do i have to replace all four tires? had a flat...

[vt_maverick]

On second thought maybe this is a more appropriate picture:

[circmand]

On second thought maybe this is a more appropriate picture:

Now you are just trying to hard.

And the post was unnecesarily wrong and proved my point. These 3 little words "or within 4/32-inch" that is a good amount of wear difference and documents that when mentioning tire size the manufacturer is really concerned about different sized tires not same sized tires with a little but different wear. I pulled a random tire off tirerack
http://www.tirerack.com/tires/TireSe...45&diameter=17

Its specs state that tread septh new is 10/32, its tread warranty is 50,000. If you divide the tread warranty 50,000 by usable tread depth 8/32 (assuming tire wear out is 2/32) you get wear ratio of 6,250 miles per 1/32 of wear. Multiply that by 4 (the allowable difference as stated in audi manual in quoted post) and you have an allowable mileage difference of 25,000 miles.

Therefore I submit for your approval that while NOT AS GOOD AS $ NEW buying 2 replacement tires is acceptable by manufacturer standards as long as the existing tires have less than 25,000 miles on them.

[etlsport]

Now you are just trying to hard.

And the post was unnecesarily wrong and proved my point. These 3 little words "or within 4/32-inch" that is a good amount of wear difference and documents that when mentioning tire size the manufacturer is really concerned about different sized tires not same sized tires with a little but different wear. I pulled a random tire off tirerack
http://www.tirerack.com/tires/TireSe...45&diameter=17

Its specs state that tread septh new is 10/32, its tread warranty is 50,000. If you divide the tread warranty 50,000 by usable tread depth 8/32 (assuming tire wear out is 2/32) you get wear ratio of 6,250 miles per 1/32 of wear. Multiply that by 4 (the allowable difference as stated in audi manual in quoted post) and you have an allowable mileage difference of 25,000 miles.

Therefore I submit for your approval that while NOT AS GOOD AS $ NEW buying 2 replacement tires is acceptable by manufacturer standards as long as the existing tires have less than 25,000 miles on them.

while the numbers presented dont mean anything because tires do not wear at a constant rate, i will absolutely agree with your logic

if the op had said that their remaining tires had 75% tread remaining there would have been no reason to replace all 4 provided an acceptable size/load/tread replacement tire was found. the question asked was at 40% should all 4 be replaced and i maintain an unwavering yes

if you read what i have been saying, my point is not that you ALWAYS have to replace 4 tires. it is that on a vehicle equipped with AWD/ 4wd, it is essential that the tires be close in tread depth. every point i have argued has been with the assumption that the tires not being replaced are below 6/32nds of tread remaining

that being said..
I have seen many vehicles in my shop with differential or transfer case damaged caused my mismatched tires. strictly my personal preference, my VX will never have more than 2/32nds tread difference

[tom4bren]

Sorry Circ, I gotta side with Eric on this one.

No logic involved ... just preference.

Well, maybe a little logic:

You drive a vehicle that can apply drive to all 4 wheels. That vehicle is designed (within reason) to do so with tires that match. If the tires don't match then obviously you are causing the drive system to work harder to compensate for the mis-match. You'll either cause a clutch system to slip more or a viscous coupling to slip more. Either way (in our case it's a clutch system) it will cause additional wear ... either clutch wear or additional heat generated by the viscous coupling. The bottom line is that if your tires don't match, it may not be unsafe, it may not damage the vehicle, it WILL cause accelerated wear of certain components.

That's all I got to say about that.

[circmand]

Etlsport however

Sorry Circ, I gotta side with Eric on this one.

No logic involved ... just preference.

Well, maybe a little logic:

You drive a vehicle that can apply drive to all 4 wheels. That vehicle is designed (within reason) to do so with tires that match. If the tires don't match then obviously you are causing the drive system to work harder to compensate for the mis-match. You'll either cause a clutch system to slip more or a viscous coupling to slip more. Either way (in our case it's a clutch system) it will cause additional wear ... either clutch wear or additional heat generated by the viscous coupling. The bottom line is that if your tires don't match, it may not be unsafe, it may not damage the vehicle, it WILL cause accelerated wear of certain components.

That's all I got to say about that.

Oh and Tom4bren your arguement fails to take into consideration the vehicle suspunsion. With the independent suspension and the weight of the vehicle all pushing downward the small diference in tire size does not affect the AWD. Its not like a solid suspenion where if you set the VX down on a flat surface the tire that is 2mm smaller in diameter (which means since the tire is attached at the center the tire is only half the diameter difference) the tire would not hover 1 mm above the surface. The suspension would still push the tire doan and contact would be solid. The radius is actual what should be measured. Heck a 1-2 mm difference would probably not be much different than driving on a paved road surface that is crowned.

[tom4bren]

Actually ... NO!!!

On paper, it's all about radius. On the road, it's all about circumference. I know that circumference is directly proportional to radius (high school was a loooooong time ago, but I remember the simple stuff). But, when the rubber meets the road, the tire can no longer be treated as a circle - it's flat where the contact is made. Believe it or not, the radius decreases (mushes down) but the circumference remains constant. A worn tire has less circumference (distance traveled per revolution of the tire) than a new one. Suspension has no bearing on the equation.