124 



" If the terms depending on k 2 be neglected in equations 

 (4), we obtain 



d x x du clz 



-j-t + N T = 2k sin X -77 + 2k cos X — ; 

 at 4 I at at 



^ + iV_^_2^cosX^. 



Eliminating N between the first two equations, we find 

 d 2 x d 2 y _..•■. ^ ( dy dx\ _ z _ dfz 



Integrating this equation, we obtain the following : 



y — - x — = k sin X (a; 2 + y 2 ) + 2k cos X \ydz. (6) 



Transforming (6) to polar co-ordinates, by the formulae 

 x = I sin sin 0, 

 y = I cos <j> sin 0, 



2 = I COS 0, 



in which <p denotes the azimuth measured from the north, and 

 the deviation of the pendulum from the vertical, we find 



<fy 7 • "v 2&COSX, • o n 7/1 ,*\ 



-£ = £ sm X . 0/1 | cos A sm 2 0c?0. (7) 



dt sm 2 J r x ' 



This equation proves, that the azimuthal velocity consists of 

 two parts ; one uniform, and equal k sin X, directed from the 

 north to the east ; the other periodic, and passing through all 

 its changes in the time of an oscillation of the pendulum, and 

 depending on the amplitude of the vibration. As the azimuth 

 $ may be considered constant during the time of an oscil- 

 lation, the second term in equation (7) may be integrated. 

 Hence we obtain, 



■j- = k sin X - %k cos X cos $ . 9; (8) 



