On Mon, 25 Jun 2018 07:50:42 +0100, Chris Hogg wrote:

On Mon, 25 Jun 2018 15:50:06 +1000, "Jeff" wrote:



Yes, but it is the distance between the axle and the road
that determines the rotation rate.

If the length of the perimeter of the tyre is say 1.5 metres,

At what state of inflation?

and the
vehicle is travelling at 150 km/hr, that means the tyre rotates 100000
times per hour, or 1666.6 rpm,

At what state of inflation?

assuming no slippage between tyre and
road or tyre and rim.

Ok.

There's no getting around that,

See above. ;-)

and there's no
mention of state of inflation,

But there should be as that's a key variable that affects the
effective 'perimeter length. ;-)

shape of the tyre or axle to road
distance.

No, but there should be as the shape of the tyre is instrumental in
the calculation.

On a solid tyre it wouldn't be because you have removed that variable.

Do you also deny the change in tyre diameter on a dragster (and
therefore rpm / mph)?

https://www.youtube.com/watch?v=YViiNxnQTr8

What about this. Imagine some lunar rover module where the 'tyre' is
made up of many completely independent segments.

With the vehicle up on jacks you might measure the effective
circumference of the wheel as the measurement made with a tape drawn
around it.

But put it under load and with it standing on one (sprung) segment.
the RW radius of that segment at that time under that load will be
less. Imagine that reduction in radius being passed round the wheel
segment by segment and you could then see (hopefully) that the
effective circumference would be calculable from the loaded radius and
it *would* be very different from the unloaded one.

Join those segments together on the outside by something plastic, like
say thin balloon rubber and nothing really changes.

Join them with something heavier but with the similar ability to
'give' and you have a pneumatic car tyre. ;-)

Cheers, T i m