"Rolling Radius" vs Circumference of Deflated Tyres and Tyre RPM.
Submitted: Saturday, Sep 25, 2021 at 16:13
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Frank P (NSW)
Being in lockdown and having plenty of time for the mind to wander I thought more about the side discussion in the space-saver spare wheel thread ....
While Zippo and others made me doubt my understanding, having thought about it more I am again convinced that deflating a tyre causes it to spin slightly faster than its fully inflated mate of the same nominal size at the same roadspeed.
Consider a tracked vehicle and its driving sprocket. The track length stays the same. For a given road speed a larger sprocket will spin slower than a smaller one to maintain the roadspeed. Ie, at a given roadspeed the RPM of the sprocket is directly related to its radius.
Now consider a tyre. The tread of the tyre is a fixed length, just like a track. As you continue to deflate a tyre, the tread with its fixed circumference becomes a little more like a track with its elongated contact patch and a little less like a perfectly round wheel . The distance from the centre of the axle to the ground is effectively the radius of the pseudo-sprocket, if I can call it that. As you deflate the tyre that radius decreases and consequently just like a tracked vehicle, to maintain a given roadspeed the wheel (sprocket) has to turn faster.
Of course the extent to which this can happen is limited by the connection of the tread to the rim by the sidewalls - they kind of get in the way (and get hot as a result) - but I think the principle is there nonetheless.
Discuss if you wish, or travel if you can. LOL
Reply By: AlbyNSW - Saturday, Sep 25, 2021 at 17:11
Saturday, Sep 25, 2021 at 17:11
I have heard both sides of the debate and have been on both sides at one point or the other
If anyone is bored during lockdown perhaps they could do an experiment on their driveway of marking a reference point on two tyres and deflating one to say 6psi . Drive the vehicle forward 10 metres or so and then see where the reference points are in relation to each other
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Follow Up By: Allan B (Sunshine Coast) - Saturday, Sep 25, 2021 at 17:48
Saturday, Sep 25, 2021 at 17:48
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Sorry Alby, can't help. My driveway is only 7 metres.
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Reply By: Allan B (Sunshine Coast) - Saturday, Sep 25, 2021 at 17:46
Saturday, Sep 25, 2021 at 17:46
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Frank,
Your tracked vehicle analagy does'nt do it for me I'm afraid. The track is no more than an interface between the sprocket and the ground. Neither its length, nor even its existence has any consequence upon the the progression of the sprocket which is simply rolling along the ground, albeit with a track interface.
A tyre has a defined circumferential length. One revolution of the wheel, and hence the tyre, moves that circumferential length along the ground by 1 unit, i.e. its length. It matters not what the shape of the tyre may be, it will retain the same circumferential length as it would if perfectly circular. Accordingly, 1 revolution should produce the same progress length regardless of a varying distance between axle and ground.
I say "should" because I know that it does not. I know that a 'flat' tyre rotates more for a given ground distance. The discrepancy, I believe, can be accounted for by compression of the flexible tyre as it is rotated by the wheel. The tyre, being more flaccid in its underdeflated state is being compressed as it is fed under the wheel thus shortening its effective length and hence its progress on the ground. Essentialy, its circumference is reduced in the same proportion as the reduction in radius.
Some of that heat you refer to is doubtless generated in this compression and recovery of the tyre as is rotates, as
well as the contribution from sidewall flexing.
Do you see any legitimacy in my thinking Frank?
AnswerID:
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Follow Up By: Frank P (NSW) - Saturday, Sep 25, 2021 at 18:27
Saturday, Sep 25, 2021 at 18:27
Allan,
We agree 100% on the tread length (circumference) being the constant.
The distance the tread will move the vehicle forward in one complete revolution will be the same on each side regardless of inflation because the treads are equal and fixed in length (circumference). On this we also agree.
The issue is what drives the tread to rotate. Think about a Caterpillar D9 or whatever with the driving sprocket at the top of a triangular arrangement of the track. A smaller sprocket will have to spin more to make the track complete one revolution, a larger sprocket less so.
Now consider the tyre. The driving "sprocket" is the rim and the sidewall. Deflate the tyre and the radius of the "sprocket" is reduced as the sidewall bulges and the axle gets closer to the ground. Therefore the "sprocket" has to spin more to make the tread do one complete revolution. (As the "sprocket's" axle gets closer to the ground it is a bit like the triangular track arrangement on the Cat D9, only inverted.)
So the wheel travels the same distance as its fully inflated mate on the other side, but the axle (and "sprocket") that drives it must spin more to make it happen because of its reduced effective radius.
Whether the "sprocket" is rolling or driving it makes no difference - with a deflated tyre it must increase RPM to keep up with or drive the fixed-length tread.
Is this more paradoxical than Aristotle's wheel? LOL
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Follow Up By: Zippo - Saturday, Sep 25, 2021 at 18:41
Saturday, Sep 25, 2021 at 18:41
Allan, in my (engineer's) view you are partly correct. The analogy I prefer is the tracked vehicle. The track has a fixed circumference. Similarly the modern tyre has a belt of definitely fixed circumference, buried under the outer casing topped with tread.
On the tracked vehicle, if you fitted a 50mm block of -say- wood to each track element, what would be the change in effective circumference? My answer is NONE.
With the tyre, the overlaying tread deforms at the point whereas it is forced into contact with the road under the belt, and a similar -but opposite- deformation as it escapes from the road surface. These cancel out, the net effect of this deformation is zero creep, so the effective circumference remains dictated by the underlying belt.
This is largely unaffected by tyre inflation except in extreme situations like a flat. Have you ever observed a flat tyre being driven on very slowly? I have, and it was interesting that the tread contact area was "interrupted" - there were TWO separate contact patches, with the belt going concave between them, not what I had expected. In that scenario you will get a departure from the "constant circumference" scenario, but reductions in tyre pressure that are less dramatic (than a flat) won't alter it.
That's my 2.2c worth.
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Follow Up By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 11:18
Sunday, Sep 26, 2021 at 11:18
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Exactly Zippo, attaching the wood blocks has raised the track but the "wheel", the sprocket, has not changed. One rotation will still propel it along the track, which is in close relation with the ground, for the same distance as before.
Bear in mind that the sprocket is a 'wheel' in itself.
On an automobile, the 'wheel' is composed of a metal centre firmly attached to an outer rubber tyre. The whole assembly is the effective 'wheel' and must be considered as such in contemplating the physics.
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Reply By: Briste - Saturday, Sep 25, 2021 at 18:25
Saturday, Sep 25, 2021 at 18:25
Also linking to that the side discussion in the space-saver spare wheel thread ... Run flat tyres were briefly mentioned. Cars with such tyres alert you to a suspected flat tyre, or at least an under-inflated one.
For those vehicles without a TPMS, my understanding (and I've unfortunately seen this alert a few times) is that the warning is shown because one of the car's systems has detected that one wheel is rolling noticeably faster than the others. I believe that this sensor also is also used for ABS and ESC.
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Follow Up By: RMD - Saturday, Sep 25, 2021 at 19:22
Saturday, Sep 25, 2021 at 19:22
Bridge
Thanks for that. It shows Zippo isn't quite on the mark and the speed difference is registered by wheel speed sensors. As mentioned before, my mates Daughter had an Audi and it flashed a light to indicate. He carefully checked tyre pressures and one was 4psi less than the others. If tyre/road radius wasn't the cause what could it be? because when inflated correctly the issue was not there. No tyre monitors.
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Follow Up By: Briste - Saturday, Sep 25, 2021 at 21:35
Saturday, Sep 25, 2021 at 21:35
False RFT alerts do rarely happen. Had one recently (pre-lockdown) driving down the Hume enroute to collect the new Prado, and thought oh no, why here!?!?!? Stopped and checked the pressures and they were fine. It was a real puzzle as to what caused caused that false alert. Drove another 300-400kms without issue. (I think it was just a coincidence that a bolt went through a tyre a few days later.)
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Follow Up By: Bob Y. - Qld - Saturday, Sep 25, 2021 at 22:15
Saturday, Sep 25, 2021 at 22:15
We have a Mazda CX-5 that has a tyre alert function, amongst other electronics warnings.
It has only activated twice in just over 60K kms. The first time was after travelling across the partly gravel road from Mt Perry to
Monto. A rear tyre had a small leak that I repaired with a plug, that did the job until we had new tyres fitted in Biloela.
The second time the light came on was travelling up the old Landsborough H'way, where a storm had fallen the previous night, and there was a degree of wheel spin involved. I understood that this function worked with ABS, to indicate a fault?
The vehicle has a space saver spare too, so on most journeys a 12v compressor & numerous plugs are always part of the kit.
Bob
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Follow Up By: Zippo - Sunday, Sep 26, 2021 at 10:43
Sunday, Sep 26, 2021 at 10:43
We had the misfortune to be in a Chevy HHR on a US trip a while back, and it had dashboard displays including tyre pressures. One time the alert triggered and the display said a front had 4psi. Got out and did the look/kick and nothing amiss, certainly nothing like 4psi would be. Drove to a nearby town and checked the tyre pressure - spot on the 34 the others had. Never did it again.
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Reply By: RMD - Saturday, Sep 25, 2021 at 19:27
Saturday, Sep 25, 2021 at 19:27
If your GPS says 100kmh and the tyre tread at the ground instant is effectively stationary, What speed is the top of the tread doing? Does the system of leverage happen? Is the axle moving at100kmh? If so there are strange things happening to the tyre!
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Follow Up By: Zippo - Sunday, Sep 26, 2021 at 10:38
Sunday, Sep 26, 2021 at 10:38
That's an easy one. The top is travelling at 200kph. And if the GPS says 100kph then surely the axle is doing the same unless the vehicle is disintegrating.
Not sure why you even bothered asking ...
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Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 08:14
Monday, Sep 27, 2021 at 08:14
Hi Zippo,
That is not correct. Think of a wind turbine which is in effect a three spoke wheel without a tyre. The axle is spinning at a relatively slow speed, while the tips of the turbine blades are spinning much faster. The speed at the tips is relative to the length of the blade. The longer the blade, the faster the tip will spin for the same axle speed.
Macca.
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Follow Up By: Michael H9 - Monday, Sep 27, 2021 at 09:43
Monday, Sep 27, 2021 at 09:43
If the speed on the ground is 100 kph then the top of the wheel is travelling at 100 kph and the axle is travelling at 100 kph. By contrast the speed of the wheel on the bottom is 100 kph in the opposite direction, that is, backwards. Velocity is speed in a given direction, so every point on the circumference is travelling at the same speed but at a different velocity because every point is continually changing direction. For example, a point that is at 3 o'çlock on the wheel is travelling 100 kph vertically upwards while also travelling at 100 kph in the direction of the car.
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Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 13:02
Monday, Sep 27, 2021 at 13:02
Michael,
I think you may be confusing speed with rotation. The forward speed of the vehicle may be 100 kph, but the outside of the tyre is rotating faster than the axle to maintain that speed. The circumference of the axle is going to be something around 80 mm where as the circumference of a 33” tyre on the wheel attached to that axle will be something just over 1.1 m. The outer edge of the tyre has to travel 1.1 m in the same time that the axle travels 80 mm. Therefore, the outer edge of the tyre has to be rotating faster for the same axle rotation in order for it to complete one rotation.
Just as in my example of a wind turbine, the tip of the turbine blade is rotating faster than the axle.
Macca.
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Follow Up By: Michael H9 - Monday, Sep 27, 2021 at 13:23
Monday, Sep 27, 2021 at 13:23
No I'm not confusing anything. The speed that the outer circumference is travelling at is the speed that the car is travelling at... unless you're doing a burn out.
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Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 13:35
Monday, Sep 27, 2021 at 13:35
Hi Michael,
Yes that is correct, but the axle will be travelling slower.
Macca.
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Follow Up By: Member - LeighW - Monday, Sep 27, 2021 at 13:39
Monday, Sep 27, 2021 at 13:39
Uuum,
The car is moving forward at 80Kmh, therefore the axle and everything in the car is moving forward
at 80Kmh an hour if fixed to the car, this is not the same a rotational speed or RPM of the tire or axle.
The road is also going by at 80Kmh and if the tyre is in contact with the road than that bit of tire inn contact with the road must also be doing 80Kmh else it would be skidding.
"Just as in my example of a wind turbine, the tip of the turbine blade is rotating faster than the axle." That would depend on how you look at it, the turbine blade is rotating at the same speed of the axle for instance if the axle is doing a 100RPM then so is the blade. The actual distance covered by the tip of the of course is a lot greater than a spot on the surface of the axle would cover in one revolution.
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Follow Up By: Allan B (Sunshine Coast) - Monday, Sep 27, 2021 at 13:47
Monday, Sep 27, 2021 at 13:47
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Gentlemen, it seems to me that you are confusing each other.
When expressing "speed" (or any motion for that matter), unless it has already been established or understood, it is necessary to nominate the reference.
For example, the speed of a car is normally understood to be in relation to the road.
But the speed of a wheel or tyre needs defining both as to what part of the tyre and in relation to what?...... the car body or the road or what else.
In the case of a car travelling at 100km/h (in relation to the road) the mass of the tyre is moving at 100km/h but an instantaneous point at the top of the tyre is moving at 200km/h in relation to the road whereas an instantaneous point at the bottom is moving at 0km/h in relation to the road.
Interestingly, a point at the very front of the tyre (having travelled 270 degrees since leaving contact with the road) is travelling forward at 100km/h at the same time as 100km/h downward in relation to the road.
Does that help to confuse the discussion?
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Follow Up By: Michael H9 - Monday, Sep 27, 2021 at 13:56
Monday, Sep 27, 2021 at 13:56
The axle has several different speeds. It's travelling at the speed of the car quite apart from the actual rotation. The outer rim of the axle is rotating such that its speed is a much slower speed than the outer circumference of the whole wheel. Take note that I said a point on the outer circumference in the 3 o'çlock position is travelling at 80kph both forward and straight up vertical, so two different directions and the same speed in both at the same time.
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Follow Up By: Michael H9 - Monday, Sep 27, 2021 at 14:05
Monday, Sep 27, 2021 at 14:05
Allan, the point on the top of the tyre isn't moving at 200kph or it would leave the tyre. That also implies that the point touching the road isn't moving at all. You can't add the speeds together similar to two cars having a head on crash each travelling at 100kph. It's not a 200kph crash it's the same as one car hitting a solid stationary object at 100kph.
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Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 14:13
Monday, Sep 27, 2021 at 14:13
Terminology: revolutions per minute vs speed. The tip of the wind turbine blade in order to make one revolution per minute has to be moving at a faster speed than the axle of the turbine. Same applies to the outer edge of the tyre vs the axle.
Macca.
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Follow Up By: Allan B (Sunshine Coast) - Monday, Sep 27, 2021 at 17:58
Monday, Sep 27, 2021 at 17:58
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Michael, the top of the tyre is most certainly moving at 200km/h relative to the ground.
You are not grasping the factors.
In a car travelling at 100km/h the axle is moving forward at 100km/h relative to the ground.
The wheel is rotating on that axle at a peripheral speed of 100. So a point on the top of the tyre is moving past the axle at 100. The top of the tyre is therefore moving at 200km/h relative to the ground.
But as that point on the tyre rotates to come into contact with the ground, which of course is stationary, that tyre point must also becomes stationary. It's forward speed is now 0km/h.
I could could draw your attention to this...... if this imaginary point on the tyre is to rise from the ground behind the axle then overtake the axle to get in front and rotate down to the ground again, it must move faster in the air than the axle is to be able to overtake it.
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Follow Up By: RMD - Monday, Sep 27, 2021 at 19:07
Monday, Sep 27, 2021 at 19:07
I wonder how many people did Physics at school and think about how physics relates to everyday life. Hard to believe what some really believe is true.
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Follow Up By: Michael H9 - Monday, Sep 27, 2021 at 20:42
Monday, Sep 27, 2021 at 20:42
Allan, when you put it like that you are correct, I misinterpreted what you meant sorry. The point on the circumference spends half its revolution going from 100kph to 200kph and back to 100kph, then spends the other half of the revolution going from 100kph to 0 kph and back to 100kph relative to the ground. Its nett forward speed is 100kph for the whole revolution.
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Follow Up By: Member - silkwood - Friday, Oct 01, 2021 at 09:44
Friday, Oct 01, 2021 at 09:44
Just looked at this thread, an interesting discussion (and all held in reasonable and polite conditions!).
Allan, I think your point here is the most relevant, the key being that measurement is only meaningful in reference. Here there is a tendency to combine two different measurements, the length of the "circumference" and the radius of the circle created from the centre of the axle to the road. Two totally different references. Reading forwards, it appears the answer was given, though in a somewhat convoluted manner!
Very enjoyable read.
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Reply By: Peter_n_Margaret - Saturday, Sep 25, 2021 at 20:31
Saturday, Sep 25, 2021 at 20:31
I have (copyright) data for the Michelin tyres that I run on the OKA.
It quotes (amongst other things) the relationship between Road Contact area (cm2), Laden Radius (mm) and Rolling Circumference (mm) when these tyres are operated at reduced pressures.
Road contact:------------- 493,,,,,541,,,,,599,,,,741.
Laden Radius:------------- 428,,,,,423,,,,,417,,, 402.
Rolling Circumference: 2820, 2810, 2800, 2760.
Conclusion? Michelin say the rolling circumference reduces with reduced pressure at the same load.
Cheers,
Peter
OKA196 motorhome
ps the extra commas and stuff was an attempt to line the numbers up :)
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Follow Up By: Frank P (NSW) - Saturday, Sep 25, 2021 at 20:55
Saturday, Sep 25, 2021 at 20:55
Thank you Peter.
That says to me that there is a correlation between reduced tyre pressure which causes a reduced radius and therefore a reduced effective circumference.
Therefore the tyre at reduced pressure must revolve more frequently to cover the same distance as a tyre at a higher pressure given the same load, nominal size, etc.
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Follow Up By: RMD - Saturday, Sep 25, 2021 at 22:31
Saturday, Sep 25, 2021 at 22:31
Peter,
That info is only from a small tyre company, what would they know eh? Anyone who thinks a deflated tyre doesn't revolve faster and hotter should go straight to
the Pool Room and stay there.
PS, my Corolla has Michelin tyres. and now they are not talking to me! Must be SUB standard.
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Follow Up By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 09:46
Sunday, Sep 26, 2021 at 09:46
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So Frank, how do you account for Boobook's video clip where it shows (before your very eyes) that the distance travelled per wheel rotation did not change with decreased tyre pressure?
p.s. Please forgive delayed response. I have been off-line. Bloody NBN!!
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Follow Up By: Briste - Sunday, Sep 26, 2021 at 10:25
Sunday, Sep 26, 2021 at 10:25
Allan - as I posted in other follow-up, "The fact that the rolling circumference doesn't change doesn't negate Frank's initial point in the original post in this thread - that an under-inflated tyre will rotate faster, because its effective radius - from the centre of the hub to the road surface - is less than the other fully-inflated tyres. That tyre tread is going to cover more distance than the others."
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Follow Up By: Frank P (NSW) - Sunday, Sep 26, 2021 at 10:51
Sunday, Sep 26, 2021 at 10:51
Allan, I never disputed the circumferential thing. The tank track analogy I used supports it. What I am trying to come to terms with is the rotational speed of the axle supporting a deflated tyre. Is there a difference between that and its mate on the other side supporting a fully inflated tyre?
Let me ask you a question.
Given the video and our agreement on same, how do those OEM built-in tyre pressure warning systems (ie, those without actual pressure sensors) work?
PS Forgiven. I, too, have been subject to the vagaries of NBN. 12 weeks Nov '20 to Jan '21 and intermittently this year.
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Follow Up By: Allan B (Sunshine Coast) - Tuesday, Sep 28, 2021 at 13:51
Tuesday, Sep 28, 2021 at 13:51
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Yes Frank, I have no doubts that a wheel not fully inflated must rotate faster than its fully inflated mate on the other end of the axle. And that is demonstrated by the successful operation of the inferential TPMS systems which operate by sensing the rotation of each tyre and performing a comparison. I suspect that the difference in effective 'rolling radius' is not great with inflation change but enough to be sensed and resolved.
Boobook's video presentation, which aims to demonstrate that an under-inflated tyre rolls the same distance as a fully inflated one, fails to convince as it is performed over one rotation only which is insufficient sampling data for the required accuracy. As I said above, the differentiation is not great and more rotations would be needed to resolve the issue. I suspect also that the significance may vary between a slowly rolling tyre and one travelling at speed with dynamic forces at play. But if so, I could not hope to predict in what way or magnitude. Just a suspicion, as dynamics often do modify the the simple geometrical reasoning.
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Reply By: GG7777 - Saturday, Sep 25, 2021 at 22:54
Saturday, Sep 25, 2021 at 22:54
Wouldn’t it be the same effect as a brand new tyre verses a bald old tyre?
This should give a different/smaller ‘rolling radius’ I would think
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Reply By: Member - Boobook - Sunday, Sep 26, 2021 at 07:34
Sunday, Sep 26, 2021 at 07:34
The circumference does not change when the tyre is flat. It can't when you think about it. The tank track analogy appears to be sound.
Here is proof. Looks like the same discussion has been around on forums. since 2009.
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Follow Up By: Member - McLaren3030 - Sunday, Sep 26, 2021 at 08:39
Sunday, Sep 26, 2021 at 08:39
That is what I suspected all along. Thanks Tony for digging that up. Some might argue that it wasn’t conducted under “controlled conditions”, but I think it is good enough to settle the argument.
Macca.
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Follow Up By: Briste - Sunday, Sep 26, 2021 at 09:30
Sunday, Sep 26, 2021 at 09:30
The fact that the rolling circumference doesn't change doesn't negate Frank's initial point in the original post in this thread - that an under-inflated tyre will rotate faster, because its effective radius - from the centre of the hub to the road surface - is less than the other fully-inflated tyres. That tyre tread is going to cover more distance than the others.
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Follow Up By: Zippo - Sunday, Sep 26, 2021 at 10:35
Sunday, Sep 26, 2021 at 10:35
Briste, that's a total contradiction. The ROLLING CIRCUMFERENCE is just that i.e. the distance the vehicle travels per revolution of the drive shaft, no more no less. The shaft-to-road height doesn't appear anywhere in that relationship.
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Follow Up By: Member - Warren H - Sunday, Sep 26, 2021 at 10:42
Sunday, Sep 26, 2021 at 10:42
I don't know about controlled conditions but close observation shows that the experiment was faulty as there was insufficient control on the the positioning of mark on the tyre as the angle doesn't appear to be at the same when each of the measurements are made. If you look at Peter's data for example, there is only 60 mm difference in the distance so everything has to be spot on. Perhaps if you set up a plumb line to control the positioning of the mark you would be able to improve the experiment.
The equation that relates the arc length (l) to the radius (r) is l=r.w, where w is the angular displacement in radians. (So for a full 360 degree rotation or 2pi radians you get the formula for the circumference 2pi.r.) So if the radius r is changed, the distance travelled must change it's not a matter of opinion.
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Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 08:25
Monday, Sep 27, 2021 at 08:25
I will go back to my original contention, and that is that when you deflate a tyre, you have not changed the circumference of the tyre, the tyre is no longer a circle, but there is still the same “length” of rubber on the road. The circumference is a measurement of length, that length hasn’t changed significantly with the tyre deflated.
Macca.
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Reply By: Michael H9 - Sunday, Sep 26, 2021 at 10:18
Sunday, Sep 26, 2021 at 10:18
Everybody take notice..... the tyre pressure monitoring in VW's, (that I know of), uses the wheel speed sensors to pick up the fact that a tyre is going flat because it starts rotating at a different speed.
How do VW tyre pressure sensors work
I don't care if it doesn't make logical sense, science is science. Things that rotate are the source of many a paradox. There's Aristotle's Wheel paradox and the Coin Rotation Paradox -
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Follow Up By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 11:06
Sunday, Sep 26, 2021 at 11:06
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That Coin Rotation demonstration is a magician's sleight-of-hand.
When the coin is being rotated and is at the half-way point the 'magician' declares that it has performed a full rotation because the image is now 'upright' as in the starting point. However that is only because it has done only a half rotation but is now inverted, which restores the image to its original upright view.
Smoke and mirrors my friend, smoke and mirrors, not science!
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Follow Up By: Michael H9 - Sunday, Sep 26, 2021 at 15:25
Sunday, Sep 26, 2021 at 15:25
No, that's incorrect Allan. For the head of the coin to be the right way up, same as when it started means that it has completed a full rotation when it reaches the bottom.
https://en.wikipedia.org/wiki/Coin_rotation_paradox
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Follow Up By: Frank P (NSW) - Sunday, Sep 26, 2021 at 16:01
Sunday, Sep 26, 2021 at 16:01
I just did it very carefully with two 20c coins on a table, no slippage. If it's a magician's trick then I'm a much better magician than I thought I was LOL.
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Follow Up By: Allan B (Sunshine Coast) - Monday, Sep 27, 2021 at 23:03
Monday, Sep 27, 2021 at 23:03
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Can you not see that, to get from the top to the bottom, the coin has 'rolled' a half turn but on a course that 'inverts' its axis at the same time? Thus it appears upright at the halfway point.
Which is why when rolled out along the straight string path, it is upside down at the halfway point and only becomes upright when it completes the full turn at the end of the string. This path does not invert the coin's axis.
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Reply By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 10:44
Sunday, Sep 26, 2021 at 10:44
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A flaw in reasoning of this subject is that the tyre is being treated as a circle when it is not. Not when distorted by a laden vehicle.
The formula of c=2*pi*r is for application to a circle, a perfect circle. A distorted tyre is a not a 'circle' so that formula cannot be applied.
A tyre attached to a wheel has a given perimeter length, even when distorted by load. If the supporting wheel is rotated by one turn, then the tyre's perimeter will describe a distance equal to its perimeter length, regardless of its deviation from a perfect circle. This surely is obvious by observation not by conjecture.
Using the formula based on 'r' is a notion based on that formula for a circle when it is not.
Frank's caterpillar track analogy is false in that the sprocket is not 'attached' to the track which is simply an interface between the sprocket and ground. One turn of the sprocket does not produce one rotation of the track as is the case with a tyre affixed to a wheel. The sprocket rolls along the track as if it were on the ground and being a 'perfect circle' the c=2*pi*r can be applied to this case.
So how does the TPMS sensing based on tyre rotation work? Dunno, I'm still working on that!
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Follow Up By: Frank P (NSW) - Sunday, Sep 26, 2021 at 11:34
Sunday, Sep 26, 2021 at 11:34
From the link in Michael H9's post immediately above about VW's indirect tyre pressure warning system:
"The tire pressure monitoring system looks at the ABS wheel speed sensors and uses that information to track the rotational speed of the tires. Any time a tire has lower pressure, it will roll at a different number of revolutions per mile than tires that have the correct tire pressure. If one tire’s rotational speed seems off compared to the rest, the TPMS will alert the driver that one of the tires has low pressure."
If we forget Dieselgate and accept this then how can it be reconciled with the fixed tread length argument?
The only variable as the tyre deflates is the radius of the rotating parts - the vertical distance from the contact patch to the axle. For the rotating parts with a smaller radius of rotation to cover the same distance, which is controlled by the fixed circumference of the tread, they MUST rotate more - a fact that VW uses in its indirect TPMS.
So if the wheel is doing more rotations and the tyre is of fixed circumference and therefore ostensibly travelling further, what is happening to the extra distance that the tyre should be travelling but is not because it is fixed to the vehicle that is being "held back" by three (or more) normally inflated tyres?
Perhaps that distance is simply lost to the tyre scrubbing or slipping on the road, which shows up as accelerated wear of the tyre.
FollowupID:
916323
Follow Up By: Briste - Sunday, Sep 26, 2021 at 12:59
Sunday, Sep 26, 2021 at 12:59
I only got into this thread because of my experience with the run-flat warnings on vehicles with RFT but no TPMS. Believe me, these systems work, that's why manufacturers use them as a poor man's TPMS. Under-inflated tyres rotate faster.
Why is that? If we had a mechanical engineer here to sort us out, like Allan_B does for electrical matters, then we wouldn't be having this classic online slugfest. Without that, everyone is going to have to come to their own understanding of why this is. I'm with Frank. Reduced effective radius. YMMV (pun intended).
FollowupID:
916325
Follow Up By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 13:08
Sunday, Sep 26, 2021 at 13:08
.
Yes Frank, I agree that 'inferential' TPMS using wheel rotation sensing does work.
Because of this, I cannot ignore that a deflated tyre with unchanged circumference rotates faster than its fully inflated mates.
But I cannot accept that the measured distance from the hub to the ground surface (rolling radius) can unconditionally be seen as a radius in simple c=pi*2r computation of a circumference.
Surely, if you were to dismantle a tyre such as to enable the tread (circumference) to be laid out straight on the ground then that would be the distance that the vehicle would travel with one revolution of a wheel.
I am missing something but cannot see it! Think I'll go and take my pills!
FollowupID:
916326
Follow Up By: RMD - Sunday, Sep 26, 2021 at 14:08
Sunday, Sep 26, 2021 at 14:08
We all recognize that the tread length remains the same if partially deflated, and many know the wheel rotates faster, to keep up, with opposite side. However, it seems many will not recognize that as deflation increases, so does the degree of tyre squirm and deformation and tread scrubbing at the road surface as the tread length is forced to slip and abrade past the road surface. With this squirm the heat increases/rises at an alarming rate. Maybe it doesn't happen with American TIRES but it happens with Australian TYRES. The tread buckles up in the centre in relation to the now lower sidewalls, that constant tread deformation has to occur for the wheel to rotate faster "cos the sidewalls are lower.
Q, why does a flatish tyre begin to make two parallel scrublines and the tread isn't touching much at all. Run a partially deflated tyre over a flat section of clay and you will see. Anyone been offroad and hit mud or clay with lower pressure would know. Perhaps they don't!
FollowupID:
916327
Follow Up By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 14:15
Sunday, Sep 26, 2021 at 14:15
.
OK, I took my pills and am now convinced that I have the answer.
As the tyre rotates and a given section makes contact with the road, it dynamically changes from being an arc (curved section of circle) and becomes a chord (straight line connecting two points on a circle). This chord is shorter than an arc joining those same two points therefore the circumference is shortened. If the circumference is shortened then the forward progress of the axle will be less per rotation of a wheel of perfect circle shape (no flat section as bottom). Or conversely, the wheel will need to rotate faster for a given progress.
This shortening of the circumference is only accomplished by compression of a section of the tyre (the flat bit in contact with the road). This constant compression/relaxation will generate heat which contributes to that being generated bu sidewall flexing.
This proposition satisfies all (sensible) argument. Yes, the 'effective radius' (dislike 'rolling radius') changes and yes the actual circumference changes. "Everybody wins a prize"!!!
FollowupID:
916328
Follow Up By: Frank P (NSW) - Sunday, Sep 26, 2021 at 14:24
Sunday, Sep 26, 2021 at 14:24
Allan,
I wrote a follow-up to yours (916326) above. It was along the lines of RMD's content (Follow-up 916327) about tyre squirm absorbing or partially absorbing the increased RPM, but you posted the above which, along with RMD's, makes my contribution redundant so I deleted it.
I had not considered the arc-chord thing and doubt that I would have got to it. As with all good engineering solutions, it is simple!
An interesting discussion, thanks for hanging in there!!
FollowupID:
916330
Follow Up By: Allan B (Sunshine Coast) - Sunday, Sep 26, 2021 at 15:08
Sunday, Sep 26, 2021 at 15:08
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Frank,
I had said in Reply 638068 above that "The discrepancy (in circumference), I believe, can be accounted for by compression of the flexible tyre as it is rotated by the wheel". I realised that could be the only explanation but had not at that time reasoned it scientifically. It really was staring us in the face.
FollowupID:
916331
Follow Up By: RMD - Sunday, Sep 26, 2021 at 16:53
Sunday, Sep 26, 2021 at 16:53
Mathematically, if you think that way, but tyres are not mathematical at all, they are flexible. To the eye it may seem as a chord at road surface but, it sure isn't and that chord line will be a fluctuating suddenly with a severe induced S bend of sorts. The rubber has to go somewhere in that instant trip under the rim. If you cut a tyre open in the tread after a deflation failure you will see the belt has been severely massaged and it was hot and delaminated.
The flapping your may hear on some isn't the sidewall, it is the tread doing Yoga under the rim. If it doesn't recover before the road hits it again then the flapping is more.
FollowupID:
916335
Reply By: Michael H9 - Sunday, Sep 26, 2021 at 15:49
Sunday, Sep 26, 2021 at 15:49
There's one inescapable fact, the velocity at the ground is equal to the perpendicular distance of the point touching the ground to the axle, multiplied by the angular velocity. The angular velocity is the rotational speed in whatever units you like eg. revs per minute. This equation is absolute if you are in a Newtonian physical framework, it's a dead set physical law. Therefore if the velocity of the car remains the same, and the distance of the axle to the ground gets smaller, then the angular velocity must increase to compensate, therefore the wheel speed must increase. Whatever forces this must exert on everything involved create a loss of energy and a heat build up as one wheel fights the other. Its why your car will start pulling to one side or another and struggle to keep straight in extreme circumstances.
AnswerID:
638084
Follow Up By: RMD - Sunday, Sep 26, 2021 at 16:58
Sunday, Sep 26, 2021 at 16:58
Michael
Thanks for the physics, that is exactly what I learnt of during my apprenticeship. therefore it was known by the auto trade a long time ago. Model T vehicles didn't flap as much because they didn't go faaaaast. Isn't it strange how some believe the laws of physics are somehow different.
PS. The vehicles which read wheel speed via a sensor system do not register while turning, because the speeds ARE different rotation rates, but they DO register rotation difference in a straight line. If the duration of the event is longer than a specific programmed time, the ECU knows a wheel isn't the correct dia within it's programmed limit and registers a fault. Simples!
Briste probably saw an alert while driving straight ahead. Last time you drove it straight it registered a similar thing and so it's memory remembers. A bit like annoying someone, after a while they react.
FollowupID:
916336
Follow Up By: Briste - Sunday, Sep 26, 2021 at 17:55
Sunday, Sep 26, 2021 at 17:55
I haven't seen an RFT alert all that often, and when I do I'm usually too busy cussing to notice fine points of detail like the direction of the wheels.
I do remember reading in the MB literature somewhere that the vehicle's systems won't immediately detect an under-inflated tyre. They take a short while to come to that conclusion, probably partly for the reasons that RMD alluded to. The calculations are based on averages over a period of time. Which makes that recent false alert on the Hume all the more puzzling.
I once seriously damaged a sidewall - a combination of bad luck and very poor judgement on my part - and I knew about it
well before the RFT alert came on. RFT are of little use in such circumstances, as the tyre definitely didn't run while flat on that occasion!
FollowupID:
916338
Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 08:41
Monday, Sep 27, 2021 at 08:41
If we think about what Michael H9 has said in regard to the vehicle pulling to one side when a tyre has deflated. Which side does it pull to? It pulls to the side that has the deflated tyre. Would this not indicate that the other side, the side with the fully inflated tyre is turning faster? If that is correct, then the under inflated tyre is rotating slower.
Macca.
FollowupID:
916342
Follow Up By: Michael H9 - Monday, Sep 27, 2021 at 09:32
Monday, Sep 27, 2021 at 09:32
Sorry, but the ground speed is the same on both the wheels because they're both on the same car after all. There's a force on the flat tyre because the tread is longer than the decreased radius allows so there's a lot of stretching and rubber deformation going on causing greater friction on the ground.
FollowupID:
916345
Follow Up By: RMD - Monday, Sep 27, 2021 at 10:35
Monday, Sep 27, 2021 at 10:35
Macca
Have you ever heard of drag and friction and resistance and equal and opposite forces? It seems not to be. Just not logical!
If the deflated tyre is on the rear, then the drag and resistance to forward motion attempts to move that wheel to the centre line of the vehicle. and steers it to the left. Flat left front also steers to left because of drag and the caster effect of lean makes it go left.
FollowupID:
916348
Follow Up By: Member - McLaren3030 - Monday, Sep 27, 2021 at 13:04
Monday, Sep 27, 2021 at 13:04
Yes, of course RMD, that would explain the pulling to one side.
Macca.
FollowupID:
916352
Reply By: Member Kerry W (Qld) - Tuesday, Sep 28, 2021 at 13:55
Tuesday, Sep 28, 2021 at 13:55
Thinking out loud here...
One thing which would be interesting to consider - In lowering the profile of the wheel, generally, by deflating the tyre. Is there a (significant) increase in the power/torque being delivered. Or rather what is the increase?
If so, is this a trade off/implication in the distance the tyre covers or doesnt cover during a single rotation compared to a fully inflated tyre?
A few of the laws of physics need to be considered but I'm guessing/hoping someone else has the time to consider from this point of view...
The maths involved is a bit over my pay grade but I can see there are more factors involved than just distance travelled.
AnswerID:
638111
Follow Up By: RMD - Tuesday, Sep 28, 2021 at 14:25
Tuesday, Sep 28, 2021 at 14:25
Kerry
All depends if it is a front wheel free rotating or a driven rear wheel providing traction force forward.
Flat but free it will soak up power, On the rear it will be using power/torque at a certain rate to create heat and noise and the additional rotation is provided by differential action, where the crownwheel and carrier is going faster than the deflated side and slower than the inflated side. The differential gives and takes, ie directly proportional.
Net feeling with one wheel deflated is, engine RPM same, but some loss of forward speed.
FollowupID:
916382
Follow Up By: Member Kerry W (Qld) - Thursday, Sep 30, 2021 at 09:46
Thursday, Sep 30, 2021 at 09:46
Yes RMD I agree RPM same but less foward speed theoretically more power available....Interesting RMD that you mentioned the freewheeling tyre - I thought I explained
well enough. The deflated wheel(s) being driven "should" or could display an increase in torque/power/gearing being delivered in proportion to the reduction in diameter. Just as changing tyre size affects gearing and torque. It is simply a question to ponder....largely irrelevant but none the less an interesting question for those above who have taken an interest in the physics involved.
thanks for your thoughts.
FollowupID:
916436
Reply By: Geoff (Newcastle, NSW) - Tuesday, Sep 28, 2021 at 16:15
Tuesday, Sep 28, 2021 at 16:15
There must be some relationship between tyre pressure and rotation.
The reason I say this is my mate bought a new VW Toureg back in the early 5cyl diesel days.
One of the dealer fitment software options was a tyre abnormality warning. You paid your money, the dealer was issued a code by VW and the dealer entered that into the Toureg's plastic brain. It could be added any time in the vehicles life.
Once done you had VW's TPMS option.
The only way my mate and I could see for this to work using the vehicles existing array of sensors was the ABS ring.
As I said, there appears some form of measurable differing relationship between a correctly inflated tyres speed of rotation and an underinflated tyres speed of rotation.
AnswerID:
638113
Follow Up By: RMD - Tuesday, Sep 28, 2021 at 19:12
Tuesday, Sep 28, 2021 at 19:12
G'day Geoff.
First hand you have seen the reality of it all. Good to see users know what is the reason. It seems though, many people have infinitely more knowledge than ALL the world car companies who use ABS rings etc to
check rotation speed and report, if difference is more than the full tread wear depth.
Definitely smart some of those Aussies.
FollowupID:
916392