I am trying to graphically simulate a series of springs in 2D. Now one of the forces I am stuck with calculating is the damping force. The given formula is $F = -k_d v$. I know that $v$ is the velocity of the vectors, but I can't seem to find how to calculate $k_d$.
[Physics] Calculate damping constant / coefficient
dragfrictionnewtonian-mechanicsoscillatorsspring
Related Solutions
Yes, one can calculate an equivalent mass. Take the following coordinates, that describe the position of the particles' centers with respect to the center of mass of the set, $G$. In the sketch that follows I have arbitrarily chosen $m_1 < m_2$, and so $G$ falls closer to $m_2$:
Since $G$ will remain fixed in the absence of external forces, it makes sense to use the coordinates above. The equations of motion of the two particles in these coordinates read $$m_1\frac{\mathrm{d}^2 x_1}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0 \\ m_2\frac{\mathrm{d}^2 x_2}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$ where $l = x_1 + x_2$ (the distance between centers). Adding these two equations and deviding through by $2$ one obtains $$\frac{m_1 + m_2}{2}\frac{\mathrm{d}^2 l}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$ which is of the form of the original equation (the one from Wikipedia) if one takes $$m = \frac{m_1 + m_2}{2}$$
Thus, the equivalent mass is just the arithmetic average of the two masses.
Let's start by plotting the damped harmonic oscillation equation:
$$x = Ae^{-\gamma t} \cos (\omega t)$$
In the above example $\gamma = 0.8$ and $\omega = 2.5\times\pi$. It should now be more clear that when $t = 0$, $x=A$ i.e. the initial displacement of the oscillator from it's equilibrium position. Here $A = 1~\mathrm{m}$. Let's assume down is the positive $x$-axis for simplicity.
Now the way one would conventionally go about finding the damping coefficient is by finding $\gamma$ form the envelope function $Ae^{-\gamma t}$ (plotted in green). You would ideally use a data logger of some fashion to sample the position of the oscillator as a function of time (hopefully with enough resolution) and then you find $x$ at the exact same point in the cycle (constant phase) for each oscillation. It is usually the easiest to locate the peaks (plotted in red) where $$\cos(\omega t) = 1 \quad \therefore \quad \omega t = 2n\pi\; (n=1,2,3,\ldots) $$
You can either fit an exponential curve to these data points or alternatively fit a linear curve to the log-log of the data.
Now in your case, it seems you only have 2 data points. The position at $t =0$, and the time $t_{1/2}$ when $x= A/2$. It's not ideal, but you have everything you need to calculate the damping coefficient now.
Hope this helps :)
Best Answer
For a viscous damper, the decay in the free oscillation amplitude is exponential (it is geometric for hysteric damping and linear for Coulomb damping). So if you have the time history of the amplitude of your decay and you know it is a viscous damper (which is the equation you gave) then you can measure the amplitude $A$ at two consecutive peaks and calculate:
$$\gamma = \ln \left(\frac{A_{t_n}}{A_{t_{n+1}}}\right) $$
you can then find the damping coefficient to give this decay as:
$$\zeta = \frac{\gamma}{\sqrt{4 \pi^2 + \gamma^2}}$$
where then of course $\zeta = k_d/(2\sqrt{k m})$.
So given a spring with unknown damping coefficient but known stiffness, you can attach a known mass to it and measure it's response to a disturbance and determine from that the damping coefficient.
Since you are just going for aesthetics, you pick your damping constants arbitrarily. I would actually recommend that you play with it and see how it influences the solution, it's actually pretty cool to visualize. All you do is pick values for $\zeta \in [0.0, 2.0]$ where the upper bound is really limitless but not much will change when it is greater than $2.0$. Then you can compute your $k_d$ based on $k$ and $m$. Depending on your time integration, you may find that $\zeta = 0$ will be unstable. You might need something nominal to stabilize the scheme. When $\zeta = 1$, it is called critically damped and you should not see much oscillation at all (it will be driven to steady state without oscillation). I say you won't see much because as a system of springs and with numerical integration, it won't be exactly critically damped.