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PROBLEM V.

19. The right ascensions and declinations of two fixed stars, observed on the same vertical given in position, together with their distance, or the arc of that vertical included between them, being known; it is required to determine the latitude of the place of observation.

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If s, s', in the annexed diagram, be the places of the two stars in the vertical zso, HR the horizon, P the elevated pole, and EQ the equator; then it is evident that the three sides of the triangle PSS' are given, as well as the angle at P. For PS, PS', are the H codeclinations of the stars (known

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R

by hypothesis), ss' their distance on the vertical given in position, and the angle SPS' the measure of the difference of their right-ascensions. Hence, by the proportionality of the sines of opposite sides and angles, we shall have,

sin ss': sin SPS:: sin s'P: sin PSS' sin PSZ

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Thus, two angles and a side opposite to one of them, will become known in the triangle Pzs, namely, the azimuth, pzs, the angle Psz, and the side PS; whence we shall have

sin PZS: sin PSZ:: sin PS: sin PZ = cos L =

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sin ps sin PS' sin SPS' cosec ss' cosec PZS. 20. Cor. 1. It is evident that the hour angle zps may be readily found from the same data; and therefore if the sun's place and his right ascension be known, the difference between it and that of the star, s, will become

known; and if from this difference the arc ED be taken, the remainder will give the arc of the equator contained betwixt the meridian of the place and that which passes through the sun, whence the moment of observation may at once be deduced. It is equally obvious that, by pursuing the computation, the altitude of each of the stars at the moment of observation may be found.

21. Cor. 2. If, instead of the azimuth of the two stars, their altitudes were given, we might readily determine the elevation of the pole; for, in that case we should know the three sides of the triangle PSS' with the angle P; and thence the angle Psz between the given sides rs, zs, of the triangle Psz.

22. Cor. 3. So again, if besides the declinations and right ascensions of the stars observed on the same vertitical, the hour angle zps were given, the latitude might be ascertained; for then, in the triangle Psz, the angles ZPS, ZSP, and the side SP, would be known, and ze would be determinable by case 3, prob. 3, of oblique spherical triangles.

23. Cor. 4. Hence, from this problem and corollaries the latitude of a place may be determined with great facility; a plumb line serving to show when two stars are in the same vertical, and the altitude being susceptible of being taken with sufficient accuracy for common purposes without employing large instruments. But correct tables of the places of the stars are indispensable.

Example.

Two stars, the right ascension of one of which s is 78° 24', and its declination 27° 25′, the right ascension of the other, s', is 104° 52′, and its declination 12° 18′, of the same kind with the former, are distant from each other 28° 30′, on a vertical whose azimuth Ezo is 73° 36. Required the latitude of the place?

Here PS 90° — 27° 25′ : = 62° 35′, ps′ = 90°, 12° 18' = 77°42′,

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26° 28',

and

ss' 28° 30', srs' = 104° 52′ 78° 24′ = PZS 180° 73° 36′ 106° 24'.

The logarithmic computation from the formula is

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Hence the latitude is 32° 23′ 18′′, of the same kind with the declinations of the stars.

PROBLEM VI.

24. Give the apparent altitudes of the moon and sun, or of the moon and a fixed star, and their apparent distance to determine their true distance.

This is an important problem in navigation, being of essential use in one of the best methods of determining the longitude at sea. The true distance of the moon and the sun, or a star at any given instant, being ascertained, we are enabled by means of the "Nautical Almanac," and the Ephemerides of different countries, to find the time at the first meridian, which corresponds with the moment of observation: the difference between these times, reduced to degrees at the rate of 15 to an hour, shows the longitude required.

Besides the error arising from the imperfection of instruments, the observed altitudes of the heavenly bodies are affected by three causes, the depression of the horizon, the refraction of rays of light in passing through the air, and the parallax, or the inclination of two visual rays passing from the celestial body, one to the earth's centre, the other to the point where the observer is placed on its surface. The principal works on nautical astronomy contain tables by which the requisite allowances may be

made for these at any place and for any heavenly body: and the purpose of making such allowances is, to reduce the observed or apparent altitude of any body at or above the surface of the earth, to the real altitude at which it would appear from the earth's centre if light were transmitted in right lines.

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25. The apparent altitude of the sun always surpasses the true; for the refraction elevates the sun more than the parallax depresses that luminary. With regard to a fixed star the parallax is evanescent. Hence, the true place of the sun, or of a fixed star, will always be below the apparent place. The moon, on the contrary, is more depressed by parallax than it is elevated by refrac- M tion. Hence the true place of the moon m will be always above its apparent place.

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26. Let, then, s be the apparent place of the sun orstar, s its true place, on the same vertical zs, m the apparent place of the moon, м its true place, on the same vertical zm, ms the apparent distance of the two bodies. Then, there are known н= 90 Zs, apparent altitude of the sun or star; H' 90° . zm, apparent altitude of the moon; D=ms, their apparent distance: also, h ・r + p = 90° zs, true altitude of the sun or star; and h'H' r' + p' 90°- ZM, true altitude of the moon; r and P, rand p', being the refraction and parallax corresponding respectively to the apparent altitude of the bodies. Consequently, in the spherical triangle zms are given all the sides to find the angle z: and then in the triangle zмs are given two sides zм, zs, and their included angle z, to find the third side мs= d the true distance required. The solution of the problem thus conducted is obviously as simple and natural as can well be wished, to men acquainted with theory and accustomed to computation: yet it has been found embarrassing to mariners; on which account most writers on the subject of navigation have investigated other

rules, several of which (especially those which depend upon subsidiary tables) are more direct and expeditious in operation than the original rules deduced without modification from the two triangles zms, ZMs.

27. The second general theorem of spherical trigonometry, when applied to these two triangles, gives

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But since cos (H + H') = COS H COS H' sin и sin ', we have sin Hsin H' COS H COS H'

and, in like manner

COS (H + H'),

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sin h sin h = cos h cos h

whence, cos z =

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COS H COS H'

cos d+cos (h + h') cos h cos h'

cos h cos h

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COS D + cos (H + H') cos d + cos (h + h')
=

COS H COS H'

cos h cos h'

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Now it is evident from the formulæ in chap. iv. that cos D+ cos (H+H') = 2 cos (H + H+ D) ×· COS (H+H D),

cos (h+h') = 2 cos2 (h + h) — 1,

cos d 1. 2 sin? d.

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These values being respectively substituted for the several quantities in equa. A, it will be transformed into

2 coś (H + H′ – D) COS § (A '+ H' + D)

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