towertro11

Even with the same size tire, rolling circumference can be different, especially with different brand/model tires. Even same brand/model tires can have different measurements due to uneven wear and inflation. Traveling at a set speed in a straight line with different rolling circumference tires will cause axle rotations per minute to be different from axle to axle. How much depends on the degree of difference. The greater the difference, the greater the strain on your drivetrain/differential/clutch pack/torsen. More strain=more risk.

Good luck,
John

charles 200TQ

Here is a quick calculation to show the difference of 3/32" on a tire. My tires are 255/40R17, so I will use the numbers for it. The results will be different for other tire and rim sizes but it illustrates the point. Results depend on the radius from center of wheel to surface of tread.

17" diameter rim is an 8.5" radius which is 215.9mm. Tire section height is 40% of section width (255mm) which is 102mm. Adding rim radius and tire section height gives total radius.
215.9mm + 102mm = 317.9mm.
317.9mm/25.4mm = 12.52"

Circumference of tire = 12.52"x2xpi = 78.67"/12" = 6.56 Foot circumference
Circ. of tire with 3/32" wear = (12.52-0.09375)x2xpi = 78.08"/12" = 6.51 foot

Number of revolutions a tire makes in 1 mile is 5280 feet/circumference

New tire: 5280 feet/6.56 feet = 804.88 revs
3/32" wear: 5280 feet/6.51 feet = 811.06 revs

Difference = 811.06-804.88 = 6.18 revolutions

In this case, a the tire with 3/32" wear will have to rotate 6 more times than the newer tire in one mile. A wheel setup with a smaller overall radius will actually see an even larger difference in revolutions (you can imagine that as a radius gets larger and larger, that taking 3/32" from its radius will eventually become negligible, while as a radius gets smaller and smaller, taking 3/32" from the radius takes a larger and larger percentage of the radius).

You can make your own conclusions as to whether or not 6 revolutions in a mile is a "strain" on the drive-train. Certainly in any turn the difference in revolutions between the inner turning wheel and outer turning wheel is far greater than that shown of the tread wear difference. However the tread wear difference is always there, and while I don't believe it puts a tremendous strain on the drive-train, it is still extra work for it and should be minimized whenever possible.

spoon2000

I would think you could induce more than a 6 revolutions per mile discrepancy by having a tire (or two or three) that is over- or under-inflated relative to the others, as I'm sure many people do. It's not like the cars are set up like pick-up trucks (and old Audis for that matter) used to be where the center differential locked and you could really make the car buck if you tried to drive it on dry pavement and turned too sharply, etc. Then again, I can imagine that taken to an extreme, there IS a problem with having an odd-size tire on the car, hence Audi's decision to supply full-size spares (I mean, if Audi had its way, I imagine they would have rather put in a lighter weight spare than to go to the trouble of putting on an aluminum front end, as they did on the 4.2).

BillyP

The math above looks correct to me but various real world variations may swamp that difference. The auto engineers are aware of that variation exists and allow for some variation in the systems they design (out of round, inflation, differing tread wear, etc.). At the same time it is also correct to say you don't want to push the limits and unncessarily put stress on your car. I have located a used tire of the same tread pattern that had 5/32 tread vs the 6/32 I have on my car now, so my current plan is to go get that tire, ride one more season and swap out for an all new set of four, unless global warming has made snow tires superfluous (not seeing it here in Boston).

Just Me

probably fine if they are intended to be direct replacement. Perfect? No. Close? Likely.

G