For all you non science loving patriots haters! The balls are inflated to between 12.5 and 13.5 psi at a temperature of 70 degrees Farenheit (294.1 K). Lets assume an average ball has a gauge pressure of 13 psi. This makes the absolute pressure of the ball 27.7 psi (gauge + atmosphere). 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 (27.7/294.1) = (P2/278). At the game time temperature, the balls would have an absolute pressure of 26.2 psi and a gauge pressure of 11.5, 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, its unlikely that anyone would inflate balls from 0, but it easily could cost another couple hundredths of a psi in pressure.
Posted on: Thu, 22 Jan 2015 00:47:28 +0000
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