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I still wonder ... the balls are inflated INDOORS ... 70 degrees (probably). They can be from 11.5 to 13.5 psi. Brady likes them on the low end, so they inflate to 11.5. They check them, and they are within range. Then they take them outside where it is almost freezing temperature and the balls sit there waiting to be used, and later measured. The psi in the balls would fall below 11.5 for sure! (It's the gas laws, folks!). When they say they were under inflated, no one is saying by how much ... 11.4 psi would be "under-inflated".

If you have a newer car that shows you tire pressure ... you'll notice it is OK in the garage (where its warmer), but when you get outside or leave the car outside, the psi in the tires drop a few pounds. Same thing with anything "inflated" with AIR.

The key here is to inflate them with nitrogen instead of air ... wouldn't change the ball, but the psi would change only barely with temperature as the expansion/contraction with nitrogen is much less than "air." Or find out what the temperature differential is between "ambient" and "game environment" ... inflate just above the 11.5 knowing it will come down to the 11.5 the quarterback wants when the ball is at game temperature.

But for Patriot-haters and Colts fans ... what difference does science make ... Brady cheated!

(Makes great TV, but about as scientifically boring as man-made global warming).

I saw this analysis at Boston.com yesterday

But here’s a poster on Reddit, supposedly a local science teacher, and his formula sounds logical to even the most removed student of Newton, describing how the elements that night may have played a role. Honestly, it’s the best breakdown I’ve seen yet this week explaining how this whole mess may have come to fruition.

Given the conditions of the game, a ball which meets specifications in the locker room could easily lose enough pressure to be considered under-inflated. Some math:
- Guy-Lussac’s Law describes the relationship between the pressure of a confined ideal gas and its temperature. For the sake of argument, we will assume that the football is a rigid enough container (unless a ball is massively deflated, it’s volume won’t change). The relationship is (P1/T1) = (P2/T2), where P is the pressure and T is the temperature in Kelvins.
- The balls are inflated to between 12.5 and 13.5 psi at a temperature of 70 degrees Fahrenheit (294.1 K). Let’s assume an average ball has a pressure of 13 psi. Since these are initial values, we will call them P1 and T1.
- The game time temperature was 49 degrees F (278 K). We are attempting to solve for the new pressure at this temperature, P2. We plug everything into the equation and get (13/294.1) = (P2/278). At the game time temperature, the balls would have a pressure of 12.3 psi, below league specifications.
Furthermore, given that it was raining all day, the air in the stadium was saturated with water vapor. At 70 degrees, water has a vapor pressure of 0.38 psi. The total pressure of the ball is equal to the pressure of the air inside the ball and the vaporized water in the ball. At 49 degrees, the vapor pressure of water is 0.13 psi. Up to 0.25 additional psi can be lost if the balls were inflated by either the team or the refs prior to the game. Granted, it’s unlikely that anyone would inflate balls from 0, but it easily could cost another couple hundredths of a psi in pressure.
- For a ball that barely meets specifications (12.5 psi), it’s pressure would drop to 11.8 psi during the game... enough to be considered massively under-inflated.
Is that enough to account for two pounds though?