Deflategate pressure drop is consistent with a ball air temperature of 72 degrees when tested initially.
I revised my original Deflategate posting after learning that it is absolute air pressure not pressure above standard sea level pressure that follows the Ideal Gas Law. I also allowed for stretching of the leather once the ball becomes wet. And for the possibility that the cold rain was was colder (45 degrees F) below the recorded air temperature at 53 degrees F. Together these adjustments make it even easier for the weather to fully explain the drop in ball pressure.
My Bottom Line: The NFL owes the Patriot Nation and Bob Kraft a big apology.
Correction #1: My initial use of the ideal gas formula did not recognize that it is absolute pressure, not pressure above the ambient air pressure that matters. Hence a ball with a pressure of 12.5 PSI is actually 12.5 PSI above the surrounding air pressure, which is about 14 PSI at sea level. So a decline from 12.5 PSI to 10.5 PSI is actually only an 8.2 percent decline in absolute pressure from 26.5 to 24.5 PSI. This makes it much easier for temperature changes to explain the difference in ball pressure. Only an 8.2 percent change in absolute temperature (approximately a 42 degree Fahrenheit drop) would be required it that were the only change needed.
Correction #2: It is well established that water allows leather to stretch. I found one site that noted that water can allow leather to stretch by 2-5% when wet. It does not specify how much force is needed to achieve this.
https://answers.yahoo.com/question/index;_ylt=A0LEVvwgfs9UP0AAr40nnIlQ?qid=20060908234923AAxt7xP
It is plausible that a new ball made of leather under pressure (scuffed up to let in the moisture quickly) might stretch 1 percent upon getting wet (such as in the rain). Since volume goes up with the cube of this stretching, this would be a (1.01)^3 -1= 3 percent increase in ball volume or decline in pressure. This amount would reduces the absolute temperature difference needed for the 2 PSI drop to only 5.2 percent (a change of only 27 degrees F.)
Correction #3: It was raining on game day, and the rain was probably much colder than the outside air temperature. So it is plausible that the game ball was as cold as 45 degrees Fahrenheit at game time when the low ball pressures were detected. This makes even lower initial testing temperatures consistent with the professed levels of underinflation.
A single formula can be used to calculate the ball temperature needed when tested initially to explain a ball pressure detected during the game that is 2 PSI lower, after getting colder (to 45 degrees F), .004 smaller (since ball volume shrinks when cold), and stretched 1% due to rain. It would be
Pregame testing temperature in F =(pressure change as a ratio)/(volume change due to cold)/(volume change due to leather stretching 1% when wet)*(45 degree ball temperature during game+460 degrees) - 460 degrees
(12.5+14)/(10.5+14)/(.996)/(1.01^3)(45+460) - 460 = 72 degrees Fahrenheit
Given this math, it would have been surprising if the ball pressure had NOT declined significantly.
Final comment #1: All of these calculations and hypotheses can be tested empirically. See the empirical analysis done by Headsmart Labs (http://www.headsmartlabs.com). They find that a rain plus a 25 degree drop is consistent with a 1.82 PSI decrease.
Final comment #2: Since the original game balls were reinflated by officials during halftime, the true ball pressures during the first half will never be known. Moreover there seems to be no documentary record of their pressures at the time they were re-inflated.
The XLIX Superbowl was a terrific game from the point of view of Patriots fans. Now it is time for the NFL to own up to its own mistake in accusing the Patriots of cheating. It was just a matter of physics.
Revised calculations
Various combinations of testing temperatures and PSI |
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A |
B |
C |
D |
E |
F |
G |
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H |
I |
J |
K |
L |
M |
N |
O |
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Adjustments for temperature only, correcting for absolute pressure at 14 PSI at sea level |
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Adjustments for changes in ball volume |
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Adjusting for temperature and football volume |
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Temperature F |
Degrees above Absolute zero |
Temperature adjustment |
Various game time or testing PSI readings |
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surface area |
sphere radius |
mean football radius |
volume |
Volume adjustment |
Various game time or testing PSI readings |
Game time temperature |
45 |
505 |
1.000 |
10.5 |
11 |
11.5 |
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189 |
3.8782 |
3.81183 |
232 |
1.000 |
10.5 |
11 |
11.5 |
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60 |
520 |
1.030 |
11.2 |
11.7 |
12.3 |
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189.2427 |
3.8807 |
3.81427 |
232.447 |
0.998 |
11.3 |
11.8 |
12.3 |
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70 |
530 |
1.050 |
11.7 |
12.2 |
12.8 |
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189.4045 |
3.8824 |
3.81590 |
232.7451 |
0.997 |
11.8 |
12.3 |
12.8 |
Possibl e testing temperatures |
80 |
540 |
1.069 |
12.2 |
12.7 |
13.3 |
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189.5663 |
3.8840 |
3.81753 |
233.0434 |
0.996 |
12.3 |
12.9 |
13.4 |
90 |
550 |
1.089 |
12.7 |
13.2 |
13.8 |
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189.7280 |
3.8857 |
3.81916 |
233.3418 |
0.994 |
12.8 |
13.4 |
13.9 |
100 |
560 |
1.109 |
13.2 |
13.7 |
14.3 |
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189.8898 |
3.8873 |
3.82079 |
233.6403 |
0.993 |
13.4 |
13.9 |
14.5 |
110 |
570 |
1.129 |
13.7 |
14.2 |
14.8 |
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190.0516 |
3.8890 |
3.82242 |
233.939 |
0.992 |
13.9 |
14.5 |
15.0 |
120 |
580 |
1.149 |
14.1 |
14.7 |
15.3 |
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190.2134 |
3.8906 |
3.82404 |
234.2378 |
0.990 |
14.4 |
15.0 |
15.6 |
130 |
590 |
1.168 |
14.6 |
15.2 |
15.8 |
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190.3752 |
3.8923 |
3.82567 |
234.5367 |
0.989 |
14.9 |
15.5 |
16.1 |
140 |
600 |
1.188 |
15.1 |
15.7 |
16.3 |
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190.5370 |
3.8940 |
3.82730 |
234.8357 |
0.988 |
15.5 |
16.1 |
16.7 |
150 |
610 |
1.208 |
15.6 |
16.2 |
16.8 |
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190.6988 |
3.8956 |
3.82892 |
235.1349 |
0.987 |
16.0 |
16.6 |
17.2 |
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160 |
620 |
1.228 |
16.1 |
16.7 |
17.3 |
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190.8606 |
3.8973 |
3.83054 |
235.4342 |
0.985 |
16.5 |
17.1 |
17.8 |
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Temperature (Fo) at which ball would pass test. |
2 PSI diff |
1.5 PSI diff |
1 PSI diff |
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88 |
77 |
67 |
Temperature only |
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86 |
75 |
65 |
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Temperature and volume change from temp |
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88 |
77 |
67 |
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temp, volume, and stretching from wetness |
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72 |
62 |
51 |
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Last row calculated as |
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(12.5+14)/(inferred test level+14)/(0.996)/(1.01^3)*(45+460)-460 |
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Notes |
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Revised calculations allow for sea level temperature to be 14 PSI, so a change from 10.5 to 12.5 PSI (above this level requires only a (12.5+14)/(10.5+14)-1=8.2 percent change in absolute temperature. |
See notes at the top, but final calculations also allow for the possiblities that ball temperature was 45 degrees, not 53 due to cold rain, and 1% stretching in leather due to rain. |
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Fields in first row and first column are input parameters, others are calculated |
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Original post
There is no mention of the temperature at which the footballs need to be stored or tested in the official NFL rule book. (Sloppy rules!)
The process of scuffing up the new balls to make them feel better no doubt warms them up. It would be impossible for it to be otherwise. An empirical question is how much did it warm them up and what temperature were they when tested?
Surface temps could have been below their internal temperature of the air, which is what matters for the pressure. Leather is a pretty good insulator (hence its use in many coats).
Anyone who took high school physics may remember that pressure and temperature satisfy
PV=nRT
Pressure*Volume=Number of moles*ideal gas constant*Temperature (Ideal Gas Law)
Temperature needs to be measured in degrees above absolute zero, which is -459.67 Fahrenheit (sorry metric readers!). The temperature at game time was 53 degrees. So the right question to ask is:At what temperature, T1, would the air in the ball have to be at the time the balls were tested such that once they cooled down to T0=53 degrees they measures two pounds per square inch (PSI) below the allowed minimum?
The lowest allowed temperature for testing was 12.5 PSI. We are told only vaguely that the balls were 2 PSI lower than this, but this is not a precise number. It could be it was rounded from 1.501 PSI. that would mean they might have been 11 pounds PSI when tested during the game. I examine 10.5, 11 and 11.5 as possible game time test PSI levels.The following tables shows possible combinations of game time testing temperature and half-time testing temperatures that would be consistent with various pressures.The right hand side of the table makes an adjustment for the fact that the leather/rubber in the ball would also have shrunk as the ball cooled down, which works against the temperature.Using the formulaPSI1=PSI0*((T1+459.67)/(T0+459.67). (See correction above!) Ignoring the volume change of the ball, it is straightforward to solve for what initial temperature the balls would have had to be for the observed game time temperatures.
Adjusting for a plausible guess at the small amount that the leather plus rubber bladder would have also changed makes only a small difference.
For a 1.5 PSI difference from testing to halftime , the air inside of them would have had to be at about 128 degrees at the time they were tested. (The leather skin could have been a lower temperature.) This would have made them feel warm but not burning hot to the hand.
Allowing the balls to be warm when tested is sneaky or perhaps accidental, but not cheating.
Go Pats!
Various combinations of testing temperatures and PSI |
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A |
B |
C |
D |
E |
F |
G |
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H |
I |
J |
K |
L |
M |
N |
O |
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Adjustments for temperature only |
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Adjustments for changes in ball volume |
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Adjusting for temperature and football volume |
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Temperature F |
Degrees above Absolute zero |
Temperature adjustment |
Various game time or testing PSI readings |
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surface area |
sphere radius |
mean football radius |
volume |
Volume adjustment |
Various game time or testing PSI readings |
Game time temperature |
53 |
512.67 |
1.000 |
10.5 |
11 |
11.5 |
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189 |
3.8782 |
3.81183 |
232 |
1.000 |
10.5 |
11 |
11.5 |
Possibl e testing temperatures |
80 |
539.67 |
1.053 |
11.1 |
11.6 |
12.1 |
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189.4368 |
3.8827 |
3.81623 |
232.8048 |
1.003 |
11.0 |
11.5 |
12.1 |
90 |
549.67 |
1.072 |
11.3 |
11.8 |
12.3 |
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189.5986 |
3.8844 |
3.81786 |
233.1031 |
1.005 |
11.2 |
11.7 |
12.3 |
100 |
559.67 |
1.092 |
11.5 |
12.0 |
12.6 |
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189.7604 |
3.8860 |
3.81949 |
233.4015 |
1.006 |
11.4 |
11.9 |
12.5 |
110 |
569.67 |
1.111 |
11.7 |
12.2 |
12.8 |
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189.9222 |
3.8877 |
3.82112 |
233.7001 |
1.007 |
11.6 |
12.1 |
12.7 |
120 |
579.67 |
1.131 |
11.9 |
12.4 |
13.0 |
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190.0840 |
3.8893 |
3.82274 |
233.9988 |
1.009 |
11.8 |
12.3 |
12.9 |
130 |
589.67 |
1.150 |
12.1 |
12.7 |
13.2 |
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190.2458 |
3.8910 |
3.82437 |
234.2976 |
1.010 |
12.0 |
12.5 |
13.1 |
140 |
599.67 |
1.170 |
12.3 |
12.9 |
13.5 |
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190.4076 |
3.8926 |
3.82600 |
234.5965 |
1.011 |
12.1 |
12.7 |
13.3 |
150 |
609.67 |
1.189 |
12.5 |
13.1 |
13.7 |
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190.5693 |
3.8943 |
3.82762 |
234.8956 |
1.012 |
12.3 |
12.9 |
13.5 |
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160 |
619.67 |
1.209 |
12.7 |
13.3 |
13.9 |
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190.7311 |
3.8959 |
3.82924 |
235.1948 |
1.014 |
12.5 |
13.1 |
13.7 |
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Temperature (Fo) at which ball would pass test. |
151 |
123 |
98 |
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159 |
128 |
101 |
Notes |
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Fields in yellow are input parameters, others are calculated |
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Column C is temperature minus absolute zero |
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Column D is the ratio of column C to the game time temp in absolute degrees and shows how much higher PSI would have been than at game time. |
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Columns E through G show possible testing PSI for three possible game time PSI levels. |
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Columns H through L show adjustments to volume which tend to reduce the PSI as a ball is heated. Calculations use rate of expansion of hard rubber per square inch per degree. |
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Columns M through O show Balll PSI after adjusting for both air temperature and football volume |
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Parameters and formulas |
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absolute zero= |
-459.67 |
fahrenheit |
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hard rubber expansion |
42.8 |
(10-6 in/(in oF))*) |
http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html |
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or |
0.0000428 |
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Used for column I expansion of surface area |
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Surface area assume to grow with the square of this proportion with temperature. |
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The approximate volume and surface area of a standard football are 232 cubic inches and 189 square inches, respectively. |
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http://www.answers.com/Q/Volume_and_surface_area_of_a_football |
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Surface of a sphere formula |
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4pr2 |
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Used to calculate radius of sphere |
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volume of sphere formula |
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4/3*pi*radius3 |
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Used to calculate volume of football. Volume adjusted downward by a fixed proportion because footballs are not spheres. |
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NFL rules
Rule 2 The BallSection 1BALL DIMENSIONSThe Ball must be a “Wilson,” hand selected, bearingthe signature of the Commissioner of the League, Roger Goodell.The ball shall be made up of an inflated (12 1/2 to 13 1/2 pounds) urethane bladder enclosed in a pebble grained, leather case(natural tan color) without corrugations of any kind. It shall have the form of a prolate spheroid and the size and weightshall be: long axis, 11 to 11 1/4 inches; long circumference, 28 to 28 1/2 inches; short circumference, 21 to 21 1/4 inches;weight, 14 to 15 ounces.The Referee shall be the sole judge as to whether all balls offered for play comply with these specifications. A pump is to befurnished by the home club, and the balls shall remain under the supervision of the Referee until they are delivered to theball attendant just prior to the start of the game.
From the Free Dictionaryideal gas lawn.A physical law describing the relationship of the measurable properties of an ideal gas, where P (pressure) × V (volume) = n (number of moles) × R (the gas constant) × T (temperature in Kelvin). It is derived from a combination of the gas laws of Boyle, Charles, and Avogadro. Also called universal gas law.