The mariners magazine, or, Sturmy's mathematical and practical arts containing the description and use of the scale of scales, it being a mathematical ruler, that resolves most mathematical conclusions, and likewise the making and use of the crostaff, quadrant, and the quadrat, nocturnals, and other most useful instruments for all artists and navigators : the art of navigation, resolved geometrically, instrumentally, and by calculation, and by that late excellent invention of logarithms, in the three principal kinds of sailing : with new tables of the longitude and latitude of the most eminent places ... : together with a discourse of the practick part of navigation ..., a new way of surveying land ..., the art of gauging all sorts of vessels ..., the art of dialling by a gnomical scale ... : whereunto is annexed, an abridgment of the penalties and forfeitures, by acts of parliaments appointed, relating to the customs and navigation : also a compendium of fortification, both geometrically and instrumentally / by Capt. Samuel Sturmy.

PROBL. XII. Having the Latitude of the Place of the Suns Declination, and his distance from the Meridian being given, to find the Suns Altitude at any Time assigned.

BY this Case may be found the Suns Altitude on all hours, and the distance of Places, in the Arch of a great Gircle; for the Suns Altitude on all hours thereby is meant, that if the hour of the Day, the Declination and Latitude be given, the Suns Altitude proper to the hour, or his Depression may be found.

Take the Chord of 60 degr. and describe the Arch HTPOD, draw the Hori∣zontal-Line HCO, and from O to P prick of the Chord of the Latitude 51 degr. 28 min. and from P to T and D set of the Complement of the Suns greatest Declination, 66 degr. 29 min. and draw the Parallel of Declination TD, and the Axis CSP, or the Meridian of the hour of Six; then draw the Radius TC, which is the Ecliptick-Line, and take off the Line of Signs, and prick down,

15	1	from 6 before it, and after it.
30	degr. 2
45	for the 3
60	Hours of 4
75	5

Then take the nearest distance from 15 degr. to CS, the Meridian of the hour of Six, or Axis, and prick it from S to 5 and 7; and likewise take the nearest distance fro 30 to CS, and lay it from S to 8 and 4; and in like manner do with the rest, then will the nearest distance from 4 5 6 7 8 9 10 11 to the Horizontal-Line HCO be the Signs of their respected Altitude;

so the Altitude	4	In Summer will be	1 deg.	34 min.
	5	9	30
for the hours	6	18	12

[illustration] geometrical diagram

And so much is the Suns Depression under the Horizon at the hours of 8, 7, and 6 in Winter; as you may soon trie by the same Division on the Parallel of the Suns greatest Declination Southward DST

The Summer Alt. for the h. of	7	are	27:23
8	36:42
9	45:42
10	53:45
11	59:42

And the Winter Altitude for

the Hours of	9	are	5 d. 13
10	10:28
11	13:48

And so much is the Sun deprest under the Horizon in Summer at the hours of 3, 2, 1, from Mid-night, as you may soon find it by the nearest distance from 9, 10, 11, in the Line LD to the Line LO; if you apply that distance to the Line of Signs, you may draw the Parallel of Altitude through each hour, as the Example is through 9 ho. e f, and measure He, or Of on the Line of Chords, and it is 45 degr. 42 min. the Suns Altitude at 9 a clock the 11 of June.

2. But admit the Latitude were 51 deg. 28 min. and his Declination 00 deg. 00 m. and suppose, for Example, seek his distance from the Meridian is 2 ho. 44 min. or 41 deg. 0′0 and the Sun upon the Equinoctial the 11 of March, and 13 of September.

Upon the Equinoctial at K is the Sun represented the 11 of March, or 13 of Septem∣ber, and distance from the Meridian 2 ho. 44, or 41 degr. 00 min. If in the Convex-Sphere you lay S ♁, from C to K the Right-Ascension 59 deg. 9 min. and through K you draw a Parallel to the Horizon, as dM, it will cut the Meridian in d, and mea∣sure the distance Hd on the Line of Chords, and it is 28 degr. 03 min. the Suns Alti∣tude required, or the nearest distance from K to the Horizon-Line HCO applyed to the Line of Signs, would have shewed the same. In the Concave-Sphere, if you take the distance CK, and draw a small Arch at r; and take the nearest distance to the Azi∣muth-Line of East and West ZC from K, and with that distance turn the other Foot over, and cross the Arch at (r,) and through K and r draw a Circle of Altitude, it will cut the Limb in E and F measure HE, or OF on the Line of Chords, sheweth the Altitude 28 degr. 03 min. as before.

In the Concave-Sphere, you may see the Triangle ZAEK; and we have given first AEZ the distance from the Zenith to the Equator, equal to the Latitude of the place, 51 degr. 28 min. Secondly, AEK the Suns distance from the Meridian 41 deg. to find the Suns Altitude from K to the Horizon.

I say, as the Radius 90 deg.	10
is to the Co-Sign of the Suns distance from the Merid. 42 d. 0′ AEK	987777
So is the Co-Sign of the distance from the Equ. to the Zen. 51 d. 28 m. AEZ	979446
to the Co Sign of ZK 61 degr. 53 min.	967223

Whose Complement is the Altitude Kp 28 deg. 03 min. which was required; you may in your practice draw a particular Figure for the Question, and shew every Tri∣angle by it self, by the Line of Chords, and half Tangents.

Secondly, when the Sun is in North Signs, ♈ ♉ ♊ ♋ ♌ ♍.

Let it be required to find the Suns Altitude at 10 a clock and 2 m. past before Noon, when the Sun is in entrance of Gemini, in Latitude 51 degr. 28 min.

First, By the Convex-Sphere, the nearest distance taken from the Suns place, to the Horizon HC, applyed to the Line of Signs, sheweth 51 degr. 12. min. the Suns Alti∣tude required; Or to find his distance from the Meridian, take ST, and prick it from C to X; then with the distance S ♁ on X as a Centre, draw the touch of an Arch at K, a Ruler laid over the Centre, over the outward edge of the Arch, cuts the Arch of the outward Meridian in (n;) then measure On on the Line of Chords, and it is 60 deg. 2 min. S ♁, whose Complement is 29 deg. 58 min. the Suns distance from the Me∣ridian OT; Wherefore in the Triangle NZ ♁, we have known, (1,) ZN the Complement of the Latitude 38 deg. 32 min. (2,) N ♁, the Complement of the Suns Declination 69 degr. 30 min. (3,) the comprehended Angle ZN ♁, the distance of the Sun from the Meridian 29 deg. 58 min. to find Z ♁, and hereby the Suns Altitude 92 ♁, I say,

As the Radius 90	10
is to the Co Tangent of the Latitude 51 deg. 28 min. NZ	990112
So is the Co-Sign of the Angle from the Meridian 29 deg. 58 m. ZN ♁	993767
to the Tangent of the fourth Ark 34 deg. 36 min. N q	983879

From the Complement of the Suns Declination N ✶ 69 deg. 30 min. Substract N q 34 deg. 36 min. there remains 34 deg. 54 min.

As the Co-Sign of the fourth Ark. 34 deg. 36 min. N q	991547
is to the Compl. Sign of the Latitude 38 deg. 32 min. ZN	989334
So is the Co-Sign of the fifth Ark 34 deg. 54 min. q ♁	991389
	1980723
to the Co-Sign of Z ♁ 38 deg. 48 min.	989176

Whose Sign is 51 deg. 12 min. p ♁ the Suns Altitude above the Horizon at 10 a clock, and 02 m past, and 1 a clock 58 min. past in the Afternoon, when he is in 1 deg. 18 m. of ♊ in Latitude of 51 deg. 28 min. North.

Now if you follow the Rules before-going, you may find the Suns Altitude by the Line of Chords, and half-Tangents by this Figure to be the same.

[illustration] geometrical diagram

Suppose the Sun in the Southern Signs ♎ ♏ ♐ ♑ ♒ ♓ in the opposite Point to the former, having South-Declination 20 degr. 30 min. and be also distant from the Meri∣dian 29 d. 58 m. take the Declination 20 deg. 30 min. and prick it from AE to B, and Q to R in both the Spheres, and draw the straight Line in the Convex-Sphere BMR, and take from the Suns place in the opposite Sign in the Parallel of Declination, his distance from the Meridian YR, and prick it on the other side of the Parallel from B to ☉; and the nearest distance to the Horizon-Line HC, applyed to the Line of Signs, shews the Altitude to be 13 d. 23 m. and in the Concave-Sphere take of the Line of half-Tangents, the Declination 20 degr. 30 min. and lay it from the Centre C to M, and the Axis CS continued; take the Complement of the Declination of the Line of Se∣cants, and place it from C on the continued Line or Axis, and that will be the Centre of the Parallel of Declination; or if you take the like Complement 69 degr. 30 min. of the Line of Tangents, and put it from M on the Axis, it will be the Centre of the Parallel of Declination; therefore draw it BMR, and it will cut the Meridian in ♁, the place where the Sun is.

Now, to find the Suns Altitude or Ark (Z ☉) or Z ♁; therefore to find how much it is; you must find the Pole of the Circle N ♁ Z, which is done after this manner.

Lay a Ruler from Z to h, and it will cut the Circle in ♃; then take 90 deg. and prick it from ♃ to ♂, then lay a Ruler over from Z to ♂, and it shall cut the Horizon in ♀, which Point ♀ is the Pole of the Circle Z ♄ n.

To measure the Ark Z ♁, you must lay a Ruler upon ♀ and ♁, which will cut the out∣ward Circle in the Point X, so shall XZ measured on the Line of Chords, give you the quantity of d. contained in the Arch XZ, which will be 76 d. 37 equal to the Comple∣ment of the Suns Altitude. I have been the larger in this precept, that it may be a Rule of Direction, to shew how the Ark of any great Circle of the Sphere; the sides of all Spherical Triangles being such, may be measured whatsoever, by his operation in the Concave-Sphere.

Observe the Figure we have given in the Oblique-Angled Triangle ZN ♁. 1. NZ as before, the complement of the Latitude 38 deg. 32 min. 2. N ♁ 110 deg. 30 min. the same distance from the North-Pole. 3. the Angle ZN ♁ 29 deg. 58′ to find the Altitude ♄ ♁.

As the Radius 90	10
is to the Tangent of NZ 38 deg. 32 min.	990112
So is the Co-Sign of the Angle, ZN ♁ 29 deg. 58′	993767
to the Tangent of Nq 34 deg. 36 min. 4 Ark, as before,	983879

From the Ark N ♁ 110 deg. 30 min. Substract the Arch Nq 34 deg. 36 min. and there remains 75 deg. 54 min.

As the Co-Sign of Nq 34 deg. 36 min.	991547
is to the Co-Sign of ZN 38 deg. 32 min.	989334
So is the Co-Sign of the fifth Ark q ♁ 75 deg. 54 min.	938670
	1928004
to the Co-Sign of Z ♁ 76 deg. 37 min.	936457

Now the Complement of Z ♁, is ♁ ♄ 13 deg. 23 min. which is the Suns Altitude required.