"NY" wrote in message
...
"Jeff" wrote in message
...
Yes, but it is the distance between the axle and the road
that determines the rotation rate.

Why?

Because that's how the physics works. And that should be obvious from
what happens with wheels with no tyre, with different diameter wheels.

Long time since 'O' level maths but I suspect the relationship between
the radius and the perimeter only works for a perfect circle.

Irrelevant to what determines the rotation rate of the wheel.

True, but the easiest way to visualise the rotation rate for a unit
distance travelled is to divide the distance travelled by the
circumference to give the number of wheel revolutions.

However... it all hinges on what the *effective* circumference is: it is
the circumference of an imaginary circle with the same radius as the
distance between centre of the axle and road surface, which will be less
than the no-load radius because the tyre is slightly flattened where it
comes in contact with the road.

I presume this means that if you took a piece of string and passed it
round the circumference of a tyre this value would be smaller than the
no-load circumference because the part that is in contact with the road
will be a flat rather than a curved profile.

I think that is unlikely given that you don't see that effect with a steel
radial although it is certainly possible that Tim's parallelogram effect
means that it isnt as rigid as you might think.

This assumes that the radius of the rest
of the tyre doesn't increase significantly when a load is applied to the
tyre - presumably this is constrained by the steel reinforcing belts in
the tyre.


I wonder how much smaller the in-contact radius is than the no-load
radius: what sort of proportion is the reduction, typically? I'll have to
measure the actual distance from the centre of the hubcap to the ground,
and then jack the wheel up until it first touches the road and measure
again. Note that you can't measure to a part of the car body because of
compression of springs :-)