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mean proportional between radius, and the sine of the maximum 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′.... 8-6400931 Multiplied by....

2

The prod. — 10, sin error 6′ 33′′ 16′′. . 7-2801862

Example III.

To find when that part of the equation of time which depends on the obliquity of the ecliptic, is the greatest possible.

Here the sun's longitude will form the hypothenuse 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. + base = 90°, that is, when sun's long. + sun's right ascension = 90°, from the equinoctial points.

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 à figure in which HO is the intersection of the horizon with the vertical plane нzo, 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 triangle zz's will be the apparent zenith distance of the body, while the hypothenus e zsz+x will be the true distance.

Now from equa. 6, art. 25, chap. vi. we have
cos zs = cos zz' cos zs: that is,
Cos I cos z = cos (z + x)

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sin z sin x.

Whence sin z sin x = cos z (cos x → Cos I)

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Dividing by sin z, there results

sin x = 2 cot z (sin2 1

sin2x).

Here since x is by hypothesis small, sin2 x is extremely minute, and may be rejected: hence

sin x = 2 cot z sin2 1.

Corol. As the zenith distance z increases, the error ≈ diminishes, the inclination I remaining the same.

Supposing the deviation from the vertical 5', and the apparent zenith distance of a star 37°, the error would be 2% of a second.

Note. If z, z', be the observed zenith distances of a circumpolar star (corrected for refraction) at the superior and inferior transits of the meridian; z, z', the true zenith distances; x, x', the corrections due to the deviation of the circle from the meridian: then will

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therefore the distance of the pole from the zenith, or, § (≈ − 2) = § (z′ + x′) — § (z + x). But in the inferior transit x' is imperceptible; therefore

lat. = 90° } (≈′ — z) = 90° — § (z′ — z) + 1x. Consequently, in the case of employing a circumpolar star, the latitude is only affected by half the error produced by the deviation of the instrument from the vertical plane.

Example V.

It is required to determine, at a given time and latitude, 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 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 cos zs cos PS + sin zs sin PS cos s. But, in the example before us, cos zs =

O, and

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B),

It may be readily shown, by multiplying together the corresponding values of cos (A + B) and cos (A (equa, c, p. 42), and substituting 1-cos2 for sin2, that COS2 PZ = COS (PZ + PS) cos (PZ — PS).

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sin2 PS Hence it follows that

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sun's diam. 15

time sun ris. = ✔[cos (lat. + dec.) cos (lat. — dec.)]

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".

Hence, diam. 15 =2.1078.

Log cos (LD).. 71° 11'.... 9.5085850
cos (LD).. 29° 13′.... 9.9409048

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So that the time required is 3.9727 minutes, or 3m 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.

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, в 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 equa tions applicable to this question are the 1st and 2d of class I, oblique spherical triangles; from which the following are at once deduced.

1. Var. dec. preces. equinox x sin obliq. eclip. > sin right ascen. from solstitial colure.

2. Var. right ascen. =

var. dec. x cot ang. of posit.

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. x cot obliq. x tan dec.

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* 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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