(9) Tan A' = cos à tan w, A' being an arc perpendicular to the meridian. II. The following formulæ of Oriani have been employed by Captain Kater in the new survey detailed in the Philosophical Transactions for 1828, to determine the difference of longitude between Greenwich and Paris.* M, the distance in feet from the perpendicular to the meridian of Greenwich. P, the distance in feet to the meridian of Greenwich. Put m = M p= P and L the latitude of Greenwich. Let a be the latitude of the foot of the perpendicular let fall from the given station on the meridian of Greenwich. (10) , the required latitude of the given station, and u, the longitude of the given station, then Though Mr Ivory's formulæ are, in general, as simple as the nature of the subject will admit, yet I have been induced to add the formulæ of Oriani in order to enable Computers to check their calculations by different methods. Explanation and Application of the foregoing Formulæ. Example 1.-The chord, or y, between Beachyhead and Dunnose, by the trigonometrical survey, is 339397.6 feet, what is the length of the arc, or A? The author is indebted to Captain Kater for a copy of this interesting survey, of which an abstract is given in the following pages. Ex. 2-Required from the new survey the difference of longitude between the Observatories of Greenwich and Paris? The length of an arc drawn from Dover perpendicular to the meridian of Greenwich is 303804 feet, or 50634 fathoms, as stated in the Philosophical Transactions, 1828, page 180. If the point where this arc intersects the meridian of Greenwich be considered as the first station in formula (6), and Dover as the second, then m = - 90°, and x'= 51° 7′ 45′′.6. Since the given distance is not a chord, but an arc on the earth's surface, may be found by reducing the given length taken as an arc on the earth's equator to degrees. Formula (1) Now a degree on the equator to an ellipticity of 0.00324 is 60856 fathoms, whence ß = A A 50634° × 60 × 60"-49′ 55′′.3. By substituting these in formula (6) we have B 0° 49′ 55′′.3 -90 0 0.0 '51 7 45 .6 log sin ẞ sin m cos x - M ε sin2 λ' = sine 8.162000 M=0.4342945 log 9.637784 sine 10.000000 ε =0.00324 log 7.510545 sec 0.202342 x 51°7′45”.6 sin 9.782589 8.364342 M. sin2 x'=0.000853 log 6.930918 0.000853 w = 1° 19′ 23′′.79 sin 8.363489, which is the longitude of Dover east from Greenwich. Ex. 3.-As no azimuth either at Dover or at Dunkirk is given, formula (7) must be employed in the following solution. General Roy makes the distance from Dover to Dunkirk 244916 feet according to his scale, and, taking the ratio of his scale to the imperial standard to be as 1.0000691 to 1, according to Captain Kater, this distance would be 40822 imperial fathoms. Now the distance being an arc on the surface of the spheroid, we 408220 have, as before, ß= = (40833) = 40′ 14′′.87. 60856 The latitude of Dover, or λ = 51° 7′ 45′′.6 Dunkirk, or x = 51 2 8.5 a+a'= 102 9 54 .1 Whence by formulæ (4) and (7), 3=1.5 log +0.176091 517 45.6 sin N. N. 0.68616 λ=51745.6 N.N. λ+x=102954.1 cos-9.323625 x=512 8.5 sin2 N.N. 0.60456 3 19.14= Sum 2 22 42 .93 = sina 2 15.930153 7.965076 -0.000852 7.964224 long. of Dunkirk E. from Dover. long. of Dunkirk E. from Greenwich. Diff. 2 20 20 .93 9m 21$.39 = longitude of Paris in time east from Greenwich. Ex. 4. Required the difference of longitude between Greenwich and Paris by the formulæ of Oriani, from the following data taken from the new survey? By Table III., page 180, Phil. Trans., 1828, M 184282.44, P = 427511.43 when referred to Notre Dame at Calais, and taking the values of a, b, and ɛ, as before stated, we have, m= p = 184282.44 20853180 sin 1′′ 20853180 sin 1′′ 427511.43 L= 51° 28′ 38".5, whence by (10,) (11,) (12,) and (13,) we have, λ = 50 58 24.8, 4 1° 10′ 12′′.1, 50° 57 32", and u = 1° 51′ 19′′. Taking the longitude of Calais to be 22′ 59′′ W. of Paris, as stated in the Connaissance des Temps, the longitude of Paris will be 2° 20′ 18′′ = 9m 21.2, or 0.19 less than before. In this manner the latitudes and longitudes in Table V. were computed by Captain Kater. Ex. 5. The latitude of Beachy Head, or a 50° 44′ 21′′ N. Dunnose, or ' 50 37 5 and the perpendicular from Beachy Head upon the meridian of Dunnose is 56020 fathoms by the Trigonometrical Survey, consequently 56020 × 3600′′ = 55′ 13′′.92, and m = 90° 60856 B = Now by formula, (6,) as in example, (1,) we have 1°27′ 6′′.65, and formula, (9,) gives A' = 55' 8".11. Whence extending this arc to a degree, or 55′ 8′′.11 : 60′ :: 56020: 60963, the length of the perpendicular degree in fathoms. Ex. 6. Again let a 50° 44′ 21′′ and x'= 50° 37′ 5′′, m = 96° 55′ 58′′, and m′ = 81° 56′ 53′′, and employing formula (8,) w will be found to be 1° 26′ 54′′-76, and A' 55' 0.59. Whence the perpendicular degree is 61102 fathoms, or 80 fathoms less than that given in the Trigonometrical Survey, and 139 fathoms greater than the result by formulæ (6) and (9) in ex. (5) stated above. In fact is found by other methods from independent data to be 1° 27′ 5′′.62, giving the perpendicular degree 60974 fathoms, exceeding the result in Ex. 5 by about 11 fathoms only, and agreeing exactly with a spheroid of 0.00324 of ellipticity. Ex. 7. The latitude of Crowborough station or a is 51° 3′ 18′′.30 N⚫ and longitude 9' 21".45 E. from Greenwich, and the latitude of Fairlight station or x is 50° 52′ 36′′.88 N. At Crowborough the observed angle between the meridian of Crowborough and Fairlight or m is 121° 4′ 58′′.36, and at Fairlight the angle between the meridian and Crowborough, or m' is 58° 33′ 26′′.14. Whence by formula (8) the difference of longitude is 27′ 49′′.90 E. Ex. 8. Again, the latitude of Blancnez or a is 50° 55′ 29′′.36 N., and at Fairlight, of which the latitude or a' is 50° 52′ 36′′ 88, the angle between the meridian and Blancnez or m is 85° 36′ 39′′.73, and at Blancnez, the angle between the meridian and Fairlight or m' is 93° 32′ 31′′.11; then by formula (8) the difference of longitude between Fairlight and Biancnez is 1° 5' 34".08 E. Hence the longitude of Crowborough being The difference between Crowborough and Fairlight 0 27 49.90 And between Fairlight and Blancnez Longitude of Blancnez Captain Kater's result by Oriani's formulæ from independent data, being ʊ, is Defect by formula (8) 0° 9' 21".45 E. 1 5 34 .08 1 42 45 .43 1 42 47 .45 2 .02 If a compression of, that adopted by Captain Kater, had been used, the difference would have been 1".78 only, which may fairly be attributed to unavoidable errors of observations, and a diversity of methods. TRIGONOMETRICAL SURVEYING AND GEODESIC 1. To apply the various formulæ for geodetical purposes, and trigonometrical surveying to practice, the greater part of the survey performed in 1821, 1822, and 1823, for determining the difference of longitude between the Royal Observatories of Greenwich and Paris, communicated by Captain Kater, to whom I have been considerably indebted, is here subjoined. 2. Triangles measured on the surface of the earth, of which the sides are small in proportion to the radius, may be considered as spherical triangles, and the sides may therefore be computed by the rules of Spherical Trigonometry. This method would in most cases be too laborious, and consequently other plans have been adopted, susceptible of greater simplicity, and of sufficient accuracy. First, the arcs may be reduced to the chords, and the sides calculated by Plane Trigonometry. Secondly, the observed angles may be reduced as has been proposed by Legendre, so that the computations may be performed by Plane Trigonometry, and of these the last is the more simple method. 3. In this case, if, from each of the observed angles of a small spherical triangle, one-third of the spherical excess be deducted, the sines of the angles thus diminished will be proportional to the lengths of the opposite sides, so that the triangle may be resolved as if it were rectilineal. The spherical excess besides being applied in this manner to prepare the angles for calculation, is useful to show the degree of accuracy of the observed angles, and from the two observed angles, to derive the third when it cannot be conveniently observed, and is therefore given in a separate column. 4. The spherical excess may be readily obtained by adding to the constant logarithm 0.373660 the logarithmic sine of the contained angle, and the logarithms of the containing sides, the sum will be the logarithm of the spherical excess in seconds; or to the same constant logarithm 0.373660, add the logarithmic cosecant of the angle opposite the given side, the logarithmic sines of the adjacent angles, and twice the logarithm of the interjacent side, the sum will be the logarithm of the spherical excess in seconds; or it may be found by any of the rules, page 193, sect. VI. 5. If, however, a plan of the triangles be constructed, to which a scale of English miles is adapted, the spherical excess may be readily found by scale and compasses, or more readily by the usual featheredged rules according to sect. VI., rule 4, by multiplying the base and perpendicular in miles together, and dividing the product by 152, the quotient will be the spherical excess in seconds.-Leslie's Geometry, p. 418. This may be very conveniently performed by the sliding rule. Set 152 on the sliding line of numbers to the base of the triangle on the fixed line in miles; then opposite the perpendicular, also in miles, on the slide, will be found the spherical excess in seconds on the fixed line. Thus in the triangle DBC, formed by Wrotham, Severndroog, and Leith Hill, the side BC will measure about 30 miles on the scale, and upon it the perpendicular from D will be about 13 miles. Now, set 152 on the sliding line of numbers, to 30 on the fixed line, then opposite 13 on the slide will be found 2".56 on |