My, my, my, I wonder if we could agree that a foot is 12 inches and that a yard is 3 feet, or 36 inches?? I doubt it!!
Elly, you may continue to calculate the wheel speed of a low tire by that “radius”. And I’ll use my method, and our results will disagree.
However, that does not mean we disagree on a foot and a yard. I don’t disagree just for the sake of disagreeing, I only disagree when I believe there to be a different truth.
Well, mountainbike, measuring the amount of travel of a low tire would be the best way to go, that should be indisputable. I don’t think I ever insisted on calculating the cirdumference by the “low tire radius”.
Why I mentioned a foot and a yard is because every other post disagrees with everyother post. It’s that way regardless of the topic.
I also figured the difference of a 13 1/2 inch and a 12 inch one and I came up with 29 inches per revolution.
In less than 10 minutes anyone can satisfy their curiosity.
Rod Knox, you proved our point - you should have seen about a 28" difference for reducing the radius from 13.5" to 12.0", and you saw only 8". (as CapriRacer pointed out) More than I’d expect, but still much less than you’d calculate if the distance from axle to ground determined the rotation speed.
The difference in the distance traveled was certainly less than I had imagined but considerably more than zero. My curiosity led me to google the subject several times and there seems to be quite a bit of heated discussion with some pseudo-scientific/mathematical “proof” offered to support each side. It seems there are quite a few who, like myself, need an early spring to get the grass and weeds growing to keep us occupied.
RK - did you happen to see what the pressure was on the deflated tire?
I was below 10 psi, texases. The gauge barely moved when I pressed the iflator/gauge on the stem.
Well, mountainbike, measuring the amount of travel of a low tire would be the best way to go, that should be indisputable. I don’t think I ever insisted on calculating the cirdumference by the “low tire rqdius”.
Why I mentioned a foot and a yard is because every other post disagrees with everyother post. It’s that way regardless of the topic.
I also figured the difference of a 13 1/2 inch and a 12 inch radius and I came up with 29 inches per revolution.
“A given spot of the tread will travel a flat line for a distance and then an arc length. The total of the flat line (call it a cord if you’d like) plus the arc length is what determines the wheel speed, not the “loaded radius”.”
A given spot on the tire in contact with the ground is always stationary relative to the ground. There’s no “flat line” (you meant to say “chord”).
A tire is not a circle or a disc. I suppose it would be a toroid of some sort, and I imagine some of the air pressure or lack thereof would go into changing the width of the sidewall.
If only someone would post what they think about this issue…
"A tire is not a circle or a disc." Mine are!
"A given spot of the tread will travel a flat line for a distance and then an arc length “” Can you explain that a little further for we who aren’t so educated??
Can a three-leged dog run as fast as a four-leged dog?? I bet that half of you will disagree with the other half!!
We will get things like; It depends on the color of the dog, or how long has it been since he was taken to the vet, or is it a male or female, or is he an indoor dog or an outdoor dog, or has he ever been hit by a car, or what does he eat, that could make a difference, or does he get a lot of rest?
This thread needs to disappear faster than a pack of smokes at an AA meeting!
“‘A tire is not a circle or a disc.’ Mine are!”
Your tires exists in only two dimensions? Interesting.
“Can a three-leged dog run as fast as a four-leged dog??”
I would suggest that your dog is alleged.
“This thread needs to disappear faster than a pack of smokes at an AA meeting!”
Yeah, I think my chances of picking up a chick with issues here are pretty much nil. But that raises the question: can you pound beers at a Smokenders meeting?
flybyyaJanuary 10 It don’t seem that you are coming back to this post. However I was thinking about your experienxce with the tank track, and I wondered if you let some air out of it and checked it??
While I can appreciate CircuitSmith’s desire to have this thread go away, it is clear that several folks don’t understand what is going on and would like to.
If you have an inflated, but unloaded tire, the shape is circular. - BUT - If you put the tire on a car and lower it to the ground, some of that shape changes. Obviously the part in contact with the road surface is flat - but the rest of the tire also deforms and no longer forms a perfect circle. This is where the tank track analogy comes in.
The tread portion of the tire - that part that determines the rolling circumference - acts very much like a tank tread in that it gets displaced in front of and behind the footprint, and the total length of the tread changes very little. While the rest of the tire may appear to have a perfectly circular shape, it is in fact distorted - and to such a small extent it’s easy to fool the eye.
Also, the axle is off center from the apparent circular center (noting that it isn’t a perfect circle) - In some respects you could say the axle is more being suspended from the top, than being supported from the bottom.
The translational velocity of a vehicle depends upon the “effective” rolling radius Re of its tires. Re is defined, somewhat tautologically, by Re = V/ω, where V denotes the translational velocity of the car and ω the angular velocity of the wheel rotation. Re lies somewhere between the unloaded tire radius Ru and the loaded radius Rℓ (which is the height of the wheel center above the ground). That is, Rℓ < Re < Ru.
A tire’s effective rolling radius decreases with loss of tire air pressure, but not by much in steel belted radial tires, a little more so in bias tires. Re varies with tire construction and perhaps even among tires of the same production line.
Below is a graph from a Japanese US patent application. Notice that a tire’s effective rolling radius only decreases on the order of a millimeter for a pressure differential of 1 bar (≈15psi).That is why data from the car’s wheel sensors must be accumulated over several miles to detect air pressure loss.
If I can find a better definition of Re other than the tautological Re = V/ω, I’ll post it here.
Aw, Mechaniker, now you’re messing things up with facts! But I can’t see the graph, would you post the link?
Not a link to the actual patent application, but I can post a link to a pdf that contains the graph:
http://www-nrd.nhtsa.dot.gov/pdf/esv/esv19/05-0082-O.pdf
I have rewritten my original post, so you might want to go back and re-read it. I don’t think the concept of “effective” tire radius was discussed in previous posts, only loaded and unloaded tire radii. But effective radius is the concept used in vehicle dynamics. I only wish I could give a better explanation of it.
Read this, Section 2.3 Rolling radius and Fig. 2.6:
http://books.google.com/books?id=yAo-6yQbg9IC&pg=PA38&lpg=PA38&dq=definition+effective+rolling+radius+pneumatic+tire&source=bl&ots=LiOshq1uaV&sig=ypVy3_8ohLjlqBWBgQTYkX4sQdI&hl=en&sa=X&ei=n-UuT8KnOovpgAfo7-37Dw&ved=0CCkQ6AEwAA#v=onepage&q=definition%20effective%20rolling%20radius%20pneumatic%20tire&f=true
Still not a satisfactory explanation of effective rolling radius.
MOUNTAINBIKE, I don’t remember insisting that a “low tire radius” would be accurate or not, but I am really curious to see the difference a low tire will make. Some warmer, dryer day I will check it.