Football Physics - Local Weatherperson "Person of Interst" in Patriots DeflateGate

Climate conditions can entirely explain the mysterious 2.0psi pressure drop in the Patriots footballs.

• -1.52 psi  Temperature -- A drop from 81^o F to 51^o F. 
• -0.16 psi  Rain  -- Wet leather stretching the ball diameter 1/32" 
• -0.47 psi  Humidity -- Using humid air from the locker room to inflate the football

The weather did it.

And without trying place blame on any particular individual, the person of interest in this case - the person the NFL most needs to talk to - is a local weatherperson.

Overview

On a cold rainy afternoon of January 18, 2015, the New England Patriots played the Indianapolis Colts in a football game to determine who gets to go to the superbowl. At the beginning of the game each side selected 12 footballs to be used by their team for the game.  By NFL rule, each ball must be inflated to a pressure between 12.5psi and 13.5psi. The refs checked the balls before the game, and confirmed the balls were within the correct pressure range.  At halftime, the refs determined that the Patriots had 11 footballs which were under-inflated by 2 psi, and re-inflated them.

At the end of the game, a major controversy erupted, with people accusing the Patriots (who won the game) of cheating.  The implication was that a lower pressure football gave an advantage to people gripping the football.  The Patriots presumably deflated the balls prior to their use in the game, and after the refs measurement.

The easier grip may be true for us amateurs.  This may or may not be true at the professional football level where speed, accuracy, and feel of the football may be paramount to the quarterback.  A computer simulation done by Popular Science of the ball with the pressure differences yielded no effective change in grip or flight speed would have occurred.

Nonetheless, if the Patriots did deflate the balls after the refs initially checked them, which the press and most stand-up comedians assume they did, they deserve to be called cheaters.

Here are the key known facts:

• The temperature on the field was 51^o F at the start of the game.  According to weather charts there was a slight change in temperature towards the end of the game, of maybe 2 degrees.  The barometric pressure was 30.16 inches.
• The balls were measured to have 10.5 psi of pressure at half-time.
• The balls were measured to have 12.5 psi of pressure 2 hours before game time.
• The balls were moved from inside to outside well before game time, and were outside for most of those 2 hours.
• It was rainy and wet for the entire game.
• The ball-boys were rubbing the balls in the Patriots locker room to make for better handling prior to their being inspected by the refs.
• The height above sea level of the game field is 257 feet.

What we do not know:

1. What was the starting temperature of the balls when measured to be 12.5 psi? 
2. Did the rubbing increase the ball temperatures by friction from the ball-boys?
3. How much does football leather stretch as it becomes wet?
4. Does leather shrink when it becomes cold?
5. What was the uncertainty in the pressure measurements?
6. What was the relative humidity of the air that was pumped into the footballs?

Our Suspicious Questions:

1. Did a nefarious ball-boy remove some pressure from the balls as they were transported outside?
2. Did Tom Brady gain an advantage by having deflated footballs?
3. Or, did the Patriots gain an advantage by having only their balls re-inflated at halftime?  And why didn’t the refs give the same opportunity to the Colts?
4. There were reports that the Colts balls were measured and "within the legal range", although we don't have precise values as we have for the Patriots.  Did the refs really measure the Colts balls?  Given a temperature drop from 76oF to 51oF is it possible for them to still be in the "legal range" without some nefarious ball-boy adding pressure to the balls?  Or perhaps the Colts were keeping their balls warm on the sideline?

And the big question:

• Can the Patriots deflation be explained innocently just using science???

Science and the Results of Calculations

The Ideal Gas Law, familiar to most high school physics students states that there is a relationship between Pressure, Volume, and Temperature for a fixed amount of gas given by the following equation:

PV = nRT 

(For details see the Appendix below on the calculations)

The Scientific Results

Applying this equation to the known conditions above, we find the following results:

1. Assuming no stretching of the football due to the rain, dry air in the ball,  and precise measurements, the starting temperature of the football when the ref measured the pressure should have been 91^o F +/- 3^o F.
    
2. TEMPERATURE: Starting at 76^o F, and 12.5psi, and otherwise dry conditions with no stretching of the football, the football pressure outside at 51^o F will be 11.23psi.
   
   Delta P (Temperature change 76 -> 51^o F) = -1.27psi
    
3.  WET LEATHER DUE TO RAIN: Assuming the football stretches when wet, it would have to stretch by .5% (about 1/32" change in 7" diameter of the football)  to give a .16psi change in the pressure of the football.  Larger stretching of the ball could entirely account for the additional drop in pressure required here.  However, it may be unlikely that the ball stretched that amount.
   
   Delta P (0.5% Stretch of leather when wet at 51^o F) = -.16psi
    
4. HUMID AIR IN THE FOOTBALL: Assuming the ball was pumped up with air at 100% humidity at 76^o F, then condensation within the ball as the temperature dropped would have resulted in an additional .47 psi loss in ball pressure at 51^o F.  So humidity of the air pumped into the ball could be a significant factor.
   
   Delta P ( 100% humidity air pumped into ball at 76^o F ) =  -.47psi
    
5. UNCERTAINTY: Assuming a .1psi uncertainty in the measurement of the football pressures, (i.e. 10.5 psi could have really been 10.6 psi, and 12.5 psi could have really been 12.4 psi), then averaging the measurement over the 11 ball measurements results in an uncertainty of temperature of +/- 2^o F.  In addition, the temperature measurement of 51oF was done at the start of the game, but the pressure measurement was at halftime. Assuming that gives us a 2^o F uncertainty, the temperature uncertainty including that due to the pressure measurements is about +/- 3^o F.
   
   Uncertainty P = +/- .1psi
   Uncertainty T = +/-  3^o F

Scientific Comments and Speculations

• The Results shown above are dependent on a 76^o F starting temperature.  If indeed the starting temperature was 88^o F, or some temperature in between, then the resulting change in pressure due to temperature may be partially or entirely accounted for. For example 81oF would increase the Temperature deflation amount from -1.27psi to -1.52psi.
   
• The Results shown above can easily account for the 2.0psi difference in ball pressure observed at the game.  However, what we don't know is how humid the air was that was pumped into the balls.  And in fact the balls come from the factory preset to about 13.0 psi, so it might be that they just let some air out of the ball, and never put any additional in.  But it does seem possible that the Patriots locker room is hot and humid.
   
• Could the ball have stretched because it got wet?  I do not know what the hydrothermal properties of the football leather are.  When surfing the net, you find people who regularly stretch leather between 2% and 5% by getting it wet.  Some even claim 10%.  It confirms my feeling about leather jackets.  An interview with the football manufacturer revealed that they initially stretch the leather using steam.  However the stretch depends on the type of leather, the grain of the leather, the tanning method, etc.  Could the football leather have been stretched 1%? (Note: The expansion/shrinkage of leather as a function of temperature is approximately known.  Calculations using this produced negligible change).
    
• Can a ball boy heat a ball using friction by rubbing it?  Sure!  The temperature of their hands can also help that.  But we don’t know if they could have added the 13^o F required to get within the uncertainty range assuming they start at 76^o F. That seems like alot of rubbing.  It might be plausible, but needs to be verified.
   
• What about the uncertainty?  My wife runs a town soccer league, and has a power pump with a meter for “precisely” inflating balls to a certain pressure.  The uncertainty in that measurement (a dial guage) is at least 2% .  Anyone who has inflated a tire probably has noted the lack of precision in the gauges which measure tire pressure.  In our estimation, however, we note that the NFL is not cheap with their equipment.  There are digital gauges with hand pumps explicitly used to pump balls and measure pressure accurately to within .1psi.  So that is what we assume their pressure gauge was. ( As an aside about soccer, the legal pressure range is wide – somewhere between 8.7psi and 15.6psi, and the refs vary it according to the skill level of the players, and climatic conditions.  Younger players generally have the lower pressure, and the more skilled professional player will be at the high end of the range.  Perhaps we should all take note of this wide range allowed for a soccer ball when we think about football.)

Some Conclusions

So where does this leave us? 

Our scientific analysis, unless given some additional information, cannot rule out the purely innocent scenario for the Patriots, and in fact it is completely plausible.  A combination of temperature change, wet balls, and humidity of the air in the ball can account for the pressure drop.

It also cannot rule out that someone released some pressure in the balls prior to the game. 


It depends a lot on the starting temperature.  For most of these calculations, we assumed a temperature of 76^o F.  That could easily have been more than 81^o F in a hot locker room or if the balls were placed near a radiator.  It also depends on the humidity of the air used to pump up the balls, and the amount of stretch that occurred on the leather as they got wet.  Was the air in the locker room humid?

Plausibility

Does it seem plausible that Tom Brady instructed someone to secretly release .5 psi of air between the time the ball left the refs in the locker room and the game field?
  
Or is it more pausible that a combination of ball-boys rubbing the balls heating them up, and the stretching of the leather because of the rain and humidity of the air, and just uncertainty in ball pressure measurement?  Or were the balls originally sitting near a heat source just before the original measurement?

My own subjective sense sees that the former is far less likely scenario than the other possibilities – however I am a Patriots fan.  Many other non-Patriot fans will want to believe the former, which is a plot led by Tom Brady working with some ball-boys to release pressure in the footballs.


I suspect that Brady's answer will ultimately be that he instructed his ball boys to work the balls over, which they did in a warm locker room, and then instructed them to also make sure the balls were at the proper 12.5psi just before the refs measured them, and in fact the refs would have adjusted the pressure to make it 12.5psi if they were not.   Then they just brought the balls out to the field and assumed they will be fine for the rest of the game.

So my answer is that we cannot prove or demonstrate an exact answer, because we do not know all the starting conditions.  Some additional investigation may determine more about some of the speculations given above.  

Recommendations for the future

In the future, if the NFL wants to regulate the pressure in the balls, they need to think about temperature.

My recommendation would be that they not change much, but that they require a pre-game filling of the balls at a temperature of 72^o F and a pressure between 12.5psi and 13.5psi using dry air.

What the teams want to do from that point is up to them.  On cold days they may want to keep the balls in a heated box.  On hot days they may want to keep the balls cool.  Or maybe they won't care.

The refs could still have the right at any point in the game to inspect a ball, measuring temperature and pressure and make sure we're not having an equipment malfunction.  The ball would have to be okay according to a chart that correctly uses the Ideal Gas Law, and not just between two fixed pressure points, but uses temperature as well as pressure measurement.

Rating the “Scientists”

Now I would like to be subjective.  Acting like a professor, I would like to grade some of the science that has gone on in this whole affair.
Starting at the top:


A+: A group of people at Carnegie Mellon who not only did the correct calculations with Ideal Gas Law, but then did the actual experiment including wetting 12 brand new NFL footballs, to get the stretch effect on the leather.  The result was convincing.  The pressure on a ball starting at 75^o F and ending at a temperature of 50^o F dropped an average 1.8psi.  By the Ideal Gas Law this should have been just 1.25psi.  The stretching of a wet ball or the humidity (not reported in this report) contributed another .55 psi.  Congratulations to this group!  Excellent job!

A-:  Bill Belichick for hypothesizing that climatic conditions including temperature and wetness can affect the pressure in a football.  He also gets an A+ for effort in presenting the findings of others about this, but it was a B presentation largely because this was such unfamiliar territory for him, and I don’t think any Bill Belichick press conference can be rated above a B.  He did mention the correct number that a 20^o F change in temperature results in a 1 psi change in the football.  So the resulting grade is an A-.

B: Tom Brady’s “I don’t know” press conference.  He may have been trying to speak the truth here, but it was not overly convincing.

C: Bill Nye the science guy.  One of my favorite guys.  But he got his physics wrong.  He gets a D- because he came up with a 6% change in psi from 80^o F to 51^o F when the correct answer is about 1.5psi / 12.5psi which is about  12%.  He made the same mistake as many news reporters and high school physics students do in the application of the Ideal Gas Law.  But he did do a good thing in trying to change the focus to global warming instead of deflategate.  That was an A+ effort on global warming, but a little off topic.  So we average it out, and give him a C.

D-: Popular Science article by Chad Orzel on the science of football pressure in which the ideal gas was mentioned, but he never presented the results of any calculations.   He then over-inflated a couple of footballs and put them in the freezer, got a pressure difference of 2 psi as they froze.  He then concluded from this experiment that because the game time conditions did not match his experiment that it must have been some devious doings that deflated the balls, and that Bill Belichick has psychological problems.  He did publish some later columns on this, explained that he did not totally understand the results of his experiment, and expressed confusion over his pressure gauge.  But at least his charts had straight lines that indicated that the Ideal Gas Law applied. Then he gave a lot of correct statements about the Ideal Gas Law and the type of computations needed, but stopped short of doing the correct computations for the Patriots case.

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Appendicitis A. The Calculations

The Ideal Gas Law, familiar to most high school physics students states that there is a relationship between Pressure, Volume and Temperature for a fixed amount of gas given by the following:

PV = nRT
 

The trick here is that the Pressure (P) is the absolute pressure which includes the weight of the atmosphere (1 atm, or about 14.696 psi at sea level, and about 14.55 psi at Foxborough stadium which is 257 feet above sea level, and 14.62psi if we adjust for the barometric pressure of 30.18 on January 18  -- I use 14.7psi in the calculations below.). This means that a measurement of 12.5 psi on the football is actually about 27.12 psi.  Likewise Temperature (T) is the absolute temperature measured from Absolute Zero, which is about -459^o F.  So a value for T at 51^o F would be 510^o.

So in all our calculations we have make the proper adjustment for the “absolute” values.

Other than that, the calculation is easy. 

We can take our initial condition T1, P1, and V1 to be the halftime measurement.  The value of n which is related to the number of gas molecules, and R which is a constant, will be constant (unless the football leaks), then we can set up two equations if we want to figure out what happens at any condition P2, T2, and V2.

P1 * V1 = n * R * T1

P2 * V2 = n * R * T2

Dividing one equation by the other, we find that:

   (P1/P2)  *( V1/V2) = (T1/T2)

The parentheses are added for emphasis to show that the ratios are what we are concerned with.  This saves us a lot of work with units.

Assuming there is no volume change (i.e. no stretching of the football), then V1/V2 is just the number 1, and:

    P1/P2 = T1/T2

Given that we know P1 and T1, we can solve for either P2 or T2 if we know one or the other numbers.  In particular if we know that P2 is the pressure measured inside the locker room, then we can solve for T2.  The correct answer here for the game conditions is 91^o F.  If you can get that answer, then you will know that you have correctly converted from absolute temperatures and absolute pressures to our usual temperature and pressure units.

So

T2 = T1 * P2/P1 = (459 + 51) * (14.7 + 10.5)/(14.7 + 12.5)  = 550  =>  91^o F

Subtracting off the 459o of absolute temperature yields 91^o F.

STARTING IN A 76 degree LOCKER ROOM

Okay, that answered one question.  Now, can we start in a 76oF locker room, and get a 2psi decrease in pressure if it is 51^o F at halftime.

First, assuming no stretching of a wet football, no problems with humidity and condensation inside the football, and perfect measurements.  Just switching around the calculation we did above we find that the temperature change alone reduces the pressure by 1.25psi.

Delta-T ( 76 -> 51 )    =>    -1.27psi

What about the stretching of football leather as it gets wet?  Well that will cause a change in volume.  Since we are only interested in the change in volume, and the ratio V1/V2, we can express the change in a percentage.

Suppose the leather stretches by 1% when it gets wet.  This stretch only refers to one dimension, not the three dimensions required by volume.  So a 1% increase in size would cause the circumference of the ball to increase by 1%.  Since the volume will be proportional to the cube of the radius (the circumference is directly proportional to the radius), we have to cube this increase.  Thus:

V1/V2 = 1.01^3 ~ 1.03

Remember that V1 refers to the volume at half-time.  If we set T2 (the temperature of the locker room measurement) to be 76^o F, then

This means the volume would increase by 3%
.

(P1/P2)  *( V1/V2) = (T1/T2)

Solving for P1 in this equation

P1 = P2 * (V2/V1) * (T1/T2) = (14.7 +12.5) * (1/1.03) * (459+51)/(459+76) = 25.174     =>  10.47 psi

P1 = P2 * (V2/V1) * (T1/T2) = (14.7 +12.5) * (459+51)/(459+76) = 25.929      =>  11.23 psi

So the

Delta-V ( 1% stretch)    =>  -0.76psi

Humidity

If a ball at 76^o F is filled with air at 100% humidity, we know that some of that water vapor will condense if lower the temperature to 51^o F.   

To solve this we can look a different form of the Ideal Gas Law, which can be written this way.

PV = NkT

where N is the number of gas molecules.  At 76^o F, that N is composed of molecules of air and molecules of water.    As the temperature is decreased some of those molecules of water will condense and become water droplets on the inside of the ball.  They are no longer part of the gas.

What we really care about here is a density of molecules in the ball, and the relative densities of the air density to that of the water.  We can look up these numbers in a chart of saturated vapor pressure versus temperature done at atmospheric pressure.  (Note: this is not strictly correct.  It would be better to get these numbers at atmospheric pressure + 12.5 psi, but in the absence of having that information, this is the best approximation we’ve got – and it should be pretty good).

Accordingly, SVP( 51 ) = 9.53, and SVP(76) = 23.16.  See these tables for info.

The ratio

N1 / N2 =  760/(760-(23.16 – 9.53)) = .982

Using the formula

P1/P2 = (N1/N2)(T1/T2)

We solve for P1, and get

P1 = P2 * (N1/N2)(T1/T2) = (14.7 +12.5) * .982 * (459+51)/(459+76) = 25.46  =>  10.76psi

Delta-N (100% humidity at 76oF)  => 10.76 – 11.23 =  -.47psi

In the event that we don’t start at 100% humidity, then condensation will not occur until we reach a temperature where 100% humidity will occur.  This needs to be taken in account here.  If we start at 50% relative humidity, then we won’t reach 100% humidity until you get to about 57^o F.  The remaining change will result in much less change in psi.  It will not be a linear scale.

The Uncertainty.

If we use a gauge with an uncertainty of .1psi, the regular uncertainty principals apply.  We made two measurements, so the resulting uncertainty is the square root of the sum of the squares, so we get .14psi. 
This means our measurement of 12.5psi – 10.5psi = 2.0psi +/- .14psi.  This results in a temperature uncertainty (going to the full 91oF ) of  +/- 5.6^o F. 
However, we made 11 such measurements (the 11 footballs), so we divide this uncertainty by the sqrt(11), and round up a bit to get an uncertainty of +/- 2oF.  The initial measurement of 51^o F also has an uncertainty, which we might guess is +/- 2^o F.  Again using the sum of the squares argument, we get

Uncertainty (T)  = +/- 3^o F