suspension dynamics (1/3)

This is the first of a 3 part series of posts on the mathematics of car suspension systems.

The purpose of this is to learn a bit more about coilover suspensions. I will assume only a basic understanding of Newtonion physics and differential equations. We will start with simple harmonic oscillation and work our way up.?

Write down the general equation of motion, Newton’s second law.

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

Assuming an idealized spring that takes advantage of hooks law, we declare the spring’s force, and then plug the force into (1). k is the spring constant. This assumption holds true in practice in the case of linear springs not undergoing extreme deflection.

\begin{equation} -kx = F \end{equation}

\begin{equation} m\frac{d^2x}{dt^2} = -kx \end{equation}

We know that sin and cos are their own second derivatives, with an additional constant term. With this information we guess that the solution to the differential equation is as follows

\begin{equation} x(t) = a cos(\omega t) + b sin(\omega t) \end{equation}

Taking the solution at time t=0, we find

\begin{equation} a = x_0 \end{equation}

Taking the first derivative of (4) yields the following

\begin{equation} v(t) = \frac{dx}{dt} = -\omega a sin(\omega t) + \omega b cos(\omega t) \end{equation}

Setting t = 0, we get

\begin{equation} b = \frac{v_0}{\omega} \end{equation}

Combining the new substitions in (4), our new equation becomes

\begin{equation} x(t) = x0 cos(\omega t) + \frac{v0}{\omega} sin(\omega t) \end{equation}

Derive this twice and factor out the constant.

\begin{equation} v(t) = -\omega x0 sin(\omega t) + \omega \frac{v0}{\omega} cos(\omega t) \end{equation}

\begin{equation} a(t) = -\omega^2 x0 cos(\omega t) - \omega^2 \frac{v0}{\omega} sin(\omega t) \end{equation}

\begin{equation} a(t) = -\omega^2 ( x0 cos(\omega t) + \frac{v0}{\omega} sin(\omega t) ) \end{equation}

Notice that the term inside of the parenthesis is equal to equation (8). Substituting x(t), we get:

\begin{equation} a(t) = -\omega^2 x(t) \end{equation}

plugging in equation (3)

\begin{equation} a(t) = -\frac{kx(t)}{m} = -\omega^2 x(t) \end{equation}

\begin{equation} \omega = \sqrt{\frac{k}{m}} \end{equation}

Now that we have all of our unknowns covered, we can write our final equation of motion

\begin{equation} x(t) = x0 cos( \sqrt{\frac{k}{m}} t ) + \frac{v0}{ \sqrt{\frac{k}{m}} } sin( \sqrt{\frac{k}{m}} t ) \end{equation}

We verify our solution by taking the second derivative of (15)

\begin{equation} a(t) = - x0 \frac{k}{m} cos( \sqrt{\frac{k}{m}} t ) - \frac{v0}{ \sqrt{\frac{k}{m}} } \frac{k}{m} sin( \sqrt{\frac{k}{m}} t ) \end{equation}

Setting that equal to (3)

\begin{equation} a(t) = -\frac{kx(t)}{m} = - x0 \frac{k}{m} cos( \sqrt{\frac{k}{m}} t ) - \frac{v0}{ \sqrt{\frac{k}{m}} } \frac{k}{m} sin( \sqrt{\frac{k}{m}} t ) \end{equation}

\begin{equation} a(t) = -x(t) = - x0 cos( \sqrt{\frac{k}{m}} t ) - \frac{v0}{ \sqrt{\frac{k}{m}} } sin( \sqrt{\frac{k}{m}} t ) \end{equation}

We see equation (15) poke its head out. Making the substitution leads to

\begin{equation} -x(t) = -x(t) \end{equation}