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EXAMPLE. If the observed altitude of the sun's upper limb, on the 10th November 1841, should be 28° 21' 24", and the height of the eye 13, what would be the true altitude?
28 0 3
Since the mean refraction is constant, and the parallax in altitude of the sun is nearly constant for a given altitude, a table is formed by combining these two corrections. As the contraction of the diameter is also sensibly constant, it might be included in the elements of the same table.
1. If the observed altitude of the sun's lower limb, on the 15th of April 1841, was 42° 10′ 15′′, the height of the eye being 25 feet, what was the true altitude of its centre, its semi-diameter being 15′ 57′′? Ans. 42° 20′ 23′′.
2. If the observed altitude of the sun's upper limb, on the 4th of June 1841, was 20° 40′ 15′′, and the height of the eye 16 feet, what was the true altitude of its centre, its semi-diameter being 15′ 47′′? Ans. 20° 22' 4".
445. PROBLEM XII.-Given the observed altitude of the upper or lower limb of the moon, to find the true altitude of its centre.
To the observed altitude apply the semi-diameter by addition or subtraction, according as the altitude of the lower or upper limb is given; from this result subtract the
*This is a contraction for upper limb, as 1. 1. is for lower limb.
depression, and the remainder is the apparent altitude of the moon's centre; and to this altitude apply the refraction and parallax in altitude, as in the preceding problem.' Or, a' — a" ± s—d, and a = ·r + p ; these letters denoting the same quantities as in the preceding problem.
EXAMPLE.-If on the 12th of July 1841, in latitude 56° 40', the observed altitude of the moon's upper limb was 57° 14' 20", the height of the eye 22 feet, the semi-diameter 15′ 35′′ and the horizontal parallax 57′ 14′′; required the true altitude of the moon's centre. Ob. alt. moon's u. 1.
57° 14′ 20′′
Moon's semi-diam. = 15 35
par. 57 4
Ap. alt. of centre a' 56 53 58
Moon's par. in alt.
P.L, 57'4" 4989
Secant a' ·2511
True altitude a
=57 25 21
446. The longitude of the place of observation, and the time of observation, must be known, in order to determine the reduced time (427), and then the semi-diameter and horizontal parallax are found for the reduced time, according to the rule in article 431.
1. At a place in latitude 36° 50' the observed altitude of the moon's lower limb was 24° 18′ 40′′, the height of the eye 17.3 feet, the moon's semi-diameter at the time of observation 15', and its horizontal equatorial parallax 55′ 2"; what was the true altitude, the corrected semi-diameter, and parallax in altitude?
Ans. The true altitude of moon's centre 25° 17′ 38′′, semi-diameter 15′ 6′′, and parallax 50".
2. If, at a place in latitude 24° 30', and longitude 23° E., on the 31st May 1796, at 5h 36m P.M., the observed altitude of the moon's lower limb was 23° 48′ 15′′, the height of the eye 17.3 feet, the moon's semi-diameter and horizontal parallax at the preceding noon and following midnight being 15′ 49′′, 15′ 56′′, 58' 1", and 58' 29"; required the true altitude of the moon's centre.
Ans. The reduced time = 4h 4m P.M., corrected semidiameter 15′ 56′′, parallax in altitude = 53′ 6′′, and altitude =24° 51' 6".
3. On 10th September 1841, in latitude 28° 40′ N., longitude 24° 45′ W., at 5h 51m P.M., the observed altitude of the moon's lower limb was 32° 40′ 15′′, height of eye 16 feet; required the true altitude of the moon's centre, having also given the moon's
At noon preceding
At midnight following Ans. Reduced time = 7h 30m P.M., corrected semi-diameter 16′ 25′′, parallax in altitude = 50′ 10′′, and altitude 33° 41' 27".
447. PROBLEM XIII.-To find the polar distance of a celestial object.
When the declination and the latitude of the place are of the same name, subtract the declination from 90°; and, when of different names, add the declination to 90°; then the difference, in the former, and the sum, in the latter case, is the polar distance.'
Let D the declination of the body, P the polar distance, or codeclination; then P90 D.
EXAMPLE. What will be the moon's north polar distance on the 12th of November 1841, at noon at Greenwich, its declination then being 6° 56′ 45′′ S.?
P=90+D=90 +6° 56′ 45′′ = 96° 56′ 45′′.
1. Find the moon's north polar distance on the 11th September 1841, at 11h P.M. at Greenwich, its declination then being 8° 19' 9" N. Ans. 81° 40′ 51′′.
2. Find the north polar distance of Mars on the 10th of December 1841, when on the meridian of Greenwich, its declination being at that time 19° 15′ 16′′.
Ans. 70° 44' 44".
448. PROBLEM XIV. To convert intervals of mean solar time to intervals of sidereal time, and conversely.
'As 1h is to 1h 0m 9s-8565, so is the given interval of mean solar time to the required interval of sidereal time; and 1h is to 0h 59m 50s.1705, as the given interval of sidereal time to the required interval of mean time; or,
Find the value of the hours of the given time by the above proportion, which is obtained by merely multiplying the second term by the number of hours; then find the equivalent of the remaining time by applying the constant number 00119 to the proportional logarithm of the remaining given time by addition or subtraction, according as sidereal or mean solar time is given, and the result is the proportional logarithm of the equivalent in the required time, which, being added to the equivalent for the hours, will give the required time.'
449. Or find the equivalents by means of a table of time equivalents; or by means of a table of accelerations and retardations.
Let m the mean time,
s = m + a, m = s — r.
Convert 10h 20m 40s of sidereal time to
the equivalent sidereal interval,
By Method I.
1h: 0h 59m 50s∙1705 = 10h 20m 40s : 10h 18m 58s.32.
Equivalent of 10h
Given sidereal time
By Method III.
1. By a Table of Time Equivalents.
9h 58m 21s-7044
0 19 56-7235
0 0 39.8908
Retardation on 10h
9h 58m 21s.704
0 0 40s
10 20 40
2. By a Table of Retardations.
0 0 40s
10 20 40
0 20 36.61
10 18 58-31
= 10 18 58.32
0 1 38.295
= 0 1 41-68
Required mean time
= 10 18 58-32
The conversion of mean time to sidereal is performed in exactly the same manner, observing that, when it is calculated by means of a table, the acceleration is to be added to the mean time.
= 10h 20m 40s
The rules depend on the facts that a meridian describes 360° in 24 sidereal hours, and 360° 59′ 8′′:3 in 24 hours of mean solar time. Hence, it describes 59' 8"-3 in 3m 55s.91 mean time, and in 3m 56s.55 sidereal time.
Hence, 24h mean time = 24h 3m 56s-55 sidereal time, 24h sidereal time 23 56 4.09 mean time. 9.856 sidereal time, 50-170 mean time.
1h mean time = 1 0 1h sidereal time = 0 59