There are two general classes of problems. The first class has two attachment points that are of equal height and the second class has uneven attachment points. The same types of questions arise from both, but the method of solution varies slightly.
Given two poles that are $2x$ meters apart, find the cable length that permits a certain amount of sag, $h.$ The cable weight is $\omega_{0}$ with units of kg/meter. This problem assumes that kilograms are a weight, not a mass and it assumes a non-extensible cable. For class 1, the attachment points are at the same level. For class two, the difference in attachment height will have to be given.
Use $x$ meters of cable between poles and equal height attachment points. For a cable weighing $\omega_{0}$ kg/m, and permitting $h$ meters of sag in the middle, how far apart must the poles be placed?
If a defined length of cable is suspended between poles that are a defined distance apart, how much sag can be expected? How does the tension change with cable weight?
Between two attachment points of unequal height, set the attachment heights so that the sag, $h,$ is centered over a specific spot between the poles. You are given $\omega_{0}$ as well as the upper attachment height, the minimum distance above ground at the lowest sag point, and the desired placement of the lower pole.
Using the Catenary Data
Figure 1: Simple catenary with equal height attachment points. The sag is $h$. The $a$ value is the distance from the minimum sag point to the origin. $s$ is the curve length from $C$ to $A.$ $A$ is the attachment point on a pole.
Let's start by supposing that we know all of the information in the drawing.
Now suppose we also know that the cable weight is $874.1$ pounds per $1000$ feet, and that its rated breaking strength is $25200$ lbs. How do we calculate the tension at point $A,$ and decide if we are close or far away from the breaking strength?
$$T_{0}=\omega_{0}a$$
If a is in feet, then $\omega_{0}=\text{0.874.1 }$ lbs/ft, and $T_{0}$ can be calculated. The curve is given by the equation $f(x)=a\cosh\left(\frac{x}{a}\right).$ Its derivative is
$$f^{\prime}(x)=\tan(\theta)=\sinh\left(\frac{x}{a}\right)$$
therefore,
$$\theta=\tan^{-1}\left(\sinh\left(\frac{x}{a}\right)\right)$$
Sometimes, a line tension and distance between poles is specified and it is asked to predict sag. When given as a single tension number, the usual intent is that it is $T_{0}$ or an average of $T$ and $T_0$ which is given, even if it is just stated as tension. Since $T_{0}=\omega_{0}a,$ and $\omega_{0}$ is a property of the cable, we have effectively been given $a.$ Remember, at low angles, $T_0\approx T$. We caculate sag as
$$h=a\cosh\left(\frac{x}{a}\right)-a$$
$$s=a\sinh\left(\frac{x}{a}\right)$$
$$\theta=\tan^{-1}\left(\sinh\left(\frac{x}{a}\right)\right)$$
So tension at the pole is
$$T=\frac{T_{0}}{\cos\theta}=\frac{\omega_{0}s}{\sin(\theta)}$$
Class 1 Problems - Equal height attachments
This class makes use of the two equations that we have derived.
$$\frac{s}{a}=\sinh\left(\frac{x}{a}\right) \tag{1} \label{1}$$
and
$$y=a\cosh\left(\frac{x}{a}\right) \tag{2} \label{2}$$
To accomodate the sag, $h,$ which was not a part of any derivation, we can subtract $a$ from the right side of $\eqref{2}.$ Since $a$ is the distance from the graph origin $(0,0)$ up to the lowest sag point, subtracting it away will put the origin coincidental to the lowest sag point. Next we can let $x$ be the distance from the origin to the base of the pole and we get
$$h=a\cosh\left(\frac{x}{a}\right)-a \tag{3} \label{3}$$
If $x$ and $h$ are given, we can solve for $a$, probably using Newton's method. With $a$ in hand, we use $\eqref{1}$ to get $s$.
If $s$ and $h$ are given, we will need some algebra to solve for $x.$ Rearrange $\eqref{3}$
$$\cosh\left(\frac{x}{a}\right)=\frac{h+a}{a}.\tag{4} \label{4}$$
We will use the trig identity $\cosh^{2}(\phi)-\sinh^{2}(\phi)=1.$ So square equations $\eqref{1}$ and $\eqref{4}$, then we see
$$\left(\frac{h+a}{a}\right)^{2}-\left(\frac{s}{a}\right)^{2}=1.$$
$$a=\frac{s^{2}-h^{2}}{2h}$$
Then from $\eqref{1}$
$$\sinh^{-1}\left(\frac{s}{a}\right)=\frac{x}{a}$$
and
$$x=a\sinh^{-1}\left(\frac{s}{a}\right).$$
Exercise: Given that $100$ meters of wire will run between two poles, we are told to hang it at a height of $30$ meters above ground. It is assumed that the wire will sag and the height above ground at the lowest point should be $27$ meters. How far apart should the poles be placed?
Answer: In this problem, we have been given $s$ and $h$. $s=100/2 \qquad s=3 \quad$ Following what we just did,
Figure 2: Pole distances apart show that horizonal distance, 100m, and line length 99.76m are nearly the same.
Rearrange $\eqref{3}$
$$\cosh\left(\frac{x}{a}\right)=\frac{h+a}{a}.$$
We will use the trig identity $\cosh^{2}(\phi)-\sinh^{2}(\phi)=1.$ So square equations $\eqref{1}$ and $\eqref{4}$, then we see
$$\left(\frac{h+a}{a}\right)^{2}-\left(\frac{s}{a}\right)^{2}=1.$$
Thus
$$a=\frac{s^2-h^2}{2h}=415.17$$
and
$$x=a\sinh^{-1}\left( \frac{s}{a} \right)=49.88$$
That is, $2x$=disance between poles $= 99.76$
In other words, the distance between poles by calculation and by wire measurement is insignificant at low angles. Also, since we were given wire length, we could ignore tension and cable weight. For high tension wire, this is a very unrealistic exercise.
Class 2 Problems - Unequal Pole Height
Figure 3: Labels for attachments of unequal height. $h_{1}$ is the sag amount measured vertically from attachment $A$ to $C.$ $h_{2}$ is the vertical distance from $B$ to $C.$ $x_{1}$ and $x_{2}$ are the horizontal distances from the base of the attachments to point $C.$ $s_{1}$ and $s_{2}$ are the respective cable lengths.
We need to address this with some additional theory to go with the foundation we have established. Although the center point is no longer in the middle, we otherwise just have two sets of the equations we previously derived. Actually, we now have at least six equations and may only need five.
$$a\sinh\left(\frac{x_{1}}{a}\right)=s_{1} \tag{2-1} \label{2-1}$$
$$a\sinh\left(\frac{x_{2}}{a}\right)=s_{2} \tag{2-2} \label{2-2}$$
$$x_{1}+x_{2}=\text{distance between poles } \tag{2-3} \label{2-3}$$
$$s_{1}+s_{2}=\text{cable length between poles} \tag{2-4} \label{2-4}$$
$$h_{1}=a\cosh\left(\frac{x_{1}}{a}\right)-a \tag{2-5} \label{2-5}$$
$$h_{2}=a\cosh\left(\frac{x_{2}}{a}\right)-a \tag{2-6} \label{2-6}$$
Game Plan 1: Given the total horizontal distance between poles, the vertical attachment points, and the maximum sag, find the distances $x_{1}$ and $x_{2}$ and the cable length, $s_{1}+s_{2}.$
Construct a drawing and list the equations to be used. In this case, we will use all but $\eqref{2-4}$.
Use the two $\cosh$ equations to solve for $x_{1}$ and $x_{2}$ as variables.
$$x_{1}=a\cosh^{-1}\left(\frac{h_{1}}{a}+1\right) \tag{2-7} \label{2-7}$$
$$x_{2}=a\cosh^{-1}\left(\frac{h_{2}}{a}+1\right) \tag{2-8} \label{2-8}$$
Solve for $a$ using $x_{1}+x_{2}=d$ where $d$ is the given total distance. Probably use Newton's method.
$$f(a)=d-x_{1}-x_{2}$$
$$f(a)=-a\cosh^{-1}\left(\frac{h_{1}}{a}+1\right)-a\cosh^{-1}\left(\frac{h_{2}}{a}+1\right)+d$$
$$f^{\prime}(a)=\frac{h_{1}}{a\sqrt{\frac{h_{1}}{a}+2}\cdot\sqrt{\frac{h_{1}}{a}}}+\frac{h_{2}}{a\sqrt{\frac{h_{2}}{a}+2}\cdot\sqrt{\frac{h_{2}}{a}}}-\cosh^{-1}\left(\frac{h_{1}}{a}+1\right)-\cosh^{-1}\left(\frac{h_{2}}{a}+1\right)$$
Then by iteration until $a_{guess}$ quits changing. If it doesn't converge, and you have checked your input carefully, then it might be that you need to make the first guess of $a$ closer to correct. In that event, graph $f(a)$ and see where it crosses the $x$-axis. That spot is $a.$ Guess accordingly.
$$\text{new }a_{guess}=a_{guess}-\frac{f(a)}{f^{\prime}(a)}$$
With $a$ in hand, we can use $\eqref{2-7}$ and $\eqref{2-8}$ to compute $x_{1}$ and $x_{2}.$
Using $x_{1}$ and $x_{2},$ we can use $\eqref{2-1}$ and $\eqref{2-2}$ to obtain $s_{1}$ and $s_{2}.$
Game Plan 2: Given the length of the cable, the attachment heights, and a maximum sag amount, find the distance between two poles that will result in the desired sag.
Construct a drawing and list the equations to be used. In this case, we will use all but $\eqref{2-3}.$
Use identity: $\cosh^{2}(\phi)-\sinh^{2}(\phi)=1$ on the pair of equations $\eqref{2-1}$ and $\eqref{2-5}.$
Algebraically isolate the $\cosh$ portion of $h_{1}=a\cosh\left(\frac{x_{1}}{a}\right)-a.$
$$\cosh\left(\frac{x_{1}}{a}\right)=\frac{h_{1}+a}{a}$$
Square both sides.
$$\cosh^{2}\left(\frac{x_{1}}{a}\right)=\left(\frac{h_{1}+a}{a}\right)^{2}$$
Algebraically isolate the $\sinh$ portion of $a\sinh\left(\frac{x_{1}}{a}\right)=s_{1}.$
$$\sinh\left(\frac{x_{1}}{a}\right)=\frac{s_{1}}{a}$$
Again square both sides.
$$\sinh^{2}\left(\frac{x_{1}}{a}\right)=\left(\frac{s_{1}}{a}\right)^{2}$$
Use the trig identity, substituting in the right and side for $\cosh^{2}(\phi)$ and $\sinh^{2}(\phi).$
$$\left(\frac{h_{1}+a}{a}\right)^{2}-\left(\frac{s_{1}}{a}\right)^{2}=1 \tag{2-9} \label{2-9}$$
Repeat step 2 with $\eqref{2-2}$ and $\eqref{2-6}$ to get
$$\left(\frac{h_{2}+a}{a}\right)^{2}-\left(\frac{s_{2}}{a}\right)^{2}=1 \tag{2-10} \label{2-10}$$
Solve $\eqref{2-9}$ and $\eqref{2-10}$ for $s_{1}$ and $s_{2}.$ Use $s_{1}+s_{2}=c,$ where $c$ is total cable length.
$$s_{1}=\sqrt{h_{1}^{2}+2ah_{1}}$$
$$s_{2}=\sqrt{h_{2}^{2}+2ah_{2}}$$
$$\sqrt{h_{1}^{2}+2ah_{1}}+\sqrt{h_{2}^{2}+2ah_{2}}=c$$
which has an algebraic solution for $a$ and that solution is much easier to get after substituting numerical values for $h_{1},h_{2}$ and $c,$ and only one of the solutions makes sense.
$$a=\frac{c^{2}h_{1}+c^{2}h_{2}-h_{1}^{3}+h_{1}^{2}h_{2}+h_{1}h_{2}^{2}-h_{2}^{3}-2c\sqrt{c^{2}h_{1}h_{2}-h_{1}^{3}h_{2}+2h_{1}^{2}h_{2}^{2}-h_{1}h_{2}^{3}}}{2h_{1}^{2}-4h_{1}h_{2}+2h_{2}^{2}}$$
With the value of $a$ in hand, we can calculate $s_{1}$ and $s_{2}$
$$s_{1}=\sqrt{h_{1}^{2}+2ah_{1}}\qquad s_{2}=\sqrt{h_{2}^{2}+2ah_{2}}$$
and using $\eqref{2-1}$ and $\eqref{2-2}$ we can get $x_{1}$ and $x_{2}.$
$$a\sinh\left(\frac{x_{1}}{a}\right)=s_{1}$$
therefore,
$$x_{1}=a\sinh^{-1}\left(\frac{s_{1}}{a}\right)$$
and similarly for $x_{2}$.
$$x_{2}=a\sinh^{-1}\left(\frac{s_{2}}{a}\right)$$
Solved Catenary Examples
Example 1: Wire type Grosbeak $(\omega_{0}=0.8741 \text{ lbs/ft})$ will be streched between poles set $1000 \text{ ft}$ apart. Cable will be attached at a height of $170 \text{ ft}$ on steel structures. Tension is planned for $4200 \text{ lbs.}$ Cable breaking strength is $25200$ lbs. What is the nominal expected sag and what is the safety factor. (The "safety factor" is a ratio, force applied divided by breaking strength.)
Answer: The specified tension is $T_{0}.\quad a=T_{0}/\omega_{0}=4805.$ The expected sag is
$$a\cosh(\frac{500}{a})-a=26\,\text{feet.}$$
The angle at the support structure is
$$\tan^{-1}\left(\sinh\left(\frac{500}{4805}\right)\right)=5.95^{\circ}=0.104rads$$
and the maximum calculated tension at the support is
$$T=\frac{T_{0}}{\cos\theta}=4223\,\text{lbs }$$
The maximum tension is only nominally greater than the specified tension and the safety factor is
$$\text{s.f. }=\frac{4223}{25200}=0.167$$
which is approximately 6 fold.
Example 2: Bluebird conductor $(\omega_{0}=2.508$ lbs/ft, rated strength $=60300$ lbs$)$ is to be run on $220$ ft tall structures. How far apart can towers be for a nominal safety factor of $0.1$ $(10$ fold$).$
Answer: This is a pretty heavy cable and we have not been told how much sag is allowed. Since $T_{0}=\omega_{0}a$ and we want $T_{0}$ to be about $(0.1)(60300)\approx6030$ lbs to assure the safety factor, we can calculate $a=6030/2.508=2404.$ Typically we might like $\theta$ to be less than about $6^{\circ}$ at the pole, so let's start there and see what we get. $6^{\circ}=0.10472\ rads$
$$T\sin\theta=\omega_{0}s$$
$$T_{0}=T\cos\theta\quad T=6030/\cos(.10472)=6063.215$$
Thus $6063.215\cdot\sin(.10472)/2.508=s=252.7$ ft.
$$s=a\sinh\left(\frac{x}{a}\right)$$
$$x=a\sinh^{-1}\left(\frac{s}{a}\right)=252.24\ \text{ft}$$
The sag would be $h=a\cosh\left(\frac{x}{a}\right)-a=13.2\ ft.$
The distance between supports, $2x=504\,\text{ft}.$ For that tall of a tower, they would be pretty close together, but the choices are
Allow more tension and thus less safety factor.
Allow more sag, and a greater incidence angle at the support.
Choose a lighter conductor.
Example 3: In the prior example we had Bluebird conductor, and the following data:
$$\omega_{0}=2.508\,\text{lbf/ft}$$
$$a=6030\,\text{ft }$$
Despite the analysis, we are told that the spans need to be $1000$ ft apart. If we maintain the safety factor and accomodate the request with more sag, what will the sag be and what is the angle at the support.
Answer: Now we have that $x=500\ \text{ft. }$ The angle $\theta$ is just $\tan^{-1}\left(\sinh\left(\frac{x}{a}\right)\right)=11.8^{\circ}$ The sag will be
$$h=a\cosh\left(\frac{x}{a}\right)-a=52.2\:\text{ft}$$
The wire length used to span the 1000 feet will be
$$s=a\sinh\left(\frac{500}{a}\right)=1007\:\text{ft }$$
Example 4 Hang a rope between two poles. Each support is to be $45$ units above ground on the poles. The lowest point of sag must be $10$ units above ground. If the poles are $30$ units apart, what is the rope length between supports?
Answer: Since the sag point is $10$ units above ground and the support is $45$ units above ground, the sag, $h=35.$ Also, $x=30/2=15.$
$$h=a\cosh(x/a)-a$$
We can solve for $a$ with Newtons method.
$$f(a)=a\cosh\left(\frac{15}{a}\right)-a-h$$
$$f^{\prime}(a)=\frac{-x\cdot\sinh\left(\frac{15}{a}\right)}{a}+\cosh\left(\frac{15}{a}\right)-1$$
$$a_{new}=a-\frac{f(a)}{f^{\prime}(a)}\quad iterate$$
$$a=5.6287$$
Since these are even, $s=a\sinh\left(\frac{x}{a}\right)=40.237$ which is half of the length and the total length between supports is $80.47.$ This is significantly greater than the $30$ units between supports and it can be verified using the summing techniques that we used for arc length.
Figure 4: Rope hanging between two supports $30$ units apart and suspended $45$ units above ground. Note that the ground in the figure is line DB, below the $x$-axis.
Example 5 A Pelican Cable $\omega_{0}=0.517$ has rated strength of $11800$ lbf and is to be run from a hill at an attachment elevation of $965$ ft, across a small lake to a pole at an attachment elevation of $290$ ft. The lake has an elevation of $200$ ft. The line must be nominally $70$ ft above the water to permit most sailboats to pass under it. Sea level distance between the attachment points is $0.53$ miles $(3854.4$ ft$).$ What are the distances from the cable low point to the two attachment points.
Answer: Begin with a sketch.
Figure 5: Cable Over Lake. Horizontal distance between the two poles is $d=3854.4\,\text{ft.}$ Attachment $A$ has a GPS elevation of $965\,\text{ft. }$ Attachment $B$ has a GPS elevation of $290\,\text{ft.}$ The lake is at GPS elevation $200\,\text{ft.}$ The line has to be about $70\,\text{ft.}$ above the water. $x_{1}$ is the horizontal distance from the sag point to a right angle with the pole attachment at $A$.
The equations that we will use are the following.
$$h_{1}=a\cosh\left(\frac{x_{1}}{a}\right)-a \tag{P1} \label{P1}$$
$$h_{2}=a\cosh\left(\frac{x_{2}}{a}\right)-a \tag{P2} \label{P2}$$
$$x_{1}+x_{2}=d \tag{P3} \label{P3}$$
The sag in the line is the distance from the minimum line height vertically to the attachment at either $A$ or $B.$ Since the lake is at $200$ ft and we need to be $70$ ft above it, $h_{1}=965-270=695,$ and $h_{2}=290-270=20.$
Solve $\eqref{P1}$ and $\eqref{P2}$ for $x_{1}$ and $x_{2}.$
$$x_{1}=a\cosh^{-1}\left(\frac{h_{1}}{a}+1\right) \tag{P4} \label{P4}$$
$$x_{2}=a\cosh^{-1}\left(\frac{h_{2}}{a}+1\right) \tag{P5} \label{P5}$$
Substitute these into equation $\eqref{P3}.$
$$f(a)=-a\cosh^{-1}\left(\frac{h_{1}}{a}+1\right)-a\cosh^{-1}\left(\frac{h_{2}}{a}+1\right)+d$$
We will solve this numerically for $a$ with Newton's method.
$$f^{\prime}(a)=\frac{h_{1}}{a\sqrt{\frac{h_{1}}{a}}\;\sqrt{\frac{h_{1}}{a}+1}}+\frac{h_{2}}{a\sqrt{\frac{h_{2}}{a}}\;\sqrt{\frac{h_{2}}{a}+1}}-\cosh^{-1}\left(\frac{h_{1}}{a}+1\right)-\cosh^{-1}\left(\frac{h_{2}}{a}+1\right)$$
which is the same as
$$\begin{aligned}f^{\prime}(a)= & \frac{h_{1}}{\sqrt{\frac{h_{1}}{a}}\;\sqrt{\frac{h_{1}}{a}+1}\;a}\\
& +\frac{h_{2}}{\sqrt{\frac{h_{2}}{a}}\;\sqrt{\frac{h_{2}}{a}+1}\;a}\\
& -\ln\left(\sqrt{\left(\frac{h_{1}}{a}+1\right)^{2}-1}+\frac{h_{1}}{a}+1\right)\\
& -\ln\left(\sqrt{\left(\frac{h_{2}}{a}+1\right)^{2}-1}+\frac{h_{2}}{a}+1\right)
\end{aligned}
$$
so use whichever derivative you like. The iterated algorithm is
$$a_{new}=a-\frac{f(a)}{f^{\prime}(a)}\qquad iterate$$
If you need an initial guess, graph $f(a)$ and see where it crosses the $x$-axis. $a\approx7909.915.$ Once we have $a,$ it can be used in $\eqref{P4}$ and $\eqref{P5}$ to get values for $x_{1}$ and $x_{2}.$
$$x_1=a\cosh^{-1}\left(\frac{h_1}{a}+1\right)=3292.03\ \text{ft}$$
$$x_2=a\cosh^{-1}\left(\frac{h_2}{a}+1\right)=562.37\ \text{ft}$$
That solves the problem.
It should be noted that $T_{0}=a\cdot\omega_{0}=4516\,\text{lbf.}$ Further, we can compute $\theta$ at the upper pole as $\theta=\tan^{-1}\sinh\left(\frac{x_{1}}{a}\right)=0.405\,rads.$ Thus the upper pole tension is $T=T_{0}/\cos\theta=4914\,\text{lbf.}$ Therefore the safety factor here is
$$s.f.=\frac{4914}{11800}=0.416.$$
$$\frac{1}{s.f.}=2.4$$
which may be OK but should at least be questioned.
Example 6 A ship has let out $250$ ft of scope and drifted back. The anchor chain is attached $45$ ft above the water which has a depth of $18$ ft. When the chain takes the form of a catenary, how far is the ship away from its anchor, as measured along the bottom of the water.
Answer: In this problem, we are given $s=250$ and the total sag is $45+18$ with one end at that height and the other at zero. We will use $s=a\sinh(x/a)$ and $h=a\cosh(x/a)-a.$ First arrange both equations to isolate the trig terms.
$$\sinh\left(\frac{x}{a}\right)=\frac{s}{a}$$
Figure 6: Catenary along anchor chain.
$$\cosh\left(\frac{x}{a}\right)=\frac{h+a}{a}$$
Square both side of both equations.
$$\left(\sinh\left(\frac{x}{a}\right)\right)^{2}=\left(\frac{s}{a}\right)^{2}$$
$$\left(\cosh\left(\frac{x}{a}\right)\right)^{2}=\left(\frac{h+a}{a}\right)^{2}$$
Use the identity: $\cosh(\phi)-\sinh(\phi)=1.$
$$\left(\frac{h+a}{a}\right)^{2}-\left(\frac{s}{a}\right)^{2}=1$$
Solve for $a.$
$$a=\frac{s^{2}-h^{2}}{2h}=\frac{250^{2}-63^{2}}{2\cdot63}=464.53$$
Now substitute back into $s=a\sinh(x/a)$ to solve for $x.$
$$x=a\sinh^{-1}\left(\frac{s}{a}\right)=239.28$$
