Not quite rocket science but as Segway has been discussed on ARocket
I'll
        venture posting there as welll. Pse yell at me offlist if this is
too too OT.

Been thinking about braking of Segway Scooter / Ginger / IT.

As someone (Peter Peres?) noted - weight is always 100% reacted vertically
through the wheels under braking.
As the scooter leans backwards against the direction of motion the
decelerating moment trying to pitch the rider forward is balanced by the
mass trying to tip the scooter backwards. Or, it is if the electronics
maintain control ! :-)

A few sums * shows the not too unexpected result that

        A = g tan(X) where X is the lean angle off vertical.

This is the forward de-acceleration which can be balanced at a given lean
angle.
At 45 degrees, tan X becomes 1 and resistable deceleration = 1g.
At backwards tip angles beyond this the resistable force rapidly increases
(as tan X goes to infinity as X approaches 90 degrees) BUT there is never
any need to exceed this angle greatly unless you can get a coefficient of
friction greater than 1 at the wheels. For practical purposes the following
lean angles are needed for given accelerations. In fact angles greater than
45 degrees or in fact greater than the available "friction angle" will
result in falling over backwards as the friction at the wheels is unable to
react the decelerating force.


DEGREES lean     / G deceleration

5      /    0.09
10    /    0. 18
15    /     0.27
20    /    0.36

25    /    0.47
30    /    0.56
35    /    0.70

As can be seen, up to 20 degrees tan(X) is linear to 2 significant figures
and braking equals about 0.02 g per degree of lean.
(or A ~= degrees/56 g deceleration).
I imagine a 20 degree angle of backwards lean would be quite unnerving until
you came to trust your control system. This is slightly more than a 1 in 3
lean - quite steep. The resultant 0.35g odd consequent braking is quite
respectable. On a dry surface at slowish speeds appropriate motorcycles can
be persuaded to straight line brake on front wheel alone - a good way to
impress the locals until you overdo it (don't try this at home). The Segway
with 2 side by side wheels and anti-rotation control should make this a
typically less painful exercise than on a motorcycle :-)



regards


        Russell McMahon


*  Derivation:

Braking moment pitching rider off = MARCosX
Backward moment due to weight = gMRSinX
Equate for balance and cancel
MARCosX=gMRSinX
A = G tanX

M = mass
A = braking acceleration
g = gravitational acceleration
R = radius of mass above pivot point
X = backwards tilt angle from vertical

Note that M can be point mass at cofg OR distributed mass taken point by
point as R for mass varies with the location of the mass and ALL masses
experience the same deceleration.

Reality may be very slightly complicated by the fact that the suspension
point is below the axle line and will rise very slightly as you lean
backwards thereby changing the geometry. In practice I imagine that this
effect would be vanishingly small.

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