Deflated Balls
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=Introduction= | =Introduction= | ||
| − | If you clicked on this link trying to gain information about your old sagging man-hood, be glad that this doesn't apply to you. The reason for this web page is to evaluate the Deflated Football from a scientific feasibility test on how a ball could become depressurized by Half Time even though | + | If you clicked on this link trying to gain information about your old sagging man-hood, be glad that this doesn't apply to you. The reason for this web page is to evaluate the Deflated Football from a scientific feasibility test on how a ball could become depressurized by Half Time even though New England met the specification at the NFL required time two hours before the game. |
=Ideal Gas Equation= | =Ideal Gas Equation= | ||
Revision as of 16:23, 23 January 2015
Contents |
Introduction
If you clicked on this link trying to gain information about your old sagging man-hood, be glad that this doesn't apply to you. The reason for this web page is to evaluate the Deflated Football from a scientific feasibility test on how a ball could become depressurized by Half Time even though New England met the specification at the NFL required time two hours before the game.
Ideal Gas Equation
PV = nRT
P = Pressure (Pa) V = Volume (m^3) n = Number of moles T = Temperature (K)
State 1
Assumptions
- Balls were inside the locker room or building at the pressure inspection time
- Balls are brand new
- Balls met the minimum required pressure test two hours prior to the game
Equation
(P1)(V1)= nR(T1)
State 2
Assumptions
- Balls were outside in the playing field all the way until half time
- Balls were used in the game at various times throughout the first half
- Balls were not tampered with in any way to change the pressure through air mass reduction
Equation
(P2)(V2)= nR(T2)
Case 1
Assumptions
- Ball remains constant Volume (V1=V2)
- nR is a constant for both states
- T1 = 82 F = 301 K
- T2 = 50 F = 283 K
- P1 = 12.5 psi = 86207 Pa
Equation
P2 = (P1)(T2/T1)
Note: Solve two equations for nR and V1 and V2 cancel
Result
P2 = (86207)(283/301) = 82143 Pa = 11.8 psi
Case 2
Assumptions
- Ball volume changes from wear and tear 5% increase in volume (V1/V2) = 0.95
- Conditions from Case 1 exactly the same
Equation
P2 = (P1)(T2/T1)(V1/V2)
Result
P2 = (86207)(283/301)(0.95) = 76999 Pa = 11.2 psi
Summary
Case 1 is a viable situation, yet the pressure drop is minimal due to temperature alone. In case 2 we are very generous on the the volume expansion. It is doubtful that the volume changes very much on a new ball, if at all. However, even with the generous volume expansion, the pressure drop still is only 1.3 psi.
The probability of a full 2 psi drop in pressure is doubtful, but cannot be completely ruled out as impossible. The part that we do not know is the exact measurements of the footballs before and after for each football and that is very important. If there was significant variation, you could have some evidence that a new football can change in volume if the lower pressure balls were also the balls used most. Probably a kicked ball being the most likely to change volume. The biggest public evidence is the fact that 11 of the 12 were low and that is a tough probability to ignore.
The end result is Plausible but Unlikely.
NFL Rules
The NFL rules matter here. If there is no temperature at time of measurement rule stated, then does the NFL really have a case against New England? I say no. A small fine at best is the worst punishment, and a rule change to prevent this exploit in the future.
