# How did Enrico Fermi calculate the classical Fermi Problem?

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.

• I wanted to add the tag: "fermi-problem", but don't have the reputation needed. Please add this or other applicable tags. Jan 16, 2012 at 2:33
• I added "order-of-magnitude". I hope that works for you. Jan 16, 2012 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... Mar 9, 2012 at 7:32
• Could someone with the appropriate reputation level add the tag 'fermi-problem'. Thanks in advance. Jul 2, 2013 at 22:28

I describe this in my book "Guesstimation 2.0" (Princeton University Press, 2012). The work done by the expanding shock wave is pressure times change in volume. The change in volume is as described by the previous answer: $$2.5\,\mathrm{m} \times 2\pi \times (16\,\mathrm{km})^2$$.

Fermi could feel the extra pressure due to the explosion, we can only estimate it. The extra force on his body must have been more than 100 N (20 lb) and less than $$10^4\,\mathrm{N}$$ (2000 lb) so we will take the geometric mean and estimate $$10^3\,\mathrm{N}$$. The extra pressure then is just $$10^3\,\mathrm{N}$$ divided by the typical frontal surface area of a person of $$1\,\mathrm{m}^2$$ or $$P = 10^3\,\mathrm{N}/\mathrm{m}^2$$.

Then $$E = P \Delta V = (10^3\,\mathrm{N}/\mathrm{m}^2)(4 \times 10^9\,\mathrm{m}^3) = 4 \times 10^{12}\,\mathrm{J} = 1\,\mathrm{kT}$$

Now multiply by a few because the energy of the bomb can go into light (photons), nuclear radiation, shock wave, ground pulse … and we get an estimate of 4 kT.

Fermi was there and he estimated 10 kT so this is in the correct ballpark.

• Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to write equations for readability. Jun 23, 2015 at 15:49

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 $2.5\textrm{ m}×2\pi×(14\textrm{ km})^2$ Multiply by 1 Atm to get energy of $3×10^{14}\textrm{ J}\simeq 80 \textrm{ kt TNT}$

• 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... Jan 16, 2012 at 4:33