Vehicle Ride

1. Wheel rate (Kw) Wheel force upon wheel travel whereas Spring rate (Ks) is nothing but the Spring force tot the Spring travel.

Kw = Ks*(MR)^2.

MR = motion ration which is usually less than 1.

Thus, Wheel rate is generally less than spring rate which means spring is not able to absorb complete wheel force or the road disturbances.

2. Wheel rate = Force / Displacement at wheel center

Ride rate = Force / Displacement at contact patch.

(If tire is infinitely stiff then ride rate would be equal to wheel rate.)

3. For the low frequency region, high damping is required & vice versa. Else will affect transmissibility.

In compression, the energy is stored in Spring in the form of KE. In rebound, the same energy is expelled in the form of heat by Damper. Thus, in the rebound condition high damping is required than in compression phase.

4. When wheel travels over the bump it undergoes through certain vibrations whose amplitude goes on dying after the pass. Due to the wheelbase, there occurs a lag of bump experience between the front & rear wheel. Hence, in order to achieve the same amplitude after the vehicle travel over the bump the rear ride frequency has to be slightly higher than the front (1.2x).

5. I will go with lower un-sprung mass leads to maintain low transmissibility ratio.

Lower damping will lead to less stiff suspension will assist the wheel hop failure.

6. From ride frequency standpoint, why are air suspensions beneficial, especially in heavy vehicles?

Ride frequency = 188/ sqrt(x), where x=static deflection.

In the air suspension system, vehicle height varies based on input load.

Thus, as load increases the height gets lowered. In heavy duty vehicles, sprung height from ground plays an important role from NVH point of view.

As the vehicle height is lowered the less jerk impact will be faced by the body & ride frequency will improve.

7.

Given:

Sprung mass of the vehicle = 2000Kg. Weight distribution = 50/50.

Thus, for quarter model Sprung mass (Ms) = 2000/4 = 500Kg

Unsprung mass on each corner (Mus) = 50Kg

Spring stiffness (Ks) = 25N/mm = 25000N/mm

Tire stiffness (Kt) = 250N/mm = 250000N/mm

Calculations:

Sprung mass natural frequency (Fs) = (1/2*pi)*(sqrt (Kr/Ms))

Where Kr = Ride rate = Kt*Ks / (Kt+Ks) = 25*250/(250+25) = 22.72 N/mm = 22720 N/m.

Thus, Fs = (1/2*pi)*(sqrt (22720/500)) = 1.07Hz.

Unsprung mass natural frequency (Fus) = (1/2*pi)*(sqrt (Keq./Mus))

Where Keq. = Kt + Ks = 250 + 25 = 275 N/mm = 275000 N/m.

(Fus) = (1/2*pi)*(sqrt (275000/50)) = 11.8Hz

8.

a. Ff = 1.2Hz & Fr = 0.8Hz.

Thus, Tf = 1/Ff = 1/1.2 = 0.83s & Tr = 1/Fr = 1/0.8 = 1.25s

Time lag = 1.25 – 0.83 = 0.42s

b. Ff = 1.2Hz & Fr = 1.45Hz

Thus, Tf = 1/Ff = 1/1.2 = 0.83s & Tr = 1/Fr = 1/1.45 = 0.69s

Time lag = 0.83 – 0.69 = 0.14s

Hence for a flat ride (Front & rear frequency in phase) Second ride frequency would be better from ride perspective as the time lag is less.

Wheelbase (X) = 98.42in = 2.5m, Speed (V)= 37.42mph = 16.72m/s.

Time lag (t) = X/V = 2.5/16.72 = 0.15s.