F150_Life

thanks brady_bot, i'll give that a shot. I think we need a new thread of this. I just want to see what everyone is running on their edge evo's.. still working on tuning mine to my liking.

BassAckwards

I get what your saying, but I still don't see how that would take away circumference from the tire like I said in my post you could squish that tire down to a very elongated oval and it's still going to have the same circumference and make the same revolutions per mile. The circumference of the tire cannot physically change.

VTX1800N1

Think of it this way, the tire "flattens" or squishes just a little when the weight of the truck is put on it. So, take that to the extreme to see if there is an effect. Take all the air out of the tire. The tire still has the same circumference everywhere but under the wheel, where it is "squished" down flat. Certainly you don't think the effective diameter and revolutions per mile are the same as an inflated tire? Essentially the effective diameter has been reduced to that of the wheel. This is also what happens with the small squish when the weight of the truck is on an inflated tire. The radius of the wheel/tire combo only matters when measured from the center of the wheel directly down to the ground. Squish height absolutely makes a difference. The circumference of the tire can absolutely change, because it is not rigid. With force (weight of a vehicle) it will deform.

BassAckwards

I think you are underestimating the rigidity of a tire. I understand how a load will change the effective radius of a tire but that still doesn't change the circumference, the tread belts are very much a rigid structure. I see what you are saying about a completely deflated tire, but even on a slightly under inflated tire, the circumference remains the same and the truck drives on that entire circumference.

I guess I'm just not seeing how any normal amount of deflection on the tire can change the circumference, the tire should always measure the same distance around.

Take a 10cm length of string and tie the ends together. No matter how you shape that string (Perfect Circle, Square, Triangle, Oval) it is still 10cm long.

VTX1800N1

I used the example of a completely deflated tire so that the concept would be obvious. The squish of the tire with a vehicle load on it is merely this same effect on a much smaller scale. When a tire is loaded with vehicle weight, it has a contact patch with the road surface. Essentially, that part of the tire is not curved, as it confirms to the road surface. That reduces the radius measurement of the tire at that point. The radius is what matters when calculating how many revs per mile the tire will make. Because of the squish height, the tire will behave as though it had the circumference based on the radius between the center if the wheel and the road surface. This is smaller than the unloaded circumference of the tire.

Again, take the example to the extreme to understand the effect. Imagine a 5" wheel with a very tall tire, such that the overall height of the wheel/tire combo is the same as the 35" on your truck now. It looks goofy, but bear with me. Mount it to the truck. Compare how many revs per mile the wheel turns with this goofy tire inflated vs deflated. The actual circumference of the tire never changes, but the revs per mile change drastically depending on whether the tire is inflated or not. That little bit of squish height affects the revs per mile in the exact same way, just on a much smaller scale.

Get it?

BassAckwards

I see where you're coming from, but I don't believe the radius does matter when calculating revs per mile. It's the tires circumference. If the tire has a circumference of 2600mm then one rotation of the wheel should equate to a distance traveled of 2600mm even under load as long as the tire isn't slipping.

Here is something I found in researching the matter:
http://www.thenxtstep.com/2007/11/wh...t-perfect.html
Now, you said yourself that the actual circumference never changes yes?
So if my tire has a circumference of 2600mm then one revolution of my wheel means I have traveled 2600mm regardless of my radius under load. Would you agree?

08silverbullet

Not sure what engine you have but on my 4.6 I am running lvl 2 with 89 octane +.50 on timing and +10 on shift firmness on all gears and I couldn't be happier I've been told I can go up to +1 on timing but I don't want to risk it

VTX1800N1

No, because there is tire slippage as the tire deforms over the contact patch and changes from a curved surface to a flat surface and back again. Most of your tire wear comes from starting, stopping, and changing directions. There is also tread wear from simply going in a straight line. That wear comes from the slippage and shape change that the tire undergoes as it meets the road surface. Do you not see how, on a very tall tire that is deflated to the point that it is almost, but not quite flat, the axle will make more revolutions per mile than if the tire were properly inflated? The entire circumference of that tire has not changed, and all of it comes into contact with the ground as the tire makes each revolution, yet the revolutions per mile will change greatly. There is massive slippage because there is massive deformation of the tire. With little deformation of the tire, there is little slippage. It's simply a question of scale. This is also why under inflated tires generate excess heat and wear out faster.

VTX1800N1

This is also why transport by rail is so much more efficient than by pneumatic tire. The train wheels and metal rails essentially do not deform (they do on a molecular scale, but you get my meaning. A locomotive certainly isn't very aerodynamic, but all that weight on non deformable wheels is many times more efficient that the same weight on deformable pneumatic tires.

BassAckwards

I understand where you are going with a completely flat tire, but I don't see that being the case on a properly inflated tire or even a slightly under inflated tire.

On the un-inflated tire there is some slipping, the tire is bending under itself and under the wheel, but on an inflated tire there isn't any slipping only deflection of the tread blocks and you still travel of the circumference of the tire for one wheel revolution.

Maybe we will just have to agree to disagree, but it's obviously a pretty highly debated topic from what little searching I have done.

TireRack.com does give a similar description to you and how it effects Rev/mile
http://www.tirerack.com/tires/tirete...jsp?techid=203

Oddly enough it states calculating the tires absolute rolling circumference isn't possible, but they use the 3% squish factor in their Rev/mile.