How did Enrico Fermi compute when the Chicago Pile-1 nuclear reactor would become critical? I'm trying to understand the first nuclear reactor, the Chicago Pile-1, specifically the math Fermi did to figure out when the reactor would go critical. There's a nice report available from Fermi, where he tracks the value of $R_{eff}^2/A$, where $R_{eff}$ is the effective radius of the pile, and A is the measured neutron intensity at the center of the pile, some screenshots below.
Fermi then claims that when $R_{eff}^2/A$ reaches 0, the pile will become critical. This is where I get lost - I read through the report and some other sources, but I don's see where this math is coming from -> why should the pile become critical when $R_{eff}^2/A=0$?


 A: This kind of plot is typically called a "1/M Plot", where M is the reactor multiplication.  To measure the multiplication, you need a neutron source and a detector.  As you add more and more material to the reactor, the detector signal increases.  When the reactor goes critical (or more precisely supercritical), the detector signal (M) will grow to infinity.  If you plot the inverse of the signal (1/M), it will become zero when the reactor goes critical.
If you plot 1/M as a function of the reactor size (or control rod position, or number of rods, etc.), you can do a curve fit to estimate where the point of criticality is.  In Fermi's plots, they were adding material (layers) and the plot shows the reactor when critical at layer 57.
A: I cannot be sure I understand it correctly, but this is how it looks to me at the moment. The report that you quote says:

In a spherical structure having the reproduction factor 1 for
infinite dimensions the activation of a detector placed at the center
due to the natural [my emphasis] neutrons is proportional to the
square of the radius.

@Adam says in a comment:

I guess the question really is why he prefers to plot $R^2_{eff}/A$
instead of just $A$

According to the quote from the report, if you just have $A\to\infty$ when the radius increases, it is possible that you just have natural neutrons. If, however, $R^2_{eff}/A\to 0$, it means that the number of natural neutrons becomes negligible compared to the number of neutrons from chain reaction, therefore, the reactor is getting critical.
