Nine geometricall exercises, for young sea-men and others that are studious in mathematicall practices: containing IX particular treatises, whose contents follow in the next pages. All which exercises are geometrically performed, by a line of chords and equal parts, by waies not usually known or practised. Unto which the analogies or proportions are added, whereby they may be applied to the chiliads of logarithms, and canons of artificiall sines and tangents. By William Leybourn, philomath.

PROPOSITIONS ASTRONOMICALL, Usefull in the Practice of NAVIGATION: Performed by the resolving of severall Sphericall Tri∣angles upon the Projection. The Sixth EXERCISE.

OF the vast number of Sphericall Questions that are resolvable by the severall Sphericall Triangles that are (by the Intersections of the great Circles of the Sphere) constitu∣ted upon the Projection, seeing there may be 30 in every Right-angled, and 60 in every Oblique-angled Sphericall Triangle resolved; I have made choice onely of such as I conceived most necessary for Sea-mens use. Of which, some of them will be assistent to them in finding the Latitude; others for finding the Sun's Altitude upon some certain Points

of the Compass, and at some particular Hours: others will help them to find the Variation of their Compass; and some of them will help them to the Hour of the Day and Night.

By the manner of Working of these Propositions upon the Pro∣jection, the ingenious Practitioner may propose divers other Questions to himself, and project the Sphere in Plano sutable unto them; in the Practice and true performance whereof he will accumulate to himself both Pleasure and Profit.

Besides the working of these Propositions upon the Projection, I have to every of them added the Analogie or Proportion, by which they may be resolved by the Canons of Artificial Sines and Tangents; so that such as require more exactness then can be ex∣pected from Instrumental performance, may by those Canons work and resolve their Questions more accurately. Many more Propositions I could have added, but these (for the Astrono∣micall part) I deemed sufficient: such as concern Geographie and Navigation shall follow in convenient place.

PROP. I. The distance of the Sun from the nearest Aequinoctial Point (either Aries or Libra) given, to find his Declination.

IN the Projection this Proposition is to be resolved upon the Right-angled Sphericall Triangle A k ♒, right-angled at ♒: which Triangle is constituted by the Intersection of the Arches of three great Circles of the Sphere; namely, of A ♒, an Arch of the Ecliptick; A k, an Arch of the Aequinoctial; and of ♒ k, an Arch of a great Circle passing through Q and R the Poles of the Ecliptick, and cutting the Ecliptick at right Angles in ♒, which is the place of the Sun at the time of the Que∣stion.

In which Triangle you have given (besides the right Angle

at ♒) (1.) the Side A ♒, 59 degr. which is the distance of the Sun from the nearest Aequinoctial Point Libra; (2.) the Angle ♒ A k, 23 degr. 30 min. which is the Angle that the Ecliptick makes with the Aequinoctial, and is alwaies equal to the greatest Declination of the Sun. Wherefore in this Tri∣angle you having given the Base A ♒, and the Angle at the Base k A ♒, you may find the Perpendicular k ♒, the Sun's Declination, by the 11. Case of Right-angled Sphericall Tri∣angles in the first Part of this Book. For which this is

The Analogie or Proportion.

As the Radius 90 degr. is to the Sine of the Sun's greatest De∣clination 23 degr. 30 min.

So is the Sine of the Sun's distance from the next Aequinoctial Point Libra 59 degr. to the Sine of the Sun's present Decli∣nation 20 degr.

To resolve the Triangle upon the Projection,

Lay a Ruler upon the Pole of the Circle R ♒ Q, (which is at ♉,) and the Point k, it will cut the Meridian in the Point l: Also lay a Ruler from ♉ to ♒, it will cut the Meridian in the Point ♑. So the distance l ♑, being measured upon your Line of Chords, will contain 20 degr. the Sun's Declination, he being in the 29. degree of ♒.

The like Declination the Sun hath when he is in 29 degr. of Taurus, in 1 degr. of Leo, or 29 degr. of Scorpio, every of which Points are distant from one of the Aequinoctial Points Aries or Libra 59 degr.

PROP. II. The Latitude of the Place, and the Declination of the Sun, being given, to find the Ascensional Diffe∣rence.

UPON the Projection this Proposition is to be resolved by finding the Side A B of the Right-angled Sphericall Triangle A ☉ B, Right-angled at B; which Triangle is com∣pounded of three Arches of great Circles, namely, of A ☉, an Arch of the Horizon, A B, an Arch of the Aequinoctial, and ☉ B, an Arch of an Hour-Circle.

In this Triangle you have given (1.) the Side ☉ B, the Sun's Declination 20 d. (2.) the Angle ☉ A B, the Complement of the Latitude 38 d. 30 m. and the right Angle at B. In this Triangle therefore you have given ☉ B, the Perpendicular, and ☉ A B, the Angle at the Base, to find the Base A B, which you may doe by the 14. Case of Right-angled Sphericall Triangles. For which this is

The Analogie or Proportion.

As the Co-tangent of the Latitude 38 degr. 30 min. is to the Tangent of the Sun's Declination 20 degr.

So is the Radius 90 degr. to the Sine of the Ascensional Dif∣ference 27 degr. 14 min.

To resolve the Triangle upon the Projection,

Lay a Ruler to P, the Pole of the World, (and also of the Aequinoctial,) and the Point B, it will cut the Meridian Circle in the Point b; the distance b S, being taken in your Compasses and measured upon your Line of Chords, will reach from the beginning thereof to 27 degr. 14 min. the Ascensional Dif∣ference; which is so much as the Sun riseth or setteth before or after Six a Clock. So these 27 degr. 14 min. being turned

into Time (by allowing 15 deg. for one Hour, and one Degree for 4 minutes of Time) is 1 Hour and 49 min. and so much doth the Sun rise or set before or after the hour of Six, ac∣cording to the time or season of the Year: for if the Sun hath North Declination, then he riseth before Six, and sets after; but if the Sun have South Declination, then doth he rise after, and set before Six.

This Ascensional Difference being added to 6 Hours will give you the Semidiurnall Arch or Half-length of the Day; and being taken from six Hours, will leave the Seminocturnall Arch or Half-length of the Night.

The Semidiurnall Arch, when the Sun hath 20 degr. of North Declination, is represented in the Projection by the Arch ♊ a ☉, and the Seminocturnall Arch by ☉ ♌. The Semidiurnall Arch, when the Sun hath 20 degr. of South Declination, is represented by the Arch ♒ E r, and the Seminocturnall by the Arch r c ♐.

Though I have shewed how these may be found by adding and subtracting the Ascensionall Difference; yet they may be found by the Projection, for the Arches are measured upon the Aequino∣ctial. Wherefore lay a Ruler to P, the Pole of the World, and the Point B, it will cut the Meridian Circle in b: So the distance b AE, being measured by your Chord, will be 117 degr. 14 min. the Semidiurnall Arch, and b ae measured will be 62 degr. 11 min. for the Seminocturnall Arch.

PROP. III. The Latitude of the Place, and the Declination of the Sun, being given, to find his Amplitude.

THERE are two Triangles upon the Projection, by the resolving of either the Proposition may be resolved, and both of them Right-angled. The one is the Triangle P☉O,

Right-angled at O. The other is the Triangle made use of in the last Proposition, A ☉ B. The first Triangle is constituted of these three Arches, viz. P O, an Arch of the Meridian, P ☉, an Arch of an Hour-Circle, and ☉ O, an Arch of the Horizon.—The second Triangle consisteth of A ☉, an Arch of the Horizon, ☉ B, an Arch of an Hour-Circle, and A B, an Arch of the Aequinoctial. In the first Triangle you have given the Perpendicular P O, the Latitude of the Place, and the Hypotenuse P ☉, the Complement of the Sun's Decli∣nation 70 degr. by which you are to find the Base ☉ O, the Sun's Amplitude from the North part of the Meridian, which may be found by the 7. Case of Right-angled Sphericall Tri∣angles.—In the second Triangle B A ☉ (which is that I will first exemplifie in) you have given (1.) the Perpendicular ☉ B, the Sun's Declination 20 degr. (2.) the Angle at the Base ☉ A B, the Complement of the Latitude 38 degr. 30 min. to find the Hypotenuse ☉ A, the Sun's Amplitude from the East or West. So having the Perpendicular ☉ B, and the Angle at the Base ☉ A B, you may find the Hypotenuse ☉ A by the 4. Case of Right-angled Sphericall Triangles. And for it this is

The Analogie or Proportion.

As the Co-sine of the Latitude 38 degr. 30 min. is to the Ra∣dius 90 degr.

So is the Sine of the Sun's Declination 20 degr. to the Sine of the Amplitude from the East or West Points of the Horizon 33 degr. 20 min.

To work the Proposition upon the Projection,

Lay a Ruler upon Z the Zenith, (which is also one of the Poles of the Horizon,) and to the Point ☉, it will cut the Meridian Circle in d; and laid from Z to A, it will cut the Nadir Point in N: So the distance N d, being taken in your Compasses and measured upon your Line of Chords, will con∣tain

33 degr. 20 min. the quantity of the Hypotenuse, which is the Amplitude of the Sun's rising or setting from the true East or West Points of the Horizon.

To perform the same Work by the other Triangle P ☉ O, Lay a Ruler to Z the Zenith, and the Point ☉, and it will cut the Meridian Circle in the Point d, as before: So the distance d O, measured upon your Line of Chords, will con∣tain 56 degr. 40 min. which is the Amplitude of the Sun's ri∣sing or setting from the North Point of the Horizon.

PROP. IV. The Latitude of the Place, and the Declination of the Sun, being given, to find the Angle of the Sun's Position at the time of his rising.

THIS Proposition is resolvable upon the Triangle P ☉ O: In which there is given (1.) the Hypotenuse P ☉, the Complement of the Sun's Declination; (2.) the Perpendi∣cular P O, the Latitude: and it is required to find the Angle P ☉ O: which may be found by the 15. Case of Right-angled Sphericall Triangles, and by the following

Analogie or Proportion.

As the Co-sine of the Declination 70 degr. is to the Radius 90 degr.

So is the Sine of the Latitude 51 degr. 30 min. to the Sine of the Angle of the Sun's Position at the time of his rising.

By the Projection.

Lay a Ruler to X, (the Pole of the Hour-Circle P ☉ S,) and the Point ☉; the Ruler so laid will cut the Meridian Circle near the Point ♌: then set 90 degr. from ♌ to x upon the Me∣ridian, and lay the Ruler from X to x; so shall it cut the Hour-Circle

S ☉ P (being continued without the Meridian Circle) in the Point Φ. Again, the Ruler laid from ☉ to Φ will cut the Meridian Circle in the Point 2. So the distance O 2, being ta∣ken in the Compasses and applied to the Line of Chords, will be found to contain 56 degr. 29 min. which is the quantity of the Angle P ☉ O.

PROP. V. The Sun's Declination, and his Amplitude from the North part of the Horizon, being given, to find the Latitude.

IN the same Triangle as in the last Proposition P ☉ O, you have given (1.) the Base ☉ O, the Sun's Amplitude from the North part of the Horizon; (2.) the Hypotenuse P ☉, to find the Perpendicular P O. So there are two Sides given to find the third, which you may doe by the 8. Case of Right-an∣gled Sphericall Triangles.

The Analogie or Proportion.

As the Co-sine of the Amplitude from the North 33 degr. 20 min. is to the Radius 90 degr.

So is the Sine of the Declination 20 degr. to the Co-sine of the Latitude 38 degr. 30 min.

By the Projection.

This is easie; for if you take in your Compasses the distance from O to P upon the Meridian, and measure it upon your Line of Chords, it will be found to contain 51 degr. 30 min. the Latitude required.

PROP. VI. The Sun's greatest Declination, with his Distance from the next Aequinoctial Point (Aries or Libra,) being given, to find his right Ascension.

THE Triangle A k ♒ in the Projection resolves this Proposition, in which you have given (1.) the Side A ♒, the distance of the ☉ from ♎; (2.) the Angle ♒ A k, the Sun's greatest Declination, and the right Angle at ♒: So you have given the Base, and the Angle at the Base to find the Hypotenuse, which you may doe by the Case of Right-an∣gled Sphericall Triangles, and this

Analogie or Proportion.

As the Radius 90 degr. is to the Co-sine of the greatest De∣clination 66 degr. 30 min.

So is the Tangent of the Sun's distance from the next Aequi∣noctial point Libra 59 d. to the Tangent of the right Ascen∣sion 56 degr. 50 min.

By the Projection.

Lay a Ruler to P and k, it will cut the Meridian Cir∣cle in s: the distance S s, taken and measured upon your Line of Chords, will contain 56 degr. 50 min. the Sun's right Ascension.

I should here shew how the right Ascension and Declination of a Star might be found; but, the Calculation Trigonometricall being very laborious, I have therefore in this place omitted it, be∣cause I have in another Treatise, now speedily to be published, fra∣med Tables of the Longitude, Latitude, Right Ascension, De∣clination,

and Semidiurnall Arches, of the most eminent Fixed Stars in the Heavens, exactly calculated for the Year of our Lord 1670; and also other Tables of the Sun's right Ascension for every Degree of the Ecliptick and Day of the Year: by help of which Tables, the Hour of the Night, the Rising, Setting, and the time of their coming to the South, may be obtained by the Rules and Directions prescribed in the forementioned Treatise, which will contain (besides the Tables here mentioned) all such others as at any time the Sea-man shall have occasion for in his Practice, and divers other things too tedious here to enumerate. But in the mean time I shall request the Reader to be satisfied with these two Tables following: These Exercises being things onely Practicall.

A Table of the Sun's right Ascension for every De∣gree of the Ecliptick.

Degr.	♈	♉	♊	♋	♌	♍
0	0 0	27 54	57 48	90 0	122 12	152 6
1	0 55	28 51	58 51	91 5	123 14	153 3
2	1 50	29 49	59 53	92 11	124 16	154 1
3	2 45	30 46	60 56	93 16	125 19	154 58
4	3 40	31 44	61 59	94 22	126 20	155 54
5	4 35	32 42	63 2	95 27	127 22	156 51
6	5 30	33 40	64 6	96 32	128 24	157 48
7	6 25	34 38	65 9	97 38	129 25	158 44
8	7 21	35 37	66 13	98 43	130 26	159 40
9	8 16	36 36	67 17	99 48	131 27	160 37
10	9 11	37 34	68 21	100 53	132 28	161 33
11	10 6	38 33	69 25	101 58	133 28	162 29
12	11 1	39 33	70 29	103 3	134 29	163 25
13	11 57	40 32	71 34	104 8	135 29	164 20
14	12 52	41 31	72 38	105 13	136 29	165 16
15	13 48	42 31	73 43	106 17	137 29	166 12
16	14 44	43 31	74 47	107 22	138 29	167 7
17	15 40	44 31	75 52	108 26	139 28	168 3
18	16 35	45 31	76 57	109 31	140 27	168 58
19	17 31	46 32	78 1	110 35	141 27	169 54
20	18 27	47 32	79 7	111 39	142 26	170 49
21	19 23	48 33	80 12	112 43	143 24	171 44
22	20 20	49 34	81 17	113 47	144 23	172 39
23	21 16	50 35	82 22	114 51	145 22	173 35
24	22 12	51 36	83 28	115 54	146 21	174 30
25	23 9	52 38	84 33	116 57	147 18	175 25
26	24 6	53 40	85 38	118 1	148 16	176 20
27	25 2	54 41	86 44	119 4	149 14	177 15
28	25 59	55 44	87 49	120 7	150 11	178 10
29	26 56	56 46	88 55	121 9	151 9	179 5
30	27 54	57 48	90 0	122 12	152 6	180 1

A Table of the Sun's right Ascension for every De∣gree of the Ecliptick.

Degr.	♎	♏	♐	♑	♒	♓
0	180 0	207 54	237 48	270 0	302 12	332 6
1	180 55	208 51	238 51	271 5	303 14	333 3
2	181 50	209 49	239 53	272 11	304 16	334 1
3	182 45	210 46	240 56	273 16	305 19	334 58
4	183 40	211 44	241 59	274 22	306 20	335 54
5	184 35	212 42	243 2	275 27	307 22	336 51
6	185 30	213 40	244 6	276 32	308 24	337 48
7	186 25	214 38	245 9	277 38	309 25	338 44
8	187 21	215 37	246 13	278 43	310 26	339 40
9	188 16	216 36	247 17	279 48	311 27	340 37
10	189 11	217 34	248 21	280 53	312 28	341 33
11	190 6	218 33	249 25	281 58	313 28	342 29
12	191 2	219 33	250 29	283 3	314 29	343 25
13	191 57	220 32	251 34	285 8	315 29	344 20
14	192 52	221 31	252 38	286 13	316 29	345 16
15	193 48	222 31	253 43	287 17	317 29	346 12
16	194 44	223 31	254 47	288 22	318 29	347 7
17	195 40	224 31	255 52	289 26	319 28	348 3
18	196 35	225 31	256 57	290 31	320 27	348 58
19	197 31	226 32	258 1	291 35	321 27	349 54
20	198 27	227 32	259 7	292 39	322 26	350 49
21	199 23	228 33	260 12	293 43	323 24	351 44
22	200 20	229 34	261 17	294 47	324 23	352 39
23	201 16	230 35	262 22	295 51	325 22	353 35
24	202 12	231 36	263 28	296 54	326 21	354 30
25	203 9	232 38	264 33	297 57	327 18	355 25
26	204 6	233 40	265 38	298 1	328 16	356 20
27	205 2	234 41	266 44	299 4	329 14	357 15
28	205 59	235 44	267 49	300 7	330 11	358 10
29	206 56	236 46	268 55	301 9	331 9	359 5
30	207 54	237 48	270 0	302 12	332 6	360 0

The right Ascension, Declination, and Ma∣gnitude of some principal Fixed Stars.

The Stars Names.	Right A∣scension.	Declina∣tion.		Magni∣tude.
	D. M.	D. M.
The Pole-Star	7 47	87 27	N	2
The Girdle of Andromeda	12 32	33 48	N	2
The former Horn of the Ram	23 38	17 37	N	4
Bright Star in the Ram's Head	26 56	21 48	N	3
The Whale's Jaw	41 3	2 42	N	2
The Head of Medusa	41 27	39 35	N	3
The Bull's Eye	64 0	15 46	N	1
The Goat	72 44	45 35	N	1
The former Shoulder of O∣rion	76 38	4 59	N	2
The latter Shoulder of O∣rion	84 7	7 18	N	2
The great Dog	97 27	16 13	S	1
The uppermost Head of the Twins	108 1	32 35	N	2
The little Dog	110 17	6 6	N	2
The lower Head of the Twins	111 0	28 49	N	2
The Crib	125 4	20 52	N	Neb.
Hydra's Heart	137 39	7 10	S	2
Lion's Heart	147 27	13 39	N	1
Lion's Loins	163 54	22 26	N	2
Lion's Tail	172 49	16 32	N	1
The Virgin's Girdle	189 32	5 20	N	3

The right Ascension, Declination, and Ma∣gnitude of some principal Fixed Stars.

The Stars Names.	Right A∣scension.	Declina∣tion.		Magni∣tude.
	D. M.	D. M.
Aliot	189 36	57 36	N	2
Vindemiatrix	191 15	12 51	N	3
The Virgin's Spike	196 44	9 17	S	1
Arcturus	209 56	21 4	N	1
The Southern Balance	217 56	14 32	S	2
The Northern Balance	224 31	8 2	S	2
In the Serpent's Neck	231 49	7 35	N	3
The Scorpion's Heart	242 4	25 34	S	1
Hercules Head	254 40	14 51	N	3
Ophiuchus Head	259 41	12 52	N	3
The Harp	276 17	38 30	N	1
The Vulture	293 27	8 1	N	2
The upper Horn of the Goat	299 30	13 32	S	3
Left Hand of Aquarius	307 10	10 43	S	4
Left Shoulder of Aquarius	318 18	7 2	S	3
Pegasus Mouth	321 49	8 18	N	3
Right Shoulder of Aquarius	326 59	1 58	S	3
Fomahant	339 29	31 23	S	1
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 Sphe∣ricall 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 De∣clination 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 con∣stituted 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 An∣gle C of your Triangle, the Ruler will cut the Meridian Cir∣cle 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 Mi∣nutes 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 Decli∣nation, 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 Per∣pendicular o C, the Sun's Declination 20 degr. (2.) the An∣gle 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 be∣ing 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 Ver∣ticall) 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 Projecti∣on from Z to N, and through that Point where the Parallel of Declination ♊ ☉ ♌, and the Axis of the World, or Hour-Cir∣cle 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 suppo∣sed 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 Hypo∣tenuse, which is the Complement of the Sun's Altitude requi∣red. This Triangle may be resolved by the first Case afore∣going.—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 Declinati∣on 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 Pro∣position 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 Ra∣dius 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 from East or West.—And r H, measured upon your Chord, will contain 102 degr. 46 min. his Azimuth from the South.

PROP. XI. The Latitude of the Place, the Declination of the Sun, and the Sun's Altitude, being given, to find the Sun's Azimuth either from the East, North or South Points of the Horizon.

ALL the foregoing Propositions have been performed by the resolving of a Right-angled Sphericall Triangle: This and some others following require the resolving of an Oblique-angled Triangle for their Solution. So this Pro∣position is performed upon the Oblique-angled Triangle Z E P, which in the Projection is constituted by the Inter∣section of P E S, an Hour-Circle, Z E N, an Azimuth Circle; and Z H N O, the Meridian: and the Arches of these Cir∣cles intersecting each other in the Points Z, E, and P, do make the Triangle Z E P; in which you have given the three Sides, (1.) Z P, the Complement of the Latitude 38 degr.

30 min. of the Place, (2.) P E, the Complement of the Sun's Declination South 110 degr. (3.) Z E, the Complement of the Sun's Altitude 78 d. to find the Angle E Z P, which may be re∣solved by the 11. Case of Oblique-angled Sphericall Triangles. But this being to resolve a particular Proposition, I shall give another Analogie or Proportion whereby to work it by the Canon.

Adde all the three Sides together into one Sum, and take the half thereof, from which half Sum subtract the Side P E, noting the difference, as here you see done.

		d.	m.
The Side	Z P	38	30
P E	110	00
Z E	78	00
The Sum	226	30
The half Sum	113	15
The difference between the half Sum and P E	3	15

Having found the Sum, the half Sum, and the difference, you may work by the following.

Analogie or Proportion.

(1.) As the Radius 90 degr. is to the Co-sine of the Altitude 78 degr.

So is the Co-sine of the Latitude 38 degr. 30 min. to the Sine of a fourth number, which is 37 degr. 30 min.

(2.) As the Sine of the fourth number 37 degr. 30 min. is to the Sine of the half Sum 113 degr. 15 min.

So is the Sine of the difference 3 degr. 15 min. to another Sine, viz. 4 degr. 54 min. Unto which seventh Sine if you adde the Sine of 90 degr. half that Sum shall be the Sine of an Arch, whose Complement being doubled is the Azimuth from the North.

To resolve this Probleme by the Projection.

That which is most intricate and difficult to perform by Numbers, is by Projection effected with the same ease as any of the rest. As in this Proposition, it is the Angle E Z P which is required.—Lay a Ruler upon the Zenith-point Z, and to the Point G, upon the Horizon; the Ruler thus laid will cut the Meridian Circle in the Point g. So the di∣stance g O, being taken in your Compasses and measured up∣on your Line of Chords, will be found to contain 146 degr. which is the Sun's Azimuth from O, the North part of the Me∣ridian.—But if you measure the distance between the g and H, it will contain 34 degr. which is the Azimuth from H, the South part of the Meridian.—And if you measure the distance g N upon your Chord-Line, you shall find that to contain 56 degr. and so much is the Sun's Azimuth from A, the East and West Points of the Horizon.

This Example of finding the Azimuth was taken when the Sun had 20 degr. of South Declination. I will now farther exemplifie this Proposition by finding the Azimuth when the Sun hath North Declination.—As let the Latitude be as before 51 d. 30 min. the Sun's Altitude 12 degr. and the Declination 20 d. North.

To work this by the Canon of Sines differeth nothing from the former, for the Analogie or Proportion is general in all Cases.

Upon the Projection it is resolved (though the same way, yet) upon another Triangle, namely, the Triangle Z P a, in which is given (1.) Z P, the Complement of the Latitude 38 d. 30 min. (2.) Z a, the Complement of the Altitude 78 degr. (3.) the Complement of the Sun's Declination North 70 degr. and you are to find the Angle P Z a, the Sun's Azimuth from the North.

Lay a Ruler upon Z unto the Point a, it will cut the Meri∣dian Circle in the Point s; the distance s O, being taken in your Compasses and applied to your Line of Chords, will there give you 72 degr. 52 m. And such is the Sun's Azimuth from the North.

If you subtract this Azimuth from the North 72 degr. 52 m. from 180 degr. the remainer 107 degr. 8 min. will give you the Azimuth from the South, which upon the Projection is the di∣stance s H.—And if from this Azimuth from the South 107 d. 8 min. you take 90 degr. the remainer 17 degr. 8 min. is the Azimuth from the East or West, which in the Projection is the di∣stance N s.

PROP. XII. The Latitude of the Place, the Sun's Declination, and the Sun's Altitude, being given, to find the Hour of the Day.

THIS Proposition is performed by the resolving of the Oblique-angled Sphericall Triangle Z P a, composed of Z P, an Arch of the Meridian, Z a, an Arch of an Azi∣muth Circle, and of P a, the Arch of an Hour-Circle: In which you have given (as in the last Proposition) the three Sides, to find the Angle Z P a, which you may doe by the 11. Case of Oblique Sphericall Triangles.

To resolve this Proposition by the Canon; Adde the three Sides together, and from the half Sum of them subtract the Complement of the Sun's Altitude, and note the difference, as you see here done.

		d.	m.
The Side	Z P, the Complement of the Latitude	38	30
Z a, the Complement of the Altitude	78	00
P a, the Complement of the Declination	70	00
	The Sum	186	30
	The half Sum	93	15
The difference between the half Sum and Z a, the Complement of the Altitude	15	15

Being thus prepared, you may resolve the Proposition by the Canon of Sines, by this

Analogie or Proportion.

(1.) As the Radius 90 degr. is to the Co-sine of the Sun's Al∣titude 78 degr.

So is the Co-sine of the Latitude 38 degr. 30 min. to a fourth. Sine, viz. 35 degr. 48 min.

(2.) As this fourth Sine of 35 degr. 48 min. is to the Sine of the half Sum 93 degr. 15 min.

So is the Sine of the Difference 15 degr. 15 min. to another Sine, viz. to the Sine of 26 degr. 40 min. Unto which Sine if you adde the Sine of 90 degr. (or Radius,) half that Sum shall be the Sine of an Arch, whose Complement being doubled is the Hour from the Meridian 95 degr. 52 min.

To resolve the Proposition by the Projection.

In the Triangle Z P a, it is the Angle at P that is to be found. Wherefore lay a Ruler from the Point P to the Point a, and it will cut the Meridian Circle in t: So the Arch t AE, being mea∣sured upon your Line of Chords, will be found to contain 95 d. 52 min. which is the Hour from the Meridian; and the Arch t ae, being measured, will contain 84 degr. 8 min. which is the

Hour from Midnight. Also the Arch t S, being measured upon the Chord, will contain 5 degr. 52 min. the Hour from Six.

		d.	m.		hours	m.
The Arch	t AE	95	52	converted in∣to Time is	6	23
t ae	84	08	5	36
t S	05	52	0	23

To convert Degrees and Minutes of the Aequinoctial into Hours and Minutes of Time: Note that 15 Degrees of the Aequi∣noctial make one Hour of Time, and one Degree 4 Minutes of Time. Therefore divide the Degrees of the Aequinoctial by 15, the Quotient is Hours; and multiply the Degrees by 4, and the Product will be Minutes of Time.—So the Hour from the Me∣ridian being 95 degr. 52 min. divide 95 by 15, the Quotient is 6 Hours, and 5 remaining, which 5 multiply by 4, and it makes 20 Minutes of Time, and the 52 min. make 3 minutes of Time and more, almost 4 minutes. So that 95 degr. 52 min. of the Aequinoctial do make in Time 6 hours and almost 24 minutes.

PROP. XIII. The Declination, Altitude, and Azimuth of the Sun, be∣ing given, to find the Hour of the Day.

THE Triangle Z E P in the Projection resolves this Problem: in which there is given (1.) E P, the Com∣plement of the Sun's Declination 70 degr. South; (2.) the Side E Z, the Complement of the Sun's Altitude 78 degr. and (3.) the Angle E Z P, the Sun's Azimuth from the North. So that you have two Sides and an Angle opposite to one of them given, to find the Angle opposite to the other, which you may doe by the 8. Case of Oblique-angled Sphericall Trian∣gles, and by the following

Analogie or Proportion.

As the Co-sine of the Declination 70 degr. is to the Sine of the Azimuth 146 degr. or 34 degr.

So is the Co-sine of the Altitude 78 degr. to the Sine of the Hour from Noon 35 degr. 36 min.

By the Projection.

Lay a Ruler to the Pole P, and upon the Point D in the Ae∣quinoctial; a Ruler thus laid will cut the Meridian Circle in e; the distance e Ae, being measured upon the Line of Chords, will give you 35 degr. 36 min. the Hour from Noon, which in Time is 2 hours and 22 minutes.

PROP. XIV. The Sun's Declination, his Altitude, and the Hour from Noon, being given, to find the Sun's Azimuth from the North part of the Meridian.

IN the same Triangle Z E P you have given (1.) the Side E P, the Complement of the Sun's Declination South 70 d. (2.) the Side Z E, the Complement of the Altitude 78 degr. and (3.) the Angle Z P E, the Hour from Noon 36 degr. 35 m. That is, you have given (as before) two Sides and an Angle opposite to one of them, to find the Angle opposite to the other, which you may doe by the 8. Case of Oblique Spheri∣call Triangles; or by this

Analogie or Proportion.

As the Co-sine of the Altitude 78 degr. is to the Sine of the Hour from Noon 36 degr. 35 min.

So is the Co-sine of the Sun's Declination 70 d. to the Sine of the Azimuth from the North part of the Meridian 146 degr. or 34 degr. from the South.

By the Projection.

Lay a Ruler to Z, and upon the Point G, which will cut the Meridian Circle in g: So g H, being measured upon your Line of Chords, will be found to contain 34 degr. the Azimuth from the South part of the Meridian, which being taken from 180 degr. the remainer will be 146 degr. equal to the Arch g O, the Azimuth from the North part of the Meridian.

PROP. XV. The Hour from Noon, the Latitude of the Place, and the Altitude of the Sun, being given, to find the An∣gle of the Sun's Position.

IN the Oblique-angled Sphericall Triangle Z P E you have given the Side Z P, the Latitude, Z E, the Complement of the Sun's Altitude, and Z P E, the Hour from Noon, to find the Angle Z E P, which is the Angle of the Sun's Position at the time of the Question: So in the Triangle Z P E you have two Sides, with an Angle opposite to one of them, given, to find the Angle opposite to the other Side, which you may find by the 2. Case of Oblique-angled Sphericall Triangles: For which this is

The Analogie or Proportion.

As the Co-sine of the Sun's Altitude 78 degr. is to the Sine of the Hour from Noon 35 degr. 36 min.

So is the Co-sine of the Latitude 38 degr. 30 min. to the Sine of the Angle of the Sun's Position at the time of the Que∣stion 21 degr. 45 min.

By the Projection.

This is the most troublesome Proposition that we have yet

met withall to be resolved by the Projection; and yet it is also thereby easily resolved in this manner.

Take in your Compasses 90 degr. of your Chords; then lay a Ruler upon Y, (the Pole of the Hour-Circle P E S,) and the angular Point E; it being so laid will cut the Meridian Circle in v. Then set 90 degr. of your Line of Chords from v to x upon the Meridian Circle, and the Ruler laid from Y to x will cut the Hour-Circle P E S in the Point y.

Again, lay a Ruler to ☉, (the Pole of the Azimuth Circle Z E N,) and to the angular Point E; it being so laid will cut the Meridian Circle in the Point M. Set 90 degr. from M to z upon the Meridian Circle, and lay a Ruler upon ☉ and z; it will cut the Azimuth Circle (it being continued without the Meridian Circle) in the Point δ.

Lastly, Lay a Ruler to the angular Point E, and this Point δ, it will cut the Meridian Circle in λ; also lay a Ruler from E to y, it will cut the Meridian in λ. The distance θ λ, being taken in the Compasses and measured upon your Line of Chords, will contain 21 degr. 45 min. and that is the quanti∣ty of the enquired Angle Z E P, which is the Angle of the Sun's Position at the time of the Question.

PROP. XVI. The Sun's Altitude, his Declination, and Azimuth from the North, being given, to find the Latitude.

IN the former Triangle Z E P, let there be given (1.) E Z, the Complement of the Sun's Altitude 78 degr. (2.) E P, the Complement of the Sun's Declination (or his distance from the North-pole) 110 degr. (3.) the Angle E Z P, the Azimuth from the North.—By the last Proposition find the Angle of the Sun's Position, Z E P, which had, the Side Z E, the Complement of the Latitude, may be found by this

Analogie or Proportion.

As the Sine of the Sun's Azimuth 146 degr. (or 34 degr.) is to the Sine of the Sun's distance from the North-pole 110 d. (or 70 degr.)

So is the Sine of the Angle of the Sun's Position 21 degr. 45 m. to the Complement of the Latitude 38 degr. 30 min.

By the Projection.

It being the Side Z P that is required, forasmuch as that is an Arch of the Meridian Circle, you have no more to doe but to take the distance Z P in your Compasses, and measure it up∣on your Line of Chords; which being done, you will find it will contain 38 degr. 30 m. the Complement of the Latitude, which taken from 90 degr. leaves 51 degr. 30 m. the Latitude it self.

By the Solution of the foregoing Propositions, both by Trigono∣metricall Calculation by the Canons of Sines and Tangents, and also by the Projecting of the Sphere, you may easily discern the facility of the one above the other, namely, that of Projecti∣on; forasmuch as those Propositions which by Calculation are most troublesome, are by the Projective way most easie.—For, here is no need of letting fall Perpendiculars to reduce the Ob∣lique Triangle into two Right-angled Triangles, and so making of two works for solving of one Problem. Besides, the Projection renders every Triangle so naturally to the eye, that they are re∣solved (as it were) by looking on them.—Many more Exam∣ples I might have given upon this one Projection, but I see the Scheme grow too full of Lines and Letters, that makes me here break off. And now I will shew how the ingenious young Sea∣man may apply these Propositions to his use at Sea.

[illustration] geometrical diagram

The Sphere Projected upon the Plain of the Meridian.

page 122.