# Football inflation

## Main Question or Discussion Point

If I take a balloon and tie it off prior to blowing it up I have a flaccid balloon with atmospheric pressure both inside and outside the balloon...say 14.7 psi.

Now if I blow up the balloon I am working against the elasticity of the rubber and it requires effort to do so. I blow more and more volume into the balloon and it requires more psi inside than outside to press outward against the elastic rubber and hold a certain size of balloon.

If the NFL's rules call for 12.5 to 13.5 psi that's less than 14.7 psi ambient atmosphere. Shouldn't the football shrink under the excessive outer pressure until there is equilibrium leaving 14.7 both inside and out?

I know that volume is the key but I don't understand. When they say 13.5 do they mean 13.5 psi greater than the 14.7 ambient or a total of 28.2?

If so does that mean that our little pump meters are reading psi above ambient?

Tex

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Simon Bridge
Homework Helper
When they say 13.5 do they mean 13.5 psi greater than the 14.7 ambient or a total of 28.2?
If so does that mean that our little pump meters are reading psi above ambient?
This is correct.

The NFL is recommending a "guage pressure", not an absolute pressure.
The pump meters are comparing the pressure inside the ball with the pressure outside the ball - whatever that happens to be - that is how all pressure guages work.

This means the ball stiffness and size does not depend much on the weather.

It's the same with pretty much all quoted pressures - i.e. tire pressure.

I'm surprised and disappointed in all the arguments regarding the football deflation.
When I whip out PV=nRT and make a few semi-reasonable assumptions - The ball doesn't leak, the ball has a constant volume, we'll ignore humidity, PV=nRT is close enough...

Then I think, what if somebody inflated the ball inside and then took it outside. Say, 75F inside. It was in the neighborhood of 50F outside...
I got a football with 11.2psig at 50F.

Then, I upped the assumptions and made the 75F room have a 47% humidity. Consulting a psychometric calculator, this gave me 100% RH at 12.3psig, ensuring 100% humidity and some condensate at 12.5psig.

At 50F, the RH in the ball is still 100%, but some of the water has condensed so the number of molecules, n, serving as gas in the ball decreased.

This is where it became difficult. I had to satisfy the ratio of n when hot versus n when cold in the psychometric chart as well as PV=nRT. So, I varied the ball pressure in each until they were but satisfied at 10.7psig.

So, I got a difference of 1.8psi, just by inflating the ball indoors and taking it out. This doesn't include the heating of the air by the compression process. There's a fair chance that the balls were warmer than the room. Then again, the balls could have been pumped in a locker room, which would likely be a tad warmer.

Anyway, I invite others to use the tools we're given to make this ball deflation event less of a mystery. Or even better, someone could perform experiments to validate these ideas

russ_watters
Mentor
Mike, thetexan; certainly the media talks a lot without facts or science, but unfortunately in this case there is a lot of room for speculation. Given Belichick's propensity for "following the letter of the law", there isn't much I'd put past him to get around the spirit of the law. Inflating the balls in a sauna or with hot/humid air so they test ok and then deflate? Certainly plausible.

Anyway, I'm getting a 36F temperature difference to generate a 2 psi pressure difference. Heat of compression of a pump alone could account for that.

My main interest came from the grilling of the QB. He was coming across as somebody worried, and treading water, but not a natural liar. I don't care for the game, but it tweaked me to have so many people attack his honor with no understanding of what might be happening in the background.

As to a business man following the letter of the law, I've lived with that all my career. Do know what you call a public health limit on rat droppings in food? A specification.

Have a good good one, Russ.

DaveC426913
Gold Member
What I don't understand is how an under-inflated ball gives one team some sort of advantage. Doesn't that door swing both ways?

I think the issue was based upon slightly deflated balls being easier to grip

PS,
I used P1V=nRT1 and PV=nRT2, the volume, amount of gas, and R are assumed constant.
P1 = nRT1/V, P2 = nRT2/V, P2 = P1T2/T1.
75F -> 297K, 50F (roughly game day temp) -> 283.2 -> T2/T1=.954.
The target pressure was 12.5psig -> 27.2psia
Thus, P2 = 27.2psia x .954 = 25.9psia or, 11.2psig

The trick with 47%RH was based upon finding a humidity that just barely caused 100% once compressed to 12.5psig. Then, I found n for the dry air and n for water.
Through iteration, I found a pressure that returned values of n air/water that returned the same pressure when the total n was placed in the PV=nRT equation. That gave 10.7psig.

Last edited:
DaveC426913
Gold Member
I think the issue was based upon slightly deflated balls being easier to grip.
Yes.

By both teams.

So...

Yes, the other team may intercept the ball, but the QB uses his team's ball and generally it's his guys catching them.
In all cases, I bet it's really bad for anyone kicking...

russ_watters
Mentor
Yes.

By both teams.
No, each team gets 12 of their own balls. 11 of the Patriot's balls were under-inflated. The 12th was presumably the kicker's ball. The Colts' balls were all inflated properly.

There are 12 Game balls from each team. Each team uses there own balls when they have the football. There is also 8 Balls for Kicking game.

Here is a link to the vague rules on the quality of the football. This came from a blog where there has been lots of chatter on this topic.

http://static.nfl.com/static/content/public/image/rulebook/pdfs/5_2013_Ball.pdf

We all get into the details of this and rules call out 12.5 to 13.5 pounds. One might think it's weight and not pressure. I guess guy that wrote rules, only had "Football Math" in school.

The Baseline issues here as Science goes.

Temperature of Balls at Inspection time., Equilibrium of Ball at measurement time.
-1- Leather Temperature from massaging and rubbing surface to improve grip.
-2- Temperature of Internal air in the Football. Was it filled from 0 psig to 12.5 psig right before it was Sent to inspection, or just checked or topped off. Might have been filled with elevated air temperature.
-3- Water Content of Air in Football.

At the Game: Balls are colder and also wet. from Rain. Wetness factor depends on the time each Ball was on the field. Were all 12 New England Balls used in 1st half. Maybe it was just 11 and 12th ball was in Ball bag. No Data...

What is the Barometric pressure at the game as the storm came in. If this was falling, then the measurement at half time should be adjusted for this.

Repeatability of the Gauges? Does the Gauge release a tad bit of Air during this process?

The Physics backs the drop in Air pressure. But what is really interesting is why did the Physic not drop the pressure in the Colt Footballs? No pressure drop in their balls is the assumption. No data out there.

The expected drop should be around ~1.5 psig. the spec is just 1.0 psig. All of the balls should be out of spec.

I think this is not really a physics problem. What if the Colts, having been preped by Ravens, and also "Bad Blood" from Jets in the politics. If you planned on bringing this up out of the blue. You sure would want all of your balls in spec. Were the Colt balls kept at a lower temperature prior to inspection, say at 13 psi and 60 F. Then they tell Refs, they like 13.5 so refs adjust them as well.

New England understanding all of this, same for Colts and Ravens and other teams.

If these NFL teams ad a few NASCAR technical staff on them, these balls would be filled with a different gas. and this would be taken to the limit.

Some Tests have been done. This are two examples of testing..

Simon Bridge