The Bounce Test
Maybe you know the bounce test in which you push downwards, for a short time, on the body of a
car and then see how the body reacts to this push.
Figure 1: Bounce Test
http://www.chamsocotokiengiang.com/tin-tuc/chi-tiet/cac-buoc-kiem-tra-va-phat-hien-loi-he-thong-treo-tren-xe-o-to/134
Bounce test:
https://www.youtube.com/watch?v=Lo_5ibeSQDc
https://www.youtube.com/watch?v=J86hrRi7jmQ
https://www.youtube.com/watch?v=BPI9beSabdI
The suspension has to be in good condition to assure good driving behaviour
Some worn shock absorbers:
https://www.youtube.com/watch?v=W8dComhoRhs
https://www.youtube.com/watch?v=Q5r64v4Yr2M
https://www.youtube.com/watch?v=5NHTLEGOgU4
Let’s analyse what is going on when this type of test is performed
When you push down, you apply a force on the mass of the car. The spring (with a certain spring
constant “k”) will be compressed by this force and reacts with a counter force. Also the damper (with
a certain damper constant “c”) will react with a counter force because of the velocity of the body.
The best way to analyse such a physical system is to make a model of it, using mathematical
equations. First we have to establish the equation of motion for this system. In general we can say
that the sum of all forces is equal to zero.
� 𝐹𝐹⃗ = 0
We use Newton’s second law of motion for describing the position of the mass.
𝐹𝐹⃗ = 𝑚𝑚 ∗ 𝑎𝑎⃗
𝑚𝑚 is a scalar quantity and 𝐹𝐹⃗ & 𝑎𝑎⃗ are vector quantity’s, meaning having a direction.
Let’s analyse what is going on by using a simplified schematic drawing as shown in Figure 2. Because
we only look at one suspension system, “m” is ⅟₄ of the mass of the car with unit [kg]. “k” is the
spring constant with unit [N/m] and “c” is the damper constant with unit [N/(m/s)] or [Ns/m]
We assume that the wheel and tire is a stiff
object and for simplicity we say that gravity
plays no role.
Further we assume that the force F(t) is
positive in the upwards direction and negative
in the downwards direction.
Also the position x(t) is positive when its
above x0 and negative when its below x0.
Figure 2: Mass-Spring-Damper System
https://commons.wikimedia.org/wiki/File:Mass_spring_damper.svg
The equation of motion for this system will be:
𝐹𝐹(𝑡𝑡) − 𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝐹𝐹𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑚𝑚 ∗ 𝑎𝑎
𝐹𝐹(𝑡𝑡) − 𝑘𝑘 ∗ 𝑥𝑥 − 𝑐𝑐 ∗ 𝑥𝑥̇ = 𝑚𝑚 ∗ 𝑥𝑥̈
→
→
Now we can calculate the acceleration:
𝐹𝐹(𝑡𝑡) − 𝑘𝑘 ∗ 𝑥𝑥 − 𝑐𝑐 ∗ 𝑣𝑣 = 𝑚𝑚 ∗ 𝑎𝑎
𝐹𝐹(𝑡𝑡) − 𝑘𝑘 ∗ 𝑥𝑥 − 𝑐𝑐 ∗
𝑑𝑑 2 𝑥𝑥
𝑑𝑑𝑡𝑡 2
𝑑𝑑𝑑𝑑
𝑑𝑑2 𝑥𝑥
= 𝑚𝑚 ∗ 2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑡𝑡
= �𝐹𝐹(𝑡𝑡) − 𝑘𝑘 ∗ 𝑥𝑥 − 𝑐𝑐 ∗
𝑑𝑑𝑑𝑑
1
�∗
𝑑𝑑𝑑𝑑
𝑚𝑚
This equation will be used for modelling in Matlab/Simulink with basic blocks.
𝑋𝑋(𝑠𝑠)
To retrieve the transfer function 𝐻𝐻(𝑠𝑠) = 𝐹𝐹(𝑠𝑠) we have to reorder the equation
𝐹𝐹(𝑡𝑡) − 𝑘𝑘 ∗ 𝑥𝑥 − 𝑐𝑐 ∗
𝑑𝑑𝑑𝑑
𝑑𝑑2 𝑥𝑥
= 𝑚𝑚 ∗ 2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑡𝑡
→
𝐹𝐹(𝑡𝑡) = 𝑚𝑚 ∗
And then perform the Laplace transform, where
𝐹𝐹(𝑆𝑆) = 𝑚𝑚 𝑠𝑠 2 𝑋𝑋(𝑠𝑠) + 𝑐𝑐 𝑠𝑠 𝑋𝑋(𝑠𝑠) + 𝑘𝑘 𝑋𝑋(𝑠𝑠)
𝑋𝑋(𝑠𝑠)
→
The transfer function will be: 𝐻𝐻(𝑠𝑠) = 𝐹𝐹(𝑠𝑠) =
𝑑𝑑 2
𝑑𝑑𝑡𝑡 2
𝑑𝑑2 𝑥𝑥
𝑑𝑑𝑥𝑥
+ 𝑐𝑐 ∗
+ 𝑘𝑘 ∗ 𝑥𝑥
2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑡𝑡
= 𝑠𝑠 2 and
𝑑𝑑
𝑑𝑑𝑑𝑑
= 𝑠𝑠.
𝐹𝐹(𝑆𝑆) = (𝑚𝑚 𝑠𝑠 2 + 𝑐𝑐 𝑠𝑠 + 𝑘𝑘) 𝑋𝑋(𝑠𝑠)
1
𝑚𝑚 𝑠𝑠2 + 𝑐𝑐 𝑠𝑠 + 𝑘𝑘