Vehicle ride
1. Wheel rate is ususally different from the spring rate because of the motion ratio/installation ratio which defines the geometric relationship between the change in position of force element i.e., spring and the change in vertical wheel center movement.
Motion ratio arises due to the fact that spring which is mounted on lower control arm will be having a offset from the wheel center.
This can be better explained with the relationship:
Kw = ks * (MR)^2
Where
Kw - wheel rate
Ks - spring rate
MR - Motion ratio (always <1)
From the above expression we can say that wheel rate is always less than spring rate.
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.
ride rate is given by,
Kr = (Kw*Ks)÷(Kw+Ks)
We know that,
Ks = Kw/(MR)^2
On substitution and simplification,
Kr = Kw ÷ [1 + (MR)^2]
Since the denominator is greater than 1, ride rate is always less than wheel rate.
3.
i.) No. We can\'t use sthe same amount of damping for low and high frequency region.
For low frequency region high damping is preferred while for high frequency region low damping is preferred.
ii.) No. We cannot use the same damping for compression and rebound.
Compression damping is less than rebound damping by a factor of 2-3.
This is because in Compression, spring absorbs energy and hence the damping needs to be lower.
While in rebound, spring releases energy and this rate at which energy is released must be controlled by damping. Hence the damping needs to be higher in rebound.
4. If both, rear and front ride frequencies are same, an out of phase motion will be caused between the front and rear suspension because of the wheelbase seperation.
This out of phase motion leads to pitching of the vehicle.
So in order to obtain a flat ride, the rear frequency must be higher than the front frequency (approximately by 1.2 times).
5. For an improved vibration isolation in the wheel hop region ( high frequency region),
i.) Higher unsprung mass is preferred.
ii.) Lower damping is preferred.
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 steel 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.
ii.) Another advantage of air springs is that, in heavy vehicles, the driver can increase the firmness when towing to account for heavier loads.
7. Given:
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. Given:
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.