Tire truing (tire shaving) on L.I.? - 4X4 -Enjoy the offroad! / CB Radio Area - Long Island Firearms

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Tire truing (tire shaving) on L.I.?


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13 replies to this topic

#1 cprstn54

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Posted February 22 2018 - 04:46 PM

If you need to replace a tire on an AWD (or even ABS) vehicle and the remaining tires are, say, 50% worn, you can have problems with the drive train if you use a new tire as the replacement. Experts say a difference in circumference as little as 1/2" and be problematic.The alternative is to spend 4X as much and replace all four tires, when three of them have maybe 25K miles of life in them.

 

There is equipment that cuts away rubber on a mounted rotating tire using a sharp disk on a router-like tool that is moved from side to side. See, for Nevada:

 

Anyone know someone on L.I. who does this?

 

Ken C

 



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#2 zzrguy

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Posted February 22 2018 - 05:33 PM

There is a couple of shops that tru up rims maybe they have the tool. But really buy a new tire cut 50% just buy 2 tires.


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#3 mrprovy

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Posted February 22 2018 - 05:41 PM

Go to a junk yard and buy used tires; why would you ruin good tires to match a worn tire?  Sounds like a waste of money to me!



#4 CJ

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Posted February 22 2018 - 06:03 PM

Do an epic AWD burnout and true them up yourself!


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#5 Camaro45th

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Posted February 22 2018 - 08:21 PM

Would rotating the tires during maintenance allow them to wear more evenly?

#6 VolkoSupply

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Posted February 22 2018 - 10:34 PM

most race shops that shave tires refuse to shape old/used tires.  They are harder, and also the last thing the shop wants is the blade on the tire shaver to hit a rock.

 

most street tires start life at 10/32 or so depth.  so basically 2/3" of depth total (or to keep the math easy 20/32.

 

You hit the wear marks around 4/32?  so 8/32 total?

 

That's a total max difference of 12/32...which is less than 1/2".

 

And yes, rotate your tires.  On a car with standard amounts of HP, the front tires will wear sooner than the rears.  The reason is they are doing more work by steering and subject to more friction.  In a FWD car this is even more so as they are both pulling the car and steering.


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#7 cprstn54

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Posted February 22 2018 - 11:16 PM

This is the article I read about the importance of keeping the tires very close together in circumference:

http://www.tirebusin...noreUserAgent=1

 

From this and others I have read, you just can't replace a 50% worn tire on an AWD car with a new one.

 

If you can find a used tire that matches in every key respect, that is great.

 

Shaving a new tire for $35  is way cheaper than replacing all four.

 

Ken C



#8 Joey232

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Posted February 23 2018 - 02:02 PM


That's a total max difference of 12/32...which is less than 1/2".

 

The 1/2" was circumference not radius.



#9 VolkoSupply

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Posted February 23 2018 - 03:49 PM

The 1/2" was circumference not radius.

you're right.  i missed that.



#10 Parashooter

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Posted February 23 2018 - 09:27 PM

There is a couple of shops that tru up rims maybe they have the tool. But really buy a new tire cut 50% just buy 2 tires.

 

a Belt sander works - hold it at 45 degrees and it spins the tire and cuts... 

 

This is the article I read about the importance of keeping the tires very close together in circumference:

http://www.tirebusin...noreUserAgent=1

 

From this and others I have read, you just can't replace a 50% worn tire on an AWD car with a new one.

 

If you can find a used tire that matches in every key respect, that is great.

 

Shaving a new tire for $35  is way cheaper than replacing all four.

 

Ken C

 

The thinking is that the larger diameter tire will have that wheel turning more slowly than the rest - the potential issues come up in the transfer case (which normally is a differential) BUT - there's also the differential on each axle - so that between the two wheels on the one axle, there would be a slight compensation in driveshaft speed/rpm - which is most likely going to be a non-issue on street tires.

 

on Offroad truck tires that have an inch or so of tread depth, I can see a new tire with 3 well-worn tires creating havoc...  but as Volko noted, we're talking about less than 1/2" (xPi, about 1-3/4" circumference) by half (7/8") from the diff...  by half from the transfer, 7/16"

 

.... not going to do the RPM math right now.... but you can see how small an error we're getting to.

 

The 1/2" was circumference not radius.

 

Nope - DIAMETER not circumference OR Radius  (it's 2r)

 

you're right.  i missed that.

 

You were closer than him though!



#11 Joey232

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Posted February 24 2018 - 08:51 AM


Nope - DIAMETER not circumference OR Radius  (it's 2r)

 

Parashooter, please read the article he linked to:

The auto dealership's invoice went on to state that proper engaging and disengaging of the four-wheel drive ``relies greatly on the equal circumference of the tires.'' A difference in tire circumference of more than 1/4-inch, the invoice said, ``will slow the process of going in and out of four-wheel-drive'' and a variance of more than 1/2-inch will ``cause damage to the drivetrain.''

 

They were talking about a 1/2" difference in circumference NOT diameter causing damage. That means that it only takes a difference in tread depth of 0.0796" or less than a tenth of an inch to cause damage. (circumference = 2 * Pi * radius)



#12 cprstn54

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Posted February 24 2018 - 01:42 PM

Found a DIY site:

 

Ken C



#13 Parashooter

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Posted February 24 2018 - 09:15 PM

 

Nope - DIAMETER not circumference OR Radius  (it's 2r)

 

Parashooter, please read the article he linked to:

The auto dealership's invoice went on to state that proper engaging and disengaging of the four-wheel drive ``relies greatly on the equal circumference of the tires.'' A difference in tire circumference of more than 1/4-inch, the invoice said, ``will slow the process of going in and out of four-wheel-drive'' and a variance of more than 1/2-inch will ``cause damage to the drivetrain.''

 

They were talking about a 1/2" difference in circumference NOT diameter causing damage. That means that it only takes a difference in tread depth of 0.0796" or less than a tenth of an inch to cause damage. (circumference = 2 * Pi * radius)

 

 

Fair enough - I didn't read the article linked, and i'll even accept what you're saying.....   But that means you're driving a car that will damage itself if you lose a few pounds of air pressure in one tire - doesn't take much to lose 1/4" in RADIUS - which multiplies out to 7/8" in Circumference.

 

What car were we talking about again? - So I can avoid buying one.


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#14 Joey232

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Posted February 24 2018 - 09:38 PM

What car were we talking about again? - So I can avoid buying one.

 

Now that I agree with 100%!






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