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425. PROBLEM I.-To convert degrees of right ascension, or of terrestrial longitude, into the corresponding time, and conversely.
'To convert degrees into time, multiply by 4, and consider the product of the degrees by 4 as minutes of time, the product of the minutes of space as seconds of time, and so on.
To convert time into degrees, reduce the hours to minutes, and consider the number of minutes of time as degrees, the seconds of time as minutes of space, and so on; and divide by 4, and the quotient will be the required number of degrees."
1. Convert 36° 12′ 40′′ to time.
(36° 12′ 40′′) × 4 = 144m 50s 40t2h 24m 50s 40t.
2. Convert 2h 24m 50s 40t to degrees.
(2h 24m 50s 40t) (144m 50s 40t)
36° 12′ 40′′. If the sun moved uniformly, it would pass over 360° of the equator or equinoctial in 24 hours of mean time, that is, 15 in 1 hour; hence, if d = the number of degrees, and h the number of hours corresponding,
1h : h = 15° : d, and therefore
= d, also d = 15h=
h = 15 and from these expressions, the rules are easily obtained.
1... 80° 32′ 40" are equivalent to 5h 22m 10s 40t
2...161 5 20
10 44 21 20
3... 98 14 48
6 32 59 12
4... 5h 22m 10s 40t are equivalent to 80° 32′ 40′′ 5... 0 28 6
7 1 30 210 18 0
6... 14 1
426. PROBLEM II.-To express civil time in astronomical time, and conversely.
'When the given time is P.M., the civil and astronomical time are the same; and when the civil time is A.M., add 12 hours to it, and the sum will be the astronomical time, reckoning from the noon of the preceding day. The rule for the converse problem is evident."
1. April 6th at 3h 12m P.M., civil time, is in astronomical time also April 6th 3h 12m.
2. June 1st at 10h 15m A.M., of civil time, is May 31st 22h 15m of astronomical time.
427. PROBLEM III.-To reduce the time under any given meridian to the corresponding astronomical time at that instant at Greenwich, and conversely.
'To find the time at Greenwich corresponding to that at another place; to the given time expressed astronomically, apply the longitude in time by addition when it is W., and by subtraction when it is E.
To find the time at a given place corresponding to a given time at Greenwich; to the given time expressed astrono
mically, apply the longitude in time by addition when it is E., and by subtraction when it is W.'
1. Find the time at Greenwich corresponding to June 20th at 9h 12m A.M., at a place in longitude 14° 2′ 30′′ W. Given time June, 19d 21h 12m 08 Longitude in time W., + 0 56 10
The reduced time at Greenwich, 19 22 8 10 2. Find the time at Greenwich corresponding to August 30th at 2h 40m 10s P.M., at a place in longitude 75° 34′ 45′′ E. Given time August, 30d 2h 40m 10s Longitude in time, 5 2 19
Astronomical time at Greenwich, 29 21 37 51 From these examples, the converse problem is evident.
1. Find the time at Greenwich corresponding to July 18th at 5h 24m A.M., at a place in longitude 40° 20′ W.
Ans. July 17a 20h 5m 20s. 2. Find the time at Greenwich corresponding to June 19th at 1h 12m 40s P.M., in longitude 90° 37′ 30′′ E.
Ans. June 18d 19h 10m 10s. 3. Find the time at a place in longitude 40° 20′ W. corresponding to July 18th at 8h 5m 208 A.M. at Greenwich. Ans. July 17 17h 24m. 4. Find the time in longitude 90° 37' 30" E., corresponding to June 19th, at 7h 10m 10s A.M. at Greenwich.
Ans. June 19d 1h 12m 40s.
428. PROBLEM IV.-To reduce the registered* declination or right ascension of the sun to any given meridian, and to any time of the day.
'As 24 hours is to the given time, so is the change of declination for 24 hours to its change for the given time. When the declination is increasing, add this proportional
*These elements of the sun's place are registered in the Nautical Almanac for noon of every day at Greenwich.
part to it; and when diminishing, subtract it; and the result will be the declination required."
By the same method, the sun's right ascension can be found for any time at a given place, only it is always increasing.
Or, if t
reduced time at Greenwich,
variation of declination in 24 hours,
24: tv':v, and v = v't, D=D' ± v. Or, if P.L. denote proportional logarithms, P.L., P.L., t + P.L., v'.
EXAMPLE. Find the sun's declination in 1828, August 30th, at 2h 40m 10s P.M., at a place in longitude 75° 34′ 45" E.
Time at Greenwich (427) August 29th = 21h 37m 518 = t
Variation of declination in 24h =0 21 28 = v' And 24h: 21h 37m 51s = 21′ 28′′ : v, and v = 0° 19′ 21′′
Hence, the required declination D'—v=D=8 58 38
P.L., t 21h 37m 51s
P.L., v 0 19 21
D=8 58 38 required declination. By changing declination to right ascension in the preceding rule, it will be adapted to the finding of the sun's right ascension.
1. Find the sun's declination in 1813, October 12th, at noon, at a place in longitude 4° 15′ W., the declination at
Greenwich at noon, 12th October, being 7° 21′ 18′′ S. and increasing, and its variation in 24 hours 22′ 33′′
Ans. 7° 21′ 34′′.
2. Find the sun's declination in 1828, June 20th, at 9h 12m A.M., at a place in longitude 14° 2' 30" W.; having given D' = 23° 27′ 22′′ N., and v' = 13", and the declination increasing. Ans. 23° 27′ 34′′. 3. Find the sun's declination in 1833, May 29th, at 2h 37m 20s P.M. in longitude 32° 40′ W.; having given D' = 21° 37′ 51′′ N., and v′ = 9′ 7′′. Ans. 21° 39′ 40′′. 4. What is the sun's right ascension in 1833, September 4th, at 4h 45m 39s P.M., at a place in longitude 72° 35′ 15′′ W., when D' = 10h 52m 20s-8 N., and v' = - 3m 37s? Ans. 10h 53m 478.6.
429. These logarithms, which are useful in calculating small quantities, as minutes of time or space, as they generally require to be carried out only to four decimal places, are obtained in this manner :
Let a, b, c,...be any quantities; assume another quantity 2, such that it exceeds any of the quantities a, b, c...then Lq-La is the proportional logarithm of a; Lq-Lỗ, that of b; and so on; also P. Lq= Lq — Lq = = 0. The quantity q so assumed is 3 hours or 3 degrees, and sometimes 24
Proportional logarithms may be used for quantities smaller than 9, like common logarithms in calculating a fourth proportional. For let a:bc:d, then
Lb LaLd-Lc, and therefore
Lq-La-(Lq-Lb) = LqLc — (Lq — Ld);
430. PROBLEM V.-To reduce the registered declination and right ascension of the moon to any given meridian, and to any time of the day.*
'Find the reduced time and the declination for the pre
*The moon's declination and right ascension are now given for every hour in the Nautical Almanac, and the variation of the former for every 10 minutes.