tire wear and all wheel drive

: >

: > Time for some math (someone double check this) : >

: > 1 mile is 63360" : > the circumfrance of a 215/60/16 is approx 82.3" : > with that, it requires that wheel to turn 770 times to go a mile : >

: > if you increase the circumfrance by 1/4 inch (.25) this wheel will : > require 767.5 : > turns to go a mile, which is a 2.5 turn delta. : >

: > At 60 MPH, the larger wheel will turn 2.5 turns less than the smaller : > wheel every minute. : >

: > Where is this difference obsorbed and will it cause damage? Is it : > different for Autos Vs Manual?

The circumference of the tire is irrelevant. Before you blast me, please consider the following. Remember the old saying "It's only flat on the bottom"? Well I think that applies here. If a tire is underinflated and the axle sits closer to the ground by, for example 1/4", then the effective rollout of the tire has been affected as if the diameter was reduced by

1/2". The length of the rubber that makes up the tread is unchanged, this is true, but since we are no longer talking about a circle, we have to consider the *effective* circumference. I think the effective circumference is

2*pi*(the distance from the center of rotation to the road)

Think of the extreme case: your tire is dead flat and you are riding on the rim. It doesn't matter how much extra rubber you have flapping around, you are riding on a shortened effective circumference.

Bayard

No kidding. Then you get tire ruined and have to throw the spare into the rotation, you know that spare has a lot less wear than the others, but there is no way you are going to buy three new tires. I mention this because I had to do it recently. If a Soob can't tolerate that much difference I don't know how impressed I am am with that engineering. I 'd be interested to hear some field experience.

-keith

Well, I was in that situation twice already. Once, I used the spare because the mileage on the others was pretty low, and I put the brand new tire I had to purchase onto the spare rim. Still measured about 1/4" difference circumference, if I recall correctly. No problems as a result _that I know of_! The second time, I had a tire belt problem, so I had to get another tire under warranty (glad it was still available even though no longer being manufactured.) I think I measured several tires on both my and my wife's Foresters, and chose those closest together for mine and for hers, maybe mixing one tire from the opposite car. We had to use the new one because at this point our spares did not match the new tires we bought when we replaced the factory tires with another type Yokohama tire. So far, we have not had any problems _that we know of_! My judgement is that the 1/4" is WAY too ridiculously tight a tolerance to abide by, and you just have to do your best to get close. If mine were above 1/2" difference, I think I would probably opt to buy another set and just try to recoup something out of these selling as used or something. (Or more likely, let them sit and rot somewhere, I'm afraid.)

I made an experiment with a bicycle : front wheel with 60 psi and rear with 30 psi. After just a few revolutions in a straight line, the mark had I made on the rear tire where it touched the ground had crept noticeably from the mark I had made on the front tire.

Radial tires have a circumferential steel belt so inflation won't change the circumference of a radial as much as it will a cross-ply tire, which I assume is what you have on your bike.

Consider the point on the tire's circumference directly below the axle. Give the wheel and tire one full revolution. The same point on the tire will again be under the axle. How far forward will the axle have moved? It will have moved a distance equal to the circumference of the tire.

If you don't believe that, consider a "wheel" that is square or rectangular and made of wood. Give it one revolution, thump, thump, thump, thump. How far has it moved? You can run this test with one of your kid's toy blocks, or a piece of 2x4 will do. It doesn't matter what the "radius" of the wooden block is; the distance moved will be the sum of the lengths of the sides of the block.

As I said in another note, if the tire is rotated at high speed the tire will stretch and deform, and differently inflated tires will have different

*effective* circumferences, or so I am led to believe.

Consider a flat tire, riding on the rim. Drive so that the wheel makes one full revolution. How far will you travel? a distance equal to the circumference of the rim. The rest of the tire just flops around. : : If you don't believe that, consider a "wheel" that is square or rectangular : and made of wood. Give it one revolution, thump, thump, thump, thump. How : far has it moved? You can run this test with one of your kid's toy blocks, or : a piece of 2x4 will do. It doesn't matter what the "radius" of the wooden : block is; the distance moved will be the sum of the lengths of the sides of : the block.

Now take your block of wood and wrap it in soft foam so that the outer surface is round. Press down on the block as you perform your experiment. After one revolution you will still have moved the sum of the four sides even though you have built up the circumference with padding. The fact that you are compressing the bottom as it hits the ground shortens the total distance traveled. This is analogous to running an underinflated tire. : : As I said in another note, if the tire is rotated at high speed the tire will : stretch and deform, and differently inflated tires will have different : *effective* circumferences, or so I am led to believe.

Agreed that there are other effects, and that the behavior will be different at different speeds. Also agreed that with an overinflated tire the effective circumference is the same as the real circumference since the weight of the car is not deforming the tire at the contact patch. : : -- : John Varela

I still stand by my original statement that a tires circumference does not change with tire pressure (or not enough to make a big difference to the circumference). I do agree that the diameter decreases with less inflation of the tire and this is what obviuosly causes a difference to the tires rpm. Tire circumference and tire diameter are two completely different things when pressure differences are brought into the equation.

Here's something that should settle some of these raging arguments.

At dmses.dot.gov/docimages/pdf68/135285_web.pdf is a US Department of Transportati "As was mentioned in the preceding section, TPMS can be divided into two main categories, Wheel-Speed Based - WSB (Indirect) and Pressure-Sensor Based - PSB (Direct). WSB systems infer tire pressures by using the vehicle's ABS hardware, specifically the wheel speed sensors, to measure tire-to-tire differences in tire rotational velocities that indicate that a tire is at a different pressure from the others."

The report also says that WSB sensors don't work very well:

"Four WSB systems installed as original equipment on 2000 or2001 vehicles were evaluated. Testing showed that all of the WSB systems would warn of a single tire being significantly underinflated (50 percent low) on a winding road course. However, on the 7.5 mile oval test track, which requires little turning input, only three of the four systems could detect tire pressure as low as 14 psi. The WSB systems also did not warn of two tires equally underinflated on the same side of the vehicle or on the same axle. Three of four systems warned of two underinflated tires in diagonal positions. All of the systems warned of three tires equally underinflated. None of the four WSB systems were able to warn of all four tires equally underinflated. The observed warning capabilities of these systems corresponded to the theoretical limitations of the sensing algorithm documented in this report."