Consequently, a plot of log 10 t as abscissa with (5/2)log 10 R as ordinate should give
a 45
straight-line whose intercept determines the energy yield, E 0 . Taylor plotted
values of R over a range from 20 to 185 meters and spanning a time interval from
about 0.24 ms to 62 ms and found that the logarithmic plot coincided almost exactly
with his theoretical predictions, these predictions as he stated “made more than four
years before the explosion took place”. According to Taylor, the straight-line plot
that appears in Fig. 1 in his article [11] represents the relation,
5
2
log 10 R À log 10 t ¼ 11:915:
By taking the density of air as 1.25 kg m
À3 and with Taylor’s first estimate of
E 0 ¼ 16,800 tons of TNT (hence, E 0 ¼ 4.2 Â 10
9
 16,800 Joules), we find, after
substituting these values for E 0 and ρ 0 in the equation above, that
5
2
log 10 R À log 10 t ¼ 6:911:
However, it should be noted that Taylor measured the shock front in cm, so that
the latter equation becomes,
5
2
log 10
R cm
100
À log 10 t ¼ 6:911,
hence,
5
2
log 10 R cm À log 10 t ¼ 11:911,
which is in very good agreement with the straight-line representing the data of the
Trinity test. Taylor points out that this estimate is based on the assumption that the
energy available for propagating the blast, and which was not radiated outside the
expanding shock front, was 16,800 tons of TNT.
He also considers an alternative possibility by taking into account the temperature
behind the shock wave in order to determine a mean value for γ in the range where
the product R
5/2 t
À1 is constant. From the plot of radius versus time he determines a
mean radius of 100 m in the range and found that the temperature behind the shock
wave at this radius to be about 2800 Kelvin. Using values of c P that were calculated
for nitrogen and oxygen at this temperature, and assuming that the relation c P ¼ c V + R
still applies at this elevated temperature, he finds that γ ffi 1.3.
We have seen that the multiplying constant in Eq. (5.24) is 1.033 when γ ¼ 1.4,
hence, to determine the multiplying constant for γ ¼ 1.3, we refer to the integrals
appearing in Sect. 5.7, in this chapter. Evaluating these integrals for γ ¼ 1.3, we
find that
5.15 Taylor’s Second Paper
251
a 45
straight-line whose intercept determines the energy yield, E 0 . Taylor plotted
values of R over a range from 20 to 185 meters and spanning a time interval from
about 0.24 ms to 62 ms and found that the logarithmic plot coincided almost exactly
with his theoretical predictions, these predictions as he stated “made more than four
years before the explosion took place”. According to Taylor, the straight-line plot
that appears in Fig. 1 in his article [11] represents the relation,
5
2
log 10 R À log 10 t ¼ 11:915:
By taking the density of air as 1.25 kg m
À3 and with Taylor’s first estimate of
E 0 ¼ 16,800 tons of TNT (hence, E 0 ¼ 4.2 Â 10
9
 16,800 Joules), we find, after
substituting these values for E 0 and ρ 0 in the equation above, that
5
2
log 10 R À log 10 t ¼ 6:911:
However, it should be noted that Taylor measured the shock front in cm, so that
the latter equation becomes,
5
2
log 10
R cm
100
À log 10 t ¼ 6:911,
hence,
5
2
log 10 R cm À log 10 t ¼ 11:911,
which is in very good agreement with the straight-line representing the data of the
Trinity test. Taylor points out that this estimate is based on the assumption that the
energy available for propagating the blast, and which was not radiated outside the
expanding shock front, was 16,800 tons of TNT.
He also considers an alternative possibility by taking into account the temperature
behind the shock wave in order to determine a mean value for γ in the range where
the product R
5/2 t
À1 is constant. From the plot of radius versus time he determines a
mean radius of 100 m in the range and found that the temperature behind the shock
wave at this radius to be about 2800 Kelvin. Using values of c P that were calculated
for nitrogen and oxygen at this temperature, and assuming that the relation c P ¼ c V + R
still applies at this elevated temperature, he finds that γ ffi 1.3.
We have seen that the multiplying constant in Eq. (5.24) is 1.033 when γ ¼ 1.4,
hence, to determine the multiplying constant for γ ¼ 1.3, we refer to the integrals
appearing in Sect. 5.7, in this chapter. Evaluating these integrals for γ ¼ 1.3, we
find that
5.15 Taylor’s Second Paper
251