Proceedings. 75 



the other side of the balance (Q), and then if a and b are 

 the arms of the balance, it is stated that 



and hence 



M= v/PQ = i(P + Q) 

 if P and Q are nearly equal. 



If the whole balance were perfectly rigid in all its parts, 

 this deduction would be correct ; but the balances as con- 

 structed shew a very appreciable bending of the beam, and 

 it is not correct to assume that the bending is the same on 

 both sides. 



"Now, how is the so-called apparent weight of the body, 

 when placed on one side of the balance, determined ? The 

 position of rest of the balance when unloaded is first obtained, 

 and then with the body in one of the pans that weight is 

 found which brings the balance back to the same position. It 

 is the object of this note to point out that the position of rest 

 of an unloaded balance has nothing to do with the behaviour 

 of the balance when it is loaded, for the centre of gravity of 

 the beam and pan may be displaced by the load. The so- 

 called zero of the balance is a matter of no importance, and 

 need therefore not be observed at all. If the practice of the 

 method gives the correct result, it is only because this zero 

 disappears from the result. Let 0o be the position of rest of 

 the unloaded balance, and let di be that position of rest to 

 which the balance will come when two equal masses M are 

 placed on it, also let W be the weight required to charge 

 the zero through one division under the given circumstances, 

 then the quantity called P above would be M-|-W(0i-0o) 

 and the quantity Q would be M-W(0i-0o). Hence W 

 would be correctly obtained by taking the arithmetical 

 mean of P and Q, but the same would be true if we 

 substitute for the value of 60 any scale division, say 

 the central one, and the labour of finding do is wasted. The 



