Vehicle Ride

1. We have the relation

Wheel Rate = (Motion ratio)^2 x Spring rate (excluding non-linearity)

When the spring is attached to the knuckle itself and it is completely vertical, the displacement of the spring is the same as the displacement in the wheel center. In that case, the motion ratio is equal to 1 and the wheel rate is the same as the spring rate.

However in most cases, the displacement in the spring is not the same as that in the wheel for two reasons: either the spring is not attached to the knucle itself or the spring is at an inclination with the vertical. This means the Motion ratio is not one and thus the wheel rate is usually different from the spring rate.

2. Wheel Rate = (Motion ratio)^2 x Spring rate (excluding non-linearity)

Ride Rate = Spring rate x Tire rate / (Spring rate x Tire rate) (for quarter car model)

Actually,

Ride Rate = Wheel rate x Tire rate / (Wheel rate x Tire rate)

Clearly from the above relations, ride rate is different from the wheel rate.

In words, wheel rate is evaluated vertical force on the wheel to the vertical displacement of the wheel center wrt sprung mass.

Ride rate is evaluated by vertical force on the wheel to the vertical displacement of the tire contact patch wrt sprung mass.

3.

I. High damping is preferred in the low frequency region and low damping is preferred in the high frequency region in order to maximize vibration isolation. So while we can use same amount of damping for all frequency range, it is preferred that we use different amount of damping for different frequency regions.

II. No we cannot. During compression, spring absorbs energy and thus the damping force needs to be lower. However during rebound, spring releases energy and the damper has to control this energy; thus we need higher amount of damping during rebound.

4. The front wheels always hit the bump first before the rear wheels do. Thus suspension oscillation in the front begins earlier than in the rear and the front and rear suspension are out of phase which causes pitching. This time lag in the suspension oscillation is compensated by a higher ride frequency in the rear which makes the suspensions more in-phase for the first few cycles (suspensions need not be in-phase for the later cycles because damping significantly minimizes the oscillation till then) and we get a flat ride. Hence, rear ride frequency needs to be higher than the front for a flat ride during bump encounters.

5. For improved vibration isolation in the wheel hop region

I. I would prefer lower unprsung mass.

II. I would prefer lower damping.

6. Ride frequency is given by f = (1/2.pi) x sqrt(k/M)

In heavy vehicles, weight distribution in higher in the rear. In addition to that, the wheel base is large and so the time lag is higher in heavy vehicles. We know that for flat ride we require higher rear ride frequency than the front by an amount that corresponds to the time lag due to wheel base.In order for that, from the above formula, stiffness of rear spring needs to be much larger than the front so we get a high frequency value than the front. Had the weight distribution been more uniform, the spring stiffness required would be smaller.

Insted of using a very stiff spring, air suspensions can be used. They simply adjust the pressure in the suspension to avoid body pitching due to wheel base lag. Thus, air suspensions are beneficial in heavy vehicles.

7. Total Sprung mass (Ms_tot) = 2000 kg

For 50/50 weight distribution, for a quarter car model, we can write, Sprung mass (Ms) = Ms_tot/4 = 500 kg

Given, unsprung mass in each corner (Mus) = 50 kg

Spring stiffness (ks) = 25 N/mm = 25000 N/m

Tire stiffness (ktr) = 250 N/mm = 250000 N/m

acceleration due to gravity (g) = 10 m/s^2

Now, ride rate (kr) = ks x ktr/(ks+ktr) = 22730 N/m

Then,

Sprung mass natural frequency = (1/2.pi) x sqrt(kr/Ms) = (1/2.pi) x sqrt(22730/500) = 1.073 Hz

And,

Unsprung mass natural frequency = (1/2.pi) x sqrt[(ks+ktr)/Mus] = (1/2.pi) x sqrt[(25000+250000)/50] = 11.803 Hz

8.

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

Time lag = 1/0.8 - 1/1.2 = 5/12 sec = 0.4167 sec (front suspension oscillates faster)

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

Time lag = 1/1.2 - 1/1.45 = 25/174 sec = 0.1437 sec (rear suspension oscillates faster)

Given,

Wheelbase = 98.42 inches; speed = 37.42 mph

Time lag due to wheelbase = Wheelbase/speed = (98.42x25.4/1000 m)/(37.42x0.44704 m/s) = 0.1494 sec

Rear suspension needs to have higher frequency in order to achieve flat ride. Only the set 2 meets this requirement. In addition to this, time lag due to wheelbase is pretty close to the time lag for the frequency set 2. Hence, for the given two sets, frequency set 2 works best from a flat ride perspective.