So you're stating, based on the example I gave above, that if one tire were to be inflated to say 5PSI and the other inflated to say 40PSI then they would BOTH make the same amount revolutions if they both traveled the same distance in a straight line <img src="/forums/images/graemlins/rolleyes.gif" alt="" />??? Then what would be the purpose of keeping the same tire pressure in both tires on the same axle if the distance they traveled was based on CIRCUMFERENCE and not ROLLING RADIUS <img src="/forums/images/graemlins/confused.gif" alt="" />???

Fred R.

Also I never once mentioned that the circumference of a tire would change with tire pressure, I said that the distanced it traveled would change with a change in tire pressure...

Fred I can see you have not GOT IT yet. Ok when one tire is severly deflated...as in SAND.. The PATCH on the ground gets wider and LONGER so it is in contact with the sand more than a fully inflated tire of the same size. Yes the bottom radius is changed (shorter) But the circumference has NOT changed! It takes the same amount of INCHES for both tires to turn one complete turn..
With your theory one could run his tires low and the speedometer would be INCORRECT!
Remember i is all about the circumference not the radius.
Big JIm <img src="/forums/images/graemlins/pfft.gif" alt="" />

Yes I do "get it", I'm not talking about the contact patch of a tire caused by it's inflation or lack there of, I'm also not talking about the bottom radius of the tire in comparison to the upper radius of the tire based on the tire's inflation, this ENTIRE time I've been talking about the [b] [color:"red"]ROLLING RADIUS [/color] [/b] (<---Click here) of the tire and how it effects how many revolutions a tire makes in a certain distance. Look it up if you don't believe me, the internet can be your friend if you let it <img src="/forums/images/graemlins/kewl.gif" alt="" />. Either way I see this is going nowhere fast so I'm officially done with this, have a good day <img src="/forums/images/graemlins/cheers.gif" alt="" />.

Fred R.

Well the arguments aside... I have had to use a one size difference spare tire in 2 situations.

A.)I was running 215's and had a flat. Put the 225 on and it did good for about 2 months until I had the cash up to buy new tires (235's).

B.)Had a slow leaking tire with the then new 235's, one morning I got up and it was flat, had to use the 225 spare. Worked till I got the truck to the tire shop that weekend.

So In both cases where I was running one tire size too big, and one too small it will do until you get the tire replaced. But don't push your luck though, get it replaced ASAP. Nice thing about my 31's now I have a same size matching spare. <img src="/forums/images/graemlins/kewl.gif" alt="" />

<img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" /> <img src="/forums/images/graemlins/lol.gif" alt="" /> I thought this thread died a while ago! Good arguments though. The new tire was on last week. She does the 6hr. trip today and back tomorrow.

I have an idea... how about if one of us gets off the computer and actually conducts a physical experiment? <img src="/forums/images/graemlins/smile.gif" alt="" />

I may have to try this with a handcart.

Michael

I have re-read the posts in this thread, and I think I know why there seems to be a paradoxical loggerhead here. That is, why the "height" crowd seems to be right and why the "diameter" crowd also seems to be right. Here's my take... We need to look at the geometric relationships involved.

As we all know, in a circle, the radius (1/2 the diameter) of a circle has a certain relationship to the circumference of a circle. Simply put, the circumference is equal to Pi diameters.

But here's the problem in this threaad --- once a tire is installed on a car, in almost all instances it is NO LONGER a true circle. That is, it's flat on the bottom. Deflate the tire a little, and the distortion is even worse!

Since the tire is no longer a true circle when mounted. the geometric relationship between diameter (or radius) and circumference ceases to exist!

So, some of you are correct -- regardless of the inflation pressure of the tire (within resonable limits), the circumference of the tire will not change. Therefore the distance the car travels per revolution of the tire does not change, despite the fact that the distance from the center of the hub to the ground has changed.

On the other hand, some others of you are also correct in stating that the effective "gear ratio" of a vehicle with deflated tires changes as the tire is deflated. Since the tires are no longer perfect circles, we can no longer use the diameter or radius of the tire when computing effective gear ratio. We must use the distnce from the center of the hub to the ground, which I'll call the "inflated height distance."

Think of it this way... torque generated by the engine is multiplied (unless in a straght gear or overdrive gear)by the transmission, and then multiplied by the rear axle ring and pinion gears. The differential applies the resultant torque to the axle shaft. The axle shaft applies that torque to the hub, and to the wheel and the tire. If I install a smaller size tire, I will have greater leverage to rotate the tire. So I have more grunt with a 31" tire than I do with a 44" tire. The same increase in grunt (change in effective gear ratio) occurs when I keep the same tire on, but just deflate the tire womewhat. By reducing the "inflated hieght distance", I decrease the length of the lever arm in the Third Class lever that the axle/tire represents.

The upshot of all this is:

(1) The deflation of a tire will increase the effective gear ratio (making it act like a numerically-higher gear ration).

(2) However, the deflation of the tire will not change the circumference of the tire, therefore the distance traveled per revolution of the tire will remain essentially the same.