diuide the lines of Solids SEing like Solids do hold in the proportion of their homologall sides triplicated if you shall finde two meane proportionalls 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 containe the diuisions required Wherefore vpon the center A Semidiameter equall to the line of Lines describe a circle and diuide it into 4 equall parts CEBD drawing the crosse diameters CB ED. Then diuide the semidiameter AC first into 10 equall parts and betweene the whole line AD AF the tenth part of AC seeke out two meane proportionall lines AI and AH againe betweene AD and AG being two tenth parts of AC seeke out two meane proportionals AL and AK and so forward in the rest So shall the line AB be diuided into 10 vnequall parts Secondly diuide each tenth part of the line AC into 10 more and betweene the whole line AD and each of them seeke out two meane proportionalls as before So shall the line AB be diuided now into an hundred vnequall parts Thirdly If the length will beare it subdiuide the line AC once againe each part into ten more and betweene the whole line AD and each subdiuision seeke two meane proportionalls as before So should the line AB be now diuided into 1000 parts But the ruler being short it shall suffice if those 10 which are nearest the center be expressed the rest be vnderstood to be so diuided though actually they be diuided into no more then 5 or 2. and this line AB so diuided shall be the line of Solids and must be transferred into the Sector But let the numbers set to them be onely 1.1.1.2.3 c. vnto 10. as in the example 5 To diuide the lines of Sines and Tangents on the side of the Sector VPon the center A and semidiameter equall to the line of Lines describe a semicircle ABCD with AB perpendicular to the diameter CD Then diuide the quadrants CB BD each of them into 90. and subdiuide each degree into 2 parts For so if streight lines be drawne parallell to the diameter CD through these 90 and their subdiuisions they shall diuide the perpendicular AB vnequally into 90. And this line AB so diuided shall be the line of Sines and must be transferred into the Sector The numbers set to them are to be 10.20.30 c. vnto 90 as in the example If now in the poynt D vnto the diameter CD we shall raise a perpendicular DE and to it drawe streight lines from the center A through each degree of the quadrant DB. This perpendicular so diuided by them shall be the line of Tangents must be transferred vnto the side of the Sector The numbers set to them are to be 10.20.30 c. as in the example If betweene A and D another streight line GF be drawne parallell to DE it will be diuided by those lines from the center in like sort as DE is diuided and it may serue 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 diagonall Scale equall 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 rootes and the line of Solids by tables of cubique rootes which I leaue to others to extract at leasure 6 To shew the ground of the Sector LEt AB AC represent the leggs of the Sector then seuering these two AB AC are equall and their sections AD AE also equall they shall be cut proportionally and if we draw the lines BC DE they will be parallell by the second Pro. 6 lib. of Euclid and so the Triangles ABC ADE shal be equiangle by reason of the common angle at A and the equall angles at the base and therefore shall haue the sides proportionall about those equall angles by the 4 Pro. 6 lib. of Euclid The side AD shal be to the side AB as the basis DE vnto the parallell basis BC and by conuersion AB shall be vnto AD as BC vnto DE and by permutation AD shall be vnto DE as AB to BC. c. So that if AD be the fourth part of the side AB then DE shall also be the fourth part of his parallell basis BC The like reason holdeth in all other sections 7 To shew the generall vse of the Sector THere may some cōclusions be wrought by the Sector euen then when it is shut by reason that the lines are all of one length but generally the vse hereof consists in the solution of the Golden rule where three lines being giuen of a known denominaton a fourth proportionall is to be found And this solution is diuerse in regard both of the lines and of the entrance into the worke The solution in regard of the lines is sometimes simple as when the worke is begun and ended vpon 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 vpon the lines of Lines finished vpon the lines of Superficies It may begin on the Sines and end on the Tangents The solution in regard of the entrance into the worke may be either with a parallell or else laterall on the side of the Sector I cal it parallell entrance or entring with a parallell when the two lines of the first denomination are applied in the parallells and the third line and that which is sought for are on the side of the Sector I call it laterall entrance or entring 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 doe stand in the parallells As for example let there be giuen three lines A B C to which I am to find a fourth proportionall let A measured in the line of lines be 40 B 50 and C 60 and suppose the question be this If 40 Monthes giue 50 pounds what shal 60 Here are lines of two denominatiōs one of months another of pounds and the first with which I am to enter must be that of 40 monthes If then I would enter with a parallell first I take A the line of 40 and put it ouer as a parallell in 50 reckoned in the line of lines on either side of the Sector from the center 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 betweene the feete of the compasses and carrying them parallell to BC I finde them to crosse the lines AB AC on the side of the
was shewed before in the second proiection And if from Z we draw circles through each of these diuisions they shall be parallels of altitude The houre circles may be here drawne as the azimuths in the third proiection For the center of EPV the houre of 6 will be found at B somewhat neare vnto N by the tangent of the latitude And if through B we draw an occult line parallell vnto EV and diuide it on each side from B in such sort as the tangent is diuided on the side of the Sector allowing 45 gr to be equall to BP and 15 gr for euery houre those diuisions shall be the centers and the distance from these diuisions vnto P shall be the semidiameters whereon to describe the rest of the houre circles The ecliptique may be drawne as the equator For the center of that halfe which hath Southerne declination shall be giuen by the tangent of the altitude which the Sun hath in his entrance into ♑ And the center of the other halfe by the tangent of his altitude at his entrance into ♋ And it may be diuided as in the former proiection or else by tables calculated to that purpose To these circles thus drawne if we shall adde the moneths of the yeare and the dayes of each moneth as we may well doe at the horizon on either side betweene the tropiques this proiection shall be fitted for the most vsefull conclusions of the globe For the day of the moneth being giuen the parallell that shooteth on it doth shew what declination the Sunne hath at that time of the yeare And where this parallell crosseth the ecliptique there is the place of the Sunne Or the place of the Sunne being first giuen the parallell which crosseth it shall at the horizon shew the day of the moneth Either of these then being giuen or onely the parallell of declination we may follâ—w it first vnto the horizon there the distance of 〈…〉 of the parallell from E or V sheweth the ampli◠〈…〉 same among the houre circles sheweth the time 〈…〉 â—â—e riseth or setteth Then hauing the altitude 〈…〉 any time of the day the intersection of the 〈…〉 on with the parallell of altitude sheweth 〈…〉 â—ay and a right line drawne from Z through this intersection to the horizon giueth the azimuth Thus in either of these proiections that which is otherwise most troublesome is easily done by the help of the tangent line and what I haue said of this line the same may be wrought by scale and numbers out of the table of Tangents CHAP. IV. 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 giuen the rest may be found by the Sector As in a Rectangle triangle 1 To finde the base both sides being giuen Let the Sector be opened in the lines of Lines to a right angle as before was shewed 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 meete in the center and marke how farre they extend For the line taken from the termes of their extension shall be the base required viz. the side opposite to the right angle Or adde 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 hauing the angles and one of the sides giuen Take the side giuen and turne it into the parallell sine of his opposite angle so the parallell Radius shall be the base 3 To find a side by hauing the base and the other side giuen Let the Sector be opened in the lines of lines to a right angle and the side giuen laid on one of those lines from the center then take the base with a paire of compasses and setting one foote in the terme of the giuen side turne the other to the other line of the Sector and it shall there shew 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 hauing the base and the angles giuen Take the base giuen and make it a parallell Radius so the parallell sines of the angles shall be the opposite sides required 5 To find a side by hauing the other side and the angles giuen Take the side giuen and turne it into his parallell sine of his opposite angle so the parallell sine of the complement shall be the side required 6 To find the angles by hauing the base and one of the sides giuen First takeout the base giuen and laying it on both sides of the Sector so as they may meete in the center and marke how farre it extendeth Then take out the laterall Radius and to it open the Sector in the termes of the base This done take out the side giuen and place it also on the same lines of the Sector from the center For the parallell taken in the termes of this side shall be the sine of his opposite angle Or take the base giuen and make it a parallell Radius then take the side giuen and carrie it parallell to the base till it stay in like sines so they shall giue the quantitie of the opposite angle 7 To finde the angles by hauing both the sides giuen Take out the greater side and lay it on both sides of the Sector so as they meete in the center and marke how farre it extendeth Then take the other side and to it open the Sector in the termes of the greater side so the parallell Radius shall be the tangent of the lesser angle The third angle is alwayes knowne by the complement 8 The Radius being giuen to find the tangent and secant of any arke 9 The tangent of any arke being giuen to find the tangent thereof and the Radius 10 The secant of any arke being giuen to find the tangent thereof and the Radius The tangent and the secant together with the Radius of euery arke do make a right angle triangle whose sides are the Radius and tangent and the base alwayes the secant and the angles alwayes knowne by reason of the giuen arkes Wherefore the solution is the same with those before In any right-lined triangle whatsoeuer 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 giuen then take out the sides of the triangle laying them the one on the one line the other on the other so as they meete in the center marke how far they extend For the line taken betweene the termes of their extension shall be the third side required 12 To find a side by hauing the other two sides and one of the adiacent angles so it be
whether the two stations be chosen at the one end of the bredth proposed or without it or within it if the line betweene the stations be perpendicular vnto the bredth as may appeare if in stead of the stations at A and H we make choise of the like stations at I and K. There might be other wayes proposed to work these Prop. by holding the Crosse euen with the distance and the Staffe parallell with the height but these would proue more troublesome and those which are deliuered are sufficient and the same with those which others haue set downe vnder the name of the Iacobs staffe CHAP. III. The vse of the Tangent lines in taking of Angles 1 To find an angle by the Tangent on the Staffe LEt the middle sight be alwayes set to the middle of the Crosse noted with 20 and 30 and then the Crosse drawne nearer the eye vntill the marks may be seene close within the sights For so if the eye at A that end of the Staffe which is noted with 90 and 180 beholding the marks K and N betweene the two first sights C and B or the marks K and P betweene the two outward sights the Crosse being drawne downe vnto H shall stand at 30 and 60 in the Tangent on the Staffe it sheweth that the angle KAN is 30 gr the angle KAP 60 gr the onâ— double to the other which is â—he râ—ason of the double numbers on this line of the Staffe and this way wil serue for any angle from 20 gr toward 90 gr or from 40 gr toward 180 gr But if the angle be lesse then 20 gr we must then make vse of the Tangent vpoÌ„ the Crosse 2 To find an angle by the Tangent of 20 vpon the Crosse Set 20 vnto 20 that is the middle sight to the middest of the Crosse at the end of the Staffe noted with 20 so the eye at A beholding the marks L and N close betweene the two first sights C and B shall seâ— them in an angle of 20 gr If the marks shâ—ll be nearer together as are M and N then draw in the Crosse from C vnto E if they be farther asunder as are K and N then draw out the Crosse from C vnto F so the quantitie of the angle shal be still found in the Crosse in the Tangent of 20 gr at the end of the Staffe and this will serue for any angle from 0 gr toward 35 gr 3 To find an angle by the Tangent of 30 vpon the Crosse This Tangent of 30 is here put the rather that the end of the Staffe resting at the eye the hand may more easily remoue th Crosse for it supposeth the Radius to be no longer then AH which is from the eye at the end of the Staffe vnto 30 gr about 22 inches and 7 parts Wherefore here set the middle sight vnto 30 gr on the Staffe and then either draw the Crosse in or out vntill the marks be seene between the two first sights so the quantitie of the angle will be found in the Tangent of 30 which is here represented by the line GH and this will serue for any angle from 0 gr toward 48 gr 4 To obserue the altitude of the Sunne backward Here it is fit to haue an horizontall sight set to the beginning of the Staffe and then may you turne your backe toward the Sun and your Crosse toward your eye If the altitude be vnder 45 gr set the middle sight to 30 on the Staffe and looke by the middle sight through the horizontall vnto the horizon mouing the Crosse vpward or downward vntill the vpper sight doe shadow the vpper halfe of the horizontall sight so the altitude will be found in the Tangent of 30. If the altitude shal be more then 45 gr set the middle sight vnto the middest of the Crosse and look by the inward edge of the lower sight through the horizontall to the horizon mouing the middle sight in or out vntill the vpper sight do shadow the vpper halfe of the horizontall sight so the altitude will be found in the degrees on the Staffe betweene 40 and 180. 5 To set the Staffe to any angle giuen This is the conuerse 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 obiect in the angle giuen 6 To obserue the altitude of the Sunne another way Set the middle sight to the middle of the Crosse and hold the horizontall sight downward so as the Crosse may be parallell to the horizon then is the Staffe verticall and if the outward sight of the Crosse do shadow the horizontall sight the complement of the altitude wil be found in the tangent on the Staffe 7 To obserue an altitude by thread and plummet Let the middle sight be set to the middest of the Crosse and to that end of the Staffe which is noted with 90 and 180 then hauing a thread and a plummet at the beginning of the Crosse and turning the Crosse vpward and the Staffe toward the Sunne the thread will fall on the complement of the altitude aboue 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 obserued betweene the two first sights there wil be such proportion between the parts of the Staffe and the parts of the Crosse as betweene the Radius and the Tangent of the angle As if the parts intercepted on the Staffe were 20 inches the parts on the Crosse 9 inches Then by proportion as 20 vnto 9 so 100000 vnto 45000 the tangent of 24 gr 14 m. But if the angle shall be obserued betweene the two outward sights the parts being 20 and 9 as before the angle will be 48 gr 28 m. double vnto the former In all these there is a regard to be had to the parallax of the eye and his height aboue the Horizon in obseruations at Sea to the Semidiameter of the Sun his parallax and refraction as in the vse of other staues And so this will be as much or more then that which hath been heretofore performed by the Crosse-staffe CHAP. IIII. The vse of the lines of equall parts ioyned with the lines of Chords THe lines of equall parts do serue also for protraction as may appeare by the former Diagrams but being ioyned with the lines of Chords which I place vpon one side of the Crosse they will farther serue for the protraction and resolution of right line triangles whereof I will giue one example in finding of a distance at two stations otherwise then in the second Cap. Let the distance required be AB At A the first statioÌ„ I make choise of a station line toward C and obserue the angle BAC by the tangent lines which may be 43 gr 20 m then hauing gon an hundred paces toward C I make my second station at D
the latitude is 50 gr For the meridian altitude of the Sunne is 30 gr by the last Prop. and the Sun hauing 10 gr of South declination the meridian altitude of the equator would be 40 gr and therefore the obseruation 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 haue been 51 gr 30 m. and so in the rest 5 To find any North latitude by the meridian altitude of the starres to the Southward Let the vpper sight be set to the starre which you intend to obserue here placed in the fore side of the Bow Then hold the North end of the Bow vpward and turning your face to the South obserue the meridian altitude as before so the lower sight shall shew the latitude of the place in the fore side of the Bow Thus if in obseruing the meridian altitude of the great Dog-starre the lower sight shall stay at 50 gr it would shew the latitude to be 50 gr For this starre 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 adde 16 gr 12 m. for the South declination of this starre it would shew 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 starres to the Northward Let the vpper sight be set to the starre which you intend to obserue here placed on the backe side of the Bow Then hold the North end of the Bow vpward and turning your face to the North obserue the altitude of the starre when he cometh to be in the meridian and vnder the pole so the lower sight shall shew the altitude of the pole in the back side of the Bow Thus the former guard coming to be in the meridian vnder the pole if you obserue and find the lower sight to stay at 50 gr such is the eleuation of the pole and the latitude of the place to the Northward For the distance betweene the two sights will shew the altitude to be 35 gr 45 m. the star is 14 gr 15 m. distant from the North pole These two doe make vp 50 gr for the eleuation of the North-pole and therefore such is the North latitude 7 To find any South latitude by knowing either the day of the moneth or the declination of the Sunne When you are come into South latitude turne both your sights to the backside of the Bow the vpper sight to the declination of the Sun or the day of the moneth at the South end and the lower sight toward the North end of the Bow Then the Sunne coming to the meridian turne your face to the North and holding the South end of the Bow vpward obserue the meridian altitude as before so the lower sight shall shew the latitude of the place in the backe side of the Bow Thus being in South latitude vpon the tenth of May if you obserue and finde the lower sight to stay at 30 gr on the backe side of the Bow such is the latitude For the declination is 20 gr Northward the altitude of the Sunne betweene 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 starres to the Northward Let the vpper sight be set to the starre which you intend to obserue here placed on the backe side of the Bow Then hold the South end of the Bow vpward and turning your face to the North obserue the meridian altitude as before so the lower sight shall shew the latitude of the place in the back side of the Bow Thus being in South latitude and the former guard comming to be in the meridian ouer the pole If you obserue and finde the lower sight to stay at 5 gr such is the latitude For this starre is 14 gr 15 m. from the North pole the altitude of the starre betweene the two sights 9 gr 15 m. the North pole depressed 5 gr and therefore the latitude 5 gr to the Southward 9 To obserue the altitude of the Sunne backward Set the vpper sight either to 60 or 70 or 80 gr as you shall find it to be most conuenient the lower sight on any place betweene the middle and the other end of the Bow and haue an horizontall sight to be set to the center Then may you turne your backe to the Sunne and the back of the Bow toward your selfe looking by the lower sight through the horizontall sight and mouing the lower sight vp downe vntill the vpper sight doe cast a shadow vpon the middle of the horizontall sight so the degrees contained betweene the two sights on the Bow shall giue the altitude required Thus if the vpper 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 backe obseruation knowing either the day of the moneth or the declination of the Sunne When you obserue in North latitude place your three sights on the fore side of the Bow the vpper sight to the declination of the Sun or the day of the moneth at the North end the lower sight toward the South end of the Bow and the horizontall sight to the center Then the Sunne coming to the meridian turne your face to the North holding the North end of the Bow vpward the South end downeward with the back of it toward your selfe obsâ—rue the shadow of the vpper sight as in the former Prop. so the lower sight shall shew the latitude of the place in the fore side of the Bow Thus being in North latitude vpon the ninth of October if you obserue 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 â—r Southward and the altitude of the Sunne betweene the two sights 30 gr the altitude of the equator 40 gr and â—herefore the latitude 50 gr as in the fourth Prop. 11 To find any South latitude by a back obseruation knowing either the day of the moneth or the declination of the Sunne When you obserue in South latitude place your three sights on the backe side of the Bow the vpper sight to the declination of the Sunne or the day of the moneth at the South end the lower sight towarâ— the North end of the Bow and the horizontall sight to the câ—nter Then the Sun coming to the meridian turne your face to the South and holding the South end of the Bow vpward with the backe of it toward your selfe obserue the shadow of the vpper sight as before so the lower sight shal shew the latitude of the place in the back side of the Bow Thus being in the South latitude vpon the tenth
a verticall plane seruing both for the meridian of the place and the houre of 12. 2 In this meridian line I make choice of a center in the vpper part of the line if it had been the Southerne 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 vpward and vpon this center I describe an occult circle representing the declining verticall belonging to this plane 4 The substylar being drawne I may inscribe the chords of the arks of the plane from the substylar and draw the houre-lines and set vp 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 vpon the plane otherwise then by chords For hauing drawne the houre-lines as in the last figure vpon paper or paist-boord we shall find the most part of them in this and such like planes that haue greater declination to fall so close together that they can hardly be discerned wherefore to draw them at large to the best aduantage of the plane I leaue out the center and draw them by tangents as in the polar plane 1 I consider the length and bredth 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 wil contain both the substylar and all those houre-lines which are required in the great square AZCQ 2 I draw two parallell lines FN GM crossing the substylar at right angles in the points F G so as they may best crosse all the houre-lines and yet the one be distant from the other as far as the plane will giue me leaue 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 Againe that the point F will fall about 6 inches below the vpper horizontall side AB and about 12 inches from the next verticall side BD for I need not here stand vpon parts 3 Because these two parallell lines are tangent lines in respect of circles drawne vpon the semidiameters CF CG and such tangents as belong to the arkes of the plane betweene the substylar and the houre-lines the proportion will hold As the tangent of 45 gr to the tangent of the arke of the plane So the length of the semidiameter to the length of the tangent line As for example the arke of the plane betweene the substylar and the houre 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 vnto the tangent of 15 gr 28 m. the same extent will reach from 115 in the line of numbers vnto 31 82 which shewes the length of the tangent line betweene F in the substylar and the houre-line of 1 to be 31 inches 82 cent or parts of 100. Againe the same extent will reach from 100 vnto 27 67 and such is the length of the lesser tangent from G to the houre of 1. The like reason holds for the length of the other tangents from the substylar to the rest of the houres as in the Table as also for the height of the style aboue these tangent lines and so the angle of the style aboue 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 obserue 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 vse the tangent of 5 gr 43 m. in stead of the tangent of 45 gr as I noted before Pag. 100. 4 Hauing found these lengths and heights and set them downe in a Table I come to the plane here resembled by the lesser square ABDE where I begin with an occult verticall FH about 12 inches from the side BD and vpon the center F about 6 inches below the side AB describe an occult arke of a circle 5 Into this arke I first inscribe a chord of 38 gr 23 m. the distance of the substylar from the meridian to make the angle HFG equall to the ZCT so the line FG shall be the substylar and then another chord of 51 gr 37 m. the complement of this distance to make vp 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 vnto G toward the center and through G draw the lesser tangent line GM parallell to the former 7 These two occult tangent lines being thus drawne I looke vnto the former Table for the houre of 1 and there finde the arke of the plane betweene the substylar and the houre 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 pricke them downe in the greater tangent from F to N and then 27 inches 67 parts and prick them downe in the lesser tangent from G to M and draw the MN for the houre of 1 which if it were produced would crosse the substylar FG in the center 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 vp the style right ouer 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 crosse the substylar FG in the center C and there make the angle FCK to be 3 gr 6 m. and so be truly placed for casting of the shadow vpon the houre-lines in this declining plane CHAP. VIII To draw the houre-lines in a meridian inclining plane ALl those planes wherein the horizontall 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 generall name of meridian planes without farther addition and are described before if they leane 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 vpper 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 aboue the plane 3 The
extent will reach in the line of numbers from 21 vnto 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 iust 12 inches as a more easie ground for the rest of the worke But if it were required to proportion the style vnto the plane so as it may cast the shadow to the ful lēgth of the substylar line at all times of the yeare you may then consider the Sunne in the tropique which is to be set nearest vnto the center and adde 66 gr 30 m. vnto 34 gr 33 m. so the remainder vnto 180 gr will be 78 gr 57 m. And if you extend the compasses from the sine of 66 gr 30 m. vnto the sine of 78 gr 57 m. the same extent will reach in the line of numbers from 21 vnto 22.47 for the length of the axis of the style 2 Hauing the length of the axis and the height of the style aboue the plane to find the length of the sides of the style The style of a plane neither equinoctiall nor polar may be either a small rod of iron set parallell to the axis of the world or perpendicular to the plane or else a thin plate of iron or brasse made in forme of a rectangle triangle BAC with the base BC parallell 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 cosine 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. vnto 12 in the line of numbers the same extent will reach from the sine of 34 gr 33 m. vnto 6.80 in the line of numbers for the length of the perpendicular side and from the sine of 55 gr 27 m. vnto 9.88 for the length of the substylar side 3 To find the distance between the center and the equator vpon the substylar line This is here represented by C ♈ and may be found by resoluing the rectangle triangle CB ♈ As the cosine 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 center Extend the compasses from the sine of 55 gr 27 m. vnto the sine of 90 gr the same extent will reach in the line of numbers from 12 vnto 14.57 Wherefore if you take 14 inch 57 cent and pricking them down on your substylar line frō C vnto ♈ draw a line through ♈ crossing the substylar at right angles the line so drawne shall be the equator 4 To find the angles contained between the equator and the houre-lines vpon your plane These angles made by B ♈ and the houre-lines are complements of those which are at C betweene BC the axis and those seuerall houre-lines and depend vpon the angles at the pole between the proper meridian and the houre-circles As the sine of 90 gr to the cosine 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 houre after noone whereupon the angle at the pole betweene this proper meridian and the circles of the houre of 1 on the one side and 3 on the other side wil be 15 gr so between this meridian and the houre-circles of 12 and 4 the angle will be 30 gr c. 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 Infinit  27 60 If then it be required to find the angle which the houre-line of 4 after noone doth make with the plane of the equator that is the angle C 4 B contained betweene the houre-line C 4 and the line B 4 drawne from the top of the style vnto the intersectiō of the houre-line of 4 with the equator Extend the compasses from the sine of 90 gr vnto the sine of 60 gr the complement of the angle at the pole the same extent wil reach from the tangent of 55 gr 27 m. the complement of the height of the pole vnto the tangēt 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 finde the angle C 9 B looke into the Table for the houre 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 vnto the tangent of 55 gr 27 m. the same extent will reach from the sine of 15 gr the complement of 75 gr vnto the tangent of 20 gr 36 m. and such is the angle C 9 B made at the equator betweene the line B 9 drawne from the top of the style and the houre-line C 9 drawne from the center The like reason holdeth for the rest which may be found and set downe in a table then may you either draw these angles at C in the former figure more perfectly and thence finish your worke or else proceed 5 To find the distance betweene the center and the parallels of declination The distances betweene the center and the parallels of declination may be found by resoluing the triangles made by the axis BC the lines of declination and the houre-lines For hauing the angles at the equator and knowing the declination of the parallell if the parallell shall fall betweene the equator and the center adde the declination vnto 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 haue the angle at the parallell Then As the sine of the angle at the parallell to the cosine of the declination So the length of the axis of the style to the distance between the center and the parallell Thus in our example the angle at the equator belonging to
20 5 SbW 13 5 10 39 SSW 1 50 1 29 Â Style 0 0 SWbS 9 25 7 38 SW 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 WNW 88 10 87 44 These angles being knowne if on the center V at the verticall 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 verticall point and the termes of those chords the lines so drawne shall be the azimuths required The like reason holdeth for the drawing of the azimuths vpon all other inclining planes wheof you haue another example in the Diagram belonging to the meridian incliner Pag. 126. Or for further satisfaction you may finde where each azimuth line shall crosse 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 vnto the line of our latitude 51 gr 30 m. the same extent will reach in the line of tangents from 10 gr vnto 7. gr 50. m. for the intersection of the equator with the azimuth of 10 gr from the meridian Againe the same extent will reach from 20 gr vnto 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 crosse the equator at 90 gr from the meridian in the same point with the horizontall line and the houre of 6. And that the azimuth of 45 gr will crosse the equator at 38 gr 2 m. from the meridian that is the line of SE will crosse the equator at the houre of 9 and 28 m. in the morning and the line of SW at 2 ho. 32 min. in the afternoone and so for the rest whereby you may examine your former worke 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 Sunne aboue the horizon haue such respect vnto the horizon as the parallels of declination vnto the equator and so may be described in like maner In an horizontall plane these parallels will be perfect circles wherefore knowing the length of the style in inches and parts and the distance of the parallell 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 parallell to the semidiameter of his circle Thus in the example of the horizontall plane Pag. 164. if AB the length of the style shall be 5 inches and that it were required to finde the semidiameter of the parallell of 62 gr extend the compasses from the tangent of 45 gr vnto 5.00 in the line of numbers the same extent will reach from the tangent of 28 gr the complement of the parallell vnto 2.65 and if you describe a circle on the center A to the semidiameter of 2 inches 65 cent it shall be the parallell required In all vpright planes whether they be direct verticals or declining or meridian planes these parallels will be conicall sections and may be drawne through their points of intersection with the azimuth lines in the same maner as the parallels of declination through their points of intersection with the houre-lines To this end you may first finde the distance betweene the top of the style and the azimuth and then the distance betweene the horizon and the parallell both which may be represented in this maner On the center B and any semidiameter BH describe an occult arke of a circle and therein inscribe the chords of such parallels of altitude as you intend to draw on the plane I haue here put them for 15. 30. 45 and 60 gr then draw right lines through the center and the termes 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 That done consider your plane which here for example is the South face of our vertical plane p. 168 wherein hauing drawne both the horizontall verticall lines as I shewed before first take out AB the length of the style pricke that downe in this horizontall line from B vnto A then take out all the distances between B the top of the style and the seuerall points wherein the verticall lines do crosse the horizontal transfer them into this horizontal line BH from the center 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 seuerall distances betweene the horizon and the lines of altitude shall giue the like distances betweene the horizontall and the parallels of altitude vpon the azimuths in your plane Vpon this ground it followeth 1 To find the distance between the top of the style and the seuerall points wherein the azimuths do crosse the horizontall line Hauing drawne the horizontall and azimuth lines as before looke into the table by which you drew them and there you shall haue the angles at the zenith Then As the cosine of the angle at the zenith is to the sine of 90 gr So the length of the style to the distance required 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 ESE 67 30 24 14 26 13 7 02 15 08 EbS 78 45 50 27 51 26 13 73 29 60 East 90 0 Infinit Infinit Infinit Infinit As in our example of the verticall 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 complement of 11 gr 15 m. the angle at the zenith belonging to SbE and SbW vnto the sine of 90 gr the same extent wil reach from 10.00 the length of the style vnto 10.20 for the distance betweene the top of the
style and the intersection of the azimuth SbE with the horizontall line which distance may be called the secant of the azimuth and may serue for the drawing of the parallell of 45 gr from the horizon The like reason holdeth for the rest of these distances here represented in the line BH 2 To finde the distance betweene the horizon and the parallels As the tangent of 45 gr to the tangent of the parallell So the secant of the azimuth to the distance required As if it were required to draw the parallell of 15 gr from the horizon vpon this verticall plane extend the compasses from the tangent of 45 gr vnto the tangent of 15 gr the same extent will reach in the line of numbers from 10. 00 the secant of the South azimuth vnto 2.68 and therefore the distance betweene the horizon and the parallell of 15 gr is 2 inches 68 cent vpon the South azimuth Againe the same extent will reach from 10.20 the secant of SbE vnto 2.73 for the like distance belonging to SbE and SbW and so for the rest which may be gathered and set downe in the table That done and the horizon and azimuths being drawne pricke downe 10 inches from the horizontall line vpon 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 seuerall azimuths then if you draw a crooked line through all these points that may make no angles the line so drawne shall be the parallell of 45 gr from the horizon In like maner may you draw the parallell of 15 gr or any other parallell of altitude vpon any verticall plane If the plane incline to the horizon after we haue found the verticall point and drawne the horizontall line we are farther to finde the length of the axis of the horizon then the angles betwixt this axis and the azimuth lines and so the seuerall distances betweene the parallels and the verticall point all which may be represented in this manner On the center B and any semidiameter describe an occult quadrant of a circle and therein inscribe the chords of such parallels of altitude as you intend to draw on the plane drawing right lines through the center and the termes 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 Or hauing the Sector you may draw an occult line VE perpendicular to the axis VB and therein prick downe the tangent of the complement of the inclination of the plane from V vnto E then draw the line EF parallell to the axis crossing the line VH produced in the point F so this line EF will be as the line of sines vpon the Sector and therein you may pricke downe the sines of the complement of the angles at the zenith from E towards F and draw the verticall lines by those points through the lines of altitude so the angles at V betweene the axis VB and those azimuth lines shall be the angles betweene the axis of the horizon and the azimuth lines on your plane and the seuerall distances between the point V and the lines of altitude shall giue the like distances betweene the verticall point and the parallels of altitude vpon the azimuths in your plane Vpon this ground it followeth 1 To find the length of the axis of the horizon The verticall point is alwayes either directly ouer or vnder the top of the style and the distance betweene them is that which I call the axis of the horizon which may thus be found As the cosine 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 complement of the inclination vnto the sine of 90 gr the same extent wil reach in the line of numbers from 6.00 vnto 7.42 such is VB the length of the axis required 2 To find the angles contained between the horizon and the verticall lines vpon your plane The angles at the vertical point betweene the axis of the horizon and the azimuth lines vpon your plane are represented in this figure by those at V betweeâ—e VB and the azimuths The angles betweene the horizon and the azimuth lines being complements to the former are represented either by those which are made by VE or by BH and the azimuth lines which are drawne from V. In our example where the inclination to the horizon is 36 gr and the angle at the zenith betweene 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 vnto the tangent of 36 gr the same extent wil reach from the sine of 65 gr 40 m. the complement of the angle at the zenith vnto 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 wil reach froÌ„ the cosine of 35 gr 35 m. the angle at the zenith belonging to SbE vnto the tangent of 30 gr 3 m. for the angle betweene the horizon and the azimuth line of SbE The like reason holdeth for the rest which may be found and set downe 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 EbS 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 SW 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