Author’s Note: What follows is not normal TAH fare. But I’ve seen so much scientifically illiterate speculation and bogus commentary on this particular issue that I could damn near hurl. And all our regular readers know how I am about running numbers to ground. (smile)
That illiteracy includes a whopper of a “rookie” mistake that appeared in the original version of this article (forgetting to change from gauge pressure to absolute pressure – hey, it’s been approaching 40 years since I took thermodynamics, so I plead the proverbial “senior moment” here). It’s corrected below.
. . .
We’ve all heard about “deflate-Gate” recently.
A quick refresher: 11 of the 12 game balls used by the NE Patriots in their conference championship game were found to be substantially underinflated. The Patriots claim that the balls were inflated to the lower end of the NFL’s specification (between 12.5 and 13.5 PSI – gauge pressure – inclusive) prior to inspection by the game’s referee. What I’ve seen indicates the underinflated balls were reportedly later all found to be approx 2 PSI below the lower limit.
Each team provides their own game balls for use on offense during the game. The balls are inspected prior to the game, and marked as compliant by the game’s referee. They are then in the possession of that team throughout the game.
Various theories have been offered. Chief among them is, “The cold temperature during the game caused the pressure to drop.” The NE Patriots claim no wrongdoing, and to have followed the NFL’s rules “to the letter”.
Well, let’s look at this.
A Bit of Science Background
Provided its chemical composition remains unchanged, gas behavior in a closed system is governed by the ideal gas law
PV=nRT
where
P is pressure – absolute, not gauge
V is volume
n is the number of gas molecules present
R is the universal gas constant
and T is temperature – Kelvin, not Celsius or Fahrenheit
This equation is equivalent algebraically to
P = nRT/V
Since R is a constant, if there’s no change in chemical composition of the gas only 3 things can change the pressure of a gas in a closed system. Those are (1) a change in temperature, (2) a change in volume, (3) removal of some of the gas molecules, or (4) a combination of the three. It really is that simple.
Change in Temperature
OK, now let’s assume the change was due solely to a change in temperature.
10.5 and 12.5 PSI (gauge) equate to 25.2 and 27.2 PSI (absolute), respectively. The balls in question were found to be inflated to (25.2 PSI) / (27.2 PSI) = 0.9265 of the minimum allowable pressure. However, they were inflated in (presumably) environmentally controlled conditions – let’s say 70F. The game conditions were in the high 40sF – let’s say 47F. Could the game temperature being 47 F have caused the discrepancy?
In a word: no.
Unlike a balloon, a football’s volume doesn’t change much if any when the internal pressure rises from 25.2 PSI to 27.2 PSI. Rather, the football’s volume remains essentially constant. And if no gas was removed, that means the number of gas molecules is constant between the two pressure measurements.
Let’s let P1 and T1 be the temperature when inspected; P2 and T2 are similarly the later pressure and temperature.
That means we have
0.9265 = P2/P1 = (nRVT2) / (nRVT1) = T2/T1
A common mistake here is to fail to convert to degrees Kelvin (yeah, it does indeed matter here). 70F is 294.26 K. Doing the math, we find that
T2 = 0.9265 x T1 = 0.84 x 294.26 K = 272.62 K
272.68 K is about 31 F. Since the game was played in conditions in the high 40’s F, um, no that isn’t the case here. This was a cool and rainy – but not exceptionally cold – winter day in New England, but the rain wasn’t freezing.
The game temperature (47 F) only explains a pressure drop of about 1.2 PSI, or somewhat more than half of the observed discrepancy. What I’ve seen indicates the 11 underinflated balls were all reportedly around 2 PSI below the lower limit – or 0.8 PSI lower than can be explained by the game temperature alone.
Change in Volume
Well, NFL teams reportedly “prepare” the surface of game balls after inspection to ensure the surface is to their quarterback’s liking after they’ve been inspected by the referee. If this “preparation” process somehow added enough volume to the ball to account for the measured discrepancy, that might explain the incident.
However, we now know the actual pressure drop due to the temperature change from 70 F to 47 F – 1.2 PSI. That makes P1’ – the expected pressure at 47 F in the absence of tampering – 26 PSI (absolute), or about 11.3 PSI (gauge). So let’s see how much additional volume at 47 F would be required to account for an additional 0.8 PSI pressure drop. Here, T1=T2=47 F. As above,
V1/V2 = 25.2/26.0 = 0.969 = P2/P1’ = (nRT2V1) / (nRT1V2)
which means that
V2 = V1/.969 = 1.032
The observed pressure of 10.5 PSI (gauge) would require an increase in volume of 3.2% at 47 F. Thus, if the “preparation” process causes a gain in volume of the football by about 3.2%, the observed low-pressure conditions in the 11 of 12 game balls would be explained.
Let’s see if that’s plausible.
Per this article, a football can be approximated by an ellipsoid of rotation with major axis a = 14.0 cm and minor axis b = 8.5 cm. Since the equation for the volume of an ellipsoid of rotation is known – V = (4/3)(PI)ab2 (where PI is the physical constant 3.1415926 . . . ) – if we assume that the proportions between a and b remain constant we can calculate how much larger the football would have to be for this to be the case.
Assuming proportionality will be retained, since 14/8.5 = 28/17, replacing a with 28/17 b yields the simplified equation
V = (4/3)(PI)(28/17)b3 = (112/51)(PI)b3
Using the necessary increase in volume of 3.2%, this yields
Vnew/V = 1.032 = [(112/51)(PI)bnew3] / [(112/51)(PI)b3] = (bnew3)/b3
Since b = 8.5cm, this yields
Vnew/V = 1.032 = (bnew3)/614.125, or bnew3 = 1.14(614.125) = 633.62
This means bnew is approximately 8.59cm.
Since proportionality is assumed retained this gives revised axis dimensions of b = 8.59cm and a = 14.14cm.
If the pre-game prep process caused the footballs to expand somewhat, that might explain the rest of the discrepancy. That’s only a slight difference (about 2mm in “fatness” and about 3mm in length) in size. Even professional athletes might not notice that small of a difference.
Still: I’d have to see that demonstrated to be the case before I’d buy it.
Change in Number of Gas Molecules
OK, that leaves a change in the number of gas molecules to explain the difference.
One way is to use gases that react chemically – like hydrogen and oxygen, for example. The problem here is that most such reactions are also exothermic (heat-producing) and fast – and would raise the football’s internal temperature and pressure to the point it would likely cause the football to explode. They also generally require something to initiate the reaction, like a spark. Those reactions that don’t generate heat generally require long times and/or some heat source. I think we can rule those out.
Inflating the footballs with gas that begins to condense at around 50 F would also be a possibility, since condensation removes gas molecules also. However, I don’t know of such a gas – and controlling that to achieve a specific pressure drop would be a nightmare. Moreover, if the weather got colder than expected, that could result in complete deflation. I think we can ignore this one also.
That leaves the possibility of gas molecules being removed.
While air does diffuse through rubber, it’s a very slow process; it’s one reason why your tires lose pressure over time. A 2 PSI drop just doesn’t happen to 11 out of 12 footballs naturally in one half of football.
A Last Possibility
There is one other possible explanation that I can think of. Suppose the footballs were intentionally inflated with heated air, then immediately inspected? Leather and rubber are thermal insulators – not great insulators, but they provide some insulation. If the time between inflation and inspection was very short, the heated air used to inflate the balls might not be apparent. This might be the case if the ball was inflated in a referee’s presence, then immediately handed to them for inspection.
Let’s see if this is feasible. How hot would the air have to be to cause a 2 PSI drop between inflation and game temp of 47 F?
From above, if volume is constant and no air is removed then P2/P1 = T2/T1. That means
P2/P1 = 25.2/27.2 = 281.483/T1, or T1 = (27.2/25.2) x 281.483 = 303.77 K, or about 87 F
So, if the footballs were intentionally inflated in the presence of the referee using air heated to around, say, 87 F to 90 F vice ambient temperature, during the game they’d drop to about 10.5 PSI (gauge).
My Guess
Based on the above, I’m guessing one of two things happened.
First, and IMO most likely, possibility: someone intentionally let some air out of the game balls in question, dropping the pressure by about 1 PSI or so. In doing that they missed one, accounting for 11 of 12 balls with low pressure.
Second: one ball was inflated previously using air at 70F, then given to the referee for his inspection. The remainder were inflated in the referee’s presence using air heated to around 87 to 90 F. This would also explain the observed conditions.
That said, it’s close. Most teams would be given the benefit of the doubt. But given the Patriots’ history, well, I’m not inclined to give them that benefit. IMO, even in 2007 NFL teams just didn’t give up a 1st round draft pick when they were alleged to have broken league rules w/o a fight if they were innocent. YMMV.
If I had to bet, I’d bet on the first. But either is possible. And IMO either is absolutely shady and unethical conduct.
The sad thing is that if this was indeed a case of tampering, like many cases we see regarding stolen valor it just simply wasn’t necessary. The way the Patriots were playing that day, as Andrew Luck – quarterback of the Indianapolis Colts – put it: “They could have used soap for balls and they’d have still won.”
Again: sorry for the rookie error in the original, but it’s been a while since I dealt with this stuff.
