Science might not be enough to save the New England Patriots from the clutches of another embarrassment. But at least it provides an opportunity to learn a little more about science, something that’s not entirely the sport’s strong suit.
NPR, for example, today explained the difference between the performance of a properly inflated football and the ones the Patriots used in whipping the Indianapolis Colts to win the AFC Championship on Sunday.
“If you reduce the mass of the ball, which happens if you let a little bit of air out, the ball can decelerate faster when you throw it,” John Eric Goff, a physicist at Lynchburg College in Virginia, told the network.
Maybe that was the problem in the first quarter of Sunday’s game when Patriots quarterback Tom Brady underthrew Shane Vereen, who had to reverse course to make a critical catch at that time in the game.
Neither of the scientists consulted by NPR think the fact 11 of 12 balls were deflated was an intentional act by the Patriots.
So what could it be? The weather? The Boston Globe has considered that theory and found it lacking.
We know that the temperature at kickoff was 51 degrees. Let’s assume that the temperature inside — where the balls were filled — was 71, and that they were already at the low end of allowable pressure: 12.5 psi.
As the balls cooled from 71 degree to 51 degrees, their pressure would actually drop from 12.5 pst to 11.5 psi. That’s quite a lot, though not enough to explain the full 2 psi decline.
For the pressure inside a football to drop 2 psi, the balls would have to get close to the freezing point, which didn’t happen Sunday, though it’s possible it has happened in some of those legendary cold-weather games — like the 1967 ice-bowl, when the temperature hit 15 below.
The weather might have played a bit of a factor, but it probably had an accomplice, the Globe concludes.
Unless, of course, there’s another theory. And, conveniently, there is.
Boston.com says the balls could’ve been heated in a very warm room, presented to officials in a cooler room, and, finally, taken to the cold field.
If the Patriots inflated, stored, and/or tested the balls by filling them to 12.5 PSI in a 70 degree environment, the PSI of the balls would drop by about 1 after half an hour outside in a 50 degree environment, according to (Martin) Schmaltz (professor of physics at Boston University).
However, the NFL found that 11 of the Patriots 12 balls were “underinflated by about 2 pounds each.” In order for a ball to register a 10.5 PSI in a 50 degree environment but register a 12.5 PSI in the testing environment, the ball would have to have been inflated, stored, and/or tested in a 91 degree environment.
Schmaltz reasons that “about 2 pounds” means that the balls were likely somewhere in the 1.6-2 PSI below standard range. In that case, the balls would have to have been inflated, stored, and/or tested in an environment that was at least 82 degrees.
Of course, this all assumes that the Patriots’ balls registered the minimum 12.5 PSI when they were tested by the NFL. It also assumes that the NFL doesn’t mandate the balls be left outside in the open air for a half hour or so to get an accurate reading on the PSI of the ball at game temperature.
Jay Busbee at Yahoo’s Shutdown Corner headed for the sporting goods shop for a football and an air pump. Science!
He found that anyone who says they noticed a difference in the footballs is probably lying, but dropping a couple of p.s.i. out of a football is easier than calculating what temperature the room has to be to heat the balls to pull off Boston.com’s theory.
So how hard is it to pull two psi out of a ball? Even easier than you’d expect. Pop a needle into the ball and you can drop two psi faster than you can say “two psi.” Granted, there are some chain-of-custody issues here; someone looking to do this would have to gain access to the balls, which are supposed to be under the referees’ control prior to the game.
Verdict: If someone were to do a Mission: Impossible-style break-in of the football vault, they could deflate all 12 footballs inside of 30 seconds … or miss one, if they’re sloppy.
So what have we learned? This: deflating footballs is a perfect cut-the-corners gambit. It’s an advantage that’s simple to execute, not easily detectable, but with measurably positive results.
What does all this look like when written out in the fashion that makes you have bad dreams forever about having a test in a class you didn’t attend all year?
It looks like this (courtesy of WUSA).
It’s called the Ideal Gas Law:
where p is pressure, v is volume, n is the number of moles of a gas, R is the Universal Gas constant, and T is temperature.
Remember, what we do to one side of the equation, we have to do to the other side as well. For example, if we increase the pressure (p), then the temperature (T) would have to increase as well. That also means that a change in volume (V) would mean a change in temperature.
We make the following assumptions, based on what we know about the procedure regarding regulation footballs in the NFL and about the Ideal Gas Law:
1) V, the volume of gas (air) in the ball should not change, since (according to procedure), no air is added to or subtracted from the ball after reaching the proper inflation,
2) n will not change for the same reason as above,
3) R does not change, since it is a universal constant.
Now, let’s just change the way the equation looks by moving all the letters to one side of the equation:
pV/nRT = 1
From here, we need to think of this as two different times: the pressure, temperature, etc. from when the balls were checked and the pressure, temperature, etc. out on the field. Let’s set those to be equal:
p1 V1 / n1 RT1 = p2 V2 / n2 RT2,
where the 1 represents the initial readings and 2 represents the readings on the field. Since the volume will not change (assuming no air is added or taken away from the ball), then V1 = V2, and those can be cancelled. For the same reason, n1 andn 2 can cancel. The R ‘s cancel, since R
is a constant. We are left with a simple equation:
p1 / T1 = p2 / T2
Now, we can start solving this puzzle quite easily!
Let’s assume that each ball was inflated to the minimum pressure required to meet the NFL rules regarding proper inflation: 12.5 psi. We convert psi (English) to pascals (Metric), which comes out to 86,184.5 Pa and assume room temperature (68ºF/20ºC) which converts to 293.15 K (Kelvin, the Metric equivalent). We now have,
86,184.5 Pa / 293.15 K = p2 / T2.
We’re down to two variables. But we also know the temperature on the field at the start of the game was reported as 51ºF/10.6ºC (283.15 K). Plug it in…
86,184.5 Pa / 293.15 K = p2 / 283.15 K
Neat! Look, we’re left with a solvable equation with one variable, p2, which is the pressure of the air inside the ball at game time! Let’s solve this riddle…
Isolate the lone variable:
(86,184.5 Pa / 293.15 K) * 283.15 K = p2
83,244.6 Pa = p2 —> 12.1 psi
83,244.6 Pa is 12.1 psi, so, according to our calculations, the balls could have been under-inflated by 0.4 psi on the field. This makes sense given the very first equation, which shows that a decrease in temperature would force a decrease in pressure, assuming the same volume of air in the football.