Vehicle Ride

1. Wheel Rate is basically the spring rate but measured at the wheel instead of the where the spring attaches to the linkage.

Wheel Rate = Spring Rate * (Motion Ratio ^ 2) * Spring Angle Correction

The motion ratio is the ratio between how much the spring is compressed compared to how much the wheel is actually moved. Thus, it is estimated by knowing the distance of the wheel to the pivot point of the suspension arm and the distance to the spring from the pivot point.The "Spring Angle Correction" is typically a small correction equal to the cosine of the spring angle from vertical. With a vertical spring angle, the cosine of 0 degrees is 1, so the correction has no affect. At 10 deg from vertical, the cosine of 10 degrees is .985, only a 1.5% effect. So with springs relatively vertical, the Spring Angle Correction can usually be ignored.

2. Ride rate is defined as vertical force per unit vertical displacement of the tire ground contact with respect to chasis whereas Wheel rate is vertical force acting per unit vertical displacement at spindle centreline of tire.

3. I cannot use same amount of damping for both the cases.

1. Low and high frequency regions : Higher damping is required at low natural frequency range (1-3Hz) whereas Lower damping is required at high natural frequency.
2. Large energy is released during rebound, thus more damping is required to be “absorbed” during rebound of suspension than during compression.

4. Rear ride frequency needs to be in phase (more the better) with the Front ride frequency of the suspension. This phase difference occurs due to the time lag between front and rear suspension after it hits the bump as function of wheelbase. The in-phase system creates flat ride during a bump.

5. For improved isolation in wheel hop region i would prefer :

1. Wheel hop region is high unsprung mass natural frequency region thus,higher unsprung mass would result in better isolation. It can be understood from the equation as follows :
2. Lower damping is required in the high frequency range.

6. From a ride frequency standpoint, air suspension is beneficial especially in heavy vehicles because:

• Suspensions with a lower natural frequency reduce the force transmitted to the vehicle, which improves ride and vehicle protection when properly engineered.
• Natural frequency includes the spring rate of the suspension and the amount of weight supported by the suspension.
• The spring rate of the air springs is a measure of the vertical stiffness of the air springs, or how much the suspension deflects for a given load being carried.
• The natural frequency of a suspension will increase if the suspension spring rate is increased. Because of a softer spring rate, air suspensions provide greater equipment protection over traditional mechanical suspensions. This is because of the air ’springs’ ability to cushion the equipment and passenger load. Outstanding ride produces superior driver and passenger comfort while helping to protect critical, lifesaving equipment and costly electronic components from excessive vibration and road inputs.
• Lower frequencies mean lower damping velocity and produce a softer suspension with more mechanical grip, however, the response will be slower in transient.

7. Given:

Sprung mass = Ms = 2000Kg

Weight distribution = 50/50 (hence, 1000kg on Front & rear axle each --> 500kg of Ms on each corner)

Unsprung mass in each corner = Mu = 50kg

Spring stiffness = Ks = 25 N/mm & Tire stiffness = Ktr = 250 N/mm

Assume g = 10 m/sec^2

To Find : Sprung mass and Unsprung mass natural freuqencies

Solution :

Ride rate = Kr = `(Ks*Ktr)/(Ks+Ktr) = (25*250)/(25+250) = 22.727`N/mm = `22.727X10^3` N/m

Thus, Sprung mass natural freuqency = fs = `(1/(2*pi))*sqrt(K_r/M_s) =`1.073 Hz

Also, Unsprung mass natural freuqency = fus = `(1/(2*pi))*sqrt((K_s +K_t)/M_u)) =`11.80 Hz

8. Given: Wheelbase = 98.42 inches = 2.5m & Speed = 37.42 Mph = 16.73 m/s

Solution : Time lag = dT = 2.5/16.73 = 0.15s

Case 1. Front ride frequency =1.2 Hz i.e. T_front = 1/1.2 = 0.833 sec

Read ride frequency = 0.8 Hz i.e. T_rear = 1/0.8 = 1.25 sec

Thus, it can be seen that T_front > T_rear + dT ; hence, Not favorable for flat ride (requires in-phase characteristics)

Case 2. Front ride frequency =1.2 Hz i.e. T_front = 1/1.2 = 0.833 sec

Read ride frequency = 0.8 Hz i.e. T_rear = 1/1.45 = 0.68 sec

Thus, it can be seen that T_front = T_rear + dT ; hence, favorable for flat ride (requires in-phase characteristics)