191 
1918-19.] The Propagation of Earthquake Waves. 
and the numbers in the third column give the ratios of the energy per 
unit surface of the nucleus at the position /3 to the energy at distance 
( b — a ) from the epicentre. 
Table VIII. — Energy Distribution. 
3. 
d. 
Energy per 
Unit Surface. 
6 
0 
' 
1-0000 
3 
7 
3 
•9836 
6 
15 
9 
•9384 
9 
23 
1 
•8638 
12 
31 
19 
•7596 
15 
40 
19 
•6279 
18 
50 
36 
•4702 
21 
63 
38 
•2797 
23 35 
90 
o-oooo 
Reducing 
Energy 
Average 
Ratio. 
in Nucleus. 
Values. 
0*994 
0-944 
•98 
•927 
•835 
•730 
•570 
*390 
•240 
•070 
•977 
•946 
•893 
•886 
•752 
•672 
•595 
0 
•961 
•888 
•771 
•673 
•472 
•316 
T66 
0 
We have now to consider the energy which passes into the nucleus 
as each of the ra 3 7 s (/3) is refracted at the boundary separating the shell 
from the nucleus. The data for making this calculation are supplied by 
Table VII. In this table, however, each incident ray, whatever its 
inclination, is assumed to have energy unity ; but in the present case 
the energy per unit surface associated with each incident ray other 
than the ray /3 = 0 is less than unity, in accordance with the short 
Table VIII just given. In order to find bow much energy passes into the 
nucleus, we must multiply each number in the third column in this table 
by the fraction appropriate to the corresponding angle of incidence as 
given in Table VII. 
From the graphs, which are shown on reduced scale in fig. 8, the 
various energies may be picked off* with sufficient accuracy for any 
required angle of incidence. When this is done for the various angles 6 
in Table VIII, certain numbers are obtained for the refracted condensational 
wave. These are tabulated under the heading Reducing Ratio, and form 
the fourth column in Table VIII. Multiplying these into the corresponding 
energies per unit surface incident on the nucleus, we obtain in the fifth 
column the corresponding energies which pass with the refracted con- 
densational wave into the nucleus. 
To find the whole energy which passes into the nucleus, we must 
evaluate the integral 
d= z 
f ^27rasin {0-/3) . ead(6 - /3) 
7(9 = 0 
where e is the energy per unit surface at position /3. 
