11
$\begingroup$

Notice that the parabola, defined by certain properties, is also the trajectory of a cannon ball. Does the same sort of thing hold for the catenary? That is, is the catenary, defined by certain properties, also the trajectory of something?

$\endgroup$
  • 1
    $\begingroup$ If you hold a necklace from its end (and don't hold it tight), gravity will make you see an upside down "parabola" which is really a catenary. In terms of buildings, the St Louis Gateway arch is an upside down catenary. The properties in physics tell us that such a building must be a catenary in order to be structurally sound $\endgroup$ – imranfat Apr 7 '16 at 2:10
  • 3
    $\begingroup$ The name comes from the Latin word for chain, as it is the shape taken by a hanging chain $\endgroup$ – Will Jagy Apr 7 '16 at 2:12
  • 5
    $\begingroup$ The parabola being the trajectory of something isn't specific to cannonballs - it's the trajectory of anything affected by a constant force. (In this case, gravity.) $\endgroup$ – Deusovi Apr 7 '16 at 8:20
  • 3
    $\begingroup$ Sure. I just drew a catenary on a sheet of paper, and so it was the trajectory of the tip of my pen. You'll have to be more specific to get an interesting answer. In any case, this is a physics question, not a math question. $\endgroup$ – Najib Idrissi Apr 7 '16 at 9:49
  • $\begingroup$ trajectory of canonball only within a very simplified physics, real trajectory with gravity varying respective height is not parabola, they make it so simple so they can teach a solution to dumb dumbs. $\endgroup$ – Arjang Jun 8 '16 at 16:43
18
$\begingroup$

From the right perspective, maybe.

enter image description here

(image from Wikipedia)

I'm not exactly sure how to frame this as a trajectory problem, but certainly there is stuff moving and a catenary is traced!

We have a square moving horizontally at a constant speed, and rotating at "the right" constant angular velocity (I'm not certain the angular velocity is fixed, but I suspect it is). Throughout a given quarter rotation starting with a vertex of the square at the bottom, the point directly below the radius will trace out an inverted catenary.

$\endgroup$
  • 1
    $\begingroup$ Wagon and Hall, in their paper, show that a straight line "wheel" can "roll" (in a sense defined in the paper) on a catenary "road"; the polygonal case is a slight modification of this. Have a look at their paper if you're interested. Nevertheless, I do not think it is an answer to the question, as the catenary is not really the locus here. $\endgroup$ – J. M. is a poor mathematician Apr 7 '16 at 9:27
  • $\begingroup$ I'm happy to take "trajectory" in a wide sense, such as, in your answer, which I am going to accept, a road giving rise to a smooth ride. I really like J.M.'s answer too, but yours is superior in the sense that it is more tied to physical reality (cannon balls, roads, etc.), whereas J.M.'s is purely a geometrical result. $\endgroup$ – user27325 Apr 7 '16 at 14:03
  • 1
    $\begingroup$ @J.M. I agree that it's not the trajectory of a single point, so on shaky grounds. But i did think it physically interesting enough to be worth mentioning. That paper is really neat, thank you! I've seen things about wheel and road shapes, but never a solid reference. $\endgroup$ – pjs36 Apr 7 '16 at 15:45
  • 3
    $\begingroup$ Then you'll want to see Stan Wagon actually riding his square-wheeled bicycle. ;) $\endgroup$ – J. M. is a poor mathematician Apr 7 '16 at 21:06
  • 1
    $\begingroup$ At the Museum of Mathematics in NYC, there is a bicycle with square wheels and an inverted catenary track on which to ride. I highly recommend. $\endgroup$ – Ron Gordon Apr 10 '16 at 16:06
15
$\begingroup$

As I've shown in a previous answer, the focus of a parabola rolling on a straight line traces a catenary. Similarly, the directrix of the same rolling parabola will envelope another catenary, a reflection of the one being traced by the focus.

Here is a modern (as in done with the current version of Mathematica) version of the cartoon I did for that previous answer:

thus a parabola rolls

$\endgroup$
  • $\begingroup$ I really like your answer, but I accepted the answer of pjs36 over yours because that of pjs36 is more tied to physical reality (cannon balls, roads, etc.), whereas your answer is purely a geometrical result. I have upvoted your answer, however. $\endgroup$ – user27325 Apr 7 '16 at 14:08
  • 4
    $\begingroup$ That's certainly your choice to make, but then "trajectory" does not mean what you imply it means. $\endgroup$ – J. M. is a poor mathematician Apr 7 '16 at 14:10
14
$\begingroup$

A freely suspended chain or string forms a catenary.

$\endgroup$
  • 3
    $\begingroup$ Hence, the name. $\endgroup$ – orion Apr 7 '16 at 9:38
4
$\begingroup$

Neglecting air resistance, and assuming constant gravity, the trajectory of anything will be a parabola. If there is air resistance, trajectories of roughly spherical objects become a lot more complicated, and are not described easily using nice geometric terms. However, the trajectories do involve the hyperbolic cosine function, which traces out the catenary curve. See this page for details. In spherical classical gravity, trajectories are conic sections, especially hyperbolas and ellipses.

It may be possible to get a true catenary if you had an object with strange aerodynamic properties, or with a very precise arrangement of objects forming a gravitational field. But in either case, the catenary trajectory would be entirely contrived.

$\endgroup$
  • $\begingroup$ Yes, so the constant gravity case you mention first is really a limiting case of the "spherical classical gravity" case you mention later. The latter must be used when we cannot neglect the fact that the Earth is not infinite in extent. So the parabola of the constant gravity approximation is really one "end" of an ellipsis from the more general classical case. However, none of this, not your answer, gives examples where the catenary is relevant. $\endgroup$ – Jeppe Stig Nielsen Apr 7 '16 at 8:32
  • 3
    $\begingroup$ @JeppeStigNielsen However, none of this, not your answer, gives examples where the catenary is relevant. I think that is the main point of Alex's answer: the catenary cannot arise as a trajectory without contrivance (e.g. a very particular air resistance), because it is not a conic section. This seems to me to be the correct answer to the OP's question. $\endgroup$ – atkins Apr 7 '16 at 9:11
  • $\begingroup$ "A failure of imagination is not an insight into necessity." -- Patrick Barrow $\endgroup$ – user27325 Apr 7 '16 at 13:55
  • 2
    $\begingroup$ @EsperantoSpeaker1 Very true. In my case, the failure is not of imagination, but rather ability to solve the particular nonlinear differential equations involved in aerodynamics and classical gravity. $\endgroup$ – Alex S Apr 9 '16 at 5:14
0
$\begingroup$

The relativistic trajectory of an object under the influence of a constant force field perpendicular to its initial direction of motion is a catenary. The trajectory reduces to a parabola in the non-relativistic limit.

(Eg : motion of electron under constant electric field)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy