IATITUDE.] 



NAVIGATION NAUTICAL ASTRONOMY. 



1089 



Z would then be given ; and this, together with Z S and 

 P S, is sufficient for the determination of Z P. 



It is more convenient to observe the bearing when the 

 sun is low than when it is high, and the result can be 

 the better depended upon ; therefore, in the problem of 

 double altitudes, the bearing of the sun is always taken 

 when the less of the two altitudes is taken. 



The method of finding the latitude here explained, 

 may of course be applied to a star as well as to the sun, the 

 interval between the observations being expressed in 

 sidereal instead of in solar time ; but as, in general, the 

 horizon increases in obscurity as the star becomes more 

 clearly visible, two observations of a star, with a sufficient 

 interval of time between them, can seldom be accurately 

 made. The following, therefore, is a more suitable 

 problem for the purpose in view. 



IiATITCDE FROM THE ALTITUDES OF TWO FlXED STARS 



OBSERVED AT THE SAME TIME. There are two advantages 

 connected with this method of deducing the latitude : the 

 first is. that as no allowance is to be made for change of 

 place in the ship, all error arising from inaccuracies in the 

 sun's bearing, and the course and distance steered, is 

 avoided ; the second is, that the risk of losing another 

 observation, from unfavourable weather, is not incurred 

 when both observations are made at the same time. 



When the altitudes of the two stars are to be taken by 

 one person, the mode of proceeding is this : The altitude 

 of one star is taken, and the time by the watch noted ; 

 the altitude of the other star is then taken, and the time 

 noted ; after a short interval, the altitude of the second 

 star is again taken, and the time noted. Thus the 

 change of altitude of the second star, in a known interval 

 of time, will be found ; and therefore the correction for 

 the time, when the first star was observed, may be found 

 by proportion. 



As to the principles of solution, they are the same as 

 in the former problem : the co-declinations or polar dis- 

 tances PS, PS' of the two stars are given ; the angle P 

 between these U also given, this angle in time being the 

 difference of the right ascensions of the two stars. We 

 may hence compute the third side S S' of the spherical 

 triangle. As the co-declinations may differ considerably, 

 this third side must not be calculated on the supposition 

 that the mean of the co-declinations may be taken for 

 each of the other two sides. It must be found from the 

 distinct co -declinations themselves. The remaining part 

 of the operation is the'same as in the former problem. 



Examples. 



1. In north latitude 52 30' by account, the corrected 

 zenith distances of Capella and Sirius were as follows : 



Capella, ZS= 29 14' 24*. 

 Pol. Dist., P S = 44 11' 39*. 

 Sirius, ZS' = 72 5' 48*. 

 Pol. dist., PS'=10628'40*. 



Difference of the two right ascensions, Ih. 33m. 43. =- 

 23 26' 11' = P. Required the latitude of the ship. 



1. In the triangU S P S' to find S S'. 



tan. P S, 44 11' 39' . . . 9-9877822 

 COB. P, 23 26' 11" . . . 9-9020072 



cot. , 48 15' 56' 

 106 28' 40" 



.+ PS'= 164 44' 36' 



cot PS 9-8555080 



sin. Ar. Comp. -1271225 

 in. ( + PS') 9-6300961 



COB. S 8', 65 47' 65" 9-6127266 

 2. In the triangle PSS'tofind the angle P S S'. 

 in. S S', G.V 47' 55" Arith Comp. '0399527 

 in. I'S', Kit; L's'.fi 

 *in. S PS', 23 26' 11' 



!i '.is 17868 

 9-5995891 



sin. P8S', 24 43' 3' 



VOL. I. 



9-6213286 



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



Z S' = 72 5' 48' 



sin. Z S' = 29" 14' 24" Arith. Comp. -3111631 

 sin. SS'=6547'54* -0399536 



8' 6" 



$ sum = 83 34' 3' 

 sin. (i sum - Z S) = 54 19' 39" 

 sin. (| sum - S S') = 17 46' 9' 



sin. 4 ZSS' = 48 14' 29" 



, 9-9097504 

 , 9-4845601 



2)19-7454272 



9-8727136 



/.ZSS'= 96 

 P S S = 24 



>' 28' 58" ) 



:43' 3") 



to be added 



121 12' 1* 



supplement = 58 47' 59' = P S Z. 

 3. In the triangle Z S P to find Z P. 



tan. PS, 44 11' 39* . . . 9 9S77822 

 cos. P S Z, 58 47 69' . . . 9-7143559 



cot. 



63 16' 3' 

 29 14' 24' 



9-7021381 



(w + Z S) = 92 30' 27' 



cos. PS 9-8555030 



sin. w Ar. Comp. -0490919 

 sin. (u+ZS) 9-9995330 



in. ZP, 53 19 23' 9-9041829 



Hence the latitude is 53 19 23" N. 



In connection with the foregoing method of solving the 

 problem of double altitudes, there are some points of 

 theoretical interest to which the learner should attend. 

 In the second step of the work the object is to find the 

 angle P S S'; the operation conducts us to the sine of this 

 angfa ; and as the angle answering to a given sine may be 

 either acute or obtuse, we have no right, independently 

 of controlling conditions, to assume it to be the one any 

 more than the other. 



In the example just solved, we have arbitrarily talcen 

 it to be acute, simply for convenience, without examining 

 into any overruling circumstances. Now, in the present 

 problem we may always take a similar liberty ; for 

 although the proper angle may really be the supplement 

 of that we put down, the subsequent step will not be 

 affected by this circumstance : the angle P S Z, which it 

 is the object of that step to find, will still be the sum or 

 difference of Z S S', PS S' or the supplement of the sum. 



But it is well to show how all doubt respecting the 

 species of the angle P S S' may be avoided. 



The fundamental formula of spherical trigonometry, 

 applied to the triangle P S S', gives 



cos. P S S' = 



cos. PS^-cos. PS cos. SS', 

 ~~sTn7P S sm7 S S' ~ 



and as it is matter of indifference which of the two stars, 

 or which of the two places of the sun we mark S or S', 

 we may always consider, in this formula, that cos. P S" is 

 numerically greater than cos. P S, in which case the 

 numerator of the above fraction, and consequently the 

 fraction itself (since its denominator is positive), will 

 have the same sign as cos. P S' ; hence taking P S' for that 

 of the two co-declinations whose sine is the leas, the opposite 

 angle P S S, will always be of the same species as P S' 

 that is, they will either be both acute or both obtuse. 



In the example just solved, cos. P S' is negative, and 

 cos. P S, cos. S S' are both positive ; consequently the 

 fraction is negative, and therefore the angle P S S' is 



A second is deducted from this arc, in order that there may be an 

 eTen number of seconds in the sum of the three, so as that, in taking tho 

 half, fractions of a second may be avoided ; the omission of a second can 

 have no sensible influence on the result. 



Cz 



