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Weight of car divided by surface area of tire on ground equals tire pressure. Also expressed as weight of car divided by tire pressure equals surface are of tire on ground. Can't have one without the other.
How do you express the weight of the car and the area of the surface and the pressure of the tire in standard units? Does this formula work in metric countries, as well? Kilograms of car weight divided by square centimeters of contact surface equal pounds per square inch, kilograms per square centimeter, bars, or pascals? Exactly how do I measure the surface area of tire contact on ground? Do I subtract the gaps between the treads? Where in my owner's manual does it state the correct surface area of tire contact?
I would even doubt your straight line formula. If you reduce the pressure from 25 to 1 psi, you will not increase the surface area by a factor of 25, because the tire doesn't have that much surface. If you reduce the pressure from 2 to 1, the increase in surface will not double, in fact it won't even be discernible.
You are correct, and even a child would intuitively know, that surface contact increases as air pressure decreases and/or vehicle weight increases, but any formula would be far too complicated to express in simple algebraic terms, and would need to involve dozens of factors.
Can you post a pic of the sticker? Offhand I think the 60 PSI must be referring to a space saver spare, assuming you have one of those rather than a full-size spare.
Typically the sticker lists a pressure for front and rear tires with a reduced vehicle load, typically just a driver and passenger, and higher pressures for full rated load, which is also posted on the sticker. Sometimes the same pressure is recommended front and rear, sometimes not.
Most performance oriented drivers, when driving an ordinary car (like me with my old Camry) will do better with *a bit* more pressure - say go from 32 to 35 or so - but maintain the same difference between front and rear tires if one is given on the sticker.
One needs to keep in mind that older rigs have tire pressure recommendations dating from when the car was new. Also keep in mind IIRC part of the Ford Explorer problem was that the OEM recommended pressures were kept low to give a soft ride.
In general keep in mind that information from OEMs, while often the best info you will get, is compromised. Marketing and accounting get some input to what's said. Case in point is certain BMW M cars, and I forget exactly which ones, but the ones with an 8000 RPM redline - that same engine in other similar cars (I think the M3 got the higher redline while the M version of the Z3 or Z4 was rated at 7500) - anyway, people who revved the cars to 8000 had rod bearing problems. Marketing got input to the redline, much to the detriment of anyone who "drank their Kool-Aid".
So what you get from the OEM corporate is not exactly what the Chief Designer of your car would tell you off the record, over a beer.
OP, if you want to talk to a real tire geek, someone who will simply try to steer you right, call Tire Rack. Alternatively, call the manufacturer of your tires, most have an 800 number, and again they have no reason to BS you.
Closing point - the guy who said look at tire wear patterns has a good point. And this is going to vary with driving style, road and ambient temperature, etc. If the center is wearing more than the edges, *you* would benefit from less pressure (not much, go in 1 or 2 PSI increments) in that tire when mounted on that position on your car. Likewise if both edges of the tire are wearing more than the center, add air. You can also on a dry day just feel of the tread, particularly on front tires, if it has a uniform temperature across the tire, you are good. This is exactly how a race car is set up, although they use a tire pyrometer rather than just the palm of the hand. But you can tell a lot just by feeling.
Thanks alot, that really helps. I looked at the sticker again and now I can see that the 60 PSI is for that spare. Although, 32 PSI seems too low when the tires can hold up to 44 PSI. I think like you mentioned before because the car is a bit older, an 2001.
Thanks alot, that really helps. I looked at the sticker again and now I can see that the 60 PSI is for that spare. Although, 32 PSI seems too low when the tires can hold up to 44 PSI. I think like you mentioned before because the car is a bit older, an 2001.
Actually that's about normal, 44 PSI is the highest safe pressure for the tire, 12 PSI worth of safety margin is not unusual or excessive.
Try them at 32, then maybe try upping it to 34 or so, if you are more interested in handling than in comfort.
At least you are paying attention to tire pressure, so many "numpties" out there don't even own a tire pressure gauge, don't know what the pressure should be, how to find that out, or what pressure is in their tires right now.
Weight of car divided by surface area of tire on ground equals tire pressure. Also expressed as weight of car divided by tire pressure equals surface are of tire on ground. Can't have one without the other.
...Huh?
Are you suggesting the weight of a car, on the tire, increases or affects the psi of the tire? I don't think so; at most it is a barely measurable increase in actual psi, and not by normal tire gauges.
Simple experiment: pump a bicycle tire to normal, off the bike. Put it on the bike and you and another person get on bike; check tire psi. It will be the same. Car is heavier but same situ. I can toss up the formula, but I am not trying to arm wrestle...just didn't understand what your post had to do with the OP's '32 psi in the tires' question, and your comment that the weight of a car has some affect on tire psi.
GL, mD
Actually that's about normal, 44 PSI is the highest safe pressure for the tire, 12 PSI worth of safety margin is not unusual or excessive.
Try them at 32, then maybe try upping it to 34 or so, if you are more interested in handling than in comfort.
At least you are paying attention to tire pressure, so many "numpties" out there don't even own a tire pressure gauge, don't know what the pressure should be, how to find that out, or what pressure is in their tires right now.
OP-your engine can run at 6000 RPMs, but that doesn't mean you want to do it all the time. As M3 says there is a safety and use margin built in. If you haul or tow often, you may want to raise the pressure. For long trips a little extra pressure can result in better MPGs. For traction in snow or sand you may want to lower the pressure, but not for long or for high speed driving. You don't want to build up a lot of unnecessary heat.
Contrary to the "thumb" method mentioned above, I take about 2 minutes a week to use a tire pressure gauge and get an accurate reading of my tires. This helps me identify a slow leak before it becomes a bigger problem. I have an air compressor at home, and I adjust pressure maybe once a month, or when I know my driving habits are going to change, such as a long trip, or a track day. If you don't have a compressor, determine how much air you need to add, then when you find air just add that amount.
How do you express the weight of the car and the area of the surface and the pressure of the tire in standard units? Does this formula work in metric countries, as well? Kilograms of car weight divided by square centimeters of contact surface equal pounds per square inch, kilograms per square centimeter, bars, or pascals? Exactly how do I measure the surface area of tire contact on ground? Do I subtract the gaps between the treads? Where in my owner's manual does it state the correct surface area of tire contact?
I would even doubt your straight line formula. If you reduce the pressure from 25 to 1 psi, you will not increase the surface area by a factor of 25, because the tire doesn't have that much surface. If you reduce the pressure from 2 to 1, the increase in surface will not double, in fact it won't even be discernible.
You are correct, and even a child would intuitively know, that surface contact increases as air pressure decreases and/or vehicle weight increases, but any formula would be far too complicated to express in simple algebraic terms, and would need to involve dozens of factors.
Quote:
Originally Posted by motordavid
OT...
...Huh?
Are you suggesting the weight of a car, on the tire, increases or affects the psi of the tire? I don't think so; at most it is a barely measurable increase in actual psi, and not by normal tire gauges.
Simple experiment: pump a bicycle tire to normal, off the bike. Put it on the bike and you and another person get on bike; check tire psi. It will be the same. Car is heavier but same situ. I can toss up the formula, but I am not trying to arm wrestle...just didn't understand what your post had to do with the OP's '32 psi in the tires' question, and your comment that the weight of a car has some affect on tire psi.
GL, mD
Let me see if I can explain my comment.
The mfg puts a sticker on the door which sets forth what they say is the correct tire pressure. This assumes a tire that is of a certain size such that the surface area of the tire at the spec'd psi is a certain area. The area of that tire can be determined by dividing the pressure per square inch into the weight of the car. So for example, a 3200 pound car with a recommended tire pressure of 32 psi one would expect a tire with 25 square inches on the ground (100/4). That tire may have a maximum psi of 40psi. But the mfg wants it filled only to 32psi because it will then spread out a little and put more area to the ground. If the pressure is lowered to 24 psi it will flatten out and present 33 square inches of area to the ground (133/4). If that tire pressure is increased to 40 psi it will become more round and less flat and only present 20 square inches to the ground. (80/4).
If the tire is even a little different in width than the OEM, the pressure must be changed to put the same rubber on the ground.
Are you suggesting the weight of a car, on the tire, increases or affects the psi of the tire? I don't think so; at most it is a barely measurable increase in actual psi, and not by normal tire gauges.
His formula didn't imply that at all. What it said was, given constant car weight, the contact surface of tire on ground varies inversely with the pressure. Change one of those, and the other changes in the opposite direction. Similarly, with constant pressure, raising the weight of the car increases the footprint (but has almost no effect on the pressure). Or, to maintain a constant surface area on the ground, raising the weight of the car requires an increase in pressure. Changing the tire pressure obviously changes the surface contact area, but obviously doesn't change the weight of the car.
All of which is true. My objection was to the formula that the proportions would remain constant and can be graphed with a straight line, as opposed to a variable slope. Due mainly to the shape of the tire, you can't create a simple formula in which a=bc. And it would not be useful anyway, because you can't meaningfully determine actual nor ideal area of surface contact.
His formula didn't imply that at all. What it said was, given constant car weight, the contact surface of tire on ground varies inversely with the pressure. Change one of those, and the other changes in the opposite direction. Similarly, with constant pressure, raising the weight of the car increases the footprint (but has almost no effect on the pressure). Or, to maintain a constant surface area on the ground, raising the weight of the car requires an increase in pressure. Changing the tire pressure obviously changes the surface contact area, but obviously doesn't change the weight of the car.
All of which is true. My objection was to the formula that the proportions would remain constant and can be graphed with a straight line, as opposed to a variable slope. Due mainly to the shape of the tire, you can't create a simple formula in which a=bc. And it would not be useful anyway, because you can't meaningfully determine actual nor ideal area of surface contact.
I think there may be some displacement based on the construction of the tire, but in general, the weight of the car will be evenly distributed across the portion of the tire that is actually on the ground. Certainly at the center of this graph it will be linear. I mean if we are talking about the range of 24-40 psi.
I drop my tire pressures for off road use to as low as 6 psi which really deforms the tire and it may be that there are little spots on the contact surface that exceed 6psi. But if you think about each of those little square inches as separate structures with a job to do (hold up 32 pounds of car weight) it makes a little more sense. (At least to me).
Oh, just to follow up, if the tire is deformed by the weight of the car, the tire may have less volume and the same amount of air. I think that would increase the pressure.
Actually, ignoring the structural stiffness of the tire itself, the tire pressure times the contact patch area is equal to the (static or dynamic) weight on any particular wheel, regardless of tire width or diameter, provided there is enough pressure to keep the wheel from contacting the road or ground.
Or so it would seem to me anyway.
So a wider tire will give you a wider but less long contact patch. This deforms the tire less, reducing heatup and in general giving better grip in dry conditions, wet conditions too so long as the tread is not overwhelmed by water.
Actually, ignoring the structural stiffness of the tire itself, the tire pressure times the contact patch area is equal to the (static or dynamic) weight on any particular wheel, regardless of tire width or diameter, provided there is enough pressure to keep the wheel from contacting the road or ground.
Or so it would seem to me anyway.
So a wider tire will give you a wider but less long contact patch. This deforms the tire less, reducing heatup and in general giving better grip in dry conditions, wet conditions too so long as the tread is not overwhelmed by water.
Right?
Yea, I think thats right. Whatever the pressure and weight (which are constants as far as this discussion is concerned) the contact area would be the same size. Just a different shape.
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...Huh?
