EXERCISE.-At Madras, 7 = 13° 4′ 9′′, and p = 39·0234; and at Melville Island, l' = 74° 47′ 12′′, and p' find e. Ans. e0033291 = = 39.207; 300 nearly. 593. PROBLEM XIV.-Given the length of a small arc on the earth's surface, and its radius of curvature, to find the number of minutes or seconds contained in it. 'Divide the arc by the radius of curvature, and multiply the quotient by 3437-75 for the number of minutes; or multiply the quotient by 206265 for the number of seconds." Let z, r = the length and radius of curvature of the arc, n°, n', n" the number of degrees, minutes, and seconds respectively in the arc, then n° = 57-29578, n' = 3437.75, and n" = 206265 Or, Ln° = 1·75812263 + L≈ Lr, - Lr, .Lr. For 2 Tr: z = 360: n, for the circumference of circle EXAMPLE.-How many minutes in an arc of 316469 feet on the earth's surface, whose radius of curvature is 20,892,000 feet? When the arc does not amount to 100', or to about 116 imperial miles, the logarithms need not be carried farther than the fifth decimal place. EXERCISE. Find the number of seconds in an arc of 423562 feet, its radius of curvature being 20,902,000 feet. Ans. n" 4179"-79. 594. PROBLEM XV.-Given the degrees contained in an arc on the earth's surface, and its radius of curvature, to find the length of the arc. Multiply the number of degrees, minutes, or seconds, by the radius of curvature, and divide the product by 57.29578, 3437-75, or 206265, according as degrees, minutes, or seconds, are given, and the quotient will be the required length of the arc.' Or, L22-24187737 + Ln° + Lr, Lz = 4·46372612 + Ln' + Lr, Lz = 6·68557487 + Ln” + Lr. The constant logarithms in these formulas are the arithmetical complements of those in the preceding problem. EXAMPLE. The number of minutes in an arc is 52.074, and its radius of curvature is 20,892,000 feet; required the length of the arc. By means of this problem, the length of a degree of the meridian, at any latitude, can be calculated. EXERCISE. What is the length of an arc of the meridian of 1° 9′ 39′′-79, in latitude 48° 50′ 48′′-59, its radius of curvature being 20,902,000, and also the length of an arc of 1° at the same place? Ans. 423562 and 364808.8. 595. PROBLEM XVI.-Given the number of degrees, minutes, or seconds in an arc, and its length, to find its radius of curvature. 'Divide the length of the arc by the number of degrees, minutes, or seconds; and multiply the quotient by 57.29578, 3437-75, or 206265, according as degrees, minutes, or seconds are given, and the product will be the radius of curvature." Or, r = 57.29578, r = 3437·757, r Or, 209 r = 206265 n EXAMPLE. If the length of an arc of 1° is 362912.2 feet, what is its radius of curvature? Lr1.7581226+ Lz-L 1° 1-7581226 +5.5598015 =7-3179241 L 20,793,300. = EXERCISES. 1. What is the radius of curvature of an arc of 57'074, its length being 316,469 feet? Ans. 20,892,000. 2. An arc of 4179.79 seconds is 423,562 feet long; find its radius of curvature. Ans. 20,902,000. 596. PROBLEM XVII.-Given the number of degrees in an arc, its length and the latitude of its middle point, and the ellipticity of the earth, to find the polar radius. Find the radius of curvature of the arc by article 595; then find the polar radius by the following formulas, in which a is the polar radius, and the latitude of the middle of the arc 1. When the given arc is a part of the meridian, a=r(1+e—3 e sin2 1), or a = 1r (2—e+3 e cos 27. 2. When the arc is perpendicular to the meridian, a = r (1. — e — e sin2 l), or a = 1r (2—3e + e cos 21). 3. When the latitude lo, or the middle of the arc is on the equator, and the arc is a part of the meridian, find, by article 595, the value of its radius of curvature r'; then, if " is that of the perpendicular arc, it follows, since l =0, a=r' (1+e), and a = r" (1 − e); that hence, Or, : = 1-e1+e=1-300:1 +300299: 301. Lr" Lr' + L 301 L 299. And, in this case, r' is just the radius of the equator; and hence "b; and, by the usual formula, a=6(1— e) 388 b, or LaLb + L 299 — L 300. EXAMPLE.—In the French survey, it was found by Delambre that the difference of latitude between Evaux and Carcassonne was 2° 57′ 48′′-24, and the length of the meridional arc 168846-7 French toises, the middle latitude being 44° 41′ 48′′; required the axes of the earth, the ellipticity being o Here / 44° 41' 48", d= 2° 57′ 48"-24177'-804, and z=168846.7. 1 F. toise: 168846·7 F. t. = 6·3946 En. f. : z in feet, LzL 168846·7 + L 6·3946 = 6·0333060. And (595) Lr3·5362739+ LzLd7-3196383; also, La = Lr (2 — e + 3 e cos 27) Lr+L 998386 73189368 L 20,841,900. = Hence, a = 20,841,900, and b = 300 a = 20,911,600. The values of a given above are also easily derived from the values of R and R' given in the next problem. EXERCISES. 1. The length of a degree is 364,535 feet, and the latitude of its middle point is 44° 51′ 2′′; find the polar radius of the earth. Ans. 20,852,000 feet. 2. In the trigonometrical survey, it was found that the distance by General Roy's standard, from Dunnose to Clifton, was 1,036,337 feet, and that the middle latitude was 52° 2' 20", and the amplitude 2° 50′ 23′′-5; find the axis. Ans. 20,848,400 feet. When this number is reduced to imperial feet by multiplying it by 1.0000691, the value of the axis in imperial feet is = 20,850,000; for a foot of General Roy's standard is 1·0000691 imperial. 597. PROBLEM XVIII.-To find the radius of curvature of a meridian of the earth at any latitude, and of the arc perpendicular to the meridian, its ellipticity and axes being known. Let R= the radius of curvature of the meridian at any point, R' the radius of curvature of the arc perpendicular to it; then a and e denoting the polar radius and ellipticity, as before (584), and the latitude. Ra(1-e+3 e sin2 7), R'a (1+e+ e sin2 7); or, Ra(2+e-3 e cos 27), R'a (2+3e-e cos 21). If r denote the radius of curvature at the given point of the arc of an oblique section, inclined at an angle to the meridian, then r= RR' or r= 2 RR' R+R' when 45°. Instead of r, half the sum of R and R' may be taken in practice. These expressions are investigated by means of the higher calculus.* In the following example and exercise, the mean values obtained for a and e are taken; namely, a and e=300 20,853,000, EXAMPLE. Find the radii of curvature R, R', a place in latitude 48° 50′ 48′′-59. And Ra(2+e-3 e cos 27) × 20853000 =20,902,000. R'a (2+3e-e cos 21)= 20,962,000. r=(R+R') nearly 20,893,500. = In this example 2190; hence, cos 27 is negative, and cos 27 becomes + cos 21. The last three figures are made cyphers, because they are uncertain (561.) * See Hymer's "Analytical Geometry," or Leroy's "Analyse Appliquée &c;" or Lacroix's "Calcul Differentiel et Integral." |