Doge

I was ignoring the centripedal force figuring v^2/r was going to be the same for all tires at the same speed and mass.
Where the resistance to deformation in that 1ms will not be the same in all tires. I don't think it is measured anywhere. We can infer it from Crr. But the comment was about contact patch.

Here is a simpler statement.
If in 1ms tire A resists full (that same as at rest) deformation more than tire B the contact patches will be different sizes at the same width and PSI.

wphamilton

pneumatic tire = spring close enough for this question. Calculating is complicated a little by the fact that force hence acceleration depends on the displacement, so it's a differential equation BUT sniff test-wise you can get the frequency of a full cycle f = (1/2pi) * sqroot(k/m). Need to find the hookes law constant for the inflated tire somewhere.

BTW I'd assume that the contact area is proportional to the vertical displacement

Doge

That would work. Then we are back to the roller diameter and smooth surface, but it would prove at least one point about contact patch.

Who's going to build it? It would have to be smooth enough we wouldn't see the patch size vary. Ground glass might work as you would see the shadow.

To calculate it start with something like HYSTERETIC EFFECTS IN DYNAMIC HERTZ TYPE CONTACT OF RUBBER BALLS.
by Florina Ciornei.

In the opening paragraph she says:

It is known that mechanical characteristic of
rubber is not linear [1, 2], but, in order to take into
account the damping, an approximation using a linear
viscoelastic model is of common use [3].
The aim of this paper is to prove that, from the
damping perspective, rubber may be modelled by a
linear viscoelastic model.

See table in picture. She is testing rubber balls between .17 Hertz and 1.4Hertz. This tire thing is 1,000Hz (1ms) - one cycle.
I wouldn't know where to start with that data.
We need a table of moving tire one vs moving tire 2 in 1ms.
Capture.jpg

noglider 

@Scarbo what is the pavement quality in your area?

Doge

The pneumatic part is a spring for this. And yes to your assumption.

I calculated - non-differential equation from the start of the deformation to the end of cycle time as being 1ms. That was algebra and arithmetic. The deformation part is actually .5ms as the rubber contacts the road until it is fully deformed. that is not much time to get full drop.

Do you, @wphamilton, think in .5ms all tires deform (drop) as much as they would at rest?

If not, do you agree the contact area's will be different?

Doge

I have and have used, and son races with from 22-28s.
The 22s are the fastest on the right road and because they are on a TT setup.

These 27.5s are light 300g. They are not as fast as the 25.5 silks @ 230g on pavement. They are used for dirt/gravel.

TarmacClearance.jpg

noglider 

Is that a 1mm clearance between tire and seat tube?!

wphamilton

I'd estimate the time the contact patch is on the ground as about .01 second, and the time for the bike tire to deform as something less than that.

Doge

Didn't measure, but that was a new 2016 Tarmac. I just wanted to show with my arm waving and all I have some experience with the different sizes. I/we choose the tire based on the event.

Doge

.01 sec is <2mph. It is .0005 seconds to total drop at 30mph. You would need the tire to totally deform in that time.
Do you think it can totally deform/drop the same amount as at rest in .0005 sec?

In English the contact patch length is about .6in. At 30mph the tire moves 528in/sec so that .6 inch is in contact with the road .6/528 sec or .0011 sec. As it much drop in the first half of that time, it takes .00057 sec to drop. The question is can it fully drop in that time.

You didn't ask for numbers, @Brian Ratliff did. My math may be wrong, but we don't have to estimate much.
I calculated it here https://www.bikeforums.net/19861110-post199.html

wphamilton

I used 3 inches for length, five times your estimation, per FLO Cyling - The Contact Patch and 20 mph, and the time it takes for the tire at front of the contact patch to rotate to the rear is about .01 second.

The frequency of the full harmonic of a bike tire which deflects 15% from a load of 30 pounds, using the hookes law formula, is about 7.7/sec, or .13 seconds for the full cycle. For the first 15% of the compression part, it would be much much less than .01 second. That would be the time it takes to compress the tire that far.

wphamilton

An easier idea, from some paper that I happened across just now, is sticking an accelerometer to the sidewall. They got a big spike in acceleration when the sidewall bulged, and along with the speed calculated the length of the bulge.

However, the data is so noisy that it was impossible to detect the effect that you're looking for.

Doge

Our math agrees. Except the time to drop is half the time it is in contact. I have under .005 sec in your parameters for full drop.
Parameters we don't agree on. 20mph vs 30mph is simple enough. 3 in long contact patch?
I was using 100PSI with 100lbs on the tire for a 1"sq contact patch. Using that if it were 3" long it would be .1 wide to get that 1"sq area ~30:1, so I don't like your contact shape, but that also means more drop, more tire deformation, just over more time.

I just tried measuring this with both brakes on and my own puddle of water on the garage floor and the foot print was way bigger than the contact patch - in width. A assume that capillary action of the water made this not work. I don't have an ink pad, but I assume that would be better. But to be correct the patch should have an area that lines up with the PSI I think. Meaning a 100PSI with 100lbs should get a 1sq in patch.

I'll agree the real patch is more elliptical than my near circle. My .6 and your 3, my 30mph and your 20mph - and the halving of the time explains the difference in the time.

Of course the more you pump the tire, the smaller that patch gets and the less it deflects, even though doing it more quickly. I was trying to be simple using a 1sq area patch.

Here is the net:
I am saying a tire that takes 20W to roll (as tested) will not obey Hooke's law (non-linear elasticity) the same as one that takes 10W to roll when the time for deformation takes milliseconds.

I think you are disagreeing with that.

Net II:
It is hard to measure.
It is hard to calculate.

I wonder if you could take a 100PSI ire and with a maching hit it on top with 100lbs between 100Hz (your number) and 1000Hz and use a high speed camera see if the tire keeps up.
Would the tire "drop"/deform fully?
Would it vary from tire to tire?
As I said, hard to measure.

HTupolev

The area of an ellipse is pi*a*b, where a and b are the semi-major axes. If the length is 3, one of the semi-major axes is 3/2, or 1.5, so:

1 = pi*1.5*b

b = 1/(pi*1.5) = .2122

The width is 2*b, so .4244 inches.

wphamilton

Yeh I'd disagree, both will obey Hooke's Law with some minor inelastic stuff added - the deflated tires will have differing constants and are not very elastic. Altogether, I don't think you'll ever be able to measure a difference in contact patch area relative to speed on the road. Nor correlate a Crr difference to a differential in contact area sizes from two tires at precisely the same pressure. Those effects will be way less than the noise IMO, and I also tend to think that neither the length nor area of the contact area can be directly correlated to Crr to begin with.

edit: BTW, measure the rim distance to ground with 10 pound load, 20 pounds, 30 etc and easily determine that the tire does or does not obey Hooke's Law.

Doge

Yup - sorry.

HTupolev

If other aspects of the setup are allowed to change, definitely not.

Scarbo

Road surfaces vary greatly. Some of the roads I ride on are glassy smooth; others are horrible--rough, with big fissures and potholes. I live relatively close to the Nevada border and I love riding in the Silver State because they seem to have all the money in the world to devote towards maintaining their roadways! If I know I am going to be riding on rougher roads I ride my steel bike that is equipped with 38mm tires.

Doge

I have Skate and Nordic skied a few times in Kirkwood.

noisebeam

Every-time I've put on new tire(s) of the exact same size and brand (25c 4000gp II) they feel faster and more supple with better handling. This may be real vs. mental given the aging and wear of the rubber which on the old tire is cracking and feels harder.

An well worn tire will also have more rev's per mile resulting in a faster speed readout (if not gps) - about 0.5%

Carbonfiberboy 

We hope the OP and arguers are all using GPS for speed. If not, the OP would certainly notice a large speed increase from the 25s. Those 28s are very tall tires.

I've been enjoying 23mm tires on my 23mm (outside) deep rims. They look fast and maybe they are. Certainly doesn't hurt on descents. It occurs to me that perhaps pros are going to 25s to match their 25mm rims. Or 28 for 28 rims if there are such animals.

wphamilton

I dunno, my bike computer says I'm going slower with the 28's, having not gotten around to changing the setting yet. About 1/2 mph less than I think I'm going.

topslop1

fastest/lightest 30's and 32's?

Doge

Slower than what?

Maelochs

Slower than he thinks he is going.