OR MOON AT THE EARTH'S SURFACE. 309 



In order to determine the effect of these forces upon a freely 

 suspended horizontal magnet at the earth's surface, we must resolve 

 X and T in the direction of the tangent, and of the radius, of the 

 parallel of latitude. The resolved forces are, respectively, 



X sin + T cos 0, X cos 6 - Y sin 0. 



Again, resolving|the forces Z and X cos - T sin in the direction 

 of the tangent to the meridian, and in that of the radius of the 

 earth, we have finally the three components, viz. : 



X sin + Y cos 0, directed eastward ; 

 Z cos A + (X cos - Y sin 0} sin X, directed northward; 

 - Z sin A + (X cos - Y sin 0) cos X, vertical, towards centre. 



Of these, the latter has no effect upon the horizontal magnet. The 

 moment of the two former to turn it is 



{X sin + Fcos 0) cos 3 - (^cos X + (X cos & - Fsin 0) sin X| sin 8, 



8 denoting the magnetic declination ; or, substituting for X, F, Z 

 their values, 



|UcosS(2P sin0- Q cos 0} - sin 8 (2P cos + Q sin 0)' sin A 



But the moment of the earth's magnetism, opposed to this, is 



JTA8 sin 1', 



in which H denotes the horizontal component of the earth's mag- 

 netic force. Wherefore 



A8= ,,*. , Jsin0(2PcoB8- QsinXsinS) 

 D H. sin 1 ( 



-cos0(2PsinAsin8 + QcosS) + R cos A sin8 j. 

 At the equator, this is reduced to 



AS - - cos 82P sin - Q cos0) + ^si 



To determine the effect of the magnetic body upon the hori- 

 zontal component of the earth's magnetic force, we must resolve 



