PROBLEM IX. Given the Height of the Eye; to find the Distance of the visible Horizon. RULE. Let the earth's diameter, in feet, be augmented by the height of the eye; then, to the logarithm thereof, add the logarithm of the height of the eye; from half the sum of these two logarithms subtract the constant logarithm 3.783904,* and the remainder will be the logarithm of the distance in nautical miles; which is to be increased by a twelfth part of itself, on account of terrestrial refraction. Example. Chimboraço, the highest part of the Andes, is said to be 20633 feet above the level of the sea: now, admitting that an observer be placed upon its summit, at what distance can he see the visible horizon, allowing a twelfth part of that distance for the effects of refraction? Dist. at which the visible horizon may be seen=165.21 miles. This is the logarithm of 6080, the number of feet in a nautical mile. PROBLEM X. Given the measured Length of a base Line; to find the Allowance for the Curvature or spherical Figure of the Earth. RULE. Let EBF represent the arc of a great circle on the earth; C, the earth's centre; CB its semi-diameter; and A B the measure of a base line, on an apparent level or horizontal plane on the earth's surface: join CA, and it will cut the arc D H of the great circle in D; then A D will be the excess of the apparent level of the horizon above its true level. = Now, in the right angled plane triangle ABC, given the perpendicular BC and the base AB; to find the hypothenuse AC: which is readily determined by Euclid, Book I., Prop. 47. Then, the difference between A C, thus found, and CD CB, will be equal to DA, or the absolute value of the true level below the apparent level: and, if this value be expressed in miles and decimal parts of a mile, it may be reduced to inches, if necessary, by being multiplied by 63360 the number of inches in an English mile. Example. Let the base line AB, in the above diagram, be 1 English mile, and the earth's semi-diameter B C = 3958. 75 miles; required the allowance for the earth's curvature answering to that base line, or the difference between the true and apparent levels on the earth's surface expressed by the measure of the line A D? Subtract CD = CB, the earth's semi-diameter, = 3958.7501263 Remainder the line A D, the allowance for curvature,= 0000.0001263 Multiply by the number of inches in an English mile, = Number of inches which the true level is below the apparent level in one mile = 63360. 8.0023680 Now, since the curvature answering to A B is known, that corresponding to any other base line on the earth's surface may be readily determined by the following proportion: As the square of AB, is to AD; so is the square of B G, to GH: whence it is manifest, that the curvature answering to any given distance, as BG, is in the duplicate ratio of that distance to AB: And, since AB is expressed by unity or 1, and that AD is a constant quantity, the proportion may be reduced to a logarithmical expression; as thus : To twice the logarithm of the given base line, expressed in miles and decimal parts of a mile, add the constant logarithm 0.903219 (the log. of 8.002368 inches); and the sum will be the logarithm of the number of inches and decimal parts of an inch which the true horizontal level at sea is below its apparent level. Example 1. Required the curvature of the earth answering to a distance of 2 miles on its surface? Hence, the curvature answering to a distance of 2 miles on the surface of the earth, is 32. 009 inches; or 2 feet, nearly. Example 2. Required the curvature of the earth answering to a distance of 15 miles? Hence, the curvature answering to a distance of 15 miles on the earth's surface, is 1800 inches; or 150 feet and half an inch. Remark. If to twice the logarithm of the given base line, in miles, the constant logarithm 9. 824037 be added, the sum (abating 10 in the index,) will be the logarithm of the excess of the apparent above the true level, in feet. Example. Required the curvature of the earth, or the excess of the apparent above the true level, answering to a base line of 15 English miles in length? Given base line = 15 miles; twice the logarithm = log. of 12 inches, = 2.352182 9.824037 Excess of the app. above the true level, in ft.,=150.044 Log. 2. 176219 Note. This problem will be found useful to land-surveyors, engineers, and others employed in the art of levelling, cutting canals, and conducting water (by means of pipes, &c.) from one place to another. PROBLEM XI. Given the measured Length of a Base Line on any elevated Level; to find its true Measure, when referred to the Level of the Sea. RULE. A B D E In the annexed diagram, let the arc AB represent the measured length of a base line, at any given elevation above the level of the sea expressed by the arc DE; let CD be the radius of the earth, or the distance from its centre to the surface of the sea; and let CA be the earth's radius referred to the level of the measured base line AB. Now, because the arcs A B and DE are concentric and similar, and that similar arcs of spheres are to each other as their radii, we have the following analogy; viz., As the radius C A, is to the radius CD; so is the arc A B, to the arc D E: that is, as the earth's semi-diameter, augmented by the height of the base line above the level of the sea, is to the earth's true semi-diameter; so is the measured length of the given base line, to the true measure of that line at the surface of the sea. Example. Given a base line of 36960 feet in length, measured on a horizontal plane which is elevated 120 feet above the level of the sea; required the measure of that base line at the surface of the sea? As CD CD 20902200 = 20902200+ DA = 120 CA 20902320=A B 36960 :: DE = 36959.787813. Hence the given base line, reduced to the level of the sea, is 36958. 787813 feet; which is about 24 inches less than the measure on the elevated horizontal plane. But, since the probable elevation of any horizontal plane on the earth, above the level of the sea, can bear but a very insignificant proportion to the earth's semi-diameter,-if, therefore, the product of the measured base line by its height above the level of the sea be divided by the earth's radius, the quotient will be the excess of the measured base above the corresponding arc at the surface of the sea. This may be reduced to a logarithmical expression, in the following manner; viz., to the constant logarithm 2.679808, add the logarithms of the base line and of its elevation above the level of the sea, both expressed in feet: the sum will be the logarithm of a natural number, which, being taken from the measured base line, will leave the measure of that line at the surface of the sea, sufficiently near the truth for all practical purposes. Thus, to work the last example, Constant log.=ar. co. of the log. of the earth's semi-diam. in ft.=2.679808 Excess of the given base line above the arc at the surface of the sea = 36960 feet. -0.212188 Log. 9.326721 Given base line, reduced to level of sea, = 36959.787812; which approximates so very closely to the true result by the direct method of computation, as scarcely to admit of any sensible difference. Remark. In consequence of the spherical figure of the earth, no two points on its surface can be situated exactly on the same horizontal plane; for it is the chord of the arc, and not the arc itself, that measures the horizontal distance between two points. Hence, when philosophical inquiries are under consideration, it becomes necessary to apply a small correction to the measured base line on a horizontal plane, so as to reduce it to the corresponding terrestrial arc; though, in general, this correction is so very inconsiderable, that, even in the most extensive trigonometrical surveys, it may be safely disregarded. If, however, it be deemed necessary to find its value, or (which amounts to the same thing) if the excess of the arc over its chord be required, it may be very readily determined by the following rule, to every desirable degree of accuracy; viz., From thrice the measured length of the base line, in feet, subtract the constant logarithm 16.020595: the remainder will be the logarithm of the excess of the arc over its corresponding chord, expressed by the given base line. |