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Here the triangle which is supposed to undergo a minute variation is oblique, one of the sides being Pz, the co-latitude, or distance from the pole to the zenith, another being Ps, the co-declination of the luminary at the time proposed, and the other zs = 90° the distance from the zenith to the horizon. It is required to ascertain the variation in the hour angle P which corresponds to any assigned variation in the opposite side Zs.

The differential equation which applies to this inquiry is, obviously, the first of class 1, oblique angled spherical triangles, which, when accommodated to the present notation, becomes

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Now, from the 2d fundamental theorem of spherical triangles, (chap. vi. p. 84), we have cos Pz = coszs cos Ps + sin zs sin Ps cos s. ... But, in the example before us, cos Zs = 0, and coS PZ sin rs'

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Let it be proposed to ascertain the time occupied by


the sun in rising from the horizon on the 25th of May, 1816, in latitude 50°12' N. It appears from the Nautical Almanac, that the sun's declination on the given day is 20° 59' N., while its apparent semidiameter is 15' 483". IIence, diam. -- 15 = 2*1078. Log cos (L + D). , 71°11'.... 9:5085850 cos (L - D). .29° 13'.... 9.9409048

Sum + 2 .............. 19-4494898

Quotient ............... 97247449
Taken from log 2-1078........ 0-3238294

Rem. log 8972, ............ oogoso

So that the time required is 3.9727 minutes, or 3" 58°.

Note. By a similar theorem we may find the time which the sun's rising or setting is affected by horizontal refraction.

Example VI. o

To determine the annual variation of the declination and right ascension of a fixed star, on account of the precession of the equinoxes.

Here, if A (in the spherical triangle ABc) be the pole of the ecliptic, B the pole of the equinoctial, and c the place of the star; the sides AB or c, and Ac or b, must be regarded as invariable. The differential equations applicable to this question are the 1st and 2d of class 1, oblique spherical triangles; from which the following are at once deduced.

1. War, dec. = preces, equinox x sin obliq. eclip.

x sin right ascen. from solstitial colure.

- * var. dec. x cotang. of posit. 2. War. right ascen. = - cos dec

Example VII.

To determine the variation in right ascension and . declination, occasioned by any variation in the obliquity of the ecliptic. In this example, the hypothenuse and the opposite angle of a right angled spherical triangle are assumed as constant. The 2d and 6th equations, class 3, right angled spherical triangles, give 1. var. dec. = var. obliq. x sin right ascen. = var, obliq. × cot obliq. x tan dec. 2, var. right ascen. = — var. obliq. x 3 tan obliq. x sin 2 right ascen.”


Miscellaneous Problems.


Problems with Solutions.


REQUIRED, the arc whose logarithmic tangent is 7-1644398. First, by rule 4, p. 55.

* For more on this curious subject the reader may consult Cagnoli's Trigonometry, chap. xix, and xxi., and Lalande's Astronomy, vol. iii. pp. 588–604.

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Difference. . . . 2981 Difference. . . . 45318

* - // 298.1 5:1/z. - - r r arc=5'1" iiji=5 l 2009 arc = decim, 9'20' = 9'29".652

... I = sexiges. 5' 1"-2032. Hence it appears that in this part of the tables Hutton's has the advantage of Borda’s in point of accuracy. Borda, however, gives a rule to approximate more nearly to the truth; while in other parts of his tables the decimal division supplies great facilities in the use

of proportional parts.

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It appears from equa, 11, chap. iv. that a” + bo - co 2ab Substituting, then, for co in this equation, its value in each of the three former, there will result, respectively,

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Corollary. In like manner it may be shown, that when

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Required a commodious logarithmic method of finding the hypothenuse of a right angled plane triangle, when the base and perpendicular are given in large numbers.

B denoting the base, P the perpendicular, and H the hypothenuse;

Find N so that 2 log P – log B = log N

and make B + N = M. Then, § (log M + log B) = log H. o

For, from the nature of logarithms, + = N;

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