# Simple mass spring damping system sizing K, and C factors

I have come across a problem, which has troubled me for some time now. What needs to be done is the following:

A mass on a rod 0.6m (mass less) has a mass of 1 kg attached at the end of it. The rod needs to be rotated 60°, within t=120 sec (see image). What I would like to do is size a rotational spring (located at the pivot point) and a damping system, such that it that will damp the spring force. Thus the rotation happens within the specified amount of time.

I have written the generic differential equation of the system:

$$J\theta'' + C\theta' + K\theta = 0$$

(typical differential equation of damped spring system)

and for a critical damped system, and for $t=0$, $\theta=0$ I have the solution:

$$\theta(t)=A t \exp(-bt)$$

where $A$ is a constant, and $b$ is the damping coefficient.

My question is how can I continue, such that I can size the damping coefficient and the spring constant?

And if I can continue from here, how should I then proceed to size my system?

Please note $g=0$, no gravity.

• When using only passive components such as springs and dampers it should theoretically always take an infinite amount of time to come to rest at its equilibrium point. But you can design the system such that it gets arbitrarily close after some finite amount of time. Also you also need to take $\theta'(0)$ into consideration. – fibonatic Oct 27 '17 at 3:00
• @fibonatic only if your model has has no stiction – joojaa Oct 27 '17 at 15:02
• Well in this case I dont want the system to have a zero velocity at 60 degrees. but rather still have some velocity such that the system can activate a latching mechanism that will hold the rod into place. Taking the velocities under consideration has not provided me with a meaning full solution. I am beginning to wonder if my initial approach is correct.. – mioumitsou Oct 30 '17 at 8:31
• The equations shown above assume the rest position of the spring to be $\theta=0$. Is this the intended behavior? – John Alexiou Dec 28 '17 at 20:34

The equation should be

$$J \, \ddot{\theta} + K \, (\theta-\theta_T) + C \, \dot{\theta} =0$$

where $\theta_T$ is the target angle and the rest position for the spring.

Things simplify wit the following substitution $$\begin{cases} K = J \Omega^2 \\ C = 2 \zeta J \Omega \end{cases}$$ where $\Omega$ is a parameter the relates to the stiffness, and $\zeta$ a parameter that relates to the damping.

The general solution of the equation $\ddot{\theta} + \Omega^2 (\theta-\theta_T) + 2 \zeta \Omega \dot{\theta} =0$ is

$$\theta = \theta_T + C_1 \exp\left( -\Omega t \left(\sqrt{(\zeta^2-1)}+\zeta \right) \right)+ C_2 \exp\left( -\Omega t \left(\sqrt{(\zeta^2-1)}-\zeta \right) \right)$$

where the coefficients $C_1$ and $C_2$ are found based on the boundary conditions. In this case, the rod is at rest when $t=0$ and thus

$$\begin{cases} C_1 =\frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}-1 \right) \\ C_2 =-\frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}+1 \right) \end{cases}$$

The fastest response is when $\zeta \rightarrow 1$ since that minimizes the value of the first exponent. With optimum damping the solution becomes

$$\theta = \theta_T \left( 1 - {\rm e}^{-\Omega t}(1+\Omega t) \right)$$

and $C = 2 J \Omega$.

But as you stated, you cannot find the correct $\Omega$ to reach the target angle at the specified time because $\Omega$ cannot be isolated from the solution.

But it can be isolated from the general solution when $\zeta>1$. From the two terms, the first one approaches zero much faster than the second one. We can find the coefficient $\varphi = \Omega t$ which makes the second exponent near zero by a value $\epsilon$ (in degrees).

$$\left. \frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}+1 \right) \exp\left( -\varphi \left(\sqrt{(\zeta^2-1)}-\zeta \right) \right) = \epsilon \right\}$$ $$\varphi = \frac{ \ln\left( 2 \frac{\epsilon}{\theta_T} ( \zeta \sqrt{(\zeta^2-1)}-\zeta^2+1)\right)}{\sqrt{(\zeta^2-1)}-\zeta} = \Omega\, t_T$$

where $t_T$ is the target time. A good compromize occurs when $\varphi = \frac{1}{\sqrt{(\zeta^2-1)}}$ which makes $\frac{{\rm d}\varphi}{{\rm d}\zeta}=0$ and hence minimizes $\varphi = \Omega \, t_T$.

This produces the solution

$$\begin{cases} K = \frac{J}{t_T^2 \gamma (\gamma+2)} \\ \zeta = \frac{ \sqrt{(J+K t_T^2)}}{t\,\sqrt{K} } \\ C = \frac{2}{t_T} \sqrt{J (J+K t_T^2)} \end{cases}$$

where $\gamma \ll 1$ is a small positive value (defined as $\zeta = 1+\gamma$).

Correct me if I'm wrong, but it looks like you have your values for $\theta$ and $t$ as 60° and 120s respectively. I'm not sure what $A$ is, it might be a design choice for you to make. For your critically damped system, you would only need to know the damping coefficient $b$. You would just rearrange the equation until it became

$$b = \frac{\ln(\theta/A)}{\ln(t) t}$$

• Hello the values are correct... Though I cannot imagine a boundary condition that will allow me to find the value of A. Would you happen to be able to think of such a condition? – mioumitsou Oct 30 '17 at 8:28
• well, what does A stand for? Generally, if you have two unknown constants, then you either have to enforce another constraint, or make a design decision. – Timothy Barrett Nov 1 '17 at 14:41