LATITUDE.] 



NAVIGATION NAUTICAL ASTRONOMY. 



1087 



Examples. 

 The two corrected zenith distances are, 



Z S = 73 54' 13", and Z S' = 47 42' 51* ; 

 the corresponding co-declinations of the sun are 

 P S = 81 4? N., and P S' = 81 45' N. ; 

 and the interval of time between the observations is 

 three hours. Required the latitude. 



For the more easy determination of S S', let it be re- 

 garded as the base of an isosceles spherical triangle, of 

 which each of the equal sides is | (P S + PS') = 81 

 43' 30", the vertical angle at P being 3h. or 45 ; then if 

 the perpendicular P M be drawn, the triangle P M S will 

 be right-angled, and we shall have given 



P S = 81 43' 30" and P = - = 22 2ff 



to find S M = S S' as follows : 



1. In the triangle P MS io find SM. 



sin. PS, 81 43' 30" 

 Bin. P, 22 W 0* 



rin. SM, 22 15' 11' '3 

 2 



9-9954547 

 9-5828397 



. 95782944 



/. SS' = 44 30' 22* -6 



2. In the triangle P S S' to find the angle P S S'. 

 sin. SS', 44 30' 22"0 Arith. Comp. -1542898 

 sin. PS', 81 45' 0* 9-9954822 

 sin. S PS' 45 0" 0* 9-8494850 



sin. PS S' 86" 38' 68' 9-9992570 

 3. In the triangle Z S S' to find the angle Z S S'. 



Z S', 47 45' 61 



sin. Z S', 73" 64' 13* Arith. Comp. -0173686 

 Bin. Sb, 443(X22 -1542898 



2)166 W 26"-G 



4 gum = 83 6'13'-3 



Rin.(isum-ZS), 9 11' 0-3 . . . 9-2030206 

 sin.(f sum-SS'),3ti 34'50"7 . . 97949179 



2)19-1695969 



sin. fJZSS', 2236'26"-4 . . . 9-6847985 

 .'. Subtract 



Z S S' - 45' 12 7 52" -8 

 fromPSS'-8638'68" 



.'. 1>SZ' = 41*26' 5 -2 



4. In the triangle Z S P to find Z P. (See the following 

 Investigation.) 



tan. PS, 81 42- 0" . - . . 10-8359917 

 C06.PSZ, 41 26' 5*2 . . . 9-8748930 



cot. u = 11 0' 41-2 

 Z S = 73" 54' 13' 



10-7108847 



*+ZS=8454'54'-2 



cos. PS 91594354 



sin. wAr. Comp. . . . 7189551 



sin. (+ZS) 9-9982874 



sin. ZP, 48' 49' 59" 7 . . 9-8766779 



Hence the latitude is 48" 50.' 



As Z P is here found by a method not always given in 

 works on Spherical Trigonometry, we shall subjoin the 

 investigation of it. Representing, as usual, the three 

 sides of a spherical triangle by a, 6, c, and the angle in- 

 cluded by the first two by C, we have, by the funda- 

 mental formula of Spherical Trigonometry (page 658), 

 cos. c = cos. a cos. b -f- sin. a sin. 6 cos. C. 



sin. a 



Or, since sin. a = cos. a = cos. a tan. a, 

 cos. a 



cos. e = cos. a (cos. I -\- tan. o sin. 6 cos. C). 



Put cot. u for tan. a cos. C ; that is, let 



cos. <a 



tan. a cos. C = cot. u = , 



sin. u 



Then we shall have 



sin. M cos. 6 -f- sin. b cos. o> 

 cos. c = cos. o : 



that is, cos. c= 



cos a sin. (01 -f- 6) 



(1) 



(2) 



The expressions (1) and (2) are those already calculated. 

 In the preceding solution more attention is paid to 

 minute quantities than is at all necessary in actual prac- 

 tice at sea, where fractions of a second are of course dis- 

 regarded. Yet in lengthy operations seconds themselves 

 ought not to be entirely neglected, except in those con- 

 fessedly approximative methods which the indirect pro- 

 cesses, by peculiar tables, generally are. The only de- 

 parture from strict mathematical rigour in the foregoing 

 work, is in the first part of it, where the arc S S' is sup- 

 posed to be equal to an arc subtending the same angle at 

 P, the sides of this angle being the mean between the 

 two slightly differing co-declinations P S, P S'. It is 

 plain, from the small amount of this difference, that the 

 fictitious arc cannot vary in length from the real arc S S' 

 by a quantity deserving of any consideration, when the 

 time between the observations is not unreasonably great. 



By any one familiar with the practical solution of 

 spherical triangles, the method here illustrated will be 

 preferred to the indirect methods adverted to above; for 

 the operation, though rather long, lays no burden upon 

 the memory, and is, moreover, free from the inaccuracies 

 small inaccuracies, no doubt which indirect methods 

 are always affected with. A distinguished practical navi- 

 gator, Captain Kater, after the example of Delambre, 

 gives the preference to the direct process : the preceding 

 Exercise U taken from the former writer ; but tho latter 

 part of the solution is conducted differently. 



2. On the 7th of February, 1840, in latitude by ac- 

 count 35" N., and longitude 47 W., at 8h. 9m. 4s. A.M. 

 mean time, the altitude of the sun's lower limb was 

 36 1W, and his bearing S. } E. ; after running N.E. 27 

 miles, the altitude of the lower limb, at llh. 30m. 18s., 

 was observed to be 41 20' ; the height of the eye was 

 20 feet. Required the latitude of the ship when the 

 second observation was made. 



/ declination. 



16 



1. For the true altitude of sun's centre. 



First alt. sun's L. L. . 36 1(X 0' 



Dip ... -4' 24") , 11 , fift , 



Semi-dL 16' 14" ' ' + U ^ 



App. alt. . 



Refraction and par. . 



True altitude 



Second alt. sun's L. L. 

 Dip. and semi-diam. 



Apparent alt. . 



Refraction and par. 



36 21' 50" 

 - 1' 10* 



36 20' 40" 



41 2V 0' 

 + 11' 50" 



41 31' 50' 

 58" 



True altitude . . . 41 30' 52' 



The angle between the sun's bearing S. ^ E. and the 

 course N.E. is 11J points, so that the ship has sailed 

 within 4} points of the direction opposite to tho sun, "a 

 distance of 27 miles. With this distance and 4| points 

 as a course, the Traverse Table gives 18' for the corre- 

 sponding difference of latitude, which is the number of 

 minutes the ship has receded from the sun during the in- 

 terval of the observations : hence these 18' must be sub- 

 tracted from the first true altitude, to reduce it to what 

 it would have been if taken by another observer at the 

 place of the second observation : consequently, the true 



