Now, in the annexed diagram, let A B ED represent the earth, sur rounded by the atmosphere, as indicated by the dotted circular space. Let the point A be the place of an observer upon the earth, and R H his visible horizon; S the place of the sun within 18 degrees of the horizon, either at his rising or setting; and the line SR, a ray of light from H that luminary falling upon the atmosphere at the point R, and making the angle SRH equal to 18 degrees, the depression of the sun below the horizon. Now, because the angle SRH is 18 degrees, its supplement, or the angle SRA, is 162 degrees.* From the centre C, draw the line CR, and it will be perpendicular to the aërial particles of the atmosphere; and, by the laws of optics, it will bisect the supplemental angle S RA, making the angle CRA equal to 81 degrees.*-Hence, in the right angled plane triangle C RA, the angles and one side are given, to find the hypothenuse CR; from which, let the semidiameter C d (=CA) be taken, and the remainder, or d R will be the height of the atmosphere. The computation may be performed by plane trigonometry, Problem II., page 172; or by contraction, as in the following Example.-The semidiameter of the earth is 3958 English miles ; required the height of the atmosphere? Angle CRA = 81 degrees Logarithmic co-secant -10.0053801 Earth's semidiameter C A 3958.75 miles Logarithm 3.5975581 = Hypothenuse, or side C R = 4008. 10 miles Logarithm= 3.6029382 Difference, or side d R = of the atmosphere, as required. 49.35 miles; which is the correct height Remark. Although the atmosphere extends to the distance of 49. 35, or about 49 miles above the earth's surface, yet it is seldom of sufficient The angle S RH, is assumed considerably more than 18 degrees, and the angle SRA less than 162 degrees, so as to render the triangle C RA clearly perceptible to the view of the young computer: for, were the proportions strictly adhered to in the projection, the triangle would be scarcely discernible on so small a scale; and for the same reason the supplemental angle SRA is not fairly bisected. density to buoy up clouds to a greater altitude than about two miles; except, indeed, those very volatile or fleecy ones, called cirrus, which sometimes float at an elevation of five or six miles.-At the height of about three miles the atmosphere becomes so attenuated, or loses so much of its density, as to be incapable of reflecting the heat of the sun, or of communicating its genial influence to the mountainous tracts at that altitude:-hence it is that, in general, at an elevation of about 16000 feet above the level of the sea, there is a total suspension of vegetation:-in that aërial region there is a perpetual congelation; and thus it is that the summits of all very high mountains, even of those within the tropics, are eternally involved in frozen snow, or crowned with impassable glacial masses, which can never be thawed or reduced to a state of fluidity by the action of the solar beams. PROBLEM XXIV. Given the Earth's Semidiameter, and the Sun's mean horizontal To the logarithm of the earth's' semidiameter, 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 disc 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 semidiameter be 3958.75 miles, its mean distance from the sun is required? Semidiameter of the earth = 3958.75 miles Log. Log.=3.5975581 Mean horizontal parallax of the sun=8765 Log. co-tang.-14. 3780860 Earth's mean distance from the sun=94546196 miles Log.-7.9756441 See the result deduced from the inverse ratio of the parallaxes of the moon and sun, page 34. PROBLEM XXV. 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 semidiameter; and the sum (abating 10 in the index) will be the logarithm of the sun's semidiameter, 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 semidiameter 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 semidiameter = Sun's true semidiameter 16:1". 65 Log. tang.=7.6685950 440797.5 miles Log.=5.6442391 True measure of the sun's diameter 881595 English miles. PROBLEM XXVI. 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 Note. The ratio of the magnitudes of the earth and a planet may be determined in the same manner; as thus :-to find how many times Jupiter is larger than the earth. Hence it appears that the planet Jupiter is about 1679 times larger than the globe of earth on which we live! PROBLEM XXVII. 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. = Given time, or Rate per hour, in miles = 1036.396 8.6197888 Log.= 4.3957369 PROBLEM XXVIII. 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 logarithm 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? Log.=. Circum. under given parallel = 15478. 45 Rate per hour, in miles, as required=644.93 Log. = . PROBLEM XXIX. 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'sapp.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 365548" 48:, as required. * See Explanatory Article 6, page 304. |