consequently to RE; and M'E will be parallel to mc, consequently to MD: also the angles PR'M and PM'R are, respectively equal to the spherical angles prm and pmr. But since, within the limits of the terrestrial triangles, the deviation of the produced normal ER from the plane MDR conceived to pass through the normal MD, is imperceptible from the station M, we may consider the planes MDR and MER as coincident; and the supposed vertical plane passing through MR to cut the parallel planes MR'D, RM'E under equal angles of inclination, so that the angle RMR' may without sensible error be considered equal to MRM': then the spheroidal azimuthal angle PRM will be as much less than the spherical angle PRM (= prm) as the spheroidal azimuthal angle PMR exceeds the spherical angle PM'R (= pmr). Consequently the sum of the spheroidal angles at R and м will, without sensible error, be equal to the sum of the spherical angles prm, pmr. P The like will be true for another triangle as PNM, in which we shall have the sum of the angles PMN, PNM equal to the sum of the angles PMN', PN'M; therefore, describing the arcs RN, R'N', we have the angles PRM + PMR = PR'M+PMR', and PNM+PMN then by addition, PN'M+PMN': R R N N PRM + PNM+ RMN=PR'M+ PN'M + R'M N'. Again in the triangles PRN, PR'N', we have as before M and subtracting this equation from the preceding, we have NRMRNM + RMN N'R'M + R'N'M + R'MN'. That is, the sum of the angles in the spheroidal triangle RMN is without sensible error equal to that of the angles in the triangle R'M N', which are those of a spherical triangle rmn, whose angular points correspond, in latitude and longitude, with the points R', M, N'. By a different investigation, Legendre has ascertained that the difference between the spherical and the spheroidal excess of the angles of a triangle, above two right angles, in the greatest triangle ever formed on the surface of the earth, does not amount to of a second; therefore, in computing the sides of the terrestrial triangles, the latter may always be considered as appertaining to the surface of a sphere. PROP. III. 414. To find the radius of curvature at a certain point a (fig. to art. 412.) in the periphery of a vertical section of a spheroid. It has been already proved that the general equation for the curve line produced by such a section is mx2 + ny` - and comparing the co-efficients of the variables x and y' with those of the same variables in the equation az2 + bx2 + • by 2 = a2b2 of a spheroid, when for x, y, z, are substituted their equivalents (art. 412.) it will be seen that also a, representing half the equatorial diameter, b half the polar diameter of the spheroid, and a2 - b2 or a2 e2 representing the square of the excentricity of the ellipse XAP, we have, by conic sections, RC= a e2 cos.l and AR = b2 a (1e2 sin.2 )a° Now ' and y' being the co-ordinates of any point in a curve, the usual formula for the radius of curvature is, when dy' is considered as constant, and differentiating the equation (1) twice successively, considering dy' as constant; also, since the radius is required for the point A where y' = 0, making y' equal to zero after each differentiation, we have But, from the preceding values of p and r, and since, at the point A, x'AR, it will be found that r=px', or r px' = 0; therefore the radius R of curvature at the point a becomes C C ; in which replacing m, n, q and x by their values above, we obtain R= (1 ab2 e2 sin.2 1)3 b2 + (a2 — b2) cos.2 l cos.2 0}° When the ellipse AB coincides with the plane of a terrestrial meridian as XAP, we have 0 = 0; and designating the radius of curvature in this case by R' (which is then in the direction of AR) we have a (1-e2 sin. 21)2 again, if the said ellipse be perpendicular to the plane of a meridian XA P, we have 90 degrees; and designating the radius of curvature in that case by R" (which is then also in the direction of AR) we have The above formulæ may be transformed into others, equiva lent to them, in which ε, the compression ( = is em a ployed instead of e, the excentricity. Thus, by the nature of the ellipse we have a2 - b2 = e2; now since the difference between a and b is very small, if we put 2a for a + b and =; then comparing this equation with that for e2, we obtain e2 = ε, or e2 = 2 ɛ. P Again, let APQ represent a meridian of the terrestrial spheroid, AQ a diameter of the equator, P one of its poles, and F the focus. Imagine PF' to be equal to the semi-transverse axis, and F'c' drawn perpendicularly to PC, to be equal to the difference between the semi-transverse and semi-conjugate axes: also let the angle FPC' be represented by I. Then, when PF', or PF, or AC, is equal to unity, F'c' (= ɛ) becomes equal to sin. I: and because the difference between the semi-axes is small, PC (= 1-8) is sometimes put for PC', and is represented by cos. I; or 1 cos. I is put for ɛ. But 1 cos. I is (Pl. Trigo., art. 36.) equal to 2 sin.21; therefore a is equivalent to 2 sin.21: or putting sin. I for sin. I because the angle I is very small, sin.21 sin.21; whence & may be represented by sin.2 1; or 2ɛ, that is, e2 by sin.2 1. PROP. IV. 415. The lengths of two meridional arcs as Aa, Bb (fig. to art. 411.), one near the pole and the other near the equator, and each subtending one degree, being given by admeasurement; to find the ratio which the earth's axes have to each other, the meridian being an ellipse. Imagine the normals a D, AD, bE, BE to be drawn from the extremities of the arcs till they meet in D and E respectively; and let AD, BE intersect CQ, the equatorial semidiameter, in N and N': also let the angles ANQ, BNQ (the geographical latitudes of the points A and B) be represented by land l' respectively. Then, by the similarity of the sectors ADa and BEb, representing AD and BE, the radii of curvature, by R and R', we have R: R': Aa: Bb, or R. arc B = R'. arc A ; or, developing 1 (A). (1-e2 sin.27) or (1 − e2 sin.2 7) —3 by the binomial theorem and neglecting powers of e above the second, R = a (1 − e2) (1 + e2 sin.21) = a (1 − e2 + e2 sin.2 1). In like manner R' = a (1 − e2 + 3 e2 sin.2 l'): therefore the above equation (A) becomes B (1 − e2 + 3e2 sin.2 1) = A (1 − e2 + § e2 sin.2 l'), and from this by transposition we obtain e2 = Thus e2 is found: ; whence a2 (1-e2) b', and a (1-e2)=b: a2 b developing the radical as far as two terms, we get 1 – 12 = 2, in which substituting the numerical value of e2 above, we have the required ratio of the semi-axes a and b. The numerical value of e being substituted in the above equation R. arc B = R'. arc A, there might from thence be obtained the value of a; and from the values of e and a, that of b might be found. Thus the equatorial and polar semidiameters of the earth would be completely determined. PROP. V. 416. To investigate the law according to which the lengths of the degrees of terrestrial latitude increase, from the equator towards either pole. It has been shown in the preceding article, that, in the plane of the meridian, But, at the equator, where = 0, the radius of curvature becomes equal to a (1-e2); therefore, by subtraction, the increment of the radius, for any latitude 7, above its value at the equator, is equal to a e2 sin.27: or the increments of the radii vary with the square of the sine of the geographical latitude of the station. Now the lengths of the degrees of latitude have been shown (art. 415. (A)) to vary with the radii of curvature in the direction of the meridian of any station; therefore, by proportion, the increments of the degrees of latitude vary with the increments of the radius; that is, with the square of the sine of the latitude. PROP. VI. 417. To determine the radius of any parallel of terrestrial latitude, and find the length of an arc of its circumference between two points whose difference of longitude is given. Let PqA be a quarter of the elliptical meridian passing through q; and let q R be the direction of a normal at A Let AC (a) be a radius of the equator, and Pc (b) be the polar ▲ semi-axis: let also qx (=sc or x) be a radius of the parallel Mq of latitude; and let ae be the excentricity of the P elliptical meridian. Then, by the nature of the ellipse, we have |