328 Prof. S. P. Thompson. On the [Feb. 14, 



value of n will give the force when 0=45° equal to -§- of the force 

 when 0=30°. Writing in the sines of these angles and equating 



3 times the force at 30° to 2 times the force at 45°, we get the value 

 n=4i'6. Or, solving similarly for the case that the force at 60° shall 

 be double the force at 30°, we get w=4'28. Here, however, the errors 

 near the intermediate angle of 45° are not inconsiderable, exceeding 



4 per cent. In any case we shall not be far wrong if we adjust the 

 balance so that w=4*5, and confine our readings within 45° of zero. 

 For purposes where great accuracy is not essential the readings might 

 be extended as far as 60°, since the curves for w=4 and n=5, show 

 how very little the straight line is departed from up to that point. 

 In the original balance of Hughes, the values of r and I are about 

 30 centims. and 7*5 centims. respectively, so that n=4i. 



[4.] I now proceed to the consideration of the action of the com- 

 pensator in affecting in different positions the sensitiveness of the 

 needle. 



In the original instrument of Hughes, the compensator turned in a 

 horizontal plane ; and therefore in every position, save only when at 

 90° to the zero of its scale or when end-on to the indicating needle, 

 the force exercised by it on the needle would have a component in the 

 magnetic meridian. As remarked above, the effect of this component 

 would be to increase or diminish the directive force of the earth's 

 magnetism upon the needle, according as the S pole of the com- 

 pensator pointed northwards, or southwards. The sensitiveness of 

 the indicating needle will, of course, be affected by this component ; 

 being a maximum when the component is such as nearly to astatise 

 it. But it is evident that if the needle be nearly astatised when the 

 compensator is at zero, it will not be so astatised when the com- 

 pensator is moved right or left. The sensibility of the needle will 

 diminish as the effective force of the compensator increases by its 

 being turned. The calculation for the component of the force at this 

 point is as follows : — 



Calling T the part of the magnetic force at B (fig. 1) resolved at 

 right angles to AX, we have — 



But 



Sill a — 



I COS 



{f+P—Zrl sin0)*' 



and 



sin /3= 



I cos 



(r» + Z8 + 2rZ sin 0)*' 



whence T=wicos0 



(r 3 + P - 2rl sin 0) I (r 3 + Z 3 + %rl sin 0) 



