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PROBLEM III.

To reduce the Right Ascension and Declination of a Planet, as given in the Nautical Almanac, to any given time under a known Meridian,

RULE,

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Turn the longitude into time, and add it to the apparent time at ship or place if it be west, but subtract it if east; and the sum, or difference, will be the corresponding time at Greenwich.

From page IV. of the month in the Nautical Almanac, take out the planet's right ascension and declination for the nearest days preceding and following the Greenwich time, and find the difference; find, also, the difference between the Greenwich time and the nearest preceding day; then,

To the proportional log. of this difference, add the proportional log. of the difference of right ascension, or declination, and the constant log. 9.9031 * ; the sum of these three logs., rejecting 10 from the index, will be the proportional log. of a correction, which being applied by addition, or subtraction, to the right ascension, or declination, (on the nearest day preceding the Greenwich time,) according as it may be increasing or decreasing, the sum, or difference, will be the correct right ascension or declination at the given time and place.

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Example.
Required the right ascension of the planet Venus, July 3, 1924, at

apparent time, at a place 75:30! west of the meridian of Green

10:20" wich?

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To find the Right Ascension :
R. A. of Venus, July 1 = 62 87

14 0 0?
7 = 6.40.

Gr. time = 3. 15. 22

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5

Ditto

Difference.
=0:32. Diff.

24 15:22
which are to be ésteemed as minutes and seconds :-hence,

63.221;

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The arithmetical complement of the proportional log. of 144 hours (6 days,) esteemed as minutes ; and, hence taken as 2 hours and 24 minutes.

Diff. bet. G. time and nearest preceding day 63: 22" prop. log. = .4534 Diff. of right ascension in 6 days .

0.32" prop. log. = .7501 Constant log.

9.9031

Correction of right ascension
Planet's R. A. on July 1, 1824 =

+14! 5" p. log. =1. 1066 6" 80:

R. A. of Venus, as required

6:22 5: To find the Declination:

Dec. of Venus, July 1 = 23:36? N. .

1 0 0 Ditto

7 = 23. 32. N. Gr. time = 3.15.22

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Difference

0 4! Diff. = 2.15.22. = 63:22; which are to be esteemed as minutes and seconds; hence, Diff. bet. G. time and nearest day preceding 63:22" prop. log. = .4534 Diff. of declination in 6 days =

0: 4! prop. log. 1.6532 Constant log.

9.9031

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Logarithmic Half-elapsed Time. This Table is useful in finding the latitude by two altitudes of the sun; and also in other astronomical calculations, as will be shown hereafter. The Table is extended to every fifth second of time under 6 hours, with proportional parts, adapted to the intermediate seconds, in the right hand margin of each page; by means of which, the logarithmic half-elapsed time answering to any given period, and conversely, may be readily obtained at sight.

As the size of the page would not admit of the indices being prefixed to the logs. except in the first column, under 0:, therefore where the indices change in the other columns, a bar is placed over the 9, or left hand figure of the log., as thus, 7, to catch the eye, and to indicate that from thence, through the rest of the line, the index is to be taken from the next lower line in the first column, or that marked 0! at top and bottom. It is to be observed, however, that the indices are only susceptible of change when the half-elapsed time is under 23 minutes,

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The logarithmic half-elapsed time corresponding to any given period, is to be taken out by entering the Table with the hours and fifths of seconds at the top, or next less fifth if there be any odd seconds, and the minutes in the left-hand column; in the angle of ineeting will be found a number, which being diminished by the proportional part answering to the odd seconds, in the right hand margin, will give the required logarithm,

Example. Required the logarithmic half-elapsed time answering to 2247"28! ?' 2:47*25: answering to which is

. 0.17572 Odd seconds .. 3. pro. part answering to which is

11 Given time = 2:47"28: corresponding log. hf.-elapsed time. 0.17561

In the converse of this, that is, in finding the time corresponding to a given Iog.;--if the given log. can be exactly found, the corresponding hours, minutes, and seconds, will be the time required :--but if it cannot be exactly found (which in general will be the case), take out the hours, minutes, and seconds answering to the next greater log.; the difference between which, and that given, being found in the column of proportional parts, abreast of where the next greater log. was found, or nearly so, will give a certain number of seconds, which being added to the hours, minutes and seconds, found as above, will give the required time.

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Example. Required the time corresponding to the logarithmic half-elapsed time 0.14964 ?

Solution. The next greater log. is 0. 14973, corresponding to which is 3:0:25: ; the difference between this log. and that given is 9; answering to which in the column of proportional parts is 3 seconds, which being added to the above found time gives 3:028: for that required.

Remark.-The numbers in this. Table are expressed by the logarithmic Ço-secants adapted to given intervals of time, the index being diminished by radius, as thus :

Let the half-elapsed time be 3.20.45: ; to compute the corresponding logarithm.

Given time = 3:20:45: in degrees = 50:11:15"; log. co-secant less radius = 0.114557; which, therefore, is the required log.; and since it is not necessary that this number should be extended beyond five decimal places, the sixth, or right hand figure, may be struck off; observing, however, to increase the fifth figure by unity or 1, when the right hand figure, so struck off, amounts to 5 or upwards :-hence, the tabular number Corresponding to 3:20*45: is 0.11456; and so of others,

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TABLE XXXI.

Logarithmic Middle Time.

This Table is, also, useful in finding the latitude by two altitudes of the sun; for which purpose it is extended to every fifth second under 6 hours, with proportional parts for the intermediate seconds, in the right-hand margin of each page ; by means of which the logarithmic middle time answering to any given period, and conversely, may be readily taken out at sight.

As the indices are only prefixed to the logs. in the first column, therefore where those change in the other columns a bar is placed over the cypher, as thus, ā, to catch the eye, and to indicate that from thence through the rest of the line, the index is to be taken from the next lower line, in the first column.

The logarithmic middle time answering to any given period is to be taken out by entering the Table with the hours and fifths of seconds at the top, or the next less fifth second (when there are any odd seconds, as there generally will be), and the minutes in the left-hand column; in the angle of meeting will be found a number, which being augmented by the proportional part answering to the odd seconds, in the compartment abreast of the angle of meeting, will give the log. required.

Example. Required the logarithmic middle time answering to 3:17*23:? 3:1720: answering to which is

6. 18099 Odd seconds

3. pro. part answering to which is

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+ 8

Given time = 3:17*23! corresponding log. middle time

6. 18107

The time corresponding to a given logarithmical number, is found by taking out the hours, minutes, and seconds, answering to the next less tabular number; the difference between which and that given, being found in the compartment of proportional parts, abreast of the said next less tabular number, will give a certain number of seconds, which being added to the hours, minutes, and seconds found as above, will be the time required.

Example. Required the time corresponding to the log. middle time, 6.01767)

Solution. The next less tabular log. is 6:01757, answering to which is 2:57.30: ; the difference between this log. and that given is 10, answering to which in the column of proportional parts is 2 seconds, which being added to the time found, as above, gives 2:5.32:, for that required.

Remark.-The logarithmic middle time may be readily computed by the following rule; viz :

To the logarithmic sine of the given time expressed in degrees, add the constant log. 6.301030, and the sum, abating 10 in the index, will be the required logarithm.

Example,
Let the middle time be 4? 1025', required the corresponding log. ?
Given time = 4:10:25:, in degs. = 62:36:15” log. sine = 9.948339
Constant log

= 6.301030

Logarithmic middle time, as required ...

= 6.249369

; and since it is not necessary that this log. should be extended beyond five places of decimals, the sixth, or right-hand figure may, therefore, be struck off ; observing, however, to increase the fifth figure by unity or 1, when the right-hand figure, so struck off, amounts to 5 or upwards; hence the tabular number corresponding to 4:10–25, is 6. 24937, and so on.

TABLE XXXII.

Logarithmic Rising.

This Table, with the two preceding, is particularly useful in finding the latitude by two altitudes of the sun; it is also of considerable use in many other astronomical calculations, such as in computing the apparent time from the altitude of a celestial object; determining the altitude of a celestial object from the apparent time, &c. &c.--The arrangement of the present Table is so very uniform with the preceding, that it is not deemed necessary to enter into its description any farther than by observing that the indices are only prefixed to the logs. in the first column :-that where those change in the other columns, large dots are introduced instead of O's to catch the eye, for the purpose of indicating that from thence through the rest of the line, the index is to be taken from the next lower line in the first column ; and that, in the general use of the Table, these dots are to be accounted as cyphers,

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