On Gravimeters. 



523 



To determine the position of unstable equilibrium, consider the 

 general statical equation 



(P+iO-^.sinfl 4-2^=^0. 



When the left-hand side is a maximum its differential with respect to 

 equals zero, therefore 



cos O +2^=O, 



whence cos ft = — — . -J^2-« 



u r P+p 



7T 37T 



Now, as in the intended position of rest, 0=-, 0=— , we have 



2 2 



Combining this with the former, it appears that 



The double wires and the single wire are no doubt of the same material, 

 but the former are thinner. Their lengths are as 4:1. Suppose their 

 thicknesses as 2 : 3. Then p l : p 2 : : 4 X 3 3 : 2 2 : : 9 : 1, which gives 



sec 6 =- 12-5 -19-64, . 

 2 



and .-. O =92° 55'. 



That is to say, the position of unstable equilibrium lies less than 3° 

 beyond the intended position of rest. [This is closer than I should 

 have thought, but it is borne out by the fact that I have not been able 

 (with the existing adjustments) to reach the intended position of rest 

 both ways, i.e., turning both to the right hand and to the left hand, 

 before reaching the point of unstable equilibrium. This is easily ac- 

 counted for by supposing the primary position to be a balanced one, a 

 slight deviation from parallelism of the mirrors being induced by a 

 slight initial torsion of the parallel wires.] To return : — 



The angle through which the loiver end of the single torsion wire is 

 turned, by the stop acting on the minor weight, is + 0. As this 

 angle approaches 360°, 0, which is the angle through which the upper 

 end and the major weight are turned, approaches, or should approach, 

 90°. As this stage is approached a very small increase of + in- 

 duces a larger and larger increase of 0, according to the proximity to 

 the value O . And the above expression for sec O shows that this 

 depends on the ratio of p 1 : p 2 . By varying the strength of the double 



2 p 2 



