Hi Bob

Sorry about the length of this one, see if you can stay awake
through it. I didn't.


On (29 Nov 98) Bob Lawrence wrote to Theo Bee...

 TB> Of course a radius change must affect the reading, simple
 TB> stuff. But on reflection it doesn't appear as simple as just a
 TB> radius change. Observing a radial shows a considerable flat
 TB> spot where it touches the road. Closer observation shows a
 TB> bulge where the radius is actually larger then the the
 TB> original one.

 BL>  It's not radius... it's the height of the axle above ground. As the
 BL> car rolls forward the axle rotates inversely proportional to its
 BL> height above ground. The axle rotation (and hence speedo reading) is
 BL> basically the linear movement on the ground divided by the axle height.

 BL>  This is just simple maths.

This is the linchpin of your argument, and it's got a huge hole in
it: it's true only for a circle!! For the umpteenth time, a
flattish tyre is NOT a circle. The axle is locked to the tyre,
*not* to a circle.

 BL>  Does a flat tyre put the car closer to the ground? If it does, then
 BL> the speedo reading changes and hence the circumference of the tyre
 BL> also changes, BY THE SAME AMOUNT.

Assuming a tyre that squeezes and stretches without resistance
(which a steel radial isn't), this argument is easily disproved.
I've given you the proof already, but you seem to have ignored it,
so here's a summary: if we take a circular tyre and squash it so
its axle is 5% closer to the ground, the flat bit on the bottom is
squashed by only 1.67%. THIS IS NOT THE SAME AMOUNT!!!!!

*THIS* is just simple maths. Do it yourself. Five minutes with a
bit of paper and a calculator. Do it, then tell me how to compress
the flat bit by 5% instead.




 BL>  If Roy's assertion is true, IF the circumference does not change,
 BL> then the only way it can do that is if the axle stays at the same
 BL> height with no air in the tyre. Is this fucking stupid, or what?

You didn't even read Theo's first para above - look at a tyre, see
the flat spot on the bottom, and see also the little bulge in
front of and behind the flat spot. That's the tread on the ground
resisting being squashed, pushing a bit of tread out in front of
and behind the grounded bit. It can do this because the sidewall
isn't rigid, and is actually bent out a bit so can be pulled in a
bit to allow the tread to move further away from the axle.







 BL> Simple my arse... it's impossible. So measure the axle height
 BL>                                                   ^^^^^^^^^^^
 BL> instead, which will reflect the speedo reading, precisely.

 BL> v = r * d0/dt as simple as that. v is the actual speed, and
 BL> d0/dt is proportional to the speedo reading. r is the radius,
 BL> axle height above ground.


 TB> Indeed, radius times angular velocity, the hitch is in the
 TB> radius, where do you measure the radius in the centre of the
 TB> tyre or the extreme contact points, something must give.

 BL>  You measure the axle height. What radius? You only care about the
 BL> axle rotation, which is where the speedo is connected, so all you care
 BL> about is axle height above ground, and movement along the ground...

Here's the flaw in your argument again. Your equation v = r * d0/dt
applies to a circle. The constantly morphing mess that is a real
tyre would need a considerably more complex equation.

See if you can concentrate long enough to poke a hole in the
following:

_____________

Look at a tyre whose axle is 5% closer to the ground than it would
be if the tyre was circular. According to your theory, the axle
speed over ground is v = 0.95 * r * d0/dt, where r is the original
radius. But this equation also applies to a circle of radius 0.95r,
centred on the same axle, and touching the ground right under the
axle.

A circle of radius r will travel 2*pi*r forwards for one axle
rotation. A circle of radius 0.95r will travel 0.95*2*pi*r for one
rotation, 5% less than the bigger circle.

So what you have is a circle of 0.95r rotating on the same axle as
the flattish 1.0r circle. According to you, the 1.0r tyre rotates
at the same rpm as the 0.95 circle, and therefore the bit of rubber
on the ground distorts exactly enough to slow the axle's forward
speed by 5%, ie it must shrink by 5% .

But the geometry shows that it only shrinks by 1.67%! For it to
shrink by 5%, it would have to pull in bits of tread from the
sections just above ground. In the real world, we don't see this,
we see the reverse. But back to the argument:

So the flattish 1.0r circle must be rotating at a rate different
from the full 0.95r circle.

_____________


If that's not enough, try this:

_____________

Take a 1.0r radius tyre, and magically make its tread totally
rigid, but leave its sidewalls flexible. Place a load on this wheel
until its axle is 5% closer to the ground. The tread stays
circular, the sidewalls bulge at the bottom and straighten out at
the top. And don't tell me this can't happen, any sidewall on an
inflated tyre bulges out a bit, so can be straightened a bit given
enough force. Let's call this the off-centre circle model.

Does this axle rotate at a different rate from the axle of an
unloaded wheel, for a given forward speed? No, because the
circumference is undistorted, so the wheel moves 2*pi*r forwards
for each axle revolution regardless of axle position.

According to your theory, the forward speed is v = r * d0/dt, where
r is the magical axle height, so for a constant forward speed v,
d0/dt has to increase as r decreases, so the axle rotates faster.
This contradicts the above para.

"Unfair!" I hear you scream. "All you've done is make a fake centre
for a real circle! Of course the v=r*d0/dt equation doesn't apply!"

But if your scream is true, then exactly where is the true centre
of a flattish circle? It isn't a real circle, so how can you define
a "centre"? It certainly can't be at the centre of the original
circle. "Yes it can," I hear you say. "The flat bit that's a bit
closer to the axle compensates by being compressed." Somehow I
doubt this - if we load the wheel up so the axle is 99% closer to
the ground than the original circle, will the d0/dt centre still be
at the axle? ie will the axle be going 100 times quicker? Somehow I
doubt it.

Any good mathematician (ie better than I am now) would be able to
derive a variant of v=r*d0/dt to suit the off-centre circle. I'm
suggesting that similar skill would be needed to get a similar
variant to suit the flattish circle model, and its result will
*not* be exactly equal to v=r*d0/dt.

_____________


And on (29 Nov 98) Bob Lawrence wrote to Niels Petersen...

NP> Roy experimented and only found a small difference on the
NP> speedo. What he forgot was the DIFF is in between. !!!!

 BL>  I doubt very much that he could measure 1.7% accurately. It's silly.
 BL> To me, a single reading of 1.7% means bugger-all.
 BL> It would have to be 10% before I'd even write it down.

It wasn't a single reading. It was two readings at full pressure,
of 126.5 and 126.3 seconds, and two readings at reduced pressure,
of 127.5 and 127.1 seconds. That's four readings within 1% of each
other. The tyres were flattened by 4.8%. Wanna explain how I got
errors in the second two readings almost exactly 5% different (and
different in the same direction) from the first two?


Cheers

--- PPoint 1.88