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counter-steering, with numbers



I'm sending this to the entire list because the analysis I found has a
rather neat example of eliminating unknown parameters to reach a
conclusion. It is a rather long posting, though. Various points which
have spun off the main thread are included in PS's at the bottom.

First, since it's taken me several weeks to work this out, let me recap.
There is agreement that when starting a turn on a bicycle or motorcycle at
low speeds, the technique is to shift the body weight into the turn so that
the bike leans into the turn. At higher speed, especially on a motorcycle,
another method is required, called counter-steering. For a rightwards
turn, you first turn the front wheel to the *left*, which causes the bike
to lean to the right, so that the right turn can then be successfully made.

The question is, how does the leftwards counter-steering cause the
rightwards lean? David Thiessen has suggested that the counter-steer
causes an upwards torque on the front wheel; since the wheel has a leftward
angular momentum, this causes the wheel to lean rightwards. I have
suggested that the lean is just a result of inertia; described in the
non-intertial frame of the bike, the leftward counter-steer results in a
(very brief) left turn in which the unbalanced centrifugal force to the
right leans the bike over. [I leave as an exercise how to describe this
without reference to the fictitous centrifugal force.]

Which of these effects is most important?

To check this, I want to compare the angle by which the bike leans to the
angle by which the front wheel turns to the left. If you push on the right
handle bar, you provide an upward torque on the front wheel. Since the
wheel has angular momentum to the left, it wants to lean to the right. But
to do that it has to push the whole bike over. So the wheel applies a
forward-pointing torque on the bicycle frame -- this is what makes the
bicycle lean. The N3 reaction forces of the frame on the wheel result in a
rearward-pointing torque of the frame on the wheel -- this is what turns
the wheel to the left.

Using:
T = magnitude of the frame-on-wheel torque and the wheel-on-frame torque.
t = time
L = the angular momentum of the front wheel
alpha = the angle by which the front wheel turns to the left
I = the moment of inertia of the bike about the pivot point where the tire
meets the road
phi = the angle by which the bike leans to the right (from vertical)

The torque T causes the front wheel to precess: alpha = (T/L) t
It also angularly accelerates the bike lean: phi = 0.5 (T/I) t^2
Eliminating the unknown T, we get: phi = 0.5 (L/I) t alpha

That relates the two angles. Now for some numbers:
For a 1 kg wheel traveling at 5 m/s (~= 11 mi/hr), L = 5 kg m/s.
For a 65 kg person ~1.2 m of the ground on a 10 kg bicycle, I ~= 75 kg m.
Plug it in, and we get phi = alpha * (t / 30 s).
Since the turn happens rather quickly, the lean phi is pretty small
compared to the wheel turn alpha. I therefore believe that the torque on
the wheel contributes a very small amount to the lean of the bicycle.

For a more complete treatment, one could add the effects on the lean (phi)
of gravity (as a function of phi) and of inertia / centrifugal force (as a
function of alpha). I think you could still eliminate the
difficult-to-measure T.

Anybody want to plug in numbers appropriate to a motorcycle? For that, the
angular momentum of the engine may be significant, but I think its effect
should be to futher stabilize the bike against leaning.

--
--James McLean
jmclean@chem.ucsd.edu
post doc
UCSD

PS: CENTRIFUGAL TORQUE?
I suggested that the lean of a bike during a turn can be viewed (in the
non-inertial bike frame) as a balance of the torques due to gravity and
centrifugal force about the wheel-road contact point. David Thiessen
pointed out that such an explanation doesn't work if treating the problem
in an intertial frame; there seems to be nothing balancing the torque due
to gravity.

He suggested using fixed point at the center of the turn, calculating
torques about that point. He found a small resulting torque, to which he
attributed the precession of the wheels as the bike goes around in a
circle. However, I don't think this analysis can be correct, since the
same analysis should apply to a skater going in circles, who has no
rotating wheels to give angular momentum.

After thinking long and hard about this, I think the answer is that there
are only two kinds of reasonable points about which to calculate the torque
on a body. One is the center of mass, which always works. The other is a
fixed point or axis on the body which you know isn't going to move. A
fixed point isn't always available, but if it is it allows you to ignore
the forces acting on the body at that joint/axle. If you calculate the
torque about any other point, then rotation of the body becomes confused
with linear acceleration.
[CAVEAT: If the forces are balanced, then I guess the torque is independant
of pivot point: translation of origin gives
Sum[r{i} x F{i}] -> Sum[(r{i}+o) x F{i}] = Sum[r{i} x F{i}] + o x Sum[F{i}]
= Sum[r{i} x F{i}]
]

For the turning bicycle, considering the torque about the center of mass
(in *either* an inertial or non-inertial frame), we see that the torque due
to the centripital frictional force is balanced by the torque due to the
normal force of the road. In the non-inertial frame, you can also take the
tire-road contact as a fixed axis, and view gravity and the fictional
centrifugal force as balancing each other. In a stationary frame, there's
no fixed point on the bike, so you have to go with the CoM calculation.

PPS: CONING?
David Thiessen also suggested that a bike turns because of 'coning', that
effect which makes a styrofoam coffee cup roll in a circle. While I agree
that a single wheel exhibits that behavior, I don't think that a bike
does. Try this experiment: holding the front wheel straight, lean your
bike and then move forward; I find that the bike does not turn.

Of course, the tires grind on the road, because each wheel wants to
'cone'. But the frame won't let them. I certainly believe that coning can
reduce this friction during a turn, thus improving performance and tire
life. But I don't think the coning is the cause of the turning.

In experimenting, remember that leaning has another effect on the radius of
the turn. The turn radius is determined by the angle between the lines of
contact between the two tires and the road. For a given angle of
handle-bar turn, the angle between the contact lines increases as the bike
leans into the turn. So leaning does affect turn radius, even if coning
does not.