The New England Patriots football team have come under recent investigation by the NFL for deflating some of the footballs they used last week in the AFC championship game. It was found that 11 of the 12 footballs the Patriots had been using in that game had been deflated by 2 PSI (pound per square inch) below the 12.5 PSI guideline designated by the NFL. A few of my friends have speculated that the cold weather on the field might have caused this deflation if the balls had been inflated inside in a warm room before the game. For some fun, here is the math on why it most likely wasn’t the weather which deflated those 11 Patriot footballs by 2 psi.
Using the ideal gas law and eliminating constants on both sides we get:
P1/T1 = P2/T2
where:
P1 is the pressure of the football before the game, inside, at room temperature (12.5 PSI)
P2 is the pressure of the football once we bring the football outside and allow the air inside to cool. (our unknown)
T1 is room temperature (72° F)
T2 is the reported temperature on the field (51° F)
Converting pressure and temperature to Pascals and Kelvin units (respectively):
P1 = 12.5 PSI = 86184.47 Pascals
T1 = 72° F = 295.37.15° Kelvin
T2 = 51° F = 283.706° Kelvin
However, as astronomer Phil Platt correctly points out to me, we must use absolute pressure (atmospheric + gauge) accounting for the the natural pressure exerted on the ball due to the atmosphere. At the game that night, atmospheric pressure was reported at 100950 Pascals. So P1 is now:
P1 = 86184.47 + 100950 = 187134.47 Pascals
Plugging into our equation:
(187134 Pascals) = unknown Pascals
(295.37.15° Kelvin) (283.706° Kelvin)
By using basic algebra, we can solve for our unknown pressure of the football on the field (P2) by multiplying both sides of the equation by T2 (283.706° Kelvin) :
(283.706° Kelvin) * (187134 Pascals) = 179743 Pascals
(295.37.15° Kelvin)
Then we subtract the atmosphere from P2 and convert back to PSI to find our unknown ball pressure:
P2 = 179743 – 100950 = 78793 Pascals
P2 = 78793 Pascals = 11.4 PSI
The pressure drop due to weather then becomes a simple subtraction problem:
ΔP = P1 – P2 = drop in pressure of the football due to weather
ΔP = 12.5 – 11.4 = 1.1 PSI
So the weather on the field can only account for a drop of 1.1 PSI in the football under the reported conditions. Perhaps you may be wondering just how cold it would need to be for a drop of 2 PSI? This is easy to find also. Just plug in 10.5 PSI for P2 (~173056 Pascals), and solve for T2.
Following this math, they could have inflated the balls in a room at 91 degrees fahrenheit. Then outside, at 50 degrees, the ball would measure 2 psi less. It seems reasonable that a locker room could be 90 degrees.
If it was 90 degrees when the players came in at halftime, or perhaps after the game at an amateur sporting facility with not enough AC, or maybe if the game were played in the summertime, then perhaps a 90 degree locker room is within the limits of reason in this situation. But 90 degrees before the game even started? In a professional multi-million dollar sporting facility in the month of January? I stand by my initial assumptions. =)