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Math . . . again
Thanks for the comment on the temp, but I'm trying to go a bit deeper, hang with me. By the way, people, the balls are inflated and then checked with gauges INDOORS!! The game is played OUTDOORS.
Someone elsewhere pointed out that
Pressure gauge= Pressure absolute - Pressure atmosphere
which means I need to do my calculations with absolute pressure to use the Ideal Gas Law
Atmospheric pressure is 14.7 psi
12.5 + 14.7 = 27.2psi = 187.54 kPA
187.54/297.04 = P2/272.04
0.29= P2/272.04
P2=171.76kPa
171.76kPa= 24.91psi
subtract atmospheric pressure and
24.91-14.7= 10.21psi
!!!!
A ball inflated at 75 degrees to 12.5 psi will be 10.2 psi at 30 degrees!!! This matches the readings by the NFL Officials!!
UPDATE REDUX
Temperature at game time was 50F, which is 282K
187.54/297.04 = P2/272.04
P2=178.07kPa
171.76kPa= 25.82psi
subtract atmospheric pressure and
25.82-14.7= 11.15psi
A ball inflated at 75 degrees to 12.5 psi will be 11.15 psi at 50 degrees!!!
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