Vehicle Ride

1.Why is the wheel rate usually different from the spring rate?

Wheel rate is the product of spring rate and (motion ratio)^2, spring rate depends on spring stiffnes alone, but wheel rate depends on motion ratio as well as spring stiffness.

2.Why is the ride rate different from the wheel rate?

Ride rate is measured by taking contact patch as reference, the vertical deviation from the sprung mass and contact patch.

whereas wheel rate is measured by taking wheel center as reference, the vertical deviation from the sprung mass and wheel centre.

3.Can you use the same amount of damping for

I.Low and high frequency regions

Nope!,

to reduce the transmissibilty ratio, for low frequency region high damping and for high frequency region low damping is required.

II.Compression and rebound

Nope!,
During compression the spring stores energy so low damping is enough, whereas during rebound spring releases energy so high damping is required.

4.Why does the rear ride frequency need to be higher than the front?

The rear ride ride frequency need to be higher than the front to avoid Pitching effect.

Because, the rear wheel hit the bump after some time delay from the front wheel hit the bump.

5.For improved vibration isolation in the wheel hop region, would you prefer:

I.A higher or lower unsprung mass?

Natural frequency is inversely proportional to the mass of the object.

So, lower unsprung mass provides higher frequency. Hence it provides better vibration isolation by avoiding resonance.

II.Higher or lower damping?

High natural frequency is acheived at low damping.

So, lower damping gives better vibration isolation by avoiding resonance.

6.From a ride frequency standpoint, why are air suspensions beneficial, especially in heavy vehicles (do some research to answer this)

Air suspension is benefecial in heavy vehicles, because depending on the load air pressure in the suspension varies. Hence it avoids pitching effect.

7.Calculate sprung mass and unsprung mass natural frequencies for a vehicle with a 2000 kg sprung mass, 50/50 weight distribution, 50 kg of unsprung mass in each corner, spring stiffness of 25 N/mm and tire stiffness of 250 N/mm. Assume gravity to be 10 m/sec^2 to simplify calculations. Report your answer in hertz

Natural frequency,

Sprung mass= 1.073 Hz


Un-Sprung mass= 11.8 Hz

8.Calculate the time lag between the front and rear suspensions for:

1.Front ride frequency of 1.2Hz, Rear ride frequency of 0.8Hz

2.Front ride frequency of 1.2Hz, Rear ride frequency of 1.45Hz

Now calculate the time lag due to wheelbase. Wheelbase = 98.42 inches, speed = 37.42 mph.

From these numbers, determine which ride frequency set (1 or 2) works best from a flat ride perspective

Ride frequency set ( 2) works best from a flat ride perspective.