Ex-Players React: Brady's Cluelessness "Unbelievable"

NY Post ^ | 1-22-15 | Howie Kussoy

[Qwackertoo]

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?

[FamiliarFace]

***Do weather conditions play any role in the ball losing PSI?***

That’s been ruled out here, because the Colts balls registered within the guidelines at halftime and after the game as well. The new substituted balls retained their pressure at the end of the game as well.

[TexasGator]

“Stop with the bullshit. If “Brady didn’t do it” then find the guy that did.”

Mother Nature?

pv = nRT

Assuming v, n, R are constant:

P/Po = T/To

p= T/To/Po

P = (50+273) / (70+273)*(14.7+13.5)= 26.6 psia

PSIG = (26.6-14.7) = 11.9

[Andy'smom]

Now the Colt’s linebacker is saying he didn’t notice.