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E90/E91/E92/E93 (2006 - 2013)
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  #1  
Old 03-05-2014, 08:48 AM
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Today's 3'er Tire Inflation Trivia Question

Your 3 Series is on a lift (hoist to you Southerners) and the mechanic inflates all four tires to door jamb pressures. He then lowers the car to the ground so that all 3400+ lbs. is sitting on the tires. He rechecks the tire pressure.
What will the tire pressure be? You don't have to be specific, what's your guess? Does the pressure rise an insignificant amount? A few pounds? More than four pounds? More than 10 pounds?

Pointandgo, no fair you answering.
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  #2  
Old 03-05-2014, 09:08 AM
jeanlucdckhard jeanlucdckhard is offline
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If I may ask this question here...

The sticker that says for speeds over 100mph.. the yellow sticker. That IS the minimum pressure I need to go over 100mph safely correct? It's autobahn time this weekend and I want to be safe and make doubly sure. Thanks!
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  #3  
Old 03-05-2014, 09:10 AM
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Seriously?

Tire pressure stays the same regardless of load or not.
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Last edited by spencers; 03-05-2014 at 09:11 AM.
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Old 03-05-2014, 09:15 AM
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My guess

My guess is that the tire pressure will increase. How much, I don't know but my guess is just a few pounds.
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  #5  
Old 03-05-2014, 09:24 AM
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Without looking at the other answers, I'll take a swing.

3400 lbs -> 850 lbs per wheel.
34 psig normal setpoint on my 225-45/17 tires. 8.9" wide / 78.5" circumference -> 700 sq.in of tire surface. 23754 lb(f) applied to inner tire surface. Increase load on tire surface from 23754 to 24604 lb(f) to pick up weight of car. 24604 / 23754 * 34 = 35.2 psig final pressure. Get the same result if you distribute the 850 pounds over the 700 sq.in. of tire surface (1.2 lb/sq.in.), so it seems to close.

So, I'd guess it goes up a psi or so. But I may be totally off base in force loading the tire.
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  #6  
Old 03-05-2014, 09:29 AM
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BiHoTTo115 BiHoTTo115 is offline
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Quote:
Originally Posted by Zooks527 View Post
Without looking at the other answers, I'll take a swing.

3400 lbs -> 850 lbs per wheel.
34 psig normal setpoint on my 225-45/17 tires. 8.9" wide / 78.5" circumference -> 700 sq.in of tire surface. 23754 lb(f) applied to inner tire surface. Increase load on tire surface from 23754 to 24604 lb(f) to pick up weight of car. 24604 / 23754 * 34 = 35.2 psig final pressure. Get the same result if you distribute the 850 pounds over the 700 sq.in. of tire surface (1.2 lb/sq.in.), so it seems to close.

So, I'd guess it goes up a psi or so. But I may be totally off base in force loading the tire.
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Old 03-05-2014, 09:43 AM
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I'm guessing it will be fractionally lower because the tech's repeated pressure checks bled an appreciable volume of air from the tires.
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  #8  
Old 03-05-2014, 10:04 AM
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Quote:
Originally Posted by Zeichen311 View Post
I'm guessing it will be fractionally lower because the tech's repeated pressure checks bled an appreciable volume of air from the tires.
I agree.
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Old 03-05-2014, 10:14 AM
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Quote:
Originally Posted by Zeichen311 View Post
I'm guessing it will be fractionally lower because the tech's repeated pressure checks bled an appreciable volume of air from the tires.
Quote:
Originally Posted by cwinter View Post
I agree.
You guys need to stop buying your pressure testers at Harbor Freight.
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  #10  
Old 03-05-2014, 10:16 AM
bbbuzzy bbbuzzy is online now
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The pressure will be related to the volume of the chamber it is in (the inside of the tire) and the temp of the air in the tire. Assuming the tire doesn't change temp, then the pressure will remain the same unless the volume of the tire changes as the weight of the car is added . I'd guess there is a minor change in volume leading to a minor change in pressure (less than 1 psi). However, now I'm skeptical, since you might have started this thread because of a bigger observed change.
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  #11  
Old 03-05-2014, 10:25 AM
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My initial reaction was to think an increase, but P=V/T says otherwise.

My vote is no change.



... Wait, how tall is the hoist? If over 10,000 ft I may want to reconsider.


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  #12  
Old 03-05-2014, 10:28 AM
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Quote:
Originally Posted by DSXMachina View Post
You guys need to stop buying your pressure testers at Harbor Freight.
Actually, in my case, I need to buy a bigger garage where I don't have to squeeze in the corners to get to the valves on the passenger side, causing me to use the pressure gauge at a less than ideal angle. But point taken.


Quote:
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However, now I'm skeptical, since you might have started this thread because of a bigger observed change.
You know what the weather is like in NH, right?
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Last edited by cwinter; 03-05-2014 at 10:30 AM.
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  #13  
Old 03-05-2014, 10:53 AM
DozerDan DozerDan is offline
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Are we north or south of the equator. What is the current ATM pressure, did he use a oiled or oilless compressor. Was he wearing a red or green shirt when he fill? Same shirt when he lowered the car. What is the tire wear currently at....
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  #14  
Old 03-05-2014, 10:54 AM
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Quote:
Originally Posted by Zooks527 View Post
Without looking at the other answers, I'll take a swing.

3400 lbs -> 850 lbs per wheel.
34 psig normal setpoint on my 225-45/17 tires. 8.9" wide / 78.5" circumference -> 700 sq.in of tire surface. 23754 lb(f) applied to inner tire surface. Increase load on tire surface from 23754 to 24604 lb(f) to pick up weight of car. 24604 / 23754 * 34 = 35.2 psig final pressure. Get the same result if you distribute the 850 pounds over the 700 sq.in. of tire surface (1.2 lb/sq.in.), so it seems to close.

So, I'd guess it goes up a psi or so. But I may be totally off base in force loading the tire.
I would say that the weight of the car is what causes a change of volume in the tire due to compression of the tire on the road surface. The weight of the car as such is meaningless to the calculation, as it is simply the cause for the volume decrease that holds the air. Of course, higher weight causes higher change in volume.

With that said, on a properly inflated tire, the dispersion of volume will be very low, I'd assume just a few percent. Since pressure and volume are inversely correlated and temperature is unchanged, the change in PSI will be inversely proportional to the volume change.

I'd assume this is also not a linear relationship. That is, the volume change on a tire with 10 PSI vs 30 PSI will not differ by a factor of 3. I'd assume a highly inflated tire will experience almost no change in volume and hence pressure.

BTW, I tried modeling the contact patch of the tire as a circle and distribute the weight across that. I found that for every 0.07 kg/cm^2 the pressure rises by 1 PSI. Using that model yielded tremendous PSI increments (10+) per tire, which seems questionable.
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Last edited by cwinter; 03-05-2014 at 12:20 PM. Reason: Fixed the correlation statement to show inverse relation, since p=t/v
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  #15  
Old 03-05-2014, 11:01 AM
Watchme Watchme is offline
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Eh, I'll take a shot.
Guess on area occupied by tire on concrete with car rested: 3x6 = 18 sq.in.
Added stress = 3400 lbs / 4 / 18 = 47.2 psi applied to the patch of the contact area.
Now we find internal pressurized area of the tire and divide by 18 sq.in. then use as a factor on 47.2 psi = my guess is less than 10% change in pressure.
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  #16  
Old 03-05-2014, 11:07 AM
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  #17  
Old 03-05-2014, 11:10 AM
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DSXMachina DSXMachina is offline
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Quote:
Originally Posted by DozerDan View Post
Are we north or south of the equator. What is the current ATM pressure, did he use a oiled or oilless compressor. Was he wearing a red or green shirt when he fill? Same shirt when he lowered the car. What is the tire wear currently at....
Finally, some pertinent questions. Obviously the answer cannot be known without this information.
1025 millibars
oilless
blue (red or green? what country are you in?)
different shirt but same color, a slush ball fenderberg fell down the back of his other shirt (typical day in a NH garage)
50% tread life
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  #18  
Old 03-05-2014, 11:14 AM
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DSXMachina DSXMachina is offline
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I'll wait until west coast lunch is over and respond. Someone got it dead right. But whom?

Where is Calwaterboy? How can we have a thread without Cal?
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  #19  
Old 03-05-2014, 11:33 AM
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galahad05 galahad05 is offline
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Quote:
Originally Posted by bocabimmer View Post
My initial reaction was to think an increase, but P=V/T says otherwise.

My vote is no change.



... Wait, how tall is the hoist? If over 10,000 ft I may want to reconsider.


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lol Absolutely true too.
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Old 03-05-2014, 11:43 AM
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galahad05 galahad05 is offline
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Oh, and the pressure will increase a tiny bit I think. You increase the pressure on a gas, you squeeze it down until it pushes back enough to stop the squeezing. But in this case the tire itself will push out a bit due to the increased pressure.

That's my story and I'm sticking to it.
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  #21  
Old 03-05-2014, 11:55 AM
Zoomie94 Zoomie94 is offline
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I agree with the "slight reduction in volume causes slight increase in pressure" camp.

Now whether or not your gauge has the precision needed to detect that slight increase in pressure is another matter.
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  #22  
Old 03-05-2014, 11:59 AM
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If any change were noted it would have to be attributed to "experimental error".


final answer: no change.
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  #23  
Old 03-05-2014, 12:03 PM
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I'm with Zooks ....... up 1 lb or so
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  #24  
Old 03-05-2014, 12:31 PM
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I cheated and looked this up. Because like most everything, this has been asked and answered before. This is what I found:

The equation you are looking for is the ideal gas law:

PV = nRT

P = Pressure
V = Volume of the container
n = The amount of gas
R = 8.314472
T = Temperature of the gas in the container

Solving for P you get:

P = (nRT)/V

In your scenario, you fill the tires and get a reading. You then put them on the car and read another reading. What has changed?

The amount of gas, R, and the temperature remain constant. (The temperature actually increases a little but that's not important right now.) However, the car slightly decreases the volume of the tire by deforming it. In terms of the equation, if V is smaller, P must be larger. Ergo, the pressure goes up.
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Old 03-05-2014, 12:32 PM
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Quote:
Originally Posted by DSXMachina View Post
I'll wait until west coast lunch is over and respond. Someone got it dead right. But whom?

Where is Calwaterboy? How can we have a thread without Cal?
I'll chime in later...
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