466 



Mr. S. G. Starling on the Equilibrium of 



It is easier for these courses to reckon <f> and <j>' from the 

 south, although in plotting the results they are considered 

 to be 180° up to 360° E. of N. 



Calculation of results. — The following Tables are obtained 

 for a position where the dip is 67°, so that 



tan d = tan 67° = 2*36. 



For values of less than 15° it is taken that <j> and </>' are 

 equal, but for values of above 15°, the value of </> is 

 calculated from 



^ tan 6 



cos0=r V. 



tan<£> 



tan d> 

 or tan = -hr . 



T COS0 



It must also be noticed that on the turns N. to E. or E. to S., 

 that is, when the north side of the card is upwards, there 

 may be a position for which x>^>, and the N. pole of the 

 magnet is below the horizontal line OQ (fig. 1). In this 

 case (<f>—a;) becomes (x — <f>). This is, in fact, generally the 

 case ; for even when = 20°, we obtain by putting <£ = # 

 in (1), that 



tan d sin /Xl , . 



n — = cos tan <p /. 



cos<£ T 



sin <£' = tan d . tan 0, 

 = 2-36 tan 20°, 

 = 0-860, 

 or <f> — 60° approx. 



For values of greater than 23°, x is always greater 

 than <j)' for 



sin<£' = 2'36tan23°, 



= 2-36x0-4245 = 1-0. 



<£' = 90°. 



Thus if is more than 23°, tan d tan must be greater 

 than unity, and there is no possible value for (f> f , such that 

 <£'=#', which means that x' must always be greater than <j>', 

 and the North polar end of the needle is below the horizontal 

 OQ from the start of the turn. This appears also from the 

 curves. As an example, the calculation for a tilt of = 70° 

 is given, and onlv a summary of the results for the other 

 angles of tilt from'lO to 85°. 



