2
$\begingroup$

I have a crane boom that is rotating and when it is stopped suddenly, the boom flexes back and fourth several times. I have measured the angles between each deflection and have $3.16 ^\circ$, $2.42 ^\circ$, $1.77 ^\circ$ and $1.31 ^\circ$, which if plotted in Excel forms a logarithmic curve equal to $$y = -1.33ln(x) + 3.2221$$ The goal now is to find how to plot this as a sin curve with the period of 3.5 seconds so that it it looks like a deflection vs time chart (or damping curve), similar to the one in red below. Can someone please explain how I can do this?

enter image description here

$\endgroup$
2
$\begingroup$

The model

The equation for a damped periodic oscillation is $$ y(t) = A e^{-\lambda t} \cos(\omega t + \phi) $$ where $A$ is the amplitude, $\lambda$ is the decay constant, $\omega = 2\pi/\tau$ where $\tau$ is the period, and $\phi$ is the phase angle at $t=0$. In your figure, $\phi = -\pi/2$, and therefore your damping curve has the form $$ y(t) = A e^{-\lambda t} \sin(\omega t) $$

Your mistake

Instead of using $$ y(t) = A e^{-\lambda t} \quad \quad \text{(Correct)} $$ you have fit the portion of the curve where $\sin(\omega t) = 1$ using the relation $$ y(t) = B \ln(t) + C \quad \quad \text{(Wrong)}\,. $$ This model predicts an infinite displacement $y$ at time $t=0$.

Data needed

You have provided angles but not displacements. We have to guess what those are from your fit.

If we fit the correct model to data extracted from your log model, we get the following curves (note that we can calculate the full deflection-time curve after fitting $A$ and $\lambda$ because $\omega$ can be calculated from the period): enter image description here

$\endgroup$
1
  • $\begingroup$ Thanks very much for this. It is exactly what I was after. $\endgroup$ – david_10001 May 26 '18 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.