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Jeff[_34_] Jeff[_34_] is offline
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Default Does a tyre change its CIRCUMFERENCE when underinflated?



"The Natural Philosopher" wrote in message
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On 28/06/18 10:24, Jeff wrote:


"The Natural Philosopher" wrote in message
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On 27/06/18 19:36, Tim Streater wrote:
You can also imagine that if you could get hold of a tire of the same
overall diameter but to go on a much smaller wheel, that the spiralling
effect when flat would be much more pronounced as the tire wall would
be bigger. But, once spiralled up, one turn

I've been thinking about this a lot trying to make it simple enough for
even a remoaner to understand.

The fallacy is in Huge's basic premise that because the axle is a given
height above te ground, that represents the 'effective/rolling radius*'
of the tyre.

It does not.


It does when considering the distance that determines the rotation rate
of the wheel.

The 'effective/rolling radius' is /always/ the circumference divided by
2 PI.


Not when it is the distance between the axle and the road that
determines the rotation rate of the wheel.

How can this be? Quite simply because the tyre is attached to the wheel
all round, and the tyre is dragged along by the sidewall such that the
tread is moving faster (angular velocity) than the rim at the point of
'flatness'.

So Huge's assumption that the tyre tread speed is actually it's distance
from the axis of rotation, times the speed of rotation of the wheel, is
simply a mistake. It is in fact *greater*, at that point, and its the
sidewall flexure that takes up the strain, and the attachment of the
tyre at other points of the rim that creates that strain.

Should the wheel deflate completely, then the strain in the sidewall and
the friction between the tread and the rim as the two models
compete...leads to rapid separation of tread and wheel.

You then end up then with a wheel rim and tyre tread rotating at
completely different angular velocity. The triumph of 'Huge's model' is
to rip the tyre sidewall apart to allow that to happen.


If you follow F1 you can see this regularly - you end up with the
sidewall disintegrating and a more or less intact tread separating from
the car wheel entirely.


This leads to the conclusions that what affects wheel RPM - in a non
slip case - apart from car speed, is change in *circumference* alone,
and that will be a function of tyre tread elasticity, tyre pressure and
tyre RPM only, due to 'centrifugal force'.

Loading won't affect it at all.

And whoever dreamt up 'effective' or 'rolling' radius deserves to be
strapped to a centrifuge.

A little knowledge is a dangerous thing.

So is relying on 'experts' who are equally ignorant.


But it is very easy to measure both a fully inflated tyre
and one that is say 20% under inflated and see if the
rotation rate of the wheel is in fact determined by the
distance between the axle and the road or the very
minimal change in the circumference of the tyre and
that must have been done when designing the ABS
system for detecting under inflated tyres.


Indeed. I see that my efforts are watsed and you havent understood a word
I said, nor even looked at the video I posted way back where someone did
precusely this and showed that within the limits of te technique,
flattening te tyre made no measureable difference


And yet we know that the TMPS system that uses the ABS sensors does
work to detect under inflated tyres very reliably.

- ceraiunly
not as much as the 'loss of radius' would account for. Nor yet te p[aper I
likned to that showed taht te differemnce with 30% deflation was of the
order of a percent or so. WAY less than would be accounted for by the
'loss of radius'



A lot of engineering is highly counter intuitive.


Doesnt matter when it is so easy measure.


Why don't you measure it then?


My car does have ABS, but doesnt show the instantaneous
rotation rate of each wheel on the OBD2 data.

I see that other people have, and that confirms my thesis.

What I wouuld ask you is how a tyre, rotating at a differentent rate from
a wheel, so that its circumference does in fact get covered on the ground
by that rotaion, stays on the wheel without ripping the tread off?