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The acceleration of sidereal on mean time in 24 sidereal hours is 3m 56.55, and the retardation of mean on sidereal time in 24 mean hours is 3m 55s.91. A table of accelerations and retardations for any number of hours, of minutes, &c., can easily be calculated.

The rule by proportional logarithms is obtained thus:— The first two terms of the proportion are either 3600s and 3609.85, or 3600s and 3590s·17, and the difference of their logarithms is 00119; then proportional logarithms may be taken for the other two terms; for if a, b, c, d, are the terms of a proportion, then (429),

and

~

La LbP.L.c~ P.L. d,
Lb-La-P.L. c- P.L. d,
P.L.d=P.L.c+ (La — Lb).

EXERCISES.

1. Convert 7h 40m 15s of sidereal time to mean time.

Ans. 7h 38m 59s.6.

2. Convert 7h 38m 59s.6 of mean time to sidereal time. Ans. 7h 40m 15s.

450. PROBLEM XV.-Given the sun's registered mean right ascension at mean noon, to find its mean right ascension at any place, and at any time of the day.

"Find the reduced time; then, as 24 hours is to the reduced time, so is 3m 56s-555 to a proportional part, which, when added to the given right ascension at the preceding mean noon, will give that required."

Let A' the registered mean right ascension at mean noon, that is, the sidereal time,

then

and

A

d'

d

the required mean right ascension,

the increase of A' in 24 mean hours
=3m 56s.555,

the proportional part for t,

t = the reduced time;

24: t=d': d, and d=d't,

A=A'+d.

451. The proportional part is just the acceleration for the reduced time, and can be found from a table of accelerations."

EXAMPLES.

1. Find the sun's mean right ascension at mean noon on the 11th April 1841, at a place in longitude 36° 15′ W.

t = longitude in time = 2h 25m.

Sun's given mean right ascension, or A' = 1h 18m 68.73 Increase in time t, or

Right ascension required, or

d

= 0 23.82

A1 18 30-55

2. What is the sun's mean right ascension at 2h 40m P.M., on the 30th April 1841, at a place in longitude 50° 20′ 30′′ W.?

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Sun's given right ascension, or A' = 2h 33m 1s.27

Increase in time t, or

Right ascension required, or

d = 0 0 59.139

A2 34 0.41

452. The principle of the rule is evident, for 3m 56s-555 is the increase of the sun's mean right ascension in 24 hours mean time, and terrestrial longitude reduced to time by the usual rule is mean time.

EXERCISES.

1. Find the sun's mean right ascension at mean noon at a place in longitude 45° 25′ W., on the 25th June 1841; its registered mean R.A. at mean noon being 6h 13m 48s.5. Ans. 6h 14m 188.34.

2. Required the sun's mean R.A. on the 20th July 1841, at 3h 20m P.M., at a place in longitude 56° 15′ W.; its registered mean R.A. at mean noon on the same day being 7h 52m 229 45. Ans. 7h 53m 32s.27.

3. What will be the sun's mean R.A. on the 14th November 1841, at 10h 40m A.M., at a place in longitude 36° 24′ 15′′ E.; its registered mean R.A. at mean noon on the 13th being 15h 29m 428.8. Ans. 15h 33m 28.3.

453. PROBLEM XVI.-To convert any given mean time on any given day to the corresponding sidereal time, and conversely.

When mean time is given, express it astronomically, and convert it into the equivalent sidereal time; then to this result add the sidereal time at the preceding mean noon, and the sum will be the required sidereal time.

When sidereal time is given, subtract from it the sidereal time at the preceding noon, and convert the remainder into its equivalent mean time, and it will be the required time." The sidereal time at the preceding noon, that is, the sun's mean right ascension, is found by the preceding problem. Let m the mean astronomical time,

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the equivalent interval of sidereal time, a = the acceleration for m,

r = the retardation for s,

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S'the registered sidereal time or sun's mean R.A. at preceding mean noon at the given place. -S' + 8.

When m is given, sm+a, and S

=

and when S is given, s = =S—S', and m = s — r.

EXAMPLES.

1. Find the sidereal time corresponding to 2h 22m 25s.62 mean time at Greenwich, January 2, 1841.

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36.77 required time.

S'

S

2. Find the mean time corresponding to 21h 10m 368-77 sidereal time, at Greenwich 2d January 1841.

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3. Find the sidereal time corresponding to 3h 40m P.M. mean time on the 11th April 1841, at a place in longitude 36° 15′ W.

The sun's mean R.A. at mean noon, that is, the sidereal time at the preceding noon at the given place, according to the first example of the preceding problem, is = 11 Ï8m 30$.55.

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454. The sidereal time at mean noon at any place is just the right ascension of its meridian at that time, that is, the sidereal interval since the transit of the first point of Aries; and this is just the right ascension of the mean sun at the mean noon. This time is given in the Nautical Almanac for Greenwich, and is easily found from the sun's right ascension at mean noon, by applying to it the equation of time; it could also be deduced from the sun's right ascension at apparent noon.

EXERCISES.

1. Convert 2h 21m 13s-08 mean solar time, 2d January 1841, at Greenwich, into the corresponding sidereal time; the sidereal time at mean noon being 18h 47m 478.76.

Ans. 21h 9m 248.04.

2. Convert 21h 9m 24s-04 of sidereal time on 2d January 1841, at Greenwich, into the corresponding mean solar time, the sidereal time at mean noon being 18h 47m 478.76. Ans. 2h 21m 13s.08.

3. Find the sidereal time corresponding to 8h 20m A.M. mean time on the 26th June 1841, at a place in longitude 45° 25′ W., the registered sidereal time at the preceding mean noon being 6h 13m 488.5. Ans. 2h 37m 388.8.

455. PROBLEM XVII.-To find the mean time of the sun's transit over the meridian of any place.

'Find the equation of time for the reduced time corresponding to the longitude, and it will be the time from mean noon, either before or after, at which the transit of the centre happens.

To the time of the meridian passage of the centre apply the time of the semi-diameter's passing the meridian, by subtraction or addition, according as the time of transit of the first or second limb is required.".

EXAMPLES.

1. Find the mean time of the transit of the sun's centre, and that of its first limb, over the meridian of Greenwich, on the 10th of January 1841.

Equation of time at apparent noon
to be added to apparent time
Time of semi-diameter's passage

Oh 7m 56s-14 = 1 10.23

0 6 45.91

Hence, the time of transit of the centre is Oh 7m 56s-14, and of the first limb Oh 6m 458.91.

2. Find the mean time of the transit of the sun's centre, and of his second limb, at a place in longitude 54° 30′ E., on the 28th April 1841. Longitude in time Reg. equation of time on 28th Reg. equation of time on 27th

= 3h 38m 0s

=

+

2m 37, 79

0

2 28-34

0 0

9.45

1.42

0 2 36-37 0 1 5.56

Increase of eq. of time in 24h
Increase of eq. of time in 3h 38m
Equation of time for reduced time =
Time of passage of semi-diameter =

Hence, the time of the passage of the 368.37, and of the second limb 3m 41s.93.

EXERCISES.

0 3 41.93 centre is Oh 2m

1. Find the time of the meridian passage of the sun's centre, and its second limb, at Greenwich on 30th April

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