
#1
Mar912, 08:31 AM

P: 70

Imagine an ordinary car wheel with rubber tire, installed in a car axle. So the weight of the car (some of the weight of course, because there are 4 wheels) is pressing on it. Let's denote this pressure of the air in the tire as [itex]P_{c}[/itex].
Now the same wheel (with its rim and tire) remove from the car axle, so it is now freely lying on the floor. Hence no car weight is pressing on it. Let's denote the pressure of the air in the tire as [itex]P_{f}[/itex]. Question: Is [itex]P_{c}[/itex] different than [itex]P_{f}[/itex] ? How to show it by using ideal gas law equation? You may assume the weight of the rim as negligible if needed. My personal guess: [itex]P_{c}[/itex] is more than [itex]P_{f}[/itex] because with the weight of the car pressing on the tire, volume of tire (hence volume of gas) reduces, thus pressure increases. In other words, if a tire pressure is 30psi while installed on car, once take it out from the car axle, its pressure will drop. 



#2
Mar912, 09:44 PM

P: 686

You are correct, but the difference is so small that you will not notice it on a tire gauge.
Suppose your car is 4000 pounds with 1000 pounds on each wheel. Also suppose your tires are 10 inches wide. At 30 psi your contact patch for each wheel will be about 33 square inches. Since the tires are 10 inches wide that means your contact patch is only 3.3 inches long. Deforming the tire with a typical 6.5+ foot circumfrence so that there is a 3.3 inch flat spot will change it's volume (and pressure) only very slightly. 



#3
Mar1012, 08:31 AM

P: 70





#4
Mar1112, 06:42 PM

P: 686

Comparing pressure of air in a tire installed in a car & a free tire
Correct.



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