Talk:Catenary/Archive 1

Archive 1

Exponential form

The cosh mathematical form of the catenery is completely correct but the cosh function itslef can be intimidating to anyone who hasn't done some college (or late highschool) maths. I suggest also putting up the exponential form of the catenary, namely: y = a*0.5*(exp(x/a)+exp(-x/a)).

As you can see from above, I haven't learn't how to write pretty maths notation on wiki yet, could someone give me some pointers please.--Commander Keane 15:51, 6 Feb 2005 (UTC)

cords

is there a difference between a real cord and a massless cord? - Omegatron 23:19, Apr 15, 2005 (UTC)

Real cords (the kind a person can touch with his hands) have mass. But why do you ask here? This article doesn't even use the words "cord" or "mass" (or "massless"). --DavidCary 06:19, 17 November 2005 (UTC)

Ah. Yeah, I guess that makes sense. I was thinking "maybe a massless cord would form a parabola instead of a catenary", but a massless cord wouldn't form anything, because it would have no inertia or be pulled by gravity. Nevermind. :-) — Omegatron 16:18, 28 November 2005 (UTC)

(Much later) I realize you may be asking if there is a difference between the 2 shapes, the shape a real cord makes and the shape a (theoretical) massless, perfectly flexible, cord makes in a constant gravitational field.

• The equations assume *some* mass. But as long as the cord is perfectly flexible and has constant mass per unit length, it makes no difference how much mass -- you get the same catenary curve, even in the limit all the way down to 0+. ("negative mass", such as a cord supported by balloons, gives a similar catenary curve that hangs *up* above its supports instead of hanging down, but the same hanging-up catenary, all the way to 0-. I suppose an exactly zero mass cord could float in any arbitrary shape you could imagine.)
• Very stiff ropes tend to make curves that are almost a perfect circle.
• Cables that support constant mass per *horizontal* distance (such as suspension bridges, and also cables that stretch a lot under tension) make curves that are perfect parabolas.
• I suppose real cables that have some stiffness, and support some not-exactly-constant mass-per-unit-length make some shape that is intermediate between the catenary, the circle, and the parabola.
• (For completeness in a list of "shapes strings make", we should mention the set of shapes a plucked string can make in a string instrument).

--DavidCary 02:20, 29 November 2005 (UTC)

Can't a plucked string make ANY shape with fixed endpoints, by Fourier's theorem?--JB Gnome (talk) 03:10, 9 October 2008 (UTC)

The chain is almost vertical near the points of suspension

Why? If you take the derivative of y = a * cosh(1/a) w.r.t. x, you get y'=sinh (x/a). This does not tend to plus/minus infinity (i.e. vertical tangent) at any finite x. Compare this with a physical situation - an empty clothes line. It's supported at two points, sags under its own weight - but by visual inspection, you see (unless it's *very* loose) that the ends are really far from vertical. Mikez 00:03, 19 November 2005 (UTC)

possible copyright issue

Some of the text is almost exactly the same as http://mathworld.wolfram.com/Catenary.html so maybe somebody can check this.

Parametric formula?

Is there a parametric forumula for this curve? -SharkD 04:14, 9 November 2006 (UTC)

Latin anagram of "abcccddeeeeeefggiiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux" ???

Can someone fix this? I'm not sure what it's supposed to be. Pleasantville (talk) 22:56, 13 December 2007 (UTC)

It was added in this diff by User:Nbarth, you might want to ask him. It seems more than a little odd. —Cronholm¹⁴⁴ 23:20, 13 December 2007 (UTC)

Fixed (I think). --Pleasantville (talk) 23:33, 13 December 2007 (UTC)

Hi Pleasantville and Cronholme144; I made the edit, which I grant is unusual. I think the contribution of Hooke to the catenary and its modern engineering applications is important to highlight.

As discussed and extensively referenced in the previous revision, the use of the catenary in construction was rediscovered by Hooke, who published the law originally as a Latin anagram, as illustrated here: [1].

He did the same thing with Hooke's law; eccentric by modern standards, but it reflects historical concerns about priority (Hooke was very concerned with this issue).

The original anagram should probably be relegated to a footnote (it's encyclopedic in that it's the original statement, but is quite distracting if it's so prominent).

I suggest we restore the content, and put the anagram in a footnote (so state the law, and put <ref>original acronym ...</ref>).

How does this sound?

Nbarth (talk) 00:17, 14 December 2007 (UTC)

I have no strong opinions about how it is presented. I was just confused. --Pleasantville (talk) 00:43, 14 December 2007 (UTC)

Ok, done. The fact that it's confusing and looks like vandalism is worth noting and clarifying—thanks!

Nbarth (talk) 00:51, 14 December 2007 (UTC)

You're already done... Oh well, I agree with the edit. Cheers all—Cronholm¹⁴⁴ 00:57, 14 December 2007 (UTC)

I found the anagram fascinating.

My high-school maths went far enough for me to remember being set the problem "derive the equation for a catenary", but I never found or saw a solution and I've been (mildly) troubled by this on-and-off ever since. My high-school Latin still allows me to make some sense of the (unravelled) anagram, though, and I come to this now because I want to build an arch. Strange how these things link up (no pun intended) in the end.

Thx

Mangodog (talk) 19:20, 16 July 2008 (UTC)

The Catenary Final Equation

I questioned my calculus teacher about this, and she believes that the equation in this article is wrong. She believes that the exponential values should be e^(x/b) and e^(-x/b). Having a different value for those exponentials causes the graph to have the same y-intercept, but a steeper slope (if the value of 'b' is higher than that of 'a', which also means a less steeper slope if the value of 'b' is smaller than that of 'a') Can this be cleared up for me? I'm quite curious as to which equation is correct. —Preceding unsigned comment added by 121.215.131.93 (talk) 09:03, 2 September 2008 (UTC)

The function a cosh(x/b) would do what you claim but it would not be the curve followed by a chain. The parameter a in a cosh(x/a) determines the "steepness" of the curve and the y-intercept can be changed by translating the curve vertically, in other words by adding a constant. You can stretch a parabola vertically or horizontally by a constant factor and still get a parabola but the same does not apply to a catenary; your function would correspond to a stretch by b/a. Hope that helps.--RDBury (talk) 03:08, 15 November 2008 (UTC)

this was disproved by Euler (1707-1783) in a work published in 1669.

I hope I don't have to say anything more. 24.238.113.229 (talk) 22:49, 6 January 2009 (UTC)

I found the correct information here (last paragraph on the page) and have now changed it. --Kri (talk) 00:28, 7 January 2009 (UTC)

catenary formulae

To the author of the catenary article:

Please use Microsoft Word 2007 to create the equations for this article and save the equations as .PNG images so that they can be more readable. Any equivalent equation writer should suffice.

Thanks,

Beekeeper596 (talk) 00:47, 2 April 2009 (UTC)

I am having no trouble reading the equations -- in fact they look quite nice. You can play with your LaTeX rendering preferences and see if they look better to you. Any specific issues? Baccyak4H (Yak!) 03:30, 2 April 2009 (UTC)

Please do not use anything other than LaTex or HTML for equations. Ever. Phancy Physicist (talk) 05:31, 4 August 2010 (UTC)

Removed image of Eifel's bridge

I removed the image of Eifel's truss arch bridge. It seems to me that an arch that's being used to support something else would follow a parabola, for the same reason that a cable used in a suspension bridge follows a parabola. Only an arch which is mainly holding up its own weight would follow a catenary. Even if I'm wrong about this, there should be a reference given that the curve of the arch in Eifel's bridge is a catenary.--RDBury (talk) 17:14, 19 June 2009 (UTC)

Reordering sections

I'm trying to put the less technical sections at the top of the article. This is going on the assumption that the average reader will be more impressed by pictures than a lot of mathematical formulas. The goal is to have a comprehensive article suitable for a general audience, followed by the mathematical material.--RDBury (talk) 20:03, 21 June 2009 (UTC)

Simplifying derivation

I'm trying to simplify the derivation section and generally make it more readable. I think it's still worth while to have two alternative ways of solving the equations since both methods are useful in solving different variations of the basic problem. I was especially able to reduce the size of the second alternative by plugging in well known formulas for arc length and using a table to solve the integrals.--RDBury (talk) 02:57, 2 August 2009 (UTC)

Archive 1