the Magnetic Compass in Aeroplanes. 



465 



For any angle of dip (d), compass course (<£'), and tilt 0, 

 the value of (<£—#) can then be found, and since <j> is 



calculated from cos 6= -. — ^-r, x then becomes known, 

 tan <J> 



In the northern hemisphere, when the aeroplane makes a 

 turn from N. to E. and continues from E. to S., the tilt of 

 the machine is downwards towards the south, and the plane 

 of the card approaches the position of being perpendicular to 

 the earth's resultant magnetic field, but the two turns N. to 

 E. and E. to S. are symmetrical, the deviations being in 

 opposite directions in these two quarter turns. 



In going from S. to W. and W. to N. to complete the 

 circle, the card is tilted towards the north, that is, its plane 

 comes more nearly into the direction of the earth's resultant 

 magnetic field, and the control of the needle is always 

 increased by the tilt, and the deviations due to tilt are again 

 symmetrical in the turns S. to W. and W. to N., and are in 

 opposite directions in the turn. 



The equation of equilibrium for this southerly half of the 

 complete circle is slightly different from the former, as the N. 



Fig. 3. 



Hm Sii <» Gw, 9 



H Sin | Cose 



H Cos 9 



polar half of the card is dipping. In fig. 4 the turn S. and 

 W. has just begun and the equilibrium equation is : 



or 



H cos </>' sin (</> + #) = V sin 6 cos ($ + #) 



+ H sin <f> cos 6 cos (<£ + a), 



tan d . sin 6 



COS0' 



tan ((/> -f x] 



+ cos #tan </>'. 



(2) 



