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Old 10-25-2006, 12:17 AM
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How to determine effective tire diameter

I'm working out transmission speedo gearing for my car. You need the tire circumference (or diameter) to do the math. I measured tire diameter of unloaded tire - but, of course, the tire changes when loaded so the effective tire circumference (or radius @ ground) changes. Is there a rule of thumb on difference between effective tire radius/circumference and unloaded circumference?
Thanks
Ed
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Old 10-25-2006, 06:38 AM
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Not enough to make a difference, I am sure.......................I think you are overanalysing this.
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Old 10-25-2006, 06:46 AM
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> Is there a rule of thumb on difference between effective tire >radius/circumference and unloaded circumference?

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
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Old 10-25-2006, 07:52 PM
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Or, you could go to the tirerack site and plug your
tire size and make in, it'll tell you how many revs
per mile...

K
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Old 10-25-2006, 08:15 PM
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Well, maybe I am over-analyzing but I'm an engineer and I can't help it! Anyway, I made some measurements. The effective radius of the tire will be the distance from the center of the axle to the ground. It's pretty easy to measure the amount of tire deflection between loaded and unloaded and you can calculate the % difference in effective circumference. For the tires on my '65 - I have low aspect ratio tires - 225/40 18s - they deflect 3/8in when loaded. This results in a 3% error. I would think 70 series tires would have larger deflection so the error would be greater - maybe 5%?
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.
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Old 10-25-2006, 10:47 PM
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My 15" 70 series Yokohamas are stated to be 26.9" diameter, or 13.45" radius. When mounted on the car, I measured from the center of the wheel to the ground and got a radius of only 12.75", an effective diameter of 25.5". That's a 5.2% reduction.

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

John
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Old 10-26-2006, 04:50 AM
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I think this is an exercise in futility.
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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Old 10-26-2006, 05:32 AM
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Quote:
Originally Posted by sabino56
The effective radius of the tire will be the distance from the center of the axle to the ground.
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.

(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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Old 10-26-2006, 10:04 AM
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Well, maybe I am over-analyzing but I'm an engineer and I can't help it!


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.
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Old 10-26-2006, 10:04 AM
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Well, I agree the best way is to do it on the road while rolling. Once I get the transmission back in the car, I'll do that for these tires and see what I get. I also agree that if you are changing tire sizes and want to know the effect on your speedo - comparing undeformed circumferences is fine. The only time it may matter is when you are doing something from scratch and trying to calcluate what speedometer gears to put in your transmission - and even then we're only talking 3-5%.
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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Old 10-26-2006, 10:21 AM
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Then car weight, tire pressure, rim width, acceleration forces will all have an effect. Not to mention tire manufacturer tolerence. These will need to be taken for consideration. Testing on said car will be the most accurate way.

God forbid the tire gets low...

Step away from the pencil....



Kidding really
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Old 10-26-2006, 11:53 AM
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Quote:
Originally Posted by sabino56
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'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.

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.
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Old 10-26-2006, 06:39 PM
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Well, as I said it was just for fun. I got interested in the math and kinematics and you can actually calculate the velocity and position of a tire element... wait - as Johngrass1 said - put down the pencil, step away from the paper...

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/help...rive/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. " Deflation Detection System (DDS). DDC identifies a loss of pressure indirectly, using data from the wheel speed sensors of the electronic braking system because when a tire loses pressure its rolling circumference decreases."

The clearest answer was from a tire expert answering the same question.
http://experts.about.com/q/Tires-235...ence-tyres.htm
Answer
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, but as a general rule a properly inflation and properly loaded tire will have a rolling circumference about 97% of the free hanging circumference - a 3% difference.
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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Old 10-26-2006, 07:01 PM
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Adjustable electronic speedometers are nice!
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Old 10-27-2006, 06:12 AM
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Again, we're talking about very small differences, but, when a couple of engineers start talking about this sort of thing, the methods of analysis become far more significant than the magnitudes of the parameters involved.

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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