Vehicle Ride
1. The spring rate is a property of the spring itself—it's the force required to compress the spring by a unit distance. However, the wheel rate is the effective stiffness felt at the wheel centre, and it's influenced by how the suspension converts the wheel's vertical motion into spring compression.
2. The vertical movement between the sprung mass and the contact patch is called the ride rate, whereas the vertical movement between the sprung mass and the wheel centre is called the wheel rate. The ride rate is different from the wheel rate because tires behave like a spring and the contact patch variation happens due to two springs (onboard springs and tires) in a series connection to the ground.
3. I. No, we cannot use the same amount of damping for low and high frequencies. To lower the transmissibility ratio, we need high damping at low frequencies and low damping at high frequencies.
II. No, we cannot use the same amount of damping for compression and rebound. During spring compression, the energy is stored in the spring so low damping is required whereas, in spring expansion, it releases the energy so high damping is required in that phase.
4. Rear ride frequency should be higher than the front to control the pitching action of the car. When the front wheel hits a bump first, it causes the vehicle to pitch. A softer rear suspension reacts slowly, increasing the pitching effect. A stiffer rear (higher ride frequency) responds faster, reducing this pitch and improving stability.
5. I. Lower unsprung mass is always better for vibration isolation, as natural frequency is inversely related to mass. A higher natural frequency reduces the chance of resonance in the wheel hop region, which improves vibration isolation.
II. Lower damping means the system’s natural frequency remains closer to its undamped value, which is higher. This higher natural frequency ensures that in the wheel hop region, it can avoid resonance.
6. Air suspensions allow adjustable spring stiffness. In heavy vehicles, the higher sprung mass naturally lowers the natural frequency as natural frequency is inversely proportional to mass.
F = (1/2*pi)* sqrt(k/m)
so to maintain the desired ride frequency, k must be higher. Adjustable air suspension can increase spring stiffness K to attain the desired frequency.
7. Ms = 2000kg/4 = 500kg, Mus =50kg, Ks = 25N/mm, Kt = 250 N/mm, g = 10m/s2
Ride rate Kr = (Ks*Kt)/(Ks+Kt) = 22.27 N/mm = 22270 N/m
Now Fs = (1/2*pi)*sqrt(Kr/Ms) = 1.07 Hz (Sprung mass natural frequency)
Ke = Ks + Kt = 275 N/mm = 275000 N/m
Now, Fus = (1/2*pi)*sqrt(Ke/Mus) = 11.8 Hz (Unsprung mass natural frequency)
8. wheel base = 98.42 inches = 2.5 m, v = 37.42 mph = 16.7 m/s
Time lag = distance/speed = 2.5/16.7 = 0.15 seconds
case 1.
Front ride frequency = 1.2 Hz = 1/1.2 Hz = 0.83s
Rear ride frequency = 0.8 Hz = 1/0.8 = 1.25s
In order to get a flat ride, the rear should be able to catch up with the front meaning
Tfront = Trear + Tlag;
0.83 < 1.25 + 0.15
Thus this is not a flat ride.
Case 2.
Front ride frequency = 1.2 Hz = 1/1.2 Hz = 0.83s
Rear ride frequency = 1.45 Hz = 1/1.45 = 0.68s
In order to get a flat ride, the rear should be able to catch up with the front meaning
Tfront = Trear + Tlag;
0.83 = 0.68 + 0.15
Thus this is a flat ride.
So case(2) is a flat ride.