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From Wikipedia: Fermi was known for his ability to make good approximate calculations with little or no actual data, hence the name. One example is his estimate of the strength of the atomic bomb detonated at the Trinity test, based on the distance travelled by pieces of paper dropped from his hand during the blast. Fermi's estimate of 10 kilotons of TNT was remarkably close to the now-accepted value of around 20 kilotons, a difference of less than one order of magnitude.

I have not been able to find any references explaining how he made this calculation. Please provide a reference or an example calculation.

See also: http://en.wikipedia.org/wiki/Fermi_problem

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I wanted to add the tag: "fermi-problem", but don't have the reputation needed. Please add this or other applicable tags. –  Halfdan Faber Jan 16 '12 at 2:33
I added "order-of-magnitude". I hope that works for you. –  Mark Eichenlaub Jan 16 '12 at 2:39
This question remains essentially unanswered. I have since wondered if Fermi perhaps did a quantified comparison to an earlier much smaller TNT test bombing. This would assume he was actually present at this earlier test, dropping pieces of paper, neither of which appears to be documented... –  Halfdan Faber Mar 9 '12 at 7:32
Could someone with the appropriate reputation level add the tag 'fermi-problem'. Thanks in advance. –  Halfdan Faber Jul 2 '13 at 22:28

1 Answer 1

Well, I'm getting an answer about an order of magnitude too large so I must be doing something different, but here's my guess:

Blast wave travels at about speed of sound - 40 seconds -> 14 km in radius at this time. The paper is moved 2.5 meters by the wave - so the effect of the bomb is to displace a hemispherical shell of air of volume 2.5m*2*pi*(14 km)^2 Multiply by 1 Atm to get energy of 3e14 J ~ 80 kT TNT

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Well, the papers were dropped from a height of 6 feet, so from the 2.5 meter displacement, we can learn something about the wind speed at Fermi's location. Clearly the displacement was more than 2.5 meter and also occurred over a much larger distance. The detonation displaced air with a density of 1.2 $kg/m^3$ and also created a local vacuum, causing a subsequent reverse air flow. I would expect the calculation to involve an integration of displacement kinetic energy, taking into account energy emitted by heat and radiation, and potential energy stored in the central vacuum... –  Halfdan Faber Jan 16 '12 at 4:33

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