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.

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.