levy@ttrdc.UUCP (Daniel R. Levy) (10/17/85)
I have a question and am open to speculation, theory, etc. concerning this. Namely, how is the pressure inside an inflated auto tire affected (if at all) by putting the car's weight on it? I seem to notice a little bit of difference between the pressure measured in a tire when it is on a loose wheel and the pressure measured in the same tire when it has been mounted on the car, but that could be measurement error. Something is supporting the quarter ton or so of force against the contact patch of the tire which is not exactly a large area either--it's obviously not all air pressure that is supporting it, yet you know that if you let out the air, the wheel will come to rest against the ground with the tire squished beneath it, so the air pressure is supporting SOMETHING, which is perhaps somehow proportional to the weight on the tire (???). So anybody who knows about this, especially who could relate the theory behind this, please either post or send me mail (I will summarize to net if warranted). Thank you. -- ------------------------------- Disclaimer: The views contained herein are | dan levy | yvel nad | my own and are not at all those of my em- | an engihacker @ | ployer or the administrator of any computer | at&t computer systems division | upon which I may hack. | skokie, illinois | -------------------------------- Path: ..!ihnp4!ttrdc!levy
levy@ttrdc.UUCP (Daniel R. Levy) (10/17/85)
I have a question and am open to speculation, theory, etc. concerning this. Namely, how is the pressure inside an inflated auto tire affected (if at all) by putting the car's weight on it? I seem to notice a little bit of difference between the pressure measured in a tire when it is on a loose wheel and the pressure measured in the same tire when it has been mounted on the car, but that could be measurement error. Something is supporting the quarter ton or so of force against the contact patch of the tire which is not exactly a large area either--it's obviously not all air pressure that is supporting it, yet you know that if you let out the air, the wheel will come to rest against the ground with the tire squished beneath it, so the air pressure is supporting SOMETHING, which is perhaps somehow proportional to the weight on the tire (???). So anybody who knows about this, especially who could relate the theory behind this, please either post or send me mail (I will summarize to net if warranted). Thank you. -- ------------------------------- Disclaimer: The views contained herein are | dan levy | yvel nad | my own and are not at all those of my em- | an engihacker @ | ployer or the administrator of any computer | at&t computer systems division | upon which I may hack. | skokie, illinois | -------------------------------- Path: ..!ihnp4!ttrdc!levy
rimey@ucbernie.BERKELEY.EDU (Ken &) (10/18/85)
>... Something is supporting >the quarter ton or so of force against the contact patch of the tire which >is not exactly a large area either--it's obviously not all air pressure that >is supporting it, yet ... Don't doubt so quick. Calculate it. Multiply the tire pressure (in pounds per square inch) by the footprint size (in square inches). You will get a quarter of the weight of your car. Ken Rimey rimey@dali.berkeley.edu ucbvax!ucbdali!rimey
levy@ttrdc.UUCP (Daniel R. Levy) (10/24/85)
In article <483@ttrdc.UUCP>, levy@ttrdc.UUCP (that's me) writes: >I have a question and am open to speculation, theory, etc. concerning this. >Namely, how is the pressure inside an inflated auto tire affected (if at >all) by putting the car's weight on it? I seem to notice a little bit of >difference between the pressure measured in a tire when it is on a loose >wheel and the pressure measured in the same tire when it has been mounted >on the car, but that could be measurement error. Something is supporting >the quarter ton or so of force against the contact patch of the tire which >is not exactly a large area either--it's obviously not all air pressure that >is supporting it, yet you know that if you let out the air, the wheel will >come to rest against the ground with the tire squished beneath it, so the >air pressure is supporting SOMETHING, which is perhaps somehow proportional to >the weight on the tire (???). So anybody who knows about this, especially >who could relate the theory behind this, please either post or send me mail >(I will summarize to net if warranted). Thank you. Thanks to all who responded. Here is a summary of the responses. ------------------------------------------------------------------------------- From: ihnp4!twitch!grt The air is supporting the tire. For example, if the pressure is 25 lbs/in2, and the load is 500 lbs, then the contact patch is 500/25 = 20 in2. The shape of the tire changes with the load, because the unloaded tire is holding the pressure with the casing. From the rim, the load pulls on the upper part of the casing via the downward force on the lower part. The bead in the edge of the casing holds it together. Tension all around the whole tire transfers the load to the bottom, which deforms until the contact patch is the right size. Low profile tires for racing are stiffer, so I assume that the sidewalls contribute some support in that case. By contrast, a racing bicycle tire holds about 100 lb/in2. The rider + bike is about 200 lbs, so the contact averages 1 in2 per tire. George Tomasevich, ihnp4!twitch!grt AT&T Bell Laboratories, Holmdel, NJ ------------------------------------------------------------------------------- From: ihnp4!oddjob!matt (Matt Crawford) When the tire is held in the air the pressure of the air within it exerts a force outward against the rubber. This force does not hold anything up against gravity because the forces, which are always perpendicular to the containing surface, are directed symmetrically about the center. When you attach the wheel to an axle the downward force exerted by the axle causes the tire to deform. The forces near the bottom, which used to point (ahem) radially outward get a greater vertical component because they are perpendicular to a more horizontal surface there. The defor- mation stops when there is enough upward force to support the load transmitted by the axle. Tires are not flexible enough to flatten at the bottom without decreasing their volume at the same time, so the pressure goes up. (You probably know that the pressure will increase with temperature also.) You can quote this if you want to. _____________________________________________________ Matt University crawford@anl-mcs.arpa Crawford of Chicago ihnp4!oddjob!matt ------------------------------------------------------------------------------- From: ihnp4!ihwld!gjphw Hello Dan, This is just a brief response to your question concerning tire pressures. To a good approximation, the ideal gas law can be used to get an intuitive understanding for what happens to the air in a tire. For the case that you describe, comparing a tire under load to one that is not loaded (not on the car), the difference in air pressure that you observe is probably due to the slight change in volume caused by the weight of the car. The car`s weight causes the tire to deform which reduces the volume of the airspace of the tire. According to the ideal gas laws, for a fixed temperature, a decrease in volume is accompanied by an increase in pressure. Examining the ideal gas law should provide the resolution to your question. Patrick Wyant AT&T Bell Laboratories (Naperville, IL) ihnp4!ihwld!gjphw ------------------------------------------------------------------------------- From: ihnp4!lanl!dxm (Douglas Miller) As a first guess, I would try to explain it like this; the axles rest on the tires, and the tires hold their shape because of the air inside them. Yes, when the car is resting on a tire, the pressure inside must go up simply because the volume decreases (the car squeezes the tire). The tire itself has some resistance to collapsing, but the main force holding the car off the ground is the air pressure that makes the tire retain its shape. Hope this helps, Doug Miller dxm@lanl ....!ihnp4!lanl!dxm Los Alamos National Laboratory, P.O.B 1663 MS J960, Los Alamos, NM 87545 ------------------------------------------------------------------------------- From: Bill Pataky <ihnp4!seismo!gymble.umd.edu!umcp-cs!pataky> Ok, Here's my interpretation (I'm no math genius by any means). The pressure exerted on the tire off the car is say X. Weight of the car is W, and is distributed evenly on all four wheels (in our case). The surface area on the inside of the tire including the steel on the wheel is SA. All units are psi and lbs. I would assume that the total pressure can be expressed as: W --- 4 TP = X + -------- (assuming no change in temperature) SA Like I said, I'm no math genius, but if this isn't it, it's got to be close since pressure is weight per area. If you come up with anything better than this, please let me know, you've got me thinking about it now. Bill Pataky Parallel Processing Lab University of Maryland pataky@gymble.umd.edu ------------------------------------------------------------------------------- From: nbires!rcd (Dick Dunn) OK, for a simple way to look at it, assume that the tire is made out of very flexible material. Fill it up with air and put the car on it. Now measure the area of the patch in contact with the ground--how many square inches? Let's say a 4x5 patch = 20 sq. in. How much of the car's weight is resting on this tire? Let's say 500 lb. Then the pressure in the tire is 25 psi (psi=pounds per sq in); it's that simple. The reason for starting with very flexible material is so that we don't have trouble figuring the effective contact area--no grooves to worry about. OK, next, with the tire still on the ground, pump it up to 50 psi. Lots less tire on the ground now--only 10 sq. in. Now to your question, but coming at it backwards--if we take the weight off the tire, what's the pressure? Depends on the tire. When we put the weight on the tire, we squashed it and reduced the volume. Pressure went up in proportion to decrease in volume--but all this depends on the shape of the tire. If you want to visualize this, think of filling up a loose bag with air--sort of a balloon that doesn't need to be stretched to fill it, or maybe a basketball that's just got enough air that it's not flat but not at all springy. Take this loose container of air--it's got essentially no difference in air pressure between inside and outside. Now sit on it, or put something (a car?) on top of it. The pressure inside goes up as you flatten it (reducing volume), and you can keep increasing the pressure until it breaks. As the pressure increases, the relationship between surface area in contact with the ground, internal pressure, and weight stays the same as I described above. Same thing with the tire--the pressure goes up as you apply weight, but just how much depends on the shape of the tire, the amount of weight, etc. --- Dick Dunn {hao,ucbvax,allegra}!nbires!rcd (303)444-5710 x3086 ...Simpler is better. ------------------------------------------------------------------------------- From: ihnp4!aoa!carl@bbncca HI. In answer to your query about auto tires: yes, the air pressure IS all that holds up the car. Consider these two points: 1) if you take an inner tube, or air mattress, or whatever, attach a pressure meter, andf squeeze the tube, the pressure reading will go up. It's a simple application of pressure*volume= constant*temperature ( which is why tire pressure goes up when the tire is hot, btw.). 2) Measure the approximate area of tire that touches the road, multiply by 4 for the four tires on a car, multiply by the measured pressure, and you get an answer in units of pounds. ( pressure in poynds per sqr inch multiplied by area in square inches) This answer should be the weight of the car. In case this sounds a little odd, remember that a tire pressure of zero is actually a pressure equal to the atmospheric pressure, NOT zero pressure. Hope this helps. -- Darwin's Dad (Carl Witthoft) ...!{decvax,linus,ima,ihnp4}!bbncca!aoa!carl @ Adaptive Optics Assoc., 54 Cambridgepark Dr. Cambridge, MA 02140 617-864-0201 " Buffet-Crampon R-13 , VanDoren B-45, and VanDoren Fortes ." ------------------------------------------------------------------------------- From: ihnp4!ucbvax!dagobah!doug i have figured that the tire pressure goes up 0.24% when you put the cars weight on it -- 29.92 to 30.00 psi. ------------------------------------------------------------------------------- -- ------------------------------- Disclaimer: The views contained herein are | dan levy | yvel nad | my own and are not at all those of my em- | an engihacker @ | ployer or the administrator of any computer | at&t computer systems division | upon which I may hack. | skokie, illinois | -------------------------------- Path: ..!ihnp4!ttrdc!levy