spherical triangle being known, that of the others may be obtained by the solution of the plane triangle. Between the latitudes 450 and 250 the spherical excess amounts to about 1" for an area of 75.5 square miles. To find the spherical excess in seconds of space therefore divide the area in square miles by 75.5. The logarithm of the mean radius of the earth in yards is 6.8427917, of r is 13.6855834. The following example shows the form* in which the above rules are applied in practice on the U. S. Coast Survey. But [Art. 19, (1) App. I.], bc, sin a' is the area of the plane triangle of which a, b, c are the sides, which area does not differ sensibly from the proposed spherical triangle. If s denote the area of either of these (8) becomes And this last is the formula of verification. The same theorem has been extended to a spheroidal triangle, the difference between the spheroidal excess and spherical excess being less than of a second in the largest triangle ever measured on the sur 8 face of the earth. To express in seconds, it must be divided by sin 1'^.t * In the form given the 1st, 3d, and 4th columns explain themselves; the 2d column contains the names of the stations at the vertices of the triangle, the 7th the spherical excess 5.67, calculated by the formula on p. 323, the 5th the difference 0.15 between the excess of the sum of the observed angles over 180°, and the spherical excess, which difference ought to be zero. This error 0.15 is equally distributed between the three angles; the 6th column contains the observed angles thus corrected; the 8th column contains the angles of a plane triangle whose sides are of the same length with the spherical one, each determined by subtracting the spheri A still more accurate formula, deduced rigorously from the spheroid, is 1+ e2 cos 2 L Excess & in which mean latitude, and a = equatorial radius. a2 This form would only become important in carrying forward azimuths in a long chain. DETERMINATION OF THE LATITUDES, LONGITUDES AND AZIMUTHS OF THE STATIONS. The latitude and longitude of a few of the stations of a chain of triangles, selected at different parts of the whole, being determined by astronomical observations, the latitudes and longitudes of the other stations may be found geodetically, as it is termed, by methods which we now proceed to explain. The problem to resolve is the following. Given the latitude and the longitude of the point s in the diagram, and the azimuth of the point s' upon the horizon of s*, to find the latitude and longitude of this second point, and the azimuth of s upon the horizon of s'. If L denote the given latitude, di the difference of latitude between the two stations to be applied as a correction to the known latitude, in order to find the required latitude, м the longitude, dм the difference of longitude of the two stations, z the azimuth, dz the difference of azimuth; then the following will be the S FORMULAS FOR COMPUTATION OF L, M, Z OF PRIMARY TRIANGLES. For the difference of latitude − dL = KB cos z + K2 c sin2 z + (dL)3 D— - h K2 E sin2 zt cal excess, or 1.89 from the corresponding spherical angles; this column also contains the sides of the plane triangle, the last two of which are computed from the first, and the angles by the sine proportion, p. 62, the logarithms for which purpose are contained in the last column. * Observed astronomically by methods to be described in the sequel. The following is the demonstration of this formula:-There are known in the triangle PSS' the colatitude PS 90° -I of the point s, the angle rss' the azimuth of s' from sz, and the side ss' K, the geodetic line or measured distance between s and s'. That is to say two sides and the included angle, to compute Ps', the colatitude, and Ps's 180°-z' the azimuth of s from s', and Pdм, the diff. of long. required. = = The determination of the latitude, especially, by this method, would be attended with considerable error. We shall deduce the formula given by Delambre, for the difference of latitude between the two stations. The formula of Art. 82, applied to the present triangle, gives or, since L+ di or [Art. 70, (3)], = COS PS' COS s sin PS sin ss' + cos PS cos ss' the latitude of s', of which rs' is the complement sin (LdL): = cos z cos L sin K+ sin L COS K sin L cos dL + cos L sin dL = cos z cos L sin K+sin L cos K or [Art. 72 (8), and Art. 71 (3)], sin L (1 2 sindL) + 2 cos L cos di sindi = cos z cos i sin K + sin L tandL = 1 Or performing the operation indicated in the second member, and rejecting the terms beyond p3, * Values of R and N for every latitude are furnished by a table, p. 44, Part I., of Lee's Tables and Formulas. See also (4), p. 366 and p. 368 App. VI. Substituting in the second member of this the value of tandL, given by the preceding it becomes } dr = p − q p2 + (1 + 2q2) p3 — } p3 rejecting the higher powers of p. Or, But, third. 2 pcos z sin K 2 q p2 = = cos z sin2 K tan L+ 2 cos z sin K sin2K tan2 L p3cos3 z sin3 K, rejecting terms containing higher powers of sin K than the and the value of p3 may be put under the form K3 cos2 z cos z, rejecting higher powers of K than к3 This last, added to the second term of (m), gives - K3 cos z (1 — cos2 z) =- - K3 cos z sin2 z so that (m) thus increased becomes K COS Z- K3 sin2 z cos z Again, putting K for its sine, being a small arc, the second term in the value of 2p becomes Thus the expression for the difference of latitude reduces to dL = K cos z— K2 sin2 z tan L— K3 sin3 z cos z (1 + 3 tan2 L) &c. Delambre observes that the meridional arcs, computed on the supposition that the earth is a sphere differ insensibly from those computed on the spheroidal hypo α= equatorial semidiameter of the earth = 6377397.15 metres, log. a = 6.8046434637, h = 1st term K⚫B COS Z. e= 0.0816967, log. e 8.9122052271. Subjoined is part of a table, from which B, C, D, E are to be taken. For an example, let us make K = 100 miles = 160900 metres, L= thesis. The consideration of the spheroidal figure is only of consequence, in converting the terrestrial arcs, measured in metres, into seconds. This may be done by dividing such ares by their radii of curvature, and this quotient by the length of 1". Our formula is deduced by considering the normal N terminating at the polar axis as the radius of curvature of the geodetic arc K, which it is very nearly; then if к and N be both expressed in metres, K N will be the length of K in terms of N as unity. If therefore K2 sin? z tan L K3 sin2 z cos z (1 + 3 tan3 L) N3 in which dr is expressed in terms of N the radius of K as unity. (g) In order that it may be expressed in terms of its own radius as unity, which we will call R, this value of di must be multiplied by N R and if the result thus obtained be divided by the length of 1" in terms of radius unity, the value of du will be expressed in seconds. N Multiplying therefore the second member of (g) throughout by R 1", and estimating z from s round by w, from 0° to 360°, which will be the supplement of the z used in the diagram, it becomes K cos Z + K2 sin2 z tan L K3 sin2 z cos z 6 No R are 1" (1 + 3 tan2 L) (a) 1 The radius of curvature at any point of a curve is the radius of a circle having the same curvature as the curve in that part of it. 2 For the demonstration of a formula for the radius of curvature of an elliptical meridian, at any point of that meridian, in terms of the latitude of that point, see the Appendix to Part VI. p. 366. |