angle OMP p the parallax in altitude, angle MPM'a the apparent altitude, OP the earth's semi-diameter; and OM or OM'd the moon's distance; then in triangle OPM, angle P = 90+ a, and sin P = cos a ; and in triangle OPM, sin P or cos a: sin p=d: r; and in triangle OPM', rad: sin p' =d: r. Hence, R: cos asin p': sin p. 437. The sun's horizontal parallax varies only about of a second, and may in practice generally be considered as invariable. The parallaxes in altitude for the sun at any given time, may therefore be considered the same for other time; and thus being constant, they are given in a table. any EXERCISES. 1. When the horizontal parallax is 54′ 16′′, and altitude 24° 29′ 30′′, what is the parallax in altitude? Ans. 49′ 22′′. 2. When the horizontal parallax is 57′ 32", and the altitude 50° 40', what is the parallax in altitude? Ans. 36′ 28′′. Reduction of the Equatorial Parallax. 438. The horizontal parallax given in the Nautical Almanac is calculated for the equatorial radius of the earth, and is the true horizontal parallax only at the equator; for the earth's radius being less the greater the latitude, the horizontal parallax will be less at any other place. If denote the latitude, and e the ellipticity of the earth, which is about of the equatorial radius, and if p' and p" denote the horizontal parallax at the given place, and at the equator, then is p'p" (1-e sino 1). For if a, r, are the radii of the earth at the equator and the given place, it is proved, in the theory of the figure of the earth, that r — a (1 — e sin2 7). Also (436) rd sin p', sin p' and ad sin p"; therefore, since? = p" sin" very nearly, p' r a = = (1 - — e sin2 1); hence, p′ = p′′ (1 — e sin2 l). A table contains the corrections, which may be considered as constant, that must be deducted from the equatorial horizontal parallax, in order to reduce it to the horizontal parallax for any given latitude. Thus, for the equatorial horizontal parallax in the preceding example, the reduction in the table under 54', and opposite to latitude 36°, is 5"; and the correct horizontal parallax for this latitude is -5"= 54' 15". 439. PROBLEM VIII.-Given the altitude of a heavenly body, to find the refraction. = 54′ 20′′. = 'The refraction for the given altitude is given in a table, and is always to be subtracted from the altitude." EXAMPLE.-Let the apparent altitude be 32° 10′, to find the true altitude. In the table, the refraction for this altitude is 1' 30", 32° 10' 0" 0 1 30 32 8 30 Hence, the true altitude Let ER be a part of the earth's surface, and ZP a portion of the upper limit of the atmosphere; B the real place of a heavenly body, B' its apparent place; O the eye of the observer; OH a horizontal, and OZ a vertical line. When a ray of light, BO, from the body B, enters the atmosphere at P, which increases in density down Z E T R B H wards, the direction of the ray approaches always nearer to that of the vertical line OZ, and thus it moves in a curved path BPO; but the direction of the body is referred to the direction of the ray when entering the eye at O, that is, to the direction OB' of the tangent to the curved path at 0; and the body thus appears at B' higher than its real position. The greater the altitude of the body, the less is its refraction, and in the zenith it vanishes. 440. The mean refraction of a body is its true refraction when the barometer stands at 29.6 inches, and Fahrenheit's thermometer at 50°. Bradley's formula for calculating the mean refraction is '56"-9 tan (z-3 r'), where r' the mean refraction, and z = the zenith dis tance. A table of mean refractions can thus be calculated; and to find the true refraction r, when the pressure of the atmosphere is h, and the temperature t, multiply the mean 400 h 400 h refraction by ·r'. 350+t29.6; that is, r — 29-6(350+t) But a table is also calculated containing the corrections that must be applied to the mean refractions when the pressure and temperature differ from 29-6 and 50.* Thus, the refraction in the preceding example, namely, 1' 30", is the mean refraction; but if the temperature and pressure were 69° and 30-35, then the correction For altitude 32° 10′, and temperature 69°, is = - - 4" And altitude 32° 10′, and pressure 30-35, is = + 2 Hence, the correction for both is Therefore the true refraction == 2 =1′30′′ =128 32° 10' 0" And the true altitude = 32 8 32 441. Unless when great accuracy is required, or when * See Woodhouse's Astronomy, p. 229. the altitude is small, the corrections for change of pressure and temperature are unnecessary. 442. PROBLEM IX.-Given the height of the eye of the observer above the surface of the earth, to find the depression of the visible horizon. The depression of the horizon HOR (fig. to preceding problem) can be calculated when the height OE of the eye and the diameter of the earth are known; for it is just the angle at the earth's centre, subtended by the arc ER, for (fig. to art. 231, Part I.) angle RBH = BCH, and the latter angle can be calculated in the same manner as angle E (fig. to art. 233, Part I.) The real depression, however, will be this angle diminished by of itself on account of refraction. EXAMPLE. Find the depression of the horizon when the height of the eye is 30 feet. Opposite to 30 in the table is 5' 18", the dip. 443. PROBLEM X.-Given the observed altitude of a fixed star, to find its true altitude. 'Correct the observed altitude by applying to it the index error of the instrument with its proper sign, and subtract the dip from the result, and the remainder will be the apparent altitude. From the apparent altitude subtract the refraction, and the remainder will be the true altitude." Then where e is to be taken positive when the index error is in defect, and negative when the error is in excess. When the observed altitude is taken by a back observation, the dip must be added to it. When great accuracy a' a" ±e = a=a" -d, and aa- -r. (d+r) e very nearly, is required, the corrections for the temperature and of the atmosphere must be applied to the refraction. EXAMPLE. The observed altitude of a star was 40° 20' 34", the height of the eye 12 feet, and the index error 2′ 25′′ in excess; find the true altitude. Observed altitude, Index error, Dip, Apparent altitude, True altitude, Let a, a' a" Then EXERCISES. 1. When the observed altitude of a the index error 1' 54" in defect, and the 20 feet; what is the true altitude? a" = e= P a' d= a' = r = 2. What is the true altitude of a star when its observed altitude is 38° 2′ 20′′, the height of the eye 18 feet, and the temperature and pressure 45° and 30-6?" Ans. 37° 57 1". pressure 444. PROBLEM XI.-Given the observed altitude of the upper or lower limb of the sun, to find the true altitude of his centre. 40° 20' 34" I Apply the sun's registered semi-diameter to the observed altitude by addition or subtraction, according as the lower or upper limb was observed, and subtract the dip, and the result will be the apparent altitude of its centre; from which subtract the refraction corresponding to it, and add the parallax in altitude, and the sum will be the required altitude. 40 18 9 40 14 46 -1 8 40 13 38 star is 25° 36′ 40′′, height of the eye Ans. 25° 32' 11". = the true and apparent altitudes of centre, the semi-diameter, the refraction, and dip, a" ± sd, and a = a' — r+p. |