Sorry, NE Pats fans...

[EricKei]

Comprehensive math/physics research aside o_O...

Guys, if you have the urge to say "This may belong on Fratching," it does belong there. Feel free to copypasta to a new thread over yonder. Thread closed.

[mjr]

Ok, now let's throw a wrench into your math. Balls don't have to be 12.5 PSI. They just have to be between 11.5 and 13.5 PSI.

Not correct. The PSI is between 12.5 and 13.5, according to NFL rules. Which is why I used 12.5.

So let's say the Patriots inflated theirs to JUST above 11.5 PSI. Then you could claim the drop in temperature is what brought the balls below legal pressure.

But not a drop by the amounts claimed. Again, even if you look at the absolute pressure (gauge pressure plus atmospheric pressure), you're still talking about less than a .5 PSI drop in gauge pressure, assuming 51 degrees F and the 1009.5 mb barometer.

[Greenday]

Ok, now let's throw a wrench into your math. Balls don't have to be 12.5 PSI. They just have to be between 11.5 and 13.5 PSI. So let's say the Patriots inflated theirs to JUST above 11.5 PSI. Then you could claim the drop in temperature is what brought the balls below legal pressure.

Or just use P1T2 = P2T1 assuming volume doesn't change (Which is silly). ANY drop in temperature will result in a drop in pressure to compensate and if they were at 11.5 PSI inside, they'll go lower outside.

[mjr]

OK, now someone update me. What are we talking about? All I know is it is something about football and 'whats-his-face' suing because of a suspension.

lordlundar gave a much more brief, more concise explanation than I would have, so I don't want to bore you.

I just find the math so beautiful. I<3Math

Then Euler's Identity will blow your mind: https://en.wikipedia.org/wiki/Euler%27s_identity

Now comes the fun part, where I get to check your figures. Oh, oh, oh and research everything too.

Nobody said I was being graded on this!!

[lordlundar]

OK, now someone update me. What are we talking about? All I know is it is something about football and 'whats-his-face' suing because of a suspension.

The Patriots were caught deflating the game balls post inspection during their championship game against the Colts. This is just another in a long line of excuses to cover it up.

[Android Kaeli]

Um...you lost me at explaining the variables. *stares* Math and I get along as much as oil and water do, I understand Greek more then I do math, ha ha.

But, in reality, it was a stupid thing to really come up with -- I don't even like the Pats (not even much of a football fan, either), but regardless I think this will still be talked about next year as well.

[Gilhelmi]

God Bless you, and your glorious beautiful math

OK, now someone update me. What are we talking about? All I know is it is something about football and 'whats-his-face' suing because of a suspension.

I honestly do not really care. I just find the math so beautiful. I<3Math

Now comes the fun part, where I get to check your figures. Oh, oh, oh and research everything too.

Thank you

[mjr]

I still have a hard time buying that argument that "the weather" could cause footballs to deflate as much as they allegedly did.

I'm going to try to avoid fratching here, and just lay out some facts, as best as I can:

Ok, first the Ideal Gas Law formula is defined as:

PV = nRT

Where P = Pressure, V = Volume, n = amount of gas, R is a constant, and T is the temperature.

1 PSI = 6.89475729 kPa
Constant R (gas constant) = 8.3144621 J/mol*K
Football volume (in Liters): 4.237
Standard Temperature: 273.15K (0º C, 32º F)
Standard Pressure: 99.973980705 kPa (14.5 PSI)

First, using the standard temperature and pressure measurements above, we have to find out how much air (n) is in the ball.

That formula is set up like this:

n = PV/RT

So in this case, let's assume a gauge pressure of 13 PSI (89.63184477 kPa). Also assuming standard temperature and using the volume of the football and the constant R, we get:

n = ( 89.63184477 * 4.237)/(8.3144621 * 273.15)

So the value for n is approximately 0.16721893

To make sure everything equals out, we plug back into the standard equation:

PV = nRT
(89.63184477 * 4.237) = (0.16721893 * 8.3144621 * 273.15)
The values are close. We end up with:
379.77012629049 = 379.77012977568510195
So if we round to 5 decimals, we get 379.77013 = 379.77013.

So, that day in Foxborough, there was a game-time temperature of 51ºF (283.706 K). For the purposes of this exercise, we will assume that we are at standard atmospheric pressure.

Using this information, we must now do a new calculation:

P = (0.16721893 * 8.3144621 * 283.706)/4.237

P = 93.09570707 kPa (or 13.50239077 PSI)

So between the two gauge pressures, there is a difference of 3.46386148 kPa, or basically an increase in a little more than ½ PSI with a 19ºF (10.556 K) increase in temperature.

Now, if we assume the balls were inflated in a room that is 72ºF (295.372 K), we get:

P = (0.16721893 * 8.3144621 * 295.372)/4.237

P = 96.92380559 kPa (14.05760952 PSI)

So assuming a 21ºF (11.666 K) change in temperature from where the balls were inflated to the outdoor ambient temperature, there's a PSI difference of 0.55521875.

Now that I've bored you completely to tears...it would appear that, assuming a 72ºF room temperature inflation and a game time temperature of 51ºF, the most the balls could have “deflated” by game time is roughly 0.6 PSI.

According to this site:

http://www.wcsh6.com/story/weather/2...roll/22065861/

The barometric pressure that day was 1009.5 mb.

So plugging that in, I got a "gauge pressure" reading of 12.08 PSI, after I started at the "low end" of 12.5 PSI.

A drop of .42 PSI.

Thoughts, everyone?