296 
MR. S. DUNKERLEY ON THE WHIRLING 
By eliiiiinating A : B : C : D : A' : B' : C' ; D' from tliese equations we obtain the 
results, that either A = 0 or 
coth mil + coth ml ,2 = cot mli + cot w4- 
First consider tlie solution A = 0. 
It follows that B, B', C, C', A' are all zei’o, and that 
D = D', D' sin == 0, D sin ml^ = 0, 
Hence, in addition to the solution 
coth mil -f coth wiA = cot mli + cot ml a, 
the equations (l)-(8) are satisfied when 
mil == I . 
> simultaneously, 
ml2 = hiT 1 
a and h being integers. Hence, if h he a, multiple of a, that is, if 4 be a multiple of 
4 , one speed of whirl is clearly that of the shorter span when the longer span 
is neglected—a result, of course, identical with the vibration of strings in segments. 
Secondly, consider the solution 
coth mil H" ^oth 7 >i 4 = cot mli + cot m4- 
If 4 = 4 cr 4 = b, we get 
coth mil = cot mil. 
The physical interpretation of this equation is that the shaft in the one case 
is horizontal at the middle bearing, and in the other at the end bearing. In other 
words we get Case IV., § 14, p. 294, which has already been solved. 
[It should be noticed that the case when comes under the first solution.] 
The solution to the general equation 
coth mil “1" ml 2 = cot mli + cot ml 2 
may be performed by putting 
ml2 = a . mil, 
where a is the ratio of the spans, being ahvays less than unity. By expanding we 
obtain the equation 
