rumb from the meridian BAC, (as was showed Cap. 2. Prop. 10.) So the degrees in the parallel between B and C, shall be found to be 5 gr. ½, the difference of longitude which was required. Wherefore the proportion holds for the Sector. As AB the Radius, to BC the tangent of the rumb from the meridian: So AB as proper difference of the latitudes, to BC the difference of longitude. According to this we may take the tangent of the Rumb which is here 33 gr. 45 m. from the meridian, out of the greater tangent on the side of the Sector; and putting it over between 90 and 90, make it a Radius: then keeping the Sector at this angle, take the proper difference of latitudes from 50 gr. to 55 gr. out of the line of Meridian's, and lay it on both sides of the Sector from the centre: so the parallel taken from the terms of this difference, and measured in the line of lines, shall show the difference of longitude to be 5 gr. ½. Or if the Rumb fall nearer the equator. As DC the tangent of the Rumb from the equator, to AD the Radius: So DC as proper difference of the latitudes, to AD the difference of longitude. According to this we may best work by parallel entrance, first taking 56 gr. 15 m. for the angle of the Rumb from the equator, out of the greater tangent, and make it a parallel Radius: then take the proper difference of latitudes out of the line of meridians, and carry it parallel to the former: so we shall find it to cross the line of lines in 5 gr. ½. And this is the difference of longitude required, the same as before. But if this difference were to be found by the common sea-chart, it should seem to be only 3 gr. 20 m. which is more than 2 gr. less than the truth. And yet this error would be greater, if either the latitude be greater, or the Rumb fall nearer the equator: as may appear by comparing the common sea-chart with the Tables following. The first Rumbe fr●m the Meridian. North and by East, South and by East, North and by West, South and by West. La Long. Dist. La Long. Dist. 〈◊〉 Long. Dist. Gr Gr. P Gr. P Gr Gr. P. Gr. P. G ●r P. Gr. P. 0 0 0 30 6 26 30 5● 60 15 01 61 18 1 20 1 02 31 6 49 31 61 61 15 41 62 20 2 40 2 04 32 6 72 32 63 62 15 83 63 21 3 60 3 06 33 6 96 33 65 63 16 26 64 23 4 80 4 08 34 7 20 34 67 64 16 71 65 25 5 1 00 5 10 35 7 44 35 69 65 17 17 66 27 6 1 20 6 12 36 7 68 36 71 66 17 65 67 29 7 1 40 7 14 37 7 92 37 73 67 18 15 68 31 8 1 60 8 16 38 8 17 38 75 68 18 67 69 33 9 1 80 9 18 39 8 43 39 77 69 19 21 70 35 10 2 00 10 20 40 8 70 40 78 70 19 78 71 37 11 2 20 11 22 41 8 96 41 80 71 20 37 72 39 12 2 40 12 24 42 9 22 42 82 72 21 00 73 41 13 2 61 13 25 43 9 50 43 84 73 21 66 74 43 14 2 81 14 27 44 9 76 44 86 74 22 36 75 45 15 3 02 15 29 45 10 04 45 88 75 23 10 76 47 16 3 22 16 31 46 10 33 46 90 76 23 90 77 49 17 3 43 17 33 47 10 62 47 92 77 24 75 78 51 18 3 64 18 35 48 10 91 48 94 78 25 67 79 53 19 3 85 19 37 49 11 21 49 96 79 26 67 80 55 20 4 06 20 39 50 11 52 50 98 80 27 76 81 57 21 4 27 21 41 51 11 83 52 0 81 28 97 82 59 22 4 49 22 43 52 12 15 53 2 82 30 32 83 61 23 4 70 23 45 53 12 47 54 4 83 31 84 84 63 24 4 92 24 47 54 12 81 55 6 84 33 61 85 62 25 5 14 25 49 55 13 16 56 8 85 35 69 86 67 26 5 36 26 51 56 13 50 57 10 86 38 24 87 69 27 5 58 27 53 57 13 86 58 12 87 41 52 88 71 28 5 80 28 55 58 14 23 59 14 88 46 15 89 73 29 6 03 29 57 59 14 62 60 16 89 54 06 90 75 30 6 26 30 59 60 15 01 61 18 90 The second Rumbe from the Meridian. North North-east. South Southeast. North Northwest South South-west La Long. Dist. La Long. Dist. La Long. Dist. Gr Gr. P Gr. P. Gr Gr. P Gr. P Gr Gr. P. Gr. P. 0 0 0 30 13 03 32 47 60 31 25 64 94 1 0 42 1 08 31 13 51 33 5● 61 32 09 66 03 2 0 83 2 16 32 14 00 34 64 62 32 96 67 11 3 1 24 3 25 33 14 49 35 72 63 33 86 68 19 4 1 65 4 33 34 15 00 36 80 64 34 79 69 27 5 2 07 5 41 35 15 50 37 88 65 35 7● 70 35 6 2 49 6 49 36 16 00 38 97 66 36 75 71 44 7 2 91 7 57 37 16 51 40 05 67 37 80 72 52 8 3 32 8 66 38 17 03 41 13 68 38 88 73 60 9 3 74 9 74 39 17 56 42 21 69 40 00 74 68 10 4 16 10 82 40 18 10 43 30 70 41 19 75 77 11 4 59 11 90 41 18 65 44 38 71 42 43 76 85 12 5 01 12 99 42 19 20 45 46 72 43 74 77 93 13 5 43 14 07 43 19 76 46 54 73 45 11 79 01 14 5 85 15 15 44 20 33 47 62 74 46 57 80 10 15 6 28 16 23 45 20 92 48 71 75 48 12 81 18 16 6 71 17 32 46 21 50 49 79 76 49 78 82 26 17 7 14 18 40 47 22 11 50 87 77 51 55 83 34 18 7 58 19 48 48 22 72 51 95 78 53 46 84 42 19 8 01 20 56 49 23 35 53 03 79 55 54 85 51 20 8 45 21 65 50 23 98 54 12 80 57 82 86 59 21 8 90 22 73 51 24 63 55 20 81 60 33 87 67 22 9 34 23 81 52 25 30 56 28 82 63 13 88 76 23 9 79 24 89 53 25 98 57 37 83 66 32 89 84 24 10 24 25 98 54 26 68 58 45 84 69 99 90 92 25 10 70 27 06 55 27 39 59 53 85 74 32 92 00 26 11 16 28 14 56 28 12 60 61 86 79 63 93 09 27 11 62 29 22 57 28 87 61 70 87 86 46 94 17 28 12 08 30 31 58 29 64 62 78 88 96 10 95 25 29 12 55 31 39 59 30 44 63 86 89 112 57 96 33 30 13 03 32 47 60 31 25 64 94 90 The third Rumbe from the Meridian. North-east by North, Southeast by South, Northwest by North, South-west by South. La Long. Dist. La Long. Dist. La Long. Dist. Gr Gr. P. Gr. P. Gr Gr. P. Gr. P. Gr Gr. P. Gr. P. 0 0 0 30 21 03 36 08 60 50 42 72 16 1 0 66 1 20 31 21 80 37 28 61 51 78 73 36 2 1 33 2 40 32 22 58 38 49 62 53 18 74 56 3 2 00 3 61 33 23 38 39 69 63 54 63 75 77 4 2 67 4 81 34 24 18 40 89 64 56 12 76 97 5 3 34 6 01 35 25 00 42 09 65 57 68 78 17 6 4 01 7 22 36 25 82 43 30 66 59 29 79 37 7 4 68 8 42 37 26 64 44 50 67 60 96 80 58 8 5 36 9 62 38 27 48 45 70 68 62 71 81 78 9 6 03 10 82 39 28 34 46 90 69 64 53 82 98 10 6 71 12 03 40 29 21 48 11 70 66 44 84 19 11 7 39 13 23 41 30 09 49 31 71 68 45 85 39 12 8 07 14 43 42 30 98 50 51 72 70 55 86 59 13 8 76 15 64 43 31 88 51 71 73 72 77 87 79 14 9 44 16 84 44 32 80 52 92 74 75 12 89 00 15 10 13 18 04 45 33 74 54 12 75 77 62 90 20 16 10 83 19 24 46 34 69 55 32 76 80 30 91 40 17 11 53 20 45 47 35 67 56 52 77 83 15 92 61 18 12 23 21 65 48 36 66 57 73 78 86 25 93 81 19 12 93 22 85 49 37 67 58 93 79 89 60 95 01 20 13 64 24 05 50 38 69 60 13 80 93 27 96 22 21 14 35 25 26 51 39 74 61 33 81 97 32 97 42 22 15 07 26 46 52 40 82 62 54 82 101 85 98 62 23 15 80 27 66 53 41 91 63 74 83 106 97 99 82 24 16 53 28 86 54 43 03 64 94 84 112 90 101 03 25 17 26 30 07 55 44 19 66 15 85 119 90 102 23 26 18 00 31 27 56 45 37 67 35 86 128 45 103 43 27 18 75 32 47 57 46 58 68 55 87 139 47 104 64 28 19 50 33 67 58 47 82 69 75 88 155 00 105 84 29 20 26 34 88 59 49 11 70 96 ●9 181 58 107 04 30 21 03 36 08 60 50 42 72 16 90 The eight Rumbe of East and West, with the Longitude answering to one degree of distance, and the distance belonging to one degree of Longitude. La Long. Dist. La Long. Dist. La Long. Dist. Gr Gr. P. Parts. Gr Gr. P. Parts. Gr Gr. P. Parts. 0 0 100 00 30 1 25 86 60 60 2 00 50 00 1 1 00 99 98 31 1 17 85 71 61 2 06 48 48 2 1 00 99 94 32 1 18 84 80 62 2 13 46 94 3 1 00 99 86 33 1 19 83 86 63 2 20 45 40 4 1 00 99 75 34 1 21 82 90 64 2 28 43 83 5 1 00 99 62 35 1 22 81 91 65 2 37 42 26 6 1 01 99 45 36 1 24 80 90 66 2 46 40 67 7 1 01 99 25 37 1 25 79 86 67 2 56 39 07 8 1 01 99 02 38 1 27 78 80 68 2 67 37 46 9 1 01 98 76 39 1 29 77 71 69 2 79 35 83 10 1 02 98 48 40 1 31 76 60 70 2 92 34 20 11 1 02 98 16 41 1 33 75 47 71 3 07 32 55 12 1 02 97 81 42 1 35 74 31 72 3 24 30 90 13 1 03 97 43 43 1 37 73 13 73 3 42 29 23 14 1 03 97 03 44 1 39 71 93 74 3 63 27 56 15 1 03 96 59 45 1 41 70 71 75 3 86 25 88 16 1 04 96 12 46 1 44 69 46 76 4 13 24 19 17 1 04 95 63 47 1 47 68 20 77 4 44 22 49 18 1 05 95 10 48 1 49 66 91 78 4 81 20 79 19 1 06 94 55 49 1 52 65 60 79 5 24 19 08 20 1 06 93 97 50 1 55 64 28 80 5 76 17 36 21 1 07 93 35 51 1 59 62 93 81 6 39 15 64 22 1 08 92 72 52 1 62 61 56 82 7 18 13 91 23 1 09 92 05 53 1 66 60 18 83 8 20 12 18 24 1 09 91 35 54 1 70 58 77 84 9 57 10 45 25 1 10 90 63 55 1 74 57 35 85 11 47 8 71 26 1 11 89 88 56 1 79 55 92 86 14 33 6 97 27 1 12 89 10 57 1 84 54 46 87 19 11 5 23 28 1 13 88 29 58 1 89 52 99 88 28 65 3 49 29 1 14 87 46 59 1 94 51 50 89 57 30 1 74 30 1 15 86 60 60 2 00 50 00 90 0 These tables are calculated for each of the Rumbs. The first seven have three columns, and of them the first ●●●●●neth the degrees of Latitude, from the Equinoctial to th● Pole: the second doth give the difference of Longitude; and the third the distance, both of them belonging to that ●●mb and latitude. As in the Table of the third Rumb; at the latitude of 50 Gr. I f●nd under the title of Longitude 38 Gr. 69 parts, and under the title of Distance 60 Gr. 13 parts. This shows that if the course held constantly on the third Rumb from the Equinoctial to the Latitude of 50 Gr. the difference of Longitude would be 38 Gr 69 parts of a 100, and the distance upon the Rumbe 60 Gr. 13 parts. For here I reckon the distance by degrees, rather than by leagues or miles, and subdivide each degree into 100 parts, rather than it to 60 minutes, for the more ease in calculation, and withal to make the calculation to agree the better, both with this, and my cross-staff, and other instruments. The use of these Tables, for the finding of the difference of Longitude, is this. Turn to the table of the Rumb, and there see what longitude belongeth to either latitude, then take the one longitude out of the other, the remainder will be the difference of longitude required. As in the former example, where the places given were A, in the latitude of 50 Gr. C in the latitude of 55 Gr. and the Rumb the third from the meridian: I look into the table of the third Rumb and there find, Latitude 50 gr. Longitude 38 gr. 69 parts. Latitude 55. Longitude 44. 19 Therefore the diff. of Longitude 5 50 There is another use of these tables, for the describing of the Rumbs both on the Globe, and all sorts of Charts. For having drawn the circles of longitude and latitude, and finding by the tables, the difference of longitude belonging to each Rumb and latitude: If we make a prick in the chart, at every degree of latitude, according to that difference of longitude, and draw lines through those pricks, so as they make no angles, the lines so drawn shall be the Rumbs required. The use of the eight Rumb is something different from the rest. For there being here no change of latitude, I have set to each latitude, the difference of longitude, belonging to one degree of distance, and the distance belonging to one degree of longitude. As if two places shall be 20 leagues, or one degree distant one from the other, in the latitude of 50 gr. the difference of longitude between them will be 1 gr. 55 parts. But if they differ one degree in longitude, the distance between them will be only 64 parts, which fall short of 13 leagues, or at the most 6428 parts, such as 10000 do make a degree. 6 By the difference of longitude, Rumb, and one latitude, to find the other latitude. As if the places given were A, in the latitude of 50 gr. C in a greater latitude but unknown, the difference of longitude 5 gr. ½, and the Rumb the third from the Meridian. In the chart let AB, D C, meridians, be drawn through A and C, according to the difference of longitude, one 5 gr. ½ from the other; and a parallel of latitude through A, crossing the meridian CD in D: then in A, with AB, make an angle of the Rumb BAC: so the degrees in the meridian between D and C, shall be found to be 5 gr. the proper difference of latitude which was required. Wherefore the proportion holds for the Sector, As AD the Radius, to DC the tangent of the Rumb from the equator: So AD as difference of longitude, to DC the proper difference of latitude. According to this, I take 56 gr. 15 m. for the angle of the Rumb from the equator, out of the greater Tangent, and make it a parallel Radius. Then I reckon 5 gr. ½ in the line of lines from the centre, for the difference of longitude. So the parallel taken from the terms of this difference, and measured in the line of meridians, shall reach from 50 gr. the latitude given, to 55 gr. which is the latitude required. Or if the Rumb fall nearer to the meridian. As BC the tangent of the Rumb from the meridian, is to AB the Radius: So BC as difference of longitude, to AD the proper difference of latitude. According to this we may best work by parallel entrance; first take 33 gr. 45 m. for the angle of the Rumb from the meridian, out of the greater Tangent, and make it a parallel Radius; then take 5 gr. ½ for the difference of longitude out of the line of lines, and carry it parallel to the former, till the feet of the compasses stay in like points: so the line between the centre and the place of this stay, being taken and measured in the line of meridians from 50 gr. forward, shall show the latitude required to be 55 gr. as in the former way. The like may be found by the tables of Rumbs. For in the table of the third Rumb, at the latitude of 50 gr. I find the longitude of 38 gr. 69 p; to this if I add 5 gr. 50 p. for the difference of longitude given, the compound longitude will be 44 gr. 19 p. and this answers to the latitude of 55 gr. But if this difference of latitude were to be found by the common sea-chart, it should seem to be 8 gr. 13 m; and so the second latitude should be 58 gr. 13 m. which is above 3 gr. more than the truth. 7 By one latitude, rumb, and distance, to find the difference of longitude. As if the places given were A in the latitude of 50 gr. C in a greater latitude but unknown, the distance upon the Rumb being 6 gr. between them, and the Rumb the third from the meridian. In the chart, let a meridian AB, and a parallel AD be drawn through A; and in A, with AB, make an angle BAC for the Rumb from the meridian; then open the compasses aceording to the latitude of the places to OF, the quantity of 6 gr. in the meridian, transferring them into the Rumb from A to C, and through C draw another meridian DC, crossing the parallel drawn through A in D: so the degrees intercepted in the parallel from A to D, shall show the difference of longitude required to be about 5 gr. ½. Wherefore the proportion holds for the Sector. As AC the Radius, is to AD, equal to BC, the sine of the Rumb from the meridian: So AC as proper distance upon the Rumb, to AD the difference of longitude. According to this I take the sine of 33 gr. 45 m. for the angle of the Rumb from the meridian, and make it a parallel Radius; then keeping the Sector at this angle, I take 6 gr. for the distance out of the meridian line, according to the estimated latitudes of both places, and lay it on both sides of the Sector from the centre: so the parallel taken from the terms of this distance, and measured in the lines of lines, shall show the difference of longitude to be about 5. gr. ½. In this, and some of the Prop. following, where there is but one latitude known, there may be sometimes an error of a minute or two, in the estimation of the proper distance, yet it may be rectified at a second operation. This proposition may also be wrought by the Tables of Rumbs. For according to the example, in the Table of the third Rumb, at the latitude of 50 gr. I find the longitude of 38 gr. 69 p. and the distance of 60 gr. 13 p. to this I a de 6 gr. for the distance given; so the compound distance will be 66 gr. 13 p. and this answers to the longitude of 44 gr. 19 p; then if I take the one longitude out of the other, the difference will be 5 gr. 50 p. as before. But if this difference were to be found by the common sea-chart, it should seem to be only 3 gr. 20 m. which is more than 2 gr. less than the truth. 8 By one latitude, Rumb, and difference of longitudes, to find the distance. As if the places given were A, in the latitude of 50 gr. C in a greater latitude but unknown, the difference of longitude between them being 5 gr. ½, and the Rumb the third from the meridian. In the chart let AB, DC, meridians be drawn through A and C, according to the difference of longitude, and a parallel of latitude through A, crossing the meridian DC in D; then in A, with AB, make an angle of the Rumb BAC: so the distance on the Rumb from A to C taken and measured in the meridian, according to the estimated latitude of the places, shall be found to be 6 gr. Wherefore the proportion holds for the Sector. As AD, equal to BC, the sine of the Rumb from the meridian, is to AC the Radius: So AD as difference of longitudes, to AC the proper distance upon the Rumb. According to this, I take the lateral Radius, and make it a parallel sine of 33 gr. 45 m. which is here the angle of the Rumb from the meridian; then I reckon 5 gr. ½ in the lines of lines from the centre, for the difference of longitude: so the parallel taken from the terms of this difference, and measured in the line of meridians, according to the latitudes of the places, shall there show the distance required to be about 6 gr. which are 120 leagues. Or if the Rumb fall nearer to the meridian, that the lateral Radius cannot be fitted over in his sine, this Prop. must be wrought by parallel entrance, and so also it gives the same distance as before. Or we may find this distance by the Table of Rumbs. For in the table of the third Rumb, at the latitude of 50 gr. I find the longitude of 38 gr. 69 p. and the distance of 60 gr. 13 p. To this longitude here found, I add 5 gr. 50 p. for the difference of longitude given: so the compound longitude will be 44 gr. 19 p. and this answers to the distance of 66 gr. 15 p. Then if I take the one distance out of the other, the remainder will be 6 gr. 02 p. for the distance required. But if this distance were to be measured on the common sea-chart, it should seem to be almost 10 gr. or at the least 197 leagues, above 77 leagues more than the truth. 9 By one latitude, distance, and difference of longitudes, to find the Rumb. As if the places given were A, in the latitude of 50 gr. C in a greater latitude but unknown, the difference of longitude between them being 5 gr. ½, and the distance 6 gr. upon the Rumb. In the chart let AB, DC, meridians, be drawn through A and C, and a parallel of latitude through A; then open the compasses according to the latitudes of the places, to OF the quantity of 6 gr. in the meridian, and setting the one foot in A, the other foot shall cross the other meridian in C; and if we draw the right line AC, the angle BAC shall show the inclination of the Rumb to the meridian to be about 33 gr. 45 m. Wherefore the proportion holds for the Sector. As AC the proper distance upon the Rumb, is to AD the difference of longitude: So AC as Radius, to AD, equal to BC, the sine of the Rumb from the meridian. According to this, I take the proper distance 6 gr. out of the line of meridians, and lay it on both sides of the Sector from the centre; then I take the difference of longitude 5 gr. ½ out of the line of lines, and to it open the Sector in the terms of the former distance: so the parallel Radius taken from between 90 and 90, and measured in the sins, doth give about 33 gr. 45 m. for the Rumb required. But if this Rumb were to be found by the common sea-chart, it should seem to be above 66 gr. and so almost the sixth Rumb from the meridian. 10 By the longitude and latitude of two places, to find their distance upon the Rumb. Let the Sector be opened in the lines of lines, unto a right angle (as was showed before Cap. 2. Prop. 7;) then take out the proper difference of latitude, and lay it on the one line, and the difference of longitude, and lay it on the other line, so as they may both meet in the centre, marking how far they extend. For the line taken from the terms of their extension, and measured in the meridian, according to their latitudes, shall show the distance required. So if the places given were A and C, A in the latitude of 50 gr. C in the latitude of 55 gr. the proper difference of latitude shall be the line AB, and let BC the difference of longitude be 5 gr. ½, we shall find that AC the distance upon the Rumb is about 6 gr. which make 120 leagues. For in the chart, let an occult meridian be drawn through A, and a parallel of latitude through C, crossing the former meridian in B, and a right line for the Rumb from A to C, so have we a rectangle triangle ABC, whose base AC, taken and measured in the meridian from E below 50 gr. to F, as much above 55 gr. doth contain the quantity of 6 gr. In the same manner the Sector being opened to a right angle, in the lines of lines: if we take the difference of latitude out of the line of meridians, in his proper place from 50 gr. to 55 gr. and place it on one of the sides from the centre, to resemble AB, then reckon the difference of longitude on the other perpendicular line from the centre to 5 gr. ½, in stead of BC, we shall have the like rectangle triangle on the Sector, to that which we had before on the chart; and if we take out the base of it, and measure it in the line of meridians from below 50 gr. to as much above 55 gr. we shall find as before, that it containeth about 6 gr. or 120 leagues. But if this distance were to be measured on the common sea-chart, it should seem to be almost 7 gr. ¼, or 145 leagues; which is 25 leagues more than the truth. 11 By the latitude of two places, and the distance upon the Rumb, to find the difference of longitude. Let the Sector be opened in the lines of lines to a right angle, then take out the proper difference of latitudes, and lay it on one of the lines from the centre, then take the proper distance with a pair of compasses, and setting one foot in the terms of the difference, turn the other foot to the other line of the Sector, and it shall there show the difference of longitude required. So if the places given were A, in the latitude of 50 gr. C in the latitude of 55 gr. with 6 gr. of distance one from another, we shall find their difference of longitude to be about 5 gr. ½. For in the chart let a meridian AB be drawn for the one, and BC, AD, parallels of latitude for them both. Then open the compasses according to the latitude of the places, to OF the quantity of 6 gr. in the meridian, and setting one foot in A, having latitude of 50 gr. turn the other to the parallel of 55 gr. and it shall there cut off the required difference of longitude BC 5 gr. ½. In the same manner, the Sector being opened to a right angle, in the lines of lines: if we take the difference of latitude out of the line of meridians in his proper place from 50 gr. unto 55 gr. and place it on one of the lines from the centre; then take 6. gr. the distance upon the Rumb out of the same line of meridians, according to the latitudes of the places, and set the one foot in the term of the former difference, turning the other foot to the other perpendicular line, we shall find that it will cross it about 5 gr. ½ from the centre: which is the difference of longitude required. But if this difference of longitude were to be found by the common sea-chart, it would seem to be only 3 gr. 20 m. which is more than 2 gr. 10 m. less than the truth. 12 By one latitude, distance and difference of longitudes, to find the difference of latitudes. Let the Sector be opened in the line of lines to a right angle and let the difference of longitude be reckoned in one of those lines from the centre; then take the proper distance with a pair of compasses, and setting the one foot in the term of the former difference, turn the other foot to the other line of the Sector, and it shall thence cut off a line, equal to the proper difference of latitude required. So if the places given were A and C, A in the latitude of 50 gr. C in a greater latitude but unknown, the difference of longitude between them 5 gr. ½, and the distance upon the Rumb 6 gr. or 120 leagues, we shall find the difference of latitude to be 5 gr. For in the chart, let occult meridians be drawn through A and C, and a parallel of latitude through A; then open the compasses according to the estimated latitudes of the places to OF the quantity of 6 gr. in the meridian, and setting the one foot in A, turn the other to the meridian drawn through C, and it shall there cut off the line DC, which is the difference of latitude required. In the same manner, the Sector being opened to a right angle, in the lines of lines, if in the one line we reckon the difference of longitude from the centre to 5 gr. ½, then taking 6 gr ●or the distance out of the line of Meridian's, according to the latitude of the places, we set the one foot in the term of the given difference, and turn the other foot to the other perpendicular line, we shall find that it cuts a line from it, which taken and measured in the line of meridians, from 50 gr. on forward, doth show the difference of latitude to be as before 5 gr. But if this difference of latitude were to be found by the common sea-chart, it would seem to be only 2 gr. 25 m. which is 2 gr. 35 m. less than the truth. Such is the difference between both these charts'. THE THIRD BOOK Containing the use of the particular Lines. THE lines of lines, of superficies, of solids, of sins, with the lateral lines of tangents and meridians, whereof I have hitherunto spoken, are those which I principally intended: that little room on the Sector which remaineth, may be filled up with such particular lines as each one shall think convenient for his purpose. I have made choice of such as I thought might be best pricked on without hindering the sight of the former, viz. lines of Quadrature, of Segments, of Inscribed bodies, of Equated bodies, and of Metals. CHAP. I. Of the lines of Quadrature. THe lines of quadrature may be known by the letter Q, and by their place between the lines of sins. Q signifieth the side of a square; 5 the side of a pentagon with five equal sides, 6 of an hexagon with six equal sides, and so 7, 8, 9, and 10. S stands for the Semidiameter of a circle, and 90 for a line equal to 90 gr. in the circumference. The use of them may be 1 To make a square equal to a circle given. 2 To make a circle equal to a square given. If the circle be first given, take his semidiameter, and to it open the Sector in the points at S: so the parallel taken from between the points at Q, shall be the side of the square required. geometric illustration If the square be given take his side, and to it open the Sector, in the points at Q: so the parallel taken from between the points at S, shall be the Semidiameter of the circle required. Let the Semidiameter of the circle giuen be AB, the side of the square equal unto it shall be found to be CD. 3 To reduce a circle given, or a square into an equal pentagon, or other like sided and like angled figure. Take the side of the figure given, and fit it over in his due points: so the parallels taken from between the points of the other figures, shall be the sides of those figures: which being made up with equal angles, shall be all equal one to the other. Let the Semidiameter of the circle giuen be AB, the side of an hexagon equal to this circle, shall by these means be found to be GH; and the sides of an octagon to be IK. Other planes not here set down, may first be reduced into a square, by the sixth Prop. Superf. and then into a circle, or other of these equal figures, as before. 4 To find a right line, equal to the circumference of a circle, or other part thereof. Take the Semidiameter of the circle given, and to it open the Sector in the points at S; so the parallel taken from between the points at 90 in this line, shall be the fourth part of the circumference: which being known, the other parts may be found out by the second and third Prop. of lines. Thus if the Semidiameter of the circle giuen be AB, the right line OF shall be found to be the fourth part of the circumference. Therefore the double of OF shall be equal to the circumference of 180 gr; and the half of OF shall be the circumference of 45 gr. and so in the rest. CHAP. II. Of the lines of Segments. THe lines of segments which are here placed between the lines of sins and superficies, and are numbered by 5, 6, 7, 8, 9, 10, do represent the diameter of a circle, so divided into a hundred parts, as that a right line drawn through these parts, perpendicular to the diameter, shall cut the circle into two segments, of which the greater segment shall have that proportion to the whole circle, as the parts cut have to 100 The use of them may be 1 To divide a circle given into two segments, according to a proportion given. 2 To find a proportion between a circle and his segments given. Let the Sector be opened in the points of an 100, to the diameter of the circle given: so a parallel taken from the points proportional to the greater segment required, shall give the depth of that greater segment. geometric illustration As if the diameter of the circle given were BL, the depth of the greater segment LO being 75, doth show the proportion of the segment OMLN to the circle to be as 75 to 100 viz. three parts of four. Hence I might show, if there were any use of it, To find the side of a square, equal to any known segment of a circle. The side of a square equal to the whole circle, may be found by the former Cap. and then having the proportion of the segment to the circle, we may dimmish the square in such proportion, by that which hath been showed Lib. 1. Cap. 3. Prop. 3. CHAP. III. Of the lines of Inscribed bodies. THe lines of inscribed bodies are here placed between the lines of lines, and may be known by the letters, D, S, I, C, O, T; of which D signifieth the side of a dodecahedron, I of an Icosahedron, C of a cube, O of an octahedron, and T of a tetrahedron, all inscribed into the same sphere, whose semidiameter is here signified by the letter S. The use of these lines may be, 1 The semidiameter of a sphere being given, to find the sides of the five regular bodies, which may be inscribed in the said sphere. 2 The side of any of the five regular bodies being given, to find the semidiameter of a sphere, that will circumscribe the said body. If the sphere be first given, take his semidiameter, and to it open the Sector in the points at S: if any of the other bodies be first given, take the side of it, and fit it over in his due points: so the parallel taken from between the points of the other bodies, shall be the sides of those bodies, and may be inscribed into the same sphere. line segments So if the semidiameter of the sphere be AC, the side of the dodecahodron inscribed shall be DE. CHAP. FOUR Of the lines of Equated bodies. THe lines of equated bodies are here placed between the l●n●s of lines 〈◊〉 solids, noted with these letters, D, I, C, S, ●, T, of whic● D stands for the side of a dodecahedron, I for the ●●de of ●n Icos●hedron, C for the side of a cube, S for the diameter of a sphere, O for the side of an octahedron, and T for the side of a tetrahed on, all equal one to the other. The use of these lin●s may be 1 The diameter of a sphere being given, to find the sides of ●he five regular bodies equal to that sphere. 2 The side of any of the five regular bodies being given, to find the diameter of a sphere, and the sides of the other bodies, equal to the first body given. If the sphere be first given, take his diameter, and to it open the Sector in the points at S: if any of the other bodies be first given, take the side of it, and fit it over in his due points: so the parallels taken from between the points of the other bodies, shall be the sides of those bodies equal to the first body given. Thus in the last diagram, if the diameter of a sphere given be BC, the side of the dodecahedron equal to this sphere, would be found to be FG. CHAP. V Of the Lines of Metals. THe lines of Metals are here joined with those before of equated bodies, and are noted with these characters ☉. ☿. ♄. ☽. ♀. ♂. ♃. of which ☉ stands for gold, ☿ for quicksilver, ♄ for lead, ☽ for silver, ♀ for copper, ♂ for iron, and ♃ for tin. The use of them is to give a proportion between these several metals, in their magnitude and weight, according to the experiments of Marinus Ghetaldus, in his book called Promotus Archimedes. 1 In like bodies of several metals and equal weight, having the magnitude of the one, to find the magnitude of the rest. Take the magnitude given out of the lines of Solids, and to it open the Sector in the points belonging to the mettle given: so the parallels taken from between the points of the other metals, and measured in the lines of Solids, shall give the magnitude of their bodies. Thus having cubes or spheres of equal weight, but several metals, we shall find that if those of tin contain 10000 D, ●he others of iron will contain 9250, those of copper 8222, those of silver 7161, those of lead 6435, those full of quicksilver 5453, and those of gold 3895. 2 In like bodies of several metals and equal magnitude, having the weight of one to find the weight of the rest. This proposition is the converse of the former, the proportion not direct, but reciprocal, wherefore having two like bodies, take the given weight of the one out of the lines of Solids, and to it open the Sector in the points belonging to the mettle of the other body: so the parallel taken from the points belonging to the body given, and measured in the lines of Solids, shall give the weight of the body required. As if a cube of gold weighed 38 l. and it were required to know the weight of a cube of lead having equal magnitude. First I take 38 l. for the weight of the golden cube, out of the lines of Solids, & put it over in the points of ♄ belonging to lead: so the parallel taken from between the points of ☉ standing for gold, and measured in the lines of Solids, doth give the weight of the leaden cube required to be 23 l. Thus if a sphere of gold shall weigh 10000, we shall find that a sphere of the same diameter full of quicksilver shall weigh 7143, a sphere of lead 6053, a sphere of silver 5438, a sphere of copper 4737, a sphere of iron 4210, and a sphere of tin 3895. 3 A body being given of one mettle, to make another like unto it, of another mettle, and equal weight. Take out one of the sides of the body given, and put it over in the points belonging to his mettle: so the parallel taken from between the points belonging to the other mettle, shall give the like side, for the body required. If it be an irregular body, let the other like sides be found out in the same manner. line segments Let the body giuen be a sphere of lead containing in magnitude 16 D, whose diameter is A, to which I am to make a sphere of iron, of equal weight: If I take out the diameter A, and put it over in the points of ♄ belonging to lead, the parallel taken from between the points of ♂ standing for iron, shall be B, the diameter of the iron sphere required. And this compared with the other diameter, in the lines of solids will be found to be 23 d. in magnitude. 4 A body being given of one mettle, to make another like unto it of another mettle, according to a weight given. First find the sides of a like body of equal weight, then may we either augment or diminish them according to the proportion given by that which we shown before in the second and third Prop. of Solids. As if the body given were a sphere of lead, whose diameter is A, and it were required to find the diameter of a sphere of iron, which shall weigh three times as much as the sphere of lead: I take A, and put it over in the points of ♄, his parallel taken from between the points of ♂, shall give me B for the diameter of an equal sphere of iron: if this be augmented in such proportion as 1 unto 3, it giveth C for the diameter required. geometric illustration CHAP. VI Of the lines on the edges of the Sector. Having showed some use of the lines on the flat sides of the Sector, there remain only those on the edges. And here one half of the outward edge is divided into inches, and numbered according to their distance from the ends of the Sector. As in the Sector of fourteen inches long, where we find 1 and 13, it showeth that division to be 1 inch from the nearer end, and 13 inches from the farther end of the Sector. The other half containeth a line of lesser tangents, to which the gnomon is Radius. They are here continued to 75 gr. And if there be need to produce them farther, take 45 out of the number of degrees required, and double the remainder: so the tangent and secant of this double remainder being added, shall make up the tangent of the degrees required. As if AB being the Radius, and BC the tangent line, it were required to find the tangent of 75 gr. If we take 45 gr. out of 75 gr. the remainder is 30 gr. and the double 60 gr. whose tangent is BD, and the secant is AD: if then we add AD to BD, it maketh BC the tangent of 75 gr. which was required. In like sort the secant of 61 gr. added to the tangent of 61 gr. giveth the tangent of 75 gr. 30 m. and the secant of 62 gr. added to the tangent of 62 gr. giveth the tangent of 76 gr. and so in the rest. The use of this line may be To observe the altitude of the Sun. Hold the Sector so as the tangent BC may be vertical, and the gnomon BASILIUS parallel to the horizon; then turn the gnomon toward the Sun, so that it may cast a shadow upon the tangent, and the end of the shadow shall show the altitude of the Sun. So if the end of the gnomon at A, do give a shadow unto H, it showeth that the altitude is 38 gr. ½, if unto D, then 60 gr. and so in the rest. There is another use of this tangent line, for the drawing of the hour lines upon any ordinary plane, whereof I will set down these propositions. 1 To draw the hour lines upon an horizontal plane. 2 To draw the hour lines upon a direct vertical plane. First draw a right line AC for the horizon and the equator, and cross it at the point A about the middle of the line with AB another right line, which may serve for the meridian and the hour of 12; then take out 15 gr. out of the tangents, and prick them down in the equator on both sides from 12: so the one point shall serve for the hour of 11, and the other for the hour of 1. Again, take out the tangent of 30 gr. and prick it down in the equator on both sides from 12: so the one of these points shall serve for the hour of 10, and the other for the hour of 2. In like manner may you prick down the tangent of 45 gr. for the hours of 9 and 3, and the tangent of 60 gr. for the hours of 8 and 4, and the tangent of 75 gr. for the hours of 7 and 5. Or if any please to set down the parts of an hour, he may allow 7 gr. 30 m. for every half hour, and 3 gr. 45 m. for every quarter. This done, you are to consider the latitude of the place, and the quality of the plane: For the secant of the latitude shall be the semidiameter in a vertical plane, & the secant of the compliment of the latitude in an horizontal plane. geometric illustration For example, about London the latitude is 51 gr. 30 m. and let the plane be vertical. If you take AV the secant of 51 gr. 30 m. out of the Sector, and prick it down in the meridian line from A unto V, the point V shall be the centre: and if you draw right lines from V unto 11, and 10, and the rest of the hour points, they shall be the hour lines required. But if the plane be horizontal, than you are to take out AH the secant of 38 gr. 30 m. for the semidiameter, and prick it down in the meridian line from A unto H: so the right lines drawn from the centre H unto the hour points, shall be the hour lines required; only the hour of 6 is wanting, and that must always be drawn parallel to the equator, through the centre V in a vertical, through the centre H in an horizontal plane. 3 To draw the hour lines on a polar plane. 4 To draw the hour lines on a meridian plane. In a polar plane the equator may be also the same with the horizontal line, and the hour points may be pricked on as before, but the hour lines must be drawn parallel to the meridian. In a meridian plane, the equator will cut the horizontal line with an angle equal to the compliment of the latitude of the place; then may you make choice of the point A, and there cross the equator with a right line, which may serve for the hour of 6: so the tangent of 15 gr. being pricked down in the equator on both sides from 6, shall serve for the hours of five 5 and 7; and the tangent of 30 gr. for the hours of 8 and 4; and the tangent of 45 gr. for the hours of 3 and 9; and the tangent of 60 gr. for the hours of 2 and 10; and the tangent of 75 gr. for the hours of 1 and 11. And if you draw right lines through these hour points, crossing the equator at right angles, they shall be the hour lines required. geometric illustration 5 To draw the hour lines in a vertical declining place. First, draw AV the meridian, and A the horizontal line, crossing one the other at right angles in the point A. 2 Then take out AV, the secant of the latitude of the place, which you may suppose to be 51 gr. 30 m. and prick it down in the meridian line from A unto V 3 Because it is a declining plane, and you may suppose it to decline 40 gr. Eastward, you are to make an angle of the declination upon the centre A, below the horizontal line, and to the left hand of the meridian line, because the declination is Eastward, for otherwise it should have been to the right hand, if the declination had been Westward. 4 Take AH, the secant of the compliment of the latitude out of the Sector, & prick it down in the line of declination from A unto H, as you did before for the semidiameter in the horizontal plane. 5 Draw a line at full length through the point A, which must be perpendicular unto AH, and cut the horizontal line according to the angles of declination, and it will be as the equator in the horizontal plane. 6 Take the hour points out of the Tangent line in the Sector, and prick them down in this equator on both sides from the hour of 12 at A. 7 Lay your ruler, & draw right lines through the centre H, & each of these hour points: so have you all the hour lines of an horizontal plane, only the hour of 6 is wanting, and that may be drawn through H perpendicular to HA. Lastly you are to observe and mark the intersections, which these hour lines do make with A the horizontal line of the plane: and then if you draw right lines through the centre V, and each of these intersections, they shall be the hour lines required. 6 To prick down the hour points another way. Having drawn a right line for the equator as before, and made choice of the point A, for the hour of 12: you may at pleasure cut of two equal lines A 10, and A 2. Then upon the distance between 10 and 2, make an equilateral triangle, and you shall have B for the centre of your equator, and the line AB shall give the distance from A to 9, and from A to 3. That done take out the distance between 9 and 3, and this shall give the distance from B unto 8, and from 8 unto 7, and from 8 unto 1: and again from B unto 4, and from 4 unto 5, and from 4 unto 11. So have you the hour points, and if you take out the distance B 1, B 3, B 5, etc. You may find the points not only for the half hours, but also for the quarters. But if it so fall out, that some of these hour points fall out of your plane, you may help yourself by the larger tangent, both in the vertical, and horizontal planes. For if at the hour points of 3 and 9, you draw occult lines parallel to the meridian; the distances DC, between the hour line of 6, and the hour points of 3 and 9, will be equal to the semidiameter AV in a vertical, and AH in a horizontal plane, and if they be divided in such sort as the line AC is divided, you shall have the points of 4, and 5, and 7, and 8, with their halves and quarters. As in the horizontal plane, take out the semidiameter AH, and make it a parallel Radius by fitting it over in the sins of 90 and 90: Then take 15 gr. out of the larger tangent, and lay them on the lines of sins, where they will reach from the centre unto the sins of 15 gr. 32 m. therefore take out the parallel sine of 15 gr. 32 m. and it shall give the distance from 6 unto 5, and from 6 unto 7, in your horizontal plane. That done take out 30 gr. out of the larger tangent, and lay them on the sins, from the centre unto the sins of 35 gr. 16 m. and the parallel sine of 35 g. 16 m. shall give you the distance from 6 unto 4, and from 6 unto 8, in your horizontal plane. The like may be done for the half hours and quarters. So also in the vertical declining plane. If you first take out the secant of the declination of the plane, and prick it down in the horizontal line from A unto E, and through E draw right lines parallel to the meridian, which will cut the former hour lines of 3 and 9, or one of them in the point C: then take out the semidiameter AV, and prick it down in those parallels from C unto D, and draw right lines from A unto C, and from V unto D; the line VD shall be the hour of 6, and if you divide these lines AC and DC, in such sort as you divided the like line DC in the horizontal plane, you shall have all the hour points required. Or you may find the point D, in the hour of 6, without knowledge either of H or C. For having pricked down AV in the meridian line, and A in the horizontal line, and drawn parallels to the meridian through the points at E, you may take the tangent of the latitude out of the Sector, and fit it over in the sins of 90 and 90: so the parallel sine of the declination measured in the same tangent line, shall there show the compliment of the angle DVA, which the hour line of 6 maketh with the meridian; then having the point D, take out the semidiameter VALERIO, and prick it down in those parallels from D unto C: so shall you have the lines DC and AC to be divided as before. The like might be used for the hour lines upon all other planes. But I must not write all that may be done by the Sector. It may suffice that I have wrote something of the use of each line, and thereby given the ingenuous Reader occasion to think of more. The conclusion to the Reader. IT is well known to many of you, that this Sector was thus contrived, the most part of this book written in latin, many copies transcribed and dispersed more than sixteen years since. I am at the last contented to give way that it come forth in English. Not that I think it worthy either of my labour or the public view, but partly to satisfy their importunity, who not understanding the Latin, yet were at the charge to buy the instrument, and partly for my own ease. For as it is painful for others to transcribe my copy, so it is troublesome for me to give satisfaction herein to all that desire it. If I find this to give you content, it shall encourage me to do the like for my cross-staff, and some other Instruments. In the mean time bear with the Printers faults, and so I rest. Gresham Coll. 1. Maij. 1623. E. G. FINIS. THE FIRST BOOK OF THE cross-staff. CHAP. I. Of the description of the Staff. THe cross-staff is an instrument well known to our Seamen, and much used by the ancient Astronomers and others, serving Astronomically for observation of altitude and angles of distance in the heavens, Geometrically for perpendicular heights and distances on land and sea. The description and several uses of it are extant in print, by Gemma Frisius in Latin, in English by Dr. Hood. I differ something from them both, in the projection of this Staff, but so, as their rules may be applied unto it, and all their propositions be wrought by it: and therefore referring the Reader to their books, I shall be brief in the explanation of that which may be applied from theirs unto mine, and so come to the use of those lines which are of my addition, not extant heretofore. The necessary parts of this Instrument are five: the Staff, the Cross, and the three sights. The Staff which I made for my own use, is a full yard in length, that so it may serve for measure. The Cross belonging to it is 26 inches ⅕ between the two outward sights. If any would have it in a greater form, the proportion between the Staff and the Cross, may be such as 360 unto 262. The lines inscribed on the Staff are of four sorts. One of them serves for measure and protraction: one for observation of angles: one for the Sea-chart; and the four other for working of proportions in several kinds. The line of measure is an inch line, and may be known by his equal parts. The whole yard being divided equally into 36 inches, and each inch subdivided, first into ten parts, and then each tenth part into halves. The line for observation of angles may be known by the double numbers set on both sides of the line, beginning at the one side at 20, and ending at 90: on the other side at 40, and ending at 180: and this being divided according to the degrees of a quadrant, I call it the tangent line on the Staff. The next line is the meridian of a Sea-chart, according to Mercators' projection from the Equinoctial to 58 gr. of latitude, and may be known by the letter M, and the numbers 1. 2. 3. 4. unto 58. The lines for working of proportions, may be known by their unequal divisions, and the numbers at the end of each line. 1 The line of numbers noted with the letter N, divided unequally into 1000 parts, and numbered with 1. 2. 3. 4. unto 10. 2 The line of artificial tangents is noted with the letter T, divided unequally into 45 degrees, and numbered both ways, for the Tangent and the compliment. 3 The line of artificial sins, noted with the letter S, divided unequally into 90 degrees, and numbered with 1. 2. 3. 4. unto 90. 4 The line of versed sins for more easy finding the hour and azimuth, noted with V, divided unequally into about 164 gr. 50 m. numbered backward with 10. 20. 30. unto 164. Thus there are seven lines inscribed on the Staff: there are five lines more inscribed on the Cross. 1 A Tangent line of 36 gr. 3 m. numbered by 5. 10. 15. ●ledge●nto 35: the midst whereof is at 20 gr; and therefore I call it ●ledge●he tangent of 20; and this hath respect unto 20 gr. in the Tangent on the Staff. 2 A Tangent line of 49 gr. 6 m. numbered by 5. 10. 15. vn●ledge●o 45; the midst whereof is at 30 gr. and hath respect unto ●ledge●0 gr. in the Tangent on the Staff, whereupon I call it the ●ledge●angent of 30. 3 A line of inches numbered with 1. 2. 3. unto 26; each inch ●ledge●qually subdivided into ten parts, answerable to the inch line ●ledge●pon the Staff. 4 A line of several chords, one answerable to a circle of ●ledge●welue inches semidiameter, numbered with 10. 20. 30. unto ●ledge●0: another to a semidiameter of a circle of six inches; and ●ledge●he third to a semidiameter of a circle of three inches; both ●ledge●umbred with 10. 20. 30. unto 90. 5 A continuation of the meridian line from 57 gr. of la●ledge●tude unto 76 gr; and from 76 gr. to 84 gr. For the inscription of these lines. The first for measure is ●ledge●qually divided into inches and tenth parts of inches. The tangent on the Staff for observation of angles, with ●ledge●e tangent of 20 and the tangent of 30 on the Cross, may ●ledge●l three be inscribed out of the ordinary table of tangents. The ●ledge●affe being 36 inches in length; the Radius for the tangent ●ledge● the Staff will be 13 inches and 103 parts of 1000: so the ●ledge●hole line will be a tangent of 70 gr. and must be numbered ●ledge● their compliments, & the double of their compliments, ●ledge●e tangent of 10 gr. being numbered with 80 and 160. The Radius for the tangent of 20 on the Cross, will be 6 inches, and the whole line between the sights a tangent 36 gr. 3 m. according as it is numbered. The Radius for the ●ledge●ngent of 30 gr. on the Cross, will be 22 inches and 695 ●ledge●rts of 1000: so the whole line between the sights will con●ledge●ine a tangent of 49 gr. 6 m. in such sort as they are num●ledge●ed. The meridian line may be inscribed out of the Table ●ledge●hich I set down for this purpose in the use of the Sector. The line of numbers may be inscribed out of the first Chiliad of Mr. Briggs Logarithmes: & the rest of the lines of proportion out of my Canon of artificial sins and tangents; and in recompense thereof this book will serve as a comment to explain the use of my Canon. CHAP. II. The use of the lines of inches for perpendicular heights and distances. IN taking of heights and distances, the Staff may be held in such sort, that it may be even with the distance, and the Cross parallel with the height: and then if the eye at the beginning of the Staff shall see his marks by the inward sides of the two first sights, there will be such proportion between the distance and the height, as is between the part intercepted on the Staff and the Cross. Which may be farther explained in these propositions. geometric illustration 1 To find an height at one station, by knowing the distance. Set the middle sight unto the distance upon the Staff the height will be found upon the Cross. For As the segment of the safr unto the segment on the Cross: So is the distance given, unto the height. As if the distance AB being known to be 256 feet, it were required to find the height BC: first I place the middle sight at 25 inches and 6 parts of 10; then holding the Staff level with the distance, I raise the Cross, parallel unto the height, in such sort, as that my eye may see from A the beginning of the inches on the Staff by the sight E, at the beginning of the inches on the Cross unto the mark C: which being done, if I find 19 inches and 2 parts of 10 intercepted on the Cross between the sights at E and D, I would say the height BC were 192 feet. Or if the observation were to be made before the distance were measured, I would set the middle sight either unto 10 inches, or 12, or 16, or 20, or 24, or some such other number as might best be divided into several parts, and then work by proportion. As if in the former example the middle sight were at 24 on the Staff, and 18 on the Cross, it should seem that the height is ¾ of the distance; and therefore the distance being 256, the height should be 192. 2 To find an height, by knowing some part of the same height. As if the height from G to C were known to be 48, and it were required to find the whole height BC: either put the third sight or some other running sight upon the Cross between the eye and the mark G. For then As the difference between the sights, unto the whole segment of the Cross: So is the part of the height given, unto the whole height. If then the difference between the sights E and F, shall be 45, and the segment of the Cross ED 180, the whole height BC will be found to be 192. 3 To find an height at two stations, by knowing the difference of the same stations. As the difference of segments on the Staff, unto the difference of stations: So is the segment of the Cross, unto the height. Suppose the first station being at H, the segment of the Cross ED were 180, and the segment of the Staff HD 300: then coming 64 feet nearer unto B, in a direct line, unto a second station at A, and making another observation; suppose the segment of the Cross ED were 180 as before, and the segment of the Staff AD 240; take 240 out of 300, the difference of segments will be 60 parts. And As 60 parts unto 64 the difference of stations: So DE 180 unto BC 192 the height required. In these three Prop. there is a regard to be had of the height of the eye. For the height measured, is no more than from the level of the eye upward. 4 To find a distance, by knowing the height. As the segment of the Cross, unto the segment of the Staff: So is the height given, unto the distance. So the segment ED being 18, and DA 24, the height CB 192, will show the distance AB to be 256. 5 To find a distance, by knowing part of the height. As the difference between the sights, unto the segment of the Staff: So is the part of the height given, unto the distance. And thus the difference between E and F being 45, and the segment DA 240; the part of the height GC 48, will give the distance AB to be 256. 6 To find a distance at two stations, by knowing the difference of the same stations. As the difference of segments on the Staff, unto the difference of stations: So is the whole segment, unto the distance. And thus the segment of the Cross being 180, the segment of the Staff at the first station 240, at the second 300, the difference of the segments 60, & the difference of stations 64, the distance AB at the first station will be found to be 256, and the distance HB at the second station 320. 7 To find a breadth by knowing the distance perpendicular to the breadth. This is all one with the first Prop. For this breadth is but an height turned sidewayes: and therefore As the segment of the Staff, unto the segment of the Cross: So is the distance unto the breadth. And thus the segment of the Staff being 24, and the segment of the Cross 18, the distance AB 256, will give the breadth BC to be 192. 8 To find a breadth at two stations in a line perpendicular to the breadth, by knowing the difference of the same stations. This is also the same with the third Prop: and therefore As the difference of segments on the Staff, unto the difference of stations: So the segment on the Cross between the two sights, unto the breadth required. And thus the difference between the stations at A and H being 64, the difference of segments on the Staff 60, the segment of the Cross 180, the breadth BC will be found to be 192. In like manner may we find the breadth GC for having found the breadth BC the proportion will hold. As DE is unto FE, so B C unto GC. Or otherwise, As H a unto HA', so FE unto GC. Neither is it material whether the two stations be chosen at the one end of the breadth proposed, or without it, or within it, if the line between the stations be perpendicular unto the breadth: as may appear if in stead of the stations at A and H, we make choice of the like stations at I and K. There might be other ways proposed to work these Prop. by holding the Cross even with the distance, and the Staff parallel with the height: but these would prove more troublesome, and those which are delivered are sufficient, and the same with those which others have set down under the name of the jacobs' staff. CHAP. III. The use of the Tangent lines in taking of Angles. geometric illustration 1 To find an angle by the Tangent on the Staff. LEt the middle sight be always set to the middle of the Cross, noted with 20 and 30, and then the Cross drawn nearer the eye, until the marks may be seen close within the sights. For so if the eye at A (that end of the Staff which is noted with 90 and 180) beholding the marks K and N, between the two first sights, C and B, or the marks K and P between the two outward sights, the Cross being drawn down unto H, shall stand at 30 and 60, in the Tangent on the Staff: it showeth that the angle CAN is 30 gr. the angle KAP 60 gr. the on● double to the other; which is ●he reason of the double numbers on this line of the Staff: and this way will serve for any angle from 20 gr. toward 90 gr. or from 40 gr. toward 180 gr. But if the angle be less than 20 gr. we must then make use of the Tangent upon the Cross. 2 To find an angle by the Tangent of 20 upon the Cross. Set 20 unto 20, that is, the middle sight to the midst of the Cross at the end of the Staff, noted with 20: so the eye at A, beholding the marks L and N, close between the two first sights, C and B, shall se● them in an angle of 20 gr. If the marks sh●ll be nearer together, as are M and N, then draw in the Cross from C unto E: if they be farther asunder, as are K and N, then draw out the Cross from C unto F; so the quantity of the angle shall be still found in the Cross in the Tangent of 20 gr. at the end of the Staff; and this will serve for any angle from 0 gr. toward 35 gr. 3 To find an angle by the Tangent of 30 upon the Cross. This Tangent of 30 is here put the rather, that the end of the Staff resting at the eye, the hand may more easily remove th' Cross: for it supposeth the Radius to be no longer then AH, which is from the eye at the end of the Staff unto 30 gr. about 22 inches and 7 parts. Wherefore here set the middle sight unto 30 gr. on the Staff, and then either draw the Cross in or out, until the marks be seen between the two first sights; so the quantity of the angle will be found in the Tangent of 30, which is here represented by the line GH; and this will serve for any angle from 0 gr. toward 48 gr. 4 To observe the altitude of the Sun backward. Here it is fit to have an horizontal sight set to the beginning of the Staff, and then may you turn your back toward the Sun, and your Cross toward your eye. If the altitude be under 45 gr. set the middle sight to 30 on the Staff, and look by the middle sight through the horizontal unto the horizon, moving the Cross upward or downward, until the upper sight do shadow the upper half of the horizontal sight: so the altitude will be found in the Tangent of 30. If the altitude shall be more than 45 gr. set the middle sight unto the midst of the Cross, and look by the inward edge of the lower sight through the horizontal to the horizon, moving the middle sight in or out, until the upper sight do shadow the upper half of the horizontal sight: so the altitude will be found in the degrees on the Staff between 40 and 180. 5 To set the Staff to any angle given. This is the converse of the former Prop. For if the middle sight be set to his place and degree, the eye looking close by the sights as before, cannot but see his object in the angle given. 6 To observe the altitude of the Sun another way. Set the middle sight to the middle of the Cross, and hold the horizontal sight downward, so as the Cross may be parallel to the horizon, then is the Staff vertical; and if the outward sight of the Cross do shadow the horizontal sight, the compliment of the altitude will be found in the tangent on the Staff. 7 To observe an altitude by thread and plummet. Let the middle sight be set to the midst of the Cross, and to that end of the Staff which is noted with 90 and 180; then having a thread and a plummet at the beginning of the Cross, and turning the Cross upward, and the Staff toward the Sun, the thread will fall on the compliment of the altitude above the horizon. And this may be applied to other purposes. 8 To apply the lines of inches to the taking of angles. If the angles be observed between the two first sights, there will be such proportion between the parts of the Staff and the parts of the Cross, as between the Radius and the Tangent of the angle. As if the parts intercepted on the Staff were 20 inches, the parts on the Cross 9 inches. Then by proportion as 20 unto 9, so 100000 unto 45000 the tangent of 24 gr. 14 m. But if the angle shall be observed between the two outward sights, the parts being 20 and 9 as before, the angle will be 48 gr. 28 m. double unto the former. In all these there is a regard to be had to the parallax of the eye, and his height above the Horizon in observations at Sea; to the Semidiameter of the Sun, his parallax and refraction, as in the use of other staffs. And so this will be as much, or more than that which hath been heretofore performed by the cross-staff. CHAP. FOUR The use of the lines of equal parts joined with the lines of Chords. THe lines of equal parts do serve also for protraction, as may appear by the former Diagrams; but being joined with the lines of Chords, which I place upon one side of the Cross, they will farther serve for the protraction and resolution of right line triangles; whereof I will give one example in finding of a distance at two stations otherwise then in the second Cap. geometric illustration Let the distance required be AB. At A the first station I make choice of a station line toward C, and observe the angle BAC by the tangent lines, which may be 43 gr. 20 m; then having gone an hundred paces toward C, I make my second station at D, where suppose I find the angle BDC to be 58 gr. or the angle BDA to be 122 gr; this being done, I may find the distance AB in this manner. 1 I draw a right line AC, representing the station line. 2 I take 100 out of the lines of equal parts, and prick them down from A the first station unto D the second. 3 I open my compasses to one of the chords of 60 gr. and setting one foot in the point A, with the other I describe an occult ark of a circle intersecting the station line in E. 4 I take out of the same line of chords a chord of 43 gr. 20 m. (because such was the angle at the first station) and this I inscribe into that occult ark from E unto F, which makes the angle FAD equal to the angle observed at the first station. 5 I describe another like ark upon the centre D, and inscribe into it a chord of 58 gr. from C unto G, and draw the right line DG, which doth meet with the other line OF in the point B, and makes the angle BDC equal to the angle observed at the second station. So the angles in the Diagram being equal to the angles in the field, their sides will be also proportional: and therefore, 6 I take out the line AB with my compasses, and measuring it in the same line of equal parts, from which I took AD, I find it to be 335, and such is the distance required. CHAP. V The use of the Meridian line. 1 THe Meridian line, noted with the letter M, may serve for the more easy division of the plane sea-chart, according to Mercators' projection. For if you shall draw parallel meridians, each degree being half an inch distant from other, the degrees of this meridian line on the Staff, shall give the like degrees for the meridians on the chart, from the Equinoctial toward the Pole: and then if through these degrees you draw straight lines perpendicular to the meridians, they shall be parallels of latitude. If any desire to have the degrees of his chart larger than those which I have put on the Staff, he may take these and increase them in a double, or triple, or a decuple proportion at his pleasure. 2 This meridian line being joined with the line of chords, may serve for the protraction & resolution of such right line triangles as concern latitude, longitude, rumb and distance in the practice of navigation. As may appear by this example. Suppose two places given, A in the latitude of 50 gr. D in the latitude of 52 gr. ½, the difference of longitude between them being 6 gr. and let it be required to know, first what Rumb leadeth from the one place to the other; secondly how many degrees distant they are asunder. 1 I draw a right line A, representing the parallel of the place from whence I depart. 2 I take 6 gr. for the difference of longitude ●●th●r out of the line of inches, allowing half an inch for e●●●y de●ree, o● out of the beginning of the meridian line; (for there the meridian degrees d ffer very little from the equinoctial degrees) and these 6 gr. I prick down in the parallel from A to E. 3 In A and E, I erect two perpendiculars, AM and ED, representing the meridians of both places. geometric illustration 4 I take the difference of latitude from 50 gr. to 52 gr. 30 m. out of the meridian line, and prick it down in the meridians from A unto M, and from E to D, and draw the right line MD for the parallel of the second place, and the right line AD for the line of distance between both places: so the angle MAD shall give the Rumb that leadeth from the one place to the other. 5 To find the quantity of this angle MAD, I may either make use of the Protractor, or else of a line of chords, and so I open my compasses unto one of the chords of 60 gr. and setting one foot in the point A, with the other I describe an occult ark of a circle, intersecting the meridian in F, and the line of distance in G; then I take the chord FG with my compasses, and measuring it in the same line of chords as before, I find it 56 gr. ¼: and such is the inclination of the Rumb to the meridian, which is the first thing that was required. 6 To find the quantity of the line of distance AD, I take it out with my compasses, and measuring it in the meridian line, setting one foot beneath the lesser latitude, and the other foot as much above the greater latitude, I find about 4 gr. ½ intercepted between both feet: and such is the distance upon the Rumb, which is the second thing that was required. But if this example were protracted according to the common Sea-chart, where the degrees of the equinoctial and meridian are both alike; the Rumb MAD would be found to be above 67 gr. and AD the distance upon the Rumb about 6 gr. ½. Suppose farther, that having set forth from A toward D, upon the former Rumb of 56 gr. 15 m. NEbE, after the ship had run 36 leagues, the wind changing, it ran 50 leagues more upon the seventh Rumb of EbN, whose inclination to the meridian is 78 gr. 45 m. And let it be required to know what longitude and latitude the ship is in, by pricking down the way thereof upon the Chart. Having drawn a blank chart as before, with meridians & parallels, according to the latitude of the places proposed. 1 I would make an angle MAD of 56 gr. 15 m. for the Rumb of NEbE, which is done after this manner: I open my compasses to one of the chords of 60 gr. and setting one foot in the point A, with the other I describe an occult ark of a circle, intersecting the meridian in F; then I take 56 gr. 15 m. out of the same line of chords, and prick them down from F unto G: so the right line AGNOSTUS shall be the Rumb of NEbE. 2 I would take 36 leagues out of the meridian line, extending my compasses from 50 gr. to 51 gr. 48 m. or rather from as much below 50 as above 51, and prick them down upon the Rumb from A unto I; so the point I, shall represent the place wherein the ship was when the wind changed. And this is in the latitude of 51 gr. 0 m. and in the longitude of 2 gr. 21 m. Eastward from the meridian AM. 3 By the same reason, I may draw the right line IK for the Rumb of EbN, and prick down the distance of 50 leagues from I unto K: so the point K shall represent the place whither the ship came, after the running of these 50 leagues and this is in the latitude of 51 gr. 30 m. and in longitude 6 gr. 16 m. Eastward from the first meridian AM, and therefore 16 m. Eastward from the second meridian ED. But if these two courses were to be pricked down by the common sea-chart, the point I would fall in the latitude of 51 gr. 0 m. and the point K in the latitude of 51 gr. 30 m. But the longitude of I would be only 1 gr. 30 m. and the longitude of K only 3 gr. 57 m. which is 33 m Westward from the meridian of the place to which the ship was bound. Such is the difference between both these charts'. CHAP. VI The use of the line of Numbers. 1 Having two numbers given to find a third in continual proportion, a fourth, a fift, and so forward. EXtend the compasses from the first number unto the second; then may you turn them, from the second to the third, and from the third to the fourth, and so forward. Let the two numbers given be 2 and 4. Extend the compasses from 2 to 4, then may you turn them from 4 to 8, and from 8 to 16, and from 16 to 32, and from 32 to 64, and from 64 to 128. Or if the one foot of the compasses being set to 64, the other fall out of the line, you may set it to another 64 nearer the beginning of the line, and there the other foot will reach to 128, and from 128 you may turn them to 256, and so forward. Or if the two first numbers given were 10 and 9: extend the compasses from 10 at the end of the line, back unto 9, then may you turn them from 9 unto 8.1, and from 8.1 unto 7.29. And so if the two first numbers given were 1 and 9, the third would be found to be 81, the fourth 729, with the same extent of the compasses. In the same manner, if the two first numbers were 10 and 12, you may find the third proportional to be 14.4, the fourth 17.28. And with the same extent of the compasses, if the two first numbers were 1 and 12, the third would be found to be 144, and the fourth to be 1728. 2 Having two extreme numbers given, to find a mean proportional between them. Divide the space between the extreme numbers into two equal parts, and the foot of the compasses will stay at the mean proportional. So the extreme numbers given being 8 and 32, the mean between them will be found to be 16, which may be proved by the former Prop. where it was showed, that as 8 to 16, so are 16 to 32. 3 To find the square root of any number given. The square root is always the mean proportional between 1 and the number given, and therefore to be found by dividing the space between them into two equal parts. So the root of 9 is 3, and the root of 81 is 9, and the root of 144 is 12. 4 Having two extreme numbers given, to find two mean proportionals between them. Divide the space between the two extreme numbers given, into three equal parts. As if the extreme numbers given were 8 and 27, divide the space between them into three equal parts, the feet of the compasses will stand in 12 and 18. 5 To find the cubique root of a number given. The cubique root is always the first of two mean proportionals between 1 and the number given, and therefore to be found by dividing the space between them into three equal parts. So the root of 1728 will be found to be 12. The root of 17280 is almost 26: and the root of 172800 is almost 56. 6 To multiply one number by another. Extend the compasses from 1 to the multiplicator; the same extent applied the same way, shall reach from the multiplicand to the product. As if the numbers to be multiplied were 25 and 30: either extend the compasses from 1 to 25, and the same extent will give the distance from 30 to 750; or extend them from 1 to 30, and the same extent shall reach from 25 to 750. 7 To divide one number by another. Extend the compasses from the divisor to 1, the same extent shall reach from the dividend to the quotient. So if 750 were to be divided by 25, the quotient would be found to be 30. 8 Three numbers being given to find a fourth proportional. This golden rule, the most useful of all others, is performed with like ease. For extend the compasses from the first number to the second, the same extent shall give the distance from the third to the fourth. As for example, the proportion between the diameter and the circumference, is said to be such as 7 to 22: if the diameter be 14, how much is the circumference? Extend the compasses from 7 to 22, the same extent shall give the distance from 14 to 44: or extend them from 7 to 14, and the same extent shall reach from 22 to 44. Either of these ways may be tried on several places of this line; but that place is best, where the seete of the compasses may stand nearest together. 9 Three numbers being given to find a fourth in a duplicated proportion. This proposition concerns questions of proportion between lines and superficies; where if the denomination be of lines, extend the compasses from the first to the second number of the same denomination: so the same extent being doubled, shall give the distance from the third number unto the fourth. The diameter being 14, the content of the circle is 154: the diameter being 28, what may the content be? Extend the compasses from 14 to 28, the same extent doubled will reach from 154 to 616. For first it reacheth from 154 unto 308; and turning the compasses once more, it reacheth from 308 unto 616: and this is the content required. But if the first denomination be of the superficial content, extend the compasses unto the half of the distance, between the first number and the second of the same denomination: so the same extent shall give the distance from the third to the fourth. The content of a circle being 154, the diameter is 14: the content being 616, what may the diameter be? Divide the distance between 154 and 616 into two equal parts, than set one foot in 14, the other will reach to 28 the diameter required. 10 Three numbers being given to find a fourth in a triplicated proportion. This proposition concerneth questions of proportion between lines and solids; where if the first denomination be of lines, extend the compasses from the first number to the second of the same denomination: so the extent being tripled, shall give the distance from the third number unto the fourth, Suppose the diameter of an iron bullet being 4 inches, the weight of it was 9 l: the diameter being 8 inches, what may the weight be? Extend the compasses from 4 to 8, the same extent being tripled, will reach from 9 unto 72. For first it reacheth from 9 unto 18; then from 18 to 36; thirdly from 36 to 72. And this is the weight required. But if the first denomination shall be of the Solid content, or of the weight, extend the compasses to a third part of the distance between the first number and the second of the same denomination: so the same extent shall give the distance from the third number unto the fourth. The weight of a cube being 72 l, the side of it was 8 inches: the weight being 9 l, what may the side be? Divide the distance between 72 and 9, into three equal parts; then set one foot to 8, the other will reach to 4, the side required. CHAP. VII. The use of the lines of artificial Sins. THis line of sins hath such use in finding a fourth proportional, as the ordinary Canon of Sines: and the manner of finding it, is always such as in this example. As the sine of 30 gr. unto the sine of 52 gr. So the sine of 38 gr. to a fourth sine. Extend the compasses in the line of sins from 30 gr. unto 52 gr; the same extent shall give the distance from 38 gr. unto 76 gr. Or extend them from 30 gr. unto 38 gr. the same extent will reach from 52 gr. unto 76 gr. which is the fourth proportional sine required. And thus may the rest of all sinical proportions be wrought two ways. The minutes which are wanting in the first degree, may be supplied by the line of Numbers. CHAP. VIII. The use of the line of artificial Tangents. THis line of Tangents hath like use, but commonly joined with the line of sins: the manner of working by it, may appear by this example. As the Tangent of 38 gr. 30 m. is to the Tangent of 23 gr. 30 m. So the Sine of 90 gr. to a fourth Sine. This Prop. and such others upon two lines, may be wrought two ways. For extend the compasses from the Tangent of 38 gr. 30 m. to the Tangent of 23 gr. 30 m; the same extent shall give the distance from the sine of 90 gr. to the sine of 33 gr. 8 m. Or else extend them from 38 gr. 30 m. in the Tangents unto 90 gr. in the line of sins; the same extent from the Tangent of 23 gr. 30 m. shall reach to the sine of 33 gr. 8 m. which is the fourth proportional sine required. And this crossework in many cases is the better, in regard the tangents which should pass on from 40 gr. to 50 gr. and so forward, do turn back at 45 gr. These two lines of Sines and Tangents, may serve for the resolution of all spherical triangles, according to those Canons which I have set down in the use of the Sector. Or if at any time one meet with a secant, let him account the sine of 80 gr. for a secant of 10 gr. and the sine of 70 gr. for a secant of 20 gr. and so take the sine of the compliment in stead of the secant. As if the proposition were, As the Radius to the secant of 51 gr. 30 m. So the sine of 23 gr. 30 m. to a fourth sine. Extend the compasses from the Radius that is the sine of 90 gr. to the sine of 38 gr. 30 m. the same extent will give the distance from the sine of 23 gr. 30 m. both to the sine of 14 gr. 22 m. and to the sine of 39 gr. 50 m. But in this case, the sine of 39 gr. 50 m. is the fourth required. For the first number being less than the second, that is, the Radius less than the secant, the sine of 23 gr. 30 m. which is the third, must also be less than the fourth. CHAP. IX. The use of the line of Sines and Tangents joined with the line of numbers. THe lines of sins and tangents have another like use joined with the line of numbers, especially in the resolution of right line triangles, where the angles are measured by degrees and minutes, and the sides measured by absolute numbers, whereof I will set down these propositions. 1 Having three angles and one side, to find the two other sides. As the side of the angle opposite to the side given, is to the number belonging to that side given: So the sine of the angle opposite to the side required, to the number belonging to the side required. As in the example of the fourth Cap. of this book, where knowing the distance between two stations at A and D to be 100 paces, the angle BAC to be 43 gr. 20 m. and the angle BDC to be 58 gr. it was required to find the distance AB. First having these two angles, I may find the third angle ABDELLA to be 14 gr. 40 m. either by substraction or by compliment unto 180. Then in the triangle BAD, I have three angles, and one side, whereby I may find both AB and DB. I know the angle ABDELLA opposite to the measured side AD to be 14 gr. 40 m. and the angle ADB opposite to the side required, to be 122 gr: wherefore I extend the compasses in the line of sins from 14 gr. 40 m. unto 122 gr. or (which is all one) to 58 gr. (for after 90 gr. the sine of 80 gr. is also the sine of 100 gr. and the sine of 70 gr. the sine of 100 gr. and so in the rest) so shall I find the same extent to reach in the line of numbers, from 100 unto 335. And such is the distance required between A and B. geometric illustration In like manner if I extend my compasses from the sine of 34 gr. 40 m. to the sine of 43 gr. 20 m. the same extent will reach in the line of numbers from 100 to 271. And such is the distance between D and B. Or in cross work, I may extend the compasses from 14 gr. 40 m. in the sins, unto 100 parts in the line of numbers: so the same extent will give the distance from 58 gr. to 335 parts, and from 43 gr. 20 m. to 271 parts. 2 Having two sides given, and one angle opposite to either of these sides, to find the other two angles and the third side. As the side opposite to the angle given, is to the sine of the angle given: So the other side given, to the sine of that angle to which it is opposite. So in the former triangle, having the two sides AB 335 paces and AD 100 paces, and knowing the angle ADB, which is opposite to the side AB, to be 122 gr. I may find the angl● ABDELLA, which is opposite to the other side AD. For if I extend the compasses from 335 to 100 in the line of numbers, I shall find the same extent to reach in the line of sins from 122 gr. to 14 gr. 40 m; and therefore such is the angle ABDELLA. Then knowing these two angles ABDELLA and ADB, I may find the third angle BAD either by subtraction or by compliment to 180, to be 43 gr. 20 m; and having three angles and two sides, I may well find the third side DB, by the former Prop. This may be done more readily by cross work. For if I extend the compasses from 335 parts, in the line of numbers, to the sine of 122 gr. the same extent will reach from 100 parts to the sine of 14 gr. 40 m. and back from 43 gr. 20 m. to 271 parts; and such is the third side DB. 3 Having two sides and the angle between them, to find the two other angles and the third side. If the angle contained between the two sides be a right angle, the other two angles will be found readily by this canon. As the greater side given, is to the lesser side: So the tangent of 45 gr. to the tangent of the lesser angle. So in the rectangle triangle AIB, knowing the side AI to be 244, and the side IB to be 230: if I extend the compasses from 244 to 230 in the line of numbers, the same extent will reach from 45 gr. to about 43 gr. 20 m. in the line of tangents; and such is the lesser angle BAJAZET, and the compliment 46 gr. 40 m. shows the greater angle ABI. The angles being known, the third side AB may be found by the first Prop. So likewise in the example of the third Cap. of this book, concerning taking of angles by the line of inches, where the parts intercepted on the Staff being 20 inches, and the parts on the Cross 9 inches, it was required to find the angle of altitude. For I may extend the compasses in the line of numbers, from 20 unto 9, the same extent will reach in the line of tangents, from 45 gr. to 24 gr. 14 m. Or in the cross work, I may extend the compasses from 20 parts in the line of numbers to the tangent of 45 gr; the same extent shall give the distance from 9 parts unto the tangent of 24 gr. 14 m. And such is the angle of altitude required. But if it be an obliqne angle that is contained between the two sides given, the triangle may be reduced into two rectangle triangles, and then resolved as before. As in the triangle ADB, where the side AB is 335, and the side AD 100, and the angle BADE 43 gr. 20 m: if I let down the perpendicular DH upon the side AB, I shall have two rectangle triangles, AND, DHB; and in the rectangle AND, the angle at A being 43 gr. 20 m. the other angle ADH will be 46 gr. 40 m; and with these angles and the side AD, I may find both AH and D H, by the first Prop. Then taking AH out of AB, there remains HB for the side of the rectangle DHB; and therefore with this side HB and the other side DH, I may find both the angle at B, and the third side DB, as in the former part of this Prop. Or I may find the angles required, without letting down any perpendicular. For As the sum of the sides, is to the difference of the sides: So the tangent of the half sum of the opposite angles, to the tangent of half the difference between those angles. As in the former triangle ADB, the sum of the sides AB, AD, is 435, and the difference between them 235; the angle contained 43 gr. 20 m; and therefore the sum of the two opposite angles 136 gr. 40 m. and the half sum 68 gr 20 m. Hereupon I extend the compasses in the line of numbers from 455 to 235, and I find them to reach in the line of tangents from 68 gr. 20 m. unto 53 gr. 40 m; and such is the half difference between the opposite angles at B and D. This half difference being added to the half sum, doth give 122 gr. for the greater angle ADB: and being subtracted, it leaveth 14 gr. 40 m. for the lesser angle ABDELLA. Then the three angles being known, the third side BD may be found by the first Prop. 4 Having the three sides of a right line triangle, to find the perpendicular and the three angles. Let one of the three sides given be the base, but rather the greater side, that the perpendicular may fall within the triangle; then gather the sum, and the difference of the two other sides, and the proportion will hold. As the base of the triangle, is to the sum of the sides: So the difference of the sides to a fourth, which being taken forth of the base, the perpendicular shall fall on the middle of the remainder. As in the former triangle ADB, where the base AB is 335, the sum of the sides AD and DB 371, and the difference of them 171. If I extend the compasses in the line of numbers from 335 unto 371, I shall find the same extent to reach from 171 unto 189.4. This fourth number I take out of the base 335.0, and the remainder is 145.6, the half whereof is 72.8, and doth show the place H, where the perpendicular shall fall, from the angle D, upon the base AB, dividing the former triangle ADB into two right angle triangles, DHA and DHB, in which the angles may be found by the former part of the third Prop. And this may suffice for right line triangles. But for the more easy protraction of these triangles, I will set down one proposition more concerning chords. 5 Having the semidiameter of a circle, to find the chords of every ark. As the sine of 30 gr. to the sine of half the ark proposed: So is the semidiameter of the circle given, to the chord of the same ark. As if in protracting the former triangle ADB, it were required to find the length of a chord of 43 gr. 20 m. agreeing to the semidiameter A, which is known to be 3 inches. The half of 43 gr. 20 m. is 21 gr. 40 m; wherefore I extend the compasses from the sine of 30 gr. to the sine of 21 gr. 40 m. and I find the same extent to reach in the line of numbers from 3.000 parts to 2.215; which shows, that the semidiameter being 3 inches, the chord of 43 gr. 20 m. will be 2 inches and 215 parts of 1000 In like manner the chord of 58 gr. agreeing to the same semidiameter, would be found to be 2 inches and 909 parts. For the half of 58 being 29; if I extend the compasses in the line of sins from 30 gr. to 29 gr. the same extent will reach in the line of numbers from 3.000. unto 2.909. Or in cross work, if I extend the compasses from the sine of 30 gr. to 3.000 in the line of numbers, I shall find the same extent to reach from 21 gr. 40 m. to 2.215 parts, and from 29 gr. to 2.909 parts, and from 7 gr. 20 m. to 765 parts; for the chord of 14 gr. 40 m. for the third angle ABDELLA. CHAP. X. The use of the line of versed sins. THis line of versed sins is no necessary line. For all triangles, both right lined and spherical, may be resolved by the three former lines of numbers, sins and tangents; yet I thought good to put it on the Staff for the more easy finding of an angle having three sides, or a side having three angles of a spherical triangle given. Suppose the three sides to be, one of them 110 gr. the other 78 gr. and the third 38 gr. 30 m. and let it be required to find the angle, whose base is 110 gr. I first add them together, and from half the sum subtract the base, noting the difference after this manner. The base 110 gr. 0 m. The one side 78 0 The other side 38 30 The sum of all three 226 30 The half sum 113 15 The difference 3 15 This done, I come to the Staff, and extend the compasses from the sine of 90 gr. to the sine of 78 gr. which is one of the sides; and applying this extent from the sine of the other side 38 gr. 30 m. I find it to reach to a fourth sine, about 37 gr. 30 m. From this fourth sine of 37 gr. 30 m. I extend the compasses again, to the sine of the half sum 113 gr. 15 m. (which is all one with the sine of 66 gr. 45 m.) and this second extent will reach from the sine of the difference 3. gr. 15 m. to the sine of 4 gr. 54 m. Over against this sine you shall find 146 gr. in the line of versed sins; and such is the angle required. THE SECOND BOOK. geometric illustration Of the use of the former lines of proportion, more particularly exemplified in several kinds. THe former book containing the general use of each line of proportion, may be sufficient for all those which know the rule of Three, and the doctrine of triangles. But for others, I suppose it would be more difficult to find either the declination of the Sun, or his amplitude, or the like, by that which hath been said in the use of the line of sins, unless they may have the particular proportions, by which such propositions are to be wrought. And therefore for their sakes I have adjoined this second book, containing several proportions for propositions of ordinary use, and set them down in such order, that the Reader considering which is the first of the three numbers given, may easily apply them to the Sector, and also resolve them by Arithmetic, beginning with those which require help only of the line of numbers. CHAP. I. The use of the line of Numbers in broad measure, such as board, glass, and the like. THe ordinary measure for breadth and length are feet and inches, each foot divided into 12 inches, and every inch into halves & quarters, which being parts of several denominations, doth breed much trouble both in arithmetic and the use of instruments. For the avoiding whereof, where I may prevail I give this counsel, that such as are delighted in measure would use several lines, first a line of inch measure, wherein every inch may be divided into 10 or 100 parts; secondly a line of foot measure, wherein every foot may be divided into 100 or 1000 parts, both which lines may be set on the same side of a two foot ruler, after this or the like manner. two foot ruler Then if they be to give the content of any superficies or solid in inches, they may measure the sides of it by the line of inches and parts of inches; but if they be to give the content in feet, it would be more easy for them to measure those sides by the foot line and his parts. For example, let the length of a plane be 30 inches, and the breadth 21 inches and 6/10 of an inch; this length multiplied into the breadth, would give the content to be 648 inches: but if I were to find the content of the same plane in feet, I would measure the sides of it by the foot line and his parts; so the length would prove to be 2 feet 50/100, and the breadth 1 foot 80/100, and the length multiplied by the breadth, cutting off the four last figures, for the four figures of the parts, would give the content to be 4.5000, which is 4 foot and 5000 parts, of a foot being divided into 10000 parts. 21.6 2.50 30.0 1.80 648.00 20000 250 4.5000 The like reason holdeth for yards and elnes, and all other measures divided into 10, 100, or 1000 parts. This being presupposed, the work will be more easy both by arithmetic and the line of numbers, as may appear by these propositions. 1 Having the breadth and length of any oblong superficies given in inch-measure, to find the content in inches. As 1 inch unto the breadth in inches: So the length in inches unto the content in inches. geometric illustration Suppose in the plane AD, the breadth AC to be 30 inches, and the length AB to be 183 inches; extend the compasses from 1 unto 30, the same extent will reach from 183 unto 5490; or extend them from 1 unto 183, the same extent will reach from 30 unto 5490. So both ways the content required is found to be 5490 inches. As 1 unto 30: so are 183 unto 5490. 2 Having the length and breadth of any oblong superficies given in inches, to find the content in feet. As 144 inches unto the breadth in inches: So the length in inches unto the content in feet. And thus in the former plane AD, working as before, the content will be found to be 38.125, which is 38 foot and ⅛ of a foot. As 144 unto 30: so are 183 unto 38.125. 3 Having the length and breadth of any oblong superficies given in foot measure, to find the content in feet. As 1 foot unto the breadth in foot measure: So the length in feet unto the content in feet. And thus in the former plane AD, the breadth will be 2 foot 50 parts, and the length 15 foot 25 parts; then working as before, the content will be found to be 38.125. As 1 unto 2.50: so are 15.25 unto 38.125. 4 Having the breadth of any oblong superficies given in inches, and the length in foot measure, to find the content in feet. As 12 inches to the breadth in inches: So the length in feet to the content in feet. So also in the former plane, the content will be found to be 38.125. As 12 unto 30: so are 15.25 unto 38.125. 5 Having the breadth of an oblong superficies given in inches, to find the length of a foot superficial in inch measure. As the breadth in inches, unto 144 inches: So 1 foot unto the length in inch measure. So the breadth being 30 inches, the length of a foot will be found to be 4 inches 80 parts. As 30 unto 144: so are 1 unto 4.80. 6 Having the breadth and length of an oblong superficies given in feet, to find the length of a foot superficial in foot measure. As the breadth in foot measure to 1 foot: So the number of feet to the length in foot measure. So the breadth being 2 foot 50 parts, the length of a foot will be found to be 40 parts, the length of 2 feet 80 parts, and the length of 3 feet 1 foot 20 parts, etc. As 250 unto 1: so are 1 unto 0.40. 7 Having the length and breadth of an oblong superficies, to find the side of a square equal to the oblong. Divide the space between the length and the breadth into two equal parts, and the foot of the compasses will stay at the side of the square. So the length being 183 inches, and the breadth 30 inches, the side of the square will be found to be almost 74 inches and 10 parts of 100 Or the breadth being 2 foot and 50 parts, the length 15 foot and 25 parts, the side of the square will be found to be about 6 feet and 17 parts. As 30 unto 74.10: so are 74.10 unto 183.027. And as 2.50 unto 6.174: so are 6.174 unto 15.247. 8 Having the diameter of a circle, to find the side of a square equal to that circle. As 10000 to the diameter: So 8862 unto the side of the square. So the diameter of a circle being 15 inches, the side of the square will be found about 13 inches and 29 parts. As 10000 unto 8862: so are 15 unto 13.29. 9 Having the circumference of a circle to find the side of a square equal to the same circle. As 10000 to the circumference: So 2821 to the side of the square. So the circumference of a circle being 47 inches 13 parts, the side of the square will be about 13 inches 29 parts. As 10000 unto 2821: so are 47.13 unto 13.29. 10 Having the diameter of a circle, to find the circumference. 11 Having the circumference of a circle, to find the diameter. As 1000 to the diameter: So 3142 to the circumference. So the diameter being 15 inches, the circumference will be found about 47 inches 13 parts: or the circumference being 47.13, the diameter will be 15. CHAP. II. The use of the line of Numbers in the measure of land by perches and acres. 1 Having the breadth and length of an oblong superficies given in perches, to find the content in perches. As 1 perch to the breadth in perches: So the length in perches to the content in perches. So in the former plane AD, if the breadth AC be 30 perches, and the length AB 183 perches, the content will be found to be 5490 perches. 2 Having the length and breadth of an oblong superficies given in perches, to find the content in acres. As 160 to the breadth in perches: So the length in perches to the content in acres. So in the former plane AD, the content will be found to be 34 acres, and 31 centesms or parts of an 100 As 160 unto 30: so are 183 unto 34.31. 3 Having the length and breadth of an oblong superficies given in chains, to find the content in acres. It being troublesome to divide the content in perches by 160, we may measure the length and breadth by chains, each chain being 4 perches in length, and divided into 100 links, then will the work be more easy in arithmetic. For As 10 to the content in chains: So the length in chains to the content in acres. And thus in the former plane AD, the breadth AC will be 7 chains 50 links, and the length AB 45 chains 75 links; then working as before, the content will be found as before, 34 acres 31 parts. 4 Having the perpendicular and base of a triangle given in perches, to find the content in acres. If the perpendicular go for the breadth, and the base for the length, the triangle will be the half of the oblong. As the triangle CED is the half of the oblong AD, whose content was found in the former Prop. Or without halfing, As 320 to the perpendicular: So the base to the content in acres. So in the triangle CED, the perpendicular being 30, and the base 183, the content will be found to be about 17 acres and 15 parts. 5 Having the perpendicular and base of a triangle given in chains, to find the content in acres. As 20 to the perpendicular: So the base to the content in acres. And so in the triangle CED, the perpendicular OF being 7.50, and the base CD 45.75, the content will be found as before to be about 17 acres 15 parts. 6 Having the content of a superficies after one kind of perch, to find the content of the same superficies according to another kind of perch. As the length of the second perch to the length of the first perch: So the content in acres to a fourth number; and that fourth to the content in acres required. Suppose the plane AD measured with a chain of 66 feet, or with a perch of 16 feet and an half, contained 34 acres 31 parts; and it were demanded how many acres it would contain if it were measured with a chain of 18 foot to the perch: these kind of propositions are wrought by the backward rule of three, after a duplicated proportion. Wherefore I extend the compasses from 16.5 unto 18.0, and the sam●redge● extent doth reach backward, first from 34.31 to 31.45, and then from 31.45 to 28.84, which shows the content to be 28 acres 84 parts. 7 Having the plot of a plane with the content in acres, to find the scale by which it was plotted. Suppose the plane AD contained 34 acres 31 centesines if I should measure it with a scale of 10 in the inch, th●redge● length AB would be 38 chains and about 12 centesmes, and the breadth AC 6 chains and 25 centesmes; and the conten●redge● would be found by the third Prop. of this Chapter, to be about 23 acres 82 parts, whereas it should be 34 acres 31 parts Wherefore I divide the distance between 23.82, and 34 31, upon the line of numbers into two equal parts; then setting one foot of the compasses upon 10, my supposed scale, I find the other to extend to 12, which is the scale required CHAP. III. The use of the line of Numbers in solid measure, such as stone, timber, and the like. geometric illustration 1 Having the side of a square equal to the base of any solid given in inch measure, to find the length of a foot solid in inch measure. THe side of a square equal to the base of a solid, may be found by dividing the space between the length and breadth into two equal parts, as in the 7 Prop. of broad mea●ledge●ure. Then As the side of the square in inches to 41.57: So is 1 foot to a fourth number; and that fourth to the length in inches. So in the solid AH, the side of the square equal to the ●ledge●as● EC, being about 25 inches 45 parts, the length of a foot ●ledge●olid will be found about 2 inches 67 parts, and the length of too foot solid 5 inches 33 parts. As 25.45 unto 41.57: so 1.00 unto 1.63: and so are 1.63 unto 2.67. 2 Having the side of a square equal to the base of any solid given in foot measure, to find the length of a foot solid in foot measure. As the side of the square in feet unto 1: So is 1 unto a fourth number; And that fourth to the length in foot measure. So in the solid AH, the side of the square equal to the base EC, being about 2 foot 120 parts, the length of a foot solid will be found about 222 parts of a foot. As 2.120 unto 1.000: so 1.000 unto 0.471: and so are 471 unto 222. 3 Having the breadth and depth of a squared solid given in foot measure, to find the length of a foot solid in foot measure. As 1 unto the breadth in foot measure: So the depth in feet to a fourth number; which is the content of the base in foot measure. Then As this fourth number unto 1: So 1 unto the length in foot measure. So in the solid AH, the breadth being 2 foot 50 parts, the depth 1 foot 80 parts, the content of the base EC will be found 4 foot 50 parts, and the length of one foot solid about 222 parts, the length of two foot solid about 444 parts of 1000 As 1.00 unto 2.50: so are 1.80 unto 4.50. As 4.50 unto 1.00: so 1.000 unto 0.222. 4 Having the breadth and depth of a squared solid given in inches, to find the length of a foot solid in inch measure. As 1 hath to the breadth in inches: So the depth in inches to a fourth number; which is the content of the base in inches. Then As this fourth number unto 1728: So 1 unto the length of a foot in inch measure. So in the solid AH, the breadth AC being 30 inches, and the d pth A 21 inches 60 parts, the content of the base EC will be found to be 648 inches, and the length of a foot solid about 2 inches 67 parts. As 1 unto 21.6: so 30 unto 648: As 648 unto 1728: so 1 unto 2.667. Or as 12 to the breadth in inches: So the depth in inches to a fourth number. As this fourth number to 144: So 1 unto the length of a foot solid in inch measure. So in the solid AH, the breadth being 30 inches, the depth 21 inches 6 parts, the fourth number will be found to be 54, and the depth of a foot solid 2 inches 67 parts. As 12 unto 21.6: so 30 unto 54. As 54 unto 144: so 1 unto 2.667. 5 Having the side of a square equal to the base of any solid, and the length thereof given in inch measure, to find the content thereof in feet. As 41.57 to the side of the square in inches: So the length in inches to a fourth number; and that fourth to the content in foot measure. So in the solid AH, the length AB being 183 inches, and the side of the square equal to the base EC about 25 inches 45 parts, the fourth number will be found about 112, and the whole solid content about 68 feet 62 parts. As 41.57 unto 25.45: so 183 unto 112: and so are 112 unto 68.62. 6 Having the side of a square equal to the base of any solid, and the length thereof given in foot measure, to find the content thereof in feet. As 1 to the side of the square in foot measure: So the length in feet to a fourth number; and that fourth to the content in foot measure. So in the former solid AH, the side of the square equal to the base A, being about 2 foot 12 parts, and the length AB 15 foot 25 parts, the conent will be found to be about 68 foot 62 parts. As 1 unto 2.12: so 15.25 unto 32.35: and so are 32.35 unto 68.62. 7 Having the side of a square equal to the base of any solid given in inch measure, & the length of the solid given in foot measure, to find the content thereof in feet. As 12 to the side of the square given in inches: So the length in feet to a fourth number; and that fourth to the content in foot measure. So in the former solid AH, the side of the equal square being 25 inches 45 parts, the content will be found to be about 68 feet 62 parts. As 12 unto 25.45: so 15.25 unto 32.35: and so are 32.35 unto 68.62. 8 Having the length, breadth and depth of a squared solid given in inches, to find the content in inches. As 1 unto the breadth in inches: So the depth in inches unto the base in inches. Then As 1 unto the base: So the length in inches unto the solid content in inches. So in the solid AH, whose breadth AC is 30 inches, the depth A 21 inches and 6 parts of 10, and length AB 183, the content of the base EC will be found 648 inches, and the whole solid content about 118500 inches. As 1 unto 21.6: so are 30 unto 648: As 1 unto 648: so are 183 to 118584. 9 Having the length, breadth and depth of a squared solid given in inches, to find the content in feet. As 1 to the breadth in inches: So the depth in inches to the base in inches. As 1728 to that base: So the length in inches to the content in feet. So in the solid AH, the content will be found to be about 68 feet 62 parts. As 1 unto 21.6: so 30 unto 648: As 1728 unto 648: so 183 to 68.62. Or as 12 to the breadth in inches: So the depth in inches to a fourth number. As 144 to that fourth number: So the length in inches to the content in feet. And so also in the same solid AH, the content will be found to be about 68 feet 62 parts. As 12 unto 21.6: so 30 unto 54: As 144 unto 54: so 183 unto 68.62. 10 Having the length, breadth and depth of a squared solid given in foot measure, to find the content in feet. As 1 unto the breadth in foot measure: So the depth in feet to the base in feet. As 1 unto that base: So the length in feet to the content in feet. And thus in the former solid AH, the breadth AC will be 2 foot 50 parts, the depth A 1 foot 80 parts, and the length AB 15 foot 25 parts; then working as before, the content of the base OF will be found 4 feet 50 parts, and the whole solid content about 68 foot 62 parts, which of all others may very easily be tried by arithmetic. As 1 unto 2.50: so 1.80 unto 4.50. As 1 unto 4.50: so 15.25 unto 68.625. 11 Having the breadth and depth of a squared solid given in inches, and the length in foot measure, to find the content thereof in feet. As 1 unto the breadth in inches: So the depth in inches unto a fourth number: which is the content of the base in inches. As 144 hath unto that fourth number: So the length in feet to the content in feet. And so in the same solid AH, the content will be found to be about 68 feet 62 parts. As 1 unto 21.6: so 30 unto 648. As 144 unto 15.25: so 648 unto 68.62. Or as 144 unto the breadth in inches: So the depth in inches unto a fourth number: which is the content of the base in feet. As 1 hath unto that fourth number: So the length in feet to the content in feet. And so in the same solid AH, the content will be found to be about 68 feet 62 parts. As 144 unto 21.6: so 30 unto 4.50: As 1 unto 4.50: so 15.25 unto 68.62. Or as 12 unto the breadth in inches: So the depth in inches unto a fourth number. As 12 unto this fourth number: So the length in feet to the content in feet. And so also in the same solid AH, the content will be found to be about 68 feet 62 parts. As 12 unto 21.6: so 30 unto 54. As 12 unto 54: so 15.25 unto 68.62. All these varieties (and such like not here mentioned) do follow upon making of the base of the solid, to be EC; there would be as many more if any shall begin with the base EH, and so likewise if they make the base to be FD. 12 Having the diameter of a cylinder given in inch measure, to find the length of a foot solid in inches. As the diameter in inches unto 46.90: cylinder So is 1 unto a fourth number: and that fourth to the length in inches. So the diameter of a cylinder being 15 inches, the fourth number will be about 3.12, and the length of a foot solid 9 inches 78 parts. As 15 unto 46.90: so 1 unto 3.127: and so are 3.127 unto 9.778. 13 Having the diameter of a cylinder given in foot measure, to find the length of a foot solid in foot measure. As the diameter in feet unto 1.128: So is 1 unto a fourth number; and that fourth to the length in foot measure. So the diameter being 1 foot 25 parts, the length of a foot solid will be found about 8.14 parts of 1000 As 1.25 unto 1.128: so 1.00 to 0.9027: and so are 9027 unto 8148. 14 Having the circumference of a cylinder given in inches, to find the length of a foot solid in inch measure. As the circumference in inches to 147.36: So is 1 to a fourth number; and that fourth to the length in inches. So the circumference being 47 inches 13 parts, the length of a foot solid will be found about 9 inches 78 parts. As 47.13 unto 147.36: so 1.00 to 3.13: and so are 3.13 unto 9.78. 15 Having the circumference of a cylinder given in foot measure, to find the length of a foot solid in foot measure. As the circumference in feet to 3.545: So is 1 to a fourth number; and that fourth to the length in foot measure. So the circumference being 3 foot 927 parts, the length of a foot solid will be found to be about 815 parts. As 3.927 unto 3.545: so 1.000 unto 0.903: and so are 903 unto 815. 16 Having the side of a square equal to the base of a cylinder, to find the length of a foot solid. The side of a square equal to the circle, may be found by the eighth Prop. of broad measure, and then this Prop. may be wrought by the first and second Prop. of solid measure. 17 Having the diameter of a cylinder, and the length given in inches, to find the content in inches. As 1.128 unto the diameter in inches: So the length in inches to a fourth number; and that fourth number to the content in inches. So the diameter being 15 inches, and the length 105, the content of the cylinder will be found to be about 18560 inches. As 1.1284 unto 15: so are 105 unto 1395.87: and so are 1395.87 unto 18555.34. 18 Having the diameter and length of a cylinder in foot measure, to find the content in feet. As 1.128 to the diameter in feet: So the length in feet to a fourth number; and that fourth to the content in feet. So the diameter being 1 foot 25 parts, and the length 8 foot and 75 parts, the content of the cylinder will be found about 10 foot 74 parts. As 1.128 unto 1.25: so 8.75 unto 9.69: and so are 9.69 unto 10.737. 19 Having the diameter of a cylinder, and the length given in inches, to find the content in feet. As 46.90 to the diameter in inches: So the length in inches to a fourth number; and that fourth to the content in feet. So the diameter being 15 inches, and the length 105, the content will be found about 10 foot 74 parts. As 46.906 unto 15: so 105 unto 33.58: and so are 33.58 unto 10.737. 20 Having the diameter of a cylinder given in inches and the length in feet, to find the content in feet. As 13.54 to the diameter in inches: So the length in feet to a fourth number; and that fourth to the content in feet. So the diameter being 15 inches, and the length 8 foot 75 parts, the content will be found about 10 foot 74 parts. As 13.54 unto 15: so 8 75 unto 9.69: and so are 9.69 unto 10.74. 21 Having the circumference and the length of a cylinder given in inches, to find the content in inches. As 3.545 to the circumference in inches: So the length in inches to a fourth number; and that fourth to the content in inches. So the circumference being 47 inches 13 parts, and the length 105 inches, the content will be found about 18560 inches. As 3.545 unto 47.13: so 105 unto 1396: and so are 1396 unto 18555. 22 Having the circumference and length of a cylinder given in inches, to find the content in feet. As 147.36 to the circumference in inches: So the length in inches to a fourth number; and that fourth to the content in feet. So the circumference being 47 inches 13 parts, and the length 105 inches, the content will be found about 10 foot 74 parts. As 147.36 unto 47.13: so 105 unto 33.58: and so are 33.58 unto 10.74. 23 Having the circumference and length of a cylinder given in foot measure, to find the content in feet. As 3.545 to the circumference in feet: So the length in feet to a fourth number; and that fourth to the content in feet. So the circumference being 3 foot 927 parts, and the length 8 foot 75 parts, the content will be found to be 10 foot 74 parts. As 3.545 unto 3.927: so 8.75 unto 9.69: and so are 9.69 unto 10.74. 24 Having the circumference of a cylinder given in inches and the length in foot measure, to find the content in feet. As 42.54 to the circumference in inches: So the length in feet to a fourth number; and that fourth to the content in feet. So the circumference being 47 inches 13 parts, and the length 8 foot 75 parts, the content will be found as before, 10 foot 74 parts. As 42.54 unto 47.13: so 8.75 unto 9.69: and so are 9.69 unto 10.74. CHAP. FOUR The use of the line of Numbers in gaugeing of vessels. THe vessels which are here measured, are supposed to be cylinders, or reduced unto cylinders, by taking the mean between the diameter at the head and the diameter at the bongue, after the usual manner. 1 Having the diameter and the length of a vessel with the content thereof, to find the gauge point. Extend the compasses in the line of numbers to half the distance between the content and the length of the vessel, the same extent will reach from the diameter to the gauge point. I put this proposition first, because these kind of measures are not alike in all places. Here at London it is said that a wine vessel being 66 inches in length, and 38 inches the diameter, would contain 324 gallons: which if it be true, we may divide the space between 324 and 66 into two equal parts, and the middle will fall about 146, and the same extent which reacheth from 324 to 146, will reach from the diameter 38 unto 17.15 the gauge point for a gallon of wine or oil after London measure. The like reason holdeth for the like measures in all other places. 2 Having the mean diameter and the length of a vessel, to find the content. Extend the compasses from the gauge point to the mean diameter, the same extent being doubled, shall give the distance from the length to the content. So the mean diameter of a wine vessel being 20 inches, and the length 25 inches, the content will be found to be 34 gallons after London measure. For extend the compasses from 17.25 unto 20, the same extent will reach from 25 unto 29.15, and from 29.15 unto 34. In like manner if the mean diameter were 16 inches, and the length 23, the content would be found to be about 20 gallons. For the same extent which reacheth back from 17.15 unto 16, will reach from 23 to 21.45, and from 21.45 unto 20. So that if the mean diameter shall be 17 inches and 15 centesmes or parts of 100, the number of inches in the length of the vessel, will give the number of gallons contained in the same vessel: if the diameter shall be more or less than 17.15, the content in gallons will be accordingly more or less than the length in inches. 3 Having the diameter and content, to find the length. Extend the compasses from the diameter to the gauge point, the same extent being doubled shall give the distance from the content to the length of the vessel. So the gauge point standing as before, if the diameter shall be 38 inches, and the content 324 gallons wine measure, the length of the vessels will be found about 66 inches. 4 Having the length of a vessel and the content, to find the diameter. Extend the compasses to half the distance between the length and the content, the same extent shall reach from the gauge point to the diameter. So the length being 66 inches, and the content 324 gallons wine measure, the gauge point standing as before, the diameter of the vessel will be found to be about 38 inches. CHAP. V Containing such Astronomical propositions as are of ordinary use in the practice of Navigation. 1 To find the altitude of the Sun by the shadow of a gnomon set perpendicular to the horizon. As the parts of the shadow are to the parts of the gnomon: So the tangent of 45 gr. to the tangent of the altitude. Extend the compasses in the line of numbers, from the parts of the shadow to the parts of the gnomon; the same extent will give the distance from the tangent of 45 gr. to the tangent of the Sun's altitude. So the gnomon being 36, and the shadow 27, the altitude will be found to be 36 gr. 32 m. Or the gnomon being 27, and the shadow 36, the altitude will be found to be 53 gr. 8 m. Or the shadow being 20, and the gnomon 9, the altitude will be found to be 25 gr. 14 m. as in the eighth Prop. of the use of the tangent line. 2 Having the distance of the Sun, from the next equinoctial point, to find his declination. As the Radius is in proportion to the sine of the Sun's greatest declination: So the sine of the Sun's distance from the next equinoctial point, to the sine of the declination required. Extend the compasses in the line of sins, from 90 gr. to 23 gr. 30 m. the same extent will give the distance from the Sun's place unto his declination. So the Sun being either in 29 gr. of ♉, or 1 gr. of ♒, or 1 gr. of ♌, or 29 gr. of ♏, that is 59 gr. distant from the next equinoctial point, the declination will be found about 20 gr. If the Sun be so near the equinoctial point that his declination fall to be under 1 gr. it may be found by the line of numbers. As if the Sun were in 2 gr. 5 m. of ♈, that is, 125 m. from the equinoctial point, the former extent of the compasses from the sine of 90 gr. to the sine of 23 gr. 30 m. will reach in the line of numbers from 125 unto 50, which shows the declination to be about 50 m. 3 Having the latitude of the place, and the declination of the Sun, to find the time of the Suns rising and setting. As the cotangent of the latitude to the tangent of the Sun's declination: So is the Radius to the line of the ascensional difference between the hour of 6 and the time of the Suns rising or setting. Extend the compasses from the tangent of the compliment of the latitude, to the tangent of the declination: the same extent will reach from the sine of 90 gr. to the sine of the ascensional difference. Or extend the compasses from the cotangent of the latitude to the sine of 90 gr. the same extent will reach from the tangent of the declination to the sine of the ascensional difference. So the latitude being 51 gr. 30 m. Northward, and the declination 20 gr. the difference of ascension will be found to be 27 gr. 14 m. which resolved into hours and minutes, doth give 1 hour and almost 49 m. for the difference between the Sun's rising or setting, and the hour of 6, according to the time of the year. 4 Having the latitude of the place, and the distance of the Sun from the next equinoctial point, to find his amplitude. As the cousin of the latitude to the sine of the Sun's greatest declination: So the sine of the place of the Sun, to the sine of the amplitude. So the latitude being 51 gr. 30 m. and the place of the Sun in 1 gr. of ♒, that is 59 gr. distant from the next equinoctial point, the amplitude will be found about 33 gr. 20 m. For extend the compasses in the line of sins, from 38 gr. 30 m. the sine of the compliment of the latitude, unto 23 gr. 30 m. the sine of the Sun's greatest declination; the same extent will reach from 59 gr. unto 33 gr. 20 m. Or extend them from 38 gr. 30 m. unto 59 gr. the same extent will reach from 23 gr. 30 m. unto 33 gr. 20 m. as before. 5 Having the latitude of the place, and the declination of the Sun, to find his amplitude. As the cousin of the latitude is to the Radius: So the sine of the declination, to the sine of the amplitude. Extend the compasses from the cousin of the latitude to the sine of 90 gr. the same extent will reach from the sine of the Sun's declination to the sine of the amplitude. Or extend them from the cousin of the latitude to the sine of the declination, the same extent will reach from the sine of 90 gr. to the sine of the amplitude. So the latitude being 51 gr. 30 m. and the declination 20 gr. the amplitude will be found to be 33 gr. 20 m. 6 Having the latitude of the place, and the declination of the Sun, to find the time when the Sun cometh to be due East or West. As the tangent of the latitude, is to the tangent of the declination: So the Radius to the cousin of the hour from the meridian. Extend the compasses from the tangent of the latitude to the tangent of the declination; the same extent will reach from the sine of 90 gr. to the sine of the compliment of the hour. Or extend them from the tangent of the latitude to the sine of 90 gr; the same extent will reach from the tangent of the declination to the sine of the compliment of the hour. So the latitude being 51 gr. 30 m. and the declination 20 gr. the Sun will be 73 gr. 10 m: that is 4 hours and 53 m. from the meridian, when he cometh to be in the East or West. 7 Having the latitude of the place, and the declination of the Sun, to find what altitude the Sun shall have, when he cometh to be due East or West. As the sine of the latitude is to the sine of the declination: So the Radius to the sine of the altitude. Extend the compasses in the line of lines from the latitude to the sine of the declination, the same extent will reach from the sine of 90 gr. to the sine of the altitude. Or extend them from the sine of the latitude to the sine of 90 gr; the same extent will reach from the sine of the declination to the sine of the altitude. So the latitude being 51 gr. 30 m. and the declination 20 gr. the altitude will be found about 25 gr. 55 m. 8 Having the latitude of the place, and the declination of the Sun, to find what altitude the Sun shall have at the hour of six. As the Radius is in proportion to the sine of the sine of the declination: So the sine of the latitude to the sine of the altitude. Extend the compasses in the line of sins, from 90 gr. to the declination; the same extent will reach from the latitude to the altitude. Or extend them from 90 gr. to the latitude, the same extent will hold from the declination to the altitude. So the latitude being 51 gr. 30 m. and the declination of the Sun 20 gr. the altitude of the Sun will be found to be about 15 gr. 30 m. 9 Having the latitude of the place, and the declination of the Sun, to find what azimuth the Sun shall have at the hour of six. As the cousin of the latitude is to the Radius: So the cotangent of the Sun's declination, to the tangent of the azimuth from the North part of the meridian. So the latitude being 51 gr. 30 m. and the declination 20 gr. the azimuth will be found to be 77 gr. 14 m. For extend the compasses in the line of sins, from 38 gr. 30 m. to 90 gr. the same extent will reach from the tangent of 70 gr. to the tangent of 77 gr. 14 m. 10 Having the latitude of the place, and the declination of the Sun, and the altitude of the Sun, to find the azimuth. First consider the declination of the Sun, whether it be toward the North or the South, so have you his distance from your pole: then add this distance, the compliment of his altitude, and the compliment of your latitude, all three together, and from half the sum subtract the distance from the pole, and note the difference. 1 As the Radius is in proportion to the cousin of the altitude: So the cousin of the latitude, to a fourth sine. 2 As this fourth sine is to the sine of the half sum: So the sine of the difference, to a seventh sine. Then find a mean proportional between this seventh sine and the Radius, this mean shall be the sine of the compliment of half the azimuth from the North part of the meridian. Suppose the declination of the Sun being known by the time of the year to be 20 gr. Southward, the altitude above the horizon found by observation 12 gr. and the latitude Northwards 51 gr. 30 m. it were required to find the azimuth. The declination is Southward, and therefore the distance from the pole 110 gr; then turning the altitude and latitude unto their compliments, I add them all three together, and from half the sum subtract the distance from the pole, noting the difference after this manner. Declin. South 20 gr. 0 m. The distance 110 gr. 0 m. Altitude 12 0 The compliment 78 0 Latitude N. 51 30 The compliment 38 30 The sum of all three 226 30 The half sum 113 15 The difference 3 15 This done, I come to the Staff, and extend the compasses from the sine of 90 gr. to the sine of 78 gr. and find the same extent to reach from the sine of 38 gr. 30 m. unto 37 gr. 30 m Or if I extend them from 90 gr. to 38 gr. 30 m. the same extent doth reach from 78 gr. unto 37 gr. 30 m. which is the fourth sine required. Then I extend the compasses again, from this fourth sine of 37 gr. 30 m. unto the sine of the half sum 113 gr. 15 m. that is to the sine of 66 gr. 45 m. (for after 90 gr. the sine of 80 gr. doth stand for a sine of 100 gr. and the sine of 70 gr. for a sine of 110 gr. and so the rest for those which are their compliments to 180 gr.) and this second extent doth reach from the sine of the difference 3 gr. 15 m. to the sine of 4 gr. 54 m. Or if I extend them from the fourth sine of 37 gr. 30 m. to the sine of the difference 3 gr. 15 m. the same extent will reach from the sine of the half sum 113 gr. 15 m. unto 4 gr. 54 m. which is the seventh sine required. Lastly, I divide the space between this seventh sine of 4 gr. 54 m. and the sine of 90 gr. into two equal parts, and I find the mean proportional sine to fall on 17 gr. whose compliment is 73 gr; the double of 73 gr. is 146 gr. and such is the azimuth required. Or having found the seventh sine to be 4 gr. 54 m. I might look over against it, in the line of versed sins, and there I should find 146 gr. for the azimuth from the North part of the meridian; and the compliment of 146 gr. to a semicircle being 34 gr. will give the azimuth from the South part of the meridian. But if it were required to find the azimuth in the same latitude of 51 gr. 30 m. Northward, with the same altitude of 12 gr. and like declination of 20 gr. to the Northward, it would be found to be only 72 gr. 52 m. though the manner of work be the same as before. Declin. North 20 gr. 0 m The distance is 70 gr. 0 m. Altitude 12 0 The compliment 78 0 Latitud. North 51 30 The compliment 38 ●0 The sum of all three 186 30 The half sum 93 15 The difference 23 15 Here as the Radius is to the sine of 78 gr: so the sine of 38 gr. 30 m. to the sine of 37 gr. 30 m. which is the fourth sine, and the same as before. Then as this fourth sine of 37 gr. 30 m. is to the sine of 93 gr. 15 m: so the sine of 23 gr. 15 m. to the sine of 40 gr. 20 m. which is the seventh sine. The half way between this seventh sine and the sine of 90 gr. doth fall at 53 gr. 34 m. whose compliment is 36 gr. 26 m; and the double of that is 72 gr. 52 m. the azimuth required. Or I may find this same azimuth in the line of versed sins, over against the seventh sine of 40 gr. 20 m. 11 Having the latitude of the place, the declination of the Sun, and the altitude of the Sun, to find the hour of the day. Add the compliment of the Sun's altitude, and the distance of the Sun from the pole, and the compliment of your latitude, all three together, and from half the sum subtract the compliment of the altitude, and note the difference. 1 As the Radius is in proportion to the sine of the Sun's distance from the pole: So the sine of the compliment of the latitude, to a fourth sine. 2 As this fourth sine is to the sine of the half sum: So the sine of the difference to a seventh sine. The mean proportional between this seventh sine and the sine of 90 gr. will be the sine of the compliment of half the hour from the meridian. Thus in our latitude of 51 gr. 30 m. the declination of the Sun being 20 gr. Northward, and the altitude 12 gr. I might find the Sun to be 95 gr. 52 m. from the meridian. Altitude 12 gr. 0 m. The compliment is 78 gr. 0 m. Declin. North 20 0 the dist. from the pole 70 0 Latitude 51 30 the compliment is 38 30 The sum of all three 186 30 The half sum 93 15 The difference 15 15 Here as the Radius is to the sine of 70 gr. So the sine of 38 gr. 30 m. to the sine of 35 gr. 48 m. As this sine of 35 gr. 48 m. is to the sine of 93 gr. 15 m. So the sine of 15 gr. 15 m. to the sine of 26 gr. 40 m. The half may between this seventh sine of 26 gr. 40 m. and the sine of 90 gr. doth fall at 42 gr. 4 m. whose compliment is 47 gr. 56 m. and the double of that, 95 gr. 52 m. which converted into hours, doth give 6 hours and almost 24 m. from the meridian. Or I might find these 95 gr. 52 m. in the line of versed sins, over against the seventh sine of 26 gr. 40 m. 12 Having the azimuth, the Sun's altitude, and the declination, to find the hour of the day. As the cousin of the declination is to the sine of the azimuth: So the cousin of the altitude to the sine of the hour. Thus the declination being 20 gr. Southward, the altitude 12 gr. and the azimuth found by the tenth Prop. 146 gr. I might find the time to be 35 gr. 36 m. that is 2 hours 22 m. from the meridian. 13 Having the hour of the day, the Sun's altitude, and the declination, to find the azimuth. As the cousin of the altitude is to the sine of the hour: So the cousin of the declination, to the sine of the azimuth. So the altitude of the Sun being 12 gr. and the declination 10 gr. Southward, and the angle of the hour 35 gr. 36 m. I should find the azimuth to be 34 gr. And so it is if it be reckoned from the South; but 146 gr. if it be taken from the North part of the meridian. 14 Having the distance of the Sun from the next equinoctial point, to find his right ascension. As the Radius to the cousin of the greatest declination: So the tangent of the distance, to the tangent of the right ascension. So the Sun being in the first degree of ♒, that is 59 gr. distant from the next equinoctial point, and the greatest declination 23 gr. 30 m. the right ascension will be found to be 56 gr. 46 m. short of the beginning of ♈, and therefore 303 gr. 14 m. 15 Having the declination of the Sun, to find his right ascension. As the tangent of the greatest declination it to the tangent of the declination given: So the Radius to the sine of the right ascension. So the greatest declination being 23 gr. 30 m. and the declination of the Sun given 20 gr. the right ascension will be found about 56 gr. 50 m. These are such Astronomical propositions as I take to be useful for Seamen. For the first and second will help them to find their latitude; the third to find the Suns rising and setting; the 4.5.6.7.8.9.10.13. Prop. to find the variation of their compass; the 11 and 12 Prop. to find the hour of the day; and the two last toward the finding of the hour of the night. For having the latitude of the place, with the declination and altitude of any star, they may find the hour of the star from the meridian, as in the 11 Prop. Then comparing the right ascension of the star with the right ascension of the Sun, they may have the hour of the night. All these propositions and such others may be wrought ●●so by the tables of sins and tangents. For where four numbers do hold in proportion; as the first to the second, so the third to the fourth; there if we multiply the second into the third, and divide the product by the first, the quotient will give the fourth required. As in the example of the last Prop. where the delination being given, it was required to find the right ascension. The tangent of 20 gr. the declination given is 3639702, which being multiplied by the Radius, the product is 36397020000000, and this divided by 4348124 the tangent of 23 gr. 30 m. the quotient is 8370741 the sine of 56 gr. 50 m. for the right ascension required. Or if any will use my tables of artificial sins and tangents, they may add the second and the third together, and from the sum subtract the first, the remainder will give the fourth required. And so my tangent of 20 gr. is 9561.0638, which being added to the Radius, makes 19561.0658; from this if they subtract 9638.3019 the tangent of 23 gr. 30 m. they shall find the remainder to be 9922.7639, which in my Canon is the sine of 56 gr. 49 m. 56 seconds; and such is the right ascension required, if it be reckoned from the next equinoctial point. The like reason holdeth for all other Astronomical propositions, as I will farther show by those two examples which I gave before for the finding of the azimuth in the 10 Prop. because they are thought to be harder than the rest, and require three operations. In the first example. Declin. South 20 gr. 0 m. The distance 110 gr. 0 m. Altitude 12 0 the compliment 78 0 Latitude Nor. 51 30 the compliment 38 30 The sum of all three 226 30 The half sum 113 15 The difference 3 15 The first operation will be to find the fourth sine; and that is done by adding the sine of the compliment of the altitude to the sine of the compliment of the latitude, and subtracting the Radius: so adding 9990.4044 the sine of 78 gr. unto 9794.1495 the sine of 38 gr. 30 m. the sum will be 19784.5539. And the Radius being subtracted, the remainder 9784.5539 is the fourth sine, and belongeth to 37 gr. 30 m. The second operation will be to find the seventh sine; and that is done by adding the sine of the half sum to the sine of the difference, and subtracting the fourth sine. So the half sum being 113 gr. 15 m. I take his compliment to a semicircle, and so find his sine to be 9963.2168, to which I add 8753.5278, the sine of the difference 3 gr. 15 m; and the sum is 18716.7446. From this I take the fourth sine 9784.5539, and the remainder will be 8932.1907, which is the seventh sine, and belongeth to 4 gr. 54 m. The third operation will be to find the mean proportional sine between the seventh sine and the Radius. This in common arithmetic is done by multiplying the two extremes, and taking the square root of the product. As in finding a mean proportional between 4 and 9, we multiply 4 into 9, and the product is 36, whose square root is 6, the mean proportional between 4 and 9 But here it is done by adding the sine and the Radius, and taking the half of them. So the sum of the last seventh sine and the Radius is 18932.1907 and the half of that 9466.0953, which is the mean proportional sine required, and belongeth to 17 gr. whose compliment is 73 gr. and the double of that 146 gr. the same azimuth as before. In the second example. Declin. North 20 gr. 0 m. The distance 70 gr. 0 m. Altitude 12 0 the compliment 78 0 Latitud. North 51 30 the compliment 38 30 The sum of all three 186 30 The half sum 93 15 The difference 23 15 The first operation will be to find the fourth sine; and ●ledge●hat is here 9784.5539, as in the former example. The second operation will be to find the seventh sine; and ●ledge●o here the sine of the half sum 93 gr. 15 m. being the ●ledge●ame with the sine of 86 gr. 45 m. his compliment to 180 gr. I ●ledge●nd it to be 9999.3009, to which I add 9596.3153 the sine ●ledge●f the difference 23 gr. 15 m. and the sum is 19595.6162. ●ledge●rom this I take the fourth sine 9784.5539, and the remain●ledge●er will be 9811.0623 for the seventh sine, and belongeth to ●ledge●0 gr. 20 m. The third operation will be to find the mean proportio●ledge●all sine between the seventh sine and the Radius. And so ●ledge●ere the Radius being added to the seventh sine, the sum ●ledge●ill be 19811.0623, and the half of that 9905.5311 doth ●ledge●iue the mean proportional sine belonging to about 53 gr. 4 m. whose compliment is 36 gr. 26 m. & the double of that ●2 gr. 52 m. the same azimuth as before. I have set down these three examples thus particularly, ●ledge●hat I might show the agreement between the Staff and the ●ledge●anon. But otherwise I might deliver both the precept and the work, for the two last, more compendiously. For generally in all spherical triangles, where three sides are known, and an angle required, make that side which is opposite to the angle required, to be the base; and gather the sum, the half sum, and the difference as before. As the rectangle contained under the sins of the sides, is to the square of the whole sine: So the rectangle contained under the sins of the half sum and the difference, to the square of the cousin of half the angle. Then for the work, we may for the most part leave out the two last figures; and if they be above 50, put an unity to the sixth place, after this manner. The second example. 70 gr. 0 m 78 0 9990 40 38 30 9794 15 186 30 19784 55 93 15 9999 30 23 15 9596 32 20000 00 39595 62 19811 07 36 26 9905 53 72 52 53 gr. 34 m. 107 8 Or for such numbers as are to be subtracted, I may take them out of the Radius, and write down the residue, and then add them together with the rest. As in the same second example, the sins of 78 gr. and of 38 gr. 30 m. being the numbers to be subtracted; if I take 9990.4044 the sine of 78 gr. out of the Radius 10000 0000, the residue is 9.5956: and so the residue of 9794.1495 is 205.8505 Wherefore in stead of subtracting those sins, I may add these residues after this manner: 70 gr. 0 m. 78 0 38 30 186 30 93 15 23 15 36 26 72 52 9 59 205 85 9999 30 9596 32 19811 06 9905 53 53 gr. 34 m. 107 8 Having these means to find the Sun's azimuth, we may compare it with the magnetical azimuth, and so find the variation of the needle. geometric illustration For let the circle AMB, drawn on the centre Z, be a plane, parallel to the horizon; A the point whereon the Sun beareth from us, M the North point of the magnetical needle, and the angle AZM the magnetical azimuth. If we find the Sun's azimuth as before, to be 72 gr. 52 m. from the North to the Westward, we may allow so many degrees from A unto N, and so we have the true North point from the meridian, and consequently the East, South, and West points of the horizon; and the distance between N and M shall be the variation of the needle. So that if the magnetical azimuth AZM shall be 84 gr. 7 m. and the Sun's azimuth AZN 72 gr. 52 m. then must NZM the difference between the two meridians, give the variation to be 11 gr. 15 m. as Mr. Bourough heretofore found it by his observations at Limehouse in the year 1580. But if the magnetical azimuth AZM shall be 79 gr. 7 m. and the Sun's azimuth AZN 72 gr. 52 m. then shall the variation NZM be only 6 gr. 15 m. as I have sometimes found it of late. Hereupon I enquired after the place where Mr. Bourough observed, and went to Limehouse with some of my friends, and took with us a quadrant of 3 foot semidiameter, and two needles, the one above 6 inches, and the other 10 inches long, where I made the semidiameter of my horizontal plane AZ 12 inches: and toward night the 13 of june 1622, I made observation in several parts of the ground, and found as followeth. Alt. ☉ AZM AZN Variat Gr. M. Gr. M. Gr. M. Gr. M. 19 0 82 2 75 52 6 10 18 5 80 50 74 44 6 6 17 34 80 0 74 6 5 54 17 0 79 15 73 20 5 55 16 18 78 12 72 32 5 40 16 0 77 50 72 10 5 40 20 10 71 2 64 49 6 13 9 52 70 12 64 25 5 47 CHAP. VI geometric illustration Containing such nautical questions, as are of ordinary use, concerning longitude, latitude, Rumb, and distance. 1 To keep an account of the ships way. THe way that the ship maketh, may be known to an old seaman by experience, by others it may be found for some small portion of time, either by the log line, or by the distance of two known marks on the ships side. The time in which it maketh this way, may be measured by a watch, or by a glass. Then as long as the wind continueth at the same stay, it followeth by proportion, As the time given is to an hour: So the way made, to an hour's way. Suppose the time to be 15 seconds, which make a quarter of a minute, and the way of the ship 88 feet: then because there are 3600 seconds in an hour, I may extend the compasses in the line of numbers, from 15 unto 3600, and the same extent will reach from 88 unto 21120. Or I may extend them from 15 unto 88, and this extent will reach from 3600 unto 21120; which shows that an hour's way came to 21120 feet. But this were an unnecessary business, to hearken after feet or fathoms. It sufficeth our seamen to find the way of their ship in leagues or miles. And they say that there are 5 feet in a pace, 1000 paces in a mile, and 60 miles in a degree, and therefore 300000 feet in a degree. Yet comparing several observations, and their measures with our feet usual about London, I find that we may allow 352000 feet to a degree; and than if I extend the compasses in the line of numbers from 352000 unto 21120, I shall find the same extent to reach from 20 leagues the measure of one degree, to 1.2, and from 60 miles to 3.6; which shows the hour's way to be 1 league and 2 tenths of a league, or 3 miles and 6 tenths of a mile. But to avoid these fractions and other tedious reductions, I suppose it would be more easy to keep this account of the ships way (as also of the difference of latitude, and the difference of longitude) by degrees and parts of degrees, allowing 100 parts to each degree, which we may therefore call by the name of centosmes. Neither would this be hard to conceive. For if 100 such parts do make a degree, then shall 50 parts be equal to 30 minutes, as 30 minutes are equal to 10 leagues. And 5 parts shall be equal to 3 minutes, as 3 minutes are equal to 1 league. And so the same extent as before, will reach from 100 parts unto 6; which shows that the hour's way required is 6 cent. such as 100 do make a degree, and 5 do make an ordinary league. This might also be done at one operation. For upon these suppositions, divide 44 feet into 45 lengths, and set as many of them as you may conveniently between two marks on the ships side, and note the seconds of time in which the ship goeth these lengths: so the lengths divided by the time, shall give the cent. which the ship goeth in an hour. Suppose the distance between the two marks to be 60 lengths (which are 58 feet and 8 inches) and let the time be 12 seconds: extend the compasses from 12 to 1, in the line of numbers; so the same extent will reach from 60 unto 5. Or extend them from 12 unto 60, and the same extent will reach from 1 unto 5. This shows that the ships way is according to 5 cent. in an hour. This may be found yet more easily, if the log line shall befitted to the time. As if the time be 45 seconds, the log line may have a knot at the end of every 44 feet; then doth the ship run so many cent. in an hour, as there are knots vered out in the space of 45 seconds. If 30 seconds do seem to be a more convenient time, the log line may have a knot at the end of every 29 feet and 4 inches; and then also the centesmes will be as many as the knots. Or if the knots be made to any set number of feet, the time may be fitted unto the distance. As if the knots be made at the end of every 24 feet, the glass may be made 24 seconds and somewhat more than an half of a second; and so these knots will show the cent. If there be 5 knots vered out in a glass, than 5 cent; if 6 knots, than the ship goeth 6 cent. in the space of an hour; and so in the rest. For upon this supposition, the proportion between the time and the feet will be as 45 unto 44. But according to the common supposition it should seem to be as 45 unto 37 ½, or in lesser terms as 6 unto 5. Those which are upon the place, may make proof of both, and follow that which agrees best with their experience. 2 By the latitude and difference of longitude, to find the distance upon a course of East and West. Extend the compasses from the sine of 90 gr. unto the sine of the compliment of the latitude; the same extent shall reach in the line of numbers from the difference of longitude to the distance. So the measure of one degree in the equator, being 100 cent. the distance belonging to one degree of longitude in the latitude of 51 gr. 30 m. will be found about 62 cent. and ¼. Or if the measure of a degree be 60 miles, the distance will be found about 37 miles and ⅓. If the measure be 20 leagues, than almost 12 leagues and ½. If the measure be 17 ½, as in the Spanish charts', then somewhat less than 11 leagues sailing upon this parallel, will give an alteration of one degree of longitude. geometric illustration 3 By the latitude and distance upon a course of East or West, to find the difference of longitude. Extend the compasses from the sine of the compliment of the latitude, to the sine of 90 gr; the same extent will reach in the line of numbers from the distance to the difference of longitude. So the distance upon a course of East or West, in the latitude of 51 gr. 30 m. being 100 cent. the difference of longitude will be found 1.60, which make one degree and 60 centesmes or 1 gr. 36 m. Or if it be 60 miles, the difference of longitude will be 96, which also make 1 gr. 36 m. as before. 4 The longitude and latitude of two places being given, to find the Rumb leading from the one to the other. Extend the compasses in the line of numbers from the difference of latitudes to the difference of longitudes; the same extent will give the distance from the tangent of 45 gr. unto the tangent of the Rumb, according to the projection of the common sea-chart. So the latitude of the first place being 50 gr. the latitude of the second 52 gr. 30 m. and the difference of longitude 6 gr. the Rumb will be found to be about 67 gr. 23 m. which is near the inclination of the sixth Rumb to the meridian. But this Rumb so found, is always greater than it should be, and therefore to be limited; which may be done sufficiently for the Seaman's use, after this manner: Extend the compasses either from the sine of 90 gr. unto the sine of the compliment of the middle latitude, the same extent will reach from the tangent of the Rumb before found, to the tangent of the Rumb limited. Or else extend them from the sine of 90 gr. unto the tangent of the Rumb before found; the same extent will reach from the sine of the compliment of the middle latitude, unto the tangent of the Rumb limited. So the middle latitude between 50 gr. and 52 gr. 30 m. being 51 gr. 15 m. and the Rumb before found 67 gr. 23 m. the Rumb limited will be found to be about 56 gr. 20 m. which is but five minutes more than the inclination of the fift Rumb to the meridian. 2 This Rumb may be found by the help of the meridian line upon the Staff. For if I take the difference of latitude out of the meridian line from 50 gr. unto 52 gr. 30 m. and measure it in his equinoctial, or at the beginning of the meridian line, I shall find it there to be equal to 4 gr. Wherefore I work as if the difference of latitude were 4 gr. and extend the compasses in the line of numbers from 4 unto 6: so shall I find the same extent to reach from the tangent of 45 gr. unto the tangent of 56 gr. 20 m. and this is the inclination of the Rumb required. 5 By the Rumb and both latitudes, to find the distance upon the Rumb. Extend the compasses from the sine of the compliment of the Rumb, unto the sine of 90 gr. the same extent in the line of numbers shall reach from the difference of latitude unto the distance upon the Rumb. So the latitude of the first place being 50 gr. the latitude of the second 52 gr. 30 m. and the Rumb the fift from the meridian. If I extend the compasses from 33 gr. 45 m. unto the sine of 90 gr. I shall find the same extent in the line of numbers to reach from 2 gr. 50 cent. to 4 gr. 50 cent. and such is the distance required. 6 By the distance and both latitudes to find the Rumb. Extend the compasses in the line of numbers from the distance unto the difference of latitudes; the same extent will reach in the line of sins, from 90 gr. unto the compliment of the Rumb. So the one place being in the latitude of 50 gr. the other in the latitude of 52 gr. 30 m. and the distance between them 4 gr. 50 cent. If I extend the compasses from 4.50 unto 2.50 in the line of numbers, I shall find the same extent to reach from the sine of 90 gr. unto the compliment of 56 gr. 15 m. and such is the inclination of the Rumb required. 7 By one latitude, Rumb, and distance, to find the difference of latitudes. Extend the compasses in the line of sins, from 90 gr. unto the compliment of the Rumb; the same extent in the line of numbers, will reach from the distance, unto the difference of latitudes. So the lesser altitude being 50 gr. and the distance 4 gr. 50 cent. upon the fifth Rumb from the meridian: if I extend the compasses from the sine of 90 gr. to 33 gr. 45 m. I shall find the same extent to reach from 4.50 in the line of numbers, unto 2.50; and therefore the second latitude to be 52 gr. 30 m. 8 By the Rumb and both latitudes, to find the difference of longitude. Extend the compasses from the tangent of 45 gr. unto the tangent of the Rumb; the same extent will reach in the line of numbers from the difference of latitudes unto the difference of longitude, according to the projection of the common sea-chart. So the first latitude being 50 gr. and the second 52 gr. 30 m. and the Rumb the fifth from the meridian: if I extend the compasses from the tangent of 45 gr. unto 56 gr. 15 m. I shall find the same extent to reach from 2.50 in the line of numbers to about 3.75, which make 3 gr. 45 m. But this difference of longitude so found, is always lesser than it should be, and therefore to be enlarged, which may be done sufficiently for the sea-mens' use, after this manner: Extend the compasses from the sine of the compliment of the middle latitude, unto the sine of 90 gr. the same extent will reach in the line of numbers from the difference of longitude before found, unto the difference of longitude enlarged. So the middle latitude in this example being 51 gr. 15 m. and the difference of longitude before found 3 gr. 75 cent. the difference of longitude enlarged will be found about 5 gr. 99 cent. which are near 6 gr. 2 This difference of longitude may be found by help of the meridian line upon the Staff. For if I take the proper difference of latitude out of the meridian line, and measure it in his equinoctial, or at the beginning of the meridian line, I shall find it to be equal to four of those degrees. Wherefore having extended the compasses as before from the tangent of 45 gr. unto the tangent of 56 gr. 15 m. the same extent will reach from 4.00 in the line of numbers, unto 5.99: which shows the difference of longitude to be about 5 gr. 99 cent. or about half a minute short of six degrees. 9 By the Rumb and both latitudes, to find the distance belonging to the chart of Mercators' projection. Take the proper difference of latitudes out of the meridian line of the chart, and measure it in his equinoctial, or one of the parallels, and it will there give the difference of latitudes enlarged. Then extend the compasses from the sine of the compliment of the Rumb unto the sine of 90 gr. the same extent will reach in the line of numbers, from the latitude enlarged, unto the distance required. Or extend them from the compliment of the Rumb to the latitude enlarged, the same extent will reach from 90 gr. unto the distance. For example, let the place given be A in the latitude of 50 gr. D in the latitude of 52 gr. 30 m. AM the difference of latitudes, and the Rumb MAD the fifth from the meridian. First I take out AM the difference of latitudes, and measure it in A one of the parallels of the equinoctial; I find it to be very near 4 gr: this is the difference of latitudes enlarged. Then if I extend the compasses from the sine of 33 gr. 45 m. the compliment of the fifth Rumb unto the sine 90 gr. I shall find the same extent to reach in the line of numbers from 4.00 unto 7.20. And this is the distance belonging to the chart. Wherefore I take out these 7 gr. 20 cent. out of the scale of the parallel A, and prick it down upon the Rumb from A unto D, where it meeteth with the parallel of the second latitude. Lastly, I measure it in the meridian line, setting one foot of the compasses as much below the lesser latitude as the other above the greater latitude, and find it to be 4 gr. 50 cent. which is the same distance that I found before in the 5. Prop. 10 By the way of the ship, and two angles of position, to find the distance between the ship and the land. The way of the ship may be known as in the first Prop. The angles may be observed either by the Staff, or by a needle set on the Staff. For example, suppose that being at A, I had sight of the land at B, the ship going East Northeast from A toward C, and the angle of the ships position BAC being 43 gr. 20 m: and after that the ship had made 10 cent. or 2 leagues of way from A unto D, I observed again, and found the second angle of the ships position BDC to be 58 gr. or the inward angle BDA to be 122 gr. then may I find the third angle ABDELLA to be 14 gr. 40 m. either by subtraction or by compliment unto 180 gr. geometric illustration In this and the like cases, I have a right line triangle, in which there is one side and three angles known, and it is required to find the other two sides and the Canon for it, is this: As the sine of the angle opposite to the known side, is to that known side: So the sine of the angle opposite to the side required, is to the side required. Wherefore I extend the compasses from 14 gr. 40 m. in the fines, to 10 in line of numbers, and this extent doth reach from 58 gr. to 33 ½, and such is the distance between A and B, and it reacheth from 43 gr. 20 m. unto 27 in the line of numbers; and such is the distance from D to B. These two distances being known, I may set out the land upon the chart. For having set down the way of the ship from A to D by that which I shown before in the use of the meridian line, I may by the same reason set off the distance AB and DB, which meeting in the point B, shall there resemble the land required. 11 By knowing the distance between two places on the land, and how they bear one from the other, and having the angles of position at the ship to find the distance between the ship and the land. If it may be conveniently, let the angle of position be observed at such time as the ship cometh to be right over against one of the places. As if the places be East and West, seek to bring one of them South or North from you, and then observe the angle of position: so shall you have a right line triangle, with one side and three angles, whereby to find the two other sides. First you have the angle of position at the ship; then a right angle at the place that is over against you; and the third angle at the other place is the compliment to the angle of position. Wherefore As the sine of the angle of position, is to the distance between the two places: So the cousin of the angle of position, to the distance between the ship and the nearer place. And so is the sine of 90 gr. to the distance from the ship to the farther place. So the places being 15 cent. or three leagues one from the other, and the angle of position 29 gr; the nearer distance will be found about 27 cent. and the farther distance about 31 cent. Or howsoever the angle of position were observed, the distance between the ship and the land may be found generally as in this example: Suppose A and D were two head lands known to be East Northeast, and West Southwest, 10 cent. or two leagues one from the other; and that the ship being at B, I observed the angle of the ships position DBA, and found it to be 14 gr. 40 m. and that D did bear 9 gr. 30 m. and A 24 gr. 10 m. from the meridian BS, this example would be like the former. For if the angle SBD be 9 gr. 30 m. from the South to the Westward, then shall NDB be 9 gr. 30 m. from the North to the Eastward. Take these 9 gr. 30 m. out of the angle NDE which is 67 gr. 30 m. because the two head lands lie East Northeast, and there will remain 58 gr. for the angle BDE, and the inward angle BDA shall be 122 gr. Take these two angles ABDELLA and BDA out of 180 gr. and there will remain 43 gr 20 m. for the third angle BAD. Wherefore here also are three angles and one side, by which I may find the two other sides, as in the last Prop. These propositions thus wrought by the Staff, are such as I thought to be useful for seamen, and those that are skilful may apply the example to many others. Those that begin, and are willing to practise, may busy themselves with this which followeth. Suppose four ports, L, N, O, P; of which L is in the latitude of 50 gr. N is North from L 200 leagues or 1000 centesmes; O West from L 1000 cent. and P West from N 1000 cent: so that L and O will be in the same latitude of 50 gr. N and P both in the latitude of 60 gr. Then let two ships depart from L, the one to touch at O, the other at N, and then both to meet at P, there to lad, and from thence to return the nearest way unto L. Here many questions may be proposed. 1 What is the longitude of the port at O? 2 What is the longitude of P? And why O and P should not be in the same longitude? 3 What is the Rumb from O unto P? 4 What is the distance from O unto P? And why the way should be more from L unto P, going by O, then by N? 5 What is the Rumb from P unto L? 6 What is the distance from P unto L? 7 What is the Rumb from N unto O? 8 What is the distance from N unto O? And why it should not be the like Rumb and distance from N unto O, as from P unto L? These questions well considered, and either resolved by the Staff, or pricked down on the chart, and compared with the globe and the common Sea-chart, will give some light to the direction of a course, and reduction of places to their due longitude, which are now foully distorted in the common Sea-charts. An Appendix concerning The description and use of an instrument, made in form of a Crossbow, for the more easy finding of the latitude at Sea. THe former Prop. suppose the latitude to be known, I will here show how it may be easily observed. Upon the centre A, and semidiameter AB, describe an ark of a circle SBN. The same semidiameter will set of 60 gr. from B unto S for the South end, and other 60 gr. from B unto N for the North end of the Bow: so the whole Bow will contain 120 gr. the third part of a circle. Let it therefore be divided into so many degrees, and each degree subdivided into six parts, that each part may be ten minutes: but let the numbers set to it be 5. 10. 15. unto 90 gr. and then again 5. 10. 15. unto 25. that 55 may fall in the middle, as in this figure. geometric illustration The Bow being thus divided and numbered, you may se● the months and days of each month upon the back, and such stars as are fit for observation upon the side of the Bow. If you desire to make use of it in North latitude, you may number 23 gr. 30 m. from 90 towards the end of the Bow at N, and there place the tenth day of june. And 23 gr. 30 m. from 90 toward S; and there at 66 gr. 30 m. place the tenth day of December. And so the rest of the days of the year, according to the declination of the Sun at the same days. The stars may be placed in like manner according to their declinations. Arcturus at 21 gr. 10 m. The Bull's eye 15 42 The Lion's heart 13 45 The Vultures heart 7 58 The little dog 6 9 from 90 toward the North end of the Bow at N. Then for Southern stars, you may number their declination from 90 toward the South end of the Bow at S. As first the three stars in Orion's girdle, The first at 0 gr. 37 m. The second 1 28 The third 2 11 The Hydra's heart 7 5 The virgin's spike 9 10 The great dog 16 12 The Scorpions heart 25 30 Fomahant 31 30 And so the South crown, the triangle, the clouds, the crosiers, or what other stars you think fit for observation. This I call the fore side of the Bow. If you desire to make use of it in South latitude, you may turn the Bow, and divide the back side of it, and number it in like manner; and then put on the months and days of the year, placing the tenth of December at the South end, and the tenth of june toward the middle of the Bow, and the rest of the days according to the Sun's declination as before. The chiefest of the Northern stars may here be placed in like manner according to their declination, Anno 1625. The pole star at 87 gr. 20 m. The first guard 75 45 The second guard 73 25 The great Bears back 63 45 In the great Bear's tail first second third 58 2 57 55 51 15 The side of Perseus 48 28 The goat 45 33 The tail of the swan 44 0 The head of Medusa 39 30 The harp 38 30 Castor 32 38 Pollux 28 52 The North crown 28 0 The Rams head 21 40 Arcturus 21 10 The Bull's eye 15 42 The Lion's heart 13 45 The Vultures heart 7 58 Orion's right shoulder 7 17 Orion's left shoulder 5 57 And so any other star, whose declination is known unto you, which being done. The use of this Bow may be 1 The day of the month being known, to find the declination of the Sun. 2 The declination being given, to find the day of the month. These two Prop. depend on the making of the Bow. If the day be known, look it out in the back of the Bow: so the declination will appear in the side. Or if the declination be known, the day of the month is set over against it. As if the day of the month were the 14 of july: look for this day in the back of the Bow, and you shall find it over against 20 gr. of North declination. If the declination given be 20 gr. to the Southward, you shall find the day to be either the eleventh of November, or the eleventh of january. 3 To find the altitude of the Sun or stars. Here it is fit to have two running sights, which may be easily moved on the back of the Bow. The upper sight may be set either to 60 gr. or to 70 gr. or to 80 gr. as you shall find to be most convenient: the other sight may be set on, to any place between the middle and the other end of the Bow. Then with the one hand hold the centre of the Bow to your eye, so as you may see the Sun or star by the upper sight, and with the other hand move the lower sight up or down until you have brought one of the edges of it to be even with the horizon (as when you observe with the cross-staff:) so the degrees contained between that edge and the upper sight, shall show the altitude required. Thus if the upper sight shall be at 80 gr. and the lower sight at 50 gr. the altitude required is 30 gr. 4 To find any North latitude, by knowing either the day of the month, or the declination of the Sun. As oft as you are to observe in North latitude, place both the sights on the foreside of the Bow, the upper sight to the declination of the Sun, or the day of the month at the North end, and the lower sight toward the South end. Then when the Sun cometh to the meridian, turn your face to the South, and with the one hand hold the centre of the Bow to your eye, so as you may see the Sun by the upper sight; with the other hand move the lower sight, until you have brought one of the edges of it to be even with the horizon: so that edge of the lower sight shall show the latitude of the place in the fore side of the Bow. Thus being in North latitude upon the ninth of October: if I set the upper sight to this day, at the fore side and North end of the Bow, I shall find it to fall to the Southward of 90 upon 80 gr. and therefore at 10 gr. of South declination. Then the Sun coming to the meridian, I may set the centre of the Bow to mine eye, as if I went to find the altitude of the Sun, holding the North end of the Bow upward, with the upper sight between mine eye and the Sun, and moving the lower sight, until it come to be even with the horizon. If here the lower sight shall stay a● 50 gr. I may well say, that the latitude is 50 gr. For the meridian altitude of the Sun is 30 gr. by the last Prop. and the Sun having 10 gr. of South declination, the meridian altitude of the equator would be 40 gr; and therefore the observation was made in 50 gr. of North latitude. By the same reason, if the lower side had stayed at 51 gr. 30 m. the latitude must have been 51 gr. 30 m. and so in the rest. 5 To find any North latitude, by the meridian altitude of the stars to the Southward. Let the upper sight be set to the star, which you intent to observe, here placed in the fore side of the Bow. Then hold the North end of the Bow upward, and turning your face to the South, observe the meridian altitude as before: so the lower sight shall show the latitude of the place in the fore side of the Bow. Thus if in observing the meridian altitude of the great Dogstar, the lower sight shall stay at 50 gr. it would show the latitude to be 50 gr. For this star being here placed at 73 gr. 48 m. if we take thence 50 gr. his meridian altitude would be 23 gr. 48 m. to this if we add 16 gr. 12 m. for the South declination of this star, it would show the meridian altitude of the equator to be 40 gr. and therefore the latitude to be 50 gr. 6 To find any North latitude, by the meridian altitude of the stars to the Northward. Let the upper sight be set to the star which you intent to observe, here placed on the back side of the Bow. Then hold the North end of the Bow upward, and turning your face to the North, observe the altitude of the star when he cometh to be in the meridian and under the pole: so the lower sight shall show the altitude of the pole in the back side of the Bow. Thus the former guard coming to be in the meridian under the pole, if you observe and find the lower sight to stay at 50 gr. such is the elevation of the pole, and the latitude of the place to the Northward. For the distance between the two sights will show the altitude to be 35 gr. 45 m. & the star is 14 gr. 15 m. distant from the North pole. These two do make up 50 gr. for the elevation of the North-pole, and therefore such is the North latitude. 7 To find any South latitude, by knowing either the day of the month, or the declination of the Sun. When you are come into South latitude, turn both your sights to the backside of the Bow: the upper sight to the declination of the Sun, or the day of the month at the South end, and the lower sight toward the North end of the Bow. Then the Sun coming to the meridian, turn your face to the North, and holding the South end of the Bow upward, observe the meridian altitude as before: so the lower sight shall show the latitude of the place in the back side of the Bow. Thus being in South latitude, upon the tenth of May if you observe and find the lower sight to stay at 30 gr. on the back side of the Bow, such is the latitude. For the declination is 20 gr. Northward, the altitude of the Sun between the two sights 40 gr. the altitude of the equator 60 gr. and therefore the latitude 30 gr. 8 To find any South latitude, by the meridian altitude of the stars to the Northward. Let the upper sight be set to the star which you intent to observe, here placed on the back side of the Bow. Then hold the South end of the Bow upward, and turning your face to the North, observe the meridian altitude as before: so the lower sight shall show the latitude of the place in the back side of the Bow. Thus being in South latitude, and the former guard coming to be in the meridian over the pole. If you observe and find the lower sight to stay at 5 gr. such is the latitude. For this star is 14 gr. 15 m. from the North pole, the altitude of the star between the two sights 9 gr. 15 m. the North pole depressed 5 gr. and therefore the latitude 5 gr. to the Southward. 9 To observe the altitude of the Sun backward. Set the upper sight either to 60, or 70, or 80 gr. as you shall find it to be most convenient, the lower sight on any place between the middle and the other end of the Bow, and have an horizontal sight to be set to the centre. Then may you turn your back to the Sun, and the back of the Bow toward yourself, looking by the lower sight through the horizontal sight, and moving the lower sight up & down, until the upper sight do cast a shadow upon the middle of the horizontal sight: so the degrees contained between the two sights on the Bow, shall give the altitude required. Thus if the upper sight shall be at 80 gr. and the lower sight at 50 gr. the altitude required is 30 gr. as in the third Prop. 10 To find any North latitude by a back observation, knowing either the day of the month, or the declination of the Sun. When you observe in North latitude, place your three sights on the fore side of the Bow: the upper sight to the declination of the Sun, or the day of the month, at the North end; the lower sight toward the South end of the Bow; and the horizontal sight to the centre. Then the Sun coming to the meridian, turn your face to the North, & holding the North end of the Bow upward, the South end downward, with the back of it toward yourself, observe the shadow of the upper sight as in the former Prop. so the lower sight shall show the latitude of the place in the fore side of the Bow. Thus being in North latitude upon the ninth of October, if you observe and find the lower sight to stay at 50 gr. on the fore side of the Bow, such is the latitude. For the declination is 10 ●r. Southward, and the altitude of the Sun between the two sights 30 gr. the altitude of the equator 40 gr. and ●herefore the latitude 50 gr. as in the fourth Prop. 11 To find any South latitude by a back observation, knowing either the day of the month, or the declination of the Sun. When you observe in South latitude, place your three sights on the back side of the Bow: the upper sight to the declination of the Sun, or the day of the month at the South end; the lower sight toward the North end of the Bow, and the horizontal sight to the c●nter. Then the Sun coming to the meridian, turn your face to the South, and holding the South end of the Bow upward, with the back of it toward yourself, observe the shadow of the upper sight as before: so the lower sight shall show the latitude of the place in the back side of the Bow. Thus being in the South latitude upon the tenth of May, if you observe and find the lower sight to stay at 30 gr. on the ●ledge●acke of the Bow, such is the altitude. For the declination ●ledge● 20 gr. Northward, the altitude of the Sun between the ●ledge●wo sights 40 gr. the altitude of the equator 60 gr. and there●ledge●re the latitude 30 gr. as in the seventh Prop. 12 To find the day of the month, by knowing the latitude of the place, and observing the meridian altitude of the Sun. Place your three sights according to your latitude; the ho●ledge●zontall sight to the centre, the lower sight to the latitude, ●ledge●d the upper sight among the months. Then when the ●ledge●nne cometh to the meridian, observe the altitude, looking ●ledge● the lower sight through the horizontal, and keeping the ●ledge●wer sight still at the latitude, but moving the upper sight ●ledge●til it give shadow upon the middle of the horizontal sight: ●ledge● the upper sight shall show the day of the month requi●ledge●d. Thus in our latitude if you set the lower sight to 51 gr. 30 ●ledge● and observing find the altitude of the Sun between ●ledge●at and the upper sight to be 28 gr. 30 m. this upper sight ●ledge●ll ●all upon the ninth of October, and the twelfth of Fe●ledge●uar●e. And if yet you doubt which o● them two is the day, ●ledge●u may expect another meridian altitude; and than if you ●ledge●d the upper sight upon the tenth of October, and the ele●ledge●nth of February, the question will be soon resolved. 13 To find the declination of any unknown star, and so to place it on the Bow, by knowing the latitude of the place, and observing the Meridian altitude of the Star. When you find a star in the Meridian that is fit for ob●ledge●uation. Set the centre of the Bow to your eye, the lower ●ledge●ht to the latitude, and move the upper sight up or down ●ledge●till you see the horizon by the lower sight, and the star by the upper sight, then will the upper sight stay at the declination and place of the star. Thus being in 20 gr. of North latitude, if you observe an●redge● find the meridian altitude of the head of the Crosier to b●redge● 14 gr. 50 m. The upper sight will stay at 34 gr. 50 m. and ther●redge● may you place this star. For by this observation the distance o●redge● this star from the South pole should be 34 gr. 50 m. and th●redge● declination from the equator 55 gr. 10 m. And so for the res●redge● The stars which I m●ntione● be●●re, do come to the meridian in this order, after the first point of Aries. Ho. Mi. The pole star at 0 29 The rams head 1 46 The head of Medusa 2 44 The sid● of Perseus 2 58 The Bull's eye. 4 15 The goat 4 49 Orion's left shoulder 5 5 Orion's girdle the first the second the third. 5 13 5 17 5 22 Orion's right shoulder 5 35 The great dog 6 29 Castor 7 10 The little dog 7 20 Pollux 7 22 The Hydra's heart 9 9 The lions heart 9 48 The great bears back 10 40 First in gr. bears tail 12 37 The Virgin's spike 13 5 Second in gr. bea● tail 13 9 Third in gr. bears tail 13 33 Arcturus 13 58 The foremost gum rd 14 52 The North crown 15 19 Th● h● dmost guard 15 25 Scorpions heart 16 7 The harp 18 24 Praetors heart 19 33 Swans tail 20 29 Fomahant 22 36 The end of the second Book. THE THIRD BOOK. Of the use of the lines of Numbers, Sins and Tangents for the drawing of Houre-lines on all sorts of Planes. THere are ten several sorts of Planes, which take their denomination from those great circles to which they are parallels, and may sufficiently for our use be represented in this one fundamental Diagram described before in the use of the Sector, and be known by their horizontal and perpendicular lines, of such as know the latitude of the place, and the circles of the sphere. 1 An horizontal plane parallel to the horizon, here represented by the outward circle ESWN. 2 A vertical plane parallel to the prime vertical circle which passeth through the zenith and the points of East and West in the horizon, and is right to the horizon and the meridian; that is, maketh right angles with them both. This is represented by EZW. 3 A polar plane parallel to the circle of the hour of 6, which passeth through the pole and the points of East and West, being right to the Equinoctial and the Meridian, but inclining to the horizon, with an angle equal to the latitude. This is here represented by EPW. 4 An equinoctial plane parallel to the Equinoctial, which passeth through the points of East and West, being right to the Meridian, but inclining to the Horizon, with an angle equal to the compliment of the latitude. This is here represented by EAW. geometric illustration 6 A meridian plane parallel to the meridian, the circle of the hour of 12, which passeth through the zenith, the pole, and the points of South and North, being right to the horizon, and the prime vertical. This is here represented by SZN. 7 A meridian plane inclining to the horizon, parallel to any great circle, which passeth through the points of South and North, being right to the prime vertical, but inclining to the horizon. This is here represented by SGN. 8 A vertical declining plane, parallel to any great circle, which passeth through the zenith, being right to the horizon, but inclining to the meridian. This is represented by BZD. 9 A polar declining plane, parallel to any great circle, which passeth through the pole, being right to the equinoctial, but inclining to the meridian. This is here represented by HPQ. 10 A declining inclining plane, parallel to any great circle, which is right to none of the former circles, but declining from the prime vertical, and inclining both to the horizon and the meridian, and all the hour circles. This may be here represented either by BMD, or BFD, or BKD, or any such great circle, which passeth neither through the South and North, nor East and West points, nor through the zenith nor the pole. Each of these planes (except the horizontal) hath two faces whereon houre-lines may be drawn; and so there are 19 planes in all. The meridian plane hath one face to the East, and another to the West: the other vertical planes have one to the South, and another to the North, and the rest one to the zenith, and another to the nadir: but what is said of the one, may be understood of the other. To find the inclination of any Plane. For the distinguishing of these Planes we may find whether they be horizontal, or vertical, or inclining to the horizon, and how much they incline, either by the usual inclinatorie quadrant, or by fitting a thread and plummet unto the Sector. For let the Sector be opened to a right angle, the lines of Sins to an angle of 92 gr. the inward edges of the Sector to 90 gr. and let a thread and plummet be hanged upon a line parallel to the edges of one of the legs, so that leg shall be vertical, and the other leg parallel to the horizon. building with angles marked If the plane seem to be level with the horizon, you may try it by setting the horizontal leg of the Sector to the plane, and holding the other leg upright: for than if the thread shall fall on his plummet line, which way soever you turn the Sector, it is an horizontal plane. If the one end of the plane be higher than the other, and yet not vertical, it is an inclining plane, and you may find the inclination in this manner. First hold the vertical leg of the Sector upright, and turn the horizontal le● about, until it lie close with the plane, and the thread fall on his plummet line: so the line drawn by the edge of that horizontal leg, shall be an horizontal line. Suppose the plane to be BGED, and that BD were thus found to be the horizontal line upon the plane; then may you cross the horizontal line at right angles with a perpendicular CF: that done, if you set one of the legs of the Sector upon the perpendicular line CF, and make the other leg with a thread and plummet to become vertical, you shall have the angle between the vertical line and the perpendicular on the Plane, as before in the use of the Sector, pag. 50. and the compliment of this angle is the inclination of the plane to the horizon. To find the declination of a Plane. The declination of a Plane is always reckoned in the horizon between the line of East and West, and the horizontal line upon the Plane. As in the fundamental Diagram, the prime vertical line (which is the line of East and West) is ECW; if the horizontal line of the plane proposed shall be BCD, the angle of declination is ECB. But because a Plane may decline diverse ways, that we may the better distinguish them, we consider three lines belonging to every Plane: the first is the horizontal line; the second the perpendicular line, crossing the horizontal at right angles; the third the axis of the plane, crossing both the horizontal line, and his perpendicular, and the plane itself at right angles. The perpendicular line doth help to find the inclination of the plane as before, the horizontal to find the declination, the axis to give denomination unto the plane. For example, in a vertical plane here represented by EZW, the horizontal line is ECW, the same with the line of East and West, and therefore no declination; the perpendicular crossing it, CZ the same with the vertical line, drawn from the centre to the zenith, right unto the horizon, and therefore no inclination. The axis of the plane is SCN, the same with the meridian line, drawn from the South to the North, and accordingly gives the denomination to the plane. For the plane having two faces, and the axis two poles, S and N; the pole S falling directly into the South, doth cause that face to which it is next to be called the South face; and the other pole at N, pointing into the North, doth give the denomination to the other face, and make it to be called the North face of this plane. In like manner in the declining inclining plane here represented by BFD, the horizontal line is BCD, which crosseth the prime vertical line ECW, and therefore it is called a declining plane, according to the angle of declination ECB or WCD. The perpendicular to this horizontal line is CF, where the point F falleth in the plane QZH perpendicular to the plane proposed, between the zenith and the North part of the horizon, and therefore it is called a plane inclining to the Northward, according to the ark FQ, or the angle FCQ. The axis of the plane is here represented by the line CK, where the pole K is 90 gr. distant from the plane, and so is as much above the horizon at H, and the other pole as much below the horizon at Q, as the plane at F is distant from the zenith: and this pole K here falling between the meridian and the prime vertical circle into the South-west part of the world, this upper face of the plane is therefore called the South-west face, and the lower the North-east face of the plane. The declination from the prime vertical may be found by the needle in the usual inclinatorie Quadrant, or rather by comparing the horizontal line drawn upon the plane with the azimuth of the Sun and the meridian line, in such sort as before we found the variation of the magnetical needle. For take any board that hath one side straight, and draw the line HO parallel to that side, and the line ZM perpendicular unto it, and on the centre Z make a semicircle HMO: this done, hold the board to the plane, so as HO may be parallel to BD the horizontal line on the plane and the board parallel to the horizon; then the Sun shining upon it, hold out a thread and plummet, so as the thread being vertical, the shadow of the Sun may fall on the centre Z, and draw the line of shadow AZ representing the common section, which the azimuth of the Sun makes with the plane of the horizon, and let another take the altitude of the Sun at the same instant: so by resolving a triangle, as I shown before pag. 65. you may find what azimuth the Sun was in when he gave shadow upon AZ. Suppose the azimuth to be (as before pag. 64.) 72 gr. 52 m. from the North to the Westward, and therefore 17 gr. 8 m. from the West, we may allow these 17 gr. 8 m. from A unto V, and draw the line ZV, and so we have the true West point of the prime vertical line: then allowing 90 gr. from V unto S, we have the South point of the meridian line ZS, and the angle HZV shall give the declination of the plane from the vertical, and the angle OZS the declination of the plane from the meridian. Or we may take out only the angle AZH, which the line of shadow makes with the horizontal line of the plane, and compare it with the angle AZV, which the line of shadow makes with the prime vertical. And so here if AZV the Sun's azimuth shall be 17 gr. 8 m. past the West, and yet the line of shadow AZ 7 gr. 12 m. short of the plane, the declination of the plane shall be 24 gr. 20 m. as may appear by the site of the plane and the circles. If the altitude of the Sun be taken at such time as the shadow of the thread falleth on BD or HO, and then a triangle resolved, the declination of the plane will be such as the azimuth of the Sun from the prime vertical. If at such a time as the shadow falleth on MZ, the declination will be such as the azimuth of the Sun from the meridian. If it be a fair Summer's day you may first find what altitude the Sun will have when he cometh to be due East or West, and then expect until he come to that altitude; so the declination of the plane shall be such as the angle contained between the line HO and the line of the shadow. Having distinguished the Planes, the next care will be for the placing of the style and the drawing of the houre-lines. The style will be as the axis of the world, sometimes parallel to the plane, sometimes perpendicular, sometimes cut the plane with obliqne angles. The houre-lines will be either parallel one to the other, or meet in a centre with equal angles, or meet with unequal angles. If the style be perpendicular to the plane, the angles at the centre will be equal; and this falls out only in the South and North face of an equinoctial plane: if the style be parallel to the plane, the houre-lines will be also parallel one to another; and this falls out in all polar planes; as in the East and West meridian planes parallel to the circle of the hour of 12, in the upper and lower direct polars parallel to the circles of the hour of 6, and in the upper and lower declining polars which are parallel to any of the other hour circles. But in the horizontal and all other planes, the style will cut the plane with an acute angle, and the hour lines will meet at the root of the style, and there make unequal angles. CHAP. I. To draw the houre-lines in an equinoctial Plane. AN equinoctial plane is that which is parallel to the equinoctial circle here represented by EAW, wherein the spaces between the hour circle's being equal, there is no need of further precept, but only to draw a circle and to divide it into 24 equal parts for the 24 hours, and subdivide each hour into halves and quarters, and then to set up geometric illustration the style perpendicular to the plane in the centre of the circle. The help which these lines of proportion do here afford us, is only in the division of the circle, which may be done readily by that which I shown before, Pag. 29. For example, suppose the semidiameter of the equinoctial circle to be six inches, and that it were required to know the distance of the houre-points each from other: here each hour being 15 gr. distant from other, I extend the compasses from the sine of 30 gr. unto the sine of 7 gr. 30 m. the half of 15 gr. and I find the same extent to reach in the line of numbers from 6.00 unto 1.56. Or in cross work I extend them from the sine of 30 gr. unto 6.00 in the line of numbers, the same extent will reach from the sine of 7 gr. 30 m. unto 1.56 in the line of numbers; which shows that in a circle of six inches semidiameter, the geometric illustration distance of the houre-points each from other will be about 1 inch and 56 centesmes or parts of 100 The like reason holds for the inscribing of all other chords in the Prop. following. CHAP. II. To draw the houre-lines in a direct polar plane. A Direct polar plane is that which is parallel to the hour of 6, here represented by EPW, wherein the style will be parallel to the plane, and the houre-lines parallel one to the other, and therefore may be best drawn by that which I have showed in the use of the Sector. They may be also drawn by the help of these lines of proportion, in this manner. First draw a right line WE for the horizon and the equator, and cross it at the point C, about the middle of the line with CB another right line, which may serve for the meridian and the hour of 12, and must also be the substylar line wherein the style shall stand. Then, to proportion the style unto the plane, consider the length of the horizontal line, and what houre-lines you would have to fall on your plane. For the distance of any one houre-line from the meridian being known, we may find both the length of the style and the distance of the rest: because As the tangent of the hour given, is to the distance from the meridian: So the tangent of 45 gr. to the height of the style. Suppose the length of the horizontal line to be 12 inches, and that it were required to put on all the houre-lines from 7 in the morning unto 5 in the evening. Here we have 5 hours and 6 inches on either side the meridian. Wherefore I allow 15 gr. for an hour, and extending the compasses the compasses from the tangent of 75 gr. the measure of 5 hours unto the tangent of 45 gr. I find the same extent to reach in the line of numbers from 6.00 to about 1.61. This shows both the height of the style, and the distance of the houre-points of 9 & 3 from the meridian to be 1 inch, 61 parts. To find the length of the Tangent between the substylar and the houre-points. As the tangent of 45 gr. to the tangent of the hour: So the height of the style to the length of the tangent line between the substylar and the houre-points. Thus having found the length of the style in our example to be 1.61, if I exend the compasses from the tangent of 45 gr. unto the tangent of 15 gr. the measure of the first hour from the substylar, I shall find the same extent to reach in the line of numbers from 1.61 unto 0.43 for the length of the tangent between the substylar and the houre-points of 11 and 1. If I extend them from the tangent of 45 gr. unto the tangent of 75 gr. the measure of the fift hour, I shall find them to reach in the line of numbers from 1.61 unto 6.00 for the length of the tangent from the substylar to the houre-points of 7 and 5. For howsoever it be the same distance in the line of tangents from 45 unto 75, as from 45 unto 15; yet because 75 are more, and 15 less than 45, the tangent lines that answer to them will be accordingly more or less than the length of the style. Ho. An. Po Tang. Gr. M. In. Par 12 0 0 0 0 11.1 15 0 0 43 10.2 30 0 0 93 9.3 45 0 1 61 8.4 60 0 2 79 7.5 75 0 6 0 6.6 90 0 Infin Again, if I extend them from 45 gr. in the tangents unto 30 gr. the measure of the second hour, I shall find them to reach in the line of numbers from 1.61 unto 0.93 for the hour of 10 and 2: if I extend them from the tangent of 45 gr. unto the tangent of 60 gr. for the fourth hour, I shall find them to reach in the line of numbers from 1.61 unto 2 79, and such is the length of the tangent line from the substylar unto the hour of 8 and 4. And the like reason holdeth for the inscribing of all other tangent lines in the propositions following. But for such tangents as fall under 45 gr. I may better use cross work, and extend the compasses from the tangent of 45 gr. unto 1.61 in the line of numbers, so shall I find the same extent to reach from 30 gr. in the tangents, to 93 parts in the line of numbers, for the distance of the second hour, and from 15 gr. in the tangents to 43 parts for the distance of the first hour from the meridian. Or if this extent from 45 gr. backward to 1.61 be too large for the compasses, I may extend forward from the tangent of 5 gr. 43 m. to 1.61 parts in the line of numbers, and the same extent shall reach from 15 gr. in the tangents, to 43 parts in the line of numbers, for the distance of the first hour; and from 30 gr. to 93 parts, for the distance of the second hour, as before. Having found the length of the tangent lines in inches and parts of inches, and pricked them in the equator on both sides of the meridian, from the centre C; if we draw right lines through each of those points, crossing the equator at right angles, they shall be the houre-lines required; and if we set a style over the meridian, so as the edge of it be parallel to the plane, and the height of it be as much above the meridian as the distance between the meridian and the houre-points of 3 or 9, it shall represent the axis of the world, and be truly placed for the casting of the shadow upon the houre-lines in a polar plane. Horae Ang. Po Tangle Gr. M. In. Pa 6 0 0 3 45 165 7 30 1 32 11 15 1 99 7 15 0 2 68 18 45 3 39 22 30 4 14 26 15 4 93 8 30 0 5 77 33 45 6 68 37 30 7 67 41 15 8 77 9 45 0 10 00 48 45 11 40 52 30 13 03 56 15 14 97 10 60 0 17 32 63 45 20 28 67 30 24 14 71 15 29 46 11 75 0 37 32 78 45 50 27 82 30 75 96 86 15 152 57 12 90 0 Infin. CHAP. III. To draw the houre-lines in a meridian plane. A Meridian plane is that which is parallel to the meridian circled here represented by SZN; it hath two faces, one to the East, and the other to the West; in each of them the style will be parallel to the plane, and the houre-lines parallel one to the other, as in a polar plane, the difference being only in the placing of the equator & in numbering of the hours. geometric illustration For supposing the length of the style CB to be ten inches, the length of the tangent line belonging to the first hour will be 2 in. 68 p. the length of the second 5 in. 77 p. as in the Table. Then the tangent of 15 gr. being pricked down in the equator on both sides from 6, shall serve for the hours of 5 and 7, and the tangent of 30 gr. for the hours of 4 and 8, and so in the rest. This done, if we draw right lines through each of these points, crossing the equator at right angles, they shall be the houre-lines required: and if we set a style over the hour of 6, so as the edge of it may be parallel to the plane, and the height of it may be equal to the distance between the hours of 6 and 9 in the equator, it shall represent the axis of the world, and be truly placed for the casting of the shadow upon the houre-lines in a meridian plane. CHAP. FOUR To draw the houre-lines in an horizontal plane. AN horizontal plane is that which is parallel to the horizon, here represented by the outward circle ESWN, in which the diameter SN drawn from the South to the North, may go both for the meridian line and the meridian circle, Z for the zenith, P for the pole of the world, and the circles drawn through P for the houre-circles of 1.2.3.4. etc. as they are numbered from the meridian. These houre-circles considered with the meridian & the horizon, do make diverse triangles, PN 1, PN 2, PN 3, in which we have known first the right angle at N, the North intersection of the meridian and the horizon; secondly the side PN, the ark of the meridian between the pole and the horizon, which is always equal to the latitude of the place; thirdly the angles at the pole, made by the meridian and the houre-circles, the angle NP 1 being 15 gr. NP 2 30 gr. each hour 15 gr. more than other, each half hour 7 gr. 30 m. each quarter 3. gr. 45. m. And these three being known, we may find the arks of the horizon between the meridian and the houre-circles N 1, N 2, N 3, etc. For As the sine of 90 gr. is to the sine of the latitude: So the tangent of the hour to the tangent of the houre-line from the meridian. Ho. Ang. Po Arc. Plam Gr. M. Gr. M. 12 0 0 0 0 3 45 2 56 7 30 5 52 11 15 8 51 1 15 0 11 50 18 45 14 52 22 30 17 57 26 15 21 6 2 30 0 24 20 33 45 27 36 37 30 31 0 41 15 34 28 3 45 0 38 3 48 45 41 45 52 30 45 34 56 15 49 30 4 60 0 53 35 63 45 57 47 67 30 62 6 71 15 66 33 5 75 0 71 6 78 45 75 45 82 30 80 25 86 15 85 13 6 90 0 90 0 geometric illustration Extend the compasses from the sine of 90 gr. to the sine of the latitude, so the same extent shall reach from the tangent of the hour, to the tangent of the houre-line from the meridian. Thus the latitude being 51 gr. 30 m. I extend the compasses from the sine of 90 gr. to the sine of 51 gr. 30 m. & find the same extent to reach from the tangent of 3 gr. 45 m. unto the tangent of 2 gr. 56 m. for the distance of the first quarter from the meridian; and from the tangent of 7 gr. 30 m. unto the tangent of 5 gr. 52 m. for the half hour; and from the tangent of 11 gr. 15 m. to the tangent of 8 gr. 51 m. for the third quarter; and from the tangent of 15 gr. 0 m. unto 11 gr. 50 m. for the first hour: and so in the rest. geometric illustration This done, I come to the Plane, and there according as the lines do fall in the fundamental diagram, 1 I draw a right line SN serving for the meridian, the hour of 12 and the substylar. 2 In this meridian I make choice of a centre at C, and there describe an occult circle representing the horizon. 3 I find a chord of 11 gr. 50 m. and inscribe it into this circle on either side of the meridian for the hours of 11 & 1; in like manner a chord of 24 gr. 20 m. for the hours of 10 and 2; and a chord of 38 gr. 3 m. for the hours of 9 and 3; and so for the rest of the hours, their halves and quarters. 4 I draw right lines through the centre and the terms of these chords, and these lines so drawn are the houre-lines required. Lastly I set up the style over the meridian, so as it may cut the plane in the centre, and there make an angle with the meridian equal to the latitude of the place, so it shall represent the axis of the world, and be truly placed for casting of the shadow upon the houre-lines in an horizontal plane. CHAP. V To draw the houre-lines in a vertical plane. A Vertical plane is that which is parallel to the prime vertical circle here represented by EZW. It hath two faces, one to the North, the other to the South; in each of them the substylar will be the same with the meridian line, and the angle of the style above the plane will be equal to ZP the compliment of the latitude. The triangles here considered are made by the vertical, the meridian, and the houre-circles, in which we know the side ZP, the angles at the pole, and the right angle at the zenith, and therefore may find the arks of the vertical, between the meridian and the houre-circles after this manner: As the sine of 90 gr. is to the cousin of the latitude: So the tangent of the hour to the tangent of the houre-line from the meridian. Extend the compasses from the sine of 90 gr. to the sine of the compliment of the latitude, so the same extent shall reach from the tangent of the hour, to the tangent of the houre-line from the meridian. Thus in the latitude of 51 gr. 30 m. I extend the compasses from the sine of 90 gr. to the sine of 38 gr. 30 m. and find the same extent to reach from the tangent of 15 gr. to the tangent of 9 gr. 28 m for the distance of the first hour from the meridian: and from the tangent of 75 gr. unto the tangent of 66 gr. 42 m. for the fift hour; and so in the rest as in the Table following. geometric illustration These arks being known, I may come to the plane, and there by help of a thread and plummet draw a vertical line serving both for the meridian and the hour of 12, and the substylar; then may I draw an occult vertical circle, and therein inscribe the chords of those former arks, and draw the houre-lines, and set up the style, as before in the horizontal plane. Ho. Ang. Po Arc. Plam Gr. M. Gr. M. 12 0 0 0 0 3 45 2 20 7 30 4 41 11 15 7 3 1 15 0 9 28 18 45 11 56 22 30 14 27 26 15 17 4 2 30 0 19 46 33 45 22 35 37 30 25 32 41 15 28 38 3 45 0 31 54 48 45 35 22 52 30 39 3 56 15 42 58 4 60 0 47 9 63 45 51 36 67 30 56 20 71 15 61 23 5 75 0 66 42 78 45 72 17 82 30 78 3 86 15 84 0 6 90 0 90 0 If it be the South face of the plane, the centre will be upward, and the style must point downward; if the North face, the centre must be in the lower part of the meridian line, and the style-point upward in all such places as are to the Northward of the equinoctial line, as it may appear by considering how the lines do fall in the fundamental Diagram. CHAP. VI To draw the houre-lines in a vertical inclining plane. ALl those Planes that have their horizontal line lying East and West, are in that respect said to be vertical; if they be also upright and pass through the zenith, they are direct verticals; if they incline to the pole-they are direct polars: if to the equinoctial, they are properly called equinoctial planes, and are described before: if to none of these three points, they are then called by the general name of inclining verticals. These may incline either to the North part of the horizon, or to the South; and each of them hath two faces, one to the zenith, the other to the nadir, in which we are first to consider the height of the pole above the plane, by comparing the inclination of the plane to the horizon, with the latitude of the place. As in our latitude of 51 gr. 30 m. if the inclination of the plane EIW shall be 13 gr. Northward, that is, if IN the ark of the meridian between the plane and the North part of the horizon shall be 13 gr. we may take these 13 gr. out of PN 51 gr. 30 m. the elevation of the pole above the horizon, and there will remain PI 38 gr. 30 m. for the elevation of the North pole above the upper face of the plane, and therefore 38 gr. 30 m. for the height of the South pole above the lower face of the plane. Or if the inclination of the plane shall be found to be 62 gr. to the Southward, we may number them in the meridian from S the Southpart of the horizon unto L, and there draw the ark ELW representing this plane; so the ark of the meridian PL shall give the height of the North pole above the upper face of this plane to be 66 gr. 30 m. and therefore the height of the South pole above the lower face of the plane is also 66 gr. 30 m. In like manner if the inclination of the plane EYW shall be 15 gr. Southward, that is, if SY the ark of the meridian between the South part of the horizon and the plane, shall be 15 gr. The height of the North pole above the upper face of the plane, and the height of the South pole above the lower face of the plane, will be also found to be 66 gr. 30 m. But if the plane shall fall between the zenith and the North pole, then will the North pole be elevated above the lower face, and the South pole above the upward face of the plane, as may appear by the projection of the sphere in the fundamental Diagram. Then in the triangles made by the plane, the meridian, and the houre-circles, we have the side which is the height of the pole above the plane, together with the angles at the pole, and the right angle at the intersection of the meridian with the plane, by which we may find the arks of the plane between the meridian and the houre-circles, after this manner: As the sine of 90 gr. is to the sine of the pole above the plane: So the tangent of the hour to the tangent of the houre-line from the meridian. Thus in the former example, where PI the height of the pole above the plane was found to be 38 gr. 30 m. if you shall extend the compasses from the sine of 90 gr. to the sine of 38 gr. 30 m. the same extent will reach from the tangent of 15 gr. unto the tangent of 9 gr. 28 m. for the distance of the first hour from the meridian, and from 30 gr. unto 19 gr. 46 m. for the second hour, and so forward as in the direct vertical. And for the two last examples, you may extend the compasses from the sine of 90 gr. unto the sine of 66 gr. 30 m: so the same extent shall reach in the line of tangents from 15 gr. unto 13 gr. 48 m. for the first hour, from 75 gr. unto 73 gr. 43 m. for the fift hour, from 30 gr. unto 27 gr. 54 m. for the second hour, from 60 gr. unto 57 gr. 48 m. for the fourth hour, and from 45 gr. unto 42 gr. 31 m. for the third hour from the meridian. These arks being known, you may first draw the horizontal line, and cross it in the middle with a perpendicular that may serve both for the meridian and the hour of 12, and the substylar; then knowing which pole is elevated above the plane, you may accordingly make choice of a fit point in the meridian for the centre of your houre-lines, and thence describe an occult ark of a circle, inscribe the chords of those former arks, and draw the hour lines, and set up the style, as I shown before in the horizontal plane. CHAP. VII. To draw the houre-lines in an vertical declining Plane. ALl upright planes whereon a man may draw a vertical line, are in this respect said to be vertical; if they shall also stand directly East and West, they are direct verticals; if directly North and South, they are properly called meridian planes, and are described before: if they behold none of these four principal parts of the world, but shall stand between the prime vertical and the meridian, they are then called by the general name of declining verticals. These have two faces, one to the South, the other to the Northward, which may be distinguished in these Northern parts of the world after this manner. If the Sun coming to the meridian shall shine upon the plane, it is the South face; if not, it is the North face of that plane. Again, if the Sun shall shine upon the plane at high noon, and yet longer in the forenoon then in the afternoon, it is the Southeast face; if longer in the afternoon then in the forenoon, it is the South-west face of the plane. But how much the declination cometh to, is best found as before. When the declination is found, there be four things more to be considered before we can come to the drawing of the houre-lines, and all four represented in the fundamental Diagram. 1 The meridian of the plane and his inclination to the meridian of the place. Let the ark EZW represent the prime vertical, and BZD a declining vertical, according to the angle of declination EZB, the meridian of the place is represented by PZS, crossing the vertical EZW at right angles at the zenith in the point Z: but the proper meridian of the plane will be PR, which is a perpendicular let down from the pole unto the declining vertical, and crossing it with right angles in the point R, so the angle RPZ shall show the inclination of the two meridians, and may thus be found. In the triangle PRZ we know the angle at R to be a right angle, and the angle at Z, for it is the compliment of the declination, and the base PZ, for it is the compliment of the latitude. And therefore geometric illustration As the sine of the latitude is to the sine of 90 gr. So the tangent of the declination to the tangent of the inclination required. 2 The height of the style above the plane. This is here represented by the perpendicular ark PR, and may be found by that which we have known in the former triangle PRZ. For As the sine of 90 gr. to the cousin of the latitude: So the cousin of the declination to the sine of the height of the style. Or if you please to make use, of the angle of the inclination of the two meridians, the proportion will hold. As the sine of 90 gr. to the cousin of the inclination of meridians: So the cotangent of the latitude to the tangent of the height of the style. 3 The distance of the substylar from the meridian. This is here represented by the ark ZR, and may be found by that which we have known in the former triangle PRZ. As the sine of 90 gr. to the sine of the declination: So the cotangent of the latitude to the tangent of the substylar from the meridian. 4 The distance of each houre-line from the substylar. The distances of the houre-lines from the substylar, are here represented by those arks of the declining vertical belonging to the plane, which are intercepted between the proper meridian of the plane and the houre-circles. To this purpose we have diverse triangles made by the declining plane, together with his proper meridian and the houre-circles. In these we have known, first the right angle at the intersection of the proper meridian with the plane; then the side which is the height of the pole above the plane; and thirdly the angles at the pole. For knowing the angle of inclination between the meridian of the plane and the meridian of the place, which is always the hour of 12, we may find the angle between the meridian of the plane and the hour of 1, by allowing in 15 gr. and the angle between the meridian of the plane and the hour of 2, by allowing in 30 gr. and so for the rest, which being known, we may find the arks of the plane from the substylar to the houre-circles, in this manner. As the sine of 90 gr. to the sine of the height of the pole above the plane: So the tangent of the hour from the proper meridian, to the tangent of the houre-line from the substylar. Thus in our latitude of 51 gr. 30 m. if the declination of an upright plane shall be found to be 24 gr. 20 m. from the prime vertical, the one face open to the South-west, the other to the Northwest, I may number these 24 gr. 20 m. in the horizon of the fundamental Diagram, from E unto B, according to the situation of the plane, and there draw the vertical BZD, which shall represent the plane proposed. Then taking the compasses into my hand, 1 I may extend them from the sine of the latitude 51 gr. 30 m. unto the sine of 90 gr. the same extent will reach in the line of tangents from 24 gr. 20 m. the declination given, to about 30 gr. and such is ZPR the angle of inclination between the meridian of the place & the meridian of the plane; and therefore the meridian of the plane will here fall upon the circle of the second hour from the meridian of the place, (as it may also appear by opening the compasses to the nearest extent, between the pole and the plane) and there I place the letter R to make this rectangle triangle PRZ. 2 I extend the compasses from the sine of 90 gr. unto the sine of 38 gr. 30 m. the compliment of the latitude, and the same extent will reach from the sine of 65 gr. 40 m. the compliment of the declination, unto the sine of 34 gr. 33 m. Or I may extend them from the sine of 90 gr. unto the sine of 60 gr. the compliment of the inclination of the meridians, and the same extent will reach from the tangent of 38 gr. 30 m. the compliment of the latitude, unto the tangent of 34 gr. 33 m. and such is the ark PR, the height of the pole above the plane. 3 I may extend the compasses from the sine of 90 gr. unto the sine of 24 gr. 20 m. the declination given, and the same extent will reach from the tangent of 38 gr. 30 m. the compliment of the latitude, unto the tangent of 38 gr. 8 m. and such is the ark ZR, the distance of the substylar from the meridian. 4 That I may find the distance of each houre-line from the substylar, I consider the angle of inclination of the meridians RPZ, and there see how that PZ the meridian of the place, which is the hour of 12, being 30 gr. distance from PR the meridian of the plane, and that one face of the plane being open to the South-west, and the other to the North-east, this meridian of the plane falleth to be the same with the hour of 2, (otherwise with the hour of 10:) therefore allowing 15 gr. for an hour, the hour of 1, RPO will be 15 gr. and RFX the hour of 11, 45 gr. distant from PR the proper meridian of the plane: and so I gather the inclination of the rest of the houre-circles toward this meridian, according to their angles at the pole, as in this Table. Then taking my compasses in my hand, I extend them from the sine of 90 gr. unto the sine of 34 gr. 33 m. the height of the pole above the plane, and find them to reach in the line of tangents from 15 gr. the inclination of the hour of 1, to 8 gr. 38 m. for the ark of 1, from the substylar, and from 30 gr. unto 18 gr. 8 m. for the hour of 12, agreeable to the third Prop. and from 45 gr. unto 29 gr. 33 m. for the hour of 11, and so the rest, which I also set down in a Table. Hor. Ang. Po Ar. Pla. Gr. M. Gr. M. 8 90 0 90 0 9 75 0 64 42 10 60 0 44 30 11 45 0 29 33 12 30 0 18 8 1 15 0 8 38 2 Merid substyl 3 15 0 8 38 4 30 0 18 8 5 45 0 29 33 6 60 0 44 30 7 75 0 64 42 8 90 0 90 0 These arks being thus found, will serve for the drawing of the houre-lines, both on the South-west face, and the Northwest face of the plane. For coming to the plane, 1 By the help of a thread and plummet I draw a vertical line, serving both for the meridian of the place and the hour of 12. 2 In this meridian line I make choice of a centre at C, in the upper part of the line, if it be the South face, as here we suppose it, that the style may have room to point downward; but in the lower part of the line, if it be the North face of the plane; for there the style must point upward: and upon this centre I describe an occult circle, representing the declining vertical belonging to the plane. geometric illustration 4 According to the Table of the arkes of the plane from the substylar, I find a chord of 8 gr. 38 m. and inscribe it into this circle, from the substylar toward the meridian, for the hour of 1. In like manner a chord of 29 gr. 23 m. for the hour of 11, and a chord of 44 gr. 30 m. for the hour of 10, and so for the rest of the hours, their halves and quarters. 5 I draw right lines through the centre and the terms of these chords, and these lines so drawn are the houre-lines required. Lastly, I set up the style over the substylar, so as it may cut the plane in the centre, and there make an angle with the substylar of 34 gr. 33 m. according to the height of the pole above the plane; so it shall represent the axis of the world, and be truly placed for casting of the shadow upon the houre-lines in this declining plane. A second example. After the like manner if in our latitude an upright plane shall decline 85 gr. from the prime vertical, the one face of it being open to the Northwest, and the other to the Southeast, we may in some sort represent it by the vertical QZH, and then working as before. 1 The angle ZPT, the inclination of the two meridians will be found to be 86 gr. 5 m. so that PT the meridian of this plane, will here fall between the houre-circles of 6 and 7 from the meridian. 2 The ark PT the height of the pole above the plane will be only 3 gr. 6 m. 3 The ark ZT the distance of the substylar from the meridian 38 gr. 23 m. 4 The Table of the angles at the pole will be also gathered, by comparing the meridian of the plane with the rest of the houre-circles. For the angle TPZ between PT the meridian of the plane, PZ the meridian of the place, and the hour of 12, being 86 gr. 5 m. allowing 15 gr. for an hour, the hour of 11 ½ will be 78 gr. 35 m. and the hour of 11 71 gr. 5 m. distant from the meridian of the plane; & so the rest of the hours, as in the second column of this Table. Hor. Ang. Po Ar. Pla. C F C G Gr. M. Gr. M. In. Par. In. Par. 12 86 5 38 23 91 08 79 21 78 35 15 3 30 92 26 89 11 71 5 9 6 18 42 16 02 63 35 6 13 12 52 10 89 10 56 5 4 36 9 25 8 05 9 41 5 2 42 5 43 4 72 8 26 5 1 31 3 05 2 65 7 11 5 0 36 1 20 1 05 Merid. Substyl 0 0 0 0 6 3 55 0 13 0 44 0 38 5 18 55 1 4 2 15 1 86 4 33 55 2 5 4 18 3 64 3 48 55 3 33 7 13 6 20 2 63 55 6 20 12 77 11 10 71 25 9 10 18 56 16 14 1 78 55 15 28 31 82 27 67 86 25 40 55 99 67 86 68 Then having the height of the pole above the plane, and these angles at the pole; the arkes of the plane, between the substylar and the houre-circles, will be found as in the third column. These arks being found, will serve for the drawing of the houre-lines on either face of this plane. 1 By the help of a thread and plummet I draw ZC a vertical plane, serving both for the meridian of the place and the hour of 12. 2 In this meridian line I make choice of a centre in the upper part of the line, if it had been the Southern face of the plane, but here in C the lower part of the line, because we supposed it to be the Northwest face of the plane, and the style must point upward; and upon this centre I describe an occult circle representing the declining vertical belonging to this plane. geometric illustration 4 The substylar being drawn, I may inscribe the chords of the arks of the plane from the substylar, and draw the houre-lines, and set up the style as in the former plane. Or the arks of the plane from the substylar being found as before, we may draw the houre-lines upon the plane otherwise then by chords. For having drawn the houre-lines as in the last figure, upon paper or paist-boord, we shall find the most part of them, in this and such like planes that have greater declination, to fall so close together, that they can hardly be discerned: wherefore to draw them at large to the best advantage of the plane, I leave out the centre, and draw them by tangents, as in the polar plane. 1 I consider the length and breadth of the plane whereon I am to draw the houre-lines, which I suppose to be a square, whose side is 36 inches, and find that the little square ABDE will contain both the substylar and all those houre-lines which are required in the great square AZCQ. 2 I draw two parallel lines FN, GM, crossing the substylar at right angles in the points F & G, so as they may best cross all the houre-lines, and yet the one be distant from the other as far as the plane will give me leave; and I find by the sight of the figure that if AB the side of the lesser square shall be 36 inches, the line CF will be about 115 inches, and the line CG about 100 inches, and therefore FG 15 inches. Again, that the point F will fall about 6 inches below the upper horizontal side AB, and about 12 inches from the next vertical side BD; for I need not here stand upon parts. 3 Because these two parallel lines are tangent lines in respect of circles drawn upon the semidiameters CF, CG, and such tangents as belong to the arkes of the plane, between the substylar and the houre-lines, the proportion will hold, As the tangent of 45 gr. to the tangent of the ark of the plane: So the length of the semidiameter to the length of the tangent line. As for example, the ark of the plane between the substylar and the hour of 1, is 15 gr. 28 m. in the former Table, the semidiameter CF 115 inches, and the semidiameter CG 100 inches: wherefore I extend the compasses from the tangent of 45 gr. unto the tangent of 15 gr. 28 m. the same extent will reach from 115 in the line of numbers unto 31, 82, which shows the length of the tangent line between F in the substylar and the houre-line of 1, to be 31 inches, 82 cent. or parts of 100 Again, the same extent will reach from 100 unto 27, 67; and such is the length of the lesser tangent from G to the hour of 1. The like reason holds for the length of the other tangents from the substylar to the rest of the hours, as in the Table; as also for the height of the style above these tangent lines; and so the angle of the style above the plane being 3 gr. 6 m. the height FK will be found to be 6 inches 23 cent. and the height GL 5 inches 42 cent. Where the Reader may observe, that if the extent from the tangent of 45 gr. to the tangent of 3 gr. 6 m. or to 115 in the line of numbers, be too large for his compasses, he may use the tangent of 5 gr. 43 m. in stead of the tangent of 45 gr. as I noted before Pag. 100 4 Having found these lengths and heights, and set them down in a Table, I come to the plane here resembled by the lesser square ABDE, where I begin with an occult vertical FH, about 12 inches from the side BD, and upon the centre F, about 6 inches below the side AB describe an occult ark of a circle. 5 Into this ark I first inscribe a chord of 38 gr. 23 m. the distance of the substylar from the meridian, to make the angle HFG equal to the ZCT; so the line FG shall be the substylar: and then another chord of 51 gr. 37 m. the compliment of this distance, to make up the right angle GFN; so the line FN shall be the greater of the two tangent lines before mentioned. 6 I set off 15 inches from F unto G, toward the centre, and through G draw the lesser tangent line GM parallel to the former. 7 These two occult tangent lines being thus drawn, I look unto the former Table for the hour of 1, and there find the ark of the plane between the substylar and the hour of 1, to be 15 gr. 28 m. and the length belonging to it in the greater tangent line to be 31 inches 82 cent. in the lesser tangent line 27 inches 67 cent: wherefore I take out 31 inches 82 parts, and prick them down in the greater tangent from F to N, and then 27 inches 67 parts, and prick them down in the lesser tangent from G to M, and draw the MN for the hour of 1, which if it were produced would cross the substylar FG in the centre C, and there make the angle FCN 15 gr. 28 m. The like reason holdeth for the drawing of all the rest of the houre-lines. Lastly I set up the style right over the substylar, so as the height FK may be 6 inches 23 cent. and the height GL 5 inches 42 cent. then shall KL represent the axis of the world, and if it were produced would cross the substylar FG in the centre C, and there make the angle FCK to be 3 gr. 6 m. and so be truly placed for casting of the shadow upon the houre-lines in this declining plane. CHAP. VIII. To draw the houre-lines in a meridian inclining plane. ALl those planes wherein the horizontal line is the same with the meridian line, are therefore called meridian planes: if they be right to the horizon, they are called by the general name of meridian planes without farther addition, and are described before: if they lean to the horizon, they are then called meridian incliners. These may incline either to the East part of the horizon, or to the West, and each of them hath two faces, the upper toward the zenith, the lower toward the Nadir, wherein knowing the latitude of the place, and the inclination of the plane to the horizon, we are to consider. 1 The inclination of the meridian of the plane to the meridian of the place. 2 The height of the pole above the plane. 3 The distance of the substylar from the meridian. 4 The distance of each houre-line from the substylar. And all these four are represented in the fundamental Diagram, as in this example. In our latitude of 51 gr. 30 m. a meridian plane inclineth Eastward 50 gr; these 50 gr. I number in the vertical circle from E unto G, according to the inclination of the plane, and there draw the ark SGN representing the plane proposed. Then I let down a perpendicular ark PV from the pole to the plane, serving for the meridian of the plane, so have I a rectangle triangle PUN, wherein the base PN is the height of the pole above the North part of the horizon, and the angle PNV the compliment of the inclination to the horizon; and these being known, 1 I may find the angle NPV of inclination of the two meridians. For As the cousin of the latitude is to the sine of 90 gr. So the tangent of inclination to the horizon, to the tangent of inclination of meridians. Extend the compasses from the sine of 38 gr. 30 m. the compliment of the latitude, unto the sine of 90 gr. the same extent will reach from the tangent of 50 gr. 0 m. the inclination of the plane to the horizon, unto the tangent of 62 gr. 25 m. and such is the inclination of the meridian of the plane to the meridian of the place; which being resolved into time, doth give about 4 hours and 10 m. from the meridian, for the place of the substylar among the houre-lines. 2 The height of the pole above the plane is here represented by the quantity of the ark of the proper meridian PV, between the pole and the plane, and may be known by that which we have given in the former triangle PUN. For As the sine of 90 gr. to the sine of the latitude: So the cousin of the inclination to the horizon, to the sine of the height of the pole above the plane. Extend the compasses from the sine of 90 gr. unto 51 gr. 30 m. the sine of the latitude, the same extent will reach from the sine of 40 gr. the compliment of the inclination of the plane to the horizon, unto the sine of 30 gr. 12 m. Or as the sine of 90 gr. to the cousin of inclination of meridians: So the tangent of the latitude to the tangent of the height of the pole above the plane. Extend the compasses from the sine of 90 gr. unto the tangent of 51 gr. 30 m. the latitude of the place, the same extent will reach from the sine of 27 gr. 35 m. the compliment of the inclination of the two meridians, unto the tangent of 30 gr. 12 m. And such is PV the height of the pole above the plane, and such must be the height of the style above the substylar. 3 The distance of the substylar from the meridian is here represented by NV the ark of the plane between the two meridians, and may be found by that which we have given at the first in the former triangle PUN. For As the sine of 90 gr. to the sine of the inclination to the horizon: So the tangent of the latitude to the tangent of the substylar from the meridian. Extend the compasses from the sine of 90 gr. unto the tangent of 51 gr. 30 m. the latitude of the place, the same extent will reach from the sine of 50 gr. the inclination of the plane to the horizon, unto 43 gr. 55 m. And such is the ark NV the distance of the substylar from the meridian. 4 The distances of the houre-lines from the substylar, are here also represented by those arkes of the plane, which are here intercepted between the proper meridian and the houre-circles, and may be found by that which we have given in the triangles made by the plane, with his proper meridian and the houre-circles. For the angle at V, between the plane and the proper meridian, is well known to be a right angle, and the side PV is the height of the pole above the plane, and the angles at the pole between the proper meridian and the houre-circles are easily gathered into a Table. The angle VPN between UP the proper meridian of the plane, and PN the general meridian of the place being 62 gr. 25 m. the angle between the proper meridian and the circle of the hour of 11, will be 77 gr. 25 m. and the angle belonging to the hour of 1, 47 gr. 25 m, and so the rest of the angles at the pole. Then As the sine of 90 gr. to the sine of the pole above the plane: So the tangent of the angle at the pole, to the tangent of the houre-line from the substylar. Hor. Ang. Po. Arc. Pla. Gr. M. Gr. M. 11 77 25 66 4 12 62 25 43 55 1 47 25 28 41 2 32 25 17 43 3 17 25 8 58 4 2 25 1 13 Merid Substyl 5 12 35 6 26 6 27 35 14 44 7 42 35 24 48 8 57 35 38 23 9 72 35 58 3 10 87 35 85 12 Wherefore I extend the compasses from the sine of 90 gr. unto the sine of 30 gr. 12 m. the height of the pole above the plane, and I find the same extent to reach in the line of tangents from 77 gr 25 m. unto 66 gr. 4 m. for the distance belonging to the hour of 11; and from the tangent of 62 gr. 25 m. to 43 gr. 55 m. for the hour of 12. as when I found the distance of the substylar from the meridian. And so for the rest of the arks of plane between the substylar and the houre-circles, as in the Table. These arks being thus found, will serve to draw the houre-lines on either side of this plane: but supposing it to be the upper side, 1 I draw the horizontal line CN, serving for the meridian and hour of 12. 2 In this line I make choice of a centre at C, and thence describe an occult ark of a circle representing the plane proposed. 3 I find a chord of 43 gr. 55 m. the distance of the substylar from the meridian, and inscribe it into this circle from N unto A, according as I find the proper meridian PV to fall in the fundamental diagram, and there I draw the line CA serving for the substylar. geometric illustration 4 The substylar being drawn, I may inscribe the chords of the arkes of the plane from the substylar, and draw the houre-lines, and set up the style, as in the former planes. CHAP. IX. To draw the houre-lines in a polar declining Plane. THose planes wherein a line may be drawn parallel to the axis of the world, are called polar planes, because that line pointeth unto the poles, and these planes are always parallel to some one of the houre-circles. If they be parallel to the hour of 6, they are called direct polar planes; if to the hour of 12, they are called meridian planes; and both these are described before: if to any other of the houre-circles, they are then called by the name of polar declining planes, because of their inclining to the pole, and declining from the vertical. These kind of planes may be known in this sort: First consider the inclination of the plane to the horizon, which in these parts of the world must always be Northward, and more than the latitude of the place. Then find the declination from the vertical. These two being known, if the proportion hold, As the sine of 90 gr. to the cousin of the declination: So the tangent of the inclination to the tangent of the latitude; it is then a polar declining plane, otherwise not. For example, in our latitude of 51 gr. 30 m. a plane is proposed declining from the vertical 65 gr. 40 m. and inclining Northward 71 gr. 51 m. the upper face being open to the Southeast, and the lower to the Northwest. If I number those 65 gr. 40 m. in the horizon of the fundamental diagram from E unto Q, and draw the line HCQ, it shall represent the horizontal line of the plane; then crossing it at right angles with the plane BZD drawn through the zenith, I number 71 gr. 51 m. for the inclination from D unto R, and there draw the circle HRQ, this circle so drawn shall represent the plane proposed; and because it also passeth through the pole, it is therefore a polar plane. But for farther trial I extend the compasses from the sine of 90 gr. to the sine of 24 gr. 20 m. the compliment of the declination, and I find the same extent to reach from the tangent of 71 gr. 51 m. the inclination proposed, unto the tangent of 51 gr. 30 m. which is the true latitude of the place, and therefore it is a polar plane. Here then the style will be parallel to the plane, and the houre-lines parallel one to the other, as in the meridian and direct polar planes. But that we may know how to draw the houre-lines, and where to place the style, we are to consider 1 The ark of the plane between the horizon and the houre-lines. In a meridian plane the ark between the horizon and the hourelines, is always equal to the latitude of the place; in a direct polar it is an ark of 90 gr; in these declining polars it is greater than the latitude, and yet less than 90 gr. and may be known in this manner. As the sine of 90 gr. to the cousin of the latitude: So the sine of the declination to the cousin of the ark between the horizon and the houre-lines. Extend the compasses from the sine of 90 gr. unto the sine of 38 gr. 30 m. the compliment of the latitude, the same extent will reach from the sine of 65 gr. 40 m. the declination proposed, unto the sine of 34 gr. 34 m. whose compliment 55 gr. 26 m. the ark of the plane between the horizon and the substylar, to which all the houre-lines must be parallel. 2 The inclination of the meridian of the plane to the meridian of the place. The substylar in a direct polar plane is always the same with the hour of 12: in a meridian plane it is the same with the houre-line of 6: in these declining polars it must be placed between 12 and 6, according to the inclination of the meridian of the plane to the meridian of the place, which is thus known. As the sine of 90 gr. to the sine of the latitude: So the tangent of the declination of the plane, to the tangent of the inclination of meridians. Extend the compasses from the sine of 90 gr. to the sine of 51 gr. 30 m. the latitude of the place, the same extent will reach from the tangent of 65 gr. 40 m. the declination proposed, unto the tangent of 60 gr. and such is the angle of inclination between the meridian of the place and the proper meridian of the plane, which resolved into time doth make four hours; and so the substylar must here be placed upon the hour of 8 in the morning. This angle being known, the rest of the angles at the pole are easily gathered. For if the hour of 12 be 60 gr. distant from the meridian of the plane, the hour of 1 will be 75 gr. and the hour of 11, will be 45 gr. distant, and the rest of the hours, as in the Table following. Then coming to the plane, 1 I draw an occult horizontal line HQ, wherein I make choice of a centre H, and describe an occult circle for the horizon of the plane. 2 I find a chord of 55 gr. 26 m. and inscribe it into this circle, from Q unto B, according to the situation of the plane; so the line HB shall be the meridian of the plane: and therefore the substylar and the line AC crossing it at right angles, shall be the equator. 3 I consider the length of the plane, and how many hours I am to draw upon it, that so I may proportion the height of the style; and I find by the fundamental diagram and the former table, that it will contain all the hours from Sun rising until 1 afternoon: and therefore the meridian of the plane falling on the hour of 8 in the morning, there will be four hours on the one side, and five on the other side of the substylar. But in all polar planes the height of the style above the substylar must be equal to the distance of the third hour from the substylar, or about 4/7 of the fourth hour, or little more than ¼ of the fift hour, and thereupon I allow the height of this style to be equal to CB, which you may suppose to be ten inches. geometric illustration 4 Because the equator AC is a tangent line in respect of the Radius BC, and the parts thereof are such as belong to the angles between the meridian of the plane and the hourelines, which angles are set down in the table following, I may find the length of each several tangent in this manner. As the tangent of 45 gr. is to the tangent of the hour: So the parts of the Radius, to the parts of the tangent line. The angle ABC between the meridian of the plane and the hour of 12, the meridian of the place is 60 gr. in the former table, and the Radius BC is supposed to be ten inches; whereupon I extend the compasses from the tangent of 45 gr. unto the tangent of 60 gr. the same extent will reach from 10 in the line of numbers, unto 17. 32, which shows the length of the tangent AC between the substylar and the hour of 12, to be 17. 32 cent. The like reason holds for the rest of the hours. Hor. An. Po. Tangent Gr. M. In. Par. 4 60 0 17 32 5 45 0 10 00 6 30 0 5 77 7 15 0 2 68 8 Merid Substyl 3 45 65 7 30 1 32 11 15 1 99 9 15 0 2 68 18 45 3 39 22 30 4 14 26 15 4 93 10 30 0 5 77 33 45 6 68 37 30 7 67 41 15 8 77 11 45 0 10 00 48 45 11 40 52 30 13 03 56 15 14 97 12 60 0 17 32 63 45 20 28 67 30 24 14 71 15 29 46 1 75 0 37 32 78 45 50 27 82 30 75 96 86 15 152 57 2 90 0 Infinite 5 These lengths being thus found and set down in the table, I take out 17 inches 32 cent. and prick them in the equator from C unto A for the hour of 12, and 37 inches 32 cent. and prick them down for the hour of 1. And so the rest of the houre-points. 6 This done, if I draw right lines through each of these points, crossing the equator at right angles, they shall be the houre-lines required: and if I set the style over the substylar, so as the edge of it may be parallel to the plane, and the height of it be ten inches equal to the former Radius BC, it shall represent the axis of the world, and be truly placed for casting of the shadow upon the houre-lines in this declining polar plane. CHAP. X. To draw the houre-lines in a declining inclining plane. IF a plane shall decline from the prime vertical, and incline to the horizon, and yet not lie even with the poles of the world, it is then called a declining inclining plane. Of these there are several sorts; for the inclination being Northward, the plane may fall between the horizon and the pole, as the circle BMD in the fundamental Diagram; or between the zenith and the pole, as BFD: or the inclination may be Southward, and so be represented by BKD, it may also fall either below the intersection of the meridian and the equator, or above it; and each of these have two faces, the upper toward the zenith, and the lower toward the nadir; wherein having the latitude of the place with the declination and inclination of the plane, we are farther to consider, 1 The ark of the meridian between the pole and the plane. 2 The inclination of the plane to the meridian. 3 The ark of the plane between the horizon and the meridian. 4 The angle of inclination between both meridians. 5 The height of the pole above the plane. 6 The distance of the substylar from the meridian. 7 The distances of each houre-line from the substylar. And all these seven may be represented in the fundamental diagram, as in this example. In our latitude of 51 gr. 30 m. a plane is proposed, declining from the vertical 24 gr. 20 m. and inclining Northward 36 gr. the upper face lying open to the South-west, the lower to the North-east. If I number these 24 gr. 20 m. in the horizon from E to B, and there draw the line BCD, it shall represent the horizontal line of the plane: then crossing it at right angles with the plane HZQ drawn through the zenith, I number 36 gr. for the inclination from Q unto M, and there draw the circle BMD, crossing the meridian in the point a; this circle so drawn shall represent the plane proposed; and because it doth not pass through the pole, is therefore no polar, but an ordinary declining inclining plane. 1 The ark of the meridian of the place between the pole and the plane, is here represented by Pa, and may be found by resolving the triangle DN a, wherein the angle at N is known to be a right angle, the angle at D is the angle of inclination, the side D N the compliment of the declination, which being known, As the sine of 90 gr. to the cousin of declination: So the tangent of inclination to the horizon, to the tangent of the meridian between the horizon and the plane. Extend the compasses from the sine of 90 gr. unto the sine of 65 gr. 40 m. the compliment of the declination, the same extent will reach from the tangent of 36 gr. the inclination proposed, unto the tangent of 33 gr. 30 m. and such is the ark of the meridian Na between the horizon and the plane. This ark Na being compared with the ark NP, which is the elevation of the pole above the horizon, and is here supposed to be 51 gr. 30 m. the difference N a cometh to 18 gr. and such is the ark of the meridian required between the pole and the plane. 2 The inclination of the plane to the meridian is here represented by the angle N add, and may be found by that which we have given in the former triangle DN a. For As the sine of 90 gr. to the sine of the declination from the vertical: So the sine of inclination to the horizon, to the cousin of the inclination to the meridian. Extend the compasses from the sine of 90 gr. unto the sine of 24 gr. 20 m. the declination of the plane, the same extent will reach from the sine of 36 gr. the inclination given, unto the cousin of 76 gr. And such is N add the angle of inclination between the plane D a, and N a the meridian of the place. Or As the sine of the ark of the meridian between the horizon and the plane, is to the sine of 90 gr. So the cotangent of the declination to the tangent of the inclination to the meridian. Extend the compasses from the sine of 33 gr. 30 m. the ark of the meridian between the horizon and the plane, unto the sine of 90 gr. the same extent will reach from the tangent of 65 gr. 40 m. the compliment of the declination unto the tangent of 76 gr. And such is the inclination of the plane to the meridian, the same as before. 3 The ark of the plane between the horizon and the meridian, is here represented by D a, and may also be found by that which we have given in the former triangle DN a. As the sine of 90 gr. to the cousin of inclination to the horizon: So the cotangent of the declination to the tangent of the plane required. Extend the compasses from the sine of 90 gr. unto the sine of 54 gr. the compliment of the inclination of the plane to the horizon, the same extent will reach from the tangent of 65 gr. 40 m. the compliment of the declination, unto the tangent of 69 gr. 54 m. And such is D a the ark of the plane, between the horizon and the meridian of the place. 4 The inclination of meridians is here represented by the angle aPb. For if I let down a perpendicular Pb from the pole unto the plane, this perpendicular shall be the meridian of the plane; and we shall have another triangle abP, wherein the angle at b is a right angle, because of the perpendicular, the angle at a is the inclination of the plane to the meridian of the place, and the side P a is the ark of the meridian between the pole and the plane, which being known, As the cousin of the ark of the meridian between the pole and the plane is to the sine of 90 gr. So the cotangent of the inclination to the meridian, to the tangent of inclination of the meridian of the plane to the meridian of the place. Extend the compasses from the sine of 72 gr. the compliment of the ark Pa, between the pole and the plane, unto the sine of 90 gr. the same extent will reach from the tangent of 14 gr. the compliment of the inclination of the plane to the meridian, unto the tangent of 14 gr. 41 m. And such is the angle aPb of inclination between the meridian of the place and the proper meridian of the plane, which resolved into time, doth make about 59 minutes, and so the substylar must here be placed near the hour of 1 after noon. 5 The height of the pole above the plane is here represented by Pb, the ark of the proper meridian between the pole and the plane, and may be found by that which we have given in the triangle abP. For As the sine of 90 gr. to the sine of the meridian of the place between the pole and the plane: So the sine of the inclination to the meridian, to the sine of the height of the pole above the plane. Extend the compasses from the sine of 90 gr. unto the sine of 18 gr. the ark Pa of the meridian of the place from the pole to the plane, the same extent will reach from the sine of baP the inclination of the plane to the meridian of the place, unto the sine of 17 gr. 26 m. Or As the sine of 90 gr. to the cousin of inclination of meridians: So the tangent of the meridian of the place between the pole and the plane, to the tangent of the height of the pole above the plane, Extend the compasses from the sine of 90 gr. unto the sine of 75 gr. 19 m. the compliment of aPb the inclination of the two meridians, the same extent will reach from the tangent of 18 gr. the ark Pa of the general meridian between the pole and the plane, unto the tangent of 17 gr. 26 m. And such is Pb the height of the pole above the plane; and such must be the height of the style above the substylar. 6 This distance of the substylar from the meridian of the place, is here represented by ab the ark of the plane between the two meridians, and may be found by that which we had given at the first in the former triangle abP. For As the sine of 90 gr. to the sine of the inclination to the meridian: So the tangent of the meridian of the place between the pole and the plane, unto the tangent of the substylar from the meridian of the place. Extend the compasses from the sine of 90 gr. unto the sine of 14 gr. the compliment of baP, the inclination of the plane to the meridian, the same extent will reach from the tangent of 18 gr. the ark of the general meridian between the pole and the plane, unto the tangent of 4 gr. 30 m. And such is the ark of the plane between the two meridians; and such must be the distance from the hour of 12 to the substylar. Hor. Ang. Po. Arc. Pla. Gr. M. Gr. M. 7 89 41 88 57 8 74 41 47 35 9 59 41 27 9 10 44 41 16 31 11 29 41 9 41 12 14 41 4 30 M●rid. Substyl 1 0 19 0 6 2 15 19 4 42 3 30 19 9 56 4 45 19 16 52 5 60 19 27 45 6 75 19 48 51 7 The distances of the houre-lines from the substylar, are here also represented by those arks of the plane, which are intercepted between the proper meridian and the houre-circles. For in these triangles the angle at b between the plane and the proper meridian is a right angle, the side P b is the height of the pole above the plane, and then the angles at the pole between the proper meridian and the houre-circles being gathered into a table. As the sine of 90 gr. to the sine of the pole above the plane: So the tangent of the angle at the pole, to the tangent of the houre-line from the substylar. geometric illustration These arkes being thus found, will serve for the drawing of the houre-lines on either side of the plane: but supposing it to be the upper side, I consider how the lines do fall in the fundamental diagram, and accordingly 1 I draw an occult horizontal line DD, wherein I make choice of the centre C, and thence draw an occult circle for the horizon of the plane. 2 I find a chord of 69 gr. 54 m. the ark of the plane between the horizon and the meridian, and inscribe it into this circle from D unto a, and there draw the line Ca for the hour of 12. 3 I find a chord of 4 gr. 30 m. the ark of the plane between the two meridians, and inscribe it into this circle from a unto b, and there draw the line Cb for the substylar. 4 The substylar being drawn, I may inscribe the chords of the arkes of the plane from the substylar, and draw the houre-lines, and set up the style as in the former planes. A second example of a Plane falling between the pole and the zenith. In like manner if in our latitude a plane be proposed declining from the vertical 24 gr. 20 m. as before, but inclining to the horizon 75 gr. 40 m. Northward, the upper face being open to the South-west, the lower to the North-east, this plane shall be here represented by the circle BFD, crossing the meridian in the point d, between the pole and the zenith, and the proper meridian of this plane, by the perpendicular ark P e. Then in this triangle DN d knowing the side DN the compliment of the declination, with the angle of inclinanation to the horizon at D, and the right angle at N, these former Canons will give Nd the ark of the meridian between the horizon and the plane to be 74 gr. 20 m; and therefore Pd the ark of the meridian between the pole and the plane will be 22 gr. 50 m. the angle DdN of the inclination of the plane to the meridian, will be found to be 66 gr. 29 m. and Dd the ark of the plane between the horizon and the meridian 83 gr. 36 m. Again, in the triangle Ped knowing the side Pd the ark of the meridian between the pole and the plane, with the angle of inclination to the meridian at d, and the right at e, the angle dPe of the inclination of the two meridians will be found to be 25 gr. 17 m. and Pe the height of the pole above the plane to be 20 gr. 50 m. and the the distance of the substylar from the meridian about 9 gr. 32 m. Declination 24 20 Inclination 75 40 Diff. meri. 83 36 dist. substy 9 32 Alti. Still. 20 50 Hor. Ang. Po. Arc. Pla. Gr. M. Gr. M. 8 85 17 76 56 9 70 17 44 47 10 55 17 27 11 11 40 17 16 43 12 25 17 9 32 1 10 17 3 41 Merid Substyl 2 4 43 1 40 3 19 43 7 16 4 34 43 13 50 5 49 43 22 46 6 64 43 37 0 7 79 43 62 58 Lastly, having found the height of the pole above the plane, and gathered the angles at the pole, the arks of the plane from the substylar to the houre-lines will be as in this table. This done, if we consider how the lines do fall in the fundamental diagram, we may there see how the North pole is elevated above the lower face, & the South pole above the upper face of the plane, and accordingly make choice of a centre, draw the horizontal, the meridian, the substylar, & the houre-lines, and set up the style as in the other planes. A third example of a Plane inclining to the Southward. If in our latitude a plane were proposed declining from the vertical 24 gr. 20 m. as before, but inclining to the horizon 14 gr. 20 m. Southward, the upper face being open to the North-east, the lower to the South-west, this plane shall be here represented by the circle BKD crossing the meridian in the point f between the equator and the horizon, and the proper meridian of this plane by the perpendicular ark Pg let down from the pole to the plane, near the hour of 11, at the North part of the horizon, as may partly appear by the nearest extent of the compasses. Then in the triangle BSf, knowing the side BS the compliment of the declination, with the angle of inclination to the horizon at B, and the right angle at S, we may find Sf the ark of the meridian between the horizon and the plane to be 13 gr. 6 m. And therefore Pf the ark of the meridian between the pole and the plane to the Southward 115 gr. 24 m. but 64 gr. 36 m. to the Northward, the angle B fS or D fN of the inclination of the plane to the meridian, will be found 84 gr. 9 m; & B f or D f the ark of the plane between the horizon & the meridian 66 gr. 20 m. Declination 24 20 Inclination 14 20 diff. merid 13 27 dist. substy 12 8 Alt. Still. 64 0 Hor. Ang. Po Arc. Pla. Gr. M. Gr. M. 6 76 33 75 6 7 61 33 58 56 8 46 33 43 30 9 31 33 28 55 10 16 33 14 58 11 1 33 1 25 Merid Substyl 12 13 27 12 8 1 28 27 25 57 2 43 27 40 23 3 58 27 55 38 4 73 27 71 41 5 88 27 88 15 Again, in the triangle Pgf knowing the side Pf the ark of the meridian between the pole and the plane, with the angle of inclination to the meridian at f, and the right angle at g, the angle fPg of the inclination of the two meridians will be found to be 13 gr. 27 m. and Pg the height of the pole above the plane, about 64 gr. and fg the distance of the substylar from the meridian 12 gr. 8 m. Having found the height of the pole above the plane, and gathered the angles at the pole, the arkes of the plane from the substylar to the houre-lines will be found as in this table. This done, if we consider how the lines do fall in the fundamental diagram, we may there see how the North pole is elevated above the upper face, and the South pole above the lower face of this plane, and accordingly make choice of the centre, draw the horizontal, the meridian, the substylar, and the houre-lines, and set up the style as in the former planes. CHAP. XI. To describe the Tropiques and other circles of declination in an equinoctial Plane. Such circles as are parallel to the equinoctial, and yet fall within the tropiques, may be described on any plane by help of these lines of proportion, but after a different manner, according as the style shall be either perpendicular, or parallel to the plane, or cut the plane with obliqne angles. In an equinoctial plane where the style is perpendicular to the plane, the tropiques and other circles of declination will be perfect circles: wherefore consider the length of the style in inches and parts, and the declination of the circle which you intent to describe in degrees and minutes, the proportion will hold, As the tangent of 45 gr. to the length of the style: So the cotangent of the parallel, to the semidiameter of his circle. geometric illustration Or if it were required to proportion the style to the plane, As the tangent of 45 gr. to the tangent of the declination: So the semidiameter of the plane, to the length of the style. As if the semidiameter of the greatest parallel upon the plane were but six inches, and that parallel should be the fift degree of declination: extend the compasses from the tangent of 45 gr. unto the tangent of 5 gr. the same extent will reach in the line of numbers from 6.00 unto about 0.53, which shows that the length of the style must be 53 parts of an inch divided into 100; then the length of the style being known, the semidiameter of the other circles will be found as before. I begin here with the fift parallel, and thence proceed unto the tropic, because the shadow of the rest near the equinoctial, would be overlong, and the equinoctial itself cannot be described. The parallels of North declination are to be set on the North face, and the parallels of South declination on the South face of the plane. Neither need these parallels to be drawn in full circles, but only to the horizontal line, which shall be described in Cap. xviij. Having by these means set up the style to his true height, and drawn the circles of declination, if we shall place the plane so as it shall make an angle with the horizon equal to the compliment of the latitude, and then turn it until the top of the style cast the shadow upon the parallel of declination belonging to the time, the meridian of the plane will show the meridian of the place, and the shadow of the style the hour of the day, without the help of a magnetical needle. CHAP. XII. To describe the Tropiques and other circles of declination in a polar plane. IN all polar planes, whether the be parallel to the meridian or to the circle of the hour of 6, or otherwise declining, the equinoctial will be a right line, but the tropiques and other circles of declination will be sections hyperbolical, and be thus described. Consider the length of the style, the declination of the parallel, and the angle at the pole between the substylar and the houre-line, whereon you mean to describe the parallel. If you would find where the parallels do cross the substylar; As the tangent of 45 gr. to the tangent of declination: So is the length of the style, to the distance of the parallel from the equinoctial. geometric illustration As in the example of the polar plane, where the length of the style BC was found to be 1 inch 61 cent. if you desire to know the distance between the equinoctial and the tropic upon the substylar line: extend the compasses from the tangent of 45 gr. unto the tangent of 23 gr. 30 m. the same extent will reach in the line of numbers from 1.61 unto 0.70; and therefore the distance required is 70 parts of an inch divided into 100 The like reason holdeth for all other parallels of declination crossing the substylar. But if you would find where the parallels do cross any other of the houre-lines, first find the distance between the axis of the style and the houre-line, than the distance between the equinoctial and the parallel, both these may be represented in this manner. On the centre B and any semidiameter BD describe an occult ark of a circle, and therein describe a chord of 23 gr. 30 m. from D unto T, with such other intermediate declinations as you intent to describe on the plane, so the line BD shall be the equator, and BT the tropic, and the other intermediate lines the lines of declination. geometric illustration That done, consider your plane, which for example may be either the meridian or the declining polar plane, wherein having drawn both the equator, and the houre-lines as before, first take out the height of the style, and prick that down in this equator from B unto C; then take out all the distances, between B the top of the style and the several points wherein the houre-lines do cross the equator, transfer them into this equator BD from the centre B, and at the terms of these distances erect lines perpendicular to the equator, crossing the lines of declination, and note them with the number of the hour from whence they were taken: so these perpendiculars shall represent those houre-lines, and the several distances between the equator and the lines of declination, shall give the like distances between the equator and the parallels of declination upon your plane. Upon this ground it followeth, To find the distance between the axis and the houre-lines. As the cousin of the hour from the substylar, is to the sine of 90 gr. So the length of the style, to the distance between the axis and the houre-line. geometric illustration As if in the former example of the meridian plane, where BC the height of the style is supposed to be 10 inches, it were required to find the distance between B the top of the style and the point wherein the hour of 11 in the morning doth cross the equator, which is here represented by B 5, because it is the fift hour from the substylar, whose angle at the pole is 75 gr. Extend the compasses from the sine of 15 gr. the compliment of the fift hour from the substylar, unto the sine of 90 gr. the same extent will reach from 10.00 in the line of numbers unto 38.64; and therefore the distance B 5 between the axis and the houre-line, is 38 inches & 64 cent. and may be called the secant of the hour. Then in the rectangle B 5 T, having the side B 5, and the angle of declination at B. To find the distance between the equinoctial and the parallel. As the tangent of 45 gr. to the tangent of the declination: So the distance between the axis and the houre-line, to the distance between the equinoctial and the parallel. Extend the compasses from the tangent of 45 gr. unto the tangent of 23 gr. 30 m. the declination of the tropic, so the same extent will reach in the line of numbers from 38.64, the distance between the axis and the fift houre-line unto 16.80; and therefore the distance is 16 inches and 80 cent. The like reason holdeth for all the rest, which may be gathered and set down in such a Table as this which followeth. Wherein I have set down these distances for several declinations, for 11 gr. 30 m. for 16 gr. 55 m. for 20 gr 12 m. for 21 gr 41 m. and for the declination of the Tropic 23 gr. 30 m. which may be applied to the like declinations in all meridian and direct polar planes. As in the former example of the polar plane, where BC the height of the style is found to be 1 inch 61 cent. if it were required to find the distance between B the top of the style and the points wherein the houre-lines of 7 in the morning or 5 after noon, do cross the equator (which distances, I called the secants of those hours.) either you may extend the compasses from the sine of 15 gr. the compliment of the hour from the substylar unto the sine of 90 gr. so the same Hor. Ang. Po Tangent Secant. 11 30 16 55 20 12 21 41 23 30 Gr. M. In. Pat. In. Pa. In. Pa. In. Pa. In. Pa In. Pa. In. Pa 0 0 0 0 0 10 0 2 3 3 4 3 68 3 98 4 35 3 45 0 65 10 02 2 04 3 05 3 69 3 99 4 36 7 30 1 32 10 09 2 05 3 07 3 71 4 01 4 39 11 15 1 99 10 20 2 07 3 10 3 75 4 05 4 43 1 15 0 2 68 10 35 2 10 3 15 3 81 4 12 4 50 18 45 3 39 10 56 2 15 3 21 3 89 4 20 4 59 22 30 4 14 10 82 2 20 3 29 3 99 4 30 4 70 26 15 4 93 11 15 2 26 3 39 4 10 4 43 4 85 2 30 0 5 77 11 55 2 34 3 51 4 24 4 60 5 02 33 45 6 68 12 03 2 44 3 66 4 42 4 78 5 23 37 30 7 67 12 60 2 56 3 83 4 64 5 02 5 48 41 15 8 77 13 30 2 70 4 05 4 89 5 29 5 78 3 45 0 10 00 14 14 2 87 4 30 5 20 5 63 6 15 48 45 11 40 15 17 3 08 4 62 5 58 6 03 6 00 52 30 13 03 16 43 3 34 5 00 6 04 6 54 7 14 56 15 14 97 18 00 3 66 5 48 6 62 7 00 7 83 4 60 0 17 32 20 00 4 07 6 08 7 36 7 95 8 70 63 45 20 28 22 61 4 60 6 88 8 32 9 00 9 83 67 30 24 14 26 13 5 31 7 95 9 61 10 39 11 36 71 15 29 46 31 11 6 33 9 47 11 45 12 37 13 53 5 75 0 37 32 38 64 7 86 11 74 14 20 15 36 16 80 78 45 50 27 51 26 10 43 15 60 18 89 20 38 22 28 82 30 75 96 76 61 15 58 23 32 28 19 30 47 33 31 86 15 152 57 152 90 31 10 46 54 56 26 60 81 66 48 6 90 0 Infin. Infin. Infin. Infin. Infin. Infin. Infin. extent will reach in the line of numbers from 1.61 the length of the style, unto 6.21, according to the former Canon. Or else you may make use of the former Table, extending the compasses in the line of numbers from 10.00 the length of the style in the Table, unto 1.61 the length of the style belonging to your plane, so the same extent shall reach from from 38.64 the secant in the Table, unto 6.21, and such is your secant required, the distance between the top of the style and the point of intersection, wherein the fift hour line from the substylar doth cross the equator. Again, the same extent will reach from 16.80 the distance in the Table belonging to the fift houre-line between the equator and the parallel of 23 gr. 30 m. declination, unto 2.70 for the like distance upon your plane; and so for the rest, which may be gathered and set down in a Table. Hor. An. Po Tangle Secant Trop. Gr. M. In. P. In. P. In P. 12 0 0 0 0 1 61 0 70 11 1 15 0 0 43 1 63 0 72 10 2 30 0 0 93 1 85 0 80 9 3 45 0 1 61 2 27 0 99 8 4 60 0 2 79 3 22 1 40 7 5 75 0 6 00 6 21 2 70 That done, and the equator drawn as before, if you would draw the tropiques in the polar plane, look into the Table, and take 70 cent. out of the line of inches, and prick them down in the substylar on either side of the equator, and so 72 cent. on the first hour, and 80 on the second hour, and 2 inches 70 cent. on the fift hour from the substylar, and the rest of these distances on their several houre-lines, and then draw a crooked line through all these points, so as it makes no angles, the line so drawn shall be the Tropic required. In like manner you may draw any other parallel of declination. CHAP. XIII. To describe the Tropiques and other circles of declination in such a Plane as is neither equinoctial nor polar. IN Planes neither equinoctial nor polar, the equator will be a right line, the tropiques and other parallels of declination will be conical sections, some of them parabolical, some elliptical, but the most of them hyperbolical. To find the points of intersection of these parallels with the houre-lines, we are to consider, first the length of the axis of the style in inches and parts of inches; secondly the height of the style above the plan●; ●hirdly the angles at the pole between the proper meridian and the houre-circles. These being known, will help us to find, first the angle between the axis and the houre-lines on the plane; and then the distance between the centre and the parallels: both these may be represented in this manner. geometric illustration Let the triangle ABC be made equal to the style belonging to your plane, AC the substylar, BC the axis of the style, AB the length of the style perpendicular to the plane. Then having drawn the line BD perpendicular to the axis on the centre B, & any semidiameter BD describe an occult ark of a circle, and therein inscribe a chord of 23 gr. 30 m. from D unto T, on either side of the line, with such other intermediate declinations as you intent to describe on the plane, so the perpendicular BD shall be the equator, and BT the tropiques, and the other intermediate lines the parallels of declination. Wherefore you may take out the distance C ♈ from the centre to the equator, and prick it down on the substylar of your plane from the centre at C unto ♈, so the line drawn through ♈ perpendicular to your substylar, shall be the equator of your plane. That done, take the distance of each houre-line between the centre and the equator of your plane, and prick them down in the equator of this figure, from the centre at C, noting the place, where they cross the equator, with the number belonging to the hour, and drawing the houre-lines from C through the lines of declination. Or having the Sector you may draw an occult line CE perpendicular to the axis BC, and therein prick down the tangent of the height of the style above the plane, from C unto E. Then draw the line OF parallel to the axis, crossing the substylar produced in the point F, this line OF will be the line of sins upon the Sector, and therein you may prick down the sins of the compliment of the angles at the pole from E toward F, and draw the houre-lines by those points through the lines of declination, so the angles at C between the axis BC and those houre-lines, shall be the angles between the axis of your style and the houre-lines on your plane, and the several distances between the point C and the lines of declination, shall give you the like distances between the centre, and the parallels of declination upon the houre-lines in your plane. Upon this ground it followeth, 1 To proportion the style unto the plane. Consider the height of the style above the plane, and the length of the substylar between the centre and the place which you intent for the tropic. If it be the tropic which is farthest from the centre, add 113 gr. 30 m: if the nearer tropic, add 66 gr. 30 m. unto the height of the style, the remainder unto 180 gr. shall give you the altitude of the Sun above the plane when he cometh to that tropic. As in our latitude the height of the style above an horizontal plane is 51 gr. 30 m. add unto this 113 gr. 30 m. the sum is 165 gr. which being taken out of 180 gr. the remainder will be 15 gr. and such is the altitude of the Sun above this plane when he cometh to be in the Winter tropic: but if you add 66 gr. 30 m. unto 51 gr. 30 m. the remainder to 180 gr. will be 62 gr. And such is the altitude of the Sun in the Summer Tropic. Then As the sine of 66 gr. 30 m. to the sine of the Sun's altitude: So the length of the substylar line, to the length of the axis of the style. geometric illustration As in the first example of the declining vertical, where the height of the style was found to be 34 gr. 33 m. and is here represented before pag. 150. by the angle BC ♋; add to this height 113 gr. 30 m. for the angle CB ♋, the sum will be 148 gr. 3 m. and the remainder to 180 gr. will be 31 gr. 57 m. and such is the angle B ♋ C of the altitude of the Sun above the plane, when he cometh to be in the tropic of ♋, which is here the farthest tropic from the centre. Then supposing the length of the substylar line between the centre and the place which is fit for the farthest tropic to be about 21 inches, extend the compasses from the sine of 66 gr. 30 m. unto the sine of 31 gr. 57 m. the same extent will reach in the line of numbers from 21 unto 12.11, and so the length of the axis of the style should be 12 inch. 11 cent. Or it may suffice to make it just 12 inches, as a more easy ground for the rest of the work. But if it were required to proportion the style unto the plane, so as it may cast the shadow to the full length of the substylar line at all times of the year, you may then consider the Sun in the tropic, which is to be set nearest unto the centre, and add 66 gr. 30 m. unto 34 gr. 33 m. so the remainder unto 180 gr. will be 78 gr. 57 m. And if you extend the compasses from the sine of 66 gr. 30 m. unto the sine of 78 gr. 57 m. the same extent will reach in the line of numbers from 21 unto 22.47 for the length of the axis of the style. 2 Having the length of the axis, and the height of the style above the plane, to find the length of the sides of the style. The style of a plane neither equinoctial nor polar, may be either a small rod of iron set parallel to the axis of the world, or perpendicular to the plane, or else a thin plate of iron or brass made in form of a rectangle triangle BAC, with the base BC parallel to the axis of the world, the side AB perpendicular to the plane, & the side AC the same with the substylar line, wherein knowing BC, and the angle BAC, As the sine of 90 gr. to the length of the axis: So the sine of the height of the style, to the length of the perpendicular side: And so the cousin of the height of the style, to the length of the substylar side. Thus in the former example, the length of the axis being supposed to be 12 inches, and the height of the style 34 gr. 33 m. Extend the compasses from the sine of 90 gr. (or else from the sine of 5 gr. 45 m.) unto 12 in the line of numbers, the same extent will reach from the sine of 34 gr. 33 m. unto 6.80 in the line of numbers for the length of the perpendicular side, and from the sine of 55 gr. 27 m. unto 9.88 for the length of the substylar side. 3 To find the distance between the centre and the equator upon the substylar line. This is here represented by C ♈, and may be found by resolving the rectangle triangle CB ♈. As the cousin of the height of the style, is to the sine of 90 gr. So the length of the axis, to the distance of the equator from the centre. Extend the compasses from the sine of 55 gr. 27 m. unto the sine of 90 gr. the same extent will reach in the line of numbers from 12 unto 14.57. Wherefore if you take 14 inch. 57 cent. and pricking them down on your substylar line from C unto ♈, draw a line through ♈, crossing the substylar at right angles, the line so drawn shall be the equator. 4 To find the angles contained between the equator and the houre-lines upon your plane. These angles made by B ♈ and the houre-lines, are compliments of those which are at C, between BC the axis and those several houre-lines, and depend upon the angles at the pole, between the proper meridian and the houre-circles. As the sine of 90 gr. to the cousin of the angle at the pole: So the cotangent of the height of the style, to the tangent of the angle between the equator and the houreline. In our example the height of the style is 34 gr. 33 m. and the proper meridian falleth to be the same with the circle of the second hour after noon, whereupon the angle at the pole, between this proper meridian, and the circles of the hour of 1 on the one side, and 3 on the other side, will be 15 gr; so between this meridian and the houre-circles of 12 and 4, the angle will be 30 gr. etc. as in the Table. Ho. An. Po Arc. Pla. An. Equ C ♈ C ♋ C ♑ Gr. M. Gr. M. Gr. M. In. P. In. P. In. P. substy 0 0 0 0 55 27 14 57 20 80 11 21 1 3 15 0 8 38 54 30 14 74 21 36 11 25 12 4 30 0 18 8 51 30 15 33 23 44 11 40 11 5 45 0 29 33 45 45 16 75 29 06 11 76 10 6 60 0 44 30 36 0 20 00 50 84 12 77 9 7 75 0 64 42 20 36 34 10 Infin. 15 82 8 8 90 0 90 0 0 0 Infinite 27 60 If then it be required to find the angle, which the houre-line of 4 after noon doth make with the plane of the equator, that is the angle C 4 B contained between the houre-line C 4 and the line B 4, drawn from the top of the style unto the intersection of the houre-line of 4 with the equator. Extend the compasses from the sine of 90 gr. unto the sine of 60 gr. the compliment of the angle at the pole, the same extent will reach from the tangent of 55 gr. 27 m. the compliment of the height of the pole, unto the tangent of 51 gr. 30 m. and such is the angle C 4 B in the diagram Pag. 150. Or in crosse-worke, if it were required to find the angle C 9 B, look into the Table for the hour of 9, and there you shall find the angle at the pole. to be 75 gr; and if you extend the compasses from the sine of 90 gr. unto the tangent of 55 gr. 27 m. the same extent will reach from the sine of 15 gr. the compliment of 75 gr. unto the tangent of 20 gr. 36 m. and such is the angle C 9 B, made at the equator between the line B 9 drawn from the top of the style, and the houre-line C 9 drawn from the centre. The like reason holdeth for the rest, which may be found and set down in a table: then may you either draw these angles at C in the former figure more perfectly, and thence finish your work, or else proceed 5 To find the distance between the centre and the parallels of declination. The distances between the centre and the parallels of declination, may be found by resolving the triangles made by the axis BC, the lines of declination, and the houre-lines. For having the angles at the equator, and knowing the declination of the parallel, if the parallel shall fall between the equator and the centre, add the declination unto the angle at the equator; or if it shall fall without the equator, take the declination out of the angle at the equator, so shall you have the angle at the parallel. Then As the sine of the angle at the parallel, to the cousin of the declination: So the length of the axis of the style, to the distance between the centre and the parallel. Thus in our example, the angle at the equator belonging to the hour of 4 after noon, was found before to be 51 gr. 30 m: if you would find the distance between the centre and the equator, extend the compasses from the sine of 51 gr. 30 m. unto the sine of 90 gr. the compliment of the declination, the same extent will reach in the line of numbers, from 12 unto 15.33, and such is the distance upon the houre-line of 4 between the centre and the equator. If you would find the distance upon this houre-line, between the centre and the inner tropic, whose declination is known to be 23 gr. 30 m. add the declination to the angle at the equator, so the angle at the parallel will be 75 gr. wherefore extend the compasses from the sine of 75 gr. unto the sine of 66 gr. 30 m. the compliment of the declination, the same extent will reach in the line of numbers, from 12 unto 11.40, and such is the length of the houre-line of 4 between the centre and the tropic of ♑. If you would find the distance upon this houre-line between this centre and the tropic of ♋, which is here the farthest from the centre, take the declination out of the angle at the equator, so the angle at the parallel will be 28 gr. unto wherefore extend the compasses from the sine of 28 gr. unto the sine of 66 gr. 30 m. the same extent will reach in the line of numbers, from 12 unto 23.44, and such is the distance between the centre and the tropic of ♋ upon this houre-line of 4. The like reason holdeth for all the rest, which may be gathered and set down in a table. That done and the equator drawn as before, if you would draw the tropic of ♋, look into the table, and there finding under the title C ♋ the distance of the substylar between the centre and the parallel of ♋ to be 20 inch. 80 cent. take 20 inch. 80 cent. out of the line of inches, and prick them down in the substylar of your plane from C unto ♋. Or if either the centre fall without your plane, or the extent be too large for your compasses, you may prick down the difference between C ♈ and C ♋. As here the distance C ♈ between the centre & the equator is 14.57, the distance C ♋ 20.80, the difference 6.23, therefore taking 6 inch. 23 cent. prick them down on the substylar from ♈ unto ♋, and you shall have the same intersection of the tropic and the substylar, as before; & the like reason holdeth for pricking down of the rest of these distances on their several houre-lines. Then having the points of intersection between the houre-lines and the parallel, you may join them all in a crooked line without making of any angles, the line so drawn shall be the tropic required. And after this manner may you draw any other parallel of declination, whereof you have examples in the most of the former Diagrams. CHAP. XIIII. To describe the parallels of the Signs in any of the former Planes. THe equator and the tropiques before described, do show the Sun's entrance into 4 of the Signs, the equator into ♈ and ♎, the one tropic into ♋, and the other into ♑, the rest of the intermediate Signs will be described in the same manner as the tropiques, if first we know their declination. The manner of finding the declination not only of the beginning of the Signs, but of all other points of the ecliptic, is before set down in 2. Prop. Astronomical, pag. 52. by which you may find the declination of the beginning of ♉, ♍, ♏, and ♓ to be 11 gr. 30 m. and of ♊, ♌, ♐ and ♒ to be 20 gr. 12 m. If then you inscribe the chords of 11 gr. 30 m. and of 20 gr. 12 m. into the former figure BDT Pag. 145. from D toward T, the lines drawn from B through the terms of those chords shall be the Signs required. And with these declinations, the height of the style, and the length of the axis, you may find the angles at the parallel, and then the distances between the centre and the parallel, which being pricked down upon their several houre-lines shall give you the points of intersection, by which you may draw the parallels of the Signs, as in the figures belonging to the polar planes. CHAP. XV. To describe the parallels of the length of the day in any of the former Planes. THe length of the day will always be 12 hours long when the Sun cometh to be in the equator, and this holdeth in all latitudes; but at other times of the year the same place of the Sun, will not give the same length of the day in another latitude; wherefore the latitude being known, we are first To find the declination of the Sun agreeing to the length of the day. Consider the difference between the length of an equinoctial day and the day proposed, and turn the time into degrees and minutes. As the sine of 90 gr. is to the sine of half the difference: So the cotangent of the latitude, to the tangent of the declination. geometric illustration If then you inscribe the chords of these arks into the former figure BDT, the lines drawn from B through the terms of these arks, shall be the lines belonging to the diurnal arkes, and the several distances between them and the point C give the like distances between the centre and the parallels of the length of the day upon the houre-lines in your plane. geometric illustration CHAP. XVI. To draw the old unequal hours in the former Planes. IT was the manner of the Ancients to divide the day into twelve equal hours, and the night into twelve other equal hours, and so the whole day and night into 24 hours. Of these 24, those which belonged unto the day, were either longer or shorter (excepting the two equinoctial days) than those which belonged unto the night; and the Summer hours always longer than the hours in the Winter, according to the lengthening of the days, whereupon they are called the old unequal (and by some the Planetary) hours. geometric illustration To express these in the former Planes: first draw the common houre-lines, the equator, and the tropiques, as before: then describe two occult parallels of the length of the day, one for 9 hours, the other for 15 hours; for so you may draw a straight line for the first unequal hour through 5 ho. 45 m. in the parallel of 15, and through 8 ho. 15 m. in the parallel of 9 This straight line shall pass directly through 7 ho. 0 m. in the equator, and so cut off a twelfth part of the arks above the horizon, both from these two parallels and the equator: and being continued unto the tropiques, it shall also cut off about a twelfth part from them, and all the rest of the parallels of declination, without any sensible error. In like manner may you draw the second unequal hour through 7 ho. in the parallel of 15, through 8 ho. in the equator, and through 9 ho. in the parallel of 9, and so in the rest, as in this Table. Horae 15 Aeq 9 Ho. M. Ho Ho. M. 0 4 30 6 7 30 1 5 45 7 8 15 2 7 0 8 9 0 3 8 15 9 9 45 4 9 30 10 10 30 5 10 45 11 11 15 6 12 0 12 12 0 7 1 15 1 0 45 8 2 30 2 1 30 9 3 45 3 2 15 10 5 0 4 3 0 11 6 15 5 3 45 12 7 30 6 4 30 And of these unequal hours you have a farther example in the diagram belonging to the polar declining plane, Pag. 130. CHAP. XVII. To draw the hours from Sun rising and sun setting in the former Planes. TO know how many hours are passed since the Sun rising, or how many remain to the Sun setting; first draw the common houre-lines, the equator, and the tropiques, as before: then describe two occult parallels of the length of the day, one for 8 hours, and the other for 16 hours. For so you geometric illustration may draw the first hour from the Sun rising through the common hours of 5 in the parallel of 16, of 7 in the equator, and of 9 in the parallel of 8. In like manner the second hour from Sun rising through the common hours of 6 in the parallel of 16, of 8 in the equator, and of 10 in the parallel of 8. And so the rest in their order. The first hour before Sun setting, or the 23 hour from the last Sun setting, may be drawn in like sort through the common hours of 3 after noon in the parallel of 8, of 5 in the equator, and of 7 in the parallel of 16. The second hour before Sun setting, or the 22 hour after the last Sun setting through the common hours of 2 in the parallel of 8, of 4 in the equator, and of 6 in the parallel of 16. And so the rest in the like order, whereof you have another example in the Diagram belonging to the declining vertical, Pag. 116. CHAP. XVIII. To draw the horizontal line in the former planes. THe common houre-lines do commonly depend on the shadow of the axis, but the parallels of the Signs, and of the length of the day, the houre-lines from Sun rising and Sun setting, with many others, depend on the shadow of the top of the style, or some one point in the axis, which here signifieth the centre of the world, and is represented by the point B. And these lines so depending, are then only useful when they fall between the two tropiques, and within the horizon. There may be several horizontal lines drawn upon every plane, as I shown before in finding the inclination of a plane; but the proper horizontal line which is here meant, must always be in the same plane with B the top of the style; so that in an horizontal plane there can be no such horizontal line, but in all other planes it may be found by applying the horizontal leg of the Sector unto the top of the style, and then working as before; and the intersection of this line with the meridian or substylar line, may be found by proportion. 1 To find the intersection of the horizon with the meridian, in an equinoctial plane. As the tangent of 45 gr. to the tangent of the latitude: So is the height of the style, to the distance between the style and the horizontal line. As in the example of the former equinoctial plane, Pag. 142. extend the compasses from the tangent of 45 gr. unto 51 gr. 30 m. the tangent of the latitude, the same extent will reach in the line of numbers, from 52 the length of the style unto 66, and such is the distance between the style and the horizontal line; wherefore I take 66 parts out of a line of inches, and prick them down in the meridian line from C unto H above the style in the upper face, but below the style in the lower face of the plane, so a right line drawn through H, parallel to the hour of 6, shall be the horizontal line. 2 To find the intersection of the horizon with the meridian, in a direct polar plane. As the tangent of 45 gr. to the cotangent of the latitude: So the length of the style, to the distance between the style & the horizontal line. As in the example of the former polar plane, Pag. 144. extend the compasses from the tangent of 45 gr. unto tangent of 38 gr. 30 m. the compliment of the latitude, the same extent will reach in the line of numbers, from 1.61 the length of the style, unto 1.28, and such is the distance upon the meridian between the style and the horizontal line. In all upright planes, whether they be direct vertical, or declining, or meridian planes, the horizontal line must always be drawn through A the foot of the style, as may appear in the examples before, Pag. 102.107.116. And generally in all planes whatsoever, the horizontal line must be drawn through the intersection of the equator with the hour of 6. Or if that intersection fall without the plane, yet if any arks of the length of the day be drawn on the plane, the horizontal line may be drawn through their intersections, with the hours of the Suns rising or setting. CHAP. XIX. To describe the vertical circles in the former Planes. THe vertical circles commonly called Azimuths, are great circles drawn through the zenith, by which we may know in what part of the heaven the Sun is, how far from the East or West, and how near unto the meridian. In all upright planes, whether they be direct verticals, or declining, or meridian planes, the semidiameter of the horizon will be the same with AB the perpendicular side of the style, and these Azimuths will be parallels one to the other, and the distance of each Azimuth, from the foot of the style upon the horizontal line, may be found in this manner. Consider the length of the style in inches and parts of inches, and the distance of each Azimuth from the style, according to the angle at the zenith in degrees and minutes. As the tangent of 45 gr. to the tangent of the azimuth: So the length of the style, to the length of the horizontal line between the style and the azimuth. geometric illustration As if it were required to draw the common azimuths on the South face of the vertical plane before described, where AB the length of the style may be supposed to be 10 inches. Here the plane having no declination, the style is in the plane of the meridian, and so pointeth directly into the South. The point of SbE is 11 gr. 15 m. distant from the style, and SSE 22 gr. 30 m. and the rest in their order: wherefore extend the compasses from the tangent of 45 gr. unto 10 in the line of numbers, the same extent will reach from the tangent of 11 gr. 15 m. unto 1.99 in the line of numbers for the length of the tangent line, between the style and the point SbE, and from the tangent of 22 gr. 30 m. unto 4.14 for SSE, and so for the rest, as in this Table. Azimuths. An. Zen. Tangen Gr. M. In. Pa. South 0 0 0 0 SbE 11 15 1 99 SSE 22 30 4 14 SEbS 33 45 6 68 SE 45 0 10 00 SEbb 56 15 14 97 EASE 67 30 24 14 Ebbs 78 45 50 27 East 90 0 Infin. In like manner in the first example of the declining plane, where the style standeth according to the declination 24 gr. 20 m. distant from the South toward the West. The next point of SbW is but 13 gr. 5 m. distant from the style; and the second of SSW only 1 gr. 50 m. and the third of SWbS is again 9 gr. 25 m. and the rest in their order. Wherefore having before found the length of the style to be 6 inches 80 parts, extend the compasses from the tangent of 45 gr. unto 6.80 parts in the line of numbers, the same extent will reach from the tangent of 24 gr. 20 m. unto 3.07 in the line of numbers for the length of the tangent line between the style & the South, and from the tangent of 13 gr. 5 m. unto 1.58 for the point of SbW; and so for the rest, as in this Table. Azimuths. An. Zen. Tangen Gr. M In. Pa. SEbE 80 35 41 00 SE 69 20 18 03 SEbS 58 5 10 91 SSE 46 50 7 25 SbE 35 35 4 86 South 24 20 3 07 SbW 13 5 1 58 SSW 1 50 0 22 The foot of the still SWbS 9 25 1 13 sweet 20 40 2 57 SWbW 31 55 4 24 WSW 43 10 6 37 WbS 54 25 9 50 West 65 40 15 02 WbN 76 55 29 26 WNWS 88 10 212 45 That done, if you take these parts out of a line of inches, and prick them down in the horizontal line on either side of the style, drawing right lines perpendicular to the horizon through these intersections, but so as they may be contained between the horizontal and the tropiques, the lines so drawn shall be the azimuths required. In an horizontal plane these azimuths are drawn more easily. For here the perpendicular side of the style is the same with the axis of the horizon, and the foot of the style is the vertical point, in which all the azimuth lines do meet as their circles do in the zenith: wherefore let any circle described on the centre A, at the foot of the style, be divided first into four parts, beginning at the meridian, and then each quarter subdivided either into eight equal parts, according to the points of the Mariner's compass, or into 90 gr. according to the Astronomical division; if you draw right lines through the centre and these divisions, the lines so drawn shall be the azimuths required. In all other planes inclining to the horizon, these vertical circles will meet in a point, but that vertical point being more or less distant from the foot of the style, the angles at this point will be unequal. 1 To find the distance between the foot of the style and the vertical point. The vertical point wherein all the vertical lines do meet, will be always in the meridian, directly under or over the top of the style; and the angle between the perpendicular side of the style and the vertical line, will be equal to the inclination of the plane to the horizon. Wherefore As the tangent of 45 gr. to the tangent of the inclination of the plane: So is the length of the style, to the distance between the foot of the style and the vertical point. Thus in the first example of the declining inclining planes, where the upper face of the plane looking South-west, the declination was 24 gr. 20 m. the inclination 36 gr; and you may suppose AB the length of the style to be 6 inches: if you extend the compasses from the tangent of 45 gr. unto geometric illustration the tangent of 36 gr. the same extent will reach in the line of numbers from 6.00 unto 4.36, for the distance AV between A the foot of the style and V the vertical point. 2 To find the distance between the foot of the style and the horizontal line. As the tangent of the inclination of the plane, is to the tangent of 45 gr. So the length of the style, to the distance between the foot of the style and the horizontal line. So the same extent of the compasses as before, will reach in the line of numbers from 6.00 unto 8.26 for the distance AH between the foot of the style and the horizontal line. Then may you take 4 inches 36 cent. and pricking them down from A the foot of the style unto V the vertical point in the meridian, draw the line VALERIO, which being produced shall cut the horizon in the point H with right angles, and be that particular azimuth which is perpendicular to the plane. Or you may take 8 inches 26 cent. and prick them down in the former line VALERIO produced from A unto H, and so draw the horizontal line through H perpendicular unto VH, which horizontal line being produced will cross the equator in the same point wherein the equator crosseth the houre-line of 6, unless there be some former error. 3 To find the angles made by the azimuth lines at the vertical point. The angles at the zenith depend on the declination of the plane, as in our example, where the style standeth according to the declination 24 gr. 20 m. distant from the South toward the West, the azimuth of 10 gr. from the meridian Eastward will be 34 gr. 20 m. the azimuth of 10 gr. Westward will be only 14 gr. 20 m. distant from the style, and so the rest in their order. Or if you would rather describe the common azimuths, the point of SbE will be 35 gr. 35 m. the point of SbW 13 gr. 5 m. distant from the style, and so the rest in their order. Then As the sine of 90 gr. to the cousin of the inclination of the plane: So the tangent of the angle at the zenith, to the tangent of the angle at the vertical point between the line drawn through the foot of the style and the azimuth required. Wherefore the inclination of the plane in our example being 36 gr. extend the compasses from the sine of 90 gr. unto the sine of 54 gr. the same extent shall reach in the line of tangents, from 24 gr. 20 m. unto 20 gr. 5 m. for the angle HVa at the vertical point, between the line VH drawn through A the foot of the style & the South. Again, the same extent will reach from the tangent of 13 gr. 5 m. unto 10 gr. 38 m. for the angle belonging to SbW; and so for the rest, as in this table. Azimuths. Ang. Ze. Ang. Ve. Gr. M. Gr. M. SEbE 80 35 78 25 SE 69 20 65 0 SEbS 58 5 52 25 SSE 46 50 40 46 SbE 35 35 30 3 South 24 20 20 5 SbW 13 5 10 39 SSW 1 50 1 29 Style. 0 0 SWbS 9 25 7 38 sweet 20 40 16 58 SWbW 31 55 26 45 WSW 43 10 37 11 WbS 54 25 48 30 West 65 40 60 48 WbN 76 55 73 58 WNWS 88 10 87 44 These angles being known, if on the centre V, at the vertical point, you describe an occult circle, and therein inscribe the chords of these angles from the line VH, and then draw right lines through the vertical point, and the terms of those chords, the lines so drawn shall be the azimuths required. The like reason holdeth for the drawing of the azimuths upon all other inclining planes, wheof you have another example in the Diagram belonging to the meridian incliner, Pag. 126. Or for further satisfaction you may find where each azimuth line shall cross the equator. As the sine of 90 gr. to the sine of the latitude: So the tangent of the azimuth from the meridian, to the tangent of the equator from the meridian. Extend the compasses from the sine of 90 gr. unto the line of our latitude 51 gr. 30 m. the same extent will reach in the line of tangents from 10 gr. unto 7. gr. 50. m. for the intersection of the equator with the azimuth of 10 gr. from the meridian. Again, the same extent will reach from 20 gr. unto 15 gr. 54 m. for the azimuth of 20 gr. And so the rest, as in these tables. Azim. Equat. Gr. M. Gr. M. 10 0 7 50 20 0 15 54 30 0 24 20 40 0 43 18 50 0 13 0 60 0 53 35 70 0 65 3 80 0 77 18 90 0 90 0 Azim. Equat. Gr. M. Gr. M. 11 15 8 51 22 30 17 58 33 45 27 36 45 0 38 2 56 15 49 30 67 30 62 6 78 45 75 44 90 0 90 0 By which you may see that the azimuth 90 gr. distant from the meridian, which is the line of East and West, will cross the equator at 90 gr. from the meridian in the same point, with the horizontal line and the hour of 6. And that the azimuth of 45 gr. will cross the equator at 38 gr. 2 m. from the meridian, that is, the line of SE will cross the equator at the hour of 9 and 28 m. in the morning, and the line of sweet at 2 ho. 32 min. in the afternoon; and so for the rest, whereby you may examine your former work. CHAP. XX. To describe the parallels of the horizon in the former planes. THe parallels of the horizon, commonly called Almicanters, or parallels of altitude (whereby we may know the altitude of the Sun above the horizon) have such respect unto the horizon, as the parallels of declination unto the equator, and so may be described in like manner. In an horizontal plane, these parallels will be perfect circles; wherefore knowing the length of the style in inches and parts, and the distance of the parallel from the horizon in degrees and minutes. As the tangent of 45 gr. is to the length of the style: So the cotangent of the parallel, to the semidiameter of his circle. Thus in the example of the horizontal plane, Pag. 164. if AB the length of the style shall be 5 inches, and that it were required to find the semidiameter of the parallel of 62 gr. extend the compasses from the tangent of 45 gr. unto 5.00 in the line of numbers, the same extent will reach from the tangent of 28 gr. the compliment of the parallel unto 2.65, and if you describe a circle on the centre A to the semidiameter of 2 inches 65 cent. it shall be the parallel required. In all upright planes, whether they be direct verticals, or declining, or meridian planes, these parallels will be conical sections, and may be drawn through their points of intersection, with the azimuth lines, in the same manner as the parallels of declination, through their points of intersection with the houre-lines. To this end you may first find the distance between the top of the style and the azimuth; and then the distance between the horizon and the parallel, both which may be represented in this manner. On the centre B and any semidiameter BH, describe an occult ark of a circle, and therein inscribe the chords of such parallels of altitude as you intent to draw on the plane, (I have here put them for 15. 30. 45 and 60 gr.) then draw right lines through the centre and the terms of those chords, so the line BH shall be the horizon, and the rest the lines of altitude, according to their distance from the horizon. geometric illustration That done, consider your plane (which here for example is the South face of our vertical plane, p. 168) wherein having drawn both the horizontal & vertical lines, as I shown before, first take out AB the length of the style, & prick that down in this horizontal line from B unto A; then take out all the distances between B the top of the style and the several points wherein the vertical lines do cross the horizontal, transfer them into this horizontal line BH, from the centre B, and at the terms of these distances erect lines perpendicular to the horizon, noting them with the number or letter of the azimuth from whence they were taken, so these perpendiculars shall represent those azimuths, and the several distances between the horizon and the lines of altitude shall give the like distances, between the horizontal and the parallels of altitude upon the azimuths in your plane. Upon this ground it followeth, 1 To find the distance between the top of the style, and the several points wherein the azimuths do cross the horizontal line. Having drawn the horizontal and azimuth lines as before, look into the table by which you drew them, and there you shall have the angles at the zenith. Then As the cousin of the angle at the zenith, is to the sine of 90 gr. So the length of the style, to the distance required. geometric illustration Azimuths. Ang Ze Tangent Secant Par. 15. Par. 30. Gr. M Inch P. Inch P. Inch. P. Inch. P. South. 0 0 0 0 10 00 2 68 5 77 SbE 11 15 1 99 10 20 2 73 5 90 SSE 22 30 4 14 10 82 2 90 6 24 SEbS 33 45 6 68 12 03 3 23 6 94 SE 45 0 10 00 14 14 3 80 8 16 SEbE 56 15 14 97 18 00 4 82 10 40 EASE 67 30 24 14 26 13 7 02 15 08 Ebbs 78 45 50 27 51 26 13 73 29 60 East. 90 0 Infinite Infinite Infinite Infinite As in our example of the vertical plane, where AB the length of the style was supposed to be 10 inches, extend the compasses from the sine of 78 gr. 45 m. (the compliment of 11 gr. 15 m. the angle at the zenith, belonging to SbE and SbW) unto the sine of 90 gr. the same extent will reach from 10.00 the length of the style, unto 10.20 for the distance between the top of the style and the intersection of the azimuth SbE with the horizontal line, which distance may be called the secant of the azimuth, and may serve for the drawing of the parallel of 45 gr. from the horizon. The like reason holdeth for the rest of these distances here represented in the line BH. 2 To find the distance between the horizon and the parallels. As the tangent of 45 gr. to the tangent of the parallel: So the secant of the azimuth, to the distance required. As if it were required to draw the parallel of 15 gr. from the horizon, upon this vertical plane; extend the compasses from the tangent of 45 gr. unto the tangent of 15 gr. the same extent will reach in the line of numbers from 10. 00 the secant of the South azimuth unto 2.68, and therefore the distance between the horizon and the parallel of 15 gr. is 2 inches 68 cent. upon the South azimuth. Again, the same extent will reach from 10.20 the secant of SbE unto 2.73 for the like distance belonging to SbE and SbW; and so for the rest, which may be gathered and set down in the table. That done, and the horizon and azimuths being drawn, prick down 10 inches from the horizontal line upon the South azimuth, & 10 inches 20 cent. on the azimuths of SbE and SbW, and 10 inches 82 cent. on the azimuths of SSE and SSW, and 12 inches 3 cent. on the azimuth of SEbS and SWbS, and so the rest of these distances on their several azimuths: then if you draw a crooked line through all these points, that may make no angles, the line so drawn shall be the parallel of 45 gr. from the horizon. In like manner may you draw the parallel of 15 gr. or any other parallel of altitude upon any vertical plane. If the plane incline to the horizon, after we have found the vertical point, and drawn the horizontal line, we are farther to find the length of the axis of the horizon, than the angles betwixt this axis and the azimuth lines, and so the several distances between the parallels and the vertical point, all which may be represented in this manner. On the centre B, and any semidiameter, describe an occult quadrant of a circle, and therein inscribe the chords of such parallels of altitude as you intent to draw on the plane, drawing right lines through the centre and the terms of these chords, so the line BH shall be the horizon, and his perpendicular BV the axis of the horizon, and the rest the lines of altitude, according to their distance from the horizon. geometric illustration Or having the Sector you may draw an occult line WE perpendicular to the axis VB, and therein prick down the tangent of the compliment of the inclination of the plane from V unto E: then draw the line OF parallel to the axis, crossing the line VH produced in the point F, so this line OF will be as the line of sins upon the Sector, and therein you may prick down the sins of the compliment of the angles at the zenith from E towards F, and draw the vertical lines by those points through the lines of altitude, so the angles at V, between the axis VB and those azimuth lines, shall be the angles between the axis of the horizon and the azimuth lines on your plane, and the several distances between the point V and the lines of altitude, shall give the like distances between the vertical point and the parallels of altitude upon the azimuths in your plane. Upon this ground it followeth, 1 To find the length of the axis of the horizon. The vertical point is always either directly over or under the top of the style, and the distance between them is that which I call the axis of the horizon, which may thus be found, As the cousin of the inclination, to the sine of 90 gr. So the length of the style, to the length of the axis of the horizon. For example in the first of the three declining inclining planes, the inclination to the horizon is 36 gr. the length of the style AB six inches, extend the compasses from the sine of 54 gr. the compliment of the inclination unto the sine of 90 gr. the same extent will reach in the line of numbers from 6.00 unto 7.42, & such is VB the length of the axis required. 2 To find the angles contained between the horizon and the vertical lines upon your plane. The angles at the vertical point between the axis of the horizon and the azimuth lines upon your plane are represented in this figure by those at V, between VB and the azimuths. The angles between the horizon and the azimuth lines being compliments to the former, are represented either by those which are made by WE or by BH, and the azimuth lines which are drawn from V. geometric illustration In our example where the inclination to the horizon is 36 gr. and the angle at the zenith between the azimuth at the style and the meridian, is according to the declination 24 gr. 20 m. extend the compasses from the sine of 90 gr. unto the tangent of 36 gr. the same extent will reach from the sine of 65 gr. 40 m. the compliment of the angle at the zenith, unto the tangent of 33 gr. 30 m. for the angle contained between the horizon and the South part of the meridian line. Again, the same extent will reach from the cousin of 35 gr. 35 m. the angle at the zenith belonging to SbE unto the tangent of 30 gr. 3 m. for the angle between the horizon and the azimuth line of SbE. The like reason holdeth for the rest, which may be found and set down in the Table. Azimuths. Ang. Ze. Ang. V. Ang. Ho Horizon 11 18 26 34 45 0 Gr. M. Gr. M. Gr. M. Inch. P. Inch. P Inch. P. Inch. P. East. 114 25 119 12 16 40 Infinite. 38 60 11 05 Ebbs 103 5 106 2 19 20 210 24 22 40 9 00 E SE 91 50 92 16 1 20 41 98 15 57 7 60 SE bE 80 35 78 25 6 47 62 82 23 44 12 07 6 68 SE 69 20 65 0 14 23 29 87 16 79 10 12 6 00 SE bS 58 5 52 25 21 0 20 70 13 61 8 99 5 79 SSE 46 50 40 46 26 25 16 68 11 90 8 31 5 53 SbE 35 35 30 3 30 35 14 58 10 90 7 90 5 42 South 24 20 20 5 33 30 13 44 10 32 7 66 5 35 SbW 13 5 10 39 35 17 12 84 10 02 7 55 5 33 SSW 1 50 1 29 35 59 12 62 9 90 7 47 5 31 Style. 0 0 36 0 12 62 9 90 7 47 5 31 SWbS 9 25 7 38 35 37 12 74 9 96 7 50 5 32 sweet 20 40 16 58 34 12 13 20 10 20 7 59 5 34 SWbW 31 55 26 45 31 40 14 13 10 67 7 81 5 39 WSW 43 10 37 11 27 55 15 85 11 50 8 15 5 49 WbS 54 25 48 30 22 55 19 05 12 94 8 73 5 66 West 65 40 60 48 16 40 25 87 15 51 9 60 5 96 W bN 76 55 73 58 9 20 45 75 20 64 11 32 6 46 W NW 88 10 87 44 1 20 318 88 33 27 14 18 7 25 NWbW 99 25 101 35 6 47 Infinite. 92 40 19 60 8 48 NWS 110 40 115 0 14 23 31 44 10 30 Then may you either draw these angles at V in the former figure more perfectly, and thence finish your work, or else proceed. 3 To find the distance between the vertical point and the parallels of the horizon. These distances may be found by resolving the triangles in the last figure made by the axis, the lines of altitude, and the azimuth lines. For having the length of the axis and the angles at the horizon, if you add the distance of the parallel from the horizon unto the angle at the horizon, you shall have the angle at the parallel. Then As the sine of the angle at the parallel, to the cousin of the altitude: So the length of the axis, to the distance between the vertical point and the parallel. Thus in our example if it were required to find the distance upon the stylar azimuth VH, between the vertical point and the horizon, you have the rectangle triangle VBH, wherein the angle at the horizon here represented by BHV is (equal to the inclination of the plane) 36 gr. and BV the axis of the horizon between the plane and the top of the style, is 7 inches 42 cent. Wherefore extend the compasses from the sine of 36 gr. unto the sine of 90 gr. the compliment of the altitude, the same extent will reach in the line of numbers from 7.42 unto 12 62, and such is the distance of the perpendicular azimuth line VH between the vertical point and the horizon. In like manner if you would find the distance upon the meridian between the vertical point and the horizon, extend the compasses from the sine of 33 gr. 30 m. the angle at the horizon, to the sine of 90 gr. the same extent will reach in the line of numbers from 7.42 unto 13.44, and such is Valerio the distance between the vertical point and the horizon upon the line of the South azimuth, that is, upon the meridian line. But if you would find the distance upon the meridian between the vertical point and any other parallel of the horizon, as upon the parallel of 26 gr. 34 m. then add these 26 gr. 34 m. unto 33 gr. 30 m. the angle at the horizon, so shall you have 60 gr. 4 m. for BDV the angle at the parallel. And if you extend the compasses from the sine of 60 gr. 4 m. unto the sine of 63 gr. 26 m. the compliment of the parallel from the horizon, the same extent will reach in the line of numbers from 7.42 the length of the axis, unto 7.66, and such is the distance ud between the vertical point and the parallel of 26 gr. 34 m. upon the meridian line. The like reason holdeth for all the rest, which may be gathered & set down in the table. That done, and the horizon drawn as before, if you would draw the parallel of 26 gr. 34 m. from the horizon, look into the table, and there finding under the title of the parallel of 26.34, the distance on the South azimuth line to be 7.66, take 7 inches 66 cent. out of a line of inches, and prick them down on the meridian of your plane, from the vertical point at V. Or if either the vertical point fall without your plane, or the extent at any time be too large for your compasses, you may prick down the distance between the horizon and the parallel. As here the distance between the vertical point and the parallel is 7.66, between the vertical point and the horizon 13.44, the difference between them 5.78 is the distance from the horizon to the parallel, which being pricked down upon the meridian, shall give the same intersection as before. And the like reason, holdeth for the pricking down the rest of these distances on their several azimuths. Having the points of intersection between the azimuths and the parallel, you may join them all in a crooked line without making of angles, the line so drawn shall be the parallel required. And upon this ground it followeth, To describe such parallels on the former planes, as may show the proportion of the shadow unto the gnomon. The proportion of a man's shadow unto his height, or other shadow to his gnomon set perpendicular to the horizon, may be showed by parallels to the horizon, if they be drawn to a due altitude, which may thus be found: As the length of the shadow, to the length of the gnomon: So the tangent of 45 gr. to the tangent of the altitude. As if it were required to find the altitude of the Sun when the shadow of a man shall be decuple to his height, extend the compasses from 10 unto 1 in the line of numbers, the same extent will reach in the tangent of 45 gr. unto the tangent of 5 gr. 42 m; which shows that when the Sun cometh to the altitude of 5 gr. 42 m. your shadow, upon a level ground, will be ten times as much as your height. In the same manner you may find that at 7 gr. 7 m. of altitude your shadow will be octuple, at 9 gr. 27 m. sextuple, at 11 gr. 18 m. quintuple, at 14 gr. 2 m. quadruple, at 18 gr. 26 m. triple, at 26 gr. 34 m. double to your height, at 33 gr. 41 m. as 3 unto 2, at 36 gr. 52 m. as 4 unto 3, at 38 gr. 40 m. as 5 unto 4, at 45 gr. equal, at 51 gr. 20 m. as 4 unto 5, at 53 gr. 7 m. as 3 unto 4, at 56 gr. 19 m. as 2 unto 3, at 59 gr. 2 m. as 3 unto 5, at 63 gr. 26 m. as 1 unto 2, etc. If then you draw a parallel to the horizon at 5 gr. 42 m. another at 7 gr. 7 m. and so the rest, when ●he shadow o● the style falleth on the parallel, you have the proportion, and thereby may you know the shadow by the height, and the height by the shadow, whereof you have examples Pag. 126. and 137. I might here proceed to show the description of th●●●●cles of position, the Signs of the Zodiac in the merid●●n, the Signs ascending and descending, with such o●●er gnomonical conclusions; but these would prove superfluous to such as understand the doctrine of the Sphere; and for others, that which is delivered may suffice for ordinary use, it being my intention not so much to explain the full use of shadows (whereof I have lately given a large example in an other place) as the use of these lines of proportion, that were not extant heretofore. An Appendix concerning The description and use of a small portable Quadrant, for the more easy finding of the hour and Azimuth. CHAP. I. Of the description of the Quadrant. Having described these standing planes, I will now show the most of these conclusions by a small Quadrant. This might be done generally for all latitudes, by a quarter of the general Astrolabe, described before in the use of the Sector, pag. 58; and particularly for any one latitude, by a quarter of the particular Astrolabe, there also described, pag. 63. which if it be a foot semidiameter, may show the azimuth unto a degree, and the time of the day unto a minute; but for ordinary use this smaller Quadrant may suffice, which may be made portable in this manner. 1 Upon the centre A, and semidiameter AB, describe the ark BC: the same semidiameter will set of 60 gr. and the half of that will be 30 gr. which being added to the former 60 gr. will make the ark BC to be 90 gr. the fourth part of the whole circle, and thence comes the name of a Quadrant. 2 Leaving some little space for the inscription of the months and days, on the same centre A, and semidiameter AT, describe the ark TD, which shall serve for either tropic. 3 Divide the line AT in the point E, in such proportion, as that AT being 10000, A may be 6556, and there draw another ark OF, which shall serve for the Equator. geometric illustration 5 This part of the ecliptic may be divided into three Signs, and each Sign into 30 A Table of right Ascensions. Gr. ♈ ♉ ♊ Gr. M. Gr. M. Gr. M. 0 0 0 27 54 57 48 5 4 35 32 42 63 3 10 9 11 37 35 68 21 15 13 48 42 31 73 43 20 18 27 47 33 79 7 25 23 9 52 38 84 32 30 27 54 57 48 90 0 gr. by a table of right ascensions, made as before pag. 60. As the right ascension of the first point of ♉ being 27 gr. 54 m. you may lay a ruler to the centre A & 27 gr. 54 m. in the Quadrant BC, the point where the ruler crosseth the Ecliptic, shall be the first point of ♉. In like manner the right ascension of the first point of ♊ being 57 gr. 48 m. if you lay a ruler to the Gr. Parts. 1 176 2 355 3 537 4 723 5 913 6 1106 7 1302 8 1503 9 1708 10 1917 11 2130 12 2348 13 2571 14 2799 15 3032 16 3270 17 3514 18 3763 19 4019 20 4281 21 4550 22 4825 23 5108 Trow 5252 centre A, and 57 gr. 48 m. in the quadrant, the point where the ruler crosseth the ecliptic, shallbe the first point of ♊. And so for the rest: but the lines of distinction between Sign & Sign, may be best drawn from the centre G. 6 The line ET between the equator and the tropic, which I call the line of declination, may be divided into 23 gr. ½ out of this Table. For let A the semidiameter of the equator be 10000, the distance between the equator and 10 gr. of declination may be 1917 more; between the equator and 20 gr. 4281; the distance of the tropic from the equator 5252. 7 You may put in the most of the principal stars between the equator and the tropic of ♋, by their declination from the equator, and right ascension from the next equinoctial point. As the declination of the wing of Pegasus, being 13 gr. 7 m. the right ascension 358 gr. 34 m. from the first point of ♈, or 1 gr. 26 m. short of it. If you draw an occult parallel through 13 gr. 7 m. of declination, and then lay the ruler to the centre A, and 1 gr. 26 m. in the quadrant BC, the point where the ruler crosseth the parallel shall be the place for the wing of Pegasus, to which you may set the name and the time when he cometh to the South, in this manner, W. Peg. * 23 Ho. 54 M. and so for the rest of these five, or any other stars. Ho. M. R. Ascen Decl. M Pegasus wing * 23 54 1 26 13 7 Arcturus * 13 58 29 37 21 10 Lions heart * 9 48 32 58 13 45 Bulls eye * 4 15 63 33 15 42 Vultures heart * 19 33 66 56 7 58 8 There being space sufficient between the equator and the centre, you may there describe the quadrat, and divide each of the two sides farthest from the centre A into 100 parts, so shall the Quadrant be prepared generally for any latitude. But before you draw the particular lines, you are to fit four tables unto your latitude. First a table of meridian altitudes for division of the circle of days and months, which may be thus made: Consider the latitude of the place and the declination of the Sun for each day of the year. If the latitude and declination be alike both North or both South, ad the declination to the compliment of the latitude; if they be unlike, one North, and the other South, subtract the declination from the compliment of the latitude, the remainder will be the meridian altitude belonging unto the day. Thus in our latitude of 51 gr. 30 m. Northward, whose compliment is 38 gr. 30 m. the declination upon the tenth day of june will be 23 gr. 30 m. Northward, wherefore I add 23 gr. 30 m. unto 38 gr. 30 m. the sum of both is 62 gr. for the meridian altitude at the tenth of june. The declination upon the tenth of December will be 23 gr. 30 m. Southward, wherefore I take these 23 gr. 30 m. out of 38 gr. 30 m. there will remain 15 gr. for the meridian altitude at the tenth of December; and in this manner you may find the meridian altitude for each day of the year, and set them down in a table. Dies. 0 5 10 15 20 25 30 more Gr. M Gr. M. Gr. M Gr. M. Gr. M. Gr. M Gr. M. january 16 31 17 24 18 26 19 37 20 57 22 24 23 58 February 24 17 25 59 27 45 29 35 31 29 33 25 March 34 35 36 33 38 32 40 30 42 27 44 22 46 15 April 46 37 48 26 50 11 51 50 53 25 54 53 56 15 May 56 15 57 29 58 35 59 33 60 22 61 2 61 31 june 61 36 61 54 62 0 61 58 61 45 61 22 60 49 july 60 49 60 6 59 14 58 13 57 4 55 48 54 24 August 54 7 52 36 50 59 49 17 47 31 45 41 43 49 September 43 26 41 30 39 33 37 36 35 38 33 41 31 46 October 31 46 29 53 28 3 26 16 24 35 22 59 21 29 November 21 12 19 51 18 39 17 36 16 43 16 0 15 28 December 15 28 15 7 15 0 15 2 15 17 15 44 16 22 The Table being made, you may inscribe the months, and days of each month into your quadrant, in the space left below the tropic. For lay the ruler unto the centre A, and 16 gr. 31 m. in the quadrant BC, there may you draw a line for the end of December and beginning of january; then laying your ruler to the centre A, and 24 gr. 17 m. in the quadrant, there draw the end of january and beginning of February, and so the rest, which may be noted with I, F, M, A, M, I, etc. the first letters of each month, and will here fall between 15 gr. and 62 gr. The second Table which you are to fit, may serve for the drawing and dividing of the horizon. For drawing of the horizon. As the cotangent of the latitude, to the tangent of the greatest declination: So the sine of 90 gr. to the sine of intersection, where the horizon shall cross the tropiques. So in our latitude of 51 gr. 30 m. we shall find the horizon to cut the tropic in 33 gr. 9 m: wherefore if you lay the ruler to the centre A, and 33 gr. 9 m. in the quadrant, the point where the ruler crosseth the tropic shall be the point where the horizon crosseth the tropic. And if you find a point at H, in the line AC, whereon setting the compasses, you may bring the point at E, and this point in the tropic both into a circle, the point H shall be the centre, and the ark so drawn shall be the horizon. Then for the division of this horizon. As the sine of 90 gr. to the sine of the latitude: So the tangent of the horizon, to the tangent of the ark in the quadrant, which shall divide the horizon. So in our latitude of 51 gr. 30 m. we shall find 7 gr. 52 m. belonging to 10 gr. in the horizon, and 15 gr. 54 m. belonging to 20 gr. And so the rest, as in this Table. Ho Gr. M Ho Gr. M Ho Gr. M. Ho Gr. M. Ho Gr. M. Ho Gr. M. 0 0 0 15 11 51 30 24 19 45 38 2 60 53 35 75 71 5 0 47 12 39 25 11 39 1 54 41 72 19 1 34 13 27 26 4 40 0 55 48 73 33 2 21 14 16 26 57 41 0 56 56 74 48 3 8 15 4 27 50 42 0 58 4 76 3 5 3 55 20 15 54 35 28 43 50 43 0 65 59 13 80 77 18 4 42 16 43 29 37 44 1 60 22 78 33 5 29 17 33 30 32 45 3 61 31 79 49 6 17 18 22 31 27 46 5 62 41 81 5 7 4 19 12 32 22 47 8 62 52 82 21 10 7 52 25 20 2 40 33 18 55 48 11 70 65 3 85 83 37 8 39 20 53 34 14 49 14 66 15 84 53 9 27 21 44 35 10 50 19 67 27 86 10 10 14 22 36 36 7 51 24 68 39 87 26 11 2 23 27 37 4 52 29 69 52 88 43 15 11 51 30 24 19 45 38 2 60 53 35 75 71 5 90 90 0 Wherefore you may lay the ruler to the centre A, and 7 gr. 52 m. in the quadrant BC, the point where the ruler crosseth the horizon shall be 10 gr. in the horizon; and so for the rest: but the lines of distinction between each fift degree, will be best drawn from the centre H. The third Table for drawing of the houre-lines, must be a Table of the altitude of the Sun above the horizon at every hour, especially when he cometh to the equator, the tropiques, and some other intermediate declinations. If the Sun be in the equator, and so have no declination. As the sine of 90 gr. to the cousin of the latitude: So the cousin of the hour from the meridian, to the sine of the altitude. Thus in our latitude of 51 gr. 30 m. at six hours from the meridian the Sun will have no altitude, at five the altitude will be 9 gr. 17 m; at four 18 gr. 8 m; at three 26 gr. 7 m; at two 32 gr. 37 m. at one 36 gr. 58 m; at noon it will be 38 gr. 30 m. equal to the compliment of the latitude. If the Sun have declination, the meridian altitude will be found as before, for the Table of days and months. If the hour proposed be six in the morning or six at night. As the sine of 90 gr. to the sine of the latitude: So the sine of the declination, to the sine of the altitude. Thus in our latitude the declination of the Sun being 23 gr. 30 m. the altitude will be found to be 18 gr. 11 m: the declination being 11 gr. 30 m. the altitude will be 9 gr. If the hour proposed be neither twelve nor six. As the cousin of the hour from the meridian, to the sine of 90 gr. So the tangent of the latitude, to the tangent of a fourth ark. So in our latitude and one hour from the meridian, this fourth ark will be found to be 52 gr. 28 m. at two 55 gr. 26 m. at three 60 gr. 39 m. at four 68 gr. 22 m. and at five hours from the meridian 78 gr. 22 m. Then consider the declination of the Sun and the hour proposed; if the latitude and declination be both alike, as with us in North latitude, North declination, and the hour fall between noon and six, take the declination out of the fourth ark, the remainder shall be your fift ark. But if either the hour fall between six and midnight, or the latitude and declination shall be unlike, add the declination unto the fourth ark, and the sum of both shall be your fifth ark: or if the sum shall exceed 90 gr. you may take the compliment unto 180 gr. This fifth ark being known: As the sine of the fourth ark, to the sine of the latitude: So the cousin of the fift ark, to the sine of the altitude. Thus in our latitude of 51 gr. 30 m. Northward, the Sun having 23 gr. 30 m. of North declination, if it shall be required to find the altitude of the Sun for seven in the morning; here because the latitude and declination are both alike to the Northward, and the hour proposed falleth between noon and six, you may take 23 gr. 30 m. the ark of the declination out of 78 gr. 22 m. the fourth ark belonging to the fift hour from the meridian, so there will remain 54 gr. 52 m. for your fift ark. Then working according to the Canon, you shall find, As the sine of 78 gr. 22 m. your fourth ark, to the sine of 51 gr. 30 m. for the latitude: So the sine of 35 gr. 8 m. the compliment of your fift ark, to the sine of 27 gr. 17 m. the altitude required. If in the same latitude and declination, it were required to find the altitude for five in the morning, here the hour falling between six and midnight; if you add 23 gr. 30 m. unto 78 gr. 22 m. the sum will be 101 gr. 52 m. and the compliment to 180 gr. will be 78 gr. 8 m. for your fifth ark. Wherefore As the sine of 78 gr. 22 m. to the sine of 51 gr. 30 m. So the cousin of 78 gr. 8 m. to the sine of 9 gr. 32 m. for the altitude required. If in the same latitude of 51 gr. 30 m. Northward, the Sun having 23 gr. 30 m. of South declination, it were required the altitude for nine in the morning; here because the latitude and declination are unlike, the one North, and the other South, you may add 23 gr. 30 m. the ark of declination, unto 60 gr. 39 m. the fourth ark belonging to the third hour from the meridian, so shall you have 84 gr. 9 m. for your fift ark. Wherefore As the sine of 60 gr. 39 m. to the sine of 51 gr. 30 m. So the cousin of 84 gr. 9 m. to the sine of 5 gr. 15 m. for the altitude required. And so by one or other of these means you may find the altitude of the Sun for any point of the ecliptic at all hours of the day, and set them down in such a Table as this. A Table for the altitude of the Sun in the beginning of each Sign at all hours of the day, calculated for 51 gr. 30 m. of North latitude. Ho. ♋ ♊ ♌ ♉ ♍ ♈ ♎ ♓ ♏ ♒ ♐ ♑ Gr. M. Gr. M. Gr. M. Gr. M. Gr. M. Gr. M. Gr. M. 12 62 0 58 42 50 0 38 30 27 0 18 18 15 0 11 1 59 43 56 34 48 12 36 58 25 40 17 6 13 52 10 2 53 45 50 55 43 12 32 37 21 51 13 38 10 30 9 3 45 42 43 6 36 0 26 7 15 58 8 12 5 15 8 4 36 41 34 13 27 31 18 8 8 33 1 15 7 5 27 17 24 56 18 18 9 17 0 6 6 6 18 11 15 40 9 0 0 0 5 7 9 32 6 50 11 37 4 8 1 32 21 40 Lastly, you may find what declination the Sun hath when he riseth or setteth at any hour. As the sine of 90 gr. to the sine of the hour from six: So the cotangent of the latitude, to the tangent of the declination. And so in the latitude of 51 gr. 30 m. you shall find that when the Sun riseth, either at five in the Summer, or seven in the Winter, his declination is 11 gr. 37 m: when he riseth at four in the Summer, or eight in the Winter, his declination is 21 gr. 40 m. which may be also set down in the Table. That done, you may there see that in this latitude the meridian altitude of the Sun in the beginning of ♋ is 62 gr. in ♊ 58 gr. 42 m. in ♉ 50 gr. in ♈ 38 gr. 30 m. etc. But the beginning of ♋ and ♑ is represented by the tropiques TD, drawn at 23 gr. 30 m. of declination, and the beginning of ♈ and ♎, by the equator EF. If you draw an occult parallel between the equator and the tropic, at 11 gr. 30 m. of declination, it shall represent the beginning of ♉, ♍, ♏, and ♓; if you draw another occult parallel though 20 gr. 12 m. of declination, it shall represent the beginning of ♊, ♌, ♐, and ♒. Then you may lay a ruler to the centre A, and 62 gr. in the quadrant BC, and note the point where it crosseth the tropic of ♋ than move the ruler to 58 gr. 52 m. and note where it crosseth the parallel of ♊; then to 50 gr. and note where it crosseth the parallel of ♉; and again to 38 gr. 30 m. noting where it crosseth the equator; so the line drawn through these points shall show the hour of 12 in the Summer, while the Sun is in ♈, ♉, ♊, ♋, ♌, or ♍. In like manner if you lay the ruler to the centre A, and 27 gr. in the quadrant, and note the point where it crosseth the parallel of ♓, then move it to 18 gr. 18 m. and note where it crosseth the parallel of ♒; and again to 15 gr. noting where it crosseth the tropic of ♑; the line drawn through these points shall show the hour of 12 in the Winter, while the Sun is in ♎, ♏, ♐, ♑, ♒ & ♓, and so may you draw the rest of these houre-lines: only that of 7 from the meridian in the Summer, and 5 in the Winter, will cross the line of declination at 11 gr. 37 m. and that of 8 in the Summer, and 4 in the Winter at 21 gr. 40 m. The fourth table for drawing of the azimuth lines, must likewise be fitted for the altitude of the Sun above the horizon at every azimuth, especially when he cometh to the equator, the tropiques, and some other intermediat declinations. If the Sun be in the equator, and so have no declination: As the sine of 90 gr. to the cousin of the azimuth from the meridian: So the cotangent of the latitude, to the tangent of the altitude at the equator. Thus in our latitude of 51 gr. 30 m. at 90 gr. from the meridian, the Sun will have no altitude; at 80 gr. the altitude will be 7 gr. 52 m; at 70 gr. it will be 15 gr. 30 m; at 60 gr. it will be 21 gr. 41 m. If the Sun have declination, the meridian altitude will be easily found as before, for the table for days and months. And for all other azimuths. As the sine of the latitude, to the sine of the declination: So the cousin of the altitude at the equator, to the sine of a fourth ark. When the latitude and declination are both alike in all azimuths from the prime vertical unto the meridian, add this fourth ark unto the ark of altitude at the equator. When the latitude and declination are both alike, and the azimuth more than 90 gr. distant from the meridian, take the altitude at the equator out of this fourth ark. When the latitude and declination are unlike, take this fourth ark out of the ark of altitude at the equator, so shall you have the altitude of the Sun belonging to the azimuth. Thus in our latitude of 51 gr. 30 m. Northward, if it were required to find the altitude of the Sun in the azimuth of 60 gr. from the meridian, when the declination is 23 gr. 30 m. Nothward, you may find the altitude at the equator belonging to this azimuth to be 21 gr. 41 m. by the former Canon, and by this last Canon you may find the fourth ark to be 28 gr. 15 m. Then because the latitude and declination are both alike to the Northward, if you add them both together, you shall have 49 gr. 56 m. for the altitude required. If the declination had been 23 gr. 30 m. to the Southward, you should then have taken this fourth ark out of the ark at the equator, which because it cannot here be done, it is a sign that the Sun is not then above the horizon. But if you take the ark at the equator out of this fourth ark, you shall have 6 gr. 34 m. for the altitude of the Sun when he is in the azimuth of 60 gr. from the North, and 120 gr. from the South part of the meridian. The like reason holdeth for the rest of these altitudes, which may be gathered and set down in a table. Lastly when the Sun riseth or setteth upon any azimuth, to find his declination. As the sine of 90 gr. to the cousin of the latitude: So the cousin of the azimuth from the meridian, to the sine of the declination. And thus in our latitude of 51 gr. 30 m. when the azimuth is 80 gr. from the meridian, the declination will be found to be 6 gr. 12 m; if the azimuth be 70 gr. the declination will be found 12 gr. 18 m; if 60 gr. then 18 gr. 8 m. And so for the rest, which may be also set down in the Table. A Table for the altitude of the Sun in the beginning of each Sign for every tenth azimuth, in 51 gr. 30 m. of North latitude. Az. ♋ ♊ ♌ ♉ ♍ ♈ ♎ ♓ ♏ ♒ ♐ ♑ Gr. M Gr. M Gr. M. Gr. M Gr. M Gr. M. Gr. M. 0 62 0 58 42 50 0 38 30 27 0 18 18 15 0 10 61 43 58 24 49 38 38 4 26 30 17 45 14 25 20 60 51 57 28 48 33 36 46 25 0 16 5 12 41 30 59 52 55 52 46 40 34 34 22 27 13 15 9 45 40 57 10 53 29 43 55 31 21 18 48 9 14 6 34 50 54 3 50 12 40 11 27 5 13 58 3 57 0 6 60 49 56 45 53 35 23 21 41 8 0 70 44 40 40 25 29 27 15 13 1 0 80 38 11 33 46 21 29 7 52 90 30 38 26 10 14 45 0 0 100 22 27 18 2 6 45 6 12 110 14 14 9 58 12 18 120 6 34 2 30 18 8 That done, if you would draw the line of East or West, which is 90 gr. from the meridian, lay the ruler to the centre A, and 30 gr. 38 m. numbered in the quadrant from C toward B, and note the point where it crosseth the tropic of ♋; then move the ruler to 26 gr. 10 m. and note where it crosseth the parallel of ♊ then to 14 gr. 45 m. and note where it crosseth the parallel of ♉ then to 0 gr. 0 m. and you shall find it to cross the equator in the point F; so a line drawn through these points, shall show the azimuth belonging to East and West. The like reason holdeth for all the rest. These lines being thus drawn, if you set two sights upon the line AC, and hang a thread and plummet on the centre A, with a bead upon the thread, the foreside of the quadrant shall be fully finished. Or in stead of the five stars before mentioned, you may place the Nocturnal (described before in the use of the Sector, pag. 60.) on the backside of the Quadrant, and so also it will be fitted both for day and night. CHAP. II. Of the use of the Quadrant in taking the altitude of the Sun, Moon, and Stars. THe Quadrant is the fourth part of a circle, divided equally into 90 gr. and here numbered by 10.20.30. etc. unto 90 gr. each degree being subdivided into 4. Lift up the centre of the Quadrant, so as the thread with the plummet may play easily by the side of it, and the Sun beams may pass through both the sights; so shall the degrees cut by the thread, show what is the altitude at the time of observation, as may appear by this example. Upon the 14 day of April, about noon, the Sunbeams passing through both the sights, the thread fell upon 51 gr. 20 m. and this was the true meridian altitude of the Sun for that day in this our latitude of 51 gr. 30 m. for which this Quadrant was made. Again, towards three of the clock in the afternoon, the thread fell upon 38 gr. 40 m. and such was the Sun's altitude at that time. CHAP. III. Of the Ecliptic. 1 The place of the Sun being given to find his right ascension. THe Ecliptic is here represented by the ark, figured with the characters of the twelve Signs, ♈, ♉, ♊, etc. each Sign being divided unequally into 30 gr. and they are to be reckoned from the character of the Sign. Let the thread be laid on the place of the Sun in the Ecliptic, and the degrees which it cutteth in the Quadrant shall be the right ascension required. As if the place of the Sun giuen be the fourth degree of ♊, the thread laid on this degree shall cut 62 degrees in the Quadrant, which is the right ascension required. But if the place of the Sun giuen be more than 90 gr. from the beginning of ♈, there must be more than 90 gr. allowed to the right ascension; for this instrument is but a quadrant: and so if the Sun be in 26 gr. of ♋, you shall find the thread to fall in the same place, and yet the right ascension to be 118 gr. 2 The right ascension of the Sun being given, to find his place in the Ecliptic. Let the thread be laid on the right ascension in the Quadrant, and it shall cross the place of the Sun in the Ecliptic, as may appear in the former example. CHAP. FOUR Of the line of declination. 1 The place of the Sun being given to find his declination. THe line of declination is here drawn from the centre to the beginning of the Quadrant, and divided from the beginning of ♈ downward into 23 gr. 30 m. Let the thread be laid, and the bead set on the place of Sun in the ecliptic; then move the thread to the line of declination, and there the bead shall fall upon the degrees of the declination required. As if the place of the Sun giuen be the fourth degree of ♊, the bead first set to this place, and then moved to the line of declination, shall there show the declination of the Sun at that time to be 21 gr. from the equator. 2 The declination of the Sun being given, to find his place in the Ecliptic. Let the thread and bead be first laid to the declination, and then moved to the Ecliptic. As if the declination be 21 gr. the bead first set to this declination, and then moved to the ecliptic, shall there show the fourth of ♊, the fourth of ♐, the 26 of ♋, and the 26 of ♑; and which of these four is the place of the Sun, may appear by the quarter of the year. CHAP. V Of the circle of Months and Days. THis circle is here represented by the ark, figured with these letters, I, F, M, A, M, etc. signifying the months january, February, March, April, etc. each month being divided unequally, according to the number of the days that are therein. 1 The day of the month being given, to find the altitude of the Sun at noon. Let the thread be laid to the day of the month, and the degrees which it cutteth in the Quadrant shall be the meridian altitude required. As if the day giuen be the 15 of May, the thread laid on this day shall cut 59 gr. 30 m. in the quadrant, which is the meridian altitude required. 2 The meridian altitude being given, to find the day of the month. The thread being set to the meridian altitude, doth also fall on the day of the month. As if the altitude at noon be 59 gr. 30 m. the thread being set to this altitude, doth fall on the 15 day of May and the 9 of july; and which of these two is the true day, may be known by the quarter of the year, or by another day's observation. For if the altitude prove greater, the thread will fall on the 16 day of May and the 8 of july: or if it prove lesser, the thread will fall on the 14 of May and the 10 of july; whereby the question is fully answered. CHAP. VI Of the Houre-lines. THat ark which is drawn upon the centre of the quadrant by the beginning of declination, doth here represent the equator: that ark which is drawn by 23 gr. 30 m. of declination, and is next above the circle of months and days, representeth the tropiques: those lines which are between the equator and the tropiques, being undivided and numbered at the equator by 6. 7. 8. 9 10. 11. 12. at the tropic by 1. 2. 3. 4. etc. do represent the houre-circles: that which is drawn from 12 in the equator to the middle of june, representeth the hour of 12 at noon in the Summer; and those which are drawn with it to the right hand, are for the hours of the day in the Summer, and the hours of the night in the Winter. That which is drawn from 12 in the equator to the middle of December, representeth the hour of 12 in the Winter; and those which are drawn with it to the left hand, are for the hours of the day in the Winter, and the hours of the night in the Summer; and of both these, that which is drawn from 11 to 1, serves for 11 in the forenoon, and 1 in the afternoon. That which is drawn from 10 to 2, serves for 10 in the forenoon, & 2 in the afternoon: for the Sun on the same day is about the same height two hours before noon, as two hours after noon. The like reason holdeth for the rest of the hours. 1 The day of the month, or the height at noon being known, to find the place of the Sun in the Ecliptic. The thread being laid to the day of the month, or the height at noon, (for one gives the other by the former proposition) mark where it crosseth the hour of 12, and set the bead to that intersection; then move the thread till the bead fall on the ecliptic, and it shall fall on the place of the Sun. As if the day giuen be the 15 of May, or the meridian altitude 59 gr. 30 m. lay the thread accordingly, and put the bead to the intersection of the thread with the hour of 12; then move the thread till the bead fall on the ecliptic, and it shall there show the fourth of ♊, the fourth of ♐, the 26 of ♋, and the 26 of ♑; and which of these is the place of the Sun, may appear by the quarter of the year, or another day's observation. 2 The place of the Sun in the Ecliptic being known, to find the day of the month, etc. Let the thread and bead be first laid on the place of the Sun in the Ecliptic, and then moved to the line of 12. As if the place of the Sun giuen be the fourth of ♊, the bead being laid to this degree, and then moved to the hour of 12, in the Summer, the thread will fall on the 15 day of May, and the 9 of july; or if it be moved to the hour of 12 in the Winter, the thread will shall on the 6 of january and the 16 of November; which of these is the day of the month required, may appear by the quarter of the year. In this and the former propositions, you have two ways to rectify the bead, by the place of the Sun, and by the day of the month; the better way is by the place of the Sun, for in the other the Leap-yeare may breed some small difference. There is yet a third way. For the Seamen having a table for the declination on each day of the year, may set the bead thereto in the line of declination. 3 The hour of the day being given to find the altitude of the Sun above the horizon. The bead being set for the time by either of the three ways, let the thread be moved from the hour of 12 toward the line of declination, till the bead fall on the hour given; and the degrees which it cuts in the Quadrant, shall show the altitude of the Sun at that time. As if the time giuen be the tenth of April, the Sun being then in the beginning of ♉, the bead being rectified, you shall find the height at noon 50 gr. 0 m. at 11 in the morning 48 gr. 12 m. at 10 but 43 gr. 12 m. at 9 but 36 gr. at 8 but 27 gr. 30 m. at 7 but 18 gr. 18 m. at 6 but 9 gr. at 5 it meeteth with the line of declination, and hath no altitude at all, and therefore you may think it did rise much about that hour. Then if you move the thread again from the line of declination toward the hour of 12, you shall find that the Sun is 8 gr. 33 m. below the horizon at 4 in the morning, & near 16 gr. at 3, and 21 gr. 51 m. at 2, and 25 gr. 40 m. at 1, and 27 gr. at midnight. 4 The altitude of the Sun being given, to find the hour of the day. The altitude being observed as before, let the bead be set for the time, then bring the thread to the altitude, so the bead shall show the hour of the day. As if the 10 of April having set the bead for the time, you shall find by the quadrant, the altitude to be 36 gr. the bead at the same time will fall upon the houre-line of 9 and 3: wherefore the hour is 9 in the forenoon, or 3 in the afternoon. If the altitude be near 40 gr. you shall find the bead at the same time to fall half way between the houre-line of 9 and 3, and the houre-line of 10 and 2: wherefore it must be either half an hour past 9 in the morning, or half an hour past 2 in the afternoon; and which of these is the true time of the day, may be soon known by a second observation: for if the Sun rise higher, it is the forenoon; if it become lower, it is the afternoon. 5 The hour of the night being given, to find how much the Sun is below the horizon. The Sun is always so much below the horizon at any hour of the night, as his opposite point is above the horizon at the like hour of the day; and therefore the bead being set, if the question be made of any hour of the night in the Summer, then move it to the like hour of the day in the Winter; if of any hour of the night in Winter, then move it to the like hour of the day in Summer; so the degrees which the thread cutteth in the Quadrant shall show how much the Sun is below the horizon at that time. As if it be required to know how much the Sun is below the horizon the 10 of April at 4 of the clock in the morning; the bead being set to his place according to the time in the Summer hours, bring it to 4 of the clock in the afternoon in the Winter hours, and so shall you find the thread to cut 8 gr. and about 30 m. in the quadrant; and so much is the Sun below the horizon at that time. 6 The depression of the Sun supposed, to give the hour of the night with us, or the hour of the day to our Antipodes. Here also because the Sun is so much above the horizon at all hours of the day, as his opposite point is below the horizon at the like hour of the night; therefore first set the bead according to the time, then bring the thread to the degree of the Sun's depression below the horizon, so shall the bead fall on the contrary houre-lines, and there show the hour of the night in regard of us, which is the like hour of the day to our Antipodes. As if the 10 of April the Sun being then in the beginning of ♉, and by supposition 8 gr. 30 m. below the horizon in the East, it be required to know what time of the night it is; first set the bead according to the day in the Summer hours, then bring the thread to 8 gr. 30 m. in the quadrant, so shall the bead fall among the Winter hours, on the line of 4 of the clock in the afternoon: wherefore to our Antipodes it is 4 of the clock in their afternoon, and to us it is then 4 of the clock in the morning. 7 The time of the year or the place of the Sun being given, to find the beginning of daybreak, and end of twilight. This proposition differeth little from the former: for the day is said to begin to break, when the Sun cometh to be but 18 gr. below our horizon in the East, and twilight to end when it is gotten 18 gr. below the horizon in the West: wherefore let the bead be set for the time, and then bring the thread to 18 gr. in the quadrant, so shall the bead fall on the contrary houre-lines, and there show the hour of twilight as before. So if it be required to know at what time the day gins to break on the tenth of April, the Sun being then in the beginning of ♉; first set the bead according to the time in the Summer hours, and then bring the thread to 18 gr. in the quadrant, so shall the bead fall among the Winter hours a little more than a quarter before 3 in the morning; and that is the time when the day gins to break upon the tenth of April. CHAP. VII. Of the Horizon. THe Horizon is here represented by the ark drawn, from the beginning of declination towards the end of February, divided unequally, and numbered by 10. 20. 30. 40. 1 The day of the month, or the place of the Sun being known, to find the amplitude of the Sun's rising and setting. Let the bead rectified for the time, be brought to the horizon, and there it shall show the amplitude required. As if the day giuen be the 15 of May, the Sun being in the fourth degree of ♊, the bead rectified and brought to the horizon, shall there fall on 35 gr. 8 m. such is the amplitude of the Sun's rising from the East, and of his setting from the West; which amplitude is always Northward when the Sun is in the Northern Signs, and when he is in the Southward Signs always Southward. 2 The day of the month, or the place of the Sun being given, to find the ascensional difference. Let the bead rectified for the time, be brought to the horizon, so the degrees cut by the thread in the quadrant, shall show the difference of ascensions. As if the day giuen be the 15 of May, the Sun being in the fourth degree of ♊, let the bead be rectified and brought to the horizon; so shall the thread in the quadrant show the ascensional difference to be 28 gr. and about 50 m. Upon the ascensional difference depends this Corollary. To find the hour of the rising and setting of the Sun, and thereby the length of the day and night. The time of the Sun's rising may be guessed at by the 3 of the last Cap. but here by the ascensional difference it may be better found, and that to a minute of time. For if the ascensional difference be converted into time, allowing an hour for 15 gr. and 4 minutes of an hour for each degree, it showeth how long the Sun riseth before six of the clock in the Summer, and after six in the Winter. As if the day giuen be the 15 of May, the Sun being in the fourth of ♊, and his ascensional difference found as before 28 gr. 50 m; this converted into time, maketh 1 ho. and somewhat more than 55 m. of an hour: wherefore the Sun at that time, in regard it was Summer, rose 1 ho. and full 55 m. before 6 of the clock; and so having the quantity of the semidiurnal ark, the length of the day and night need not be unknown. CHAP. VIII. Of the five Stars. I Might have put in more Stars, but these may suffice for the finding of the hour of the night at all times of the year: and first I make choice of Ala Pegasi, a star in the extremity of the wing of Pegasus, in regard it wants but 6 minutes of time of the beginning of ♈; but because it is but of the second magnitude, and not always to be seen, I made choice of four more, one for each quarter of the Ecliptic, as of Oculus ♉ the Bull's eye, whose right ascension converted into time, is 4 ho. 15 m; then of Cor ♌ the Lion's heart, whose right ascension is 9 ho. 48 m; next of Arcturus, whose right ascension is 13 H. 58 m; and lastly of Aquila, or the Vulture's heart, whose right ascension is 19 H. 33 m. These five stars have all of them Northern declination; and if any others, some of these will be seen at all times of the year. The use of them is, The altitude of any of these five Stars being known, to find the hour of the night. First put the bead to the star which you intent to observe, take his altitude, and find how many hours he is from the meridian by the fourth Prop. of the sixth Chap; then out of the right ascension of the star, take the right ascension of the Sun converted into hours, and mark the difference; for this difference being added to the observed hour of the star from the meridian, shall show how many hours the Sun is gone from the meridian, which is in effect the hour of the night. As if the 15 of May, the Sun being in the fourth of ♊, I should set the bead to Arcturus, and observing his altitude should find him to be in the West about 52 gr. high, and the bead to fall on the houre-line of 2 afternoon, the hour would be 11 bo. 50 m. past noon, or 10 m. short of midnight. For 62 gr. the right ascension of the Sun, converted into time, makes 4 ho. 8 m. which if we take out of 13 ho. 58 m. the right ascension of Arcturus, the difference will be 9 ho. 50 m. and this being added to 2 ho. the observed distance of Arcturus from the meridian, shows the hour of the night to be 11 ho. 50 m. Another example will make all more plain. If the 9 of july the Sun being then in 26 gr. of ♋, I should set the bead to Oculus ♉, and observing his altitude should find him to be in the East about 12 gr. high, and the bead to fall on the houre-line of 6 before noon, which is 18 ho. past the meridian, the hour of the night would be better than a quarter past 2 of the clock in the morning. For 118 gr. the right ascension of the Sun, converted into time, makes 7 ho. 52 m; this taken out of 4 ho. 15 m. the right ascension of Oculus ♉, adding a whole circle, (for otherwise there could be no substraction) the difference will be 20 ho. 23 m. and this being added to 18 ho. which was the observed distance of Oculus ♉ from the meridian, shows that the Sun (abating 24 ho. for the whole circle) is 14 gr. 23 m. past the meridian, and therefore 23 m. past 2 of the clock in the morning. CHAP. IX. Of the Azimuth lines. THose lines which are drawn between the equator and the tropiques, on that side of the quadrant which is nearest unto the sights, and are numbered by 10. 20. 30. etc. do represent the azimuths, the uttermost to the left hand representeth the meridian, that which is numbered with 10 the tenth azimuth from the meridian, and that which is numbered with 20 the twentith, and so the rest. Those lines which are drawn from the equator to the left hand, do show the azimuth in the Summer; and those other to the right hand, do show the same in the Winter. The use of them is, 1 The azimuth whereon the Sun beareth from us being known, to find the altitude of the Sun above the horizon. First let the bead be set for the time, as in the former Chapter, then move the thread until the bead fall on the azimuth; so the degrees which the thread cutteth in the quadrant, shall show the altitude of the Sun at that time. Where you are to observe, that seeing the azimuths are drawn on the right side of the quadrant, you are also to begin to number the degrees of the Sun's altitude from the right hand toward the left. As if the sights had been set on the line AB, and you had turned your right hand towards the Sun in observing of his altitude, contrary to our practice in the former Chapter. As if the time given were the 2 of August, when the Sun hath about 15 gr. of North declination, you may set the bead for the time, so you shall find the height at noon when the Sun is in the South, to be 53 gr. 30 m. when he is 10 gr. from the South 53 gr. 10 m. when 20 gr. then about 52 gr. 8 m. when 30 gr. then 50 gr. 20 m. when 40 gr. then 47 gr. 48 m. when 50 gr. then 44 gr. 12 m. when 60 gr. then 39 gr. 35 m. when 70 gr. then 33 gr. 50 m. when 80 gr. then 27 gr. when he is in the East or West 90 gr. from the meridian, then is the height near 19 gr. 20 m; when he comes to be 100 gr. then 11 gr. 15 m. when 110 gr. then 3 gr. 20 m; and before he cometh to the azimuth of 120 gr. he hath no altitude. For the Sun having 15 gr. of North declination, will rise and set at 114 gr. 34 m. from the meridian. 2 The altitude of the Sun being given, to find on what azimuth he beareth from us. Let the bead be set for the time, and the altitude observed as before; then bring the thread to the compliment of that altitude, so the bead shall show the azimuth required. As if the second of August, having set the bead for the time, you shall find the altitude of the Sun to be 19 gr. 20 m. remove the thread unto 70 gr. 40 m. the compliment of the altitude; or, which is all one, to 19 gr. 20 m. from the right hand toward the left, and the bead will fall on the line of 90 gr. from the meridian. And therefore the point whereon the Sun beareth from us, is one of these two, either due East or due West. And which of these is the true point of the compass, may be soon known by a second observation: for if the Sun rise higher, it is the forenoon; if it be lower, it is the afternoon. By knowing the azimuth or point of the compass whereon the Sun beareth from us, it is easy to find, A meridian line; and thereby The coasting of the Country. The site of a building. The variation of the Compass. As if the second of August in the afternoon, I should find by the height of the Sun that he bears from me 60 gr. from the meridian toward the West; then there being 90 gr. belonging to each quarter, the West will be 30 gr. to the right hand, the East is opposite to the West, the North and South lie equally between them. CHAP. X. Of the Quadrat. THe Quadrat hath two sides divided, the other two sides next the Centre may be supposed to be divided, each of them into 100 equal parts: of the sides divided, that which is next the horizontal line contains the parts of right shadow, the other next the sights, the parts of contrary shadow. The use of the Quadrat is, 1 Any point being given, to find whether it be level with the eye. Lift up the centre of the quadrant, so as the thread with the plummet may play easily by the side of it; then look through the sights to the place given: for now if the thread shall fall on AB the horizontal line, then is the place given levelly with the eye: but if it shall fall within the said line on any of the divisions, than it is higher; if without, than it is lower than the level of the eye. 2 To find an height above the level of the eye, or a distance at one observation. Look through the sights to the place, going nearer or farther from it, till the thread fall on 100 parts in the quadrat or 45 gr. in the quadrant, so shall the height of the place above the level of the eye, be equal to the distance between the place and the eye. geometric illustration measuring shadow cast by a tower And on the contrary, As the parts cut by the thread are to 100: So the height unto the distance. But when the thread shall fall on the parts of contrary shadow: if it fall on 50 parts, the height is double unto the distance; if on 25, it is four times as the distance. For as oft as the thread falleth on the parts of contrary shadow, As the parts cut by the thread are unto 100: So is the distance unto the height. And on the contrary, As 100 are unto the parts cut by the thread: So is the height unto the distance. And what is here said of the height and distance, the same may be understood of the height and shadow. 3 To find a height or a distance at two observations. As if the place which is to be measured might not otherwise be approached, & yet it were required to find the height BC, and the distance: first if I make choice of a station at A, where the thread may fall on 100 parts in the quadrat, and 45 gr. in the quadrant, the distance AB will be equal to the height BC; then if I go farther in a direct line with the former distance, and make choice of a second station at D, where the thread may fall on 50 parts of right shadow, the distance BD would be double to the height BC: wherefore I may measure the difference between the two stations A and D, and this difference AD will be equal both to the distance AB and the height BC. Or if I cannot make choice of such stations, I take such as I may, one at D, where the thread falleth at 50 parts of right shadow; the second at E, where it falleth on 40 parts: and supposing the height BC to be 100, I find that As 50 parts are unto 100, the side of the quadrat: So 100 the supposed height, unto 200 the distance BD. And as 40 parts, at the second station, unto 100: So 100 the supposed height, unto 250 the distance BE. Wherefore the difference between the stations D & E should seem to be 50; and than if in measuring of it, I should find it to be either more or less, the proportion will hold, as from the supposed difference to the measured difference, so from height to height, and from distance to distance. As if the difference between the two stations D and E being measured, were found to be 30. As 50 the supposed difference, unto 30 the true difference: So 100 the supposed height, unto 60 the true height. And 200 the supposed distance, unto 120 the true distance: And 250 at the second station, unto 150 the distance BE. The like reason holdeth in all other examples of this kind: and if an Index with sights were fitted to turn upon the Centre, it might then serve by the same reason for the finding of all other distances. FINIS.