## Catenary versus parabolic curves: math on the fly

I’ve been tied up this week finishing out the semester as well as assisting with some calculations regarding some equipment to be used in the cleanup efforts in the Gulf of Mexico.  In relation to those efforts, I spent some time studying catenary and parabolic curves.  Mind you, I was doing this in “crisis mode” at a manufacturing site without many of my reference books and, ultimately, without reliable Internet access.

Specifically I was looking at these from the point-of-view of load calculations and was interested in trying to predict the behavior theoretically (of course, the eventual real-world test didn’t go as planned and my friend’s boat nearly sank because he forgot to put the drain plugs back in after the winter, but, hey…).

So considering some flexible object (like a cable) hanging from two secured points, not necessarily at equal height (e.g. a tow rope connecting a boat to a parasail).  Mechanically, the general deflection curve for something like this is given by

$y=\frac{1}{H}\int \left[\int w(x) dx\right] dx + C_{1}x + C_{2}$

where H is the horizontal component of the tensile force in the deflected object and w(x) is a loading function given per unit length.  The constants of integration must be determined from boundary conditions.

Suppose we now consider a cable (like a telephone wire) connected at the same height and subjected to a uniform load (e.g. a line of birds sitting on it).   The loading function in this case is a constant, the constants of integration can be made to be zero, and the deflection curve turns out to be

$y=\frac{w}{2H}x^2$

which is a parabolic curve.  If we assume the greatest sag h occurs in the middle, we set $y=h$ for $x=l/2$ where $l$ is the straight-line distance between the two connection points (i.e. it’s a chord), and thus

$h=\frac{w}{2H}\frac{l^2}{4}$.

We may use this to solve for H and substituting back in we can find the deflection curve entirely in terms of h and l which are both easily measurable:

$y=4\frac{hx^2}{l^2}$.

On the other hand, if we consider a cable loaded only by its own weight, we get a deflection curve

$y=\frac{H}{w}\left(\textrm{cosh}\frac{w}{H}x-1\right)$.

This is actually a catenary curve and not a parabola!  To solve for H in this instance, we again set $y=h$ for $x=l/2$ to get

$h=\frac{H}{w}\left(\textrm{cosh}\frac{wl}{2H}-1\right)$.

Unfortunately this is not easy to solve (many mechanics texts simply advise using trial and error on this).  Obviously, then, in practice the parabolic curve is much easier to work with.  So, given the parameters of the problem under which I was working, I set out to figure out where the two diverged.  Using values for w and H from a known model, I overlaid plots of the two curves and found that they began to diverge roughly at a ratio of $y/x \approx 0.46$.  This meant that as long as the ratio $h/l \le 0.23$, I could use a parabola instead of the catenary.

In my real-world model, I knew the value of l.  Solving for h from this again required trial and error (at least given that I was doing this in crisis mode on the fly without any of my reference books).  So I used a series approximation for a sine function that popped up in the equation and after a few more calculations found that for my real-world application, $h/l \approx 0.154$.  Thus I was safe modeling it as a parabola which meant finding other parameters would be much easier!

And that’s “math on the fly,” as it were…

### 17 Responses to “Catenary versus parabolic curves: math on the fly”

1. You could use a Taylor series to approximate cosh.

2. quantummoxie Says:

Actually, I did that in one instance – I ran it out about three or four terms – but it still made it difficult to solve in the field (meaning out on the water with diesel exhaust blowing in one’s face, etc.).

3. […] Catenary versus parabolic curves: math on the fly « Quantum Moxie […]

4. BlackGriffen Says:

It’s easier to solve than you think – you just need to use the hyperbolic equivalent of the $\latex \sin^2(x) = \frac{1 – \cos(2x)}{2}$. That is: $2 \sinh^2\left(\frac{x}{2}\right) = \cosh(x) - 1$. Thus if $\latex z = \frac{wl}{2H}$ then $z = \frac{l}{h} \sinh^2(z/2)$. It doesn’t look like we’ve made much progress, but the form of this equation is amenable to my favorite technique for solving such equations because it’s so brain-dead simple – recursion. Whether or not it will converge to the desired answer depends on the slope of the function at the intersection with $y=x$, ie the slope of the recursively evaluated function must be less than 1 at the desired solution or recursion will cause divergence instead of convergence (the linear approximation suffices to show this). That sounds like a chicken and the egg problem, but in this case there is no problem because if repeated application of $f(x) = \frac{l}{h] \sinh^2(x/2)$ causes the answer to diverge or converge to zero, we can easily invert the function to go the other way (that was actually the main reason for the trig substitution).

Thanks for helping out down in the gulf, by the way.

5. Great article, but i suggest a change in the colors of the background and the text. Actual selection is very difficult to read.

6. quantummoxie Says:

Yeah, I’m waiting for WordPress to come up with something better. Even if I had a light background, the equations still appear on the dark background.

@Sean: try thinking that out while on a boat that is sinking (my friend forgot to put the drain plugs back in his boat after the winter!) 🙂 On the other hand, thanks for the pointer since I might turn this into a paper (what the hell, why not?).

• BlackGriffen Says:

It’s actually the first thing I thought of because I had to solve this problem before for a class I TAed. That and the 1-cosine to sine squared trick is one I’ve learned pops up all the time when trying to compute small arc lengths accurately (ie the Haversine formula).

• quantummoxie Says:

Ah! I haven’t dealt with this stuff in ages so I’d forgotten all the tricks.

7. BlackGriffen Says:

It’s actually the first thing I thought of because I had to solve this problem before for a class I TAed (freshman honors mechanics, if you can believe it). The 1-cosine to sine squared identity comes up all the time in spherical trig when you want to calculate the arc length between two points on a sphere that are close to each other accurately.

8. BlackGriffen Says:

Sorry for the duplicate reply – ran into some problems with the submission.

9. quantummoxie Says:

No problemo.

10. I had this situation years ago. This engineer had designed a concrete box culvert replacement project and he made an error in his calcs and the culvert was set too low which nullified all his fill requirements for the road project (culvert was set at a low point in the road where a stream passed).

I was the construction inspector and when the error was noted I called the engineer and told him we needed new Item fill numbers for the bottom of the stream because the culvert was going in too deep. He told me I “must be wrong” (Hey.. a Gurl can’t POSSIBLY know) and to set the culvert boxes anyway.

After the contractor set the culvert they went home for the day and the Engineer Ilicensed too!) came out to the site (I called him again) and he was aghast at the depth (this was going to mess up the hydraulic opening too.. but there it was and the Crane was gone so we had what we had.

So he gets out his calculator to refigure the fill over the culvert to meet the vertical curve he had calculated for the center of the road. I did not mind.. I was into over time and time and a half always feels good.

He was obviously struggling so I said, “We can do this that way OR we can peg a line at the beginning of the fill and peg the line at the other end of the fill length and pull it tight and measure from the line to the existing for Center line and subtract 3 inches for the edges ()to make a road crown) and calculate the fill.” I then told him, “I know it is not a TRUE vertical curve as we are now dealing with a catenary.. but it will be close and if we do this smoothly it will work as the speed of the cars here is limited to 45MPH.”

He was red faced and said, “NO!! I can figure this..” So I shut up and he fiddled with his calculator another hour (the deepest part of the curve was offset from the ends so things were not symetrcal to the center).

He eventually looked up and growled at me, “Where are the PK nails? Get the line…” and we set up the catenary and in about 45 minutes had staked and marked or painted the center line fill on stations and subtracted the 3inches for edge of road to make a crown and were done.

The point of this is that the difference in deflection for a parabola and a catenary can be minor enough in some situations to allow for the one to be substituted for another.

I reiterated this story today to co workers and then decided to check out the math again and came upon this and postings and thought you all might be interested in how a catenary was used to fix a parabolic problem.

11. quantummoxie Says:

Cool! Thanks for the post! Sounds like that engineer was a whopper of a jerk. Oh well. There’s nothing like pissing someone off by being right. 🙂

12. I am pretty “old” compared to the posters here (in my 50’s) and while working in that job I went back to school (after.. oh… 30 years…) and retook calculus and chemistry (and then went past where I had before in college as I did not need this for the degree I hold).

I found great pleasure in returning to mathematics with an application. I am pleased to have found this site and this discussion. I am pretty old school in that I do not much enjoy putting math thru a computer program (tho I did use a Beam Design software for Steel Bridge beams once). I like to “feel” what the math tells me and the ‘puter can get in the way of that (spend so much time making the ‘puter do the thing I lose the math). I am not the sharpest tack in the shed, but I try to maintain a reasonable edge.

Sadly I no longer work for the County and no longer design bridges/roads (under a PE) (needed a job that paid more and this job is largely economics, not applied engineering math).

I do work as a math checker for an ME/PE who does Boiler recertification for Steam Locomotives and sometimes I have to rework something if I find a mistake. Part of the formulation (based on UT of the boiler, crown sheet etc. and a 4 safety factor) involves use of the Quadratic Formula. when I get to that part I always smile like you do meeting an old friend.. something I never thought would be possible when I learned it in High School.

I will keep coming back to this site to see what all you young whipper snappers are up to. This Old dog is up to learning a few new tricks. Maybe. 🙂

13. quantummoxie Says:

LOL, young whipper-snappers? 🙂 I know for a fact that at least one commenter on here (other than yourself) is over 50. And these days, that’s young. I play softball and most of the folks on my team are over 50. At 36, I’m one of the younger ones. As my grandmother once said, you’re only as old as you feel. She lived to 97 and spent the last 29 years of her life living alone (after my grandfather died) – never went into a nursing home.

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