Discussion on UK Fatbike Club last night around whether lower pressures affect gear ratios.

My thinking was that a soft tyre would have a smaller effective radius (the effective bit being at the bottom where the tyre is flattened) reducing the circumference. In essence a smaller wheel size.

This was argued against as like a tank track- it doesn’t matter on the shape, the perimeter will be the same.

I seem to have lost the argument but still can’t picture why. Help!

If you reduced the pressure to atmospheric then the rim would be much closer to the ground. One revolution of the wheel would therefore result in less forward distance travelled.

Tank tracks are on solid wheels. You’ll very slightly affect gearing ratios with a lower pressure tyre than a higher pressure tyre.

At the point the tyre touches the floor, it will be a smaller diameter.

It’s the internet, so someone is always right, and they’ll dig up all sorts of dodgy research.

Ie.

Yes exactly my thoughts. But now watch this:
Blew my mind

How does that work??

I seem to have lost the argument but still can’t picture why

My theory: With a theoretically perfectly round, perfectly hard wheel (on a perfectly smooth, perfectly hard road), the wheel is effectively pivoting about the point where the wheel contacts the road. If the tyre deforms, the wheel will be pivoting about a point below the road surface, so the effective diameter of the wheel will still be the same, even though the axle is closer to the road.

RE the video… i guess over a long enough distance it would show?
I mean, if the radius of the EARTH was 1m bigger, the equator would only be 6 and a bit m longer…
So the tine difference of that tyre isn’t gping to show that much i guess?

also..does having 3 other NORMAL wheels sort of ‘push’ the flatter wheel forward????

Dunno!

DrP

As above, guessing the wheel on the other side and a differential would mean if you keep the steering wheel straight, it should be the same, but the flatter wheel would be going slightly faster.

does having 3 other NORMAL wheels sort of ‘push’ the flatter wheel forward????

No, it’s rolling freely.

Another way of thinking about it is that the key dimension is the diameter where the tyre first contacts the road surface (i.e. the forward most point on the contact patch). This will change only slightly if you reduce pressure.

The apparent paradox is because you are thinking that the circumference is determined by the vertical distance from the axle to the ground, when it is actually determined by the distance from the axle to the point where the tyre first contacts the ground.

Differential innit.

Do the same experiment with the fat bike and stick it on youtube

Explains why friends Strava distances are sometimes so different on the same ride.

Down to tyre pressure innit. 👍

I’m not sure that video is a great test.

First of all, as other people have commented, vehicle mechanics (specifically the differential) come into play, and there’s no discussion of that.

There are no measurements of the wheel dimensions when inflated/deflated, nor indeed any tyre pressure measurements offered.

There are no predictions or calculation of what you might expect the difference in rolling circumference to be based on the change in dimensions from inflated to deflated, so its impossible to know whether a single revolution would show a detectable difference…

… particularly as the accuracy of the measurement approach isn’t addressed. Given the nature of the test and measurements (rolling vehicle being stopped when some bloke bangs on the side, and the distance measured using a tape and thick white marks on the ground) I’d be surprised if they were accurate to within 5mm…

… but that’s even something of a moot point because they only make an actual measurement once in the video, and the rest is done by the camera and the white marks, and there’s no accounting for any parallax error.

Finally, if you step the video frame by frame from 1:50 (https://youtu.be/_2l5bOhHNxU?t=110) there’s a clear difference between the white mark on the tyre and the white mark on the floor — this is mentioned in the YT comments but isn’t mentioned in the video I don’t think.

So all in all I don’t think the video tells you very much, and given the mechanical differences between a van and a bike I especially don’t think it tells you much about bike tyres 🙂

vehicle mechanics (specifically the differential) come into play

The differential has no effect on the front wheel. That’s a two-wheel drive vehicle without a locked differential between the front and rear wheels. The front wheel is completely independent of the others.

There are no measurements of the wheel dimensions when inflated/deflated, nor indeed any tyre pressure measurements offered.

It doesn’t matter. The axle is clearly much closer to the ground because the tyre is half flat. If the tyre was deflated by an inch, the naive expectation would be that a full revolution would be about 3 inches different. That obviously didn’t happen.

they only make an actual measurement once in the video, and the rest is done by the camera and the white marks,

The white marks are measures. Just because they aren’t calibrated against a metric or imperial scale doesn’t prevent them from functioning as measures.

Take an elastic band (or oring etc) mark a point on its surface and place it on your desk at a marked point, roll it along your desk whilst “circular” mark where it finishes 1 full revolution, repeat with the band flattened.

It’ll finish in the same place.

The shape is irrelevant the only thing which matters is point a on the circumference is always the same distance (pi*d in a perfect circle) away from its self, no matter how you deform it, without adding or removing material that distance remains constant. If you made the wheel square each side would be pi*d/4, the travel would be the same but the axle would be a varying distance from the ground as it rotated.

#caveat being at high enough pressure the tyre may stretch, in practice I don’t imagine that’s by much mind.

The differential doesn’t matter (unless it’s locked, therefore not a differential.).

Lower presssure still might have given a more notable difference, but essentially any difference is minor so it’s not worth changing gearing or anything.

The differential has no effect on the front wheel. That’s a two-wheel drive vehicle without a locked differential between the front and rear wheels. The front wheel is completely independent of the others.

Fair enough. I don’t know enough about vans to be able to make that judgement, and the video doesn’t say. I was perhaps thrown by the intro which refers to “a 4×4 forum”.

It doesn’t matter. The axle is clearly much closer to the ground because the tyre is half flat. If the tyre was deflated by an inch, the naive expectation would be that a full revolution would be about 3 inches different. That obviously didn’t happen.

That’s sort of what I was getting at. What exactly is the naive expectation? It’s not hard to call it out, but the video didn’t. So you’ve no idea what to expect.

The white marks are measures. Just because they aren’t calibrated against a metric or imperial scale doesn’t prevent them from functioning as measures.

Fair point, and I agree.

I should point out that I don’t really know what to expect from this sort of test, and I’m not saying that the conclusion of the YT video is wrong. It might well be right. But it’s still not a convincing experiment, and that’s all I was really getting at.

In cars there are two types of tyre pressure monitoring. Direct uses a pressure sensor on the wheel. Indirect is linked to ABS/ESP and measures the speed of rotation of the wheel. The latter only works because the speed of rotation changes as pressure varies.

What exactly is the naive expectation?

That the rolling distance will decrease proportionately with the apparent radius of the tyre (i.e. the distance from the axle to the ground). Circumference is 2 x radius x pi, so dropping the pressure enough to lower the vehicle by 1 inch intuitively seems like it would reduce the rolling distance by about 6 inches (1 inch radius reduction x 2 x 3.14). That 6 inch difference would easily be visible using the fairly simple methodology in the video (i.e. the effect size would be an order of magnitude greater than the measurement error).

The only difference changing the pressure has is changing the circumference slightly. More pressure means more tension and some extension of the material.

Ignoring that if the tyre doesn’t slip clearly we travel the same distance per revolution. Some lovely explanations above about centres of rotation

actually..the elastic band idea helped me..

The actual CIRCUMFERENCE of the tyre is always the same – it doesn’t stretch or shrink – it just flattens out…

In a similar but different vain..I always thought cars ‘figured out’ if your tyre was flat/losing air, but noting one wheel spinning faster than the other..I guess that isn’t the case, no??

DrP

.I always thought cars ‘figured out’ if your tyre was flat/losing air, but noting one wheel spinning faster than the other..I guess that isn’t the case, no??

No, they have a small microphone on the rim which listens out for the hiss, and can also pick up people shouting ‘Oi! Mate! You’ve got a flat!’

Yes, but the gearinches reduction will be massively offset by the effect of the extra friction involved in running squishy tyres.

(instant answer, ive not looked at the discussion above yet)

I get the point about the circumference being constant therefore the distance/revolution is the same but if you think of gearing as mechanical leverage then the distance between the hub center and the ground is reduced so the leverage/gearing must reduce also. Therefore the work done to get that full revolution has changed.

In the square example surely the gearing must change? It would take a lot more work to get it to turn.

Just to add fuel to the fire…

Would it make a difference if the ground was rocky rather than flat? I think it might, as then subsequent pointy bits of rock would in the extreme case be driven at the lower radius, i.e. in line with the “naive” expectation of the reduction in rolling diameter corresponding to a lower gear.

No idea what magnitude of difference it makes, but I’d expect my 4″ JJ at its normal ~25PSI+ for tarmac to roll further on one revolution than the ~6-10PSI some of you use for JJs offroad.

The actual CIRCUMFERENCE of the tyre is always the same – it doesn’t stretch or shrink – it just flattens out…

That’s how I see it and the radius doesn’t tell you anything once the tyre is no longer a circle.

I disagree that the circumference remains the same. I think its slightly smaller with lower pressure.

I.E a firm tyre you could measure the circumference of the centre of the tread (the bit that is touching the road on a round profile tyre) and it would be, for example, 1000mm.

If you ran the tyre soft so the shoulder blocks of the tyre now touch the road, you, would measure the effective circumference around the shoulder of the tyre, which could be 975mm, no?

I remember watching Guy Martin and him explaining why the revs on his bike increase when he is cornering. Because when he has his knee down, the tyre is rolling on its shoulder, effective circumference is smaller than the centre of the tyre, so the engine has to rev higher (or the tyre slow down but that would result in loss of grip and a crash)

P.S I haven’t watched the video linked earlier (yet) because surely the only way to settle this is with a treadmill?

Edit: Just watched the vid. Now I’m doubting myself 🙂

Because when he has his knee down, the tyre is rolling on its shoulder, effective circumference is smaller than the centre of the tyre

A motorbike tyre has a very different profile though to a car tyre, when he leans it over then the effective circumference reduces but that’s a different effect than lowering pressure.

Circumference does not remain the same. There’s a flat bit across the bottom of the tyre. If you were to run around in a big circle, then take a short cut straight across the last bit, you’d cover slightly less distance would you not? Same thing for a tyre with a flattened section. The lower the pressure, the longer the flat bit so the less the circumference. This matches up with the principle of smaller effective wheel size. If you were to do this experiment with a circular piece of string then yes flattening one part would make the rest of the circle larger. But a tyre is a 3D object – by flattening one part the extra material is bulging out sideways – this is how it’s possible.

So the answer to the OP is yes, and there is no contradiction here.

Let’s say you’d marked the circle you are running around with a long rope and joined the two ends. No you flatten the bottom of the circle and it’s now a shorter distance which is what you are saying. So in which case what has happened to the bit of rope that represents the shortening

Perhaps the sketch above could be different; rather than having the bottom face as being flat where it contacts the ground, imagine it arcs upwards towards the hub so that the overall circumference remains the same. The real life tyre deformations make it look less like this up to a point, but if you think of how an underinflated car tyre wears on the outsides but not the centre it will be because the edges of the tyre are more in contact with the ground than the centre (as per the middle portion arcing upwards). Hence, the idea already mentioned about measuring as per the radius at the points it first & last contacts the ground would make sense.

So in which case what has happened to the bit of rope that represents the shortening

As I said – if you do the experiment with rope then you have this problem, because the rope is not compressible. But your tyre is. The extra material splays out sideways. If you get a flat tyre on your car, rim is obviously closer to the floor but the rest of the tyre stays the same size, it does not get bigger. The section on the floor just gets wider as well as flatter.

Circumference does not remain the same. There’s a flat bit across the bottom of the tyre. If you were to run around in a big circle, then take a short cut straight across the last bit, you’d cover slightly less distance would you not? Same thing for a tyre with a flattened section.

That’s an interacting point. I still think that the physical length of the tyre doesn’t change. So if we now have a short cut where has the extra length gone?

On reflection I think that the radius goes up at the edge of the flat bit. So that’s how the circumference stays the same. I was hoping to find ac sortable picture to test my theory

Oh another thought. At lower pressure the tyre gets less wide and taller, in cross section. So the radius increases ever where except the flat bit?

the distance between the hub center and the ground is reduced

No, the distance from the hub center to the point where the tyre first comes into contact with the ground stays the same. With a perfectly round wheel, this will be the vertical distance from the ground to the axle. With a half flat tyre, this point is at the front of the contact patch, so considerably in front of the axle. The wheel is trying to pivot about that point, like a pole vaulter pivoting about the point where the pole contacts the ground.

Molgrips….
But with a flat tyre, the bottom is flat, but the rest of the tyre isn’t still a perfect circle is it…. It bulges out, i. E is deformed beyond just the flat bit… Because of the bulge…

DrP