may be constructed the unit of measurement, without again having reference to an arc of the meridian. Messrs. Mechain and Delambre conducted these difficult enquiries. Nothing but an actual perusal of their works on these subjects can convey an adequate idea of the difficulty of the undertaking. 1 Before concluding this article, it is necessary to mention that the French commissioners, in their calculations, employed 3 for the earth's compression at the poles, which is now well known not to be correct; therefore the metre can no longer be considered as strictly a standard unit of measurement. Mr. Baily gives 3 as the value of the earth's compression, making the polar axis to the equatorial diameter as 304 to 305. Doctor Pearson varies it from 넓이 to 325. 1 1 300 After this, Mechain repaired to Spain, anxious to realise his early project of extending the meridian as far as the Balearic Isles; but having caught an epidemic fever, of which he died in 1805, the remaining part of the work was committed to Biot, who brought it to a conclusion. In one of the great triangles connecting the coast of Spain with the Balearic Isles, one side exceeds one hundred miles in length. At such a vast distance, the stations, however elevated, could not be observed during the day; but they were rendered visible at night, by a combination of Argand's lamps and powerful reflectors. Before we illustrate the subject by practical examples, it may not be out of place to insert the following propositions relating to spherical triangles, as bearing immediately upon the subject before us. The proof is given in a work which the author has prepared for the press. The proof may, indeed, be seen in almost all the modern works on spherical trigonometry. 1. A spherical triangle, whose sides are very small when compared with the radius of the sphere, being proposed: if from each of its angles you subtract one third of the excess of the sum of its three angles above 180°, the angles thus diminished may be taken for the angles of a rectilinear triangle, whose sides are equal in length to those of the proposed spherical triangle. The preceding proposition is illustrated by the following example, taken from the trigonometrical survey of England. Stations. Observed Angles. Distance of Hundred Acres from Hundred Acres, 53° 58′ 35′′.75 Hanger-hill, 71932.8 feet St. Ann's-hill 57 36 39.5 St. Ann's-hill, 79209.7 feet 179 59 59.25 Let A, B, C represent the three angles of a plane triangle, of which the opposite sides a, b, c, are small, relatively to the radius of the sphere; and let d be indefinitely small. Then (A+d)+(B+d)+(C+d)-180°=3 d= a. b. sin C =the area of a plane triangle, whose sides are a and b, and contained angle C. Here a=71932.8, b=79209.7, and C=53° 58′ 35′′.75 a=35966.4, log=4.5558969 b=79209.7, log=4.8987783 log sin of C=log sin of 53° 58′ 35′′.75=9.9078287 log of the area of the triangle=19.3625039 log of the area of the triangle=19.3625039. log of the spherical excess= .0357302 The number corresponding to which is 1.0857 seconds =3d: hence d=.36" nearly. But the observed angles are spherical angles; therefore C=53° 58′ 35′′.75 .36" 53° 58′ 35′′.39. Now with this, as an included angle, and the two sides given above, find the angles and the third side of a plane triangle by the rules of plane trigonometry. The following formula will answer : tang = rad, and (AB) = log b=4.8987783 log a=4.8569269 cot C, tang ( — 450) rad 0.418514 log rad=10. tang =10.0418514 Hence Φ=47° 45′23′′.1 Ф-45°- 2 45 23.1 tang (Φ-45°)=tang 2° 45′23′′.1= 8.6825576 cot C=cot 26° 59′ 17′′.695=10.2930544 sum (rejecting rad=10)= 8.9756120 Hence (AB)= 5° 24' 2".327 Therefore B=68°24′ 44′′.627 The third side may be found by the formula c= 9.9078281+4.8569269-9.9265646= This solution is similar to that given by Legendre, and by adding to each of the angles one-third of the spherical excess, a complete solution will be given to the spherical triangle. Angles of spherical triangle. C+d=53° 58′ 35′′.75 180 00 1.07 Sides of spherical triangle. c=68895.43 Delambre and General Mudge, in the trigonometrical surveys of France and England, reduced the observed angles to the angles formed by the chords of arcs; they also considered an arc of the meridian as formed by the chords of curves. 2. If from the sum of the three angles of any spherical triangle two right angles be deducted, the remainder will be to two right angles, as the area of the triangle is to one-fourth of the surface of the sphere. 3. The measure of the surface of a spherical triangle, is the difference between the sum of its three angles and two right angles. (A+B+C)-180° surface. The measure of the spherical angle contained between any two arcs of a great circle of the sphere, is the same as that contained between two tangents to those arcs. In a trigonometrical survey, the sides which connect the successive stations may be considered as arcs of great circles, whose radii are equal to the radius of the sphere. 4. The sum of the three angles of every spherical triangle being greater than 180°, the sum of every three observed angles of every triangle on the surface of the earth, ought to exceed 180°. To find this excess, which is called the spherical excess: From the logarithm of the area of the triangle taken as a plane one, in feet, subtract the constant logarithm 9.3267737, and the remainder is the logarithm of the excess above 180° in seconds, nearly. Stations. Observed Angles. Buster-hill 76° 12′22′′ 180 00 1.25 Distance from Dunnose from Buster-hill 140580.4 feet. Dean-hill 183436.2 feet. In this example we have two sides of a plane triangle, and the angle contained between them given, to find the area. a=140580.4, b=183496.2, and > C=55° 43′ 7′′, then by the rule given in the preceding part of the work, we have a×b×nat.sin. of > C=area of the triangle log. sin. C=log. sin. 55° 43′ 7"=9.9171279 10.0276497 R |