OT: Football "Physics"

I think the Pats know if you fill them in a hot room they'll be soft when exposed to outside temps.

Cheers

Guy-Lussac law using absolute pressure and including the compressiblity of nitrogen shows that physics could be responsible for a large portion of the drop

defeat-deflategate.html

It also depends on the temp and location of the compressor.

Place five numbered balls into five refrigerators at five temps, and make one a chamber that sits 10 degrees above normal ambient.

Use 72F as a std. ambient point.

Place the compressor tanks in each fridge as well, and let them all settle out for some hours.

Fill each to exact calibration center point. Make measurements.

Cycle the balls through each fridge, settle time and fill and measure events. That way, you rule out ball to ball differences.

Collect data which can then be used as a baseline expected rule of thumb.

On game day, all balls should be stacked into a vertical, clear plexiglass box for ALL to see at ALL times. The ball boy gets balls from there feeding out of the bottom of the box. Once filled and inspected, they should get their fill valves SEALED, and get placed into the sideline boxes. Easy Greasy.

Howard Hughes proved that liquids are compressible.

Gas most certainly is.

Any idiot understands the operational metrics of a bourdon tube.

Any others should likely remain out of the conversation(s).

I don't know the dynamics of the bladder of a football, but that much pressure drop doesn't seem to me to fit "natural" deltas. ...Jim Thompson

Oh! Are they nitrogen-filled? ...Jim Thompson

Cheers

Maybe they fell for that tire inflation with N2 scam ?>:-} ...Jim Thompson

By my math, the pressure drop (for 30F delta) should be about 0.72 PSI, actually less since the ball isn't rigid. Less than half of what the Prominent Boston Physicist got. Somebody could check this.

I get an 11.4 PSI drop. Go thee unto: Click on Solve for P2 T1 = 273 + 70C = 343K T2 = 273 + 40C = 313K Click on Calculate and it produces an 11.4 PSI drop in pressure.

Oops. I made the same mistake as the Prominent Boston Physicist. P1 = 12.5 PSI as the initial pressure. The calculated P2 = 11.4 PSI. Pressure drop is 12.5 - 11.4 = 1.1 PSI

I'm not gonna mess with PVT calcs, but just thought I would let you know that no pro football game has ever been played at 70C, and no pro football game has ever been at 40C, at least during the winter.

--- Joe

Oops 2.0. This is what happens when I rush and don't check my work.

P1 = 12.5 PSI T1 = 273 + 21C = 294K T2 = 273 + 4.4C = 277K

Calculated P2 = 11.8 PSI Pressure drop = 12.5 - 11.8 = 0.70 PSI

Looks like the other JL is correct.

Thanks.

You guys are getting closer. Now there is one more thing that you are forgetting.

Dan

You need to use absolute pressure.

Right. That makes the drop more like 1.56 PSI.

ESPN reports that 11 of 12 balls were 2 PSI below the legal limit of

Should the pressures be gauge or absolute?

Sigh. I can't believe I screwed it up this badly. This is basic stuff. 4th try is the charm (I hope).

Just for fun, converting back to gauge pressure: 25.6 PSIA = 10.9 PSIG Pressure drop, as measured by the gauge on the pump: 12.5 - 10.9 = 1.6 PSI

Double checking: P1/T1 = P2/T2 P2 = P1*T2/T1 = 27.2 * 277/294 = 25.6 PSIA Ok, that works.

Enough. Gone sulking...

defeat-deflategate.html

Yes. Being elastic, the bladder/ball/entire containment actas as a spring for a small period of the deflation. Once past a certain point, however, it also relaxes, because it is not like a deflated balloon, where the expansion is several tens of percentage points.

Nice catch. The effect may be less than you equated due to the relaxation point not being very far from a mere couple pounds off "full". IOW, you may have given it more spring than it actually has.