Thanks

Ed

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Thanks

Ed

Tire manufacturers often specify revolutions per mile at a specific speed in their spec charts. If your tires are a readily available popular size, that could be of help. I just looked at B F Goodrich's website and they give the revs per mile at 45 mph. If your size isn't listed, try doing a "roll out" measurement. Mark the tire and move the car until the tire has made about 10 revolutions and divide the distance by number of revs to get an average circumference. Divide by pi to get the effective diameter.

Bob

Looking on the tire rack site for same tire dimension the revs/mile calculates to be 4% greater than what you would get using the diameter of the tire.

At speed they might expand a little from centrifugal force, but I don't know if it is significant.

John

I think the circumference of the tire doesn't change enough to matter. I really doubt that it "grows" like a dragster tire, and even with a flat spot where it contacts the ground, I think the circumference stays essentially the same.

A 225/40 18 tire should have a diameter in the 25" range.

(225 X.40 X 2 divided by 25.4 + 18 = 25.086")

25.086 X 3.1416 = 78.80" circumference.

Jack a tire off the ground and measure the circumference and see how close it is.

Just my opinion,

JA

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This is the effective radius when calculating acceleration, but it is not the radius you need for speedometer calculations. Consider the tire as it is often pictured in, say, a Walter Lantz cartoon. It can be terribly deformed, but, as the wheel makes a complete revolution, all that rubber has to pass under the axle. And, so it is with a real tire. The circumference you measure as the tire is mounted on the wheel at the tire shop is essentially the same circumference you experience as you drive down the road. Now, there is some expansion with speed. With the old bias ply tires, it was considerable. As I recall, it amounted to 0.2 wheel revolutions per mile per mile per hour. With radials, however, it is almost inconsequential. So, with radials, the "rollout" method (as has been previously described) is probably the best method for determining that which you need.sabino56 said:The effective radius of the tire will be the distance from the center of the axle to the ground.

(In the late fifties, my 8 to 5 job was sitting in front of a mechanical Friden calculator, calculating miles per gallon and acceleration for Chrysler Corporation products. Surprisingly, I didn't find it all that boring.)

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I hear that, but if no one else cares enough and the numbers stated are only close to apporoximation, then your precise measurement's are still wrong.

But, since we are having fun and debating angels on pins...

The tangential velocity of any point about a center of rotation is the product of the distance from the center and the angular velocity. That is the radius x rotation rate in rad/sec. The tangential acceleration is radius x angular accel in rad^2/sec. The radial accel, that is the acceleration required to keep it moving in a circle is radius x angular accel squared. (This is the term which results in drag tires ballooning out when spinning fast.

When a tire is deflected by load the perimeter length of the tire does decrease from the undeflected perfect circle. The rubber deforms as it goes through the contact patch. The tangential velocity of a tread block going around the undeflected part is actually a little faster than it's speed when it's directly underneath the axle. The reason is the distance from where the tire touches the pavement at the front straight back to where it leaves the pavement is a little shorter than arc of an undeflected tire (draw an arc and connect ends with straight line - the straight line is shorter than the arc). All this deformation takes energy and it ends up as heat. Which I understand is why under-inflated tires degrade gas mileage. I've also heard that this generated heat is one of the reasons tires can fail at very high speeds as the tire construction can't disapate the heat and it weakens. I assume radial acceleration forces also play a part in that.

I'm sure more than any of you wanted - but long story short. It looks like the tire deformation does matter a litte and I assume that this is why tire manuf. spec both a circumference and a revs/mile - and they don't match doing straight math.

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God forbid the tire gets low...

Step away from the pencil....

Kidding really

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I'm afraid you're still missing the point. I wouldn't expect the manufacturer's circumference and wheel revs per mile information to correlate EXACTLY. But, this has nothing to do with that which you're describing. You're describing a radius change which will, most definitely, affect the thrust delivered by the tire patch, but has nothing to do with the tire circumference which must pass under the axle with every revolution.sabino56 said:The reason is the distance from where the tire touches the pavement at the front straight back to where it leaves the pavement is a little shorter than arc of an undeflected tire (draw an arc and connect ends with straight line - the straight line is shorter than the arc).

I suppose you could check this out by comparison rollouts with 10 psi and 50 psi. Again, they might not come out EXACTLY the same, but they'll be a whole lot closer than road surface to axle centerline differences would dictate.

In short, give this a lot of thought before you decide to disagree with the way the industry handles its performance calculations.

Billyshope - I think I do understand your point, I just didn't agree and was attempting to explain why. And believe me, I would never think of disagreeing with the industry. In fact, thought I was just discussing it with you!

Regardless, I did what all good nerds do when faced with a problem - I Googled it. This is not a new question and people have made careers out of tire mechanics and dynamics. (Check out http://www.mathworks.com/access/helpdesk/help/toolbox/physmod/drive/index.html?/access/helpdesk/help/toolbox/physmod/drive/tire.html) The net is that "Rolling Circumference" (ie. the distance a loaded tire rolls in one rev) is not the same as the unloaded tire circumference. The difference is due to the tire deflection under load. There are plenty of references and papers on this. Alot of patents, papers on estimating tire pressure by monitoring changes in rolling circumference.

This one from Continental Tire. "

The clearest answer was from a tire expert answering the same question.

http://experts.about.com/q/Tires-2359/Rolling-circumference-tyres.htm

Herman,

There are a couple of things that complicate the "circumference" of tires.

1) There are "calculators" that will calculate the circumference (diameter) of a tire based on the tire size. Use a search engine with the key words "tire calculator"

2) These "calculators" give you an answer based on the "size", which is different than the actual physical dimensions. Said another way, a P205/65R15 does not

actually have to be 205 mm wide and have an aspect ratio of 65%. There is quite a bit of variability in the market

3) The actual circumference (diameter) of a free hanging (not touching anything) tire is different than the rolling circumference (Rolling diameter) because the tire deflects

under load. Different inflation pressures and different loads will affect the rolling circumference,

4) The difference in circumference between tires will have a minor effect compared to other factors. Rolling resistance greatly affects fuel economy, so acceleration is also affected and rolling resistance varies from tire to tire. However, a change in rolling circumference of a tire acts in a similar way as a change in final drive gearing.

Hope this helps.

About Barry Smith

Expertise

I have over 30 years experience in the design, manufacturing, and testing of tires. I have served as the technical advisor to the "800" number. I have authored or co-authored many publications - usually without credit. I can answer almost any technical question, but please don`t ask me to compare brands. I have prejudices because of my work experience.

Experience

Member SAE (Society of Automotive Engineers) Member Tire Society (Tire Technical Organization) SCCA Regional Competiton License holder Authored many training manuals on tires, their care and use.

So, if you really care - you need to adjust unloaded circumference by about 3%. However, now that I know - I don't care anymore!

Thanks for the responses!

Ed aka "Bob Goodyear"

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Yes, loading and tire pressure affect the rolling circumference of a tire. This is why I said the measured circumference of a tire on the tire changing machine is "essentially the same" as it is in operation. As Barry points out, that "essentially the same" is expressed as a 3% difference. The strains involved in load and air pressure would also affect the rolling circumference. This is why I said that the rollout results with 10 and 50 psi would not be "EXACTLY" the same.

But, the point I was trying to get across is that these differences are NOT the result of the very obvious flattening of the tire at the footprint. Again, the 10 and 50 psi rollout tests would clearly show this. While I would expect a small difference in results (from the sources discussed in the preceding paragraph), it would be nowhere near the difference expected if the rolling radius was assumed to be the vertical distance from roadway to axle centerline.

This vertical distance from roadway to axle centerline comes into play, however, when calculating acceleration. Barry mentions the gearing effect of a change in rolling circumference ("However, a change in rolling circumference of a tire acts in a similar way as a change in final drive gearing."), but this is a small effect compared to the deformation of the tire at the footprint. In other words, the radius change from the change in rolling circumference could never account for the deflection at the footprint.

A common error...yes, even in "the industry"...is to use the wheel revolutions per mile data from the proving grounds and then calculate the radius for use in acceleration calculations. This is commonly found in SAE papers. Well, I shouldn't call this an "error." We're talking about such small differences that the familiar term "accurate within engineering accuracy" is more than sufficient.

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I couldn't help but make a comment on the discussion, for those interested in some graphic footage of a tyre under deflection in race conditions, get a hold of Bathurst 2006, V8 Supercars Australia, where a camera was mounted under the Ford of winner Craig Lowndes. Bathurst is a twisty, mountain circuit where an annual pilgrimage takes place in October for the 1,000 Km event, which is renowned as a heart breaker. My experience is limited compared to most who have commented herein, however for what it's worth, there are as many variables as types of use, eg drag racing, street and circuit racing pressures all vary, especially with the width of rim and vehicle weight. Loaded radius is a starting point, low profile tyres require much higher operating pressures than is the norm, 38-42 psi is not unusual on a street driver. A particular tyre size will require different pressures under different conditions, wheel width, vehicle weight/load, operating conditions and in the case of the above mentioned race clip, driving conditions, operating temps, etc. Regards, Bob

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Should be rad/sec^2sabino56 said:The tangential acceleration is radius x angular accel in rad^2/sec.

Should be radius x angular velocity squared.sabino56 said:The radial accel, that is the acceleration required to keep it moving in a circle is radius x angular accel squared.

(Yes, I realize these were careless typos, sabino56, and I probably should have sent you a PM and given you a chance to make the corrections yourself.)

Thanks! you are correct. I appreciate you catching the errors.BillyShope said:Just to avoid confusion among the young readers who follow these threads:

Should be rad/sec^2

Should be radius x angular velocity squared.

(Yes, I realize these were careless typos, sabino56, and I probably should have sent you a PM and given you a chance to make the corrections yourself.)

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