PROBLEMS FOR MAKING VARIOUS ASTRONOMICAL CALCULATIONS. PROBLEM I. To work, by logistical logarithms, a proportion, the terms of which are minutes and seconds of a degree, or of time, or hours and minutes. With the minutes at the top and seconds at the side, or if a term consists of hours and minutes, with the hours at the top and minutes at the side, take from table LXXX., the logistical logarithms of the three given terms, and proceed in the usual manner of working a proportion by logarithms. The quantity, in the table, corresponding to the resulting logarithm, will be the fourth term. Note 1. The logistical logarithm of 60′ is 0. 2. The student will easily perceive that proportions that are worked by logistical logarithms, may also be worked by the common rule in arithmetic. As 12h. EXAM. 1. When the moon's hourly motion is 31′ 57′′, what is its motion in 39m. 22sec.? Ans. 20′ 58′′. As 60 m. : 39 m. 22 sec. :: 31′ 57′′ : 20′ 58′′ 4567 2. If the moon's declination change 2° 29′ in 12 hours, what will be the change in 8h. 21m.? Ans. 1° 44′. : 8h. 21m. :: 2° 29' 0 1830 2737 6990 8565 13831 22396 : 1° 44′ 15406 3. When the sun's hourly motion is 2′ 31′′, what is its motion in 17m. 18sec.? Ans. 0′ 44′′. 4. When the sun's declination changes 22′ 14′′ in 24 hours, what is its change in 19h. 25m.? Ans. 17′ 59′′. PROBLEM II. From a table in which quantities are given, for each Sign and Degree of the circle, to find the quantity corresponding to Signs, Degrees, Minutes, and Seconds. Take out, from the table, the quantity corresponding to the given signs and degrees; also take the difference between this quantity and the next following one. Then 60': odd minutes and seconds :: this difference: a fourth term. This fourth term added to the quantity taken out, when the quantities in the table are increasing; but subtracted, when they are decreasing, will give the required quantity. Note 1. When the quantities change but little from degree to degree, the required quantity may frequently be estimated, without the trouble of making a proportion. Note 2. The given quantity with which a quantity is taken from a table, is called the Argument. Note 3. In many tables, the argument is given in parts of the circle, supposed to be divided into 100, 1000, or 10,000, &c., parts. The method of taking quantities from such tables is the same as is given in the above rule; except that, when the argument changes by 10, the first term of the proportion must be 10, and the second, the odd units; when the argument changes by 100, the first term must be 100, and the second, the odd parts between hundreds; and so on. EXAM. 1. Given the argument 1s 9° 31′ 26′′, to find the corresponding quantity in table XLIV. Ans. 11° 13′ 57′′. 1s 9° gives 11° 11′ 15′′. The difference between 11° 11′ 15′′ and the next following quantity in the table is 5′ 9′′. To As 60′ : 31′ 26′′ : : 5′ 9′′ : 2′ 42′′. * Add 11 13 57 2. Given the argument 10s 13° 16′ 54′′, to find the corresponding quantity in table XLVII. Ans. 93° 32′ 37′′. 10° 13° gives 93° 33′ 40′′. * The student can work the proportion, either by common arithmetic, or by logistical logarithms, as he may prefer. The difference between 93° 33′ 40′′ and the next following quantity in the table, is 3' 43". As 60': 16′ 54′′ : : 3′ 43′′ : 1′ 3′′ Take 93 32 37 3. Given the argument 4° 11° 57' 10", to find the corresponding quantity in table XXVII. Ans. 3° 24′ 6′′. 4. Given the argument 3721, to find the corresponding quantity in table XXXVII. Ans. 4′ 52′′. PROBLEM III. To convert Degrees, Minutes, and Seconds of the Equator into Time. Multiply the quantity by 4, and call the product of the seconds, thirds; of the minutes, seconds; and of the degrees, minutes. EXAM. 1. Convert 72° 17′ 42′′ into time. 72° 17′ 42′′ 4h. 49m. 11sec. nearly. 4h. 49m. 10sec. 48". 2. Convert 117° 12′ 30′′ into time. 3. Convert 21° 52′ 27′′ into time. Ans. 7h. 48m. 50sec. PROBLEM IV. To convert Time into Degrees, Minutes, and Seconds. Reduce the time to minutes, or minutes and seconds; divide by 4, and call the quotient of the minutes, degrees; of the seconds, minutes; and multiply the remainder by 15, for the seconds. EXAM. 1. Convert 5h. 41m. 10sec. into degrees, &c. h. m. sec. 5 41 10 60 4) 341 10 85° 17′ 30′′ 2. Convert 7h. 48m. 50sec. into degrees, &c. 3. Convert 11h. 17m. 21sec. into degrees, &c. Ans. 117° 12′ 30′′. Ans. 169° 20′ 15′′. PROBLEM V. The Longitudes of two Places, and the Time at one of them being given, to find the corresponding Time at the other. Express the given time astronomically. Thus, when it is in the morning, add 12 hours, and diminish the number of the day by a unit. When the given time is in the afternoon, it is already in astronomical time. Find the difference of longitude of the two places, by subtracting the less longitude from the greater, when they are both of the same name, that is, both east or both west; but by adding the two longitudes together when they are of different names. When one of the places is Greenwich, the longitude of the other is the difference of longitude. Then, if the place, at which the time is required, is to the east of the other place, add the difference of longitude, in time, to the given time; but if it is to the west, subtract the difference of longitude from the given time. The sum or remainder is the required time. Note. The longitudes of the places mentioned in the following examples, are given in table VI. EXAM. 1. When it is August 8th, 2h. 12m. 17sec. A. M. at Greenwich, what is the time as reckoned at Philadelphia? Time at Greenwich, August, Time at Philadelphia, 7 9 11 37 P. M. 2. When it is April 11th, 3h. 15m. 20sec. P. M. at New York, what is the corresponding time at Greenwich? Longitude of Paris, do. Diff. of Long., d. h. m. sec. 11 3 15 20 4 56 4 11 8 11 24 P. M. 3. When it is Sept. 10th, 3h. 20m. 35sec. P. M. at Paris, what is the time as reckoned at New Haven ? Time at New York, April, Diff. of Long., Time at Greenwich, d. h. m. sec. 7 14 12 17 5 0 40 of New Haven, h. m. sec. 0 922 E. 4 51 51 W 5 1 13 Time at Paris, September, Diff. of Long., Time at New Haven, d. h. m. sec. 10 3 20 35 5 1 13 9 22 19 22 Or, Sept. 10th, 10h. 19m. 22sec. A. M. 4. When it is January 15th, 9h. 12m. 10sec. P. M. at Washington, what is the corresponding time at Berlin? Ans. Jan. 16th, 3h. 13m. 52sec. A. M. 5. When it is Oct. 5th, 7h. 8m. A. M. at Quebec, what is the time at Richmond? Ans. Oct. 5th, 6h. 43m. 18 sec. A. M. 6. When it is noon of the 10th of June at Greenwich, what is the time at Philadelphia? Ans. June 10th, 6h. 59m. 20sec. A. M. PROBLEM VI. For a given mean time, to find the Sun's Longitude, Semidiameter, Hourly Motion, the apparent Obliquity of the Ecliptic and the Earth's Radius Vector; also the Sun's Right Ascension and Declination and the Apparent Time. For the Longitude. When the given time is not for the meridian at Greenwich, reduce it to that meridian by the last problem. * With the mean time at Greenwich, take from Tables XXII., XXIII., and XXIV., the quantities corresponding to the year, month, day, hour, minute, and second, and find their sums. The sum in the column of mean longitudes will be the tabular mean longitude of the sun; the sum in the column of perigee, will be the tabular longitude of the perigee; and the sums in the columns I., II., III., and N., will be the arguments for the small equations of the sun's longitude, and for the equation of the equinoxes, which forms one of them. Subtract the longitude of the perigee from the sun's mean longitude, borrowing 12 signs when necessary; the remainder is the sun's mean anomaly. With the mean anomaly, take the equation of the sun's centre * In adding quantities that are expressed in signs, degrees, &c., reject 12 or 24 signs, when the sum exceeds either of these quantities. In adding any arguments, expressed in 100, 1000, &c., parts of the circle, when they are expressed by two figures, reject the hundreds from the sum: when by three figures, the thousands; and when by four figures, the ten thousands. Y |