Found this, may prove interesting...
So how does the vertical position of the CG affect turning
performance?...
Well, to achieve a 1g lateral acceleration in a turn, one would
expect to have to lean the bike over to 45 deg from
the vertical. But to counter the gyroscopic forces one has to lean it
over even more. Below is a table showing the
lean angle required to achieve a 1g turn as a function of the bike's
vertical CG position, for a particular case, namely
my bike at 100kph.
% -------Results-------
% ZCG (m) Bank (deg)
% 0.0000 90.0000
% 0.1000 82.4798 ZCG - height above ground of CG
% 0.2000 65.2291 Bank - angle of lean of bike
% 0.3000 59.0005
% 0.4000 55.7331
% 0.5000 53.7108 Note : Ignoring Gyroscopic forces
% 0.6000 52.3332 gives a bank angle of 45 deg.
% 0.7000 51.3334
% 0.8000 50.5742 Radius of turn Tr = 78.65m
% 0.9000 49.9780 Time to 360 deg 1/O*2*pi = 17.79sec
% 1.0000 49.4973
Note: The only variables that affect this angle are the speed of the
bike (100kph/62mph) and the ratio of the mass of
the wheels to the bike (11% per wheel).
So one can see from this that lowering the bike's CG increases the
required lean angle for a particular turn, thus
reducing the bikes turning performance. Answering the initial
question.
But as we all know, it is not practical to steer a bike by shifting
one's weight, one has to counter-steer. The reason
why this works is again because of the gyroscopic forces, and one
neat trick called precession. What precession
means is that if you apply a moment to a gyroscope (counter steer on
your front wheel) your force applied will be
rotated through 90 deg in the direction of rotation. This means that
any turning force applied to your handle bars is
then applied to your bike, tipping it over in the opposite direction.
Note that the steering does not turn but the bike is
pushed over. This is then opposed by the gyroscopic force due to the
turning of the bike through the turn. You can
either stop applying the counter-steer at this point and use the
bike's mass to hold you in the turn, or hold the turn
with continual counter-steer thus reducing the bike's required lean
angle (interesting idea! don't try this at home
kids). So back to the initial question, a bike with a low CG will
require a larger force on the handle bars to turn it.
Now back to the inertia debate, a quick calculation of the force on
the handle bars to a achieve a rapid lean of the
bike (45 deg/s in one second - more than the average) taking into
account the inertia (guess) of the bike, gives a
force of about 41Nm (8kg - for a 0.5m handlebar). Probably a more
realistic value would be half of this.
Interestingly the righting force (gyroscopic) for the above 1g case
is 288Nm (58kg at the bars - obviously further
leaning is necessary).
So how does the vertical position of the CG affect turning
performance?...
Well, to achieve a 1g lateral acceleration in a turn, one would
expect to have to lean the bike over to 45 deg from
the vertical. But to counter the gyroscopic forces one has to lean it
over even more. Below is a table showing the
lean angle required to achieve a 1g turn as a function of the bike's
vertical CG position, for a particular case, namely
my bike at 100kph.
% -------Results-------
% ZCG (m) Bank (deg)
% 0.0000 90.0000
% 0.1000 82.4798 ZCG - height above ground of CG
% 0.2000 65.2291 Bank - angle of lean of bike
% 0.3000 59.0005
% 0.4000 55.7331
% 0.5000 53.7108 Note : Ignoring Gyroscopic forces
% 0.6000 52.3332 gives a bank angle of 45 deg.
% 0.7000 51.3334
% 0.8000 50.5742 Radius of turn Tr = 78.65m
% 0.9000 49.9780 Time to 360 deg 1/O*2*pi = 17.79sec
% 1.0000 49.4973
Note: The only variables that affect this angle are the speed of the
bike (100kph/62mph) and the ratio of the mass of
the wheels to the bike (11% per wheel).
So one can see from this that lowering the bike's CG increases the
required lean angle for a particular turn, thus
reducing the bikes turning performance. Answering the initial
question.
But as we all know, it is not practical to steer a bike by shifting
one's weight, one has to counter-steer. The reason
why this works is again because of the gyroscopic forces, and one
neat trick called precession. What precession
means is that if you apply a moment to a gyroscope (counter steer on
your front wheel) your force applied will be
rotated through 90 deg in the direction of rotation. This means that
any turning force applied to your handle bars is
then applied to your bike, tipping it over in the opposite direction.
Note that the steering does not turn but the bike is
pushed over. This is then opposed by the gyroscopic force due to the
turning of the bike through the turn. You can
either stop applying the counter-steer at this point and use the
bike's mass to hold you in the turn, or hold the turn
with continual counter-steer thus reducing the bike's required lean
angle (interesting idea! don't try this at home
kids). So back to the initial question, a bike with a low CG will
require a larger force on the handle bars to turn it.
Now back to the inertia debate, a quick calculation of the force on
the handle bars to a achieve a rapid lean of the
bike (45 deg/s in one second - more than the average) taking into
account the inertia (guess) of the bike, gives a
force of about 41Nm (8kg - for a 0.5m handlebar). Probably a more
realistic value would be half of this.
Interestingly the righting force (gyroscopic) for the above 1g case
is 288Nm (58kg at the bars - obviously further
leaning is necessary).
