the Magnetic Compass in Aeroplanes. 



463 



For simplicity the pole of the needle is supposed to be at 

 the edge of the card and the strength m, and the radius of 

 the card Z, but this assumption does not affect the result. 



Fig. 1. 



Then 

 and 



TW = RW.tan</>', 



SW = RW.tan<£. 



TW tan<£' , n TW 



SW = tan-^ ;butC0S ^ = SW 



COS0: 



tan <f>' 

 tan cf> ' 



a result which will be required shortly. 



Vertical component of eartli s field. — There will be a vertical 

 force Vm acting downwards upon the N. pole of the needle 

 and, of course, an equal and opposite upward force upon the 

 S. pole. 



This has a component E'Z perpendicular to the plane of 

 the card, which has no turning effect upon the needle, and a 

 component E'Y = V»?sin# in the plane of the card, and 

 perpendicular to OQ. 



Horizontal component of ear tKs field. — The horizontal force 

 Hra on the pole E' may be resolved into two horizontal 



. \ 



