Optical Calibration Demo and Three Hours Working on a Bug

Technical Details
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I referenced the wikipedia page here: https://en.wikipedia.org/wiki/Catenary. Go to the section labeled “Analysis”. It goes through the whole derivation. Toward the end of the “Analysis” section, there is a “Determining Parameters” subsection. This covers the equation we want to use.

Here is the content from Wikipedia:

I couldn’t get it to display very well. Sorry. You’ll have to follow the link to view

Determining parameters

In general the parameter a is the position of the axis. The equation can be determined in this case as follows:[53] Relabel if necessary so that P 1 is to the left of P 2 and let h be the horizontal and v be the vertical distance from P 1 to P 2. Translate the axes so that the vertex of the catenary lies on the y-axis and its height a is adjusted so the catenary satisfies the standard equation of the curve

{\displaystyle y=a\cosh \left({\frac {x}{a}}\right)}

and let the coordinates of P 1 and P 2 be ( x 1, y 1) and ( x 2, y 2) respectively. The curve passes through these points, so the difference of height is

{\displaystyle v=a\cosh \left({\frac {x_{2}}{a}}\right)-a\cosh \left({\frac {x_{1}}{a}}\right),.}

and the length of the curve from P 1 to P 2 is

{\displaystyle s=a\sinh \left({\frac {x_{2}}{a}}\right)-a\sinh \left({\frac {x_{1}}{a}}\right),.}

When s 2 − v 2 is expanded using these expressions the result is

{\displaystyle s^{2}-v^{2}=2a^{2}\left(\cosh \left({\frac {x_{2}-x_{1}}{a}}\right)-1\right)=4a^{2}\sinh ^{2}\left({\frac {h}{2a}}\right),,}

so

{\displaystyle {\sqrt {s^{2}-v^{2}}}=2a\sinh \left({\frac {h}{2a}}\right),.}

This is a transcendental equation in a and must be solved numerically. It can be shown with the methods of calculus[54] that there is at most one solution with a > 0 and so there is at most one position of equilibrium.

For Maslow:

We know the parameter “a” a-priori because it is a function of the horizontal tension and the chain-weight. So, we need to re-formulate the equation to solve for “s”, the catenary length.

How to use the Equation:

image

However, it has to be manipulated to solve for s.

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