Vehicle Ride

1.wheel rate- The measure of effective stiffness of the suspension system measured at the wheel center.

spring rate- the amount of weight required to compress the spring one inch,
and we all so know that

wheel rate = spring rate* (motion ratio)^2

2. ride rate- vertical deviation from sprung mass and contact patch

where as wheel rate is measured by taking wheel center as reference, i.e., vertical deviation from the sprung mass and wheel centre.

3. no.

to reduce the (force applied/ force transmitted) transmissibility ratio, for low frequency high damping and for high frequency low damping.

no, on compression energy is stored in the spring so low damping is enough, on rebound spring releases energy so damping is required more.

4. for avoiding pitching effect,

5. lower sprung mass and lower damping as they provide higher frequency and better vibration isolation.

6. high natural frequency is achieved at low damping. So, lower damping gives better vibration isolation by avoiding resonance. Air suspension is beneficial in heavy vehicles, because depending on the load air pressure in the suspension varies. hence it avoids pitching effect.

7. M(s) = 2000kg

M(us)= 50kg (each corner)

M(s) = 500kg (each wheel)

K(s) = 25N/mm

K(t)= 250N/mm

for sprung mass, K(s)= [k(s) * K(t)]/[k(s) + k(t)] = 22.73*10^3 N/m

for unsprung mass, K(us) = [k(s) + k(t)] = 275*10^3N/m

Natural frequency,

for SM, f(s) = {1/2pi}* root of[ k(s)/ m(s)] =1.073hz

for USM, f(us) = {1/2pi}*root of[k(us)/m(us)] = 11.8hz

8.WB= 98.42 inch= 2.49m

speed= 37.42miles/hr= 16.73m/s

time lag- wheel base/ speed = .015 sec

frequency,

f(f)= 1.2hz f(r)= .8hz

time peiod= T(f)= 1/ f(f) = .83 T(r)= 1/f(r)= 1.25sec

T(f) < Tr + time lag ----- flat ride not possible

f(f) = 1.2 ; f(r) 1.45 hz

T(f) = 0.83 sec ; T(r) = 0.69 sec

T(f) >= T(r) + time lag ------ so flat ride is achieved.