G. I. Taylor famously published rather accurate estimates of the Trinity test’s yield using only the series of time-stamped still images published by Life magazine (example image at that link). The yield was still classified at the time.

Taylor’s calculation was rather involved, but you can get a pretty good answer using the single image in the Life magazine link above and some dimensional analysis. To wit…

Things we know: At time *t*=0.025 seconds after detonation the shockwave reaches a radius *r*=130 meters, give or take. En route it was pushing air out of the way, and air has a density *d*=1.2 kg/m[sup]3[/sup]. (Instead of the density of air you might find it more natural to think in terms of the total mass of the air displaced, but this mass is simply the air’s density times the volume under that hemispherical shockwave, both of which we already know. So, we can work with density or total mass displaced.)

We want to calculate the energy released by the bomb. Energy has SI units of kg/m[sup]2[/sup]/s[sup]2[/sup], so we need an expression involving *t*, *r*, and *d* that gives us the right units(*). There is only one way to do so:

*E* = *d* *r*[sup]5[/sup] / *t*[sup]2[/sup] .

Assuming any missing numerical (unitless) factors are of order unity (*i.e.*, not too much bigger or smaller than 1, a common – and commonly good – assumption), we can just run the numbers. Google can do the unit conversions for you to get to “kilotons TNT”:

Google: (1.2 kg/m^3)*(130 m)^5/(0.025 s)^2 in kilotons TNT = 17 ktons TNT.

The actual yield is estimated at 21 kton TNT.

(*) self-nitpick: we need an expression that has the correct “dimensions”. Units are a matter of convention.