An error in my math - potentially significant.
P pressure in the ideal gas law is
absolute pressure. Our pressure gauges read
relative pressure, that is, relative to atmospheric pressure of 14.7 psi. So a football for which the gauge reads "12.5 psi" is actually at 27.2 psi absolute ( = 14.7 +12.5). Reduce that by 4% by moving the football from 75 F to 55 F, the pressure goes to 26.11 psi absolute (= 27.2 x 0.96). Subtract atmospheric pressure, and the gauge will read 11.4 psi (= 26.1 - 14.7). So, potentially
a full 1.1 psi drop on the pressure gauge (= 12.5 - 11.4). Or a bit less, since the football has some tiny bit of elasticity.
Looks like the Patriots have done some experiments. I assume the NFL has too.
http://espn.go.com/boston/nfl/story/_/id/12222029/bill-belichick-says-new-england-patriots-followed-every-rule-preparation-footballs
Quote:
Originally Posted by jyl
The temperature factor could explain the balls being under inflated during the game if they were inflated to the bare legal minimum before the game, in a warm room. Maybe some equipment guy figured that was okay - get them as soft as possible, as long as they pass inspection.
But the magnitude of pressure drop wouldn't be very large. 75 F is 297 K, 55 F is 285 K, 297 to 285 is a 4% drop, that means the pressure in pascals would drop about 4%. Converting to psi, if the pressure was 12.5 psi at 75 F, it would drop to 12.0 psi at 55 F.
Of course we don't know how under inflated the balls were during the game.
Edit: FYI, the formula is PV = nrT, the "ideal gas law" familiar to every physics student
P is pressure in pascals
V is volume
T is temperature in kelvin (K)
nr are constants
Assume V volume is constant which is approximately true, a football isn't very elastic (not like a balloon), at least at the range of pressures we're talking about. Then P pressure will change proportional to T temperature - if you are using the correct units, which is why you have to convert Farenheit to Kelvin and pascals to psi.
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