The people who calculate the various errors in electronic distance measurement in regard to tire pressure deltas are the people who express interest in this ad

We don't care what the rest of the tire is doing.

No, this isn't the "circumference" because the loaded and low-pressure tire is not a circle!

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Let's try it this way.

1) Remove the tire. Do the rollout. That's the smallest possible radius or circumference (*). This represent a wheel without any pressure and/or a very large load.

2) Put a completely rigid tire on (one that does not deform). Do the rollout. That's the largest possible radius or circumference. This number represents a tire that at maximum pressure without any load.

A real tire with any particular pressure and load (weight of the cyclist and bike) will yield some radius or circumference between the two above values.

If you have less pressure or more weight (more deformation), the radius or circumference will be closer to the smallest radius or circumference.

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* (This number is actually too small because it doesn't include the thickness of a completely uninflated tire.)

sure we do - because that determines the distance a tire will cover in one revolution between one point on the outside of the tire and that same point again.

You're right - it's not a circle. And that's why the radius/diameters don't matter anymore. A square has a circumference, for what it's worth (technically, it has to be a closed curve - and the sharp corners make a square not a curve, but shapes other than a circle have a circumference, so the example works, and in the case of the square the outside perimeter is equivalent to the circumference). And it can rotate. The distance covered if it does one rotation and doesn't slip is circumference times rotations (n this case, one). That's a relationship that holds for simple 2 dimensional shapes with an area. Like our tire under deflection.

JB

PS - you're confusing the impact of inflating the tire (which most definitely can stretch the rubber) and weighting it down (which I'm arguing doesn't, at least not enough to matter in the real world). You also are consistently using radius when we have a non-round shape. Doesn't work anymore.

The difference is that the a rigid square wheel would cause the hub to bob up and down.

The tire deforms in a way that the distance between the road and the center of the hub is constant. That constant is the radius of the virtual circle that the rollout measures.

The perimeter measurement (the circumference) of the tire is a constant but the rollout measurement will be smaller for low pressure and large low and larger for high pressure and low load.

That is, the real rollout measure varies depending on pressure and load.

The perimeter measurement can't be the correct number. It's going to be larger than the real rollout number (*).

If there was no errors in the measurement, measuring the rollout is no different than measuring the height of the hub (the two numbers are exactly related by pi()*2r).

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(*) Practically, the perimeter measurement might be close to the real rollout number but it will still be larger than the real roll out.

Ok, substitute an ellipse for the square in my example. It has a circumference. if it rotates slowly enough it won't leave the ground (the hub bouncing up and down is irrelevant). And if you want to calculate the circumference, you can actually use the radii, just in a far more complicated equation. The roll out is still equal to the circumference. Non-round shapes have a circumference. And that is equal to the roll out. And trying to use the calculations from a circle (even a virtual one) probably won't get you the right answer.

Like I said, it's easy to verify. do the rollout measurements one way (inflated but not bearing weight) and the other (bearing weight). I've done it both, and the error trying to ensuring you are measuring from the same point on the outside of the tire is greater than the difference between the averages of measuring each way 10x.

One final example - I flat spotted my tire when my chain broke last week and locked up my rear wheel. The roll out still is equal to the circumference (the flat spotting definitely changed the amount, but not the fact that they are equivalent), but it won't be equal to your virtual circle. Yes the hub goes up and down when i get to the flat spot. It doesn't change the distance I'm travelling or the number of revolutions to do it. And your equation doesn't work (indeed it will give different answers, depending on if the flat spot is on the ground or not).

JB

edit - none of this is helping the OP one bit, beyond pointing out that he really doesn't have to worry about whether his rollout is measured with him on the bike or not!

No, the rollout would only be the same if the tire was rigid and did not deform.

The measured circumference defines the upper bound of the real rollout measurement.

The amount of deformation changes the rollout measurement and it depends on the load and pressure. The real rollout value will be smaller for lower pressure and higher load. (We aren't talking about huge differences: they might be on the order of 1%.)

Assuming an even deformation and no error, measuring the height of the hub (and using pi()*2r) would yield the same value as the rollout.

1% is significant in terms of cycling pissing matches.

It is possible that mounting a wireless sensor close to the hub will put it marginally within or out of range from the computer. Although I am hesitant to contradict a reported Sheldon opinion, since the sensor operates via a magnetically induced electrical pulse, wouldn't a rapid pass of magnet by the sensor be at least as effective in inducing current as a slow pass?

I dunno, some (many?) sensors are just reed switches where the magnet open/closes the switch. If the field is too weak and/or fast it may not open/close the switch.

OK, you're focused on the shape of the tire when looking at it from the side. That deformation is sidewall deformation, and does not change what you'd measure taking a tape measure around the outside of the tire. The contact patch changes - it gets longer and wider. the longer contact patch is not taken into account in your virtual circle model, but it is still part of the circumference and needs to be accounted for.

The tire doesn't need to stretch or shrink when it deforms - within reasonable limits. And if it doesn't stretch or shrink, the circumference didn't change. All that changed is you no longer have a circle, you have a different closed shape. You can still measure the circumference - which is exactly what the rollout measure does, directly and with no calculations needed. And if the circumference didn't change, the # of rotations needed to go a certain distance haven't changed. All the rest is trying to fit a non-circular curve back into your model for a circle.

If you have some data that shows that deformation due to carrying a load makes a difference in rollout, I'd be very interested to see it.

JB

OK, to simplify:

You claim that what matters is not the circumference, but the radius of some imaginary circle.....that you are calculating from the rollout (which whether you like it or not is by definition the circumference)

You claim that because the radius is different, the rollout will be different.

I say that's all a red herring - calculating the radius is completely unneccesary because we can measure the rollout, which unless the tire is slipping or the tread is somehow shrinking or stretching under very slow rotations is by definition the circumference. (not my definition, it's the definition in every geometry text out there)

Now the real point of contention is whether the circumference can change on a tire within reasonable limits when you place a load on it. I say no, it really can't - at least in any meaningful sense. Not only that but unless folks are very careful the noise from trying to measure rollout will swamp any difference.

You say the effective radius of the virtual circle got smaller - implying that the tire circumference is shrinking because a load is place on it. That seems problematic to me - I say the circumference didn't really change, the contact patch lengthened and that's the piece missing in your model. In reality, I wouldn't be surprised if the tire actually stretched under those conditions, since the rest of the tire would be facing higher effective pressure with the load - but I still think it would be too small to be measured the way most of us are measuring rollout.

Not data, but expert (see credentials on sidebar and see his webpage, but yes agreed still some dude on the internet) provided info about car tires and no load vs. load circumference:
https://en.allexperts.com/q/Tires-235...ence-tyres.htm

"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."

makes the same mistake - uses a change in a circular dimension (diameter) to assume that a dimension that applies to more than just circles (circumference) also changes in the same way even though the shape is changing from circular to some other shape - thus invalidating the relationship between the measurements.

JB

Are you also disputing that psi doesn't make a difference? The unloaded circumference of a tire at 20psi is very very close the the circumference of same tire at 120psi.

If the circumference doesn't change, then the rollout won't change. They are by definition the same measurement.

Pumping up a tire should stretch it out/change the shape, so it should increase the circumference. By how much, i don't know. And I'm speculating.

JB

OK, I have reams of data that shows when psi drops the measured miles increases.

Then the circumference changed.

JB

I'm a bit confused here with the circumference thing. If I take a wheel with a tire mounted and nominally inflated that is exactly 700mm in diameter and roll it out unloaded, I would get a circumference of 2199.115mm. If I load the tire such that it deflects 1mm, and do a roll out, I should get a measurement of 2192.83mmm or a wheel diameter of 698mm. The circumference of the tire did not change, but by loading the wheel, I made a shorter radius from hub to contact point. This is the distance that is measured during a roll out? I am displacing the sidewall and that allows for the hub to get closer to the ground. Yes the tire will 'squirm' to adjust to the new contact radius, but it won't be that noticeable. Consider the case of a fully flat tire. One should get the roll out number for the rim diameter plus the tire thickness.

Am I correct in this line of thinking?

You are still measuring the circumference when you measure rollout, but in those cases it is no longer a circle so it doesn't have a diameter. Unless there is slippage, shrinkage or stretching going on, the circumference/rollout shouldn't change - and it should be equal to what you would get if you cut the tire and then flattened it out. FWIW, inflating a tire should stretch it out.

In your example, the circumference along the tread would have had to shrink for that to work. I don't know of any mechanism that could give that result. If you told me that the tire got bigger, I could actually understand that since the deformation would force more air into the rest of the tire, creating higher effective pressure and possibly stretching it a bit.

What's confusing people is that circumference is a measurement that applies to shapes other than a circle. However, it is only in the case of a circle that you can calculate the circumference using diameter (or radius) and a constant. When a tire deflects, it becomes less circular. At some point, radius (or diameter) become useless, even for estimating what the circumference is. The shape is complex enough that equations to calculate the circumference from easily observable measurements would be interesting.

However, the circumference doesn't necessarily change just because the shape changed, unless there is shrinkage or stretching. And, by definition, circumference equals rollout unless there is slippage going on.

JB

Way too many people with way too much time on their hands here...by the way, the "censor" is still making me laugh. A computer with a censor...that's a first.

Nice analysis. This never occurred to me -- I've always been a "load it down" guy -- but when you think about it, it makes sense, so long as there is at least one point (fixed w.r.t. the hub, I suppose) that doesn't slip.

I understand the circumference of the tire is essentially fixed. But in a roll out I am measuring one wheel revolution. If the center of the wheel is closer to the ground, I have to get a smaller number. I can't explain the loss of distance from the nominal circumference of the tire.

Which circumference? At the center of the tread, or the side of the center, closer to the sidewall, where the circumference is slightly lower? On a partially-squished tire, both the center of the tread and the side tread are contacting the ground. But these have two different circumferences, so slippage must occur.

This is why there's a discrepancy between the tire circumference and the real rollout measurement.

The contact patch is flat from the front to the back of the tire. The contact patch is longer with low-pressure and high load.

It's as if the tire is a track on a tank with two rollers on the ground close together.

With low-pressure/high-load, the two rollers are farther apart. The circumference of the tire hasn't changed but the points on the edge take a detour (they no longer take a circular path). With less-pressure/more-load, these points take even less of a direct path.

https://gunpoint-3d.com/art-aw-T-3.html

On a tank, measuring the rollout would entail measuring the distance traveled through one rotation of a roller (pick one of the ones on the ground). That measurement is clearly much, much smaller than the tread circumference.

That measurement would be pi()*2r of the height of the center of the roller.

For a nondeformable tire, the circumference of the tire will be the real rollout. In the real world, it will be smaller and will match pi*2r of the center of the height of the hub.

The circumference around the midline of the tire. No (appreciable) slippage occurs.

Too bad it isn't correct!