How taut must a stretchable, horizontally-oriented string be in order for a straight line to approximate the string to within a given margin of error? [closed]

Question

My question deals with a string that can stretch due to its own weight. If the string is allowed to stretch then I'd assume there would always be a bit of a bulge due to gravity.

The only progress I've made is first modeling a string that cannot stretch. In that special case, I've taken note of the fact that if you increase the distance between the endpoints to exactly the length of the string, then that particular taut string can be modeled by a perfect horizontal line segment.

For the sake of dealing with the non-stretching string, I assume that an idealized non-stretching string with uniform mass distribution along its length will form a catenary when its endpoints are held some distance apart (less than the length of the string) at a height at least half the length of the string.

Assuming no stretching, I've come up with the following:

I believe the following equation can be used to model the position of a string of constant length as you move the endpoints away horizontally (one endpoint is fixed to the origin while the other moves away in the positive x direction)

Let $d$ be the distance between the endpoints. Let $L$ be the length of the string. Define $D$ such that $d = 2D\sinh^{-1}\left(\frac{L}{D}\right)$. Since $d(-D)=d(D)$, let $D∈(0,\infty)$.

Then the catenary representing the string of length L whose endpoints are a distance d apart is:

$f(t)=\left(D\sinh^{-1}\left(\frac{L}{D}\right)+D\sinh^{-1}\left(\frac{t}{D}\right),D\cosh\left(\sinh^{-1}\left(\frac{t}{D}\right)\right)-D\cosh\left(\sinh^{-1}\left(\frac{L}{D}\right)\right)\right)$ on the domain $t∈[-L,L]$

Answer 1

I believe you've lost a factor of two somewhere in your calculations. For a catenary of length $\ L\ $ with endpoints at $\ (0,0)\ $ and $\ (0,d)\ $ with $\ d>0\ $, I get the following parametric equations \begin{align} x&=\frac{g}{2}\sinh^{-1}\frac{L}{g}+\frac{g}{2}\sinh^{-1}\frac{t}{g}\\ y&=\frac{g}{2}\cosh\sinh^{-1}\frac{t}{g}-\frac{g}{2}\cosh\sinh^{-1}\frac{L}{g} \end{align} for $\ t\in[-L,L]\ $, where $\ g\ $ is the (unique) positive solution to the equation $$ d=g\sinh^{-1}\frac{L}{g}\ . $$ Note that $\ \lim_\limits{u\rightarrow\infty}u\sinh^{-1}\frac{a}{u}=a\ $ and consequently also $\ g\rightarrow\infty\ $as $\ d\rightarrow L\ $. Therefore, in the above parametric equations $$ \lim_{d\rightarrow L}x=\frac{L+t}{2} $$ and \begin{align} \lim_{d\rightarrow L}y&=\lim_{g\rightarrow\infty}\left(\frac{g}{2}\cosh\sinh^{-1}\frac{t}{g}-\frac{g}{2}\cosh\sinh^{-1}\frac{L}{g}\right)\\ &=\lim_{\delta\rightarrow0}\frac{\cosh\sinh^{-1}t\delta}{2\delta}-\lim_{\delta\rightarrow0}\frac{\cosh\sinh^{-1}L\delta}{2\delta}\\ &=\frac{d\cosh\sinh^{-1}tx}{dx}\Bigg|_{x=0}-\frac{d\cosh\sinh^{-1}Lx}{dx}\Bigg|_{x=0}\\ &=t\sinh\sinh^{-1}0-L\sinh\sinh^{-1}0\\ &=0\ . \end{align} Thus, in the limit as $\ d\rightarrow L\ $, the above parametric equations do indeed approach those \begin{align} x&=\frac{L+t}{2}\\ y&=0 \end{align} of the line segment from $\ (0,0)\ $ to $\ (0,L)\ $.