1907-8.] Professor C. G. Knott on Seismic Radiations. 
221 
0 = 0-0 
dx j jq 
l xW a?-x 
dx 
i x J , 
a A — x i 
-4> 2 
( 2 ) 
(3) 
The integrals of these are 
& 
R 
4>0 + | sin-'^a + 4 > 2 ) - 4> 2 - + 4 > 2 ) - 4» 2 [ 
4 { sin 
in "VS _ 
sin x 0 */a 2 - 1 
} 
o= e+ jfy{ sin -VS 1+ ^ )_ ^ _sin " , v / i (1+ ^ ) "^} 
+ 
{ sin_i VS 
sin 1 <j>\J a ‘ 2 - 1 
i 
(4) 
(5) 
Multiply (5) by 0 and subtract from (4) and we get a very simple 
expression for the time, namely, 
For each possible value of 0 equation (5) gives a ray, and the least 
value of x for this ray is determined by the condition 
^ 2= a 2 0 2 _ 1-20 2 
1 + 4,2 1+4,2 ' ■ ■ ' 
Also, since a 2 0 2 cannot exceed 1 -I- 0 2 , we find for the greatest possible 
value of 0 2 the expression 
0 2 = 
1 
a? - 1 
In the case to be discussed, u 2 = T2; hence 0 2 cannot exceed 5, and 0 
cannot exceed 2*236. 
When x 2 has the limiting value just given, equation (6) becomes 
T /xT 
574 ~ R 
1 
J 1 + ^ 2 
sin' 
V+} 
( 8 ) 
and gives the time of describing half the ray. 
Again, to find half the arc subtended at the centre by a complete ray, 
we put & 2 (1 + 0 2 ) = a 2 0 2 in equation (5) and obtain 
0 = 0 + 
-^=1 
ji+4,2 
sin 1 0 - sin 1 
+ sin. 1 1 - sin-^^/O'S 
(9) 
