If, - (3.) When I is nothing, or when one of the degrees is at the equator r° (d'—d) Therefore, the excess of the degree in any latitude above this degree at the equator, when divided by the square of the sine of the latitude, should always give the same quotient; or the excess of the degrees of the meridian above the degree at the equator, should be as the squares of the sines of the latitudes. Since e= sin (+1) po (d'-d) (5.) If d' and d are two contiguous degrees, so that l'=1+1°, then 3 e d'-d= sin (21+1°) sin 1°, and since the sine of one degree is 3 e x 0.017453 0.017453, d'-d= sin (27+1°) (6.) The contiguous degrees therefore differ by a quantity proportional to the sine of twice the middle latitude. The difference is a maximum when 21+1° = 90°, or when the middle latitude is 45°. From the foregoing five different measures combined so as to pro duce the most accurate result, I found = 0.003227 = 310 nearly, and the equation representing the degrees of the meridian setting out from 45°, will be Hence, e = 11252.12= (7.) (8.) Miles. t = 3486863. 0 = 3962.344 The radius of curvature for the parallel of 45° = t fath. 3955.906 miles. The circumference of the meridian is therefore equal to the product of the mean degree at 45° by 360 = 24856 miles; and the circumference of the equator is 24896 miles, or about 40 miles more than the preceding. A geographical mile is therefore 1012.6 fathoms, or 6075.6 feet. The semidiameter or distance from the centre to the surface, at any latitude l, or r = t (1— sin 2 1+ § & 2 sin 2 l cos 2 1.) (9.) If d be a degree of the meridian at any point of which the latitude is l, and Ď a degree of the curve perpendicular to the meridian Ex. 1.-In the Philosophical Magazine for 1828, D the degree perpendicular to the meridian is given equal to 60973 English fathoms; d= 60819, and 7 = 50° 40′ N. By formula (12.) Ex. 2.-The length of a degree in latitude 52° 2′ 20′′ N. is 57074 toises, that in 11° 0 N. is 56755 toises; required the ellipticity by formula (2.)? l' = 52°21′ 20′′ const. log 1=11 0 0 1+1=63 2 20 cosecant, 9.522879 0.049969 1-141 22 0 cosecant, 0.182718 If L = the length of a degree of longitude, then If the value of the degree is wanted in toises, fathoms, or feet, the second member of this equation must be multiplied by the semitransverse axis in the same measure. Ex. 3.-Required the length of a degree of latitude at Edinburgh, in 56° N.? By formula (7), D = 60759.06 + 294.58 × sin. 22° = 60759.06 + 294.58 × 0.374607 = 60869.4 fathoms. Ex. 4.-Required the length of a degree of longitude in latitude 56° N., the ellipticity being ? 1 300 0.559193 By formula (13), L = {1+ 57.2957795 1 300 ×0.68694} ; or L= 0.009760 × 1.00229 × 20921200=204651 feet, or, taking in the second term mentioned in the note, it is 204664 feet. These formulæ are useful for fixing the latitude and longitude of a particular point when * If great accuracy is not required, 2 sin. 1. may be omitted in the quantity within the parenthesis. Indeed any small error in the ellipticity or radius of the equator will produce an equal or greater effect. referred to some object whose situation has been well determined, such as many places in Britain are by the trigonometrical survey. In this case any amateur observer may verify the latitude and longitude of his observatory deduced from his own observations, by a comparison with some point well settled in that work, when properly connected by trigonometrical operations. Even by taking a few angles with great care, the situation of a particular point may be well determined by spherical trigonometry. If p be the length of a degree perpendicular to the meridian, t the equatorial radius, c the semipolar axis, t-cd the difference of these, r the length of an arc in degrees equal to radius, or 57°.2957795, t+d sin. 21 and the latitude, then no p = nearly. (14.) Ex. 5.-If t-3486850 fathoms, d= 11160 fathoms, and = 56°, 3486850+11160 × ⚫68694 then P= 57.29578 =60992 fathoms. If p be the measure of a degree of a great circle perpendicular to a meridian at a certain point, m that of the corresponding degree on the meridian itself, and o the length of a degree on an oblique arc, making an angle a with the meridian, then Ex. 6.-Ifp=60973 fathoms, m = 60819 fathoms, and a = 81° 56' 53", therefore length of the oblique degree in fathoms. For an extension of this subject, see Mr Ivory on the properties of a line of the shortest distance traced on the surface of the oblate spheroid, in the sixty-seventh volume of the Philosophical Magazine, and vol. IV., new series, for a correction of the trigonometrical survey. The papers are rather too long and difficult to be inserted in this place. Edinburgh, 4th July, 1827. Ex. 8.-Determination of the latitude and longitude of the Observatory on the Calton-hill. The observations were made with a repeating theodolite placed near the old entrance from the east into the grounds surrounding the Observatory. The angle between North Berwick Law, in latitude 56° 3′ 8′′ N., longitude 2° 42' 11" W., and the Isle of May Light, in latitude 56° 11' 22" N., longitude 2° 32′ 47′′ W., was 13° 48′ 48′′; and the angle between the Isle of May and the West Lomond, in latitude 56° 14' 57" N., longitude 3° 17′ 4′′ W., was 68° 40′ 46′′; what is the latitude and longitude of the Observatory? Let P be the North Pole, C the Calton-hill, L the West Lomond, M the Isle of May Light, only yo and B North Berwick Law; also ECFG a pa-ids l rallel of latitude passing through C, and PE, PC, PF, and PG meridians passing through L, C, B, and M. Then, by the problem originally proposed by Townley, let the marginal figure be constructed, (page 47) and apply the principles of spherical trigonometry, since the lines employed are arcs on the surface of the earth, which in this case may be considered as a sphere. Now by the trigonometrical survey we have West Lomond latitude 56° 14' 57" N. longitude Isle of May Light 56 11 22 North Berwick Law 56 3 8 E FG 3° 17' 4" W. 2 32 47 1. In the triangle LPB there are given the sides PL, PB, and the angle LPB, to find LB = 22′ 44′′.48 26.17 English miles. = 2. In the triangle LPM are given PL, PM, and the angle LPM, to find LM = 24′ 52′′.88 = 28.632 English miles. 3. In the triangle BPM are given PB, PM, and the angle BPM, to find BM = 9′ 45′′.56 = 11.23 English miles. 4. In the triangle PLB are given all the sides to find the angles PLB and PBL 121° 3′ 52", and 58° 27' 10" respectively. 5. In the triangle LBM are given all the sides now found by computation to determine the angles LBM 90° 51' 40", and BML 66° 2' 54". Also LBM - LBD 90° 51' 40" 68° 40′ 46′′ DBM 22° 10' 54"; and 180° (BLD+LBD) = 180° 82° 97° 30′ 26′′ BLM will also be found to be 23° 23° 5' 28" 13° 48′ 48"-9° Likewise LD will be found = 21′ 22′′.08, LDM = 128° 28′ 28′′, LMD = LMC = 42° 14′ 52". But LMC+LCM = 110° 55′ 38", and consequently CLM 69° 4' 22", CLM BLM 69° 4′ 22′′23° 5' 28" 45° 58' 54" BLC. But BLC+PLB = PLC = 167° 6. It is now necessary to compute the sides LC, MC, and BC, which will be found to be LC17 57".46, MC = 24' 56".84, and BC 16' 29" 70. Hence LC 20.66 English miles, MC28.71, and BC= 18.98. 7. In the triangle PLM are given the sides PL, PM, and LM, to find the angles PLM 97° 58' 24", PML = 81° 24′ 48′′. If Napier's analogies be applied to the triangle LPM, these will be found to be 97° 58′ 34", and 81° 24′ 38′′ respectively. Also PLM+BLM +CLB PLC = 167° 2′ 46′′, as before. In like manner PML+ LMC PMC 123° 39' 40". 8. In the triangle LBC are given the sides to find the angle LBC= 51° 31' 32", and consequently LBC+PBL = 109° 58′ 42′′ – PBC. 9. In the triangle PLC are given the two sides, PL, LC, and the contained angle PLC, to determine PC the colatitude = 34° 2′32′′.6, and the latitude = 55° 57' 27".4 N. deduced from the West Lomond. 10. In the triangle PMC are given the sides PM, MC, and the contained angle PMC, to deduce PC the colatitude, and thence the latitude, = 55° 57′ 26′′.8 obtained from the Isle of May. 11. In the triangle PBC are given the sides PB, BC, and the contained angle PBC to find PC, the colatitude, and thence the latitude, 55° 57 28", deduced from North Berwick Law. The mean of these three is Error in the latitude of Greenwich from which these were deduced Correct latitude 55 57 27.4 N. 1.6 55 57 25.8 N. 12. In the triangle PLC there are given the three sides to find the angle PLC, the difference of longitude between the West Lomond and the Calton-hill, = 7′ 16′′.5 E. Hence 3° 17′ 4′′ — 7′ 16′′.5 = 3° 9′ 47′′.5 W. by West Lomond: 13. Again in the triangle PMC there are given all the sides to find the angle MPC, the difference of longitude between the Isle of May and the Calton-hill, 37' 5".6 W. = Hence 2° 32′ 47′′+375′′.63° 9′ 52".6 W., the longitude of the Calton-hill deduced from the Isle of May. 14. Lastly, in the triangle PCB, the sides are given to find the angle CPB, the difference of longitude between North Berwick Law and the Calton-hill, = 27′ 42′′.4 W. Whence 2° 42′ 11′′+27′ 42′′.4 — 3° 9′ 52′′.6 W. The mean of these three gives 3° 9′ 51′′.2=12m 39$.5 W. Ex. 9.-Given the latitude of the staff on North Berwick Law, 56° 3′ 8′′ N., longitude 2° 42'11" W., and the latitude of the Isle of May Light 56° 11′ 22′′ N., longitude 2° 32′ 47′′ W.; the angle at North Berwick Law, between the Isle of May and Dunglass Tower, was observed by Captain Hall, R. N., to be 87° 41' 1", that at Dunglass, between the Isle of May and North Berwick Law, being 37° 20′ 13′′; required the latitude and longitude of Dunglass Tower? Ans.-Latitude 55° 56′ 31′′.7 N., longitude 2° 21′ 42′′ W. Ex. 10.-Determination of the latitude and longitude of Makerston Observatory by comparison with the trigonometrical survey communicated by Lieut. General Sir Thomas Makdougal Brisbane. K. C. B. By means of angles taken at the various stations, A, B, C, D, &c. and the latitudes, longitudes, and distances, obtained from the trigonometrical survey, the point D, from which the other stations could be seen, was accurately fixed, and the distance DO correctly ascertained, whence. O, the position of the Observatory, was determined. N B |