Almanac; the sum, rejecting 10 in the index, will be the logarithm of the approximate time T in seconds. 3. Enter Table XLV., with the arc B at the top, and the time T at the side, and find the corresponding correction; to the logarithm of which add the two first logarithms above found; the sum, rejecting 10 in the index, will be the correction of the approximate time, to be applied with the same sign as the arc B, and the correct value of T will be obtained, which will express the longitude of the place of observation, if it be west from Greenwich; but if the longitude be east, we must subtract this value of T from 12h. to obtain the true longitude in time east from Greenwich. EXAMPLE II. Suppose that, in a place in west longitude, on the 16th of September, 1836, the moon's bright limb passed the meridian in 3m. 20s.65, sideral time, before the star Antares. Required the longitude of the place of observation. In the Nautical Almanac, column 4, the star Antares or a Scorpii's right ascension, Subtract the observed difference of the transits in sideral time...... 16 16 16h. 19m. 23.07 9m. 40s.71 The remainder is the right ascension of the moon's bright limb at the transit...... The four right ascensions to be taken from the Nautical Almanac, are those corresponding to September 15, l. c., September 16, u. c., September 16, l. c., and September 17, v. c., being the same as those in the preceding example, where we have found A=30m. 50s.74= 1850s.74, B+61s.10. The rest of the calculation is as follows: T=3h. 48m. 288.3=the longitude of the place of observation. This longitude agrees, within a fraction of a second, with the value of the longitude assumed in Example I.; observing that the computed right ascension in Example Lis 16h. 16m. 02s.42, being the same as that which is supposed to be observed in the present example. When the difference of meridians is small, we may compute their difference from the observed difference of the times of the moon's transit, by means of the arc H, deduced from column 6 of the table of moon-culminating stars, by the following rule :— RULE. To compute the difference of meridians by means of the arc H. 1. To the constant logarithm 3.55630 add the arithmetical complement of the logarithm of the arc H, and the logarithm of the difference of the times of the moon's transit over the two meridians in sideral time; the sum, rejecting 10 in the index, will be the logarithm of the difference of meridians expressed in seconds of time. EXAMPLE III. Suppose that, in a place west from Greenwich, Sept. 16, 1836, the moon's bright limb passed the meridian in 20m. 02s.30, sideral time, after the star Antares. Required the longitude. It appears by column 4 of the table of moon-culminating stars, that, on September 16, the right ascension of Antares was, 16h 19m. 23s.07. Adding this to 20m. 02s.30, we get 16h. 39m. 258.37 for the right ascension of the moon's bright limb at the time of its transit over the meridian of the place of observation. Subtracting from this the time of transit at Greenwich, 16h. 37m. 12s.45, taken from column 4 of the table of moon-culminating stars, we get 2m. 128.92=132s.92, for the difference of the times of the transits, to be used in the above rule. Moreover, the arc H, corresponding to the time of the transit at Greerwich, is, by column 6 of the table, H=156s.77. Then we have, In strictness, the value of H, here used, ought to be increased a little; for, by column 6 of the preceding table, its value for Greenwich is 151s.62, and for a place in the longitude of 12h. west, is 156s.77. The difference between these two values of H, is 58.15, which repre sents its increment corresponding to a change of 12h. in the longitude, being at the rate of Os.429 for a change of 1h. in the longitude; and at this rate the increment for the longitude, 1 second differences depending on the arc B, and compute the arc H as in Example 1. Hence the value of the arc H, corresponding to the meridian of the place of observation, is 156s.77+0s.38157s.15. If we take the mean of the values of H at Greenwich, 1568.77,* and at the place of observation, 157s.15, it becomes H = 156s.96, and with this we may repeat the above calculation, and obtain a corrected result. Corrected arc H=1568.96.. Difference of times of transit 1328.92......... Constant Log. 3.55630 .Arith. Comp. Log. 7.80421 ...Log. 2.12359 Correct difference of longitude 50m. 488.630488.6.......... ..Log. 3.48410 In general, the longitudes of places where such observations are made, are known, within a few seconds, so that it will be easy to find at once the value of the arc H, corresponding to the estimated meridian which falls midway between the meridians of the two places of observation; the meridian of Greenwich being used as one of these places, when the times of transit given by the Nautical Almanac are used as if they were actual observations. We shall give the following example of this method :-- EXAMPLE IV. In a place whose longitude was known to be 3h. 38m. 298. W. from Greenwich, it was found by observation, on September 16, 1836, that the moon's bright limb passed the meridian 3m. 46s.2, sideral time, before the transit of the star Aldebaran; and in another place, estimated to be 20m. in longitude west from the first place, or in 3h. 58m. 298. W., the observed difference of the transits was 2m. 55s.0. Required the difference of longitude which results from this observation. The mean of these two longitudes is 3h. 48m. 29s., and we have found in Example I., that the arc H, corresponding to this meridian on that day, was 153s.30. Moreover, the difference of the two times of transit, 3m. 46s.2, and 2m. 55s.0, is 51s.2; then we have, as in the last example, Gives the longitude of the second place 3h. 58m. 31s.3 W., as it is deduced from this observation. PROBLEM XVII. Given the longitudes of the sun and moon, and the moon's latitude. to find their distance. RULE. Find the difference of the two longitudes, and to its log. cosine add the log. cosine of the moon's latitude; the sum, rejecting 10 in the index, will be the log. cosine of the sought distance, which will be of the same affection* as the difference of longitude. EXAMPLE. July 20th, 1836, at noon, mean time at Greenwich, by the Nautical Almanac, the sun's longitude was 117° 42′ 31", the moon's longitude 193° 46′ 05′′, and the latitude 2° 47' 16" N. Required their distance. 's longitude.......... 117° 42′ 31 D's longitude.......... 193 46 05 Difference of longitudes 76 03 34..........Cosine 9.38186 Distance.... 76 04 35..........Cosine 9.38135, as in the Nautical Almanac This is calculated by another method in Example III. of Problem XVIII. In this rule, the sun's latitude is neglected, being only a fraction of a second. The distances being calculated from noon and midnight by this (or by the following) problem, they may be interpolated for every three hours, by Problem I. The following example will serve for an illustration : EXAMPLE. Given the distances of the sun and moon, in July, 1836, 19d. 12h., 20d. Oh., 20d. 12h. and 21d. Oh., respectively 70° 02′ 35′′, 76° 04′ 35′′, 82° 11′ 29′′, and 88° 23′ 32′′. Required the distances, July 20d. at 3h., 6h., and 9h. Two arcs are said to be of the same affection when they are both greater than 90°, or both less than 90°, but of different affection when the one is greater and the other less than 90°. Given the longitudes and latitudes of the moon and a star, to find their distance. RULE. To the log. secant of the difference of longitude of the moon and star, add the log. tangent of the greatest latitude; the sum, rejecting 10 in the index, will be the log. tangent of an arc A, of the same affection as the difference of longitude. Take the sum of the arc A, and the least latitude, if the latitudes are of a different name, but their difference if of the same name, and call this sum or difference the arc B. Then add together the log. secant of the difference of longitude, the log. secant of the greatest latitude, the log. cosine of the are A, and the log. secant of the arc B; the sum, rejecting 30 in the index, will be the log. secant of the distance of the moon and star, which will be of the same affection as B. EXAMPLE I. Required the distance of the moon from the star a Pegasi, at noon, mean time at Greenwich, July 9d. 1836, when, by the Nautical Almanac, the moon's longitude, counted from the mean equinox, was 59° 40′ 32", and her latitude 0° 59' 15" N.; the longitude of the star, computed † as in Problem XIX., being 351° 12′ 29′′, and its latitude 19° 24′ 29′′ N. This distance agrees with the calculated value given in page 146 of the Nautical Almanac. We may observe, that the log. secant of the distance is also equal to the sum of the log. cosecant of the greatest latitude, the log. sine of the arc A, and the log. secant of the are B, rejecting 20 in the sum of the indices; but the above rule is in general most convenient, on account of the smallness of the greatest latitude, except when the difference of longitude is nearly equal to 90°. We may use the same method for finding the distance of the moon from the sun, star, or a planet, when their right ascensions and declinations are given, instead of their longitudes and latitudes. The rule is the same as that we have given above, changing longitude into right ascension, and latitude into declination. To exemplify this, we shall compute the same example by this second method. EXAMPLE II. Required the distance of the moon from the star a Pegasi, at noon, mean time at Green wich, July 9d. 1836, when, by the Nautical Almanac, the moon's right ascension was 57° 15' 01" from the mean equinox, the moon's declination 21° 3′ 55′′ Ñ.; the star's right ascension from the same equinox 344° 9′ 20′′, and the star's declination 14° 19′ 32′′ N. D's right ascension.... 57° 15′ 01′′ *'s right ascension.... 344 09 20 Difference............ 73 05 41..........Secant.. 10.53642.. Greatest declination.. 21 03 55 N........Tangent 9.58566.. .......Secant 10.03004 ... 10.53642 .Cosine 9.77998 ....Secant 10.10720 Difference is arc B... 38 37 24... Distance 69° 23′ 58"..........Secant 10.45364 We have preferred these computed values, as being rather more accurate than the numbers in Table XXXVII. The sum must be used if the latitudes are of different names. This differs 2" from the former method, from the neglect of the tenths of a second in the angles, and from not taking the logarithms to 6 or 7 places of figures. EXAMPLE III July 20, 1836, at noon, mean time at Greenwich, by the Nautical Almanac, the sun's right ascension was 119° 47′ 35", the sun's declination 20° 38′ 23′′ N., the moon's right ascension 193° 45′ 07′′, and the moon's declination 2° 52′ 03′′ S. Required their distance This agrees with the distance marked in the Nautical Almanac. PROBLEM XIX. Given the right ascension and declination of a celestial object, with the mean obliquity of the ecliptic E, to find its longitude and latitude. RULE. To the log. tangent of the declination add the log. cosecant of the right ascension of the object; the sum, rejecting 10 in the index, will be the log. tangent of the arc A, to be taken out less than 90°, and called north or south, as the declination is. If the right ascension is less than 180°, call the obliquity of the ecliptic south; if above 180°, north. If A and E are of the same name, take their sum, otherwise their difference, which call B, and mark it with the same name as the greater number, whether N. or S. Then add together the log. secant of A, the log. cosine of B, and the log. tangent of the right ascension; the sum, rejecting 20 in the index, will be the log. tangent of the longitude in the same quadrant as the right ascension, unless B be greater than 90°, in which case the quantity found in the same quadrant as the right ascension, subtracted from 360°, will be the longitude. To the log. sine of the longitude add the log. tangent of B; the sum, rejecting 10 in the index, will be the log. tangent of the latitude, of the same name as B. Remark. As the Tables of this collection are not marked above 180°, you must subtract 180° from the right ascension, when it exceeds that quantity, and find the log. tangent and log. cosecant of the remainder; and then the arc, corresponding to the log. tangent of the longitude, is to be taken of the same affection as this remainder, and 180° added thereto; the sum will be the longitude, unless B is greater than 90°, in which case the supplement of that sum to 360° is to be taken, as observed above. EXAMPLE. From the Nautical Almanac we find, that, on the 9th of July, 1836, the right ascension of a Pegasi was 22h. 56m. 37s.35=344° 9' 20", its declination 14° 19' 32" N., and the mean obliquity of the ecliptic 23° 27′ 38". Required its longitude and latitude. Declination.. 14° 19' 32" N.. A. ..Tang.. 9.40717 Right ascens. 344 09 20 ..........Cosec. 10.56380........Tang. 9.45303 E........... B. ..Cosine 9.59988........Tang. 10.36268 *'s longitude 351° 12′ 29′′..........Tang. 9.18939........ Sine.. 9.18425 This longitude is counted from the mean equinox, July 9d. 1836, and if we wish to reduce it to the apparent equinox, we must apply to the preceding longitude the equation of the equinoxes deduced from Table XL., which is nearly-12"; so that the longitude, counted from the apparent equinox, is 351° 12′ 17", and the apparent latitude 19° 24′ 29′′ N. We have, in this example, taken the right ascension and declination of the star from the Nautical Almanac, where they are given to fractions of a second; which is more accurate than Table VIII., where the declinations are given to the nearest minute. We may, however, use the numbers in Table VIII., when great accuracy is not required, correcting for the aberration, as in the precepts to Table XLI. The numbers computed in this problem agree nearly with the results obtained from Table XXXVII. †The difference is to be used if the declinations are of the same name. PROBLEM XX. The longitude and latitude of a celestial object being given, with the mean obliquity of the ecliptic E, to find the right ascension and declination, RULE. To the log. tangent of the latitude add the log. cosecant of the longitude; the sum, rejecting 10 in the index, will be the log. tangent of the arc A, which is to be called north or south, as the latitude is. If the longitude is less than 180°, call the obliquity E north; if above 180°, south. If A and E are of the same naine, take their sum, otherwise their difference, which call B, marking it with the same name as the greater number. Then add together the log. secant of A, the log. cosine of B, and the log. tangent of the longitude; the sum, rejecting 20 in the index, will be the log. tangent of the right ascension in the same quadrant as the longitude, unless B be greater than 90°, in which case the quantity found in the same quadrant as the longitude, subtracted from 360°, will be the right ascension. To the log. sine of the right ascension add the log. tangent of B; the sum, rejecting 10 in the index, will be the log. tangent of the declination, of the same name as B. Remark. If the longitude exceeds 180°, you must subtract 180° from it, and find the log. tangent and log. cosecant of the remainder. The arc corresponding to the log. tangent of the right ascension is to be taken of the same affection as this remainder, and 180° added thereto, will be the right ascension, unless B is greater than 90°, in which case the supplement of that sum to 360° is to be taken, as was observed before. EXAMPLE. On the 9th of July, 1836, the apparent longitude of the star a Pegasi was 351° 12′ 29′′, counted from the mean equinox, the star's apparent latitude 19° 24′ 29′′ N., and the mean obliquity of the ecliptic 23° 27' 38". Required its right ascension and declination. E....... ..Tang. 10.36268........Secant 10.40012 .Cosine 9.86352..........Tang. 9.97097 23 27 38 S. [This is N. when the longitude *'s right ascension 344° 09′ 20′′.....Tang. 9.45303.. .......Sine 9.43820 ........Tang. 9.40717 The assumed longitudes in this example are the same as those computed in Problem XIX, by means of the right ascension and declination taken from the Nautical Almanac; being rather more accurate than the results of Table XXXVII. The right ascension and declination computed in this example, agree with those assumed in Problem XIX., which serves as a proof of the correctness of the calculation. If the given longitude, latitude, and obliquity, are the mean values, the resulting right ascension and declination will be the mean values; but if the proposed quantities are corrected for aberration and nutation, the resulting quantities will also be corrected. This remark is equally applicable to the preceding problem. SPHERIC TRIGONOMETRY. Most of the rules given in the preceding problems may be easily demonstrated by Spheric Trigonometry. As, for example, that of Problem XVII. may be investigated as follows:-In Plate XIII., fig. 1, let A be the place of the moon, C that of the sun, CP an arc of the ecliptic, and AP a circle of latitude passing through the moon, and cutting the ecliptic at right angles, at P. Then the difference of longitude of the sun and moon is equal to the arc CP, and the moon's latitude is AP; whence the distance AC may be found by the rule of Napier, radius X cos. AC=cos. AP X cos. CP. This in logarithms gives log. cos. AC=log. cos. AP+ log. cos. CP-log. radius, which is the formula made use of. Want of room prevents the insertion of the demonstrations of the methods of calculating the other problems. The celebrated rules given by Lord Napier, for solving the problems of right-angled spheric trigonometry, being very easily remembered, are much made use of by mathemati cians. In a paper communicated by the author of this work to the American Academy of Arts and Sciences, and published in the third volume of the first series of the Memoirs of that society, a method was given for the more easy application of those rules to oblique spheric trigonometry, and as the tables of this collection may sometimes be made use of in solving various problems of spherics besides those given in the former part of this work, it was thought proper to insert this improved method, with the formulas most frequently made use of, to enable any person acquainted with spheric trigonometry, to make use of the tables, without the trouble of referring to another work for the rules. |