Talk:Catenary

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Please add useful comments here--Cronholm144 09:47, 23 May 2007 (UTC)

1 Gaudi, derivation by forces
2 Exponential form
3 cords
4 The chain is almost vertical near the points of suspension
5 Gaudi's Casa Mila
6 possible copyright issue
7 Parametric formula?
8 Towed cables - clarification needed
9 Physics of the Catenary
- 9.1 Explanation
10 Latin anagram of "abcccddeeeeeefggiiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux" ???

[edit] Gaudi, derivation by forces

This problem has been open for quite a well, but I checked my father's guide book from when we visited Barcelona (there's a fair section on Gaudi), and there is a very famous story of him hanging ropes and measuring the distances to produce the curves shown. As you probably know, that produces a catenary.

Derivation by forces: Suppose A to be the gravitation acceleration vector, λ the linear density, T to be the tension, φ to be the tangent angle, and s to be the arc length. Then

$-A\lambda=\frac d{ds}(\cos\phi,\sin\phi)T$

Assuming Aλ=(0,-1) we get

$(0,1)=(\cos\phi,\sin\phi)\frac{dT}{ds}+(-\sin\phi,\cos\phi)T\frac{d\phi}{ds}$

$\frac{dT}{ds}=\begin{vmatrix}0&-T\sin\phi\\ 1&T\cos\phi\end{vmatrix}/ \begin{vmatrix}\cos\phi&-T\sin\phi\\ \sin\phi&T\cos\phi\end{vmatrix}=\sin\phi$

$\frac{d\phi}{ds}=\begin{vmatrix}\cos\phi&0\\ \sin\phi&1\end{vmatrix}/ \begin{vmatrix}\cos\phi&-T\sin\phi\\ \sin\phi&T\cos\phi\end{vmatrix}=\frac{\cos\phi}T$

$\frac{dT}{d\phi}=\frac{dT/ds}{d\phi/ds}=T\tan\phi$

k T = secφ

(k arbitrary)

$\frac{d\phi}{ds}=k\cos^2\phi$

tanφ = C + k s

(C arbitrary)

which is a known Whewell equation. (Got a suggestion how to bridge that? I think it'll convert into the y' DE that follows...)

PS: that last eq'n is sort of obvious; one can handwave that the horizontal component of tension is constant and the vertical obviously has to equal the weight which is proportional to arc length. And that's probably how Bubbaloo first did it. But it's nice to know the top, general eq'n (eg if you're doing the skipping rope sometimes used for vertical wind turbines, or the constant-strain catenary...It'd've saved me hours on another tinker, a chain hanging in a plane rotating on a vertical axis.) 142.177.169.142 03:37, 18 Aug 2004 (UTC)

Duh. Converting that to (x,y) is easy. Roughly:

cos 2 φ = 1 / (1 + tan 2 φ)

$dx/ds=1/\sqrt{1+s^2}$

x = arcsinh s

sin 2 φ = tan 2 φ / (1 + tan 2 φ)

$dy/ds=s/\sqrt{1+s^2}$ $y=\sqrt{1+s^2}$

$y=\sqrt{1+\sinh^2x}=\cosh x$

142.177.169.142 04:42, 18 Aug 2004 (UTC)

Derivation by minimal energy: Minimise $\int U\lambda ds$ where U is the gravitational potential, λ the linear density, and s the arc length. Assume Uλ=y, let ' denote d/dx, and change variables

minimise $\int y\sqrt{1+y'^2}dx$

The Euler characteristic $\frac{\partial I}{\partial y}=\frac d{dx}\frac{\partial I}{\partial y'}$ gives

$\sqrt{1+y'^2}=\frac d{dx}\frac{yy'}{\sqrt{1+y'^2}}$

$1+y'^2=y'^2+yy''-\frac{yy'^2y''}{1+y'^2}$

1 + y' 2 = y y''

which is satisfied by $y = cosh x$ (Got a suggestion how to bridge that magic?) 142.177.169.142 23:26, 17 Aug 2004 (UTC)

S'pose one could do the usual guessing until

y = a e α x + b e β x

comes up. Then

α + β = 0

and

a b (α - β) 2 = 1

which gives

y = cosh(α x + ln(2 a α)) / α

. Vertical translations require adding a constant to the potential (!). 142.177.24.163 14:19, 18 Aug 2004 (UTC)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] Gaudi's Casa Mila

Gaudi's arches are described in this article as catenaries--which may well be true. However, they are also used (in fact, an identical photograph of them is used) in the article entitled "Parabola" sn an example of THAT shape, which unfortunately means that one of these claims must be wrong. Anybody know the answer? (I'm leaving basically this exact post on the discussion page of that other article, in the hope that someone will more likely come across this issue and clear it up.) --Buck

The article on Gaudi states he was fascinated by paraboloids and parabolic arches. That article also identifies the arches as parabolic. It would seem that the arches probably are parabolic. Either that or that article should be disputed as well.

I've been to Barcelona and seen the stringed contraptions he used to design the buildings. Since they are always three dimentionally curved I would presume that none of the arches are exact catenaries or exact parabolas. Maybe we should precede each mention with "approximate"

The photo mentioned below doesn't appear in the Parabola article anymore. I guess thats a good thing . . . Lansey (talk) 23:44, 24 November 2007 (UTC)

The foto used in the Parabloa article is named "parabola", and in the description is said to be describing catenarys. The confusion seems to be total: anybody has access to a real book about Gaudi?Mossig 16:12, 30 November 2006 (UTC)

[edit] 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.

[edit] Parametric formula?

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

[edit] Towed cables - clarification needed

The section about towed cables could perhaps become a separate article.

The subject of the section must be properly introduced. Does it talk about a cable being towed from one end? From both ends? A cable that floats on a water surface and is towed from one end, assumes a straight line. A cable hanging down from a boat in motion, or from a tower in wind, is a different scenario. Please tell the reader what she should have in mind when reading.

What is "the incident angle"? The angle between the tension forces and the drag forces at a given point of the cable? The direction of a water or air flow with respect to the orientation of a section of the cable? Without a stated scenario, it is hard to know. It should be said explicitly if we are talkning about quantities at a point on the cable, or about quantities defined for the whole cable.

So a critical angle is one that does not change? Does not change along the cable? Does not change with time? I do not understand what this is. Under what conditions does this fenomenon arise? "Far from significant point forces" does not cut it for me. Most hanging cables are subject to point forces only at the ends, towed or not. What angle does not change for most of the length of such cables? What shape does a cable assume when it hangs in steady wind with one free end? A straight line? A straight line only if the the ratio of drag forces to weight satisfies some condition? This is not obvious.

g=9.81m/s^2 - sounds like the accelleration of gravity?

And 'a', what does this letter stand for in the equations? Accelleration? So the equation is a dynamic one? Perhaps I could infer that from the equation for dT/ds, since this becomes zero if 'a' is zero. But then I must first figure out what 's' is, which is the next question. But even if 's' is 'slope', is it reasonable that dT/ds be zero in absense of accelleration? I don't know.

"'s' is the cable scope". Slope, perhaps?

It would probably be a good idea to state explicity what coordinate system is being used. The x axis horizontal, the y axis vertical, and the cable lying in the x-y plane? Gravity acting in the negative y direction, etc. What about a cable suspended horizontally and subject to a horizontal wind or water flow perpendicular to the line through the ends? Such a cable would lie in a non-vertical plane.

This brings up the question, if 's' is 'slope', what slope? Assume a coordinate system O-xyz, with z pointing up, and the ends having y=0. The midle of the cable can have non-zero 'y' coordinates. $dz/\sqrt{dx^2+dy^2}$ is one possible slope.

Regards, PerezTerron 21:11, 14 December 2006 (UTC)

[edit] Physics of the Catenary

Maybe someone who understands it could explain how $a \cdot {\cosh \left ({x \over a} \right )}$ comes from the mass of a hanging chain. The less mass causing less of a curve makes sense, but the derivation of the expression would be nice.

Dmbrown00 03:40, 15 December 2006 (UTC)

[edit] Explanation

It depends on what maths you want to use. You can't avoid calculus on this, and usually it is done via undergraduate ordinary differential equations. However, it is possible to do it just using standard integral calculus, with a bit of knowledge / intuition of how parametric curves work. (This intuition is usually informed by physics, so we are OK here.) Here goes:

Assumptions: Suppose the chain has linear density ρ and that the tension at the lowest point is τ.
Parameterisation: Suppose curve is described parametrically as (x(t),y(t)), where t is the distance from the lowest point. Since t is the distance along the curve (i.e. (x(t),y(t)) moves along the curve with "constant speed" wrt t) we have that (x'(t),y'(t)) is a unit vector. So we have:
- (x'(t))² + (y'(t))² = 1
- (x'(t),y'(t)) is a tangent vector to the curve. No harm in assuming x'(t)≥0 and y'(t)≥0.
Tension: Between the lowest point, where the tension vector is (τ,0), and another point (x(t),y(t)) is chain of length t, mass ρt, which is subject to vertical gravitational force ρgt. By resolving three vectors,
- the tension vector at (x(t),y(t)) is (τ,ρgt)
Flexibility: since the chain is perfectly flexible at (x(t),y(t)), the tension vector (τ,ρgt) is aligned with tangent vector (x'(t),y'(t)). Hence
- y'(t) / x'(t) = ρgt / τ, so y'(t) = t x'(t) / a
- Here a = τ / ρg is a constant, which you can interpret as some sort of relative horizontal tension.
Combining equations: 1 = (x'(t))² + (y'(t))² = (x'(t))² + ( t x'(t) / a )² = ( 1 + t² / a² ) (x'(t))².
Solve for x'(t) and y'(t):
- x'(t) = 1 / ( 1 + t² / a² )^1/2
- y'(t) = t x'(t) / a = t / ( a² + t² )^1/2
Integrate to find x(t) and y(t):
- x(t) = a sinh^-1(t/a) + C. (From a standard table of integrals.) Assuming x(0)=0 gives C=0.
- y(t) = ( a² + t² )^1/2 + K. (Using integration by substitution.) Since the height y(0) is arbitrary, we may as well use K=0.
Eliminate t: t = a sinh( x / a ) and so y = a cosh ( x / a )

Voila! Andrew Kepert 10:30, 4 November 2007 (UTC)

[edit] 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)