okay well you obviously don't know basic geometry and probably have no real life experience if you honestly think you're correct. you have me and one other person saying you're wrong. both of us have actually been there and done it instead of just thinking we know what we're talking about. If you had some sort of prof or logical reasoning behind your statement i would take you more serious. circumference does have something to do with tire wear when you have two dissimilar tires mounted together. the small one will be dragging to make up for the difference in circumference.

 

You have to measure the radius as the distance from the center of the wheel to the ground, and not from the center of the wheel to the top of the tire.

If both wheels are touching the ground then their effective radii are equal. Therefore, there effective circumferences are equal. (from basic geometry C=2 pi r)

There is no difference in effective circumference and therefore one tire is NOT dragging as a result of it.

For the purpose of this experiment - If you never turn:

The tire with the larger diameter will be carrying a larger share of the load and will therefore deform more and wear more... But more than what?

More than the smaller tire... Yes.
More than itself if the smaller tire was completely removed... NO!

 

Okay well think what you want. I know I'm right.

 

I want to say this respectfully but you're full of it

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What tiny fraction of my LOGICAL argument do you disagree with. Show me ONE fact I have wrong or one conclusion that I got wrong. You see, I have proven my case using Logic. Can you find one error? If you can point out where you are having difficulty, perhaps I can localize on that one specific aspect and explain it better.

I have been very careful to only take exception to the following statement.

The smaller tire spins at the same rate as the larger tire. It is measured as function of angular displacement per unit of time. RPM, degrees per hour, radians per blue moon, whatever you want.

Perhaps you are thinking of the speed of a spot on the surface of the tire tread? (relative to the axis)
That would be rotational speed(rpm) times the radius. Do you not belive that the radius is the same where both tires hit the ground? Do you not believe that the radius of the tire changes as it contacts the ground?

Can you precisely pinpoint one relevant flaw in my argument?

 

the radius may change but circumference doesnt meaning the larger tire will always roll farther per every revoluting meaning the smaller tire (which doesn't roll as far) scuffs the ground to make up the distance.

you're wrong. part of being a big boy is realizing when to accept that fact.

 

I miss spoke. The small tire spins slower. The larger tire has a bigger footprint on the ground giving it more traction. The smaller tire will just scruff the ground as it passes over it. But don't take my word for it or experience. Go buy yourself some different sized tires and see which ones wear out faster. The smaller tires I had on my dually went bald in about 4000 miles. They were on the inside so I didn't notice until I went to get them rotated. I think we're way past geometry and probably into physics, but what do I know.

 

OBSPowerstroke7.3
Circumference = 2 pi r
You have just stated that this is not true.
If the radius changes, but the circumference does not then:
Circumference (does not =) 2 pi r?
Are you sure you want to say I am and therefore Archimedes was wrong and that if you divide the circumference of a circle by it's diameter you get something other than Pi?

jamesrett... ahhh physics (the F word in my house)

Yes! Yes! Yes! I have no reason to doubt that a smaller tire on a dual would wear out very quickly. but not because it is traveling further. The larger tire bears more weight and therefore has more traction...yes this is true, I never said it wasn't. In fact It makes perfect sense that that is why the small one wears out quickly.

When A vehicle turns, the outside wheel on any axle travels farther than wheel on the opposite end of the same axle. This is also true of a set of duals mounted together on the same end of an axle... This is also true of the right side and the left side of one single wheel.

One side of that single tire travels farther than the true average distance and therefore scuffs that side of the tire in a braking fashion (like dragging the vehicle with the brakes locked) and the other side of that same tire is traveling less far than that same average and is therefore scuffing in the opposite direction (like spinning while trying to tow an immovable object). Somewhere in the middle of that tire (in a turn) is a neutral line that is not scuffing in either direction and that represents the true distance traveled. The further you get from that neutral line, the worse the scuffing.

In a dual set-up, that neutral line is normally between the duals and one whole wheel scuffs one direction and it's mate in the other. (dual wheels scuff in corners more than singles). If you mismatch tire sizes on a dual, you move that neutral line away from the center towards the tire with better traction (the larger tire). So the larger tire will scuff less to some degree (more like a single wheel)and the smaller tire will scuff way more.

Let's take this experiment to an extreme. single wheels mounted on a welded posi-traction rear-end. In effect one giant dual wheel. How much scuffing goes on when you try to turn one of those?

So yes I can believe that mismatched dual tires would be hell on the smaller tire...Because of the turning.

 

I think your assuming that just because both tires are touching the ground that they have the same radii. Now one tire being bigger has a larger rolling circumference. So if I understand your logic the bottom of the tire spins at the same speed but the upper part of the tire spins faster? How is that possible. Ill go back to my original analogy of the 20' rope. I would rather be hit a ten pound weight on the end of a one foot rope then a twenty foot rope. That sucker would be moving fast. My head hurts.

 

The radius does change. the top of the tire is moving faster. the radius changes when the tire is deformed by the road. Since the radius changes, the circumference must also change and that is why you must use the radius measured to the ground. It is what matters. that is why you must calculate the circumference from the radius measurement and not measure the circumference directly. If you had a mechanical device that pushed against the inside of the tire every time the device got to it's apex it would increase the radius measured there and therefore increasing the speed of the top of the tire and if it released before coming down, Your distance traveled does not change Because the radius measured to the ground did not change. Even a single tire decreases it's own radius when it contacts the road. A radius measured from anywhere else is irrelevant.

The top of the tire is moving faster than the bottom. Not because the angular speed(RPM) is changing, but because the radius is changing. Speed of the surface of the tire = angular speed (RPM) times distance. Decrease the distance (by altering the radius as it contacts the road) and the speed must change.