Frabjous said:
You can calculate them at the shock front.
I did it with overpressure and I'm getting some odd results:
dR/dt transforms with the chain rule like this:
$$\frac{dR}{dt} = \frac{dR}{dz}\frac{dz}{d\tau}\frac{d\tau}{dt}$$
And our variables are these:
$$
\begin{align*}
R_{0}z &= R \\
\tau &= \frac{a_{0}t}{R_{0}}
\end{align*}
$$
So their final derivatives and the final value of dR/dt is as follows:
$$
\begin{align*}
\frac{dR}{dz} &= R_{0}\\
\frac{d\tau}{dt} &= \frac{a_{0}}{R_{0}}\\
\frac{dR}{dz}\frac{dz}{d\tau}\frac{d\tau}{dt} &= R_{0}\frac{dz}{d\tau}\frac{a_{0}}{R_{0}} = a_{0}\frac{dz}{d\tau}
\end{align*}
$$
We know what ##\frac{dz}{d\tau}## is as that is the final result of the paper and earlier in the paper we get this expression $$P(R) = (\frac{2 \gamma M_{s}^{2}-(\gamma-1)}{\gamma+1})P_{0}$$ for the pressure at the shock front where $$M_{s} = a_{0}^{-1}\frac{dR}{dt}$$ this means $$M_{s} = \frac{dz}{d\tau}$$ The rest of what I did is just agebraic manipulation to get it into units that aren't dimensionless.
$$
\begin{align*}
\frac{dz}{d\tau} &= (1+\frac{1}{K_{0}z^{3}})^{\frac{1}{2}}\\
z &= \frac{RP_{0}^{\frac{1}{3}}}{E_{0}^{\frac{1}{3}}} \\
\frac{dz}{d\tau} &= (1+\frac{E_{0}}{K_{0}R^{3}P_{0}})^{\frac{1}{2}} = M_s\\
P(R) &= (\frac{2 \gamma(1+\frac{E_{0}}{K_{0}R^{3}P_{0}})-(\gamma-1)}{\gamma+1})P_{0}\\
&= (\frac{\frac{2 \gamma E_0}{K_{0}R^{3}P_{0}}+\gamma+1}{\gamma+1})P_{0}\\
&= (\frac{2 \gamma E_0}{K_{0}R^{3}P_{0}(\gamma+1)}+1)P_0\\
&= \frac{2 \gamma E_0}{K_{0}R^{3}(\gamma+1)}+P_0
\end{align*}
$$
Overpressure is just ##P-P_0## so the expression for overpressure is $$\frac{2 \gamma E_0}{K_{0}R^{3}(\gamma+1)}$$ which implies that overpressure increases linearly with explosion size, which looking at nukemap doesn't seem to be true
