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mean proportional between radius, and the sine of the marimum error on the theodolite.
Suppose that the face of the theodolite, instead of
being horizontal, was inclined in an angle of 5 degrees. Then log tan #B. . . .2° 30'.... 86400931 Multiplied by.......... 2
To find when that part of the equation of time which
depends on the obliquity of the ecliptic, is the greatest ossible,
p Here the sun's longitude will form the hypothemuse of a right angled spherical triangle, his right ascension will be the base, and the obliquity of the ecliptic is supposed constant. It is required to find when hyp. – base is a maximum; which is, evidently, as in the preceding example, when hyp. 4- base = 90°, that is, when sun’s long. + sun’s right ascension = 90°, from the equinoctial points. e
This happens in the year 1816, about May 7 and November 8.
• Example IV.
To ascertain the error that may be committed in the observation of zenith distances, by any conceivable deviation of the instrument from the vertical plane.
Draw a figure in which Ho is the intersection of the horizon with the vertical plane Hzo, and Hz’so the position of the plane of the instrument, z’ being the apparent zenith upon that plane, and s the place of the heavenly body when it has arrived at the plane of the instrument. The arc zz' will measure the inclination I of the plane at H or o, and the base z's = z of the right angled spherical trianglezz's will be the apparentzenith distance of the body, while the hypothenuse zs = z + = will be the true distance.
duced by the deviation of the instrument from the vertical plane.
It is required to determine, at a given time and lati
tude, how long an interval is taken by the body of the sun to rise from the horizon.
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 . 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 *zs - - *zs + = sins sin Ps; or, & P = −.
&P sin S sin Ps
Now, from the 2d fundamental theorem of spherical
triangles, (chap. vi. p. 84), we have cos Pz = cos Zs cos Ps + sin zs sin Ps coss. But, in the example before us, cos Zs = 0, and
Let it be proposed to ascertain the time occupied by - 7
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' 48}". Hence, diam. -- 15 = 2-1078. Log cos (L + D) .. 71°11' . . . . 9:5085850 cos (L – D). .29° 13'.... 9.9409048
Sum + 2 .............. 194494898
Quotient . . . . . . . . . . ..... 97247449
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.
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.
To determine the variation in right ascension and declination, occasioned by any variation in the obliquity of the ecliptic. In this example, the hypothentise 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. x cot obliq. x tan dec. 2. var. right ascen. = — var. obliq, x * tan obliq. x sin 2 right ascen.”
Problems with Solutions.
PRob LEM I.
REQUIRED the arc whose logarithmic tangent is 7-1644,398. 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. ~