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Fast & Dirty
2009 Forester XT 4 speed auto
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Discussion Starter #1
Sorry but could not find consistent answer.

For all 4 wheels on a Subaru:

What has to be within a 1/4" for the AWD system to function properly?
Does the diameter need to be within a 1/4" on all 4 wheels?
Does the circumference need to be within 1/4" on all 4 wheels?

You assistance would be appreciated.
 

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2008 Forester AE
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1,726 Posts
from subaru.com

4. Do I have to replace all four tires on my AWD Subaru?
All of the tires on your AWD Subaru must be within 1/4 of an inch of rolling circumference (part that touches the road). This is because of our All Wheel Drive System.

Proper rotation of the tires at the appropriate service intervals will increase the life expectancy of your tires. This will also ensure that all four tires stay relatively equal in their tire tread wear. When vehicles are serviced, tires should be routinely checked to ensure that the alignment and tires are in good working condition.
 

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2005 Impreza RS Wagon Auto
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Yes it's circumference by technicality. In real though I doubt it is that small but that is another can of worms we shouldn't open...
 

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Fast & Dirty
2009 Forester XT 4 speed auto
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Discussion Starter #7
Thanks for the link. My thinking is to have a full size spare with the same rim as the other 4. This would allow me to rotate all 5. Thus, if I blow a tire while off roading I will run 4 until something happens or the tread wear means I need new tires.
 

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2015 Highlander AWD XLE 6AT
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4,259 Posts
Depends on the size of the original tire.
The relationship is actually independent of the tire size. Increasing the diameter of any tire by ~0.08" (~5/64ths; radius by ~0.04", or ~5/128ths) will increase the circumference by 0.25".

Examples:

Circumference of 24" diameter tire = 75.4"
Circumference of 24.08" diameter tire = 75.65", or +0.25"

Circumference of 36" diameter tire = 113.1"
Circumference of 36.08" diameter tire = 113.35", or +0.25"​

The same circumferential difference between the two wheel sizes would, however, result in different changes in the angular velocities of various drive train components (i.e., center differential output shafts), and it's those differences that Subaru is trying to minimize.

I have a sneaking suspicion that the 0.25" difference spec -- if valid -- is a convenient number based on typical wheel (rim + tire) sizes used by Subaru.

HTH,
Jim / crewzer
 

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2005 Impreza RS Wagon Auto
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The relationship is actually independent of the tire size. Increasing the diameter of any tire by ~0.08" (~5/64ths; radius by ~0.04", or ~5/128ths) will increase the circumference by 0.25".

Examples:

Circumference of 24" diameter tire = 75.4"
Circumference of 24.08" diameter tire = 75.65", or +0.25"

Circumference of 36" diameter tire = 113.1"
Circumference of 36.08" diameter tire = 113.35", or +0.25"​

The same circumferential difference between the two wheel sizes would, however, result in different changes in the angular velocities of various drive train components (i.e., center differential output shafts), and it's those differences that Subaru is trying to minimize.

I have a sneaking suspicion that the 0.25" difference spec -- if valid -- is a convenient number based on typical wheel (rim + tire) sizes used by Subaru.

HTH,
Jim / crewzer
I must be doing my math wrong cuz circumference of 24 doesn't give a diameter of 75.65...

Circumference of 24", radius (better tells tread depth difference) tire = 3.81972"
Circumference of 24.08" radius tire = 3.85951", or +.03979 tread depth

Circumference of 36" diameter tire = 5.72958"
Circumference of 36.08" diameter tire = 5.76937", or +.03979" (the depths would be the same, that is where you were right)


But now that doesn't quite seem right because that means the tire can only have a difference of ~1/32"...
 

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Premium Member
2015 Highlander AWD XLE 6AT
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I must be doing my math wrong cuz circumference of 24 doesn't give a diameter of 75.65...
Right. But a diameter of 24" results in a circumference of 75.4".

Reworded:

24" diameter x Pi (3.14159) = 75.4" circumference
or
12" radius x 2 x 3.14159 = 75.4" circumference

vs.

24.08" diameter x 3.14159 = 75.65" circumference
or
12.04" radius x 2 x 3.14159 = 75.65" circumference​

HTH,
Jim / crewzer
 

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Super Moderator
2018 2.5i Premium CVT
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17,882 Posts
Subaru did once specify the 1/4" max difference in rolling circumference, but the general view here is that they have now backed away from that. The original spec did indeed work out to somewhere between 1/32" and 2/32" difference in tread depth, and that was an awfully tight spec. It was probably written by the lawyers.

Quite coincidentally the spec seems to have gone away, and at about the time when Subaru introduced the donut spare, which is a major violator of the 1/4" difference! Subaru is now counseling moderation. Keep them close, if they aren't close it's OK to temporarily drive gently, and so forth.

Yes, the math says that the difference in tread depth is dependent on the difference in circumferences, and it does not vary by actual circumference. So 16", 17", 18", rolling circumference, or measured circumference—it's all the same.

There's a counter-intuitive puzzler that engineers and the like sometimes throw at each other. You're given a very long length of magic string, which has the ability to float in the air when its two ends are tied together. Its length is supposed to be the exact circumference of the earth, along a specified route (plus just a little bit more to account for the knot). You're also given a pair of magic shoes that let you walk along that entire route, never deviating from it, across oceans, through buildings, and over mountains, paying out the string as you go. However, it turns out that the string wasn't measured quite right and is about six feet too long. "It won't matter," you think. "Stretched out over about 25,000 miles the difference will just get lost." Then you get back to your starting point, tie the two ends together, and the string snaps tight, assuming its ability to float in the air—about a foot above the earth, all the way around! Moral of the story: Radius still equals Circumference/(2*pi), even for large values of circumference.
 

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2018 2.5i Premium CVT
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Here's the math that shows independence of circumference.

Radius = C/2Pi

Therefore, difference in radius (tread depth) at max spec = (C+1/4)/2Pi - C/2Pi

= ((C+1/4) - C))/2Pi The Cs cancel!

= (1/4)/2Pi

= .25/6.28

= .04​

1/32 = .031, and 2/32 = .062

So it's right in between, and just a tad closer to 1/32. Even if you have those monster rims from Lexani Wheels!
 

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2005 Impreza RS Wagon Auto
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Right. But a diameter of 24" results in a circumference of 75.4".

Reworded:

24" diameter x Pi (3.14159) = 75.4" circumference
or
12" radius x 2 x 3.14159 = 75.4" circumference

vs.

24.08" diameter x 3.14159 = 75.65" circumference
or
12.04" radius x 2 x 3.14159 = 75.65" circumference​

HTH,
Jim / crewzer
Ah, I read it as the other way around. It was confusing how you made it seem that the diameter was larger than the circumference.

Here's the math that shows independence of circumference.

Radius = C/2Pi

Therefore, difference in radius (tread depth) at max spec = (C+1/4)/2Pi - C/2Pi

= ((C+1/4) - C))/2Pi The Cs cancel!

= (1/4)/2Pi

= .25/6.28

= .04​

1/32 = .031, and 2/32 = .062

So it's right in between, and just a tad closer to 1/32. Even if you have those monster rims from Lexani Wheels!
Makes sense mathematically. I highly agree though that it is all legality to cover Subaru's buns. That tolerance either way is extremely small!
 

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Super Moderator
2018 2.5i Premium CVT
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This one is even better, though there's apparently an error in it somewhere. I have not attempted to chase that down.

 
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