B
Bret Cahill
Guest
Check this out, especially the forcing function that increases with
mass.
To keep things simple assume a motor vehicle consists of 2 masses
connected by a spring with const. k and a shock with const c. Call
the body, frame, engine etc. one mass, m1 and the suspension is the
spring + shock and the wheels are m2.
When there is a forcing function P(t) on m2, which increases linearly
with m2, the equation of motion, derived from summing all the forces
on each mass is:
m1x1''(t) + c(x1'(t) - x2'(t)) + k(x1(t) - x2(t)) = 0
m2x2''(t) + c((x2'(t)-x1'(t)) + k(x2(t) - x1(t)) = m2P(t)
x(t), x'(t) and x"(t) are displacement, velocity and acceleration of
each mass respectively.
Adding both equations:
m1x1''(t) + m2x2''(t) = m2P(t)
or
x1"(t) = [m2P(t) - m2x2"(t)] / m1 = m2 (P(t) - x2"(t)) / m1
x1''(t) is the acceleration on the chassis.
The more you increase m2 [heavy wheels] and decrease m1, the more you
increase x1''(t), the acceleration on m1.
The short qualitative answer:
The harder you hit the pot hole the harder pothole hits back.
Why duke it out with a pothole?
That's one argument for shifting weight from the wheels to above the
suspension.
The other argument is heavy wheels increase the energy necessary to
get what is basically a flywheel up to speed.
Go to a bike shop and try a really nice $700 racing bicycle. Compared
to a beach cruiser, when you step down on the pedal it almost leaps
ahead.
Bret Cahill
mass.
To keep things simple assume a motor vehicle consists of 2 masses
connected by a spring with const. k and a shock with const c. Call
the body, frame, engine etc. one mass, m1 and the suspension is the
spring + shock and the wheels are m2.
When there is a forcing function P(t) on m2, which increases linearly
with m2, the equation of motion, derived from summing all the forces
on each mass is:
m1x1''(t) + c(x1'(t) - x2'(t)) + k(x1(t) - x2(t)) = 0
m2x2''(t) + c((x2'(t)-x1'(t)) + k(x2(t) - x1(t)) = m2P(t)
x(t), x'(t) and x"(t) are displacement, velocity and acceleration of
each mass respectively.
Adding both equations:
m1x1''(t) + m2x2''(t) = m2P(t)
or
x1"(t) = [m2P(t) - m2x2"(t)] / m1 = m2 (P(t) - x2"(t)) / m1
x1''(t) is the acceleration on the chassis.
The more you increase m2 [heavy wheels] and decrease m1, the more you
increase x1''(t), the acceleration on m1.
The short qualitative answer:
The harder you hit the pot hole the harder pothole hits back.
Why duke it out with a pothole?
That's one argument for shifting weight from the wheels to above the
suspension.
The other argument is heavy wheels increase the energy necessary to
get what is basically a flywheel up to speed.
Go to a bike shop and try a really nice $700 racing bicycle. Compared
to a beach cruiser, when you step down on the pedal it almost leaps
ahead.
Bret Cahill