PROBLEM XII. Given the Earth's Semi-Diameter, and the Sun's mean horizontal Parallax; to find the Earth's Distance from the Sun. RULE. To the logarithm of the earth's semi-diameter, add the logarithmic co-tangent of the sun's mean horizontal parallax; and the sum (abating 10 in the index,) will be the logarithm of the sun's mean distance from the earth. Example. By the transits of Venus over the sun's disk in the years 1761 and 1769, the sun's mean horizontal parallax appears to be about 8. 65 seconds of a degree; now, if the earth's semi-diameter be 3958.75 miles, its mean distance from the sun is required? = Semi-diameter of the earth. 3958.75 miles Log, 3958.75 miles Log, 3.5975581 Mean horizontal parallax of the sun=8".65 Log. co-tang.=14.3780860 Earth's mean distance from the sun=94546196 miles Log. = 7.9756441 PROBLEM XIII. Given the Sun's mean Distance from the Earth, and his apparent SemiDiameter, at a mean Rate; to find the true Measure of his Diameter, in English Miles. RULE. To the logarithm of the sun's mean distance from the earth, add the logarithmic tangent of his semi-diameter; and the sum (abating 10 in the index,) will be the logarithm of the sun's semi-diameter, in English miles; the double of which will be the measure of his whole diameter. Example. If the sun's mean distance from the earth be 94546196 English miles, and his mean apparent semi-diameter 16:1". 65, the true measure of his diameter is required? Sun's mean distance from the earth-94546196 miles Log. = 7.9756441 Sun's apparent semi-diameter = 16:1". 65 Log. tangent=7.6685950 Sun's true semi-diameter = 440797.5 miles Log. 5. 6442391 = True measure of the sun's diameter=881595 English miles. PROBLEM XIV. Given the Diameters of the Earth and the Sun; to find the Ratio of their Magnitudes. RULE. Since the magnitudes of all spherical bodies are as the cubes, or triplicate ratio, of their diameters (Euclid, Book XII., Prop. 18),—therefore, from thrice the logarithm of the sun's diameter subtract thrice the logarithm of the earth's diameter, and the remainder will be the logarithm of the ratio of their magnitudes. Example. If the earth's diameter be 7917.5 English miles, and that of the sun 881595 such miles, required the ratio of their magnitudes ? Sun's diameter, in English miles, =881595 Thrice its log. = 17.8358076 Earth's diameter, in ditto,=. 7917.5 Thrice its log. 11. 6957643 Ratio of the magnitudes of the earth and sun=1380522 Log. 6. 1400433 PROBLEM XV.. Given the Circumference of the Earth; to find the Rate, per Hour, at which the Inhabitants under the Equator are carried, in consequence of the Earth's diurnal Motion round its Axis. RULE. To the arithmetical complement of the logarithm of 24 hours, add the logarithm of the earth's circumference, and the logarithm of 1 hour: the sum of these three logarithms (abating 10 in the index,) will be the logarithm of the rate per hour at which the inhabitants under the equator are carried by the earth's diurnal motion on its axis. Example. Let the circumference of the earth be 24873.5 miles; required the rate per hour at which the inhabitants under the equator are carried, in consequence of the earth's diurnal motion? One day, or 24 hours, Arith. comp. of its log. = 8.6197888 24873.5 miles Log. 4.3957369 1036.396. Log. = 3.0155257 PROBLEM XVI. To find the Rate at which the Inhabitants under any given Parallel of Latitude are carried, in consequence of the Earth's diurnal Motion on its Axis. RULE. The circumference of the earth under the equator is 24873.5 miles; and since the circumference under any parallel of latitude decreases in proportion to the co-sine of the latitude of such parallel,-therefore, to the logarithm of the earth's circumference, under the equator, add the logarithmic co-sine of the latitude of the given parallel; and the sum (abating 10 in the index,) will be the logarithin of the earth's circumference under that parallel: with which proceed as directed in the last problem. Example. Let the circumference of the earth be 24873.5 miles; required the rate per hour at which the inhabitants under the parallel of London are carried by the earth's motion on its axis? Circumference of the earth = 24873.5 miles Log. =. 4.3957369 Latitude of the parallel of London = 51:31 Log. co-sine Circumference under given parallel=15478.45 Log. Rate 9.7939907 per hour, in miles, as required, 644. 93 Log. PROBLEM XVII. To find the Length of the tropical or solar Year. Rule. It has been found, by observation, that the sun apparently advances in the ecliptic 59'8". 33 of a degree every day at a mean rate; that is, from the time of his leaving any given meridian to the time of his returning to the same meridian. Now, since the ecliptic is a great circle of 360 degrees, -therefore, as the sun's apparent diurnal motion in the ecliptic, is to 1 day, or 24 hours; so is the great circle of 360 degrees, to the true length of the tropical or solar year; that is, to the time of the sun's periodical revolution round the ecliptic from any equinoctial or solstitial point to the same point again. Hence, by logarithms, Example. The sun's daily motion in the ecliptic is 59'8". 33 in every natural day, or 24 hours, at a mean rate; required the length of the tropical or solar year? Sun's app. diur. motion 59'8". 33, in secs.=3548. 33 Log.ar.co.6. 4499760 Length of the tropical year, in seconds=31556928 Log. = 4.9365137 6.1126050 7.4990947 Hence the tropical or solar year consists of 36554848, as required. PROBLEM XVIII. To find the Rate at which the Earth moves in the Ecliptic during the Time of its annual or periodical Revolution round the Sun. RULE. Since the earth's mean distance from the sun is 94546196 miles (Problem XII., page 614), the diameter of the orbit in which it moves round that great luminary is 189092392 miles; and since the diameter of a circle is to its circumference in the ratio of unity or 1, to 3.14159265, the circumference of the earth's orbit is 594051320 miles. Now, as the earth describes this circumference in 365 548 48: (last problem), or 8766 hours nearly, we have the following computation by logarithms: = As the length of the year, in hours, 8766 Log. ar. comp.-6.0571985 Log.=4.8310224 To the earth's hourly motion in its orbit = 67768 miles Log. 4. 8310224 PROBLEM XIX. Given the Moon's mean Distance from the Earth, and her apparent Semi-diameter, at a mean Rate; to find the true Measure of her Diameter, in English Miles. RULE. It is shown in page 9, under the head "Augmentation, of the Moon's Semi-diameter," that the moon's mean distance from the earth is 236692.35 miles. Now, since her apparent semi-diameter is 15:43" at a mean rate, therefore, to the logarithm of her mean distance from the earth, add the logarithmic tangent of her apparent semi-diameter; and the sum (abating 10 in the index,), will be the logarithm of the moon's semi-diameter in English miles: the double of which will be the measure of her whole diameter. Example. Let the moon's distance from the earth be 236692.35 miles, and her semi-diameter 15:43"; required the true measure of her diameter in English miles? Moon's mean distance from the earth-236692. 35 miles Log.5.3741842 Moon's apparent semi-diameter = 15:43" Log. tangent 7.6600896 Moon's true semi-diameter = . 1082.1 miles Log.=3.0342738 True measure of the moon's diameter=2164. 2 English miles. PROBLEM XX. Given the Diameters of the Earth and the Moon; to find the Ratio of their Magnitudes. Note. This is performed by Problem XIV., page 615. Example. If the earth's diameter be 7917.5 English miles, and that of the moon 2164.2 such miles, required the ratio of their magnitudes? Ratio of the magnitudes of the earth and moon=48.96 Log.=1.6898721 PROBLEM XXI. To find how much larger the Earth appears to the lunar Inhabitants than the Moon appears to the terrestrial Inhabitants. RULE. Since the distance between the earth and the moon is such as to cause their opposing hemispheres to appear, reciprocally from each other, like flat |