ceding hour; then, as 10 minutes is to the time past that hour, so is the variation in 10 minutes to the variation in the past time, which being applied, by addition or subtraction, according as the declination is increasing or diminishing, to the given declination, will give that required.

The same rule applies for finding the right ascension, only 1 hour or 60 minutes must be used for 10 minutes; and as the right ascension is always increasing, the variation is always to be added."

Let D', D the earlier given and required declination,

R', R = the earlier given and required right ascension, t time past the hour preceding the reduced time,

:v,

for the declination, 10m: tm = v′ :·
and
to'; hence, D = D′ ± v ;
and for the right ascension, 60m: tm = v′ : v,
and
to'; hence, R = R' + v.

v=

v', vthe variations, for 60m and the past time of the right ascension, and for 10m and for the past time of declination; then

EXAMPLE. What will be the declination and right ascension of the moon on the 15th November 1841, at 0h 30m A.M., at a place in longitude 36° 45′ W?

Given time on 14th Nov., = 12h 30m
Longitude in time,
= +236

Reduced time, (427)

Declination.

Right Ascension. On 14th at 15h, R′ = 16h 53m 37s-35, D' = 26° 18′ 47′′-5 On 14th at 16h, 16 55 57.69, 26 20 25.9

Change in 60m

0 1 38-4

v = 0 2 20$.34 Diff. dec. in 10m 16"-4.

For declination, v' in 10m = 16"-4; hence

v = 1 tv' = 1 × 16·49′′·8,
D D'+= 26° 18' 57"-3.

and
For the right ascension,

and

15 6 and t6m

II.

v = √ tv = X 140834148-03,
R=R'+v=16h 53m 518.38.

60

From the variation of declination 1' 38"-4 in 60m, its change in the past time 6m could be found in the same manner as that for right ascension.

EXERCISES.

1. What will the moon's right ascension and declination be at Greenwich on the 18th of October 1841, at 10h 40m P.M., from these data?

October 1841.

Right Ascension. On 18th at 101, R′=17h 2m 18s·88, On 18th at 11h, 17 4 38.13 Ans. R17h 3m 518.71, and D =26° 35′ 7′′-8. 2. Required the moon's right ascension and declination on the 10th May 1841, at 1h 43m P.M., at a place in longitude 40° 15' E., from these data:

May 1841.
On 9th at 23h,

On 10th at Oh,

Declination.

D'= 26° 34′ 19′′·7 26 35 31.9

Right Ascension.

Declination.

R' = 19h 4m 28.62, D' = 25° 2′ 25′′-6.
19 6 11.19,
24 56 50.1.

Var. dec. in 10m 55"-92.
Ans. Reduced time = 9d 23h 2m, R = 19h 4m 68.9, and
D= 25° 2' 14"-4.

431. PROBLEM VI.-To reduce the registered semi-diameter or horizontal parallax of the moon to any meridian, and any time of the day.*

"Find the civil time at Greenwich; then, as 12 hours is to the reduced time, so is the change in either of these elements in 12 hours to its change for the reduced time, which is to be applied by addition or subtraction to the earlier given element, according as it is increasing or decreasing.'

Let s', s the earlier given and required semi-diameter, p', p the earlier given and required horizontal parallax,

t = the reduced time in hours,

v', vthe change in either of these elements for 12h and for the reduced time; then

for the semi-diameter,

12h: th=v' :v, v = 11⁄2 tv', and s=s' + v ; and for the horizontal parallax,

12h: th = v′ : v, v = tv', and p =p' + v.

* These elements are registered for every noon and midnight.

EXAMPLE. Find the semi-diameter and horizontal parallax of the moon, at a place in longitude 4° 20′ 15′′ W., in 1825, March 20th, at 7h 42m 39s A.M.; having given the registered elements for the preceding noon and midnight. The reduced time is 20th, 8h A.M.

March 20th, 1825.

At noon,
At midnight,

Semi-diameter. s' 14' 43" 14 44

=

1841, May 2.
At noon,
At midnight,

0.7

Horizontal parallax.

p' = 54′ 1′′

54 6

05 3.3

14 43.7

54 4.3 2 × 1′′=0′′-7, and s = s' + v, x 5" 3"-3, and p=p'+v.

=

EXERCISES.

1. Find the semi-diameter and horizontal parallax of the moon on 2d May 1841, at 10h 30m at Greenwich, from these elements:

Semi-diameter.

s=15′ 39′′-7
15 35.7

Horizontal parallax.
p=57′ 28′′-3
57 13.9

=

Ans. s 15′ 36′′-2, and p = 57′ 15′′-7. 2. Find the semi-diameter and horizontal parallax of the moon on the 10th of November 1841, at 4h 7m 10s P.M., at St Helena, in longitude 5° 42′ 30′′ W., from these ele

ments:

Semi-diameter.

1841, Nov. 10.

At noon,
s'15' 49"-3
At midnight, 15 45.8

Horizontal parallax.
p' = 58′
3"-8
57 50-8

Ans. Reduced time = 4h 30m, s = 15′ 48′′, and p = 57′ 58′′-9.

Augmentation of the Moon's Semi-diameter.

432. Since when the moon is in the zenith, it is nearer to the observer than when in the horizon by the radius of the earth, its apparent magnitude is consequently increased, and at intermediate altitudes its augmentation will be intermediate. The amount of this augmentation for any given

altitude is sensibly constant, and is given in a table, and can be easily applied.

For the altitude of 15°, this augmentation is 4"-5, so that the semi-diameter found above must be augmented by this quantity for this altitude, and would then be

= 14′ 43′′"·6 + 4′′·5 = 14′ 48′′.

The semi-diameter of the moon given in the Nautical Almanac is that which it would appear to have when seen from the centre of the earth. If this semi-diameter be denoted by s, and its apparent semi-diameter at the given place by s', and a its altitude; then s' can be calculated from the equation s's+ms2 sin a.

h

Where m = k sin 1", k = 3.6697, and k =

where h and s are the registered horizontal parallax and semi-diameter. The value of m is '00001779, for the ratio of h to s is constant.* When a = 0, then s′ = s.

Contraction of the Moon's Semi-diameter.

433. The lower limb of the moon is apparently more elevated by refraction than its upper limb, as its altitude is less, and consequently every diameter of the moon, except the horizontal one, is contracted, and the vertical one is subject to the greatest contraction. This contraction is greater the less the altitude, and is sensibly constant for a given altitude, and is therefore conveniently applied by means of a table.

The contraction for an altitude of 15° is 4"; so that the semi-diameter 14' 48", previously found, now becomes 14' 48"-4" 14' 44".

=

This semi-diameter, neglecting these corrections, was found to be 14' 43"-6; so that, in this instance, these two corrections very nearly compensate each other.

The Sun's Semi-diameter.

434. The sun's daily change of distance from the earth is so small compared with its distance, that its semi-diameter does not sensibly change in apparent magnitude in the

* See Astronomie Pratique, Francœur, p. 58.

course of a day, so that its registered semi-diameter may be considered as constant for at least one day. Its distance also is so great compared with the earth's radius, that its semi-diameter is not subject to an apparent augmentation dependent upon altitude.

The sun's diameter, however, like that of the moon, is subject to an apparent contraction by the unequal refraction of its upper and lower limb, and its amount is sensibly the same as for the moon.

435. PROBLEM VII.-Given the horizontal parallax of a celestial body, and its altitude, to find its parallax in altitude.

Radius is to the cosine of the apparent altitude, as the sine of the horizontal parallax to the sine of the parallax in altitude."

Let p' p

a

the horizontal parallax,
the parallax in altitude,
the apparent altitude;

=

L, R
L, cos a
L, sin p'

then,.

rad: cosa

sin p': sin p, sin psin p' cos a, when rad = 1. Or, by proportional logarithms,

or,

P.L,p=P.L, p'+L sec a-10.

EXAMPLE.-When the horizontal parallax of the moon is 54' 20", and its altitude 36° 45', what is its parallax in altitude?

Here a = 36° 45′, and p′ = 54′ 20′′.

By Logarithms.
= 10.

= 9.903770

=

8.198758

By Proportional Logarithms.

= 10

= 10.0962

•5202

L, R

L, sec a
P.L, p'

=

8.102528

P.L, p

43' 32".

Hence, p

L, sin p Hence, p = If, instead of sin p', the value of p' in seconds be taken, namely, 3260, and L, 3260 instead of L, sin p' in the first method above by common logarithms; then the result would be, Lp instead of L, sin P, and then p would be 2612 seconds, and five decimal places in the logarithm would be sufficient.

=

.6164

= 43' 32".