CO which is also equal to PI. 13. Theor. To divide a right line in two parts so that the right angled figure made of the whole line and one part shall be equal to the square of the other part The right line given is AB upon the same line AB make a square as ABCD and divide the side AD in two equal parts the midst is M from M draw a line to B and produce AD to H so that MH be equal to MB and upon AH make a square as AHGF. Then extend GF to E and then is the right angled figure FC being made of the whole line FE which is equal to AB and the part BF equall to the square of the other part AF that is to the square AHGF. Forasmuch as by the last aforegoing the right angled figure comprehended of HD and HA or the right angled figure of HD and HG as the figure GHED with the square of AM are together equal to the square of HM being equall to BM it followeth that if we take away the square â—f AM common to both that the square â—f AB that is the square ABCD is equal â—o the right angled figure HGED and the â—ommon right angled figure AE being taken from them both there shall remain the right angled figure FC equal to the square â—HFG which was to be proved 14 Theor. To divide a right line given by extream and mean proportion A right line is said to be divided by an extream and mean proportion when the whole is to the greater part as the greater is to the lesse And thus a right line being divided as the right line AB is divided in the preceding Diagram in the point F it is divided in extream and mean proportion that is As AB is to AF so is AF to BF Demonstration Forasmuch as the right lined figure included with AB and FB as the figure FBCE is equal to the square of AF that is to the square AFGH it followeth by the eleventh Theorem of this Chapter that the line AB is divided in extream and mean proportion that is As AB is to AF So is AF to FB 15 Theor. In all plain Triangles a line drawn parallel to any of the sides cutteth the other two sides proportionally As in the plain Triangle ABC KL being parallel to the base BC it cutteth off from the side AC one fourth and also it cutteth off from the side AB one third part the reason is because the right line EH cutteth off one third part from the whole space DGFB therefore it cutteth off one third part from all the lines that are drawn quite through that space And hereupon parallel lines bounded with parallels are equal as the parallels ED and GH being bounded with the parallels DG and HE are equal for since the whole lines DB and GF are equall DE and GH being one fourth part thereof must needs be equal also 16 Theor. Equiangled Triangles have their sides about the equall angles proportionall and contrarily Let ABC and ADE be two plain equiangled Triangles so as the angles at B and D at A and A and also at C and E be equal one to the other I say their sides about the equal angles are proportionall that is 1 As AB is to BC So is AD to ED. 2 As AB is to AC So is AD to AE 3 As AC is to CB So is AE to ED. Demonstration Because the angles BAC and DAE are equal by the Proposition therefore if A + B be applied to AD AC shall fall in AE and by such application is this figure made In which because that AB and AD do meet together and also that the angles at B and D are equall by the Proposition therefore the other sides BC and DE are parallel and by the last aforegoing BC cutteth the sides AD and AE proportionally and therefore As AB to AD So is AC to AE Moreover by the point B let there be drawn the right line BF parallel to the base AE and it shall cut the other two sides proportionally in the points B and F and therefore 1. As AB to AD so is EF to ED Or thus As AB to AD so is CB to ED because that FE and BC are equal by the last aforegoing 1. Theor. In all right angled plain Triangles the sides including the right angle are equal to the the third side In the right angled plain triangle ABC right angled at B the sides AB and BC are equal in power to the third side AC that is the squares of the sides AB and BC to wit the squares ALMB and BEDC added together are equal to the square of the side AC that is to the square ACKI Demonstration 18. Theor. The three angles of a right lined Triangle are equal to two right angles As in the following plain Triangle ABC the three angles ABC ACB and CAB are equal to two right angles Let the side AB be extended to D and let there be a semicircle drawn upon the point B and let there be also dawn a line parallel unto AC from B unto G. Demonstration I say that the angle GBD is equal to the angle BAC by the 9 th hereof and the angle CBG is equal to the angle ACB by the same reason and the angles CBG and GBD are together equal to the angle CBD which is also equal to the angle ABC by the 18 th of the first and therefore the three angles of a right lined Triangle are equal to two right angles which was to be proved 19. Theor. If a plain Triangle be inscribed in a Circle the angles opposite to the circumference are halfe as much as that part of the Circumference which is opposite to the angles As if in the circle ABC the circumference BC be 120 degrees then the angle BAC which is opposite to that circumference shall be 60 degrees The reason is because the whole circle ABC is 360 degrees and the three angles of a plain triangle cannot exceed 180 degrees or two right angles by the last aforegoing therefore as every arch is the one third of 360 so every angle opposite to that arch is the one third of 180 that is 60 degrees Or thus From the angle ABC let there be drawn the diameter BED and from the center E to the circumference let there be drawn the two Radii or semidiameters AE and AC I say then that the divided angles ABD and DBC are the one halfe of the angles AED and DEC for the angles ABE and BAE are equall because their Radii AE and EB are equall and also the angle AED is equal to the angles ABE and BAE added together for if you draw the line EF parallel to AB the angle FED shall be equal to the angle ABE by the 9 th hereof and by the like reason the angle AEF is also equal to the angle BAE and therefore the angle AED is equal to the
base is to the summe of the sides So is the difference of the sides to the difference of the segments of the base Let BCD be the triangle CD the base BD the shortest side upon the point B describe the circle ADFH making BD the Radius thereof let the side BC be produced to A then is CA the summe of the sides because BA and BD are equal by the work CH is the difference of the sides CF the difference of the segments of the base Now if you draw the right lines AF and HD the triangles CHD and CAF shall be equiangled because of their common angle ACF or HCD and their equal angles CAF and HDC which are equall because the arch HF is the double measure to them both and therefore as CD to CA so is CH to CF which was to be proved Consectary Therefore the three sides of a plain oblique angled triangle being given the reason of the angles is also given For first the obliquangled triangle may be resolved into two right angled triangles by this Axiome and then the right angled triangles may be resolved by the first Axiome As in the plain oblique angled triangle BCD let the three sides be given BD 189 paces BC 156 paces and DC 75 paces and let the angle CBA be required First by this Axiome I resolve it into two right angled triangles thus As the true base BD 189 co ar 7.7235382 Is to the sum of BC DC 231 2.3636120 So the difference of BC DC 81 1.9084850 To the alternate base BG 99 1.9956352 As the the hypothenusal BC is to Radius So is the base AB 144 to the sine of the angle at the perpendicular whose complement is the angle at the base inquired In like manner may be found the angle at D and then the angle BCD is found by consequence being the complement of the other two to two right angles or 180 degrees CHAP. VII Of Sphericall Triangles A Sphericall Triangle is a figure described upon a Sphericall or round superficies consisting of three arches of the greatest circles that can be described upon it every one being lesse then a Semicircle 2. The greatest circles of a round or Spherical superficies are those which divide the whole Sphere equally into two Hemispheres and are every where distant from their own centers by a Quadrant or fourth part of a great circle 3. A great circle of the Sphere passing through the poles or centers of another great circle cut one another at right angles 4. A spherical angle is measured by the arch of a great circle described from the angular point betwixt the sides of the triangle those sides being continued to quadrants 5. The sides of a Spherical triangle may be turned into angles and the angles into sides the complements of the greatest side or greatest angle to a Semicircle being taken in each conversion It will be necessary to demonstrate this which is of so frequent use in Trigonometry In the annexed Diagram let ABC be a sphericall triangle obtuse angled at B let DE be the measure of the angle at A. Let FG be the measure of the acute angle at B which is the complement of the obtuse angle B being the greatest angle in the given triangle and let HI be the measure of the angle at C KL is equal to the arch DE because KD and LE are Quadrants and their common complement is LD LM is equall to the arch FG because LG and FM are Quadrants and their common complement is LF KM is equal to the arch HI because KI and MH are Quadrants and their common complement is KH Therefore the sides of the triangle KLM are equal to the angles of the triangle ABC taking for the greatest angle ABC the complement thereof FBG And by the like reason it may be demonstrated that the sides of the triangle ABC are equal to the angles of the triangle KLM For the side AC is equall to the arch DI being the measure of the angle DKI which is the complement of the obtuse angle MKL. The side AB is equall to the arch OP being the measure of the angle MLK And lastly the side BC is equal to the arch FH being the measure of the angle LMK for AD and CI are Quadrants so are AP and OB BF and CH. And CD AO and CF are the common complements of two of those arches Therefore the sides of a spherical triangle may be changed into angles and the angles into sides which was to be demonstrated 6. The three sides of any spherical triangle are lesse then two Semicircles 7. The three angles of a spherical triangle are greater then two right angles and therefore two angles being known the third is not known by consequence as in plain triangles 8. If a spherical triangle have one or more right angles it is called a right angled spherical triangle 9. If a spherical triangle have one or more of his sides quadrants it is called a quadrantal triangle 10. If it have neither right angle nor any side a quadrant it is called an oblique spherical triangle 11. Two oblique angles of a spherical triangle are either of them of the same kinde of which their opposite sides are 12. If any angle of a triangle be neerer to a quadrant then his opposite side two sides of that triangle shall be of one kinde and the third lesse then a quadrant 13. But if any side of a triangle be nearer to a quadrant then his opposite angle two angles of that triangle shall be of one kinde and the third greater then a quadrant 14. If a spherical triangle be both right angled and quadrantal the sides thereof are equall to the opposite angles For if it have three right angles the three sides are quadrants if it have two right angles the two sides subtending them are quadrants if it have one right angle and one side a quadrant it hath two right angles and two quadrantal sides as is evident by the third Proposition But if two sides be quadrants the third measureth their contained angle by the fourth proposition Therefore for the solution of these kindes of triangles there needs no further rule But for the solution of right angled quadrantall and oblique spherical triangles there are other affections proper to them which are necessary to be known as well as these general affections common to all spherical triangles The affections proper to right angled and quadrantal triangles we will speak of first CHAP. VIII Of the affections of right angled Sphericall Triangles IN all spherical rectangled Triangles having the same acute angle at the base The sines of the hypothenusals are proportional to the sines of their perpēdiculars As in the annexed diagram let ADB represent a spherical triangle right angled at B so that AD is the sine of the hypothenusal AB the sine of the base and DB is the perpendicular Then is DAB the angle at the base and IH the sine
line TP being the difference of the two given tangents CG and BP is double to the right line BK being the tangent of the difference of the two given arches or which is all one I say that the right line TP is equal to the right line MK Demonstration Then that the right line MT is equall to the right line KA is thus proved the right line MA is equall to the right line KA by the work but the right line MT is equal to the right line MA and therefore it is also equall to the right line KA That the right line MT is equal to the right line MA doth thus appear for that the angles MAT and MTA are equall and therefore the sides opposite unto them are equal for equall sides subtend equall angles and the angles MTA and MAT are equal because the angle MTA is equal to the angle TAC by the like reason that the angle KPA is equal to the angle DAC and the angle MAT is equall to the angle TAC by the proposition for the arches CS and SO are put to be equal therefore it followes that they are also equal one to another Generally therefore the difference of the tangents of two arches making a Quadrant is double to the tangent of the difference of those arches which was to be demonstrated And by consequence the tangents of two arches being given making a Quadrant the tangent of the difference of those arches is also given And contrarily the tangent of the difference of those two arches being given together with the tangent of one of the arches the tangent of the other arch is also given Example Let there be given the Tang. of 72 de 94 m. And the Tang. of its complement that is of 17 6 Halfe the difference of these two arches is 27 94 Tangent of 72 de 94 m. is 32586438 Tangent of 17 6 306â—761 Their difference is 29517677 The halfe whereof is 14758838 The Tangent of 55 de 88 min. Or let the tangent of the greater arch 72 d. 94 m. be given with the Tangent of the difference 55 de 88 m. and let the lesser arch 17 de 6 m. be demanded Tangent of 72 de 94 m. is 32586438 Tang. of 55 de 88 m. doubled is 29517676 Â Â Their difference is 03068762 The Tangent of 17 de 6 m. Or lastly let the lesser arch be given with the Tangent of the difference and let the greater arch be demanded Tang. of 55 de 88 m. the diff is 14758838 Â Â Which doubled is 29517676 To which the tang of 17 d. 6 m. ad 3068761 Their aggregate is 32586437 the tangent of 72 degrees 94 minutes Theor. 2. The tangent of the difference of two arches making a Quadrant with the tangent of the lesser arch maketh the secant of the difference The Reason is Because the tangent of the difference BL or BO that is the right line BK or BM with the tangent of the lesser arch BS that is with the right line BT maketh the right line MT which is equall to the Secant AK by the demonstration of the first Theorem Therefore the tangent of the difference of two arches making a Quadrant and the tangent of the lesser arch being given the secant of the difference is also given And contrarily For example Let the tangent of the former difference 55 degrees 88 minutes and the tangent of the lesser arch 17 degrees â—â— minutes be given I say the secant of this difference is also given Tang. of the diff 55 de 88 m. is 14758838 The tangent of 17 06 is 3068762 Â Â Their sum is the secant of 55 88 17827600 Theor. 3. The tangent of the difference of two arches making a Quadrant with the secant of their difference is equal to the tangent of the greater arch Because the tangent of the arch BL being the difference of the two arches BC and DC making a Quadrant with the secant of the same arch BL that is the right line BK with the right line AK is equal to the right line BP by the demonstration of the first Theorem therefore the tangent of the difference of two arches making a Quadrant being given with the secant of their difference the tangent of the greater arch is also given For example Let the tangent of the difference be the tang of the arch of 55 de 88 m. viz. Â 14758838 The secant of this difference is 17827600 Their sum is the tang of 72 94 32586438 the greater of the two former given arches And now by the like reason these Rules may be added by way of Appendix Rule I. The double tangent of an arch with the tangent of half the complement is equall to the tangent of the arch composed of the arch given and half the complement thereof For if the arch BL be put for the arch given the double tangent thereof shall be TP by the demonstration of the first Theorem And the complement of the arch BL shall be the arch LC whose half is the arch LD or DC whose tangent is the right line GC or BT but TP added to BT maketh BP being the tangent of the arch BD composed of the given arch BL and half the complement LD therefore the double tangent c. Rule II. The tangent of an arch with the tangent of half the complement is equal to the secant of that arch For if you have the arch BL or BO for the arch given the tangent of the arch given shall be BM the tangent of half the complement shall be BT which two tangents added together make the right line MT but the right line MT is equal to the right line AK by the demonstration of the first Theorem which right line AK is the secant of the arch given BL by the proposition Therefore the tangent of an arch c. Rule III. The tangent of an arch with the secant thereof is equal to the tangent of an arch composed of the arch given and half the complement For if you have the arch BL for the arch given BK shall be the tangent and AK the secant of that arch But the right line AK and KP are equal by the demonstration of the first Theorem therefore the tangent of the arch given BL that is the right line BK with the secant of the same arch that is AK is equall to the right line BP which is the tangent of the arch BD being composed of the given arch BL and LD being half the complement These rules are sufficient for the making of the Tables of natural Sines Tangents Secants The use whereof in the resolution of plain spherical triangles should now folow but because the Right Honourable John Lord Nepoir Baron of Marchiston hath taught us how by borrowed numbers called Logarithmes to perform the same after a more easie and compendious way we will first speak
like triangles by the 22 of the second and their sides proportional that is AB  AB BC  AO    AC  DE CF  CF And therefore the oblongs of BC×AC AO × DE and AB × CF are equal and the sides of equal rectangled figures reciprocally proportional that is as BC AO ∷ DE AC or as AO BC ∷ AC DE. If therefore you multiply AO the half Radius by DE the sine of the arch given and divide the product by BC the sine of half the arch given the quotient shall be AC the sine complement of half the given arch Or if you multiply BC the sine of an arch by AC the sine complement of the same arch and divide the product by AO the half Radius the quotient shall be DE the sine of the double arch And therefore the sines of 45 degrees being given or the Logarithmes of those sines the rest may be found by the rule of proportion For illustration sake we will adde an example in naturall and artificiall numbers Naturall As BC 28 46947 Is to AO 30 50000 So is DE 56 82903 To AC 62 88294 Logarith As BC 28 9.671609   Is to AO 30 9.698970 So is DE 56 9.918574   To AC 62. 9.945935 18. § The composition of the naturall Tangents and Secants by the first and second of the fourth are thus to be made 1. As the sine of the complement is to the sine of an arch So is the Radius to the tangent of that arch 2. As the sine of the complement is to the Radius so is the Radius to the Secant of that arch and by the same rules may be also made the artificiall but with more ease as by example it will appear Let the tangent of 30 degrees be sought  Logarith As the co-sine of 60 degrees 9.937531   Is to the sine of 30 9.698970 So is the Radius 10.000000   To the tangent of 30 9.761439 And thus having made the artificiall Tangents of 45 degrees the other 45 are but the arithmeticall complements of the former taken as hath been shewed in the eighth rule of the fifth Chapter Again let the secant of 30 degrees be sought As the co-sine of 60 degrees 9.937531   Is to the Radius 10.000000 So is the Radius 10.000000    20.000000   To the secant of 30 10.062469 And thus the Radius being added to the arithmetical complement of the sine of an arch their aggregate is the secant of the complement of that arch And this is sufficient for the construction of the naturall and artificiall Canon How to finde the Sine Tangent or Secant of any arch given in the Canon herewith printed shall be shewen in the Preface thereunto here followeth the use of the naturall and artificiall numbers both first in the resolving any Triangle and then in Astronomy Dialling and Navigation CHAP. VI. The use of the Tables of natural and artificial Sines and Tangents and the Table of Logarithmes In the Dimension I. Of plain right angled Triangles THe measuring or resolving of Triangles is the finding out of the unknown sides or angles thereof by three things known whether angles or sides or both and this by the help of that precious gemme in Arithmetick for the excellency thereof called the Golden Rule which teacheth of four numbers proportional one to another any three of them being given to finde out a fourth and also of these Tables aforesaid Of Triangles as hath been said there are two sorts plain and sphericall A triangle upon a plain is right lined upon the Sphere circular Right lined Triangles are right angled or oblique A right angled right-lined Triangle we speak of first whose sides then related to a circle are inscribed totally or partially Totally if the side subtending the right angle be made the Radius of a Circle and then all the sides are called Sines as in the Triangle ABC Partially if either of the sides adjacent to the right angle be made the Radius of a circle and then one side of the Triangle is the Radius or whole Sine the shorter of the other two sides is a Tangent and the longest a secant Now according as the right angled Triangle is supposed whether to be totally or but partially inscribed in a circle so is the trouble of finding the parts unknown more or lesse whether sides or angles for if the triangle be supposed to be totally inscribed in a circle we are in the solution thereof confined to the Table of Sines onely because all the sides of such a triangle are sines but if the triangle be supposed to be but partially inscribed in a circle we are left at liberty to use the Table of Sines Tangents or Secants as we shall finde to be most convenient for the work In a right angled plain Triangle either all the angles with one side are given and the other two sides are demanded I say all the angles because one of the acute angles being given the other is given also by conâ—â—quence Or else two sides with one angle that is the right angle are given and the other two angles with the third side are demanded In both which cases this Axiome following is well nigh sufficient The first AXIOME In all plain Triangles the sides are in portion one to another as are the sines oâ— the angles opposite to those sides As in the triangle ABC the side AB is in proportion to the side AC as the sine of the angle at B is in proportion to the sine of the angle at c and so of the rest Demonstration The circle ADF being circumscribed about the Triangle ABC the side AB is made the chord or subtense of the angle ACB that is of the arch AB which is opposite to the angle ACB The side AC is made the subtense of the angle ABC and the side BC is made the subtense of the angle BAC and are the double measures thereof by the 19 Theorem of the second Chapter therefore the side AB is in proportion to the side AC as the subtense of the angle ACB is in proportion to the subtense of the angle ABC but half the subtense of the angle ACB is the sine of the angle ACB and half the subtense of the angle ABC is the sine of the angle ABC now as the whole is to the whole so is the half to the half Therefore in all plain Triangles c. The first Consectary The angles of a plain triangle and one side being given the reason of the other sides is also given The second Consectary Two sides of a plain Triangle with an angle opposite to one of them being given the reason of the other angles is also given by this proportion If the side of a Triangle be required put the angle opposite to the given side in the first place If an
proportion to his opposite side the base So is Radius to his opposite side the hypothenusal and thus you see that the hypothenusal may be found without the trouble of squaring the sides and thence extracting the square root And hence also all the cases of a right angled plain triangle may be resolved several wayes that is to say 1. In a plain right angled triangle the angles and one side being given every of the other sides is given by a threefold proportion that is as you shall put for the Radius either the side subtending the right angle or the greater or lesser side including the right angle 2. Any of the two sides being given either of the acute angles is given by a double proportion that is as you shall put either this or that side for the Radius to make this clear we will first set down the grounds or reasons for varying of the termes of proportion and then the proportions themselves in every case according to all the variations The reasons for varying of the termes of proportion are chiefly three The first reason is because the Radius of a circle doth bear a threefold proportion to a sine tangent or secant and contrariwise a sine tangent or secant hath a threefold proportion to Radius by the second Axiome of this Chapter For As sine BC to Rad. AC in the 1. triangle So Rad. BC to secant AC in the 3d. tri So tang BC to secant AC in the 2d tri contra Again As tang BC to Rad. AB in the 2d triang So Rad. BC to tang AB in the 3d. trian So sine BC to sine BA in the first triang contra Lastly As secant AC to Rad. BC in the 3d. tri So Rad. AC to sine BC in the first trian So secant AC to tang BC in the 2d tri contra Hence then As the sine of an arch or ang is to Rad. So Rad. to the secant comp of that arch so is the tang of that arch to his sec contr Also As the tang of an arch or ang is to Rad. So is Rad. to the tangent compl thereof And so is the sine thereof to the sine of its complement contra Lastly As the secant of an arch or ang to Rad. So is Radius to the sine compl thereof And so is secant complement to tangent complement thereof contra Example Let there be given the angle at the perpendicular 41 degrees 60 minutes and the base 768 paces to finde the perpendicular First by the natural numbers As the secant of BAC 41 d. 60m 13372593 Â Â Is to Radius 10000000 So is the base AB 768 Â Â To the perpendicular BC â—74 574 By the Artificiall As the secant of BAC 41.60 10.1262157 Â Â Is to Radius 10.0000000 So is the base 768 2.8853612 Â Â Â 12.8853612 To the perpendicular 574 2.7591455 Secondly by the naturall numbers As the Radius 10000000 To the co-sine of BAC 41.60 7477981 So is the base AB 768 To the perpendicular BC 574 By the Artificiall As the Radius 10.0000000 To the co-sine of BAC 41.60 9.8737843 So is the base AB 768 2.8853612 To the perpendicular BC 574 2.7591455 Thirdly by the natural numbers As the co-secant of BAC 41.60 15061915 Is to the co-tang of BAC 41.60 11263271 So is the base AB 768 To the perpendicular BC 574 By the artificiall As the co-secant of BAC 41.60 10. 1778802 Is to the co-tang of BAC 41.60 10.0516645 So is the base AB 768 2.8853612 To the perpendicular BC 574 2.7591455 COROLLARY Hence it is evident that Radius is a mean proportional between the sine of an arch and the secant complement of the same arch also between the tangent of an arch and the tangent of the complement of the same arch The second Reason The sines of several arches and the secants of their complements are reciprocally proportional that is As the sine of an arch or angle is to the sine of another arch or angle So is the secant of the complement of that other to the co-secant of the former For by the foregoing Corollary Radius is the mean proportional between the sine of any arch and the co-secant of the same arch Therefore whatsoever sine is multiplied by the secant of the complement is equall to the square of Radius so that all rectangles made of the sines of arches and of the secants of their complements are equal one to another but equall rectangles have their sides reciprocally pro portional by the tenth Theorem of the second Chapter Therefore the sines of several arches c. The third Reason The tangents of severall arches and the tangents of their complements are reciprocally proportional that is As the tangent of an arch or angle is to the tangent of another arch or angle so is the co-tangent of that other to the co-tangent of the former For by the foregoing Corollary Radius is the mean proportionall between the tangent of every arch and the tangent of his complement Therefore the Rectangle made of any tangent and of the tangent of his complement is equall to the square of Radius so that all rectangles made of the tangents of arches and of the tangents of their complements are equall one to another but equal rectangles c. as before To these three reasons a fourth may be added For in the rule of proportion wherein there are alwayes four termes three given the fourth demanded It is all one whether of the two middle terms is put in the second or third place For it is all one whether I shall say As 2 to 4 so 5 to 10 or say as 2 to 5 so 4 to 10 and from hence every example in any triangle may be varied and thus you see the reasons of varying the termes of proportion we come now to shew you the various proportions themselves of the severall Cases in right angled plain triangles Right angled plain triangles may be distinguished into seven Cases whereof those in which a side is required viz. three may be found by a triple proportion and those in which an angle is required viz. three may be found by a double proportion CASE 1. The angles and base given to finde the perpendicular First As sine the angle at the perpendicular is to the base so is sine the angle at the base to the perpendicular Or secondly thus As Radius to the base so tangent the angle at the base to the perpendicular Or thirdly thus As the tangent of the angle at the perpendicular is to the base so is Radius to the perpendicular CASE 2. The angles and base given to finde the hypothenusal First As the sine of the angle at the perpendicular is to the base so is Radius to the hypothenusal Or secondly thus As Radius is to the base so the secant of the angle at the base to
and LM the tangent thereof Also DF is the sine of the perpendicular DB and KB is the tangent thereof I say then As AD is to FD So is AI to IH by the 16 th Theoreme of the second Chapter And because it is all one whether of the mean proportionals be put in the second place therefore I may say As AD the sine of the hypothenusal is in proportion to AI Radius So is FD the sine of the perpendicular to IH the sine of the angle at the base 2. In all rectangled spherical triangles having the same acute angle at the base The sines of the bases and the tangents of the perpendiculars are proportional For as AB to KB so is AM to ML by the 16 th Theorem of the second Chapter or which is all one As AB the sine of the base is in proportion to AM Radius so is BK the tangent of the perpendicular to ML the tangent of the angle at the base 3. If â— circles of the Sphere be so ordered that the first intersect the second the second the third the third the fourth the fourth the fift and the fift the fift at right angles the right angled triangles made by their intersections do all consist of the same circular parts As in this Scheme let IGAB be the first circle BLF the second FEC the third GAD the fourth HLEI the fift Then do these five circles retain the conditions required The first intersecting the second in B the second the third in F the third the fourth in C the fourth the fift in H the fift the angle we mark or note intersections at B F â— to a quadrant As angles therefore I sayâ—nt as the completriangles made by the inteâ—â—or AD we write circles namely ABD Dâ— write compl EGI and GCA do all coâ—â—d AB besame circular parts for the circuâ—â—◠〈◊〉 in every of these triangles are as hâ—â—d by peareth In ABD are AB BD c BDA c AD c DAâ— DHL c HLD c LD c LDH DH HL LFE coÌ„ ELF LF FE coÌ„ FEL c EL EGI IG coÌ„ IGE c GE coÌ„ GEI IE GCA c GA c AGC GC CA c CAG Where you may observe that the side AB in the first triangle is equal to compl HLD in the second or compl ELF in the third or IG in the fourth or com GA in the fift and so of the rest To expresse this more plainly AB in the first triangle is the complement of the angle HLD in the second or the complement of the angle ELF in the third or the side IG in the fourth or the complement of the hypothenusal GA in the fift And from these premises is deduced this universall proposition 4. The sine of the middle part and Radius are reciprocally proportional with the tangents of the extreams conjunct and with the co-sines of the extreams disjunct Namely As the Radius to the tangent of one of the extreames conjoyned so is tangent of the other extream conjoyned to the sine of the middle part And also As the Radius to the co-sine of one of the extreams dis-joyned so the co-sine of the other extream dis-joyned to the sine of the middle part Therefore if the middle part be sought the Radius must be in the first place if either of the extreams the other extream must be in the first place For the better Demonstration hereof it is first to be understood that a right angled Spherical Triangle hath five parts besides the right angle As the triangle ABD in the former Diagram right angled at B hath first the side AB secondly the angle at A thirdly the hypothenusal AD fourthly the angle ADB fifthly the side DB. Three of these parts which are farthest from the right angle we mark or noâ—e by their complements to a quadrant As the angle BAD we account as the complement to the same angle For AD we write comp AD and for ADB we write compl ADB But the two sides DB and AB being next to the right angle 〈…〉 are not noted by their complements Of these five parts two are alwayes given to finde a third and of these three one is in the middle and the other two are extreams either adjacent to that middle one or opposite to it If the parts given and required are all conjoyned together the middle is the middle part conjunct and the extreams the extream parts conjunct If again any of the parts given or required be dis-joyned that which stands by it self is the middle part dis-joyned and the extreames are extream parts dis-joyned Thus if there were given in the triangle ABD the side AB the angle at A to finde the hypothenusal AD there the angle at A is in the middle and the sides AD and AB are adjacent to it and therefore the middle part is called the middle conjunct and the extreames are the extreames conjunct but if there were given the side AB the hypothenusal AD to finde the angle at D here AB is the middle part dis-junct because it is dis-joyned from the side AD by the angle at A and from the angle at D by the side DB for the right angle is not reckoned among the circular parts and here the extreams are extreams dis-junct These things premised we come now to demonstrate the proposition it self consisting of two parts first we will prove that the sine of the middle part and Radius are proportional with the tangents of the extreams conjunct The middle part is either one of the sides or one of the oblique angles or the hypothenusal CASE 1. Let the middle part be a side as in the right angled spherical triangle ABD of the last diagram let the perpendicular AB be the middle part the base DB and comp A the extreame conjunct then I say that the rectangle of the sine of AB and Radius is equal to the rectangle of the tangent of DB and the tangent of the complement of DAB for by the second proposition of this Chapter As the sine of AB is in proportion to Radius so is the tangent of DB to the tangent of the angle at A. Therefore if you put the third term in the second place it will be as the sine of AB to the tangent of DB so is the Radius to the tangent of the angle at A. But Radius is a mean proportional between the tangent of an arch and the tangent of the complement of the same arch by the Corollary of the first reason of the second Axiome of plain Triangles and therefore as Radius is to the tangent of the angle at A so is the tangent complement of the same angle at A unto Radius Therefore as the sine of AB is in proportion to the tangent of DB so is the co-tangent of the angle at A to Radius and therefore the rectangle of AB Radius is equall to the rectangle of the
the angle at the base to the co-tangent of the angle at the perpendicular Then by the third Consectary of this Chapter the proportion is 〈◊〉 the co-sine of the first angle at the base to the co-sine of the second so is the sine of the first angle at the perpendicular to the sine of the second which being added to or substracted from the first arch found according to the â—iâ—ectâ—on following their summe or difference is the angle sought If the perpendicular fall Within the triangle adde both arches together Without and the angle opposite to the given side acute substract the first from the second arch Without and the angle opposite to the given side obtuse substract the second from the first 1. Example In the oblique angled Triangle ABC let there be given the angle BAC 56 deg 44 min. and ACB 37 deg 92 min. and the side AB 38 deg 47 min. to finde the angle ABC First let fall the perpendicular FB then in the right angled triangle AFB we have known the hypothenusal AB and the angle at A to finde the angle ABF for which I say As Radius 90 deg 10.000000 To co-sine of AB 38.47 9 893â—26 So the tangent of BAF 56. â—4 10.178229 To the co-tangent of ABF 40.28 10.071955 Secondly to finde FBC I say As the co-sine of BAF 56.44 0.257424 To the co-sine of ACB 37.92 9 8970â—5 So is the sine of ABF 40.28 9.810584 To the sine of FBC 67.32 9.965013 Now because the perpendicular fals within the Triangle I adde the first arch found ABF 40 deg 28 min. to the second arch found FBC 67 deg 32 min. and their aggregate is 107 deg 60 min. the angle ABC required 2. Example In the same Triangle let there be given the angle ACB 37 deg 92 min. and ABC 107 deg 60 min. and the side AB 38 deg 47 min. to finde the angle BAC First let fall the perpendicular AE and let the side BC be continued to E then in the right angled triangle AEB we have known the Hypothenusal AB and the angle at B 72 deg 40 min. the complement of ABC to finde EAB I say then As the Radius 90 10.000000 To the co-sine of AB 38.47 9.893726 So is the tangent of ABE 72.40 10.498641 To the co-tangent of EAB 22. â—6 10,392367 Secondly to finde EAC I say As the co-sine of ABE 72.40 0.519462 To the co-sine of ACB 37.92 9.897005 So is the sine of EAB 22.6 9.574699 To the sine of EAC 78.49 9 99â—166 Now because the perpendicular falls without the triangle and the angle opposite to the given side acute I substract the first angle found E. AB 22 deg 6 min. from the second arch found 78 deg 49 min. and their difference 56 deg 43 min. is the angle BAC required 3 Example In the same triangle ABC let there be given the angles ACB 37 deg 92 min. and ABC 107 deg 60 min. and the side AC 74 deg 84 min. to finde the angle BAC Let fall the perpendicular AE and then in the right angled triangle AEC we have known the hypothenusal AC and the angle ACB to finde the angle EAC As the Radius 90 10.000000 To the co-sine of AC 74.84 9.417497 So is the tangent of ACE 37.92 9.891559 To the co-tangent of EAC 78.49 9 3090â—6 Secondly to finde EAB I say As the co-sine of ACE 37.92 0 102â—95 To the co-sine of ABE 72.40 9.480538 So is the sine of EAC 78.49 9.991177 To the sine of EAB 22.6 9.574790 Now because the perpendicular falls without the Triangle and the angle opposite to the given side obtuse therefore I substract the second arch found EAB 22 deg 6 min. from the first arch found EAC 78 deg 49 min. and their difference 56 deg 43 min. is the angle BAC required CASE 10. Two angles and a side opposite to one of them being given to finde the side between them First by the 7th Case of right angled Sphericall Triangles I say As Radius to the co-sine of the angle at the base so is the Tangent of the Hypothenusal to the Tangent of the Base Then by the second Consectary of this Chapter the proportion is As the co-tangent of the first angle at the base to the co-tangent of the second so is the sine of the first base to the sine of the second which being added to or substracted from the first arch found according to the direction of the 9th Case giveth the side required Example In the oblique angled triangle ABC let there be given the two angles BAC 56 deg 44 min. and ACB 37 deg 92 min. with the side BC 57 deg 53 min. to finde the side AC Let fall the perpendicular BF then in the right angled triangle BCF we have known the Hypothenusal BC and the angle FCB to finde the base FC say then As the Radius 90 10.000000 Is to the co-sine of FCB 37.92 9.897005 So is the tangent of BC 57.53 10.196314 To the tangent of FC 51.11 10.093319 Secondly to finde AF I say As co-tangent FCB 37.92 co ar 9.891559 To co-tangent of BAC 56.44 9.821771 So is the sine of FC 51.11 9.891176 To the sine of AF 23.72 9.604506 Now because the perpendicular falls within the Triangle I adde the first arch FC 51 deg 11 min. to the second arch AF 23 deg 72 min. and their aggregate is 74 deg 83 min. the side AC required CASE 11. The three sides given to finde an angle The solution of this and the Case following depends upon the Demonstration of this Proposition As the Rectangular figure of the sines of the sides comprehending the angle required Is to the square of Radius So is the Rectangular figure of the sines of the difference of each containing side taken from the half summe of the three sides given To the square of the sine of half the angle required Let the sides of the triangle ZPS be known and let the vertical angle SZP be the angle required then shall ZS the one be equal ZC In like manner PS the base of the vertical angle shall be equal to PH or PB then draw PR the sine of PZ and CK the sine of CZ or ZS Divide CH into two equal parts in G draw the Radius AG and let fall the perpendiculars Pâ— and CN which are the sines of the arches PG and CG The right line EV is the versed sine of a certain arch in a great circle and SC the versed of the like arch in a less then if you draw the right line NF parallel to SH bisecting CH in N it shall also bisect the versed sine SC in F by the 15th of the second and RM bisecting TP in R and drawn parallel to TX shall for the same reason bisect PX in M and the triangles SCH and FNC shall be like as also the triangles TPX and RPM are like and ZG shall be equal to the half summe of the three sides
fractions is in the calculation very tedious besides here no fractions almost are exquisitely true therefore the Radius for the making of rhese Tables is to be taken so much the more that there may be no errour in so many of the figures towards the left hand as you would have placed in the Tables and as for the numbers superfluous they are to be cut off from the right hand towards the left after the ending of the supputation Thus to finde the numbers answering to each degree and minute of the Quadrant to the Radius of 10000000 or ten millions I adde eight ciphers more and then my Radius doth consist of sixteen places This done you must next finde out the right Sines of all the arches lesse then a Quadant in the same parts as the Radius is taken of whatsoever bignesse it be and from those right Sines the Tangents and secants must be found out 21. The right Sines in making of the Tables are either primary or secondary The primarie Sines are those by which the rest are found And thus the Radius or whole Sine is the first primary Sine the which how great or little soever is equall to the side of a six-angled figure inscribed in a circle that is to the subtense of 60 degrees the which is thus demonstrated Out of the Radius or subtense of 60 degrees the sine of 30 degrees is easily found the halfe of the subtense being the measure of an angle at the circumference opposite thereunto by the 19 of the second if therefore your Radius consists of 16 places being 1000.0000.0000.0000 The sine of 30 degrees will be the one half thereof to wit 500.0000.0000.0000 22. The other primary sines are the sines of 60 45 36 and of 18 degrees being the halfe of the subtenses of 120 90 72 and of 36 degrees 23. The subtense of 120 degrees is the side of an equilateral triangle inscribed in a circle and may thus be found The Rule Substract the Square of the subtense of 60 degrees from the Square of the diameter the Square root of what remaineth is the side of an equilateral triangle inscribed in a circleâ— or the subtense of 120 degrees The reason of the Rule The subtense of an arch with the subtense of the complement thereof to 180 with the diameter make in the meeting of the two subtenses a right angled triangle As the subtense AB 60 degrees with the subtense AC 120 degrees and the diameter CB make the right angled triangle ABC right angled at A by the 19 of the second And therfore the sides including the right angle are equal in power to the third side by the 〈◊〉 of the second Therefore the square of AB being taken from the square of CB there remaineth the square of AC whose squar root is the subtense of 〈◊〉 degrees or the side of an equilateral triangle inscribed in a circle Example Let the diameter CB be 2000.0000 0000.0000 the square thereof is 400000. 00000.00000.00000.00000.00000 The subtense of AB is 100000.00000.00000 The square thereof is 100000.00000.00000 00000.00000.00000 which being substracted from the square of CB the remainder is 300000.00000.00000.00000.00000.00000 whose square root 173205.08075.68877 the subtense of 120 degrees CONSECTARY Hence it followeth that the subtense of an arch lesse then a Semicircle being given the subtense of the complement of that arch to a Semicircâ—e is also given 24. The Subtense of 90 degrees is the side of a square inscribed in a circle and may thus be found The Rule Multiply the diameter in it self and the square root of half the product is the subtense of 90 degrees or the side of a square inscribed in a circle The reason of this Rule The diagonal lines of a square inscribed in a circle are two diameters and the right angled figure made of the diagonals is equal to the right angled figures made of the opposite sides by the 20 th of the second now because the diagonal lines AB and CD are equal it is all one whether I multiply AC by it self or by the other diagonal CD the pâ—oduct will be still the same then because the sides AB AC and BC do make a right angled triangle right angled at C by the 〈◊〉 of the second that the 〈◊〉 AC and â—B are equal by the work the half of the square of AB must needs be the square of AC or CB by the 17 th of the second whose square rootes the subtense of CB the side of a square or 90 degree Example Let the diameter AB be 200000.00000 00000 the square thereof is 400000.00000 00000.00000.00000.00000 the half whereof is 200000.00000.00000.00000.00000 00000. whose square root 14142â— 356â—3 73095. is the subtense of 90 degrees or the side of a square inscribed in a Circle 25. The subtense of 36 degrees is the side of a decangle and may thus be found The Rule Divide the Radius by two then multiply the Radius by it self and the half thereof by it self and from the square root of the summe of these two products substract the half of Radius what remaineth is the side of a decangle or the subtense of 36 degrees The reason of the rule For example Let the Radius EB be 100000.00000.00000 then is BH or the half thereof 500000. 00000.00000 the square of EB is 100000 00000.00000.00000.00000.00000 and the square of BH 250000.00000.00000.00000 00000.00000.00000 The summe of these two squares viz 125000.00000.00000 00000 00000. 00000 is the square of HE or HK whose square root is 1118033â— 887â—9895 from which deduct the halfe Radius BH 500000000000000 and there remaineth 618033988749895 the right line KB which is the side of a decangle or the subtense of 36 degrees 26 The subtense of 72 degrees is the side of a Pentagon inscribed in a circle and may thus be sound The Rule Substract the side of a decangle from the diameter the remainer multiplied by the Radius shall be the square of one side of a Pentagon whose square root shall be the side it self or subtense of 72 degrees The Reason of the Rule In the following Diagram let AC be the side of a decangle equal to CX in the diameter and let the rest of the semicircle be bisected in the point E then shall either of the right lines AE or EB represent the side of an equilateral pentagon for AC the side of a decangle subtends an arch of 36 degrees the tenth part of a circle and therefore AEB the remaining arch of a semicircle is 144 degrees the half whereof AE or EB is 72 degrees the fift part of a circle or side of an equilateral pentagon the square whereof is equal to the oblong made of DB and BX Demonstration Draw the right lines EX ED and EC then will the sides of the angles ACE and ECX be equal because CX is made equal to AC and EC common to both and the angles themselves are equal because they are in equal segments
of the same circle by the 19 of the second and their bases AE and EX are equal by the 23 of the second and because EX is equal to AE it is also equal to EB and so the triangle EXB is equicrural and so is the triangle EDB because the sides ED and DB are Radii and the angles at their bases X and B E and B by the 24th of the second and because the angles at B is common to both therefore the two triangles EXB and EDB are equiangled and their sides proportional by the 18 th and 16 th Theoremes of the second Chapter that is as DB to EB so is EB to BX and the rectangle of DB in BX is equal to the square of EB whose square root is the side EB or subtense of 72 degrees Example Let AC the side of a decangle or the subtense of 36 degrees be as before 618033988749895 which being substracted from the diameter BC 200000.00000 00000. the remainer is XB 1381966011151105 which being multiplied by the Radius DB the product 1381966011251105 00000.00000.0000 shall be the square of EB whose square root 1175570504584946 is the right line EB the side of a Pentagon or subtense of 72 degrees CONSECTARY Hence it followes that the subtense of an arch lesse then a semicircle being given the subtense of half the complement to a semicircle is given also Thus much of the primarie Sines the secondary Sines or all the Sines remaining may be found by these and the Propositions following 27. The subtenses of any two arches together lesse then a semicircle being given to finde the subtense of both those arches The Rule Finde the subtense of their complements to a semicircle by the 23 hereof then multiply each subtense given by the subtense of the complement of the other subtense given the sum of both the products being divided by the diameter shall be the subtense of both the arches given The reason of the Rule Example Let AI the side of a square or subtense of 90 degrees be 141421.35623.73059 And EO the side of a triangle or subtense of 120 degrees 173205.08075.68877 the product of these two will be 2449489742783 77659465844164315. Let AE the side of a sixangled figure or the subtense of 60 degrees be 100000 00000.00000 And IO the side of a square or subtense of 90 degrees 141421.35623.73059 the product of these two will be 141421.35623.73059 00000.00000.00000 the summe of these two products 3863703305156272659465844164315 And this summe divided by the diameter AO 200000.00000.00000 leaveth in the quotient for the side EI or subtense of 150 degrees 1931851652578136. the half whereof 965925826289068 is the Sine of 75 degrees 28 The subtenses of any two arches lesse then a Semicircle being given to finde the subtense of the difference of those arches The Rule Finde the subtenses of their complements to a semicircle by the 23 hereof as before then multiply each subtense given by the subtense of the complement of the other subtense given the lesser product being substracted from the greater and their difference divided by the diameter shall be the subtense of the difference of the arches given The Reason of the Rule Let the subtenses of the given arches be AE and EI and let the subtense sought be the right line EI then because the right angled figure made of the diagonals AI and EO is equal to the right angled figures made of their opposite sides by the 20 of the second therefore if I subtract the right angled figure made of AE and IO from the right angled figure made of AI and EO the remainer will be the right angled figure of AO and EI which being divided by the diameter AO leaveth in the quotient EI. Example 29. The sine of an arch lesse then a Quadrant being given together with the sine of half his complement to finde the sine of an arch equal to the commplement of the arch given and the half complement added together The Rule Multiply the double of the sine given by the sine of half his complement the product divided by the Radius will leave in the quotient a number which being added to the sine of the half complement shall be the sine of the arch sought The reason of the Rule That is as AI is to AM so is CP to PM and so is PS to PN and then by composition as AI AM so is CS to MN Now then let ES be the arch given and SI the complement thereof to a Quadrant then is CG or IB being equal to EY the half of the said complement SI and AM is the Sine thereof and the Sine of ES is the right line HS and the double CS MN is the difference between AM the Sine of CG or IB and AN the Sine of SB and AI is the Radius and it is already proved that AI is in proportion to AM as CS is to MN therefore if you multiply AM by SC and divide the product by AI the quotient will be NM which being added to AM doth make AN the Sine or the arch sought Example Let ES the arch given be 84 degrees and the Sine thereof 9945219 which doubled is 19890438 the Sine of 3 degrees the halfe complement is 523360 by which the double Sine of 84 degrees being multiplied the product will be 104098â—9 631680 which divided by the Radius the quotient will be 10409859 from which also cutting off the last figure because the Sine of 3 degrees was at first taken too little and adding the remainer to the Sine of 3 degrees the aggregate 1564345 is the Sine of 6 degrees the complement of 84 and of 3 degrees the halfe complement added together that is it is the sine of 9 degrees 30. The subtense of an arch being given to find the subtense of the triple arch The Rule Multiply the subtense given by thrice Radius square and from the product substract the cube of the subtense given what remaineth shall be the subtense of the triple arch The reason of the Rule Now then we have already proved that the square of AO divided by Radius is equal to OX and also that OX is equal to SA and therefore SN is less then twice Radius by the right line AS or thus NS is twice Radius less by AO square divided by Radius and NS multiplied by SA is the same with twice Radius lesse by AO square divided by Radius multiplied into AO square divided by Radius and NS multiplied by SA is equal to SC multiplied by OS and therefore twice Radius less AO square divided by Rad. multiplied by AO square divided by Radius is equal to SC multiplied by SO or thus 2 Radius less AO square divided by Radius multiplied into AO square divided by Radius and divided by AO or SO is equal to SC. All the parts of the first side of this Equation are fractions except AO and the two Radii as will plainly appear
10 must now stand between 2 and 3 of the afternoon hours And lest there should be yet any doubt conceived I have drawn to the South declining East 25 the North declining West as much from which to make the South declining West and North declining East you need to do no more then prick these hour lines through the paper and draw them again on the other side stile and all so shall they serve the turn if you place the morning hours in the one where the afternoon were in the other APPENDIX To draw the hour lines upon any plane declining far East or West without respect to the Center THe ordinary way is with a Beamcompasse of 16 18 or 20 foot long to draw the Diall upon a large floor and then to cut off the hours stile and all at 10 12 or 14 foot distance from the center but this being too mechanical for them that have any Trigonometrical skill I omit and rather commend the way following by help whereof you may upon half a sheet of paper make a perfect model of your Diall to what largenesse you please without any regard at all to the Center Suppose the wall or plane DZG on which you would make a Diall to decline from N to C that is from the South Easterly 83 degrees 62 min. set down the Data and by them seek the Quaesita according to the former directions The Data or things given are two 1. PS the poles elevation 51 degrees 53 minutes 2. SA the planes declination southeast 83 deg 62 min. The Quaesita or things sought are three 1. PR the height of the stile 3 degrees 97 minutes 2. ZR the distance of the substile from the meridian 38 deg 30 min. 3. ZPR the angle of the meridian of the plane with the meridian of the place 85 degrees which being found according to the former directions the substile line must fall within five degrees of six of the clock because 85 degrees wanteth but 5 of 90 the distance of 6 from 12. Now therefore make a table according to this example following wherein set down the houres from 12 as they are equidistant from the meridian and unto them adjoyn their Equinoctial distances and write Meridian and substile between the hours of 6 and 7 and write 5 degrees against the hour of 6 10 degrees against the hour of 7 and to the Equinoctial distances of each hour adde the natural tangents of those distances as here you see So is the Table prepared for use by which you may easily frame◠the Diall to what greatnesse you will after this manner Hours Equ dist Tang. 4 8 35 0 700 5 7 20 0 364 6 6 5 0 087   Meridiā Substile 7 5 10 0 176 8 4 25 0 166 9 3 40 0 839 10 2 55 0 1.428 11 1 70 0 2.747 12 12 85 0 11.430 The Geometricall projection Proportion the plane BCDE whereon you will draw the Diall to what scantling you think fit Let VP represent the horizontal line upon any part thereof as at P make choice of a fit place for the perpendicular stile though afterwards you may use another forme neer about the upper part of the plane because the great angle between the two Meridians maketh the substile which must passe thorow the point P to fall so near the 6 of clock hour as that there may be but one hour placed above it if you desire to have the hour of 11 upon the plane which is more useful then 4 let P be the center and with any Chord the greater the better make two obscure arches one above the horizontal line the other under it and with the same Chord set off the arch of 51.70 which is the angle between the substile and horizon and is the complement of the angle between the substile and meridian and set it from V to T both wayes then draw the streight line TPT which shall be the substile of this Diall Which is 5 inches and 66 hundred parts for the distance of 11 a clock from the point H and will be the same with those points set off by the natural tangents in the Table Having done with this Equinoctiall you must do the like with another to finde the place whereof it will be necessary first to know the length of the whole line from H the Equinoctial to the center of the Diall in parts of the perpendicular stile PO if you will work by the scale of inches or else the length in natural tangents if you will use a diagonall Scale first therefore to finde the length thereof in inch-measure we have given in the right angled plain triangle HOP the base OP and the angle at O to finde HP and in the triangle OP center We have given the perpendicular OP and the angle PO center the complement of the former to finde H center wherefore by the first case of right angled plain triangles As the Radius 90 10.000000 Is to the base OP 206 2.313867 So is the tang of HOP 3.97 8.841364   To the perpendicular PH14 1.155231 Again As the Radius 90 10.000000 Is to the perpend OP 206 2.313867 So is the tang PO center 86.3 11.158636   To the base P center 2972 3.472403 Adde the two lines of 014 and 2972 together and you have the whole line H center 2986 in parts of the Radius PO viz. 29 inches and 86 parts out of this line abate what parts you will suppose 343 that is 3 inches and 43 parts and then the remainer will be 2643. Now if you set 343 from H to I the triangle IO center will be equiangled with the former and I center being given to finde LO the proportion is As H center the first base 2986 co ar 6.524911 Is to HO the first perpend 206. 2.313867 So is I center the 2d base 2643 3.422097   To IO the 2d perpend 182 2.260875 Having thus found the length of IO to be one inch and 82 parts make that the Radius and then NT4 shall be a tangent line thereunto upon which according to this new Radius set off the hour-distances before found and so have you 2 pricks by which you may draw the height of the stile OO and the hour-lines for the Dial. The length of H center in natural tangents is thus found HP 069 is the tangent line of the angle HOP 3 deg 97 min. and by the same reason P center 14421 is the tangent line of PO center 86.3 the complement of the other and therefore these two tangents added together do make 14490 the length of the substile H center that is 14 times the Radius and 49 parts out of which substract what number of parts you will the rest is the distance from the second Equinoctial to the center in natural tangents suppose 158 to be substracted that is one radius and 58 parts which set from H to T in proportion to the Radius HO and from the point
R of the substile both wayes set off the hour distances by help of the Chord for 8 of the clock 2 degr 91 min. and so of the rest and draw streight lines from the center G through those points which shall be the true hour lines desired Last of all the height of the stile PR 26 degr 69 min. being set from R to P draw the streight line GP for the axis of the stile which must give the shadow on the diall Erect GP at the angle RGP perpendicularly over the substile line GR and let the point P be directed to the North pole GO12 placed in the Meridian the center G representing the South and the plane at EF elevated above the horizon 55 degrees so have you finished this diall for use onely remember because the Sun riseth but a little before 4 and setteth a little after 8 to leave out the hours of 3 and 9 and put on all the rest And thus you have the projection of four Dials in one for that which is the West recliner is also the East incliner 〈◊〉 you take the complements of the recliners â—ours unto 12 and that but from 3 in the afternoon till 8 at night again if you draw the same lines on the other side of your â—â—per and change the houres of 8 7 6 c. into 4 5 6 c. you have the East recliner and the complement of the East recliners hours from 3 to 8 is the West incliner onely remember that as the stile in the West recliner beholds the North and the plane the Zenith so in the East incliner the stile must behold the South and the plane the Nadir Probl. 9. To draw the hour-lines upon any direct South reclining or inclining plane AS the base of East and West reclining or inclining planes do alwayes lie in the meridian of the place or parallel thereunto and the poles in the prime vertical so doth the base of South and North reclining or inclining planes lie in the prime vertical or azimuth of East and West and their poles consequently in the Meridian Now if you suppose the circle of position which Astrologically taken is fixed in the intersection of the meridian and horizon to move about upon the horizon till it comes into the plane of the prime vertical and being fixed in the intersection thereof with the horizon to be let fall either way from the Zenith upon the meridian it shall truly represent all the South and North reclining and inclining planes also of which there are six varieties three of South and three of North reclining for either the South plane doth recline just to the pole and then it becommeth an Equinoctial because the poles of this plane do then lie in the Equinoctiall some call it a polar plane or else it reclineth more and less then the pole and consequently the poles of the plane above and under the Equinoctiall somewhat differing from the former In like manner the North plane reclineth just to the Equinoctial and then becometh a polar plane because the poles of that plane lie in the poles of the world some term it an Equinoctiall plane Or else it reclineth more or lesse then the Equinoctial and consequently the poles of the plane above and under the poles of the world somewhat differing from the former Of the Equinoctiall plane And because this Diall is no other but the very horizontall of a right Sphere where the Equinoctial is Zenith and the Poles of the world in the Horizon therefore it is not capable of the six of clock hour no more then the East and West are of the 12 a clock hour which vanish upon the planes unto which they are parallel and the twelve a clock hour is the middle line of this Diall because the Meridian cutteth the plane of six a clock at right angles which the Sun attaineth not till he be perpendicular to the plain And this in my opinion besides the respect of the poles is reason enough to call it an Equinoctiall Diall seeing it is the Diall proper to them that live under the Equinoctiall This Diall is to be made in all respects as the East and West were being indeed the very same with them onely changing the numbers of the hours for seeing the six of clock hour in which this plane lieth crosseth the twelve of clock hour at right angles in which the East and West plane lieth the rest of the hour-lines will have equall respect unto them both so that the fifth hour from six of the clock is equal to the fift hour from twelve the four to the four and so of the â—est These analogies holding the hour distances from six are to be set off by the natural tangents in these Dials as they were from twelve in the East and West Dials The Geometricall Projection Draw the tangent line DSK parallel to the line EZW in the Scheme crosse it at right angles with MSA the Meridian line make SA the Radius to that tangent line on which prick down the hours and that there may be as many hours upon the plane as it is capable of you must proportion the stile to the plane as in the fifth Problem after this manner let the length of the plane from A be given in known parts then because the extream hours upon this plane are 5 or 7 reckoning 15 degrees to every hour from 12 the arch of the Equator will be 75 degrees and therefore in the right angled plain triangle SA ♎ we have given the base A ♎ the length of the plane from A and the angle AS ♎ 75 degrees to finde the perpendicular SA for which as in the fifth Chapter I say As the Radius 90 10.000000 Is to the base A ♎ 3.50 2.544068 So is the tangent of A ♎ S 15 9,428052   To the perpendicular AS 94 1.972120 At which height a stile being erected over the 12 a clock hour line and the hours from 12 drawn parallel thereunto through the points made in the tangent line by setting off the natural tangents thereon and then the Diall is finished Let SA 12 be placed in the meridian and the whole plane at S raised to the height of the pole 51 degr 53 min. then will the stile shew the hours truly and the Diall stand in its due position 2. Of South reclining lesse then the pole This plane is represented by the prickt circle in the fundamental Diagram ECW and is intersected by the hour circles from the pole P as by the Scheme appeareth and therefore the Diall proper to this plane must have a center above which the South pole is elevated and therefore the stile must look downwards as in South direct planes to calculate which Dials there must be given the Poles elevation and the quantity of reclination by which to finde the hour distances from the meridian and thus in the triangle PC 1 having the poles elevation 51 degr 53 min. and the reclination