Gillette might want to check its wallpaper in the press room

Or effective?

He just said the press was barking up the wrong tree.

Throwing him under the bus would be like "Tom came to me and said 'coach, I can't win unless we deflate the balls illegally' and I told him to go for it".

If he never even looks at a football during the game, then what answer do you want him to give? The Superman Super-shield answer? "I'm responsible for everything and everyone and if I fart I always double-check that it didn't change the external pressure on any of our game balls so when they came back under-inflated I know exactly why...but I'm not telling any of you"?

I'm revoking your man card if you think that's being "thrown under the bus".

I think he is setting up Brady to take the hit for this. Someone has to fess up, Krafty must be lurid beyond belief. Someone has to have control of the ball situation, the cast of characters is limited. But the timing is suspect. How where the balls ok at check in , then some how in violation by half time.

You know, about seeking emergency treatment if you have an excessive lead for more than four hours ... um ... quarters?

Ignoring changes in external barometric pressure, and using the simple relationships between temperature and pressure*, the ball would lose/gain about 0.0425 psi for every degree Kelvin (Celsius) change in temperature.

That's based on the ball having 12.5 psi at 294K (70F).

So, a ball that had 12.5psi at 70F would have 12.03 psi at 50F(283K). That's without letting any air out (and assuming constant external barometric pressure).

Handy Chart (for a ball inflated to 12.5psi at 70F):
Temperature (F) = Pressure (psi)
-10 (Green Bay day) = 10.5
50 (Sunday's game) = 12.03
70 (indoor temp) = 12.50
90 (hot room/day) = 12.97
100 (Texas day) = 13.22

No wonder Aaron Rodgers likes his ball inflated as full as he can get away with! For a given mass and pressure at 70F, there will be a nearly 3psi range in game ball pressure over the range of temps in a given season!

(*note: I rearranged PV=nRT to get P=T(constant) because volume, mass, and the gas constant are, well, constant; and plugged and chugged)

(note 2: yes, working with both unit systems at once hurts my brain, but I needed absolute temps!)

By 'weather' I assume you mean temperature difference between air output from a compressor vs outside air temp? The ideal gas law can answer this question.

PV=nRT
P = pressure (in pascals)
V = volume (in cubic meters)
n = the amount of gas (in moles)
R = the ideal gas constant
T = tempterature (in degrees Kelvin)

Let's see what change in temperature T is necessary for a change in pressure P of 2 lbs (aka ~13,800 pascals)...

We have beginning and ending states:

P₁V = nRT₁ - ball when filled

P₂V = nRT₂ - ball during game

Isolate the constants:

P₁/T₁ = nRT/V

P₂/T₂ = nRT/V

Set the two equations equal:

P₁/T₁ = P₂/T₂

Isolate the unknown term (original air temp T₁):

T₁ = T₂(P₁/P₂)

P₁/P₂ is expressed in pascals. 1 psi = 6895 pascals, so 13.5 psi = 93,000 pascals and 11.5 psi = 79,250 pascals:

P₁/P₂ = 93,000/79,250 pascals = 1.17

The temperature T₂ will be the outside air temp (in Kelvin), 50°F = 283°K so:

T₁ = 283°K * 1.17 = 331°K = 136°F

So, the compressed air would have had to been at least 136° Farhenheit. Is this reasonable? Well, compressed air obeys the same ideal gas law we've been playing with - squeeze air into a smaller volume and and it gets hotter. And a quick google indicates that air compressors output air at much higher temps than they take in - so much so (as much as a couple hundred degrees Farhenheit) that large industrial compressors must be fitted with 'aftercoolers'.

So yeah, it seems pretty reasonable to this non-football fan science nerd that the pressure discrepancy can be accounted for by temperature difference.