arguments; he found that the radius R of the shock front was proportional to t
2/5
where t is the time following detonation. Using the equations of hydrodynamics in
Eulerian form together with the Rankine-Hugoniot equations he went on to derive a
set of coupled ordinary differential equations for the pressure, gas velocity and
density behind the shock front. He solved these equations numerically and then
derived analytical approximations to the numerical solution which turned out to be
remarkably accurate.
5.4 Taylor’s Analysis of Very Intense Shocks
In the following sections we will follow Taylor’s analysis of the self-similar solution
and adopt Taylor’s notation (however, here we use E 0 for the energy of the explosion
rather than E and c 0 rather than a for the sound speed in air). In the case of a nuclear
explosion we can regard the energy being liberated almost instantaneously and
coming from essentially a point source after the shock wave has traversed a distance
that is large in comparison to the dimensions of the nuclear device. The pressure
generated by the resulting disturbance in the early stages of the explosion amounts to
several tens of thousands of atmospheres and is so large that the normal atmospheric
pressure p 0 is negligible in comparison. Consequently, the only dimensional parameters appearing as inputs are the energy of the explosion E 0 and the undisturbed
density ρ 0 of the air. Based on the similarity solution, Taylor, Sedov and von
Neumann showed independently that any dimensionless function of r and t, such
as, pressure, velocity or density depend only on the dimensionless combination r(ρ 0 /
E 0 t
2 )
1/5 or some function of it. In fact, the disturbance generated in the air is headed
by a shock front at r ¼ R(t), where t is the time since the explosion. R(t) can be
inferred from dimensional arguments and, as the only other dimensional parameters
are E 0 and ρ 0 , we can write,
R / E
α
0 t
β
ρ
δ
0 ,
and expressing these quantities in terms of their dimensions we have
L / M
α L
2α T
À2α T
β M
δ T
À3δ
:
Equating dimensions on both sides gives
α þ δ ¼ 0, 2α À 3δ ¼ 1 and À 2α þ β ¼ 0,
hence,
α ¼ 1=5, β ¼ 2=5 and α ¼ À1=5:
5.4 Taylor’s Analysis of Very Intense Shocks
219
2/5
where t is the time following detonation. Using the equations of hydrodynamics in
Eulerian form together with the Rankine-Hugoniot equations he went on to derive a
set of coupled ordinary differential equations for the pressure, gas velocity and
density behind the shock front. He solved these equations numerically and then
derived analytical approximations to the numerical solution which turned out to be
remarkably accurate.
5.4 Taylor’s Analysis of Very Intense Shocks
In the following sections we will follow Taylor’s analysis of the self-similar solution
and adopt Taylor’s notation (however, here we use E 0 for the energy of the explosion
rather than E and c 0 rather than a for the sound speed in air). In the case of a nuclear
explosion we can regard the energy being liberated almost instantaneously and
coming from essentially a point source after the shock wave has traversed a distance
that is large in comparison to the dimensions of the nuclear device. The pressure
generated by the resulting disturbance in the early stages of the explosion amounts to
several tens of thousands of atmospheres and is so large that the normal atmospheric
pressure p 0 is negligible in comparison. Consequently, the only dimensional parameters appearing as inputs are the energy of the explosion E 0 and the undisturbed
density ρ 0 of the air. Based on the similarity solution, Taylor, Sedov and von
Neumann showed independently that any dimensionless function of r and t, such
as, pressure, velocity or density depend only on the dimensionless combination r(ρ 0 /
E 0 t
2 )
1/5 or some function of it. In fact, the disturbance generated in the air is headed
by a shock front at r ¼ R(t), where t is the time since the explosion. R(t) can be
inferred from dimensional arguments and, as the only other dimensional parameters
are E 0 and ρ 0 , we can write,
R / E
α
0 t
β
ρ
δ
0 ,
and expressing these quantities in terms of their dimensions we have
L / M
α L
2α T
À2α T
β M
δ T
À3δ
:
Equating dimensions on both sides gives
α þ δ ¼ 0, 2α À 3δ ¼ 1 and À 2α þ β ¼ 0,
hence,
α ¼ 1=5, β ¼ 2=5 and α ¼ À1=5:
5.4 Taylor’s Analysis of Very Intense Shocks
219