the errors of the observations; find the sum of the squares of the errors, and divide twice this sum by the square of the number of observations, and the quotient is the reciprocal of the weight.' Let u, v, w, be the reciprocals of the weights of the mean observed values A1, B1, and C1. Let n = the number of observations by which the mean value A ̧ was determined, and e1, 2, 3,...en, the errors of these observations, then 2 2 2 2 • 222 (e12 + €22 + € 32 + ... en2) ; e and in a similar manner the reciprocals v and w are found. u = EXAMPLES. 1. Let three observed values of an angle A be 54° 20' 32"-5, 54° 20′ 33′′4, and 54° 20′ 33′′-7; required the reciprocal of the weight of the mean value. The mean value, or A1 = (163° 1′ 39′′-6) = 54° 20′ 33′′-2. Hence, e1 =0"7, e2 =0"-2, and es―0"-5. = 2 2 And u = (·49+04+25) = ×·78=0·173. 32 = 2. The mean observed values of three angles are A1 62° 40′ 20′′-34, B1 = 58° 26' 32"-82, and C1 = 58° 53' 13"-05; and the reciprocals of their weights are found respectively to be 25, 32, and ∙18; also the spherical excess is 2"-71; what are the true angles? Here E2"-71, also E' 6"-21; hence, E, =-3"-5. Now, E, is to be distributed among A1, B1, C1, respectively, as the numbers 25, 32, and 18. •75: ·25 = 3"-5: 1"-17 •25 1.49 3.5: 0.84 Hence, .32 •18 ·75: 32 =— •75: 18 1′′-17 1.49 0.84 62° 40′ 19′′-17 58 53 12.21 180 0 2.71 FIGURE OF THE EARTH. 584. The figure of the earth is that of an oblate spheroid, whose equatorial diameter exceeds its polar by about part of the latter. The earth's diameter has been determined with such accuracy, that the error can amount at most to only about 15000 part of the whole, or about half a mile. In stating the length of its diameter in feet, therefore, the three last figures may be made cyphers. Let 2 a, 2b, e, denote the polar and equatorial diameters of the earth, and its ellipticity, that is, the ratio of the difference of these axes to the polar axis; then it has been found that = 7898-87 miles, 2 a 41,706,000 feet b α 1 =·00333 = nearly. 300 α 26.32 miles. Hence, 26-2a = 139000 feet And the difference of the polar and equatorial radii, that is, the compression of each pole =69500 feet 13.16 miles. b Since a:b300: 301. P 585. The radius of curvature of any plane curve on the surface of the earth, is the radius of a circle whose curvature is equal to that of the given curve at that point. Thus, the radius of curvature of a meridian at the point M is a line MR, such that the circle described with it from the centre R has a more intimate contact with the elliptic meridian at M than any other circle. The point R is called the centre of curvature for the point M; and the centres of curvature, for all the points in the arc PE, lie in a curve aRb, called the evolute of PE, and the radii of curvature, as RM, are tangents to the evolute; so cra is the evolute of PQ. P' and e= a = a 1 = 300 nearly; therefore 300 b = 301 a, and r с R a R M b M m E L 586. The radius of curvature for any point is perpendicular to the tangent at that point; or RM is perpendicular to the tangent at M. The normal to any point is that part of the radius of curvature, produced if necessary, which is intercepted between the point and the axis. Thus, se is the normal at the point s. 587. The true or astronomical latitude is the angle contained by the normal and the plane of the equator. Thus, MRL is the true latitude of M. The geocentric latitude of a point on the earth's surface, called also the reduced latitude, is the angle contained by the earth's radius at that point and the plane of the equator. Thus, MCE is the geocentric latitude of M. 588. The true difference of latitude of two places is the inclination of their normals. The inclination of the normals MR, M'R', is the difference of the latitudes of M and M'. The normal produced above the earth is called the vertical; and the difference between the true and geocentric latitude is called the angle of the vertical, or the reduction or correction of the latitude. This angle is just the inclination of the normal and the earth's radius at the place; thus, angle CMR is the angle of the vertical at M. If l, l', denote the true and geocentric latitudes of a place, l' may be found from the formula 12 tan l' = a2 tan 1, or L tan l' =Ï·9971098 + L tan l. Then the reduction is vll. When /= 54° 30', is v = 10′ 49′′-6.* then 589. The length of an arc of any number of degrees is greater the greater the latitude. If the normals MR, mR, contain the same angle as M'R' and m'R', then, since the latter normals are longer than the former, and the arcs M'm', Mm, are nearly arcs of circles, whose centres are R' and R, the arc M'm' must exceed Mm. 590. PROBLEM XII.-To determine the ellipticity of the ' earth by means of the lengths of two arcs of the meridian measured at different latitudes, and also the number of degrees in the arcs. *For the value of this angle at different latitudes, see Galbraith's useful collection of " Mathematical Tables," or the "Nautical Almanac for 1826." See also Francœur's "Geodesie," p. 175. Let 7, the latitudes of the middle points of the arcs, = the lengths of the arcs in the same denomi nation, d, d' the number of degrees, minutes, or seconds in the arcs, and di a, a' a1 = e and L also and or e = 3 · the ellipticity, 1-a, di then *e= 3 (sin2 l'— sin2 ) cos 21 EXAMPLE. -By measurements in Lapland, Svanberg found that an arc of 1°, the middle of which was in latitude 66° 20′ 10′′, was 111488 metres long; and in Peru, Bouguer found that the length of 1°, the middle of the arc being in latitude 1° 31', was 110582; required the ellipticity of the earth. α a a a' = Here /= 1° 31' 0", a = 110582, d = 1° = 3600", l' = 66° 20' 10", a' = 111488, d' = 1° 3600". a d' In this example d = d'; hence, a a d a -La-La'-5-043685-5.047229-I-996456 Therefore, e =}• =L.99187. 1-99187 998599+677803 1 003233 = 309-3 Since here 2'90, cos 2l' is negative. 1 307-4 = cos 2/ 1-a, d1 α, = ༔ • EXERCISES. 111108, d = 1°; 1. Given / = 44° 51' 2", a = l' 66° 20' 10", a' 111488, d' = 1o. = The former data are obtained from the measurements of Delambre, and the latter from Svanberg. The value of e will be ⚫00813 1-676402 * For a demonstration of this formula, see Airy's "Mathematical Tracts," p. 185. 2. Given 7 9° 34′ 44′′, a = 1029100·5 feet, d = 10210"-5. Also, l' = 66° 20′ 10′′, a′ = 593277·5 feet, d' = 5837"-6. The former data are obtained from Lambton's measurements in India, and the latter from Svanberg's in Lapland. 1 The value of e= 306-3 591. The ellipticity of the earth can be determined with greater facility by observing, by means of a pendulum, the intensity of gravity at two places, considerably distant, on the earth's surface. The length of the seconds' pendulum is greater the greater the latitude, both because the intensity of gravity is greater the higher the latitude, and because the centrifugal force, which counteracts a part of gravity, is less. 592. PROBLEM XIII.-Given the lengths of the seconds' pendulum at two places, whose latitudes are considerably different, to find the ellipticity of the earth. Let 1, l' the latitudes of the two places, p, p' the lengths of seconds' pendulum at them, P1 = p÷p', m and 289 ratio of centrifugal force at the equator to equatorial gravity; 1 -P1 1-Pi cos 2/-cos 27" then, if n = p EXAMPLE. The length of the seconds' pendulum at London, in latitude 51° 31′8′′, is 39·13929 inches; and at Melville Island, in latitude 74° 47′ 12′′, it is 39-207; required the ellipticity of the earth. and n = Hence, e= Lp-Lp'1·5926129-1.5933636-9992493 = L·998273, = .0027102. =0032301 = -001727 6367546 1 309.6 and |