Deflate-Gate: Could weather have played a role in deflated balls?

(NEWS CENTER) -- By now, you've probably heard that member(s) of the Indianapolis Colts have claimed that balls used during the AFC Championship Game at Gillette Stadium on Sunday were under-inflated. The claim is perhaps under even more public scrutiny because of the televised ball swap made by an official at the start of the second half of the game.

NFL regulation balls are inflated to a pressure of between 12.5 and 13.5 psi (pounds per square inch). The footballs are handled by several different NFL officials and a group of the balls are inspected prior to each game. Given this procedure, it would seem unlikely that a ball could make it through to the line of scrimmage without an official noticing a ball that may have been tampered with and/or not meeting NFL regulations.

But could the weather have played a role in "Deflate-Gate?"

Editor's Note: It has been brought to our attention that the calculations used in an earlier version of this article did not use the proper pressure values (the pressure of the ball measured by a gauge versus the "absolute pressure," which should be used in the equations below). The text below has been edited from its original version to reflect the correct pressure values.

To answer this question, let's begin with a little equation you may or may not remember from chemistry or physics, back in high school. It's called the Ideal Gas Law:

pV=nRT

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! But before we do, we also have to know the atmospheric pressure during the game, since p in this case is the absolute pressure; the pressure inside the ball plus the pressure of the atmosphere (which exerts a force on the ball as well).

At 6pm, the atmospheric pressure at nearby Norwood Airport was 1009.5 mb (1009.5 hPa or 100950 Pa).

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 a room temperature of 68ºF (20ºC) which converts to 293.15 K (Kelvin, the Metric equivalent). We now have,

(86,184.5 Pa + 100950.0 Pa) / 293.15 K = (p2 + 100950.0 Pa) / 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 + 100950.0 Pa) / 293.15 K = (p2 + 100950.0 Pa) / 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 + 100950.0 Pa) / 293.15 K] * 283.15 K} - 100950.0 Pa = p2

79,800.9 Pa = p2 ---> 11.8 psi

83,244.6 Pa is 11.8 psi, so, according to these calculations, the balls could have been under-inflated by 0.7 psi on the field, just due to the change in temperature from inside to outside. 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.

If we use an indoor temperature of 80º, we would get a final pressure of 11.0 (10.99) psi. Using a 90º indoor temperature, we get a final pressure of 10.5 psi.

So, according to NFL reports, either the indoor temperature when the footballs were inflated was around 90ºF, or something else happened after the footballs were inspected.

The question then becomes why this hasn't been an issue of this magnitude before, especially in games played at colder weather stadiums like Lambeau Field or Soldier Field? No one complained about the ball when the Giants squared off against the Packers at Lambeau in January of 2008, when the game time temperature was -1ºF.

Just sayin'...


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