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I have a cable suspended between two poles, both $1086.4$ ft high, that are $4444.7$ ft apart. The sag at the middle of the cable is $197.8$ ft. Given this, the constant should be $c = 1086.4 - 197.8 = 888.6$. However, when I try to use that $c$ in any equations, my results come out completely wrong.

Ex: $y=c \cosh\frac{x}{c}$

For the y at the pole/end,

$c \cosh\frac{x}{c} = 5454$

which is obviously very off from the pole's actual height of $1086.4$.

What am I doing wrong?

(I've looked through the following resources for help but haven't been able to find a formula/definition for c that comes close to my actual values. http://www.ce.siue.edu/examples/Worked_examples_Internet_text-only/Data_files-Worked_Exs-Word_&_pdf/cables_catenaries.pdf

http://euclid.trentu.ca/aejm/V4N1/Chatterjee.V4N1.pdf

http://www.stem2.org/je/catenary.pdf)

Andrew Chin
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  • That equation applies when the $x$-axis coincides with the catenary’s directrix. The one in your question appears to be translated vertically from this. – amd Sep 17 '19 at 00:05
  • On your first link, Example 2 is similar to your task here, the sag is known (and they could put it at any height), but they do not just subtract the sag from the height. Instead on page 3, they calculate $c$ by solving the equation $c + 40 = c \cosh (200/c)$ numerically. I.e. $y$ and $c$ are not heights from an arbitrary ground level. They are heights from a particular elevation relative to the curve. – Paul Sinclair Sep 17 '19 at 01:39

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