Sir Geoffrey Ingram Taylor. The Formation of a Blast Wave by a Very Intense Explosion (PDF)

The Formation of a Blast Wave by a
very Intense Explosion
Part I and Part II
Sir Geoffrey Ingram Taylor (* 7 March 1886, † 27 June 1975)

Proceedings of the Royal Society A
1950

THE ROYAL SOCIETY
6-9 Carlton House Terrace, London SW1Y 5AG

2

Contents
1 The formation of a blast wave by a very intense explosion I. Theoretical discussion
1.1 Summary and Introduction . . . . . . . . . . . . . . . . . . . . . .
1.2 Shock-Wave Conditions . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Numerical Solution for γ = 1.4 . . . . . . . . . . . . . . . . . . . .
1.5 Approximate Formulae . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Blast Wave Expressed in Terms of The Energy of the Explosion
1.7 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 Velocity of Air and Shock Wave . . . . . . . . . . . . . . . . . . . .
1.9 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.10 Heat Energy Left in the Air After it has returned to Atmospheric
Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.11 Comparison with high Explosives . . . . . . . . . . . . . . . . . .
1.12 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 The formation of a blast wave by a very intense explosion. II. The
atomic explosion of 1945
2.1 Summary and Introduction . . . . . . . . . . . . . . . . . . . . . .
2.2 Comparison with Photographic Records of the First Atomic Explosion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Calculation of the Energy Released by the Explosion . . . . . . .
2.4 Energy of the First Atomic Explosion in New Mexico . . . . . . .
2.5 An Alternative Possibility . . . . . . . . . . . . . . . . . . . . . . .
2.6 Some Dynamical Features of the Atomic Explosion . . . . . . . .
2.7 The initial rate of rise of air from the seat of the explosion . . . .
2.8 Distribution of Air Density after the Explosion . . . . . . . . . .
2.9 Calculation of the Rate of Rise of the Heated Air . . . . . . . . .
2.10 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 The formation of a blast wave by a very
intense explosion I. Theoretical discussion
1.1 Summary and Introduction
This paper was written early in 1941 and circulated to the Civil Defence Research Committee of the Ministry of Home Security in June of that year. The
present writer had been told that it might be possible to produce a bomb in
which a very large amount of energy would be released by nuclear fission-the
name atomic bomb had not then been used-and the work here described
represents his first attempt to form an idea of what mechanical effects might
be expected if such an explosion could occur. In the then common explosive bomb mechanical effects were produced by the sudden generation of a
large amount of gas at a high temperature in a confined space. The practical
question which required an answer was: Would similar effects be produced
if energy could be released in a highly concentrated form unaccompanied by
the generation of gas? This paper has now been declassified, and though it
has been superseded by more complete calculations, it seems appropriate to
publish it as it was first written, without alteration, except for the omission of
a few lines, the addition of this summary, and a comparison with some more
recent experimental work, so that the writings of later workers in this field
may be appreciated.
An ideal problem is here discussed. A finite amount of energy is suddenly
released in an infinitely concentrated form. The motion and pressure of the
surrounding air is calculated. It is found that a spherical shock wave is propagated outwards whose radius R is related to the time t since the explosion
started by the equation
2

1

−1

R = S(γ)t 5 E 5 ρ0 5 ,
where ρ0 is the atmospheric density, E is the energy released and S(γ) a calculated function of γ, the ratio of the specific heats of air.
The effect of the explosion is to force most of the air within the shock front
into a thin shell just inside that front. As the front expands, the maximum
pressure decreases till, at about 10 atm., the analysis ceases to be accurate.
At 20 atm. 45 % of the energy has been degraded into heat which is not
available for doing work and used up in expanding against atmospheric pres-

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
sure. This leads to the prediction that an atomic bomb would be only half as
efficient, as a blast-producer, as a high explosive releasing the same amount
of energy.
In the ideal problem the maximum pressure is proportional to R−3 , and comparison with the measured pressures near high explosives, in the range of
radii where the two might be expected to be comparable, shows that these
conclusions are borne out by experiment.
The propagation and decay of a blast wave in air has been studied for the case
when the maximum excess over atmospheric pressure does not exceed 2 atm.
At great distances R from the explosion centre the pressure excess decays as
in a sound wave proportionally to R−1 . At points nearer to the centre it decays
more rapidly than R−1 . When the excess pressure is 0.5 atm., for instance,
a logarithmic plot shows that it varies as R−1.9 . When the excess pressure
is 1.5 atm. the decay is proportional to R−2.8 . It is difficult to analyse blast
waves in air at points near the explosion centre because the initial shock
wave raises the entropy of the air it traverses by an amount which depends
on the intensity of the shock wave. The passage of a spherical shock wave,
therefore, leaves the air in a state in which the entropy decreases radially so
that after its passage, when the air has returned to atmospheric pressure,
the air temperature decreases with increasing distance from the site of the
explosion. For this reason, the density is not a single-valued function of the
pressure in a blast wave. After the passage of the blast wave, the relationship between pressure and density for any given particle of air is simply the
adiabatic one corresponding with the entropy with which that particle was
endowed by the shock wave during its passage past it. For this reason, it is
in general necessary to use a form of analysis in which the initial position of
each particle is retained as one of the variables. This introduces great complexity and, in general, solutions can only be derived by using step-by-step
numerical integration. On the other hand, the great simplicity which has
been introduced into two analogous problems, namely, the spherical detonation wave (Taylor 1950) and the air wave surrounding a uniformly expanding
sphere (Taylor 1946), by assuming that the disturbance is similar at all times,
merely increasing its linear dimensions with increasing time from initiation,
gives encouragement to an attempt to apply similar principles to the blast
wave produced by a very intense explosion in a very small volume.
It is clear that the type of similarity which proved to be possible in the two
above mentioned problems cannot apply to a blast wave because in the latter
case the intensity must decrease with increasing distance while the total energy remains constant. In the former, the energy associated with the motion
increased proportionally to the cube of the radius while the pressure and
velocity at corresponding points was independent of time. The appropriate
similarity assumptions for an expanding blast wave of constant total energy

5

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
are
pressure, p/p0 = y = R−3 f1 ,
density, ρ/ρ0 = ψ ,
3

radial velocity, u = R 2 φ1 ,

(1.1)
(1.2)
(1.3)

where R is the radius of the shock wave forming the outer edge of the disturbance, p0 and ρ0 are the pressure and density of the undisturbed atmosphere.
If r is the radial co-ordinate, η = r/R and f1 , φ1 and ψ are functions of η. It
is found that these assumptions are consistent with the equations of motion
and continuity and with the equation of state of a perfect gas.
The equation of motion is
∂u
∂u
p0 ∂y
+u
=−
.
∂t
∂r
ρ ∂r

(1.4)

Substituting from (1.1), (1.2) and (1.3) in (1.4) and writing f10 , φ01 for




3
p0 f10
0
− 52 dR
−4
0
φ1 + ηφ1 · R
+ R · φ1 φ1 +
= 0.
−
2
dt
ρ0 ψ

∂
f, ∂φ,
∂η 1 ∂η 1

(1.5)

This can be satisfied if
3
dR
= AR− 2 ,
dt

(1.6)

where A is a constant, and


3
p0 f10
0
−A ·
φ1 + ηφ1 + φ1 φ01 +
= 0.
2
ρ0 ψ

(1.7)

The equation of continuity is
∂ρ
∂ρ
+u +ρ·
∂t
∂r



∂u 2u
+
∂r
r


= 0.

Substituting from (1.1), (1.2), (1.3) and (1.6), (1.8) becomes


2
0
0
0
−Aηψ + ψ φ1 + ψ · φ1 + φ1 = 0 .
η
The equation of state for a perfect gas is




∂
∂
+u
· pρ−γ = 0 .
∂t
∂r

(1.8)

(1.9)

(1.10)

6

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
where γ is the ratio of specific heats.
Substituting from (1.1), (1.2), (1.3) and (1.6), (1.10) becomes
A · (3f1 + ηf10 ) +

rf1 0
ψ · (−Aη + φ1 ) − φ1 f10 = 0 .
ψ

(1.11)

The equations (1.7), (1.9) and (1.11) may be reduced to a non-dimensional
form by substituting
f = f1 · a2 /A2 ,
φ = φ1 /A ,

(1.12)
(1.13)

where a is the velocity of sound in air so that a2 = γp0 /ρ0 . The resulting equations, which contain only one parameter, namely, γ, are
1 f0 3
− φ,
γψ
2
0
0
ψ
φ + 2φ/η
=
ψ
η−φ

φ0 (η − φ) =

3f + ηf 0 +

γψ 0
f (−η + φ) − φf 0 = 0 .
ψ

(1.7a)
(1.9a)
(1.11a)

Eliminating ψ 0 from (1.11a) by means of (1.7a) and (1.9a) the equation for
calculating f 0 when f , φ, ψ and η are given is





1
2
0
2
f (η − φ) − f /ψ = f · −3η + φ 3 + γ − 2γφ /η .
(1.14)
2
When f 0 has been found from (1.14), φ0 can be calculated from (1.7a) and
hence ψ 0 from (1.9a). Thus if for any value of η, f , φ and ψ are known their
values can be computed step-by-step for other values of η.

1.2 Shock-Wave Conditions
The conditions at the shock wave η = 1 are given by the Rankine-Hugoniot
relations which may be reduced to the form

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1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

ρ1 γ − 1 + (γ + 1)y1
=
,
ρ0 γ + 1 + (γ − 1)y1
U2
1
· {γ − 1 + (γ + 1)y1 } ,
=
2
a
2γ
u1
2(y1 − 1)
=
,
U
γ − 1 + (γ + 1)y1

(1.15)
(1.16)
(1.17)

where p1 , u1 and y1 represent the values of ρ, u and y immediately behind the
shock wave and U = dR/dT is the radial velocity of the shock wave.
These conditions cannot be satisfied consistently with the similarity assumptions represented by (1.1), (1.2) and (1.3). On the other hand, when y1 is large
so that the pressure is high, compared with atmospheric pressure, (1.15),
(1.16) and (1.17) assume the approximate asymptotic forms
ρ1 γ + 1
=
,
ρ0 γ − 1
2γ
U2
=
y1 ,
2
a
γ+1
u1
2
=
.
U
γ+1

(1.15a)
(1.16a)
(1.17a)

These approximate boundary conditions are consistent with (1.1), (1.2), (1.3)
and (1.6); in fact (1.15a) yields, for the conditions at η = 1,
ψ=

γ+1
,
γ−1

(1.15b)

f=

2γ
,
γ+1

(1.16b)

φ=

2
.
γ+1

(1.17b)

(1.16a) yields

And (1.17a) yields

1.3 Energy
The total energy E of the disturbance may be regarded as consisting of two
parts, the kinetic energy

8

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

R

Z
K.E. = 4π
0

1 2 2
ρu r dr ,
2

and the heat energy
Z
H.E. = 4π
0

R

pr2
.
γ−1

In terms of the variables f , φ, ψ and η



Z 1
Z 1
p0
1
2 2
2
2
ψφ η dη +
f η dη
,
ρ0 ·
·
E = 4πA ·
2
a2 · (γ − 1) 0
0
or since p0 = a2 ρ0 /γ, E = Bρ0 A2 , where B is a function of γ only whose value is
Z 1
Z 1
4π
2 2
B = 2π
ψφ η dη +
·
f η 2 dη .
(1.18)
γ · (γ − 1) 0
0
Since the two integrals in (1.18) are both functions of γ only it seems that for
a given value of γ, A2 is simply proportional to E /ρ0 .

1.4 Numerical Solution for γ = 1.4
When γ = 1.4 the boundary values of f , φ and ψ at η = 1 are from (1.15a),
(1.16a), (1.17a), 76 , 56 and 6. Values of f , φ and ψ were calculated from η = 1.0
to η1 = 0.5, using intervals of 0.02 in η. Starting each step with values of f 0 ,
φ0 , ψ 0 , f , φ and ψ found in previous steps, values of f 0 , φ0 and ψ 0 at the end of
the interval were predicted by assuming that the previous two values form a
0
geometrical progression with the predicted one; thus the (s + 1)th term, fs+1
in
0
0
0 2
0
a series of values of f was taken as fs+1 = (fs ) /fs−1 . With this assumed
value


0
the mean value of f 0 in the sth interval
was taken as 12 fs+1
+ fs0 and the in

0
0
0
crement in f was taken as (0.02) 21 (fs+1
+ fs0 ). The values of fs+1
, φ0s+1 and ψs+1
were then calculated from formulae (1.14), (1.7a) and (1.9a). If they differed
appreciably from the predicted values, a second approximation was worked
0
out, replacing the estimated values of fs+1
by this new calculated value. In
the early stages of the calculation near η = 1 two or three approximations
were made, but in the later stages the estimated value was so close to the
calculated one that the value of f 0 calculated in this first approximation was
used directly in the next stage.
The results are given in table 1.1 and are shown in the curves of figure 1.1.
These curves and also table 1.1 show three striking features: (a) the φ curve
rapidly settles down to a curve which is very nearly a straight line through
the origin, (b) the density curve ψ rapidly approaches the axis ψ = 0, in fact at

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1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
η = 0.5 the density is only 0.007 of the density of the undisturbed atmosphere,
(c) the pressure becomes practically constant and equal to 0.436/1.167 =
0.37 of the maximum pressure. These facts suggest that the solution tends
to a limiting form as η decreases in which φ = cη, φ0 = c = constant, f = 0.436,
0
f 0 , ψ and ψ 0 become small. Substituting for γ1 fψ from (1.7a), (1.14) becomes


f0
3
1
2γφ2
2
0
· (η − φ) = γφ (η − φ) + γφ − 3η + 3 + γ φ −
.
(1.19)
f
2
2
η
Dividing by η − φ (1.19) becomes
f0
2γφ
· (η − φ) = γφ0 − 3 +
.
f
η

(1.20)

If the left-hand side which contains f 0 /f be neglected the approximate solution of (1.20) for which φ vanishes at η = 0 is
(1.21)

φ = η/γ .

The line φ = η/γ is shown in figure 1.1. It will be seen that the points
calculated by the step-by-step method nearly run into this line. The difference
appears to be due to the accumulation of errors in calculation.

1.5 Approximate Formulae
The fact that the φ curve seems to leave the straight line φ = η/γ rather rapidly
after remaining close to it over the range η = 0 to η = 0.5 suggests that an
approximate set of formulae might be found assuming
φ = η/γ + α · η n ,

(1.22)

where n is a positive number which may be expected to be more than, say, 3
or 4. If this formula applies at η = 1,
2
1
+α=
γ
γ+1

or α =

γ−1
;
γ · (γ + 1)

(1.23)

inserting φ = η/γ + αη n , φ0 = 1/γ + nα · η n−1 in (1.20), the value f 0 /f at η = 1 is
f 0 /f = αγ · (n + 2)(γ + 1)/(γ − 1). From (1.14) and (2.15), (2.16), (2.17) the true
2 +7γ−3
value of f 0 /f at η = 1 is 2γ γ−1
. Equating these two forms,
n=

7γ − 1
.
γ2 − 1

(1.24)

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1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

Figure 1.1
– · – curves f and ψ (step-by-step calculation); − − + − −, curve f (approximate
formulae). In the other curves the small dots represent the steps of the calculations, the larger symbols represent approximate formulae for: 4, curve φ; ,
curve φ = η/γ; •, curve ψ

11

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

Table 1.1
Step-by-step calculation for γ = 1.4
η

f

φ

ψ

1.00

1.167

0.833

6.000

0.98

0.949

0.798

4.000

0.96

0.808

0.767

2.808

0.94

0.711

0.737

2.052

0.92

0.643

0.711

1.534

0.90

0.593

0.687

1.177

0.88

0.556

0.665

0.919

0.86

0.528

0.644

0.727

0.84

0.507

0.625

0.578

0.82

0.491

0.607

0.462

0.80

0.478

0.590

0.370

0.78

0.468

0.573

0.297

0.76

0.461

0.557

0.239

0.74

0.455

0.542

0.191

0.72

0.450

0.527

0.152

0.70

0.447

0.513

0.120

0.68

0.444

0.498

0.095

0.66

0.442

0.484

0.074

0.64

0.440

0.470

0.058

0.62

0.439

0.456

0.044

0.60

0.438

0.443

0.034

0.58

0.438

0.428

0.026

0.56

0.437

0.415

0.019

0.54

0.437

0.402

0.014

0.52

0.437

0.389

0.010

0.50

0.436

0.375

0.007

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1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
The values of α and n have now been determined to give the correct values
of f 0 /f ; φ and φ0 at η = 1, ψ 0 is determined by (1.9a) so that all the six correct
values of f , φ, ψ, f 0 , φ0 , ψ 0 are consistent with (1.22) at η = 1. Substituting for
φ from (1.22) in (1.20),
f0
(n + 2)αγ 2 · η n−2
=
.
(1.25)
f
γ − 1 − γα · η n−1
The integral of (1.25) which gives the correct value of f at η = 1 is




2γ 2 + 7γ − 3
γ + 1 η n−1
2γ
−
· log
−
log f = log
γ+1
7−γ
γ
γ

(1.26)

At η = 0.5 this gives f = 0.457 when γ = 1.4. The value calculated by the
step-by-step integration is 0.436, a difference of 5 %.
The approximate form for ψ might be found by inserting the approximate
forms for φ and φ0 in (1.9a). Thus

 Z 1
γ+1
3 + (n + 2)αγ · η n−1
log ψ = log
dη .
(1.27)
−
γ−1
(γ − 1)η − αγ · η n
η
Integrating this and substituting for α from (1.23),




γ+1
3
(γ + 5)
γ + 1 − η n−1
log ψ = log
+
· log η − 2
· log
.
γ−1
γ−1
7−γ
γ

(1.28)

When η is small, this formula gives
ψ = D · η 3/(γ−1) ,

(1.29)

where

log D = log

γ+1
γ−1



(γ + 5)
−2
· log
7−γ



γ+1
γ


.

(1.30)

When γ = 1.4 (1.30) gives D = 1.76 so that
ψ = 1.76 · η 7.5 .

(1.29a)

At η = 0.5 this gives ψ = 0.0097; the step-by-step calculation gives ψ = 0.0073.
At η = 0.8 formula (1.28) gives ψ = 0.387, while table 1.1 gives ψ = 0.370. At
η = 0.9 formula (1.28) gives ψ = 1.24, while the step-by-step solution gives 1.18.
Some points calculated by the approximate formulae are shown in figure 1.1.
In the central region of the disturbance the density decreases proportionally
to r3/(γ−1) ; the fact that the pressure is nearly constant there means that the
temperature increases proportionally to r−3/(γ−1) . At first sight, it might be
supposed that these very high temperatures involve a high concentration of

13

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
Table 1.2
Approximate calculation for γ = 1.666
η

f

φ

ψ

1.00

1.250

0.750

4.00

0.95

0.892

0.680

2.30

0.90

0.694

0.620

1.14

0.80

0.519

0.519

0.63

0.70

0.425

0.445

0.29

0.50

0.379

0.300

0.05

0.00

0.344

0.000

0.00

energy near the centre. This is not the case, however, for the energy per unit
volume of a gas is simply p/(γ −1) so that the distribution of energy is uniform.
Values of f , φ and ψ for γ = 53 calculated by the approximate formulae are
given in table 1.2.

1.6 Blast Wave Expressed in Terms of The Energy of
the Explosion
It has been seen in (1.18) that E /ρ0 A2 is a function of γ only. Evaluating
the integrals in (1.18) for the case for γ = 1.4, and using the step-by-step
calculations, it is found that
Z 1
Z 1
2 2
η φ ψdη = 0.185 and
η 2 f dη = 0.187 .
0

0

The kinetic energy of the disturbance is therefore
K.E. = 2π(0.185)ρ0 A2 = 1.164 · ρ0 A2 ,

(1.31)

while the heat energy is
H.E. =

4π
(0.187)ρ0 A2 = 4.196 · ρ0 A2 ;
(1.4)(0.4)

(1.32)

the total energy is therefore
E = 5.36 · ρ0 A2 .

(1.33)

14

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

1.7 Pressure
The pressure p at any point is
p = p0 R−3 f

2
A2
−3 ρ0 A
= 0.133 · R−3 E f .
=
R
f
a2
γ

(1.34)

The maximum pressure at any distance corresponds with f = 1.166 at r = R.
This is therefore
pmax. = 0.155 · R−3 E .
(1.35)

1.8 Velocity of Air and Shock Wave
The velocity u of the gas at any point is
3

3

1

1

u = R− 2 Aφ = R− 2 E 2 (Bρ0 )− 2 φ .

(1.36)

The velocity of radial expansion of the disturbance is, from (1.6),
3
3
1
1
dR
= AR− 2 = R− 2 E 2 (Bρ0 )− 2 ,
dt
so that, if t is the time since the beginning of the explosion,
1
1
5 1
1
2 5
t = R 2 (Bρ0 ) 2 E − 2 = 0.926 · R 2 ρ02 E − 2 ,
5

(1.37)

(1.38)

when γ = 1.4.
The formulae (1.34) to (1.38) show some interesting features. Though the
pressure wave is conveyed outwards entirely by the air the magnitude of the
pressure depends only on E R−3 and not on the atmospheric density ρ0 . The
1

time scale, however, is proportional to ρ02 . It is of interest to calculate the
pressure-time relationship for a fixed point, i.e. the pressure to which a fixed
object would be subjected as the blast wave passed over it. If t0 is the time
since initiation taken for the wave to reach radius R0 the pressure at time t
at radius R0 is given by
 3
p
R0
f
=
·
,
(1.39)
p1
R
[f ]η=1
where p1 is the pressure in the shock wave as it passed over radius R0 at the
time t0 , R is the radius of the shock wave at time t and η = R0 /R. [f ]η=1 is the
maximum value of f , namely, 1.166 when γ = 1.4. η is related to t/t0 through
(1.38) so that

15

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

Figure 1.2
Pressure-time curve at a fixed point

η = (t0 /t)

2
5

and

p
1
=
p1
[f ]η=1

  65
t0
[f ]η=(t0 /t)2/5 .
t

(1.40)

Values of p/p1 calculated by (1.40) for γ = 1.4 are shown in figure 1.2.

1.9 Temperature
The temperature T at any point is related to the pressure and density by the
relationship
pρ0
1.33 · E R−3
T
=
=
,
T0
p0 ρ
p0 ψ

when γ = 1.4 .

(1.41)

Since f tends to a uniform value 0.436 in the central region (r < 12 R) and ψ
tends to the value ψ = 1.76η 7.5 , T tends to the value

16

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

T
E R−3 (0.133)(0.436) −7.5
T0
=
η
= 0.033 · E R−3 η 7.5 .
T0
p0
1.76
p0

(1.42)

Thus the temperature near the centre is very high; for instance, when the
wave has expanded to such a distance that the pressure in the central region
is reduced to atmospheric pressure, p0 = (0.133)(0.436)E R−3 , then (1.42) gives
T /T0 = η −7.5 /1.76 and at η = 0.5, η −7.5 = 0.181 so that T /T0 = 103. If T0 = 273◦ ,
T = 27, 000◦ . The temperature left behind by the blast wave is therefore very
high, but the energy density is not high because the density of the gas is
correspondingly low.

1.10 Heat Energy Left in the Air After it has returned to
Atmospheric Pressure
The energy available for doing mechanical work is less than the total heat
energy of the air. The heated air left behind by the shock wave can in fact only
do mechanical work by expanding down to atmospheric pressure, whereas
to convert the whole of the heat energy into mechanical work by adiabatic
expansion, the air would have to be expanded to an infinite extent until the
pressure was zero. After the blast wave has been propagated away and the
air has returned to atmospheric pressure it is left at a temperature T1 which
is greater than T0 , the atmospheric temperature. The energy required to raise
the temperature of air from T0 to T1 is therefore left in the atmosphere in a
form in which it is not available for doing mechanical work directly on the
surrounding atmosphere. This energy, the total amount of which will be denoted by E1 , is wasted as a blast-wave producer.
The energy so wasted at any stage of the disturbance can be calculated by
finding the temperature T1 to which each element of the blast wave would be
reduced if it were expanded adiabatically to atmospheric pressure. If T is the
temperature of an element of the blast wave
T
=
T1



p
p0

(γ−1)/γ


=

A2 −3
fR
a2

(γ−1)/γ
.

Also
T
pρ0
f A2 −3
=
=
R ,
T0
p0 ρ
ψ a2

17

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
hence
f 1/γ
T1
=
·
T0
ψ



A2 R−3
a2

1/γ
.

(1.43)

The total heat energy per unit mass of air at temperature T1 is
T1
T1 p0
· (gas constant) =
.
γ−1
(γ − 1)ρ0 T0
The increase in heat energy per unit mass over that which
 contained be the air
p0
T1
fore the passage of the disturbance is therefore (γ−1)(ρ0 ) · T0 − 1 . The increase


p0 ψ
T1
per unit volume of gas within the disturbed sphere is therefore (γ−1) · T0 − 1 .
Hence from (1.43) the total energy wasted when the sphere has expanded to
radius R is
)
 2 −3 1/γ
Z 1(
AR
1/γ
3 p0
·
f ·
− ψ η 2 dη .
(1.44)
E1 = 4πR
2
γ−1 0
a
This expression may conveniently be reduced to non-dimensional form by
dividing by the total energy E of the explosion which is related to A by the
formula (1.18). After inserting a2 /γ for p0 /ρ0 , this gives
#

 " 2 −3 1/γ Z 1
Z 1
4π
a2
E1
AR
=
·
·
·
f 1/γ η 2 dη −
ψη 2 dη . (1.45)
E
B(γ − 1) · γ
A2 R−3
a2
0
0
4
πρ0 R3
3

is the total mass of air in the sphere of radius R.
R1
This is also 4πR3 ρ0 · 0 ψη 2 dη, so that
Z 1
1
ψη 2 dη = .
3
0

(1.46)

The quantity A2 R−3 /a2 is related to the maximum pressure p1 at the shock
wave by the equation
y1 =

p1
A2 R−3
=
[f ]n=1 ,
p0
a2

where y1 is the pressure in the shock wave expressed in atmospheres. (1.46)
therefore reduces to
"
#
1/γ
Z 1
E1
4π
f
1
1/γ
2
.
(1.47)
=
[f ]η=1 · y1 ·
η dη −
E
B(γ − 1) · y1
fη=1
3
0

18

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
Table 1.3

y1 (atm. at shock 10,000
wave)

1,000

100

50

20

10∗

5∗

E1 /E (proportion
of energy wasted)

0.069

0.132

0.240

0.281

0.325

0.333∗

0.28∗

(E1 + E2 )/E

0.096

0.189

0.337

0.393

0.455

–

–

∗

Formulae inaccurate when y1 < 10.

For γ = 1.4 numerical integration gives
Z
B = 5.36 see (1.18) , fn=1 = 1.166 ,

1

f 1/γ η 2 dη = 0.219 .

0

(1.47) reduces therefore to
i
1 h
E1
1/1.4
=
· 0.958 · y1
− 1.63 .
E
y1

(1.48)

Some values of E1 /E are given in the second line of table 1.3.
It is clear that E1 /E must continually increase as R increases, and y1 decreases because the contribution to E1 due to the air enclosed in the shockwave surface when its radius is R2 , say, remains unchanged when this air
subsequently expands. A further positive contribution to E1 is made by each
subsequent layer of air included within the disturbance. The fact that formula
(1.48) gives a value of E1 /E which increases till y1 is reduced to 10 and then
subsequently decreases is due to the inaccuracy of the approximate boundary
conditions (1.15a), (1.16a) and (1.17a), which are used to replace the true
boundary conditions (1.15), (1.16) and (1.17).
When γ = 1.4 and y1 = 10 the true value of ρ1 /ρ0 is 3.8 instead of 6.0 as is
assumed, the true value of U 2 /a2 is 8.7 instead of 8.6 and the true value of
u1 /U is 0.74 instead of 0.83.
When γ = 5 the errors are much larger, namely, ρ1 /ρ0 is 2.8 instead of 6.0,
U 2 /a2 is 4.4 instead of 4.3, and u1 /U is 0.64 instead of 0.83. The proportion
of the energy wasted, namely, E1 /E , is shown as a function of y1 in figure 1.3,
y1 being plotted on a logarithmic scale.
It will be seen that the limiting value of E1 /E is certainly greater than 0.32, its
value for y1 = 20. It is not possible to find out how much greater without tracing the development of the blast wave using laborious step-by-step methods
for values of η, less than, say, 10 or 20.

19

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

Figure 1.3
Heavy line, (E1 + E2 )/E ; thin line E1 /E . (E1 + E2 )/E is the proportion of the initial
energy, which is no longer available for doing work in propagation. E2 /E is the
work done by heated air expanding against atmospheric pressure (see note
added October 1949 (p. 172))

20

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

1.11 Comparison with high Explosives
The range within which any comparison between the foregoing theory and the
blast waves close to actual high explosives can be made is severely limited.
In the first place the condition that the initial disturbance is so concentrated
that the mass of the material in which the energy is originally concentrated
is small compared with the mass of the air involved in the disturbance at
any time, limits the comparable condition during a real explosion to one in
which the whole mass of air involved is several times that of the explosive. In
the second place the modified form of the shock-wave condition used in the
analysis is only nearly correct when the rise in pressure at the shock-wave
front is several -say at least 5 or 10- atmospheres. In a real explosive this
limits the range of radii of shock wave over which comparison could be made
to narrow limits. Thus with l0 lb. of C.E.∗ the radius R at which the weight
of explosive is equal to that of the air in the blast wave is 3 ft., while at 3.8 ft.
the air is only double the weight of the explosive. The pressure in the blast
wave at a radius of 6 ft. was found to be 9 atm., while at 8 ft. it was about
5 atm. It seems, therefore, that in this case the range in which approximate
agreement with the present theory could be expected only extends from 3.8
to 6 ft. from the l0 lb. charge.
Taking the energy released on exploding C.E. to be 0.95 kcal./g. the energy released when 10 lb. is exploded is 1.8 · 1014 ergs. If this energy had
been released instantaneously at a point as in the foregoing calculations the
maximum pressure at distance R given by (1.35) is
pmax. R3 = (0.155)(1.80 · 1014 ) = 2.79 · 1013 ergs.

(1.49)

Expressed in terms of atmospheres pmax. is identical with y1 . If R is expressed
in feet, (1.49) becomes
y1 R 3 =

2.79 · 1013
= 9.9 · 102 .
(30.45)3 · 106

(1.50)

The line representing this relationship on a logarithmic scale is shown in
figure 1.4. Though no suitable pressure measurements have been made, the
maximum pressure in the blast from l0 lb. of C.E. has been found indirectly
by observing the velocity of expansion of the luminous zone and, at greater
radii, the blast-wave front. These values taken from a curve given in a report
on some experiments made by the Road Research Laboratory are given in
table 1.4. The observed values of U in ft./sec. given in column 2 of this table
and the values of y1 (in atmospheres) found from the shock-wave formulae
are given in column 3, where they are described as observed values though
they were not observed directly. The “observed” values are shown in figure

21

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

Figure 1.4
Blast pressures near 10 lb. charge of C.E. compared with calculated blast pressures due to instantaneous release of energy of 10 lb. C.E. at a point. The
numbers against the points on the curve give distances in feet
1.4. The values of y1 calculated from (1.50) are given in column 4.
Though the observed values are higher than those calculated, it will be noticed that in the range of radii 3.8 to 6 ft., in which comparison can be made,
the observed curve is nearly parallel to the theoretical line y1 R3 = 990. In this
range, therefore, the intensity of the shock wave varies nearly as the inverse
cube of the distance from the explosion. The fact that the observed values are
about twice as great as those calculated on the assumption that the energy
is emitted instantaneously at a point may perhaps be due to the fact that the
measurements used in table 1.4 correspond with conditions on the central
plane perpendicular to the axis of symmetry of the cylindrical charge used.
The velocity of propagation of the luminous zone is greater on this plane and
on the axis of symmetry than in other radial directions so that the pressures
deduced in column 3 of table 1.4 are greater than the mean pressures at the
corresponding radii.
On the other hand, it has been seen that by the time the maximum pressure
has fallen to 20 atm., 32 % of the energy has been left behind in the neighbourhood of the concentrated explosive source, raising the air temperature
there to very high values. The burnt gases of a real high explosive are at a
very much lower temperature even while they are at the high pressure of the
detonation wave. Their temperature is still lower when they have expanded

22

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
Table 1.4
Calculated values y1 from equation (1.50)
y1 (atm.)
R (ft.) U (ft./sec.) observed
with C.E.

y1 calculated for
concentrated
explosion by (1.50)

8

2350

6.2

.

6

3100

9.3

4.6

5

3800

14.0

7.9

4

4820

22.6

15.5

3

6200

37.5

.

2

8540

71.8

.






range of
comparison




Table 1.5
Pressure y1 p0 at distance R from explosion of weight W of T.N.T. - R.D.X. mixture
1

1

R/W 3 (ft./lb. 3 )
U (thousands ft./sec.)
y1 (atm.)
1
3

0.5

1.0

1.5

2.0

2.5

3.0

3.5

14.3

11.0

8.4

6.6

5.1

4.0

3.3

68.3

42.0

25.1

15.4

10.5

16.2

21.5

27.0

32.4

37.8

198
4

1
3

R/E · 10 (cm./ergs. )

5.39

117
10.8

adiabatically to atmospheric pressure, so that little heat energy is left in them.
To this extent, therefore, a real high explosive may be expected to be more efficient as a blast producer than the theoretical infinitely concentrated source
here considered.
Note added, October 1949. The data on which the comparison was based
between the pressures deduced by theory and those observed near detonating explosives were obtained in 1940. More recent data obtained at the Road
Research Laboratory using a mixture of the two explosives R.D.X. and T.N.T.
have been given by Dr. Marley. These are given in table 1.5, which shows
1
the values of U observed for various values of R/W 3 . R the distance from the
explosive is expressed in feet and W its weight in pounds. The third line in
table 1.5 shows the result of deducing y1 from U using γ = 1.4 in (1.16) and
a = l, l00 ft./sec. in (1.16).
For comparison with the concentrated point-source explosion, the value of

23

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
1

1

RE − 3 expressed in cm. (erg.)− 3 is found by multiplying the figures in line 1,
1
1
−3
table 1.5, by 30.41 ·
. The first factor converts ft. (lb.)− 3
1 = 1.078 · 10
7
(454) 3

(1,200·4.2·10 ) 3

1

to cm. (g.)− 3 , and the second replaces 1 g. by the equivalent energy released
1
by this explosive, namely, 1,200 cal. The values of RE − 3 are given in line 4,
1
table 1.5. In figure 1.5 values of log10 y1 are plotted against log10 RE − 3 , and
the theoretical values for a point source of the same energy as the chemical
explosive are plotted in the same diagram. Comparing figures 1.4 and 1.5
it seems that the more recent shock-wave velocity results are qualitatively
similar to the older ones in their relation to the point-source theory. The
range of values of y1 for which comparison between theory and observation
might be significant, is marked in figure 1.5.
It will be seen that the chemical explosive is a more efficient blast producer
than a point source of the same energy. The ratio of the pressures in the
range of comparison is about 3 to 1. This is more than might be expected in
view of the calculation of E1 /E as a function of y1 which is given in table 1.3.
E1 is the heat energy which, is unavailable for doing mechanical work after
expanding to pressure p0 . Of the remaining energy, E − E1 a part E2 is used in
doing work against atmospheric pressure during the expansion of the heated
air. The remaining energy, namely, E − E1 − E2 , is available for propagating
the blast wave.
To find E2 , the work done byunit volume
of the gas at radius ηR in expanding

T1 p
to atmospheric pressure is T p0 − 1 p0 . From (1.43)
T1
=
T



A2 −3
R f
a2

(1−γ)/γ
and

p
A2 R−3
=f
,
p0
a2

hence
E2 = 4πR p0 ·
3

Z

1

0

but

A2
a2 R3

=

y1
,
(f )η=1




2
−3 A
R f 2 − 1 η 2 dη ,
a

(1.51)

so that

4πR3 p0
E2
=
·
E
E

"

y1
fη=1

1/γ Z
·
0

1

f 1/γ η 2 dη −

Z

1

#
η 2 dη .

(1.52)

0

The first integral has already been calculated and found to be 0.219 when
γ = 1.4 (see (1.47) and (1.48)). Substituting for pmax. from (1.35),
"
#
(1−γ)/γ
E2
0.219 · y1
1
= 4π · (0.155) ·
−
.
(1.53)
E
(1.166)1/γ
3y1

24

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion
Values of (E2 + E1 )/E have been added as a third line in table 1.3 and a corresponding curve to figure 1.3.

1.12 References
Taylor, Sir Geoffrey 1946 Proc. Roy. Soc. A, 186, 273.
Taylor, Sir Geoffrey 1950 Proc. Roy. Soc. A, 201, 175.

25

1 The formation of a blast wave by a very intense explosion I. Theoretical
discussion

Figure 1.5
Blast pressures near a chemical explosive (R.D.X. + T.N.T.) compared with theoretical pressure for concentrated explosion with same release of energy. Heavy
line (upper part) is taken from shock-wave velocity measurements. Heavy line
(lower part) is from piezo-electric crystals. Thin line, y1 = 0.155 · E /(p0 R3 ). The
figures against the points represent the ratio of the mass of the air within the
shock wave to the mass of the explosive

26

2 The formation of a blast wave by a very
intense explosion. II. The atomic explosion
of 1945
2.1 Summary and Introduction
Photographs by J.E. Mack of the first atomic explosion in New Mexico were
measured, and the radius, R, of the luminous globe or “ball of fire” which
spread out from the centre was determined for a large range of values of t,
the time measured from the start of the explosion. The relationship predicted
5
in chapter 1, namely, that R 2 would be proportional to t, is surprisingly
5
accurately verified over a range from R = 20 to 185 m. The value of R 2 t−1 so
found was used in conjunction with the formulae of chapter 1 to estimate the
energy E , which was generated in the explosion. The amount of this estimate
depends on what value is assumed for γ, the ratio of the specific heats of air.
Two estimates are given in terms of the number of tons of the chemical
explosive T.N.T. which would release the same energy. The first is probably
the more accurate and is 16,800 tons. The second, which is 23,700 tons,
probably overestimates the energy, but is included to show the amount of
error which might be expected if the effect of radiation were neglected and
that of high temperature on the specific heat of air were taken into account.
Reasons are given for believing that these two effects neutralize one another.
After the explosion, a hemispherical volume of very hot gas is left behind and
Mack’s photographs were used to measure the velocity of rise of the glowing
centre of the heated volume. This velocity was found to be 35 m./sec.
Until the hot air suffers turbulent mixing with the surrounding cold air, it may
be expected to rise like a large bubble in water. The radius of the “equivalent
bubble” is calculated
√ and found to be 293 m. The vertical velocity of a bubble
of this radius is 23 · g 29, 300 or 35.7 m./sec. The agreement with the measured
value, 35 m./sec., is better than the nature of the measurements permits one
to expect.

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.2 Comparison with Photographic Records of the
First Atomic Explosion
Two years ago, some motion picture records by Mack (1947) of the first atomic
explosion in New Mexico were declassified. These pictures show not only the
shape of the luminous globe which rapidly spread out from the detonation
centre, but also gave the time, t, of each exposure after the instant of initiation. On each series of photographs, a scale is also marked so that the
rate of expansion of the globe, or “ball of fire”, can be found. Two series of
declassified photographs are shown in figure 2.5, plate 7.
These photographs show that the ball of fire assumes at first the form of a
rough sphere, but that its surface rapidly becomes smooth. The atomic explosive was fired at a height of 100 ft. above the ground and the bottom of the
ball of fire reached the ground in less than 1msec. The impact on the ground
does not appear to have disturbed the conditions in the upper half of the
globe which continued to expand as a nearly perfect luminous hemisphere
bounded by a sharp edge which must be taken as a shock wave. This stage of
the expansion is shown in figure 2.6, plate 8 which corresponds with t = 15
msec. When the radius R of the ball of fire reached about 130m., the intensity
of the light was less at the outer surface than in the interior. At later times,
the luminosity spread more slowly and became less sharply defined, but a
sharp-edged dark sphere can be seen moving ahead of the luminosity. This
must be regarded as showing the position of the shock wave when it ceases
to be luminous. This stage is shown in figure 2.7, plate 9, taken at t = 127
msec. It will be seen that the edge of the luminous area is no longer sharp.
The measurements given in column 3 of table 2.1 were made partly from
photographs in Mack (1947), partly from some clearer glossy prints of the
same photographs kindly sent to me by Dr. N. E. Bradbury, Director of Los
Alamos Laboratory and partly from some declassified photographs lent me by
the Ministry of Supply. The times given in column 2 of table 2.1 are taken
directly from the photographs. To compare these measurements with the
analysis given in chapter 1 of this paper, equation (1.38) was used. It will be
seen that if the ball of fire grows in the way contemplated in my theoretical
5
analysis, R 2 will be found to be proportional to t. To find out how far this
prediction was verified, the logarithmic plot of 52 log R against log t shown in
figure 2.1 was made. The values from which the points were plotted are given
in table 2.1. It will be seen that the points lie close to the 45◦ line which is
drawn in figure 2.1. This line represents the relation
5
log10 R − log10 t = 11.915 .
2

(2.1)

28

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Table 2.1
Radius R of blast wave at time t after the explosion
authority

5
2

t (msec.)

R (m.)

log10 t

0.10

11.1

4.0

3.045

7.613

0.24

19.9

4.380

3.298

8.244

0.38

25.4

4.580

3.405

8.512

0.52

28.8

4.716

3.458

8.646

0.66

31.9

4.820

3.504

8.759

0.80

34.2

4.903

3.535

8.836

0.94

36.3

4.973

3.560

8.901











strip of declassified 



photographs lent
by Ministry of Sup-



ply










1.08

38.9

3.033

3.590

8.976

1.22

41.0

3.086

3.613

9.032

1.36

42.8

3.134

3.631

9.079

1.50

44.4

3.176

3.647

9.119

1.65

46.0

3.217

3.663

9.157

1.79

46.9

3.257

3.672

9.179

1.93

48.7

3.286

3.688

9.220














3.26

59.0

3.513

3.771

9.427

3.53

61.1

3.548

3.786

9.466

3.80

62.9

3.580

3.798

9.496

4.07

64.3

3.610

3.809

9.521

4.34

65.6

3.637

3.817

9.543

4.61

67.3

3.688

3.828

9.570

106.5

2.176

4.027

10.068

130.0

2.398

4.114

10.285

145.0

2.531

4.161

10.403

175.0

2.724

4.243

10.607

185.0

2.792

4.267

10.668
















strip of small
images MDDC221 













strip of small
images from
MDDC221

large single
photographs
MDDC221















15.0






 25.0
34.0




53.0



 62.0

log10 R

log10 R

29

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945
The ball of fire did therefore expand very closely in accordance with the theoretical prediction made more than four years before the explosion took place.
This is surprising, because in those calculations it was assumed that air
behaves as though γ, the ratio of the specific heats, is constant at all temperatures; an assumption which is certainly not true.
At room temperatures γ = 1.40 in air, but at high temperatures γ is reduced
owing to the absorption of energy in the form of vibrations which increases
Cv . At very high temperatures, γ may be increased owing to dissociation. On
the other hand, the existence of very intense radiation from the center and
absorption in the outer regions may be expected to raise the apparent value
of γ. The fact that the observed value of R5 t−2 is so nearly constant through
the whole range of radii covered by the photographs of the ball of fire suggests
that these effects may neutralize one another, leaving the whole system to
behave as though γ has an effective value identical with that which it has
when none of them are important, namely, 1.40.

2.3 Calculation of the Energy Released by the
Explosion
The straight line in figure 2.1 correspond with
R5 t−2 = 6.67 · 102 (cm.)5 sec.−2 .
The energy, E , is then from equation (1.18) of chapter 1


4π
2
I2 ,
E = ρ0 A · 2πI1 +
γ · (γ − 1)

(2.2)

(2.3)

where
1

Z

ψφ2 η 2 dη

I1 =

(2.4)

0

and
Z
I2 =

1

f η 2 dη ,

(2.5)

0

where f , φ and ψ are non-dimensional quantities proportional to pressure,
velocity and density which are defined in chapter 1.
A is found by integrating equation (1.6) of chapter 1, so that

30

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Figure 2.1
5
Logarithmic plot showing that R 2 is proportional to t

31

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2 5
A = R 2 t−1 .
5

(2.6)

Writing


4
4π
K=
· 2πI1 +
I2 ,
25
γ · (γ − 1)
E = Kρ0 R5 t−2 .

(2.7)
(2.8)

It was shown in chapter 1 that I1 and I2 are functions of γ only. For γ =
1.40, their values were found to be I1 = 0.185, I2 = 0.187, using step-by-step
methods for integrating the equations connecting f , φ and ψ. Using the
approximate formulae (1.22) to (1.30) of chapter 1, values of f , φ and ψ were
calculated in chapter 1 for γ = 1.667 and are there given in table 1.2 of chapter
1. Further calculations have now been made for γ = 1.20 and γ = 1.30 using
the approximate formulae. The results for γ = 1.30 are given in table 2.2 and
are shown in figure 2.2. The corresponding values of I1 , I2 and K are given in
table 2.3; K is shown as a function of γ in figure 2.3.

2.4 Energy of the First Atomic Explosion in New
Mexico
Having determined R5 t−2 and assuming that ρ0 may be taken as 1.25 · 10−3
g./cm.3 , the figures in table 2.3 were used to determine E from equations
(2.8) and (2.2). Different values are found for different assumed values of
γ. These are given, expressed in ergs, in line 5 of table 2.3. It has become
customary to describe large explosions by stating the weight of T.N.T. which
would liberate the same amount of energy. Taking 1 g. of T.N.T. as liberating
1,000 calories or 4.18 · 1010 ergs, 1 ton will liberate 4.25 · 1016 ergs. The T.N.T.
equivalents found by dividing the figures given in line 5 of table 2.3 by 4.25·1016
are given in line 6.
It will be seen that if γ = 1.40, the T.N.T. equivalent of the energy of the
New Mexico explosion, or more strictly that part of the energy which was not
radiated outside the ball of fire, was 16,800 tons.

32

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Table 2.2
Calculation for γ = 1.30 using formulae 1.22 to 1.30 of chapter 1
η

f

φ

ψ

T /T1

1.0

1.130

0.869

7.667

1.00

0.98

0.896

0.833

4.534

1.34

0.96

0.772

0.801

2.989

1.75

0.94

0.666

0.772

2.067

2.19

0.92

0.606

0.745

1.454

2.73

0.90

0.563

0.721

1.058

3.61

0.85

0.499

0.669

0.509

6.76

0.80

0.468

0.623

0.255

12.5

0.75

0.453

0.580

0.128

24.0

0.70

0.445

0.540

0.063

48.0

0.65

0.441

0.501

0.029

103

0.60

0.439

0.462

0.013

229

0.55

0.438

0.423

0.006

.

0.50

0.438

0.386

0.002

.

0.45

0.438

0.347

0.001

.

0.40

0.438

0.309

0.000

.

Table 2.3
Calculated constants used in determining the Energy E of the explosion with a
range of assumed values for γ
γ

1.20

1.30

1.40

1.667

I1

0.259

0.221

0.185

0.123

I2

0.175

0.183

0.187

0.201

K

1.727

1.167

0.856

0.487

14.4

9.74

7.14

4.06

34, 000

22, 900

16, 800

9, 500

E · 1020 erg.
T.N.T. equivalent (tons)

33

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Figure 2.2
Distribution of radial velocity φ, pressure f , density ψ and temperature T /T1 for
γ = 1.30 expressed in non-dimensional form

34

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Figure 2.3
Variation of K with γ
35

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.5 An Alternative Possibility
If the effect of radiation, which cannot be estimated, is disregarded, a mean
value of γ might be taken which is appropriate to the temperature calculated
5
to correspond with the mean value of R in the range over which R 2 t−1 is nearly
constant. The least value of R which lies near the line in figure 2.1 is R = 20
m. and the greatest is R = 185 m. The mean value is therefore approximately
100 m. It will be found that the value of γ appropriate to the temperature
behind the shock wave at 100 m. is about 1.3. The pressure at any point is
from equations (1.1), (1.6) and (1.12) of chapter 1
p = p0 R−3 f1 = p0 R−3

A2
4 ρ0 −3
R f (R5 f −2 ) .
f=
2
a
25 γ

For γ = 1.3, the value of f behind the detonation front is 1.13, so that
p = 9.3 · 1022 ρ0 R−3
The temperature T1 at that point is given by
p ρ0
T1
=
,
T0
ρ p0
where T0 is the undisturbed atmospheric temperature. When
γ = 1.3, ρ0 /ρ = 1/ψ = 1/7.667,
Thus

and if ρ0 = 0.00125, p/p0 = 7.43 · 1013 R−3 .

T1
= 0.97 · 1013 R−3 .
T0

(2.9)

At R = 100 m., T1 /T0 = 9.7, so that if T0 = 15◦ C = 288◦ K, T1 = 2,800◦ K.
Values of Cp at temperatures up to 5,000◦ K have been calculated for nitrogen
by Johnston & Davis (1934) and for oxygen by Johnston & Walker (1935).
At 2,800◦ K, Cp is given for nitrogen as 8.82 and for oxygen as 9.43, so that
for air Cp = 8.92. Since there is little dissociation at that temperature, it
seems that Cv = Cp − R = 6.92 and γ = 8.92
= 1.29. Thus the use of γ = 1.30 for
6.92
calculating the temperature at R = 100 m. is justified when effects of radiation
are neglected.
Using the curve, figure 2.3, it seems that the value of K appropriate to γ =
1.29 is 1.21. Using ρ0 = 0.00125 and R5 t−2 = 6.67 · 1023 , equation (2.8) gives
E = 1.01 · 1021 ergs and the T.N.T. equivalent is 23,700 tons.
Of these two alternative estimates, it seems that the first, namely, 16,800
tons, is the more likely to be accurate.

36

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.6 Some Dynamical Features of the Atomic
Explosion
It will be seen in figure 1.1 of chapter 1 that if γ = 1.4 the air density is a maximum at the shock wave front where it reaches six times atmospheric density.
Within the shock wave, the density falls rapidly till at a radius of about 0.6 · R
it is nearly zero. Within the radius 0.6 · R the gas has a radial velocity which
is proportional to the distance from the centre, a very high temperature, and
a uniform pressure about 0.43 time the maximum pressure.
The maximum pressure at the shock front is found by inserting the value of
E from line 5 of table 1.3 in the formula (1.35) of chapter 1. The pressure
-expressed in atmospheres- is
p
= 0.155R−3 (7.14 · 1020 ) · 106 = 1.11 · 1014 R−3 .
p0

(2.10)

At R = 30 m. this is 4,100 atm., or 27 tons/sq.in. At R = 100 m. the pressure
would be 1 ton/sq.in. These pressures are much less than would act on rigid
bodies exposed to such blasts, but the pressures on obstacles depend on their
shape so that no general statement can be made on this subject.
The temperature rises rapidly as the centre is approached, in fact the ratio
T /T1 , T1 being the temperature just inside the shock wave, is equal to

 
 

f
ψη=1
p1 ρ 1
p p0 ρ1 ρ0
=
=
·
.
(2.11)
p ρ
p0 p1 ρ0 ρ
fη=1
ψ
The values of T /T1 for γ = 1.3, namely
table 2.2, and are shown in figure 2.2.

7.66f
,
1.13ψ

are given in the fifth column of

37

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.7 The initial rate of rise of air from the seat of the
explosion
When the shock wave had passed away from the ball of fire, it left a cloud
of very hot air, which then rapidly rose. Mack (1947, p. 37) gives a rough
picture of the process in a series of diagrams representing the outlines of
the boundary of the heated region, so far as his photographs could define
them, at successive times from t = 0.1 to t = 15.0 sec. It is not possible to
know exactly what these outlines represent, though in the later numbers of
the series they seem to show the limits of the region to which dust thrown off
from the ground and sucked into the ascending column of air has penetrated.
This dust rapidly expands into a roughly spherical shape owing, to turbulent
diffusion or convection currents in the central region. The radius of the outer
edge of the glowing region is not the same in all photographs taken nearly
simultaneously, but the height of its apparent centre seems to be consistent
when photographs taken simultaneously from different places are compared.
The heights, h, of the top of the illuminated column and their radii, b, were determined, so far as was possible, from Mack’s published photographs. These
rough measurements are given in table 2.4. The height of the centre of the
glowing area is taken as h − b, and points corresponding with those given
in table 2.4 are plotted in figure 2.4. It will be seen that the centre of the
glowing volume seems to rise at a regular rate. The line drawn in figure 2.4
corresponds with a vertical velocity of
U = 35 m./sec.

(2.12)

Table 2.4
Height, h, and radius, b, of the glowing region from 3 12 to 15 sec. after the
explosion

authority
Mack MDDC221,
sketches on p. 37

photographs on p. 38

t (sec.)
3.5
8.0
10.0
15.0
14.8

b
(radius)
(m.)
160
240
300
360
550

h
(height
of top)
375
688
810
1060
1200

h−b
(height
of centre)
215
448
510
700
650

38

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Figure 2.4
Height of centre of glowing region from 3 12 to 15 sec. after the explosion.
from diagram p. 37,  from photograph p. 38 of MDDC221

39

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Figure 2.5
Succession of photographs of the “ball of fire” from t = 0.10 msec. to 1.93 msec.
40

2 The formation of a blast wave by a very intense explosion. II. The atomic explosion of 1945

Figure 2.6
The ball of fire at t = 15 msec., showing the sharpness of its edge

41

2 The formation of a blast wave by a very intense explosion. II. The atomic explosion of 1945

Figure 2.7
The glowing volume t = 127 msec. showing an indefinite edge and hemispherical shape

42

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.8 Distribution of Air Density after the Explosion
To give a dynamical description of the rise of hot air from the seat of the
explosion it is first necessary to know the distribution of density immediately
after the shock wave has passed away and left hot air at atmospheric
pressure. Formulae are given in chapter 1 for the temperature ultimately
attained by air, which passed through the shock wave when its pressure
was y1 p0 , but the position at which this air comes to rest when atmospheric
pressure was attained was not discussed. If the ground had not obstructed
the blast wave, the distribution of temperature would evidently be spherical.
It has been pointed out, however, that the shape of the upper half of the
blast wave has not been affected by the presence of the ground. The same
thing seems to be approximately true of the temperature distribution, for
Mack publishes a photograph showing the luminous volume at t = 0.127 sec.
when the shock wave had moved well away from the very hot area. This is
reproduced in figure 2.7, plate 9. It will be seen that the glowing air occupies
a nearly hemispherical volume, the bottom half of the sphere being below
the ground. It seems that it may be justifiable to assume that most of the
energy associated with the part of the blast wave, which strikes the ground
is absorbed there. In that case, we may neglect the effect of shock waves
reflected from the ground and consider the temperature distribution as being
that calculated in part I for an unobstructed wave. With this assumption, the
distribution of density will be calculated.
The following symbols will be used:
T0 , p0 , ρ0 , the temperature, pressure and density in the undisturbed air,
T , y1 p0 , ρ, R, the temperature, pressure, density and radius at the shock
wave,
T1 , ρ1 , r, the temperature, density and radius after the pressure has become
atmospheric.
From (1.43), chapter 1,
T
(γ−1)/γ
= y1
=
T1



A2 f R−3
α2

(γ−1)/γ
(2.13)

and
f A2 −3
T
=
R ,
T0
ψ α2

(2.14)

so that
T1
T T0
1
=
= ·
T0
T1 T
ψ



A2 f R−3
α2


,

(2.15)

43

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945
and since
A2
4E
,
=
2
α
25Kγp0
 −3/γ
R
T1
=β ·
,
T0
r0

(2.16)

where
1
β= ·
ψ



4f
25Kγ

1/γ
(2.17)

,

and r0 is the length defined by
r03 = E /p0 .

(2.18)

r0 is introduced in order to make the equations non-dimensional, ψ and f
have their values at the shock wave front. When γ = 1.40, f = 1.167, ψ = 6.0;
K = 0.856 (see table 2.3), so that
(2.19)

β = 0.044
(2.16) may be written
 γ
 γ
 3
T0
T0
R
γ
=β ·
= 0.1265 ·
r0
T1
T1

when

γ = 1.40 .

(2.20)

To find the values of r and T1 /T0 corresponding with a given value of y1 the
equation of continuity must be used. This is
ρ0 R2 dR = ρ1 r2 dr ,

(2.21)

and since T0 /T1 = ρ1 /ρ0 , (2.21) can be integrated after substitution from (2.20),
thus
 3

  γ−1
r
γ
T0
γ
=β ·
·
,
(2.22)
r0
γ−1
T1
and from (1.35) of chapter 1 and (2.20)


R
r0

3

0.155
=
= βγ ·
y1



T0
T1

γ
.

(2.23)

Eliminating T0 /T1 from (2.22) and (2.23) and using values appropriate to
γ = 1.40

44

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945
Table 2.5
Density ρ1 at radius r expressed as a proportion of ρ0 , the undisturbed density
y1 (atm.)

r/r0

ρ1 /ρ0
uncorrected

C

ρ1 /ρ0
corrected

R/r0

10

(0.360)

1.155

0.753

0.873

0.249

20

0.337

0.704

0.858

0.605

0.198

30

0.325

0.527

0.900

0.475

0.173

40

0.316

0.430

0.923

0.397

0.157

60

0.304

0.321

0.947

0.305

0.137

100

0.289

0.223

0.968

0.216

0.116

500

0.247

0.071

0.071

0.068

5000

0.200

0.013

0.013

0.031



r
r0

3

= 0.0906 · y1−0.2857

(2.24)

ρ0
T1
=
= 0.167 · y10.7143 .
ρ1
T0

(2.25)

and

Values of R/r0 , ρ1 /ρ0 , r/r0 calculated for a range of values of y, are given in
table 2.5.
It has been pointed out that T1 /T0 contains two factors, T /T0 and T1 /T . T /T0
represents the temperature change through the shock wave and is equal
γ−1
to ρ0 y1 /ρ. In calculating this the approximate expression ρρ0 = γ+1
was used
instead of the true value
ρ0
γ + 1 + (γ − 1)y1
=
.
ρ
γ − 1 + (γ + 1)y1

(2.26)

The proportional error in calculating T /T0 for a given values of y1 is therefore
equal to the proportional error in using (γ − 1)/(γ + 1) instead of the correct
expression for ρ0 /ρ. The second factor, T1 /T , which represents the reduction
in temperature for given expansion ratio is correct, so that correct values of
ρ/ρ0 for a given value of y1 can be found by multiplying the figures given in

45

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

Figure 2.8
Distribution of density after the explosion. The upper scale corresponds with
calculations using the value of E given in table 2.3 for γ = 1.40
column 3 of table 2.5 by a correcting factor
C=

γ − 1 + (γ + 1)y1 γ − 1
.
·
γ + 1 + (γ − 1)y1 γ + 1

(2.27)

Values of C are given in column 4, table 2.5, and the corrected values of ρ1 /ρ0
in column 5. It must be pointed out that though the figures in column 5 are
correct, the values obtained for r/r0 when y1 is less than about 40 are subject
to an appreciable error owing to using approximate values of f , φ and ψ at
the shock front.
The variation of ρ1 /ρ0 with r/r0 is shown in figure 2.8. It will be seen that
the density is very small when r/r0 < 0.2, that it begins to rise steeply at
about r/r0 = 0.28, and that it has nearly attained atmospheric density when
r/r0 = 0.36.

46

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.9 Calculation of the Rate of Rise of the Heated Air
Though it would be difficult to calculate the effect of buoyancy on a fluid
with the density distribution shown in figure 2.8, the rise of bubbles in water,
when the change from very light to heavy fluid is discontinuous, has been
studied. It has been shown both experimentally and theoretically (Davies &
Taylor 1950) that the vertical velocity U of a large bubble is related to a, the
radius of curvature of the top of the bubble, by the formula
2 √
· ga .
(2.28)
3
It seems worth while to compare the observed rate of rise of the air heated
by the New Mexico explosion with that of a large bubble in water, and for
that purpose it is necessary to decide on the radius of a sphere of zero
density which might be expected to be comparable with air having the density
distribution of figure 2.8. The simplest guess is to take the radius as the
value of r for which ρ/ρ0 = 12 ,and in figure 2.8 this corresponds with the
broken line for which
U=

r/r0 = 0.328 .

(2.29)

In the New Mexico explosion the best estimate of E was that corresponding
with the measured rate of expansion of the ball of fire, assuming γ = 1.40.
This is given in table 2.3, namely E = 7.14 · 1020 ergs.
Using this value and p0 = 106 dynes/sq.cm.
1

r0 = (7.14 · 1014 ) 3 = 8.9 · 104 = 894 m. .

(2.30)

The radius chosen for comparison with a bubble rising in water is therefore
r = 0.328 · 894 = 293 m. .

(2.31)

The predicted velocity of rise is therefore
2 p
· (981 · 2.93 · 104 ) = 35.7 m./sec.
(2.32)
3
Comparing this with the observed value of the vertical velocity of the centre
of the glowing volume, namely, 35 m./sec., it will be seen that the agreement
is better than the nature of the measurements would justify one in expecting.
A far less good agreement would justify a belief that the foregoing dynamical
picture of the course of events after the atomic explosion is essentially correct.
U=

47

2 The formation of a blast wave by a very intense explosion. II. The atomic
explosion of 1945

2.10 References
Davies, R. M. & Taylor, Sir G. 1950 Proc. Roy. Soc. A, 200, 375.
Johnston, H. L. & Davis, C. O. 1934 J. Amer. Chem. Soc. 56, 271.
Johnston, H. L. & Walker, M. K. 1935 J. Amer. Chem. Soc. 57, 682.
Mack, J. E. 1947 Semi-popular motion picture record of the Trinity explosion.
MDDC221. U.S. Atomic Energy Commission.

48