Wednesday, January 21, 2015

Balls lose pressure in the cold

Regarding deflate-gate

Amounton's Law says

"The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas's absolute temperature."

So P/T = Constant

For comparing the same substance under two different sets of conditions, the law can be written as:

P1/T1 = P2/T2

I'm also going to do this in metric to avoid other potential errors. And I apologize that I am 30 years from doing this sort of work

12.5 psi = 86.18 kPa ( kiloPascals)
75F=23.89C = 297.04K
30F=-1.11C = 272.04K


86.18/297.04 = P2/272.04
0.29= P2/272.04
78.93kPa= 11.45PSI

Temperature change from 75F to 30F degrees would result in a football ball losing ~1 pound of pressure per square inch.

Now this does not account for ALL of the claimed 2lbs PSI that was measued as having been lost -- but we don't know that the ball began at 12.5 -- what if they began at 12? What if they were inflated in a 78 degree room?

I think early calls dismissing temperature as a force are hasty.

Someone 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
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!!


Temperature at game time was 50F, which is 282K

187.54/297.04 = P2/272.04
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!!!  


Unknown said...

The end-game temperature was in the upper 40s, so the numbers would have changed again.

Not sure by how much, but i've heard rumors of 47/48F at agme end.

Unknown said...

Glad to see that you corrected for absolute pressure. We do not know a lot of things about the method and conditions of the measurements. Sticking to just the temperature issue for a moment, the rain falling through very cold air higher in the atmosphere would be much colder than ground air conditions.Not below freezing but possibly close. If you assume 10F colder rain (38F) then you would get an expected 2 Deg F drop.

Unknown said...

Minor correction to the above post. You would get "an expected 2 psi drop"

Anonymous said...

Also worth mentioning that leather expands when wet, increasing volume and thus decreasing pressure even further.