Deflate-Gate: Here We Go Again

Dr Pain said:

I expect the NFL will announce something like, all gameballs were inspected pregame, the Colts made a complaint about them and as a result the NFL took 1 ball for examination. It was found unaltered from the examination. I think the NFL has to say what prompted them to look. I suspect it was the Colts and the Colts had Kravitz malign the Patriots. This all reminds me of the accusation made against Edleman years ago. Thank god for that video but even being proven it was a false accusation, I still see people bring up this accusation.

Screw the tuck, where would Ray Lewis be now if the bloody white suit he wore did not disappear?

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For every N.F.L. game, each team has 12 to 20 balls that it has meticulously groomed and prepared according to the needs of its starting quarterback. The balls, brushed and primed using various obvious and semisecret techniques, bear the team logo and are switched out from sideline to sideline depending on which team is on offense.

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Triumph said:

Yeah, alleged taping walkthroughs of the Super Bowl opponent and a deflated football are equal.

NOT

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shmessy said:

......doesn't look like too many people agree with you nonchalance on this.


.

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shmessy said:

Question: Why would it take the NFL investigation of something so particular and precise (and simple) to take several weeks through and after the Super Bowl??

Why are they taking that strategy? It only hovers the cloud over the Patriots franchise.

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• Guy-Lussac's Law describes the relationship between the pressure of a confined ideal gas and its temperature. For the sake of argument, we will assume that the football is a rigid enough container (unless a ball is massively deflated, it's volume won't change). The relationship is (P1/T1) = (P2/T2), where P is the pressure and T is the temperature in Kelvins.
  
  
• The balls are inflated to between 12.5 and 13.5 psi at a temperature of 70 degrees Farenheit (294.1 K). Let's assume an average ball has a pressure of 13 psi. Since these are initial values, we will call them P1 and T1.
  
  
• The game time temperature was 49 degrees F (278 K). We are attempting to solve for the new pressure at this temperature, P2. We plug everything into the equation and get (13/294.1) = (P2/278). At the game time temperature, the balls would have a pressure of 12.3 psi, below league specifications.

• For a ball that barely meets specifications (12.5 psi), it's pressure would drop to 11.8 psi during the game... enough to be considered massively underinflated."

Jackson 2 said:

Surprise, surprise: on Comcast just now Felger believes it's true and Merloni thinks it could be. Then the clincher: Felger says they'll have Borges and Buckley coming up to see what they think. Click...

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NENGFAN said:

Kraft needs to tell the Commish to get his guys investigate and make a statement asap. It's us fans that are suffering the most.

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Dr Pain said:

*Furthermore, given that it was raining all day, the air in the stadium was saturated with water vapor. At 70 degrees, water has a vapor pressure of 0.38 psi. The total pressure of the ball is equal to the pressure of the air inside the ball and the vaporized water in the ball. At 49 degrees, the vapor pressure of water is 0.13 psi. Up to 0.25 additional psi can be lost if the balls were inflated by either the team or the refs prior to the game. Granted, it's unlikely that anyone would inflate balls from 0, but it easily could cost another couple hundredths of a psi in pressure.

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Dr Pain said:

I set up racing carts and cars in the past and tire pressure is very dependent on temperature. This is a better explanation than I can give
http://www.reddit.com/r/nfl/comments/2sxul5/deflategate_could_the_weather_have_an_effect_on/

"Science teacher here. Given the conditions of the game, a ball which meets specifications in the locker room could easily lose enough pressure to be considered under-inflated. Some math:

• Guy-Lussac's Law describes the relationship between the pressure of a confined ideal gas and its temperature. For the sake of argument, we will assume that the football is a rigid enough container (unless a ball is massively deflated, it's volume won't change). The relationship is (P1/T1) = (P2/T2), where P is the pressure and T is the temperature in Kelvins.
  
  
• The balls are inflated to between 12.5 and 13.5 psi at a temperature of 70 degrees Farenheit (294.1 K). Let's assume an average ball has a pressure of 13 psi. Since these are initial values, we will call them P1 and T1.
  
  
• The game time temperature was 49 degrees F (278 K). We are attempting to solve for the new pressure at this temperature, P2. We plug everything into the equation and get (13/294.1) = (P2/278). At the game time temperature, the balls would have a pressure of 12.3 psi, below league specifications.

*Furthermore, given that it was raining all day, the air in the stadium was saturated with water vapor. At 70 degrees, water has a vapor pressure of 0.38 psi. The total pressure of the ball is equal to the pressure of the air inside the ball and the vaporized water in the ball. At 49 degrees, the vapor pressure of water is 0.13 psi. Up to 0.25 additional psi can be lost if the balls were inflated by either the team or the refs prior to the game. Granted, it's unlikely that anyone would inflate balls from 0, but it easily could cost another couple hundredths of a psi in pressure.

• For a ball that barely meets specifications (12.5 psi), it's pressure would drop to 11.8 psi during the game... enough to be considered massively underinflated."

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Dr Pain said:

I set up racing carts and cars in the past and tire pressure is very dependent on temperature. This is a better explanation than I can give
http://www.reddit.com/r/nfl/comments/2sxul5/deflategate_could_the_weather_have_an_effect_on/

"Science teacher here. Given the conditions of the game, a ball which meets specifications in the locker room could easily lose enough pressure to be considered under-inflated. Some math:

• Guy-Lussac's Law describes the relationship between the pressure of a confined ideal gas and its temperature. For the sake of argument, we will assume that the football is a rigid enough container (unless a ball is massively deflated, it's volume won't change). The relationship is (P1/T1) = (P2/T2), where P is the pressure and T is the temperature in Kelvins.
  
  
• The balls are inflated to between 12.5 and 13.5 psi at a temperature of 70 degrees Farenheit (294.1 K). Let's assume an average ball has a pressure of 13 psi. Since these are initial values, we will call them P1 and T1.
  
  
• The game time temperature was 49 degrees F (278 K). We are attempting to solve for the new pressure at this temperature, P2. We plug everything into the equation and get (13/294.1) = (P2/278). At the game time temperature, the balls would have a pressure of 12.3 psi, below league specifications.

*Furthermore, given that it was raining all day, the air in the stadium was saturated with water vapor. At 70 degrees, water has a vapor pressure of 0.38 psi. The total pressure of the ball is equal to the pressure of the air inside the ball and the vaporized water in the ball. At 49 degrees, the vapor pressure of water is 0.13 psi. Up to 0.25 additional psi can be lost if the balls were inflated by either the team or the refs prior to the game. Granted, it's unlikely that anyone would inflate balls from 0, but it easily could cost another couple hundredths of a psi in pressure.

• For a ball that barely meets specifications (12.5 psi), it's pressure would drop to 11.8 psi during the game... enough to be considered massively underinflated."

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