Radius \(R\) depends on time \(t\), energy \(E\), density of the atmosphere \(\rho_0\), and the specific heat of the atmosphere \(\gamma\), so \(R = f( E, t, \rho_0, \gamma )\) for some unknown function \(f\). The fireball is moving so fast, the atmospheric density has no chance to change in the short term. The specific heat \(\gamma\) has units of joules per kilogram per degree Kelvin.
\(E\) | \(t\) | \(R\) | \(\gamma\) | \(\rho _ 0\) | |
distance | 2 | 0 | 1 | 2 | -3 |
time | -2 | 1 | 0 | -2 | 0 |
mass | 1 | 0 | 0 | 0 | 1 |
temp | 0 | 0 | 0 | -1 | 0 |
We can save ourselves some work in the row-reduction by re-arranging the rows and columns of the matrix ...
t | R | E | \(\rho _ 0\) | \(\gamma\) | |
time | 1 | 0 | -2 | 0 | -2 |
distance | 0 | 1 | 2 | -3 | 2 |
mass | 0 | 0 | 1 | 1 | 0 |
temp | 0 | 0 | 0 | 0 | -1 |
Now, by row-reduction,
t | R | E | \(\rho _ 0\) | \(\gamma\) | |
time | 1 | 0 | 0 | 2 | 0 |
distance | 0 | 1 | 0 | -5 | 0 |
mass | 0 | 0 | 1 | 1 | 0 |
temp | 0 | 0 | 0 | 0 | 1 |
So, there is a 1-dimensional column nullspace with the spanning basis \(\{ [-2,5,-1,1,0]^T \}\).
After dimensional analysis,
\[F(t^{-2} R^{5} E^{-1} \rho_0^{1}, \gamma) = 0\]
\[R = E^{1/5} t^{2/5} \rho_0^{-1/5} S(\gamma)\] \[\log R = \frac{1}{5}\left( \log E + 2 \log t - \log \rho_0 \right) + \log S(\gamma)\] \[\frac{5}{2} \log R = \log t + \frac{1}{2} \left(\log E - \log \rho_0 \right) + \frac{5}{2} \log S(\gamma)\]
From this, we can determine the y-intercept, which implies
Knowing something about specific heats and air density, G I Taylor was able to solve for energe E, and estimate the explosion size at 16.8 kilotons. The actual strength was estimated with other methods as equivalent to 20 kilotons of TNT.