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1841, the equation of time at apparent noon being 2m 55s 19, to be subtracted from apparent time, and the mean time of the semi-diameter's passing the meridian 1m 58.72.
Ans. For centre 29d 23h 57m 4s-81, and for the second limb 29d 23h 59m 108.53.
2. Required the time of the meridian passage of the sun's centre, and that of its first limb, at Edinburgh, in longitude 12m 43s W., on 16th November 1841, the equation of time at apparent noon (to be subtracted from apparent time), on the 16th and 17th, being 15m 18.05, and 14m 49s-21, and the mean time of the semi-diameter's passage being 1m 88.47.
Ans. For centre 15d 23h 44m 59s.04, and for the first limb 15d 23h 43m 508.57.
456. PROBLEM XVIII.-To find the mean time of a star's culmination at any given meridian.
'Find the sun's mean right ascension at mean noon at the given place, and subtract it from the star's right ascension, increased if necessary by 24 hours; and the remainder, which is a sidereal interval, being converted into mean time, will be the mean time of culmination."
Let A' the star's apparent right ascension at given time, A the sun's mean right ascension at mean noon preceding the transit at given place, (450) Tthe sidereal time of transit after mean noon, Tthe mean time of transit after mean noon; then TA' A, and T = T' — r by 449.
1. When will Arcturus culminate at Greenwich on 1st April 1841 ?
R.A. of Arcturus, or
R.A. of sun at mean noon,
Sidereal time of cul. after noon,
0 2 12.66
13 27 33.56
Mean time of transit,
2. Find the time of the passage of Arcturus over the
meridian of a place in longitude 62° 15′ W., on the 7th December 1841.
Longitude in time 62° 15′ W. = 4h 9m
Sun's registered mean R.A. on 7th = 17h 4m 206-18 Increase or acceleration for 4h 9m, or a = + 0 0 40·90
Mean time of transit,
1. At what time will Sirius culminate at Greenwich on the 17th December 1841, its right ascension being 6h 38m 128.6, and sun's mean right ascension or the sidereal time, at mean noon, being 17h 43m 45s.76?
Ans. At 12h 52m 50s.17.
2. When will Aldebaran culminate at New York, in longitude 73° 59′ W., on the 17th November 1841, its right ascension being 4h 26m 53s-42, and the registered sidereal time at mean noon being 15h 45m 29s-03?
Ans. At 12h 38m 31s.17.
457. PROBLEM XIX.-To find the mean time of the moon's culmination at any given meridian.
I. 'Find the sun's mean right ascension for the reduced time corresponding to the longitude, and find also the moon's right ascension for the same time; subtract the former from the latter, increased if necessary by 24 hours, and the remainder will be an approximate time.
Then as 1 hour, diminished by the difference between the hourly variations in right ascension of the sun and moon, is to the approximate time, so is this difference to a fourth term, which, added to the approximate time, will give the true time.
the moon's R.A. at the reduced time,
Then TA' — A.
1-(vv): Tv-v: r, and TT' + r.
458. The interval of time between two successive meridian passages of the moon, is called the moon's daily retar
II. If the transit is required only to about a minute of accuracy, it can easily be found by the following rule :—
"Find the difference between the times of the preceding and succeeding meridian passages, that is, the moon's daily retardation; then find the proportional part for the longitude in time; add this part to the first registered time, and the sum will be the required time.'
registered time of preceding passage,
the variation for the given longitude in time, t = the longitude in time.
24: tv': v, and v=v't. P.L. v=P.L. v' + P.L. t. P.L. t being taken from P.L. for 24 hours.
EXAMPLE. Find the time of the moon's culmination in longitude 40° 45′ W., on 2d May 1841.
By the second Method.
Longitude 40° 45′ = 2h 43m.
Time of reg. meridian passage on 2d May = + 9h 47m.3 Time of reg. meridian = passage on 3d May
By the first Method.
Sun's m. R.A. at m. noon on 2d May =
Sun's R.A. at noon at given place, A =
Increase in 1h or
Moon's R.A. at noon at place, A’
P.L. 18m 85.2
= +12 14
2h 40m 54s-39 26.78
v = 1
9 34 19.73 v= Om 98.856 v = 2 0.06
1m 50s-2 : r.
T=9 52 27·93 = mean time of transit.
459. The second method is somewhat simplified by means of a table of retardations, which contains at the top of the columns the daily retardations of the moon's meridian passage at Greenwich, and in the side column the longitude; under the former, and opposite the latter, is the retardation that must be added to the time of the preceding passage at Greenwich to obtain that at the given place. In the second method, the retardation by the table* under 46m.2, and opposite to 40° 45′ W., is + 5·1 as found above.
* See Table XVII., Riddle's Navigation.
Let the meridian be at S at mean noon at the given place, and the moon then at M'; then, if M be the position of the same meridian of the earth at the culmination of the moon, the meridian will have moved over the arc SM, while the moon has moved over M'M. Now, if arc SM'T' in sidereal time = A'— A, SMT" in sidereal time,
T= the mean time corresponding to T",
24 h' — (v' —v): 24 h' T': T.
By this proportion, the required time T can be found.
1 — (v' — v): v' — v = T': T—T.
Or, if r
T-T. This proportion is the rule, and the calculation can be made by proportional logarithms, taking those for 24 hours for the first two terms.
The first of the following exercises is to be solved by the second, and the second by the first method.
1. Find the time of the moon's meridian passage at a place in longitude 68° 30′ W., on the 5th November 1841, the time of its passage, on the 5th and 6th, over the meridian of Greenwich, being 18h 6m-4 and 18h 57m.2.
Ans. At 18h 16m.1.
2. Find the time of the moon's meridian passage at the same place, on the same day, the registered sidereal time, on the 5th, being 14h 58m 10s-35, and the moon's registered right ascension on the same day at 4h, being 8h 35m 18.64, and at 5h, 8h 37m 228.13. Ans. At 18h 17m 158.
460. From the irregularity in the increase of the moon's right ascension in 24 hours, the result is 11 minutes less in