BY the Precepts before in this Treatise delivered you have the whole Doctrine both of plain Right-lined Triangles and also of Sphericall both Right and Oblique-angled wrought by a way not usually practised which if it be carefully performed you may by a Line of Chords of 3 or 4 inches Radius come within a few minutes of Calculation which will be of sufficient use for any Practice either in Astronomicall Geographicall or Nauticall Questions wrought thereby And if in the working of any of the Cases there be any difficulty it is in the moving of the Compasses along one Line while the other foot being turned about doth only touch another Line But this is not a real but a seeming difficulty for in 3 or 4 times trial you will finde it very expeditious and easie and although it be not so Geometricall as the way by drawing of Parallels yet it is altogether as exact more ready and speedy in performance and avoids the drawing of multiplicity of Lines which would much cumber the Diagrams to no purpose If the Schemes that are here drawn seem to be cumbred with multiplicity of Lines the reason is because many of them serve for 3 4 or 5 Cases but if a particular Diagram had been drawn for every Case then none of them would have consisted of above 5 Lines besides the Quadrant as you will see by trying of any Case And here note that every one of these Cases both in Right-lined and Sphericall Triangles is applicable to some one thing or other in the Practice of the Mathematicks As by plain Right-lined Triangles all Propositions are resolved that concern the taking of Heights Depths and Distances Measuring of Land Sailing by the plain Sea-Chart and also by Mercator's and divers other particulars The Solution of Sphericall Triangles resolves Questions in Astronomy Geography Dialling Sailing by the Arch of a great Circle with many other particulars too tedious here to be recited of some of which especially those that concern the taking of Distances in Astronomy Geography Navigation c. I shall in the following Treatises exemplifie their Vse And because the Doctrine of Sphericall Triangles is more difficult then that of Plain Triangles I will in this next Treatise following shew you how by your Line of Chords onely you may draw or project the Globe or Sphere upon a Plain by which means you may see before you how the Sphericall Triangles lie in the Sphere it self which will be a great help to the understanding of the nature of a Sphericall Triangle The Sphere being thus projected upon a Plain I will shew how upon that Plain to measure both the Sides and Angles of a Sphericall Triangle and in so doing resolve some Questions in Astronomy by which you may discern in that particular how subservient Trigonometry is to Astronomy and as the Proverb is judge of Hercules by his Foot for to instance in the Variety of things that the Doctrine of Triangles is assistent to were an endless Work To project the SPHERE Upon the Plain of the MERIDIAN BY A LINE OF CHORDS Whereby the Sides and Angles of Sphericall Triangles are naturally laid down in Plano as they are in the Sphere it self By which the nature of them is discovered and their Sides and Angles measured with speed and exactness The Fourth EXERCISE BEing to treat of the Projection of the Sphere in Plano I suppose the Reader to be acquainted with the Doctrine of the Sphere or Globe and with the Circles thereof the nature of them how they are there situate one from another in respect of distance and to what use each of them is appropriate But if there be any that hath a desire to make use of this Treatise and is ignorant of the Sphericks let them reade the Books of such as have written of the Sphere or Globe In Latin there are divers as Theodocius Orontius Clarius c. In English there is Record's Castle of Knowledge Hill's School of Skill Blundevil's Exercises Hews of the Globes Newton Moxon c. But that this Treatise may not be accounted deficient in that which is so absolutely necessary for the understanding and practice of what is herein contained I will in this place give you a brief and plain Description of the Names Properties Distances c. of such Circles of the Sphere as in this Book we shall have occasion to project for the delineating of Sphericall Triangles in Plano and that in a correspondent Position to their situation on the Globe or Sphere it self The Circles therefore chiefly made use of in this Projection are these 1. The Meridian 2. The Horizon 3. The Aequinoctial 4. The Ecliptick 5. The Prime Verticall or Circle of East and West 6. The Hour-Circles 7. Azimuths or Verticall Circles 8. The two Tropicks 9. Parallels or Circles of Declination 10. Circles or Parallels of Altitude Of these Circles all but the three last are great Circles of the Sphere which divide it into two equal parts and the two Tropicks the Circles of Altitude and Declination are smaller Circles and divide the Sphere into two parts unequally Besides these great and small Circles there are severall Points of note upon the Globe as 1. The Zenith which is the Point in the Heavens directly over our heads in what part of the Earth soever we be 2. The Nadir which is the Point directly under our feet 3. The Poles of the World about which the Heavens are moved 4. The Poles of the Ecliptick 5. The Poles of all other Circles I. Of the MERIDIAN THE Meridian is a great Circle of the Sphere which passeth through both the Poles of the World and also through the Zenith and Nadir Points and crosseth the Horizon in the North and South Points thereof Unto this Circle any Day in the year when the Sun cometh it is Noon or Mid-day and when the Moon Stars or Planets in the Night come to touch this Circle they are then said to be upon the Meridian or at the highest they will be that Night This Circle in the Scheme of this Projection is noted by the Letters Z H N O. II. Of the HORIZON THE Horizon also is a great Circle of the Sphere and it is that Circle which divideth the visible part of the Heavens which we see from the not visible that is it divideth the Sphere into two Hemispheres the lower and the higher To this Circle when either the Sun Moon Stars or Planets come on the East part they are then said to rise and when they have passed from the Easterly Point by the Meridian and descended to the Western part of this Circle they are then said to set This Circle is represented in the Projection by the right Line H A O. III. Of the AEQVINOCTIAL THE Aequinoctial is a great Circle and in the Sphere it is elevated above the Horizon upon the Meridian Circle so much as is the Complement of the Latitude of the Place As at London where the Latitude
is 51 degr 30 min. there the Aequinoctial is elevated 38 degr 30 min. which is so much as 51 degr 30 min. wants of 90 degr and it cutteth the Horizon in the Points of East and West Unto this Circle when the Sun cometh which is twice every year namely about the 10. of March and the 12. of September it causeth the Daies and Nights to be of equal length all the World over This Circle is noted in the Scheme with AE A ae and cuts the Horizon in the Point A which represents both the East and West Points thereof IV. Of the ECLIPTICK THIS also is a great Circle of the Sphere and in the Northern Hemisphere where the North Pole is visible above the Horizon and the South Pole not visible is elevated above the Aequinoctial Circle so much as is the Sun 's greatest Declination which is 23 degr and about 30 min. and is as much depressed below the Aequinoctial in the Southern Hemisphere This Circle is called by some The Way of the Sun for that the Sun in his motion never swerveth or goeth out thereof and so his Longitude or Place is counted in this Line It cutteth the Horizon in the East and West Point A as the Aequinoctial did It is represented in the Scheme by the Line ♋ A ♑ and hath charactered upon it the 12 Signs of the Zodiack the six Northern Signs ♈ ♉ ♊ ♋ ♌ and ♠being on that half which is above the Horizon and the six Southern Signs ♎ ♠♠♑ ♒ and ♓ on the other half which is below the Horizon V. Of the PRIME VERTICALL THE Prime Verticall or Circle of East and West generally called the Aequinoctial Colure and then as the Sphere is here projected the Meridian representeth the Solstitial Colure is a great Circle passing through the Zenith and Nadir Points and also through the East and West Points of the Horizon Unto this Circle when the Sun Moon Stars or Planets do in their Motions arrive they are then due East or West It is in the Projection signified by the right Line Z A N passing through Z the Zenith N the Nadir and A the East and West Point of the Horizon and also cutteth the Aequinoctial in the Points ♈ and ♎ VI. Of the hour-HOVR-CIRCLES THE hour-Hour-Circles are great Circles of the Sphere meeting together in the Poles of the World and crossing the Aequinoctial at right Angles dividing it at every 15 degrees and then every of those Divisions is one Hour of time but if they pass through other parts of the Aequinoctial dividing it unequally then do those Hour-Circles represent unequal Spaces of time according to the distance they are from the Meridian or one from another Of these Circles in the Scheme of the Projection there are four thus noted P B S P A S P C S and P D S. VII Of the AZIMVTH CIRCLES THE Azimuth or Verticall Circles are great Circles of the Sphere meeting together in the Zenith and Nadir Points as the hour-Hour-Circles do in the Poles of the World and divide the Horizon at right Angles either equally or unequally as the Hour-Circles did the Aequinoctial In the Scheme of the Projection there are four of these Verticall Circles thus noted Z O N Z F N Z A N and Z G N. VIII Of the TROPICKS THE Tropicks are lesser Circles of the Sphere dividing it unequally and are drawn parallel to the Aequinoctial at 23 degr 30 min. distance therefrom equal to the Sun 's greatest Declination on either side That Tropick which is on the North-side is called The Tropick of Cancer to which when the Sun cometh which is but once in a year about the 10. of June it maketh the longest Daies to all the Northern Inhabitants of the World and the shortest Nights The other Tropick which is on the South-side of the Aequinoctial is called The Tropick of Capricorn to which when the Sun cometh which is about the 11. of December it maketh the shortest Daies and the longest Nights to all Northern Inhabitants and the contrary to all the Southern Inhabitants of the World In the Projection the Tropick of Cancer is signfied by ♋ I ♋ and the Tropick of Capricorn by ♑ K ♑ IX Of the CIRCLES or PARALLELS of DECLINATION THESE also are smaller Circles of the Sphere and are drawn parallel to the Aequinoctial towards both the Tropicks and up to them Those that are on the North-side of the Aequinoctial are called Parallels of North Declination and those that are on the South-side of the Aequinoctial are called Parallels of South Declination Of these Parallels there are in the Scheme of the Projection two one towards the Tropick of Cancer the other towards the Tropick of Capricorn and either of them 20 degrees distant from the Aequinoctial The Northern Parallel of Declination is noted with ♊ ☉ ♌ and the Southern with ♒ E ♠X. Of the CIRCLES or PARALLELS of ALTITVDE THE Circles of Altitude are likewise small Circles of the Sphere and are drawn parallel to the Horizon as the Circles of Declination were to the Aequinoctial These Parallels are drawn from the Horizon towards the Zenith Point and upon occasion in many Cases quite up unto it By these Parallels are measured the Altitude or Height of the Sun Moon and Stars In the Scheme there is onely one of them and that is expressed by the Letters M E L. Thus have I given you a brief and plain Description of the Circles both great and small which we shall have occasion to use in this following Treatise And here note that every Circle of the Sphere both great and small hath his proper Poles which Poles of all the great Circles are 90 Degrees or a Quadrant of a Circle distant from the Circle it self The Poles of the Circles in this Projection are as followeth Z and N Are the Poles of H A O the Horizon P and S Are the Poles of AE A ae the Aequinoctial O and H Are the Poles of Z A N the Prime Verticall Q and R Are the Poles of the Ecliptick AE and ae Are the Poles of P A S the Axis of the World The Poles of these five Circles are all in the Meridian and so there needeth no farther Precept for the finding of them and the Pole of the Meridian is the Centre thereof But for the three Azimuth Circles they fall in several Points of the Horizon and the three hour-Hour-Circles in certain Points in the Aequinoctial How to finde which Points shall be shewed afterwards in due place A Is the Pole of the Meridian Z H N O. T Is the Pole of the Azimuth Circle Z F N. G Is the Pole of the Azimuth Circle Z ☉ N. ☉ Is the Pole of the Azimuth Circle Z G N. X Is the Pole of the hour-Hour-Circle P B S. Y Is the Pole of the hour-Hour-Circle P D S. V Is the Pole of the hour-Hour-Circle P C S. The Poles of the World P and S are also the Poles of the
Tropicks and of all the Parallels of Declination And The Zenith and Nadir Z and N are the Poles of all the Parallels of Altitude Having sufficiently acquainted the Reader with the several Circles Lines Points and Poles belonging to every Circle I will now proceed to my intended purpose namely to project or lay down in Plano all these Circles Lines Points and Poles in their true Positions How to project the Sphere upon the Plain of the Meridian FIrst take 60 degr of your Line of Chords and with that distance upon the Point A. as a Centre describe the Circle Z H N O representing the Meridian within which Circle all the rest are to be projected and cross it with the two Diameters H A O the Horizon and Z A N the Prime Verticall Secondly because the Latitude of the place for which you draw your Projection viz. London is 51 degr 30 min. take 51 degr 30 min. from your Line of Chords and set them upon the Meridian from Z to AE and from N to ae and draw the Line AE A ae for the Aequinoctial Also set 51 degr 30 m. from O to P and from H to S and draw the Line P A S representing the Axis of the World and the Hour-Circle of 6 a clock Thirdly take 23 degr 30 min. the quantity of the Sun 's greatest Declination and also of the distance of the two Tropicks from the Aequinoctial and set them upon the Meridian from AE to ♋ above the Aequinoctial and also from AE to ♑ below the Aequinoctial In like manner set the same distance of 23 degr 30 min. from ae to ♋ above the Aequinoctial and from ae to ♑ below it This done lay a Ruler upon the Points AE and ♋ and it will cut the Axis of the World P A S in the Point I. So a Circle drawn which shall pass through these three Points ♋ I ♋ shall be the Tropick of Cancer Again lay a Ruler to AE and ♑ and it will cut the Axis in the Point K. So a Circle drawn through ♑ K ♑ shall be the Tropick of Capricorn But to shew how you may find the Centres upon which these Tropical Circles are to be described I must make this Diversion But for those Parallels of Altitude which fall near the Horizon those Circles or Parallels of Declination which fall near to the Aequinoctial those hour-Hour-Circles which fall near to the Axis of the World or Hour of Six and those Azimuth Circles which are near to the Prime Verticall or Azimuth of East and West those that make Mathematicall Instruments have an Instrument called a Bow which by the help of one or more Screws according to the length of the Bow may be extended to touch any three Points which lie near in a straight Line by the edge of which Bow you may draw your Hour-Circles Azimuths Parallels of Declination and Altitude as easily as you may draw a right Line by the edge of a Ruler But to return again to our Projection Fourthly draw a right Line ♋ A ♑ between the two Tropicks touching the Tropick of Cancer above the Horizon at ♋ and the Tropick of Capricorn below the Horizon at the Point ♑ This Circle hath upon it the Characters of the 12 Signs of the Zodiac which are to be put on in this manner Take 23 d. 30 min. out of your Line of Chords and set them from P to Q and from S to R which Points Q and R are the two Poles of the Ecliptick Then take 60 degr from your Line of Chords and set them from Q to 1 and from Q to 3. Also set the same distance from ♋ to 2 and from ♑ to 4. This done lay a Ruler to the Pole R and the figure 1 it will cut the Ecliptick in the Point ♊ and ♌ the Ruler laid to R and 2 will cut it in the Point ♉ and ♠and laid to R and 4 in ♠and ♓ and laid to R and 3 in ♠and ♒ So have you the true Points for the Sun's entrance into every Sign And if you would have every tenth degree of each Sign divide every of the Spaces ♋ 1 12 2 Q Q 4 4 3 and 3 ♑ into three equal parts so will each part contain 10 d. and a Ruler laid to each of them and the Point R shall give you the Points upon the Ecliptick answering to the 10. degr of every Sign And in the same manner may you if your Projection be large put on every Degree Fifthly for the putting on of the Hour-Circles consider how far the Circle you are to put on is distant from the Meridian and set so many degrees upon the Meridian from the Aequinoctial a Ruler laid from Z to those degrees will cross the Aequinoctial and through that Point in the Aequinoctial where the Ruler so crosseth the Hour-Circle will pass Example The Hour-Circle P B S in this Projection is distant from the Meridian 62 d. 46 m. wherefore take 62 d. 46 m. from your Chords and set them from a to b then laying a Ruler from Z to b it will cut the Aequinoctial in B through which Point the Hour-Circle of 62 d. 46 m. must pass To find the Centre of this Hour-Circle and so of any other repair to the former Scheme for finding of the Centres of the Parallels of Altitude and Declination and because this Hour-Circle is distant from the Meridian 62 degr 46 min. take 62 degr 46 min. from your Line of Chords and set them upon the Quadrant A B C from C to l and draw the Line A l m. So shall the Line A l m be the Semidiameter of the Hour-Circle P B S which being taken in your Compasses and set upon your Projection from B upon the Line B AE being extended shall there give you the Centre of that Hour-Circle And in the same manner may the Centres of all the rest be found Sixthly the Azimuth Circles are to be drawn upon the Projection and the Centres of them found in all respects as the Hour-Circles were So the Azimuth Circle Z ☉ N being 56 degr 40 min. from the Meridian take 56 degr 41 min. out of your Line of Chords and set them upon the Meridian of your Projection from O to d then laying a Ruler unto Z and d it will cut the Horizon in the Point ☉ through which the Azimuth of 56 degr 41 min. Z ☉ N must pass Then to find its Centre repair to the former Scheme for finding of Centres and upon the Quadrant A B C set 56 degr 41 min. of your Chords from C to n and draw the Line A u o so shall the Line A u o be the Semidiameter of the Azimuth Circle Z ☉ N which being taken in your Compasses and set upon your Projection from ☉ upon the Line ☉ H being extended shall there give you the Centre of the Azimuth Circle Z ☉ N. And in this manner may the Centre of any other Azimuth Circle be found
And here note I. That the Centres of all Azimuth Circles fall in the Horizon H A D being extended where need is The Centres of all the hour-Hour-Circles fall in the Aequinoctial Line AE A ae being extended The Centres of the Tropicks and Parallels of Declination fall in the Axis of the World P A S extended And the Centres of the Circles of Altitude fall in the Prime Verticall Circle Z A N. Note II. That if the middle Point of any Hour-Circle do not fall just in the Aequinoctial or any Azimuth Circle just in the Horizon but on either side of them then you may find the Centres by the Geometricall Propositions at the beginning of this Book though there be other waies to find the Centres upon the Projection it self which I omit for that I would not cumber the Scheme with unnecessary Lines Seventhly Every Circle in the Projection hath its proper Pole as was before intimated Now for the finding of them you are to note that the Pole of every great Circle is 90 degr or a Quadrant of a Circle distant from the Circle it self upon that Line which cutteth the Circle at right Angles Thus the Poles of all the Hour-Circles are upon the Aequinoctial and the Poles of all the Azimuths upon the Horizon Now if you would find the Pole of the Hour-Circle P D S lay a Ruler upon P and D and it will cut the Meridian Circle in e then take 90 degr of your Line of Chords and set them from e to f a Ruler laid from P to f will cut the Aequinoctial in Y so is Y the Pole of the Hour-Circle P D S. Lastly The finding of the Poles of the Azimuth Circles is the same with the Hour-Circles So if you would find the Pole of the Azimuth Circle Z G N lay a Ruler upon Z and G it will cut the Meridian Circle in g then set 90 degr of your Chord from g to d so a Ruler laid from Z to d will cut the Horizon H A O in the Point ☉ which Point ☉ is the Pole of the Azimuth Circle Z G N. And thus have you found the Poles of one of the Hour and one of the Azimuth Circles And by the same manner of Work you may find the Poles of all the rest As The Pole of the Hour-Circle P D S will be found at Y The Pole of the Hour-Circle P C S will be found at V The Pole of the Hour-Circle P A S will be found at AE or ae The Pole of the Hour-Circle P B S will be found at X The Pole of the Azimuth Circle Z G N will be found at ☉ The Pole of the Azimuth Circle Z A N will be found at H or O The Pole of the Azimuth Circle Z F N will be found at T The Pole of the Azimuth Circle Z ☉ N will be found at G The Poles of the Horizon H A O are Z and N the Zenith and Nadir The Poles of the Aequinoctial AE A ae are P and S the Poles of the World The Poles of the Ecliptick ♋ A ♑ are Q and R. Thus have I given you at large a plain and easie method how to project the Sphere upon the Plain of the Meridian Circle by help of the Line of Chords onely Vpon which Projection by the intersection or crossing of the severall Circles thereof are constituted divers Sphericall Triangles some Right-angled and others Oblique-angled By the resolving of which Triangles variety of Questions appertaining to Astronomie Geographie and Navigation may with speed and exactness be resolved But before I come to shew the manner of working particular Questions of any kind it will be expedient that I shew you 1. how to measure or find the quantity of the Sides and Angles of a Sphericall Triangle as they are here projected and 2. how to project or lay down an Angle or Side of any quantity that shall be required I. A Sphericall Triangle being projected how to find the quantity of any Angle thereof LAY a Ruler to the angular Point and the extremity of the Sides containing the Angle they being continued to Quadrants and note where the Ruler cuts the Meridian or outward Circle at both which places make marks upon the Meridian the distance between those two marks being measured upon your Line of Chords shall give you the quantity of the Angle required Example I. IN the Triangle P ☉ O in the Projection let it be required to find the quantity of the Angle ☉ P O. First lay a Ruler upon the angular Point P and to the extreme ends of the Sides P ☉ and P O they being extended to Quadrants which is to that Circle which measures that Angle as the Aequinoctial measures all the Angles at P the Pole of the World the Horizon all the Angles at Z the Zenith c. So the Ruler laid from P to ae will cut the Meridian in ae and being laid from P to B it will cut the Meridian in the Point b. The distance b ae being taken in your Compasses and measured upon your Line of Chords will be found to contain 62 degr 46 min. which is the quantity of the Angle ☉ P O. But if upon the Point P you were to project an Angle to contain 62 degr 46 min. then take 90 degr of your Chords and set them from P to ae and through the Centre A draw the Line AE A ae then take 62 degr 46 m. out of your Line of Chords and set them from ae to b and laying a Ruler from P to b it will cut AE A ae in the Point B the Circle P B S being drawn the Angle at P will contain 62 degr 46 min. Example II. LET it be required to find the quantity of the Angle Z E P. Lay a Ruler to ☉ the Pole of the Circle Z E N and the Point E it will cut the Meridian Circle in M from M set 90 degr to z a Ruler laid from ☉ to z will cut the Circle Z E N it being extended beyond the Zenith Z at the Point δ. Again Lay a Ruler upon Y the Pole of the Circle P E S and it will cut the Meridian Circle in v set 90 degr from v to x upon the Meridian a Ruler laid from Y to x will cut the Circle P E S in y. This done lay a Ruler from E to δ and it will cut the Meridian in θ also lay the Ruler from E to y it will cut the Meridian in λ the distance θ λ being taken in your Compasses and applied to your Line of Chords will be found to contain 21 degr 45 min. And such is the quantity of the Angle Z E P. These two sorts of Angles are the most troublesome to find their quantities and therefore I have instanced in them There are other Angles in the Projection which render their measures to the eye without farther Instructions for finding their quantities II. A Sphericall Triangle being projected to find the quantity of
Upper Wing of Pegasus 341 53 13 21 N 2 In the tip of Pegasus Wing 358 52 13 15 N 2 PROP. VII The Latitude of the Place and the Sun's Declination being given to find at what Hour the Sun will be upon the true East or West Points UPON the Projection there are two Right-angled Sphericall Triangles by either of which this Proposition may be solved The one is the Triangle Z P o made by the Intersections of Z o an Arch of the Prime Verticall P o an Arch of an Hour-Circle and Z P an Arch of the Meridian In which Triangle there is given Z P the Perpendicular the Complement of the Latitude of the Place 38 degr 30 min. and the Hypotenuse P o the Complement of the Sun's Declination 70 degr to find the Angle at the Perpendicular Z P o which you may doe by the 14. Case of Right-angled Sphericall Triangles The other Triangle is o C A right-angled at C and is constituted of o C an Arch of an Hour-Circle C A an Arch of the Aequinoctial and o A an Arch of the Prime Verticall In which Triangle you have given 1. the Perpendicular O C the Sun's Declination 2. the Angle at the Base C A o the Latitude 51 degr 30 min. to find the Base C A. Thus having the Perpendicular and the Angle at the Base you may find the Base C A as followeth this being The Analogie or Proportion As the Tangent of the Latitude 51 degr 30 min. is to the Tangent of the Sun's Declination 20 degr So is the Radius 90 degr to the Co-sine of the Hour from Noon To resolve the Proposition by the Projection Lay a Ruler upon P the Pole of the World and the Angle C of your Triangle the Ruler will cut the Meridian Circle in the Point g So g Ae being taken in your Compasses and measured upon your Line of Chords will be found to contain 73 degr 10 min. which converted into Hours and Minutes will be 4 hours and about 53 min. So that the Sun when he hath 20 degr of Declination will come to the East Point at 7 min. past 7 in the Morning and will be due West 53 min. after 4 in the Afternoon PROP. VIII Having the Latitude of the Place and the Sun's Declination given to find what Altitude the Sun shall have when he is upon the true East or West Points THIS Proposition may be resolved by either or both of the Triangles mentioned in the last Proposition For in the Triangle Z P o you have given P Z the Perpendicular and P O the Hypotenuse to find Z o the Base by the 8. Case of Right-angled Sphericall Triangles But in the Triangle o C A you have given 1. the Perpendicular o C the Sun's Declination 20 degr 2. the Angle at the Base o A C the Latitude of the Place 51 degr 30 m. to find the Hypotenuse o A for which this is The Analogie or Proportion As the Sine of the Latitude 51 degr 30 min. is to the Sine of the Declination 20 degr So is the Radius 90 degr to the Sine of the Sun's Altitude being due East or West 25 degr 55 min. To resolve the Proposition by the Projection Lay a Ruler upon O one of the Poles of the Prime Verticall and to the Angle o of the Triangle a Ruler thus laid will cut the Meridian Circle in the Point p So the distance H p being taken and measured upon the Line of Chords will be 25 degr 55 min. and such height will the Sun have when he is either East or West PROP. IX The Latitude of the Place and the Declination of the Sun being given to find what Altitude the Sun shall have at Six of the Clock FOR finding of the Triangles upon the Projection which will resolve this and the following Propositions you must suppose another Azimuth Circle to be drawn in the Projection from Z to N and through that Point where the Parallel of Declination ♊ ☉ ♌ and the Axis of the World or Hour-Circle of Six P A S do cross each other The drawing of which Azimuth Circle I purposely omitted chiefly because the Scheme in that place is more cumbred with Lines and Letters then any other part thereof But you may well enough for the solving of these two Propositions imagine it to be drawn the Pole whereof is at * This Azimuth Circle being supposed to be drawn you have upon the Projection two Triangles like-angled which will perform the Work of resolving this Proposition In one of which you have given the Base which is the Complement of the Declination and the Perpendicular which is the Complement of the Latitude to find the Hypotenuse which is the Complement of the Sun's Altitude required This Triangle may be resolved by the first Case aforegoing In the other Triangle there will be given the Hypotenuse which is the Sun's Declination and the Angle at the Base which is the Latitude to find the Perpendicular which is the Sun's Altitude at Six a Clock To find which this is The Analogie or Proportion As the Radius 90 degr is to the Sine of the Sun's Declination 20 degr So is the Sine of the Latitude 51 degr 30 min. to the Sine of the Sun's Altitude at Six 15 degr 30 min. To resolve the Proposition by the Projection Lay a Ruler upon the Point * and that Point where the Parallel of Declination ♊ ☉ ♌ crosseth the Axis or Hour of Six the Ruler thus laid will cut the Meridian Circle in the Point g. So O g being measured upon the Chords will give you 15 degr 30 min. And such Altitude will the Sun have at the Hour of Six in the Latitude of 51 degr 30 min. when he hath 20 degr of Declination PROP. X. The Latitude of the Place and the Declination of the Sun being given to find what Azimuth the Sun shall have at Six a Clock THE two Triangles that were supposed in the last Proposition to be drawn upon the Projection will resolve this Proposition also but seeing the Triangles are not drawn but supposed I will onely give you the Analogie and then the way of working it upon the Projection The Analogie or Proportion As the Co-sine of the Latitude 38 degr 30 min. is to the Radius 90 degr So is the Co-tangent of the Sun's Declination 70 degr to the Tangent of the Sun's Azimuth counted from the North part of the Meridian 77 degr 14 min. To resolve the Proposition upon the Projection Lay a Ruler to the Zenith-point Z and upon the Point where the Parallel of Declination cuts the Hour-Circle of Six the Ruler thus laid will cut the Meridian Circle in r So the distance r O being measured upon the Line of Chords will contain 77 degr 14 min. which is the Azimuth from the North part of the Meridian The distance N r measured upon the Chords will give you 12 degr 46 min. which is the Azimuth