PROBLEMS AND TABLES. I. TO CONVERT DEGREES, ETC., INTO TIME. RULE 1.--Divide the degrees by 15, for hours; and multiply the remainder, if any, by 4, for minutes. 2. Divide the odd minutes and seconds in the same manner by 15 for minutes, seconds, &c., and multiply each remainder by 4, for the next lower denomination. EXAMPLE 1.-Convert 32° 34′ 45′′ into time. Ans. 32° 34′ 45′′-2h. 10′ 19′′ EXAMPLE 2.-If it is 12 o'clock at this place, what is the time 20° east of us? II. TO CONVERT TIME INTO DEGREES, ETC. RULE.-Multiply the hours by 15, and to the product add one-fourth of the minutes, seconds, &c., observing that every minute of time makes 4o, and every second of time '. EXAMPLE 1.-In 2 hours, 10 minutes, and 19 seconds, how many degrees? Ex. 2.-When it is 12 o'clock at Hartford, it is 4 hours, 51 minutes, and 20 seconds past noon at Greenwich; how many degrees is Hartford west of Greenwich? Thus: 15 times 4 is 60-added to this, increased by of 20, is 72° 50′. of 51, is 72° 45′′, and Ans. Ex. 3.-The moment of greatest darkness, during the annu lar eclipse of 1831, took place at New Haven, 10 minutes after 1 o'clock. A gentleman reports that it happened precisely at 1, where he observed it; and another, that it was 5 minutes after 1 where he saw it: Query. How far east or west were these gentlemen from each other, and how inany degrees from New Haven ? III. TO FIND THE EARTH'S MEAN DISTANCE FROM THE SUN. RULE.-As the Sun's horizontal parallax is to radius, so is the semi-diameter of the Earth to its distance from the Sun. By Logarithms.-As tangent of the Sun's horizontal parallax is to radius, so is the Earth's semi-diameter to her mean distance from the Sun. 8.5776: 206264".8:: 3962: 95,278,869 miles. As tangent of the Sun's horizontal parallax, 8.5776== 5.6189407 So is the Earth's semi-diameter, To the Earth's distance, =10.0000000 8962 8.5979143 95,278,869 7.9789788 IV. TO FIND THE DISTANCE OF ANY PLANET FROM THE SUN, THAT OF THE EARTH BEING KNOWN. RULE. Divide the square of the planet's sidereal revolution round the Sun, by the square of the Earth's sidereal revolution, and multiply the cube root of the quotient by the Earth's mean distance from the Sun. By Logarithms.-From twice the logarithm of the planet's sidereal revolution, subtract twice the logarithm of the Earth's sidercai revolution, and to one-third of the remainder add the logarithm of the Earth's mean distance from the Sun. EXAMPLE-Required Mercury's mean distance from the Sun, that of the Earth being 95,273,869 miles. Mercury's sidereal revolution is 87.969258 days, or 7600548.8912: the Earth's sidereal revolution is 365.256874417 days, or 995916962096952.25 by which divide 57768267575827.21 and the quotient will be 0.052005106713292, the cube root of which is 0.8870977, and this multiplied by 94,881,891 gives 86,727,607 miles, for Mercury's distance from, the Sun. This problem may be performed by logarithms in as many minutes as the former method requires hours. Mercury's Sid. Rev. 7600543.9 log.=6.8808447 × 2 Earth's Sid. Rev. 81558151". log.=7.4991302 × 2 མ་ ༣ང 13.7616894 14.9982604 TO FIND THE HOURLY MOTION OF A PLANET IN ITS ORBIT. RULE.-Multiply the planet's mean distance from the Sun by 6.2831853, and divide the product by the time of the planet's sidereal revolution, expressed in hours, and the deci mals of an hour. By Logarithms.-Add 0.7981799 to the logarithm of the planet's mean distance from the Sun, and from the sum subtract the logarithm of the planet's revolution, expressed in hours. EXAMPLE-Required the Earth's hourly motion in its orbit. Log. of Earth's distance 7.9780738+0.7981799= Gives Earth's horary motion, 68,288 miles, 8.7771557 8.942S090 4.8848447 VI. TO FIND THE HOURLY MOTION OF A PLANET ON ITS AXIS. RULE.-Multiply the diameter of the given planet by 3.14159, and divide the product by the period of its diurnal rotation. By Logarithms.-Add 4.0534524 to the logarithm of the planet's diameter, and from the sum subtract the logarithm of its diurnal rotation, expressed in seconds. VII. TO FIND THE RELATIVE MAGNITUDE OF THE PLANETS. RULE.-Divide the cube of the diameter of the larger planet by the cube of the diameter of the less. By Logarithms.-From three times the logarithm of the larger, subtract three times the logarithm of the less. EXAMPLE.-How much does the size of the Earth exceed that of the Moon? Earth's diameter, 7912 log. 3.8982863 × 3= The Earth exceeds the Moon, 49.1865 times. Ans. 11.6948589 10.0030128 1.6918461 In this example, 7912 miles is assumed as the mean between the Earth's equato rial and polar diameter: the former being 7924, and the latter 7898 miles. VIII. TO FIND THE PROPORTION OF SOLAR LIGHT AND HEAT AT EACH OF THE PLANETS. RULE.-Divide the square of the planet's greater distance from the Sun, by the square of the less. Or, subtract twice the logarithm of the greater distance from twice the logarithm of the less. EXAMPLE.-How much greater is the Sun's light and heat at Mercury, than at the Earth? Log. of Earth's distance 66 of Mercury's Ana 6.6786 times greater= 7.9789738 x 2=15.9579476 IX. TO FIND THE CIRCUMFERENCE OF THE PLANETS. RULE.-Multiply the diameter of the planet by 3.14159, or, add the logarithm of the planet's diameter to 0.4971499. X.-TO FIND THE CIRCUMFERENCE OF THE PLANETARY ORBITS. RULE.-Multiply the planet's mean distance from the Sun by 6.2831853; or, to the logarithm of the planet's mean distance add 0.7981799, and the sum will be the logarithm of the answer. XI. TO FIND IN WHAT TIME ANY OF THE PLANETS WOULD FALL TO THE SUN, IF LEFT TO THE FORCE OF GRAVITATION ALONE. RULE.-Multiply the time of the planet's sidereal revolution by 0.176776; the result will be the answer. By Logarithms.-From the logarithm of the planet's sidereal revolution, subtract 0.7525750, and the remainder will be the logarithm of the answer, in the same denomination as the sidereal revolution. Required the times, respectively, in which the several planets would fall to the Sun by the force of gravity. |