http://mmqb.si.com/2015/01/23/nfl-deflategate-new-england-patriots-deflated-footballs-super-bowl-49/
This was actually in a Peter King article from a Pats fan
Nerd alert: What I have summarized below is the comprehensive, correct gas law predictions and real-world analysis for “deflategate”. It is explained from the basic ideal gas law, but anyone can hopefully follow. I go through it very clearly, I believe. While a great many people have posted the gas law calculations on the internet, most have been wrong in that they confused absolute pressure with relative pressure. Absolute pressure takes barometric pressure into account.
This post has been corrected after its first posting to include the actual reported barometric pressure in the first half and also to focus only on the pressure/temperature issue, since the "expanding wet leather issue" is harder to model.
The ideal gas law: PV = nRT
The ideal gas law modeled for two conditions 1 and 2: P1V1 = nRT1 and P2V2 = nRT2
Solving for Volume V in each case: V1 = nRT1/P1 and V2 = nRT2/P2
Now we will assume that V1 = V2; that is, the volume of the football will not change during the game; therefore; nRT1/P1 = nRT2/P2;
Dividing each side by the constants n and R; leaves: T1/P1 = T2/P2
Now let‘s determine the real values based upon:
T1 = initial temperature of inflation (indoors) = room temperature,
likely equals 72 degrees F = 22.22 degrees C = 295.37 Kelvin;
T2 = final temperature = game time temperature (published value at kickoff) = 51 degrees F = 10.56 degrees C = 283.71 Kelvin
P1 = absolute pressure at the onset, equals relative pressure + atmospheric pressure, and since reported atmospheric pressure at the time point nearest kickoff time, as reported by Weather Underground, equals 29.75 in = 14.61 psi at 6:53 PM;
P1 equals 12.50 psi + 14.61 psi, = 27.11 psi
All that we need to do is solve for P2:
Since T1/P1 = T2/P2; thereciprocal is true and P1/T1 = P2/T2
Multiplying both sides by T2 solves for P2:
P2 = P1T2/T1
Plugging in the actual values for the known quantities P1, T1 and T2;
P2 = (27.11) x (283.71) / (295.37) = 26.04 = absolute pressure of the football at the end condition (halftime).
Relative pressure = absolute pressure – atmospheric pressure = 26.04 – 14.70 = 11.34 psi
Therefore, the drop in relative pressure P2 – P1 thus equals 12.50 – 11.34 =
1.16 psi
Plain English: By the ideal gas law, a football inflated to 12.50 psi at 72 degrees F and cooled to 51 degrees F will have a final pressure of 11.34 psi, thus a loss of 1.16 psi.
Carnegie Mellon Finding #1: Footballs inflated to 12.50 psi at 75 degrees F and cooled to 50 degrees F had a final pressure of 11.4 psi, a loss of 1.1 psi (summarized in the pdf document at
http://www.headsmartlabs.com/
Conclusion #1: Experiment seems to match ideal gas law prediction rather closely. Note that they used a slightly larger temperature drop, 25 degrees, not 21 degrees. We do not know room temperature in the ref’s room, though, anyway.
Carnegie Mellon Finding #2: Footballs inflated to 12.50 psi at 75 degrees F and cooled to 50 degrees F and then soaked with water had a final pressure of 10.7 psi, a loss of 1.8 psi.
Conclusion #2: A second factor, the expansion of a football as it gets wet, also leads to a drop in psi. This factor contributes another 0.7 psi in pressure drop. This in essence shows that the “constant volume assumption” of the ideal gas law is not fully valid since a football is not infinitely rigid.
One important caveat on the Carnegie Mellon Experiment #2: In this experiment they immersed the football in water for a time. Is that analogous to heavy rain, or is it overkill? That criticism concerns me a bit. The real world effect of a heavy rain is probably between their dry ball result (1.1 psi) and the soaked ball result (1.8 psi).
But to err on the side of caution, let’s ignore the water effect for now and go with the dry ball result: 1.1 psi drop at Carnegie Mellon, in agreement with the 1.16 psi calculation.
Plain English ultimate conclusion for the Patriots footballs: It would be reasonable to expect, based on both experimental results and ideal gas law calculations, for a pressure drop of at least ~1.2 psi to have occurred within the Patriots footballs in the first half of the AFCCG, based on the known game time conditions and the observation that the footballs were inflated to 12.5 (relative) psi at room temperature.
Aha though- what about the Colts footballs?
We don’t know their initial pressure, unfortunately, but if we assume that they were at the maximum legal pressure of 13.5 psi relative pressure (since they apparentlyknew that ball pressure loss would be monitored), we can calculate the expected pressure drop.
T1 = 295.37 Kelvin, as before
T2 = 283.71 Kelvin, as before
P1 = absolute pressure at the onset = 13.50 psi + 14.61 psi, = 28.11 psi
P2 = (28.11) x (283.71) / (295.37) = 27.00
Relative pressure = 27.00 – 14.70 = 12.30 psi
Thus the Colts footballs should have been a final pressure of 12.3 psi. The legal lower limit is 12.5 psi. The Colts footballs should have been illegal by 0.2 psi.
Question: Would a referee call a reading of 12.3 rather than 12.5 to be clearly out of specifications and illegal?
Maybe yes, maybe no. It certainly depends on both the accuracy and precision of the pressure gauge. A digital readout often shows significant drift/fluctuation in the last digit. If in real time the ref saw values pop up such as these: 12.3, 12.4, 12.5, 12.4, 12.3, 12.4; he would likely say: It looks to be about 12.5; pass! Similarly at the beginning, if he saw in real time 13.5, 13.6, 13.5, 13.6, 13.5, 13.6; he would likely say: It looks to be about 13.5, pass!
Final conclusion: It is not unreasonable at all to assume that the Patriots balls would fail the inspection and the Colts balls would (barely) pass or (barely) fail, based upon logical assumptions of inflation levels and inflation temperatures in concert with the issues of temperature-related gas expansion, and the human-element: deciding when (and if) you are sure about that last digit on the pressure gauge.
Not taken into account at all in this analysis is the ball preparation (rubbing) procedure. Thus in essence I am assuming that the ball preparation procedure does not affect psi in the least, which is Bill Nye’s assertion. To me, a 1 psi drop based on friction, even severe friction, seems quite a bit high, because it would correspond to a 25 C temperature rise. I disagree, though, that the effect must be zero. It is probably non-zero, but unknown and unknowable unless we knew the exact ball prep procedure. Still, it is not needed to explain a pressure drop of the magnitude seen. Also ignored is the expanding wet football concept. This is likely real but harder to quantify.
A key piece of the puzzle to nail this is the actual ref-recorded data for the Colts footballs, very clearly.
45-7 
