1148 
530 
AS BE AR ONS AND (Di. Ro YEN NTE 
Fy | | |p ptrotat eNercy Fux 
re] 1 
E 4 
- 
0078 3 La 
napa 
Y ezee 
sat MBG 
ws # 4 
it 
200 cli 80) IAS 
| 
i 
AB (Zee 
=: EAE 
LI | 
IN-LB. 
wz Lels 
[ 
E| 
ENERGY FLUK 
yen 
ENERGY FLUX 
[ 
° 
° 
° 
2 8 10 1s 
REOUCED TIME 
20 26 30 
T+ % 
Fic. 8. Energy flux, impulse, and total particle velocity versus time for initial positive pressure phase 
of shock wave. Explosive: TNT; charge depth: 500 ft:; distance from center of charge: R= W?/0.352. 
@=time constant of initial shock-wave decay. 
The total volume flow through a spherical 
surface may be expressed as 
t 
AV=4rR? i udt. 
0 
In the case under consideration, the total volume 
flow up to the time of bubble maximum is 
tM, Ap 
avaaer' [ dt 
0 poCo 
tM, t Ap 
ster? f"[ fala, 
0 0 poR 
and since the first term on the right is very small, 
i) A f “apai at 
Since AV is nearly independent of the radius, 
Eq. (35) should give us the total flow through 
the surface at the maximum bubble radius, i.e., 
the volume of the bubble at tyy1. This volume may 
also be calculated independently from experi- 
mental knowledge of the maximum radius. 
From high speed photographic work carried 
4rR 
AV= 
(35) 
Po 
out at Woods Hole by J. C. Decius and E. Swift, 
the following empirical equation relating maxi- 
mum bubble radius to charge size and depth has 
been obtained : 
Am=Ji(W/Z,)}. (36) 
Am is maximum bubble radius in feet, W is 
charge weight in pounds, Zo is the total hydro- 
static head in feet (depth +33 ft.), and J; is a 
nearly constant factor which has a value of 12.6 
for TNT over the range of depths under con- 
sideration. ‘ 
At a depth of 500 ft. Eq. (35), utilizing the 
pressure-time curve of Fig. 6, gives a volume of 
17 cu. ft. per lb. of explosive, while Eq. (36) 
gives 15.7 cu. ft. per lb. At 250 ft. the respective 
figures are 33.6 cu. ft. per lb. and 29.6 cu. ft. 
per Ib. In each case the integrated particle 
velocity gives a greater volume change than the 
direct measurement of the radius by 8 and 13 
percent, respectively. The error in the radius 
formula (36) is of the order of 2 percent, which 
could amount to an 8 percent error in the 
volume. The error in the double integration in 
Eq. (35) is of the order of 5 percent, because of 
base line inaccuracies, etc. (A base line shift of 
about 5 lb./in.* in the pressure-time curve would 
