324 Mr. Galbraith on the Deviation of Falling Bodies. 



By dynamics s = ^gt^, or in this case a = ^g f, there- 

 fore t ^ J-^ : (6) 



The earth performs a rotation about its axis in SS** 56™ 4* 

 or 86,164 seconds. Let this be represented by q ; whence 



/ a ^ 2t /a 2cr/cf3 



g:2Tra:: \/ j-^: = a ^ — — V T7 • • • • ( ' ) 



the difference between the arcs in the time of fall. But the 

 body in falling describes a small portion of an ellipse, which 

 on account of its minuteness may be considered parabolic, as 

 shown in our ordinary treatises on Natural Philosophy, 

 (Leslie's, vol. i. p. 128.) of which the contained area is two- 

 thirds of the circumscribing rectangle ; wherefore, 



D = §x^xV^ (8) 



This formula gives the deviation at the equator. \n a given 

 latitude \ from what precedes, 



D = I X — X cos XV 7— = --,— cos X V -T—— (9) 



4 T 



Let - — = k, and formula (9) becomes 



« / n3 



T> = k cos X V T- (^) 



As it requires a fall of a considerable number of feet to pro- 

 duce a sensible change in the deviation, ^g may be taken at 

 16 feet, which introduced into formula (9) gives 



D = -r- cos A « # 



3s ^ 



Now, calling -^ = K, we shall have, finally, 



D = K cos A a I (B) 



an expression remarkably simple. 



This formula is well adapted for logarithmic calculation, in 

 which case it will become 



Log. D = const, log. 5-084702 +log. cosX + | log. a ... (C) 

 Let Emerson's example be solved by this formula, in which a 

 is 5280 feet, and X is 51° 30'. 



Constant logarithm 5'084702 



A = 51° 30' N. log. cosine . . . 9*794150 

 a = 5280 feet logarithm .... 3*722634 



half same log 1-861317 



D= 2-9026 feet (Sum) 0-462803 



This result differs in a slight degree from Emerson's, which 

 is 2-88 feet ; but the difference may be accounted for by his 



taking 



