suspension dynamics (2/3)

F=ma, yo

This is part 2 of a 3 part series of posts about car suspensions, and more specifically - solving the motion equations for a coilover suspension system.

Again, we start with Newtons second law

\begin{equation} F = ma \end{equation}

Now instead of just one force, we have 2

\begin{equation} F = -kx -c \frac{dx}{dt} \end{equation}

c in this case is the damping constant. The damping force is proportional to the velocity of the mass and acts in the direction opposite of motion. This has been experimentally shown for viscous friction, which is how shock absorbers function. [1]

\begin{equation} m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \end{equation}

Following in the footsteps of Euler, let’s assume that

\begin{equation} x(t) = e^{rt} \end{equation}

\begin{equation} m r^2 e^{rt} + c r e^{rt} + k e^{rt} = 0 \end{equation}

Because of the properties of exponentials, we know that

\begin{equation} e^{rt} \neq 0 \end{equation}

So, dividing by the exponential is allowed, and it gives us

\begin{equation} mr^2 + cr +k = 0 \end{equation}

\begin{equation} r = \frac{-c \pm \sqrt{c^2 - 4 m k}}{2m} \end{equation}

Notice that the inside of the square root can be negative depending on the parameters of our initial conditions. This leads to a piecewise system of solutions for (3). If r is real and greater than zero, the system is over-damped. If r is zero, the system is critically damped. If r is imaginary, the system is under-damped. Luckily for us, we care about the under-damped situation. Hooray for complex numbers!

Lets simplify our variables a bit before we move on.

\begin{equation} r = - \frac{c}{2m} \pm \frac{ \sqrt{c^2 - 4 m k}}{2m} \end{equation}

\begin{equation} - \frac{c}{2m} = a \end{equation}

\begin{equation} \frac{\sqrt{c^2 - 4 m k}}{2m} = ib \end{equation}

\begin{equation} r = a \pm ib \end{equation}

Our solutions now looks like:

\begin{equation} x(t) = e^{(a \pm ib) t} \end{equation}

Separating exponents in preparation of the big event

\begin{equation} x(t) = e^{at}e^{\pm ibt} \end{equation}

and BAM! Euler’s formula.

\begin{equation} x(t) = e^{at}( cos(bt) \pm i sin(bt) ) \end{equation}

Next up, we need to get rid of those pesky i’s.

My limited experience with abstract linear algebra really showed during my research of this. With the help of a few college professors, I was able to get to a point of understanding that I am comfortable with. Props to Mark Lyon and Adam Boucher.

Since the differential equation is linear and homogeneous, the superposition principle states that any linear combination of solutions is itself a solution.

Combining both equations in (16) with the coefficients of 1 and 1, we get the following.

\begin{equation} x(t) = e^{at}(cos(bt) + i sin(bt) ) + e^{at}(cos(bt) - i sin(bt) ) \end{equation}

\begin{equation} x(t) = e^{at}( 2 cos(bt) ) \end{equation}

Choosing the coefficients of i and -i, we get:

\begin{equation} x(t) = e^{at}(i cos(bt) + i^{2} sin(bt) ) - e^{at}(i cos(bt) - i^{2} sin(bt) )\end{equation}

\begin{equation} x(t) = e^{at}( 2 i^{2} sin(bt) ) \end{equation}

\begin{equation} x(t) = - e^{at}( 2 sin(bt) ) \end{equation}

(17) and (20) are particular solutions for the differential equation. Because we are free to do what we want, let’s take another stab at guessing the general solution to the differential equation. This time choosing the following.

\begin{equation} x(t) = A e^{at}( cos(bt) ) + B e^{at}( sin(bt) ) \end{equation}

to be finished when I get time…