1. Hanging Cable Problem

In my engineering class we were learning about the hanging cable under its own weight.

If you have coordinate system centered with the hanging cable, the equation of the curve is,
$y=c\cosh (x/c)$
Where, $c$ is the konstant to be determine.
If you know the span $L$ and it sag $h$.
We need to find $c$.

We know that when $x=L/2$ then we have, $y=c+h$

Thus,
$c+h=c\cosh (L/2c)$
Thus,
$1+h/c=\cosh (L/2c)$
The way the professor taugh us to solve this is to guess through values until it works.

That approach is too primitive for me.
I told him that it might be possible to use the Lambert function to solve it, he slaped me across my face.
---
So I have been thinking,
Let $x=1/c$.

Then,
$1+hx=\frac{e^{.5Lx}+e^{-.5Lx}}{2}$
Thus,
$2+2hx=e^{.5Lx}+e^{-.5Lx}$

It seems I have been around engineers too much these days, and I learned this trick from them.
$e^{-.5Lx}\approx 0$
(I cannot believe that I did that).

Thus,
$2+2hx=e^{.5Lx}$
Thus, the problem redues to solving,
$a+bx=e^x$
Which I think is possible using a proper variable substition, right

Thus, we would have an approximate solution through the Lambert function for the solution of konstant $c$.
---
I had another approach, more mathematical, is to create error bounds on the solution.
Since,
$0
We solve the equation as if,
$e^{-.5Lx}=0$
And then,
$e^{-.5Lx}=1$
Those would create an interval on which the solution exists.
Which is good because it leads to,
$a+bx=e^x$

Anyone, has any other ideas, how to solve this equation.

2. Originally Posted by ThePerfectHacker
In my engineering class we were learning about the hanging cable under its own weight.

If you have coordinate system centered with the hanging cable, the equation of the curve is,
$y=c\cosh (x/c)$
Where, $c$ is the konstant to be determine.
If you know the span $L$ and it sag $h$.
We need to find $c$.

We know that when $x=L/2$ then we have, $y=c+h$

Thus,
$c+h=c\cosh (L/2c)$
Thus,
$1+h/c=\cosh (L/2c)$
The way the professor taugh us to solve this is to guess through values until it works.
I have no suggestion about a closed form solution, but I do have a
suggestion to simplify the equation. Make the substitution:

$
x=h/c
$

Then the equation becomes:

$1+x=\cosh (Lx/h)$

Then let $K=L/h$ to give:

$1+x=\cosh (Kx)$

RonL

3. This might work
If you have, if $b\not = 0$
Working with linear exponential equation,
$a+bx=e^x$
Let,
$x+a/b=y$
Thus,
$a+b(y-a/b)=e^{y-a/b}$
Thus,
$by=e^{y}e^{-a/b}$
Thus,
$bye^{-y}=e^{-a/b}$
Thus,
$-ye^{-y}=-(1/b)e^{-a/b}$
Lambert function!
$-y=W(-(1/b)e^{-a/b})$
Thus,
$x=-a/b-W(-(1/b)e^{-a/b})$
The only question I have am I allowed to do that, maybe I am not in the domain of the Lambert function?

4. Hi,

I don't know if this post is of any use for you, but ...

I send you a copy of the chapter about the catenary (literally translated it means chain line) from Hairer, Wanner, Analysis by Its History, New York, 1996

Maybe the equation number (7.7) is of some use for you.

EB

5. Originally Posted by earboth
Hi,

I don't know if this post is of any use for you, but ...
It does not help, but it does look interesting. Does it imply that Leibniz was the first person to guess the shape of the caternary.

6. Originally Posted by ThePerfectHacker
It does not help, but it does look interesting. Does it imply that Leibniz was the first person to guess the shape of the caternary.

"The catenary is the shape of a perfectly flexible chain suspended by its ends and acted on by gravity. Its equation was obtained by Leibniz, Huygens and Johann Bernoulli in 1691. They were responding to a challenge put out by Jacob Bernoulli to find the equation of the 'chain-curve'. "

RonL