Blitzer88
Active member
So much sciencey stuff..........................my brain hurts...................
Blitzer88":2u8ed3y8 said:So much sciencey stuff..........................my brain hurts...................
tacomahawk":166q6x1o said:They did check Indy's balls, and they were correct pressure
bmorepunk":t6zu7v9s said:Vancanhawksfan":t6zu7v9s said:Maybe somebody smarter than me can check these calculations, but it seems to me that there is a possibility that the ball pressure discrepancy could be explained if the referees originally tested the balls @ 12.5 psi in a heated locker room of 69.57 F, but were re-checked at half-time at 11 psi at a recorded outside temperature of 52 F...in other words... if the referees F'd up.
Here are my calculations...
Assumptions:
1. Let's assume that all balls tested were done pregame in the referees locker room just a little before 2hrs before gametime (4:53pm or 16:53) at 12.5 psi;
(NFL rules state that balls are turned over to teams 2 hrs 15 mins before gametime)
2. And then let's also assume all "deflated" balls were tested outside and found to be 11 psi at 7:53 pm EST (19:53)
Weather Observations January 18, 2015 - Foxboro, Massachusetts
http://www.weatherforyou.com/reports/in ... &icao=KOWD
@ 16:53
Temp in referees locker room: to be determined
Atmospheric air pressure: 29.83 in
@ 19:53
Outside temperature: 52F = 284.61K
Atmospheric air pressure: 29.68 in
Ideal Gas Law
PV = nRT
rewritten as P/T = nR/V = constant
Therefore:
P(1)/T(1) = P(2)/T(2)
(29.83 in + 11 in)/284.61K = (29.68 in + 12.5 in)/T(2)
T(2) = 295.10K = 69.57 F = possible referee locker room temperature
It seems to me that it is very realistic that the ref's locker room temperature was 70 F. Let's face it - refs aren't physics majors and they may very well not have considered that the balls would have had to be warmed to the ref's locker room temperature before being retested again.
Adding barometric pressure to the calculation certainly takes into effect that the kind of pressure gauge used to check the balls are relative. However, you have made a huge mistake in adding barometric pressure (which is measured in inHg) and psi together. You have to convert inHg to psi first.
29.83 inHg = 14.65 psi
29.68 inHg = 14.58 psi
Difference in measurement due to barometric pressure: 14.65 psi - 14.58 psi = 0.07 psi (and I'm spotting you the round up here). The barometric pressure's effect is two orders of magnitude below that of a 2 psi measurement, and really not worth putting into the calculation for something like this.
The temperature drop to account for 2 psi is:
Code:P1/T1 = P2/T2 T2 = (T1*P2) / P1 = (295.37 Kelvin * 72,394.95 pascals) / 86,184.47 pascals = 248.11 Kelvin
which is about -13 F.
The resulting pressure, if started at 12.5 psi, going from 72 F (295.37 K) to 52 F (284.26) would be:
Code:P2 = (P1*T2) / T1 P2 = (86,184.47 pascals * 284.26 K) / 295.37 K P2 = 82942.74 pascals
That's about 12.0 psi.
Vancanhawksfan":s91ji006 said:bmorepunk":s91ji006 said:Vancanhawksfan":s91ji006 said:Maybe somebody smarter than me can check these calculations, but it seems to me that there is a possibility that the ball pressure discrepancy could be explained if the referees originally tested the balls @ 12.5 psi in a heated locker room of 69.57 F, but were re-checked at half-time at 11 psi at a recorded outside temperature of 52 F...in other words... if the referees F'd up.
Here are my calculations...
Assumptions:
1. Let's assume that all balls tested were done pregame in the referees locker room just a little before 2hrs before gametime (4:53pm or 16:53) at 12.5 psi;
(NFL rules state that balls are turned over to teams 2 hrs 15 mins before gametime)
2. And then let's also assume all "deflated" balls were tested outside and found to be 11 psi at 7:53 pm EST (19:53)
Weather Observations January 18, 2015 - Foxboro, Massachusetts
http://www.weatherforyou.com/reports/in ... &icao=KOWD
@ 16:53
Temp in referees locker room: to be determined
Atmospheric air pressure: 29.83 in
@ 19:53
Outside temperature: 52F = 284.61K
Atmospheric air pressure: 29.68 in
Ideal Gas Law
PV = nRT
rewritten as P/T = nR/V = constant
Therefore:
P(1)/T(1) = P(2)/T(2)
(29.83 in + 11 in)/284.61K = (29.68 in + 12.5 in)/T(2)
T(2) = 295.10K = 69.57 F = possible referee locker room temperature
It seems to me that it is very realistic that the ref's locker room temperature was 70 F. Let's face it - refs aren't physics majors and they may very well not have considered that the balls would have had to be warmed to the ref's locker room temperature before being retested again.
Adding barometric pressure to the calculation certainly takes into effect that the kind of pressure gauge used to check the balls are relative. However, you have made a huge mistake in adding barometric pressure (which is measured in inHg) and psi together. You have to convert inHg to psi first.
29.83 inHg = 14.65 psi
29.68 inHg = 14.58 psi
Difference in measurement due to barometric pressure: 14.65 psi - 14.58 psi = 0.07 psi (and I'm spotting you the round up here). The barometric pressure's effect is two orders of magnitude below that of a 2 psi measurement, and really not worth putting into the calculation for something like this.
The temperature drop to account for 2 psi is:
Code:P1/T1 = P2/T2 T2 = (T1*P2) / P1 = (295.37 Kelvin * 72,394.95 pascals) / 86,184.47 pascals = 248.11 Kelvin
which is about -13 F.
The resulting pressure, if started at 12.5 psi, going from 72 F (295.37 K) to 52 F (284.26) would be:
Code:P2 = (P1*T2) / T1 P2 = (86,184.47 pascals * 284.26 K) / 295.37 K P2 = 82942.74 pascals
That's about 12.0 psi.
Thank you for addressing the question at hand. It is much more appreciated than the name calling I'm getting from some of the others here.
So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?
bmorepunk":3u1xurf5 said:Vancanhawksfan":3u1xurf5 said:bmorepunk":3u1xurf5 said:Vancanhawksfan":3u1xurf5 said:Maybe somebody smarter than me can check these calculations, but it seems to me that there is a possibility that the ball pressure discrepancy could be explained if the referees originally tested the balls @ 12.5 psi in a heated locker room of 69.57 F, but were re-checked at half-time at 11 psi at a recorded outside temperature of 52 F...in other words... if the referees F'd up.
Here are my calculations...
Assumptions:
1. Let's assume that all balls tested were done pregame in the referees locker room just a little before 2hrs before gametime (4:53pm or 16:53) at 12.5 psi;
(NFL rules state that balls are turned over to teams 2 hrs 15 mins before gametime)
2. And then let's also assume all "deflated" balls were tested outside and found to be 11 psi at 7:53 pm EST (19:53)
Weather Observations January 18, 2015 - Foxboro, Massachusetts
http://www.weatherforyou.com/reports/in ... &icao=KOWD
@ 16:53
Temp in referees locker room: to be determined
Atmospheric air pressure: 29.83 in
@ 19:53
Outside temperature: 52F = 284.61K
Atmospheric air pressure: 29.68 in
Ideal Gas Law
PV = nRT
rewritten as P/T = nR/V = constant
Therefore:
P(1)/T(1) = P(2)/T(2)
(29.83 in + 11 in)/284.61K = (29.68 in + 12.5 in)/T(2)
T(2) = 295.10K = 69.57 F = possible referee locker room temperature
It seems to me that it is very realistic that the ref's locker room temperature was 70 F. Let's face it - refs aren't physics majors and they may very well not have considered that the balls would have had to be warmed to the ref's locker room temperature before being retested again.
Adding barometric pressure to the calculation certainly takes into effect that the kind of pressure gauge used to check the balls are relative. However, you have made a huge mistake in adding barometric pressure (which is measured in inHg) and psi together. You have to convert inHg to psi first.
29.83 inHg = 14.65 psi
29.68 inHg = 14.58 psi
Difference in measurement due to barometric pressure: 14.65 psi - 14.58 psi = 0.07 psi (and I'm spotting you the round up here). The barometric pressure's effect is two orders of magnitude below that of a 2 psi measurement, and really not worth putting into the calculation for something like this.
The temperature drop to account for 2 psi is:
Code:P1/T1 = P2/T2 T2 = (T1*P2) / P1 = (295.37 Kelvin * 72,394.95 pascals) / 86,184.47 pascals = 248.11 Kelvin
which is about -13 F.
The resulting pressure, if started at 12.5 psi, going from 72 F (295.37 K) to 52 F (284.26) would be:
Code:P2 = (P1*T2) / T1 P2 = (86,184.47 pascals * 284.26 K) / 295.37 K P2 = 82942.74 pascals
That's about 12.0 psi.
Thank you for addressing the question at hand. It is much more appreciated than the name calling I'm getting from some of the others here.
So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?
The temperature change alone is not enough to explain a supposed 2 psi difference. The other things that have come up about this seem improbable, but I'm not in the profession of handling footballs.
randomation":z8avi3tc said:It was 2 psi btw not sure why you keep saying 1.5.
Vancanhawksfan":2uy60zy1 said:So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?
KiwiHawk":fwlgdb4h said:Vancanhawksfan":fwlgdb4h said:So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?
The difference between 295.37K and 248.11K is 47.26K. The ratio of units between Kelvin and Fahrenheit is same as the units between Celsius and Fahrenheit, which is 9/5. Therefore the temperature change required is 85.09 degrees F. If the game was played at 50F, the temperature in the referee's room would need to have been 135F. Definitely not plausible.
RunTheBall":2zzbfi64 said:This guy sure seems pretty defensive for a "Seahawks" fan. It's crazy all these accounts that were made a day or two after the NFC/AFC Title Games that blindly defend a known cheating franchise, QB, and coach.
Vancanhawksfan":35i8fu50 said:Sarlacc83":35i8fu50 said:Already been calculated at ~80 degrees F for the temp change.
Your calculation is wrong because you didn't take into account the existing air pressure in Foxboro.
Vancanhawksfan":246vtk27 said:RunTheBall":246vtk27 said:This guy sure seems pretty defensive for a "Seahawks" fan. It's crazy all these accounts that were made a day or two after the NFC/AFC Title Games that blindly defend a known cheating franchise, QB, and coach.
What does me discussing Tom Brady/Belichek/Patriots have anything to do with whether or not I'm a Seahawks fan? Stop being a knob, check my post history (even though its only a week) and then tell me if you still think I am.
The difference between you and I is that I don't "blindly" defend or criticize anybody which is something that you are doing to me right now. What...do you think that only a "true" Seahawks fan would only trash the Brady and company? How stupid is that?
Talk about hypocrisy.
Vancanhawksfan":1cg99ny4 said:randomation":1cg99ny4 said:It was 2 psi btw not sure why you keep saying 1.5.
I understood that the balls were measured at 11 psi at halftime.
Regulation balls are to be between 12.5 - 13.5 psi. Brady has stated that he prefers them at 12.5 psi.
I have not heard an announcement stating what the refs measured them at. But if Brady likes them at the lower end of the scale then its safe to assume that they were at 12.5 psi when the refs originally measured them.
No that is a temperature converter between degrees Fahrenheit and degrees Kelvin, not between the units of measure. The two scales don't start at the same point.Vancanhawksfan":2ipgcw7a said:No. The ratio between Farenheit and Celsius is 9/5. Here is a conversion calculator for you for Farenheit to Kelvin:
https://www.google.ca/?gws_rd=ssl#q=con ... +to+kelvin
SPIRITOF12":1luvl0ks said:Love this portion of Belichick's presser.....
BELICHICK: There is this possibility. Now if those footballs were picked up by, let’s say swallows, and carried up several thousand feet they would be directly affected by the changing barometric pressure and temperatures of the upper atmosphere.
ESPN REPORTER: What? A swallow carrying a football?
BELICHICK: It could grip it by the stitching.
ESPN REPORTER: It's not a question of where he grips it! It's a simple question of weight ratios! A five ounce bird could not carry a one pound football.
BELICHICK: Well, it doesn't matter. Will you please put this whole thing to rest?
ESPN REPORTER: Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?
BELICHICK: Please!
ESPN REPORTER: Am I right?