Sort of.

Edit: note that "patch" below refers to the contact patch (the part of the tire that is touching the road).


Note that the result you are looking for is the horizontal distance the center of the axle moves with a full rotation of the wheel.

This is related to the patch radius (the distance of the axle to the ground directly under it). With a load, the patch radius is smaller than the tire radius.

The axle is rotating around an "imaginary" circle that has the patch radius. That is, this circle would be the size of the wheel if the wheel was perfectly rigid (no patch deformity).

The (C) distance the axle moves horizontally in a full rotation is equal to the (C) circumference of the patch circle.

(Imagine the patch radius below rotating full circle.)




The only number that matters is the vertical patch radius. You really only need to measure this number (accurately). That's kind of hard to do. So, people measure the roll-out distance instead (which is the circumference of the patch-radius circle).

What is going on elsewhere with the tire (the tire radius) doesn't matter at all. (The tire radius isn't strictly constant anyway.)

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Note that the actual radius to use is the "rolling" (or effective) radius but it's hard to explain what is going on with that. (This was somewhat clarified below.)

The rolling radius would be forward of the contact patch and (generally?) would be larger than the patch radius but smaller than the tire radius.

If the wheel was completely rigid, the wheel would have this radius (the patch radius). It's the circumference of this imaginary wheel that the roll-out is measuring.

If the radius changes (due to different loads, mostly), the circumference changes.

With a higher load, the bicycle moves a shorter distance with each full rotation of the wheel.

The "circumference" (the path of the outer edge) of the tire doesn't change but it doesn't matter what that number is (as long as the patch radius is the same).

Your theory is incorrect. The effective radius of a tire is less than the unloaded radius, but greater than the loaded ("patch") radius that you show.
Actually, measuring the "patch" radius in your diagram is trivial -- it's the distance from the center of the hub to the ground. If you measure it and the roll-out distance of a tire, you'll quickly find that the effective radius of the tire is greater than the loaded radius, and the difference increases as the tire is more heavily loaded.

Huh???

"Tire radius" (in my illustration) is the radius of the part of the tire that isn't at the patch on a loaded wheel.

Tire radius (loaded) > patch radius (loaded). As I said and as the illustration shows.

This tire radius (loaded) is going to be very close to tire radius (unloaded). But we don't care about these numbers at all (we only care about the patch radius),

??? It seems it would be harder to load the bike and take that measurement. You'd probably need two people.

?? I said this.

??? This is what the illustration is showing. The effective radius of the tire is irrelevant. Only the patch radius matters.

I think we're in agreement based on with what you're saying, but your diagram shows something different.

????

No, it doesn't (the illustration is correct).

Tire radius (loaded) > Patch radius (loaded). <<== what we are both saying.

The circle (with the tire radius) is larger than the circle with the patch radius.

The illustration shows the two circles (the patch circle is smaller and inside the tire radius circle).

Quite honestly, I can't follow your reasoning, because I can't figure out what your definition of "patch radius" is.

???

Look at the illustration.

It's the vertical distance from the center of the wheel to the ground. It's pointing directly at the part of tire that is being compressed.

You "figured it out" here.

You "figured it out" here too.

With that definition, your theory is incorrect. The effective radius of the tire is larger than what you're calling the "patch" radius, and it is related to the tire circumference (or roll-out) by the familiar L = 2 * pi * radius.

Unloaded radius > effective radius > loaded (patch) radius

Let us agree on a few things, one complete revolution of the rim equals one complete revolution of the tire.
If you cut the tire circle at one point and lay it flat on the ground in a straight line, like a belt, one complete revolution of the circle equals
one full length of this belt, this does not change, ever, squish or no squish. This distance does not change if it gets squished under load or no squish under no load.
One rev of the wheel equals one circumference distance. Assume there is not any slip during squish load footprint.
I fully realize there other variables going on, at the squish load spot, compression, temperature, ect, but all these are irrelevant
until the understanding of the belt theory is understood. Am def willing and able to hear all opinions about this subject.

Let us agree on a few things, one complete revolution of the rim equals one complete revolution of the tire.
If you cut the tire circle at one point and lay it flat on the ground in a straight line, like a belt, one complete revolution of the circle equals
one full length of this belt, this does not change, ever, squish or no squish. This distance does not change if it gets squished under load or no squish under no load.
One rev of the wheel equals one circumference distance. Assume there is not any slip during squish load footprint.
I fully realize there other variables going on, at the squish load spot, compression, temperature, ect, but all these are irrelevant
until the understanding of the belt theory is understood. Am def willing and able to hear all opinions about this subject.

I said this multiple times! The illustration clearly shows it!

Your "effective radius" isn't clear. (The actual effective radius is the patch radius.)

I suspect it's more like:

"Tire radius (loaded) approximately equal to tire-radius (loaded) > loaded (patch) radius."

(But we don't care about either tire radius. We only care about the patch radius.)

No one is saying it does. The perimeter of the tire doesn't change.

It's important to keep in mind that the loaded tire isn't a circle. It's a deformed circle (there's a flat part at the bottom). It's also important to be clear that circumference is an attribute of an actual circle (or ellipse).

This is where you are making a mistake.

There are two circumferences to keep mind of: (1) the circle with the tire radius and (2) the circle with the "patch radius" (the vertical distance between the center of the wheel and the ground). (The tire perimeter is a third thing to keep in mind.)

The circumference that matters is the second one. This is the 2*(vertical distance) * Pi circle. The vertical distance varies with the load (it's shorter with heavier loads). Thus. the circumference varies with load (as does the horizontal distance travelled). (The tire perimeter is irrelevant.)

Look at my illustration.

The distance travelled in one rotation on a loaded wheel is less than the length of the perimeter of the tire (the length covered by the cut tire laid flat).

What the "belt" is doing is irrelevant. The only thing that matters is the distance of the center of the axle and the ground.

Tanks are an extreme example of what is going on. The distance travelled in a rotation is the circumference of the circle with the radius of the wheel axis to the round (not the track length). The tire perimeter is like the track: it doesn't matter how long it is.



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(1) Consider what the deal would be if you removed the tire and just ran on a raw rim. Note that this would be what would happen if you had an extreme load (or no air in the tire).

In this case, the horizontal distance would be equal to the circumference of the rim.

(2) Now, imagine running on a raw rim that was as big as the wheel with a fully-inflated tire. Or on the same rim but with a solid tire that doesn't compress.

In this case, circle and circumference is larger, which means the horizontal distance would be larger than case (1).

So, the circumference of a real wheel is going to be between case (1) and case (2). The circumference will be smaller with more load or less pressure.

Again, a simple roll out test, one loaded and one unloaded will give the answer. Of course that might not be as fun as several pages of physics arguments.

I’m going to start a new thread as a tangent to this one: How many Angels can dance on the contact patch of a 28mm tire, loaded, or unloaded.

Does the tyre compress less at speed? There is work done deforming the tyre.

Is the contact patch compressed horizontally due to the torque exerted by the rider?

The following diagram shows the relationship between the three radii, where Ru = unloaded radius, Rl = loaded radius, and Re = effective radius. Your illustration shows the effective radius and the "patch" radius as being the same. They are not, in general.

Geometry arguments

OK this is physics arguments

Can just about read that. A higher quality image would have helped.
What’s the definition of the “effective radius” there? Cos based on the image alone it’s “distance from hub to an arbritrary dotted line”. Does PC=CA? And if so, why?

And the doppler shift from the satellite signal.

Indeed, one of the biggest improvements in going from cyclocomputer + cue sheet to GPS for long distance navigation, is eliminating the need for math in your head. Cue sheet says turn right at 47.18 miles, from which you have to subtract 0.62 in your head due to error, and remember the result before the turn comes. Then let's say there's another turn at 47.29, which is pretty quick so you have to decide ahead of time whether to remember 0.11 and add that upon turning, or subtract 0.62 again and remember two results, or just hope there's only one road in close proximity. Keeping in mind the 0.62 becomes 0.63 after a bit, now try that short on sleep on day two or three. Then add rain, with required efforts to keep the cue sheet dry. The best is when the cue says turn at T in the road, in which case you give the brain a break and ride to the T.

An errorless cyclocomputer would have been a game changer.

With GPS you just obey the GPS. Every turn is easy like a T. Admittedly something is lost when navigation becomes effortless, but alas sometimes GPS fails and one is forced to a backup method. Which can be fun in its own way.

Ok I think I understand what tomato coupe is saying but I think he’s wrong for weights in bike territory as opposed to cars.
Figure 8.12 of this suggests the effective radius actually being smaller than the patch radius for weights below 1200N (about 120kilos)
https://www.sciencedirect.com/topics...rolling-radius

It's too hard to read. You should provide a link to it.

https://www.sciencedirect.com/topics...%20the%20wheel.

You've finally managed to almost get your point across! You still aren't being clear enough.

So, you are saying the actual radius to use is somewhat larger than the patch radius. So, sure, what I said was not quite correct. It's still explains closely-enough why the unloaded radius is wrong (too large).

So, you are correct but explained it very poorly.

How different is this effective radius from the patch radius?

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So, to be clear...

The actual radius to use for measuring distance per rotation is called the "effective" or ("rolling") radius. You can't really measure it directly, which is why one does a roll-out measurement.

https://www.tractorbynet.com/forums/...2#post-4599241

Talking about the loaded (patch) radius is a simplification. I think it's a useful simplification: it would be impossible to talk about the effective (rolling) radius without being clear about the loaded radius.

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Note that the effective radius is different between "free rolling" and "driven" wheels.

So, there are situations where the effective radius can be smaller than the loaded (patch) radius. (This might not be the case when bicycling.)

https://liu.diva-portal.org/smash/get/diva2:1762057/FULLTEXT01.pdf

The diagram only shows the relationship between the radii -- it's not meant to be a means of determining the actual value of the effective radius.

Auto tires are obviously designed to support higher loads than bicycle tires, but the same principle of effective radius under load applies to both.
I don't see figure 8.12, but the first available figure in the link shows the same relationship between the unloaded, loaded, and effective radii as the diagram I posted.

Your initial post was about calibrating a speed sensor. If your concern is about accurately determining distances during a ride, the speed sensor is not offering any advantage -- just use your GPS computer without it.