Good morning, sir.

I offer the following theoretical experiment for your consideration
and contemplation.

Take a 1 metre circumference wheel, radius r1 = 159.155mm, and poke
100 small nails into its tread, 10mm apart, so they'll leave nail
head marks in the road as we roll the wheel along. The nail
directly under the axle at time zero we'll call Spike.

Every time we roll the wheel one revolution, there will be 99 nail
marks between Spike's first mark (Spike 0) and Spike's second mark
(Spike 1). If the wheel was a perfect circle, the nail marks would
be exactly 10mm apart, and Spike's two marks would be 1 metre
apart. Let's call the internailmark distance NM. For a rigid wheel,
NM = 10 mm.

If the wheel is a flattish tyre, then one revolution of the wheel
will still produce 99 nail marks and two Spike marks. At the
beginning of the revolution, Spike will be directly under the axle.
At the end of the revolution, Spike will be directly under the
axle.

What is in dispute is the distance between Spike 0 and Spike 1 when
the wheel is a flattish tyre, instead of a perfect circle. The
change in this distance must be reflected in the distance NM, and
it's this distance that my entire argument is based on.

Let's define a "flatness ratio" FR, where FR = (axle to ground
distance for a flattish tyre) / (axle to ground distance for a
rigid circular tyre), or r2/r1.

(Picky bit:

 This definition is in trouble already, because it's a bit hard to
 measure the radius of a perfectly circular tyre. Any tyre, fully
 inflated and not supporting any load, will not have a dead flat
 tread. The tread will bulge out in the middle, just like a gut
 with too much beer, or a head with too much air. This also pokes a
 bit of a hole in my own theory, because it shows that the steel
 reinforced tread can indeed be stretched, at least to some extent.
 But if the tread was infinitely flexible, an unsupported tyre
 would look like a sausage balloon tied in a circle, ie it would
 have a near circular cross section. My argument says it will
 resist this tendency somewhat.

End picky bit.)

Now let's deflate the perfectly circular tyre so that FR = 0.95,
and assume that the tread is infinitely flexible, so what we have
is a circle with its bottom sliced off and replaced with a straight
line. The flat section in contact with the road will subtend an
angle Theta = 2 * acos(0.95) = 36.390 degrees at the wheel centre.
The length of the flat section will be d1 = 2 * r1 * sin (Theta/2)
= 99.393 mm. The length of the original circumference that subtends
Theta is d2 = 1000 * Theta / 360 = 101.083 mm.

The piece of tyre in contact with the road has been compressed from
101.083 mm to 99.393 mm, a factor of 1.67 percent. So the nail head
marks made on the road as we roll this flattish tyre forwards will
be 9.833 mm apart (on average: the flat bit of the tyre will be
compressed more near the centre than at the edges, so the nail head
marks will be a bit smeared. The distance between the centres of
the smears, though, will still be 1.67% shorter than 10mm, because
that's the *average* distance between nail heads in this straight
bit.)

One rotation of the wheel will mark on the road Spike 0, 99
intermediate marks 9.833 mm apart, and Spike 1. The distance
between Spike 0 and Spike 1 will be 983.3 mm, 1.67% shorter than
the original 1000 mm. So for a given speedo indication, the wheel
will move 1.67% less distance, so the speedo will be reading approx
1.67% high.

If the theory that the speedo accuracy is directly proportional to
FR was correct, then the nail head marks should be 9.5mm apart,
which means the length d1 should be 0.95 * d2, or 96.029 mm. This
would leave a gap of 3 mm in the flattened circle. We might try to
compensate for this by saying that the bits of tyre very near the
flat bit are also compressed a bit, but that would only reduce the
required compression on the flat bit and increase the NM distance
even more.

Conclusion:

If the tyre was infinitely flexible, we could expect a 1.67% speedo
error for a 5% reduction in axle to ground distance. Since the tyre
is not infinitely flexible, the flat bit will be compressed by
something less than 1.67%, and we should expect a lower speedo
error.

Biased footnote:

My experiment's result was that a 4.7% change in axle to ground
distance (*NOT* radius!!!) had a 0.7% change in speedo reading. The
only criticism of this experiment was Dorkbrain's suggestion that
at 70 kph, centrifugal force would make the tyre a lot less "flat".

This idea can be easily disproved by a simple experiment that I've
already done: (a) find a tyre that you can see from your driver's
seat while you're driving, or bribe a passenger (b) check that the
tyre is at normal pressure, then observe it at rest and at 70kph.
(c) deflate the tyre to 95% of normal axle-ground distance at rest,
observe how flat it looks at rest (d) drive at 70 kph, and see if
the fat bulge at the bottom resembles the faint bulge at normal
pressure. If you're really picky, do this experiment with a lightly
loaded rear wheel, like my Mazda van. When I did this, I could see
that a low tyre at 70 kph was still a low tyre, considerably lower
than the same tyre at normal pressure.

---The Truth Is Out There - it sure ain't in here...


--- PPoint 1.88