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which is the same with the formula, given in the preface to the Navigator, for calculating Table XVI.

128. Refraction, by elevating the stars in the horizon, will affect the times of their rising and setting; and the star will not set until its zenith distance is

90° + horizontal refraction,

and the corresponding hour angle is easily found by solving the triangle PZB (fig. 35).

129. Another astronomical phenomenon, connected with the atmosphere, and dependent upon the combination of reflection and refraction, is the twilight, or the light before and after sunset, which arises from the illuminated atmosphere in the horizon. This light begins and ends when the sun is about 18° below the horizon; so that the time of its beginning or ending is easily calculated from the triangle PZB (fig. 35).

130. EXAMPLES.

1. Find the dip of the horizon, when the height of the eye is twenty feet.

Ans. 264" = 4′ 24′′.

Dip.

2. Find the dip of the sea at the distance of 3 miles, when the height of the eye is thirty feet.

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3. Find the dip of the sea at the distance of 2 miles, when the height of the eye is forty feet.

Ans. 10'.

4. Find the dip of the sea at the distance of of a mile, when the height of the eye is thirty feet.

Ans. 68'.

Parallax in altitude.

CHAPTER XI.

PARALLAX.

131. The fixed stars are at such immense distances from the earth, that their apparent positions are the same for all observers. But this is not the case with the sun, moon, and planets; so that, in order to compare together observations taken in different places, they must be reduced to some one point of observation. The point of observation which has been adopted for this purpose, is the earth's centre; and the difference between the apparent positions of a heavenly body, as seen from the surface or the centre of the earth, is called its parallax.

132. Problem. To find the parallax of a star.

Solution. Let O (fig. 53) be the earth's centre, A the observer, S the star, and OSA, being the difference of directions of the visual rays drawn to the observer and the earth's centre, is the parallax. Now since SAZ is the apparent zenith distance of the star, and SOZ is its distance from the same zenith to an observer at O, the parallax

OSA = P

is the excess of the apparent zenith distance above the true zenith distance. If, then,

Parallax in altitude.

z= SAZ, R= OA the earth's radius,

rOS

we have

=

the distance of the star from the earth's centre,

r : R = sin. z: sin. p,

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133. Corollary. If P is the horizontal parallax, we have

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which agrees with (604) and Tables X. A., XIV, and XXIX, are computed by this formula, combined, in the last table, with the refraction of Table XII.

134. Corollary. In common cases, the value of the horizontal parallax can be taken from the Nautical Almanac; but, in eclipses and occultations, regard must be had to the length of the earth's radius, which is different for different places. The earth is not a sphere, but a spheroid slightly compressed at the poles; the polar radius being less than the equatorial one by about th part. The spheroid may be obtained from the sphere by such a compression over

Reduction of parallax.

the whole surface parallel to the polar axis, that each place is brought nearer to the plane of the equator byth part.

Thus, if OEAP (fig. 54) is a section of the earth through the polar axis OP and OEA'P', the section of the sphere of which the equatorial semidiameter OE is the radius; and if A'AM, B'BN, are drawn parallel to OP, each of the distances A'A, BB, PP, &c., will be th part of the distances A'M, B'N, PO, &c.

135. Problem. To find the reduction of parallax.

The horizontal parallax is, by (747), proportional to the earth's radius, so that it diminishes at the same rate, from the equatorial value which is given in the Nautical Almanac. -Hence, if AR is drawn perpendicular to OA,

we have

L" = A'OL,

&R the diminution of R for the latitude L,

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A'MOA' sin. A'OM= R sin. L"

AAm R sin. L" m R sin. L nearly

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