Vehicle Ride

1)

Wheel rate is usually different from spring rate because the wheel and the spring are mounted at different distances from the pivot point. Thus, a concept of motion ratio comes into picture. Motion ratio is a geometric measure that relates the change in displacement of force element to a change in vertical wheel center movement.

The relation of wheel rate and spring rate is given by

K_w = K_s * MR²

where,

K_w = wheel rate

K_s = spring rate

MR = motion ratio = a/b ………(from the figure)

If the wheel and the spring distances are same from the pivot point i.e. a = b as in McPherson Strut, and if the spring is vertical, then the wheel rate and spring rate will be equal.

But, packaging wise this is very difficult to achieve in a vehicle and hence, the wheel rate is usually different from the spring rate.

2)

The wheel rate is given by

K_w = K_s * MR² ......... (a)

where,

K_w = wheel rate

K_s = spring rate

MR = motion ratio

This expression shows that the wheel rate depends only upon the spring rate as it forms the part of unsprung mass.

The ride rate depends upon the spring rate as well as the wheel rate as the ride rate is for sprung mass and it is the equivalent of spring stiffness and tire stiffness attached in series with each other.

K_r = K_w * K_s / (K_w + K_s)

From (a), we have,

K_s = K_w/MR²

Therefore,

K_r = K_w/(1 + MR²).

Since, the denominator will always be greater than 1,

Hence, the ride rate will always be lesser than wheel rate.

3)

I. No we cannot use same amount of damping at low and high frequency regions.

For low frequency regions, the acceleration levels are quite high at the passenger seat and hence, we require high damping in the low frequency regions. As the frequency increases, the acceleration levels decrease and hence, at higher frequencies we need to provide lower damping.

II. No we cannot use same amount of damping for compression and rebound.

Compression damping is lower than rebound damping (by a factor of around 2) because during compression, the spring absorbs energy and hence, the damping force needs to be lower.

In rebound, the energy is released from the spring. The damper has to control this energy release and therefore rebound damping needs be higher.

4)

There is an out of phase motion of front and rear because the rear hits the bump a little later than the front due to wheelbase separation.

If the front and the rear frequencies are same, the front and the rear suspension are out of phase and this causes pitching.

To avoid this pitching, that is to get a flat ride the rear frequency needs to be higher than the front frequency.

Typically, the rear frequency is kept 1.2 times the front frequency.

5)

In a wheel hop region or the region of unsprung mass natural frequency, for better vibration isolation,

I. Lower unsprung mass is preferable

II. Lower damping is preferable.

6)

In heavy vehicles, the change in the height of sprung mass from the ground (ride height) varies to a greater extent with the increase in load as compared to a light vehicle where the variation in load and height from ground is not much.

This leads to a large amount of variation in natural frequency of the sprung mass when the heavy vehicle is empty and when the vehicle is fully loaded.

The conventional metal springs have a constant spring stiffness in all conditions whether the vehicle is unladen or loaded. Hence, the natural frequency is much higher in unladen conditions which deteriorates the ride.

The main advantage of air suspension is that it maintains the ride height of the sprung mass constant in all loading conditions which is near the static deflection height, by changing the amount of air in the bellow. The relation of natural frequency with the static deflection is given by

ω = 188/sqrt (x)

where, ω = natural frequency (cycles/min)

x = static deflection (in.)

From this relation, it can be seen that if the ride height remains same, the natural frequency will also remain same in all conditions.

Thus, the natural frequency of an air suspension remains almost constant in all conditions.

This natural frequency is a function of pressure in the air bellow in air suspension and it can be so adjusted that the frequency can be kept low for a softer ride and lower shock inputs.

Thus, the natural frequency of an air suspension is low and remains almost constant throughout the normal load range of a heavy vehicle improving the ride quality.

7)

Sprung mass, ms = 2000 kg

As there is 50/50 weight distribution, at each wheel, ms = 2000/4 = 500 kg

Unsprung mass, m_us = 50 kg (each wheel)

Spring stiffness, K_s = 25 N/mm

Tire stiffness, K_w = 250 N/mm

Gravity, g = 10 m/s²

Solution:-

f_us = 1/2Π √((K_s + K_w)/m_us) = 1/2Π √((25 + 250)/50) = 1/2Π √(275*1000 /50) = 11.8 Hz

f_s = 1/2Π √((K_w * K_s /(K_w + K_s))/m_s) = 1/2Π √((25 ∗ 250/(25 + 250))/2000) = 1/2Π √(22.73 ∗ 1000/500) = 1.07 Hz

Answer:-

Sprung mass natural frequency = 1.07 Hz

Unsprung mass natural frequency = 11.8 Hz

8)

Given:-

Wheel base = 98.42 = 2.5 m

Speed = 37.42 mph = 37.42 * 1.6 * (5/18) = 16.7 m/s

Now, time lag due to wheel base,

Time = Distance / Speed

Therefore, Time (lag) = 2.5 / 16.7 = 0.15 s

Therefore, Time lag = 0.15 s due to wheel base separation

i) For the 1st case,

Front ride frequency = 1.2 Hz

Front time period = 1/1.2 = 0.83 s

Rear ride frequency = 0.8 Hz

Rear time period = 1/0.8 = 1.25 s

In order to get a flat ride, front and rear should come to the same point after striking a bump in shortest interval of time.

Thus, T (front) = T (rear) + T (lag)

But, in this case, 0.83 < 1.25 + 0.15

In the 1st case, the rear frequency will not be able to catch up with the front frequency and hence, will not result in a flat ride, it will continue to pitch.

ii) For the 2nd case,

Front ride frequency = 1.2 Hz

Front time period = 1/1.2 = 0.83 s

Rear ride frequency = 1.45 Hz

Rear time period = 1/1.45 = 0.68 s

In order to get a flat ride, front and rear should come to the same point after striking a bump in shortest interval of time.

Thus, T (front) = T (rear) + T (lag)

In this case, 0.83 = 0.68 + 0.15

Thus, in the 2nd case, the rear frequency will be able to catch up with the front frequency resulting in a flat ride.

In conclusion, the 2nd ride frequency set works best from a flat ride perspective.