Help with this diagram
- Thread starter Pingu27
- Start date
topsquark
Senior Member
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- Aug 27, 2012
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There is very little help I can give except to give you some definitions.
The curve represented here is a catenary (See the pink panel on the left.), not a parabola as many might think. The equation of the curve (with an origin at the midpoint of the curve (where the 20 m measurement is taken from) is the equation [math]y = a ~ cosh \left ( \dfrac{x}{a} \right )[/math].
You can look up the rest but it's a bit of a slog. I hate to just post it but there's really no other way.
[math]d = \left ( \dfrac{L^2 - h^2}{2 h} \right ) ~ sinh^{-1} \left ( \dfrac{2hL}{L^2 - h^2} \right )[/math]where d is half the width of what you are looking for, h is the distance from 20 m above the ground to a height of 50 m (ie. h = 30 m), and L is the half the length along the wire (L = 40 m.)
This will be easy or hard to calculate depending on how comfortable you are with your calculator. That [math]sinh^{-1}(.)[/math] might be a bit to work with if you don't have a programmable calculator. If you run into problems with this, just let us know.
-Dan