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1° +1°. a {1 + h. cos. 2 (p+6)} sin.2+4°.ah.tan..cos.2(+6) (16)

If we had retained the term u" = h. cos. 2 (p+6), which gives

'du"

ddu".

-) = 0 ; (dd, ') = — 4 h. cos 2 (p+6) it would produce the cor

rection

=

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1

and since =1+tan. 2. this may be put under the form

cos.2

1o. ah. { -3-4 tan. 2}. cos. 2 (p +6)

which being added to the expression (16) we get the corrected value of the length of this degree equal to

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1° +1°. a. ↓2. {1 + h. cos. 2 (p+6)}. sin. 42 —3°. ≈ h. cos. 2 (p +6).

IV.

Method of correcting the apparent distance of the Moon from the Sun, or a Star for the effects of Parallax and Refraction.

BY NATHANIEL BOWDITCH, LL. D.

THE method of correcting the apparent distance of the Moon from the Sun, or a Star for the effect of parallax and refraction, which was published by me in a work, entitled the "Practical Navigator," has one great advantage over other methods of approximation; namely, that all the corrections are additive; which renders it peculiarly well adapted for the use of mariners. It was given without demonstration, and two of the smaller corrections were neglected, as is usual in such methods. I have here drawn up a general demonstration of the formula, with tables for computing the neglected terms, from which it will appear that they are generally insensible,

M

B

C

A

S

Let Z be the zenith, M the true, and L the apparent place of the Moon; S the true, and A the apparent place of the Sun or Star. Upon the arch LA (continued if necessary) let fall the L perpendiculars MK, SB, join MB, and continue it towards C to meet the perpendicular SC let fall thereon. Then in the case of the present figure, in which all the angles and sides of the triangle AZ L are acute, we shall have the following expression of the true distance M S.

K

Bowditch on correcting the apparent distance of the Moon &c. 51

=

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MS (LA-2)+(60′ —SA) + (59′ 42′′ — LM) + (SA + AB) +(LM—LK)+(18′′ + BM — BK)+BC + (MS — MC) Because the terms — 2° + 60′ + 59′ 42′′ + 18′′, -SA+SA, - LM+ LM, LA+AB-LK-BK, BM+ BC-MC, which occur in this expression mutually destroy each other, leaving the identical equations MS= MS. Now ABSA.cos A, neglecting the third power of SA. Hence SAAB SA. (1 + cos. A) 2. SA. cos. A2, and if we put ZL 90°-m, ZA 90°-8, LA = d, 2S=d+s+m, Snd=f, S-8=g, we shall have, by the noted theorem for finding an angle of a spherical triangle when the three sides are given, namely

=

=

=

=

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+

tional logarithms Prop. log. (SA + AB) = Prop. log.

log. sin. d + log. cosec. g + log. sec. f.

2. SA

COS. S

To simplify this calculation, 1 have computed and published in the “Practical Navigator" the tables, numbered XVII (or XVIII), in which by a single entry may be found the quantity 60'-SA, and the Prop. log and then by the preceding formula the quantity SA + AB, called the first correction, is to be found.

2. SA

COS. S

=

LK LM-LM. cos. L

In a similar manner we have LM - LK =2. LM. sin. L' =

2LM.

44

sin. § (ZA + LA —ZL) . sin. § (ZA + ZL — LA) — 2 LM. sin. g. cos. S

sin. ZL. sin. LA

and by using Proportional logarithms, we shall get

2. LM

COS. m sin. d'

Prop.log.(LM-LK)=Prop.log. +log. sin. d+log. cosec. g+ log. sec. S.

cos. ገዜ

This is also simplified by a Table numbered XIX, containing

the value of the quantity 59′, 42′′- LM for all altitudes, and at

2. LM

the same entry the Proportional Logarithm of by means of

cos. m

which the quantity LM-LK called the second correction is computed. This requires only one additional logarithm; because log. sin. d and log. cosec. g were found in making the computation of the first correction.

The next term in the values of MS is 18" + BM — BK. Now it is well known that when the arch MK is very small in comparison with BK, the difference BM-BK will be very nearly equal to ± MK2. cot. BK, or ± . MK2. cot. d: so that the term now treated of is 18′′ MK2. cot. d. The upper sign being used if d90, the lower if d> 90°. This is found by means of Table XX, which contains two vertical columns corresponding to each value of d, the arguments at the side being the quantites 60'-LM and 60'-LK, or in other words the correction of Tab. XIX, and corr. Tab. XIX + second correction; the tabular numbers corresponding when d90°, being respectively

18′′ + + LM2. cot. d and LK2. cot. d, whose difference is 18′′ + 1. (LM2— LK2). cot. d or 18" + MK2. cot. d. When d> 90°, the tabular numbers are 18" + LK2. cot. d, and ÷ LM2. cot. d,

whose difference is 18" 14 MK. cot. d, as above. This is called the third correction.

Therefore if we neglect the two very small terms BC + (MS - MC) in the value of MS, it will become

MS=(app.dist.-2°)+corr.Tab.XVII÷corr.Tab.XIX+1acorr.+2acorr.+3acorr. (A) and all the terms of this expression will have the affirmative sign. As an example of this formula, let us take the following, which is the first in the Practical Navigator, Pag. 155, Edit. 4. The apparent distance being 38° 52', Star's app. alt. 43° 14', 's app. alt. 53° 4. D's Horiz. Parallax 54′ 35′′

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2S=d+s+m=135 10 Tab.xvii.log. 1.8112 Tab. xix log. 2266

S

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67.35 1st Corr. P.L. 2.0506 2d, corr. P.L. 8277

Cor. 2 26 46

Sid=f
S-s=g

28.43
24.21

Tab xx 29

True distance

38 47 27

Thus we see that this method is quite short, and it has the inestimable advantage of being free from a variety of cases in the application of the corrections, since all the terms are additive.

The two neglected terms BC + (MSMC) may be computed in the following manner. We have BCBS. sin. BSC =

BS. sin MBK nearly; and sin. MBK

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MK

sin. d nearly, hence BC

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= 2. cot. d. cosec. 2 d, or sind

=

2. cot. d. cosec 2 d, which being substituted in BC, it becomes 言

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BC= 2. BS. MK. cot. d.cosec. 2 d, and this may be calculated by means of the Tables E, F, G subjoined. In Table E the argument at the top is 60'-SA, at the side, the first correction SA+ AB, the corresponding number is the value of BS =

SA-AB2 in minutes = E. In Table F the argument at the top is E, and at the side the third correction 18" ± MK2. cot. d, the tabular number corresponding being

F

= √{BS√√. MKP. cot. d} = √E. √ MK2. cot d. expressed in minutes. Then in Table G, the argument at the top is d, and at the side F, corresponding to which is the expression of

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