ride theory

Question:

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

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

3.Can you use the same amount of damping for

I.Low and high frequency regions

II.Compression and rebound

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

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

I.A higher or lower unsprung mass?

II.Higher or lower damping?

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

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

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

Answers:

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)

1. No we cannot use same amount of damping at low and high frequency regions. for low and high-frequency regions we will have to use the different type of dampings. For a low-frequency region, we have to use high damping whereas for high-frequency regions low damping is preferred. The idea here is to keep the transmissibility ratio low as shown in the graph below:
2. No we cannot use same amount of damping for compression and rebound.

For compression in the spring, the energy is going into the spring and because of that, the spring is restricting the further compression of it so in this case, we will need a lower amount of damping whereas, in the case of rebound the spring is releasing its energy and so for that we will need a high amount of damping to restrict its further expansion.

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)

The answer to Question 5:

(i) A lower unsprung mass will provide better vibration isolation in the wheel hop region.

(ii) less damping increases the natural frequency of the unsung mass.

Hence, less damping is required to achieve a high natural frequency of the unsprung mass so that for resonance to occur at wheel hop region than it could happen at a high-frequency point.

hence, lesser damping is preferred for good vibration isolation.

6)

The advantage with Air Suspension is that we can change the stiffness of it by pumping in or out some gas.

Another advantage is that we can change the ride height too.

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

1. 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.

1. 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.