In the right-angled spherical triangle AGS.

1. AG the time from 6 o'clock when the sun is due east or west. Or it is the complement of the time from a star's culminating.

2. As the altitude of the sun, or a star, above the horizon, when on the prime vertical zSAN.

3. Gs the sun's or star's declination, of the same name as the latitude.

4. The angle GAS = the latitude of the place; for the elevation HÆ of the equinoctial is always equal to the complement of the latitude.

Any two of these quantities being given, the rest may be found.

The triangle SN Zenith, right-angled at Zenith, is complemental to the triangle AGS.

EXAMPLE I.

In latitude 51°.32' North, when the sun's declination is 19°.39' North; required his altitude, and the time when he will appear due east or west?

1. To find the altitude As.

*

Rad x sine Gs=sine GAS X sine as.

Sine GAS 51°.32′ : sine Gs=19o.39′ :: rad : sine As—25°.26'.

2. To find the hour from 6, AG.

Rad x sine AG tang GSX cot GAS.

Rad: tang Gs=19°39′:: cot GAS=51°32′ : sine AG=16°.28′48′′. Now 16°.28'.48′′=1.5′.55′′ the time from 6, hence the sun will be east at 7.5.55" in the morning, or west at 4.54′.5′′ in the afternoon.

(T) As the declination increases, the altitude and hour from 6 increase, and when the declination vanishes, the sun appears in the horizon upon the prime vertical. When the latitude is equal to 90°, the sun's altitude upon the prime vertical is equal to his declination.

When the latitude of the place is less than the declination, the sun never appears on the prime vertical.

At what time will Arcturus appear upon the prime vertical, or be due east or west from Greenwich, on the first of April 1813; and what will be its altitude, its right ascension being 14.7.8" and declination=20°.9′.48′′ North?

The sun's right ascension on the same day at noon being= ●1.41′.59′′, and on April 2d=0.45′.28′′.

1. Te

1. To find the altitude As.

*

Rad X sine Gs=sine GAS X sine as.

Sine GAS 51°.28′.40′′ : rad :: sine Gs=20°.9′.48" : sine As 26°.0'.50".

2. To find AG, the complement of the time from the star's culminating.

*

Rad X sine AG tang GSX cot GAS.

Rad: tang Gs=20°.9′.48" :: cot GAS 51°.28.48′′ : sine AG= 16.59.40". Hence 90°-16°.59′.40′′ 73°.0.20", which reduced to time=4".52′.1′′, the time from the star's culminating. The star comes to the meridian at 13.23.16" (see Example II. Prob. II.); therefore 13h.23.16" - 4.52'.1"-8.31.15" the time in the evening, when the star will appear due east; and 13".23′.16′′+4.52′.1′′=18.15′.17′′; or 6.15.17′′ next morning, when the star will appear due west.

(U) The height of the same star upon the prime vertical in any place, is always nearly the same, for the reasons already assigned (R. 255.); and the difference of the times of its coming to the prime vertical, will be equal to the difference of the times of its culminating, which is nearly equal to the diurnak difference of the sun's right ascension.

ERACTICAL EXAMPLES.

1. The sun's declination being 19°.39′ North, and having 25°.26' altitude upon the prime vertical; required the latitude of the place, and the hour of the day?

=

Answer. Latitude 51°.32′ N., time 7.5'.55" or 4.54.5", according as the observation was made in the forenoon or after

noon.

2. At what time will Aldebaran appear due east or west at Greenwich, latitude 51°.28.40" N. on the 20th of November 1813; and what will its altitude be at that time, its right ascension being 4.25.12", and declination 16°.7.27" N?

The sun's right ascension at noon 20th November = 15h.42′12′′, and on the 21st 15h.46'.24".

Answer. Aldebaran culminates at 12h.40′.47′′ at night, is due east at 7.34', due west at 17.47.34", and its altitude=20°.47.30′′. The arc AG=13°.18′.17′′.

3. The sun's declination being 19°.39′ N., and he was observed to be due west at 4.54′ in the afternoon; required his altitude, and the latitude of the place?

Answer. Altitude -25°.26', latitude 51°.32′ N.

4. At what time will Regulus appear due east or west at

Greenwich,

Greenwich, latitude 51°.28′.40" N., on the 6th of February 1813; and what will be its altitude at that time, its right ascen sion being 9.58'.24" and declination 12°.52'.41" N.?

The right ascension of the sun at noon on the 6th of February being=21.19′.25′′, and on the 7th=21".23′.25′′.

Answer. Regulus culminates at 12.36.52", is due east at 7.18'.49", due west at 17.54.55", and its altitude 16°.35.3". The arc AG-10°.29′.10′′.

5. In latitude 51°.32′ North the sun's altitude when on the prime vertical was 25°.26'; required his declination, and the hour of the day?

Declination 19°.39′ North

Answer. Time 7.5'.55" or 4.54'-5" according as the observation was made in the forenoon or afternoon.

Y PROBLEM V. (Plate III. Fig. 1.)

(W) Given the latitude of the place, and the sun's altitude, when on the equinoctial, to find his azimuth and the hour of the day.

In the right-angled spherical triangle adc.

1. AG the complement of AG the hour from noon, being an arc of the equinoctial, on which G represents the sun's place between noon and 6 o'clock.

2. Ad the complement of Hd the sun's azimuth from the meridian, or south point of the horizon.

3. dg the sun's altitude.

4. Gad the complement of the latitude, being measured by the arc HÆ the elevation of the equator above the horizon. Any two of these quantities being given, the rest may be found.

This triangle is complemental to the triangle GA zenith; for Gz is the complement of do the altitude; AG is the hour from noon; AZ the latitude; and the angle EzG, measured by the arc Hd, is the sun's azimuth from the meridan.

EXAMPLE.

In latitude 51°.32' North, when the sun was on the equinoctial, his altitude was 22°.15'; required his azimuth, and the hour of the day?

1. To find ad the complement of the azimuth.

*

Rad x sine ad tang dGx cot Gad.

Rad: tang do 22°.15' :: cot Gad=38°.28′ : sine ad➡30°.59′.

L1 2

Hence

Hence the azimuth=59°.1' from the south towards the east or west, according as the time is before or after noon.

2. To find AG, the hour from 6.

37°.29'.45",

Sine GAD 38°.28': rad :: sine dc=22°.15' :sine AG which reduced into time=2h.29′.59"; this added to, and subtracted from, 6 hours gives 8h.29.59" or 3.30'.1" according as the sun is in the eastern or western semi-circle of the globe.

PRACTICAL EXAMPLES.

1. In latitude 51°.32′ N. when the sun has no declination, what is his altitude, and azimuth at three hours and a half from noon?

Answer. The altitude 22o,15′. Azimuth = S. 59°.1' E or W.

2. At the time of the equinox the sun's altitude was found to be 22°.15', and his azimuth S. 59°. E; required the hour of the day, and the latitude of the place?

Answer. Time 8h.30'. Latitude 51°.32' North.

3. In latitude 51°.32′ North, the sun being in the equinox, there is given the angle (AGd) which the vertical circle passing through the sun forms with the equinoctial 57°.46'; to find the hour of the day and the altitude, and azimuth of the sun? Answer. Time 8.29'.56" or 3.30.4". Altitude 22°.15', azimuth S. 59o. E. or W.

(X) PROBLEM VI.

(Plate III. Fig. 2.)

The difference of longitude between two places, both in one parallel of latitude, being given, to find the distance between them.

Before we give any examples to this problem, it will be necessary to remark, that all the meridians on the terrestrial globe are great circles meeting each other in the poles of the equator. Therefore the meridional distances of places vary as the latitudes; that is, the distance between the meridians PE and pw is greater on the equator woQE (supposing woQE to be a part of the equator) than it is on the parallel of latitude MN, and greater at MN than at /L, &c.

Longitude is always counted on the equator, where a degree is reckoned 60 geographical miles, hence a degree of longitude in the parallel MN must be less than 60 miles. This shews that what is called meridional distance in navigation, is always less than the longitude, except the ship sail on the Equator.

Since all the meridians PW, PO, PQ, &c. cut the equator at

right angles (I. 130.) they must necessarily cut all the parallels of latitude at right angles.

The arcs of the equator wo, oq, &c. are the measures of the spherical angles WPO, OPQ, &c. at the pole (D. 129.)

If wm or on represent the latitude of two places on the same parallel, then mp or nP will be their co-latitudes.

The arcs wo and mn are similar (P. 131 and R. 132.). hence mrcv: mn :: cw: wo. But my is the sine of the latitude wm, cv its cosine, and cw the radius of the sphere, Therefore,

any arc on the

Radius cosine of : any latitude :: the length of equator the length of a similar arc in that latitude.

PRACTICAL EXAMPLES.

1. Suppose a ship, in latitude 40° North, sail due east until her difference of longitude be 60°, required the distance sailed in that parallel?

Rad: cos 40°:: diff. long-3600 miles: the distance 2757.7 miles.

EXAMPLE II.

2. Suppose a ship, in latitude 40° north, sail due east 2757.7 miles, what is her difference of longitude?

Answer. 3600 miles.

(Y) By the first of these examples, geographers have calculated tables shewing how many miles answer to a degree of longitude for each degree of latitude; thus in latitude 45°, a degree of longitude is 42-43 miles; in the latitude of London 51°.32′, the length of a degree is 37.32 miles, &c.

3. If a ship sail 2757.7 miles directly eastward or westward, so as to alter her longitude 60°, what latitude is she in?

Answer. 40 north or south.

4. Suppose a ship to sail from latitude 40° North to latitude 10° North, till her difference of longitude be 60°, and it be required to find her departure, course, and distance sailed.

Here IL the parallel of 40° is the latitude left, Aa the parallel of 10° is the latitude arrived at, and MN the parallel of 25° is the middle latitude between them; hence MP is the complement of the middle latitude, and MN is the meridional distance, or the true departure nearly*; Bw is a quadrant or 90° and we is the difference of longitude.

(Z) Hence the following CONSTRUCTION (Plate III. Fig. 3.): 1. With the chord of 60°, which is equal to the sine of 90°,

See the Theory of Navigation, Book IV. Chapter II.

and