PROBLEM III. To reduce the Moon's Right Ascension and Declination, as given in the Nautical Almanac, to any given time under a known Meridian. RULE. To the mean time at ship, always reckoned from the preceding noon, add the longitude in time if it be west, but subtract it if east: the result will be the corresponding mean time at Greenwich. As the moon's right ascension and declination are now given in the Nautical Almanac for every hour, between pages V. and XII. of the month; therefore, enter those pages and take out, under the given day, the right ascensions and declinations answering to the hours which are next less and next greater than the Greenwich time, and find their difference then, to the proportional logarithm of the difference of right ascension, or of declination, add the proportional logarithm of the minutes and seconds in the Greenwich time, and the constant logarithm 9.5229 (the prop. log. ar. comp. of one hour); the sum, abating 10 in the index, will be the proportional logarithm of a correction, which is always to be added to the right ascension at the next less hour; but to be applied by addition or subtraction to the declination, at that hour, according as it may be increasing or decreasing. Example. Required the moon's right ascension and declination, August 20th, 1836, at 9:40:36: mean time, in longitude 36:30: east? Mean time at ship or place Mean time at Greenwich = Moon's right ascension at 7 hours = 16:27":38: 0 231 prop. log. = 1.8544 prop. log. = 1.0909 = 9.5229 0:37: prop. log. = 2.4682 16.27.38 16 2815 as required. Moon's correct declination =.23:56:21 south, as required. PROBLEM IV. 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. Turn the longitude into time, and add it to the mean time at ship, if it be west, but subtract it if east; the sun or difference will be the corresponding mean time at Greenwich. As it is the four bright planets, viz. Venus, Mars, Jupiter, and Saturn, that are of any importance to the mariner; the places of which are given, in the order thus named, for every noon between pages 278 and 302, and tween 323 and 346 of the Nautical Almanac: therefore, enter those pages and take out the planet's right ascension and declination for the noons immediately preceding and following the Greenwich mean time, and find their difference; then, To the proportional logarithm of the difference of right ascension, or of declination, add the proportional logarithm of the Greenwich time, (esteeming the hours as minutes, and the minutes as seconds,) and the constant logarithm 9. 1249*; the sum, abating 10 in the index, will be the proportional logarithm of a correction, which being applied by addition or subtraction to the right ascension or declination at the noon preceding the Greenwich time, according as those elements may be increasing or decreasing, the sum, or difference, will be the planet's correct right ascension, or declination, at the given time and place. Example. Required the right ascension, and the declination of the planet Venus at 10:20 mean time, on the 3rd of July, 1836; the longitude being 75:30! west of the meridian of Greenwich? This is the prop. log. ar. comp. of 24 hours, esteemed as minutes. Venus's declination July 3rd. 16: 3:19 north. 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, 9, 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. 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 meeting 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 24728!? 2:47 25 answering to which is Odd seconds Given time 3. pro. part answering to which is .0.17572 11 24728: corresponding log. hf.-elapsed time. 0.17561 In the converse of this, that is, in finding the time corresponding to a given log. ;—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. 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 3:025; 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 3028: for that required. Remark. The numbers in this Table are expressed by the logarithmic co-secants adapted to given intervals of time, the index being diminished by radius, as thus: Let the half-elapsed time be 32045; to compute the corresponding logarithm. Given time 320" 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. 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 cy pher, as thus, 0, 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. 3:1723: ? 3:1720: answering to which is . 6.18099 Given time 3:17 23: corresponding log. middle time 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 2:530; 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 2:532:, for that required, |