Vehicle Ride
1. Spring Rate refers to the amount of weight it takes to compress a spring a certain distance. The higher the spring rate, the stiffer the spring whereas , Wheel rate is basically the Spring Rate multiplied by the motion ratio squared.
2. Wheel rate is defined as the vertical force acting on the wheel per unit vertical displacement of wheel along the spindle relative to the chassis.
On the other hand, ride rate is defined as the vertical force acting on the wheel at the tire contact patch per unit displacement of the sprung mass relative to the chassis.
Therefore the difference between these two is that wheel rate arises due to the dispacement of the wheel alone where as the ride rate is due to the displacement of both wheel as well as the spring.
3. * High damping is preferred in low frequency region and low damping is preferred in mid and high frequency region.
* High damping ratio at low speed and Low damping ration at high speed.
4. Rear ride frequency must be higher than the front to compensate the time delay in compression and rebound which is caused due to the distance between the front and the rear wheels.
5. * Lower Unsprung mass gives good isolation.
* Lower Damping is preferred in High frequency range.
6. From ride frequency standpoint, air suspension is beneficial in the following ways:
i.) Air suspensions have better compliance than steel suspensions or we can say that air suspensions have lower stiffness compared to spring suspensions.
Ride frequency is given by,
fs = (1/2π)*√(Kr/Ms)
Since the air suspension stiffness is of lower value, Ride rate(Kr) value goes down.
Due to this ride frequency (fs) will also be low.
This results in a more smoother ride and also keeps the NVH levels low.
7. Sprung mass, Ms=2000 kg
50/50 weight distribution
On each corner, Ms = 2000/4 = 500 kg
Unsprung mass, Mus = 50kg(each corner)
Spring stiffness, Ks = 25 N/mm
Tire stiffness, Kw = 250 N/mm
g=10 m/s^2
i.) Sprung Mass natural frequency (fs),
fs= (1/2π)*√(Kr/Ms)
Ride rate,
Kr = Kw*Ks/(Kw+Ks)
Kr = 25*250/(25+250) = 6250/275
Kr = 22.72 N/mm = 22727.27 N/m
fs = 0.1591*√(22727.27/500)
fs = 1.07 Hz
ii.) Unsprung mass natural frequency (fus),
fus = (1/2π)*√(Keq/Mus)
Equivalent stiffness,
Keq = Kw + Ks = 250 + 25
Keq = 275 N/mm = 275000 N/m
fus = 0.1591*√(275000/50)
fus = 11.8 Hz.
8. Wheelbase = 98.42 inches = 2.5meters
Speed = 37.42 mph = 16.67 m/s
Now time lag due to wheelbase,
T(lag) = wheelbase/speed
T(lag) = 2.5/16.67
T(lag) = 0.15 s
To get a flat ride, the front and rear of the car should reach the same point after 1 cycle in the shortest interval of time.
T(front)= T(rear) + T(lag)
Considering the two given cases,
i.) Front frequency, f1= 1.2 Hz
Time period, T(front)= (1/1.2) = 0.833s
Rear frequency, f2 = 0.8 Hz
Time period, T(rear) = (1/0.8) = 1.25s
Requirement for flat ride:
T(front) = T(rear) + T(lag)
But here,
0.833 < 1.25 + 0.15
ii.) Front frequency, f1= 1.2 Hz
Time period, T(front) =0.83 s
Rear frequency, f2= 1.45 Hz
Time period, T(rear) = 0.68 s
Requirement for flat ride:
T(front) = T(rear) + T(lag)
0.83 = 0.68 + 0.15
The equation holds.
Thus we can conclude that 2nd case works best for flat ride perspective.