Talk:Catenary

Archive 1
Old posts are archived here. Phancy Physicist (talk) 05:05, 4 August 2010 (UTC)

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, &lambda; the linear density, T to be the tension, &phi; to be the tangent angle, and s to be the arc length. Then
 * $$-A\lambda=\frac d{ds}(\cos\phi,\sin\phi)T$$

Assuming A&lambda;=(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$$
 * $$kT=\sec\phi$$ (k arbitrary)
 * $$\frac{d\phi}{ds}=k\cos^2\phi$$
 * $$\tan\phi=C+ks$$ (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:
 * $$\cos^2\phi=1/(1+\tan^2\phi)$$  $$dx/ds=1/\sqrt{1+s^2}$$   $$x=\mbox{arcsinh}s$$
 * $$\sin^2\phi=\tan^2\phi/(1+\tan^2\phi)$$  $$dy/ds=s/\sqrt{1+s^2}$$   $$y=\sqrt{1+s^2}$$
 * $$y=\sqrt{1+\sinh^2x}=\cosh x$$
 * 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, &lambda; the linear density, and s the arc length. Assume U&lambda;=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
 * $$\sqrt{1+y'^2}=\frac d{dx}\frac{yy'}{\sqrt{1+y'^2}}$$
 * $$1+y'^2=y'^2+yy-\frac{yy'^2y}{1+y'^2}$$
 * $$1+y'^2=yy''$$

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 $$y=ae^{\alpha x}+be^{\beta x}$$ comes up. Then $$\alpha+\beta=0$$ and $$ab(\alpha-\beta)^2=1$$ which gives $$y=\cosh(\alpha x+\ln(2a\alpha))/\alpha$$. Vertical translations require adding a constant to the potential (!). 142.177.24.163 14:19, 18 Aug 2004 (UTC)

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

Read the first paragraph again for the first time: The author asserts that the parabola and the catenary look similar but have different mathematical formulas.StevenTorrey (talk) 17:46, 5 September 2011 (UTC)

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)


 * I found a reference that the arches are centenaries and added it to the image caption.--RDBury (talk) 20:07, 5 August 2009 (UTC)

Did you mean 'centenaries' or 'catenaries'? In the dictionary, I find no definition of 'centenary' related to suspension cables. Please, let's not confuse people more than necessary! The mistake can be corrected by the author.StevenTorrey (talk) 17:46, 5 September 2011 (UTC)
 * I meant 'catenaries', slight case of dyslexia.--RDBury (talk) 18:39, 5 September 2011 (UTC)

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)

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)

If the mass is uniformly distributed then it doesn't affect the shape. A light string and a massive chain will follow the same curve if they are the same length and assumed to have no stretch and no stiffness. Nophead (talk) 14:32, 16 September 2020 (UTC)

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: Voila! Andrew Kepert 10:30, 4 November 2007 (UTC)
 * 1) Assumptions: Suppose the chain has linear density ρ and that the tension at the lowest point is τ.
 * 2) 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:
 * 3) *(x'(t))2 + (y'(t))2 = 1
 * 4) *(x'(t),y'(t)) is a tangent vector to the curve. No harm in assuming x'(t)≥0 and y'(t)≥0.
 * 5) 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,
 * 6) * the tension vector at (x(t),y(t)) is (τ,ρgt)
 * 7) 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
 * 8) * y'(t) / x'(t) = ρgt / τ, so y'(t) = t x'(t) / a
 * 9) * Here a = τ / ρg is a constant, which you can interpret as some sort of relative horizontal tension.
 * 10) Combining equations: 1 = (x'(t))2 + (y'(t))2 = (x'(t))2 + ( t x'(t) / a )2 = ( 1 + t2 / a2 ) (x'(t))2.
 * 11) Solve for x'(t) and y'(t):
 * 12) * x'(t) = 1 / ( 1 + t2 / a2 )1/2
 * 13) * y'(t) = t x'(t) / a = t / ( a2 + t2 )1/2
 * 14) Integrate to find x(t) and y(t):
 * 15) * x(t) = a sinh-1(t/a) + C.  (From a standard table of integrals.)  Assuming x(0)=0 gives C=0.
 * 16) * 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.
 * 17) Eliminate t: t = a sinh( x / a ) and so y = a cosh ( x / a )

Who discovered the equation?
The 1911 Encyclopedia Britannica says the equation was found by James Bernoulli in 1691. However the MacTutor History of Mathematics archive has the same year but says the equation of the curve was found by Leibniz, Huygens and Johann Bernoulli (James' brother) responding to a challenge by Jacob (=James) Bernoulli.

Is there any way to resolve this? Until it is I vote for applying the maxim, "When in doubt, be vague," and just list all the names without trying to assign credit to who did it first.--RDBury (talk) 15:20, 12 September 2008 (UTC)

Artistic and Practical Applications of Approximate Catenary Curves
The caternary curve, in addition to being a mathematical topic is also a practical one. As can be seen by the architectural examples, some forms of the curve, or near visual approximations to it, have an aesthetic appeal ( and engineering properties ) that has been used in a number of areas by designers and artists.

The hang of chain necklace around the wearer's neck approximates a catenary curve (see the initial diagram in the article), and is a matter of aesthetic interest to jewelry designers. One wonders how the better designs are created -- whether it is on the basis of sketching from life or imagination, or from more explicit knowledge of the mathematics of caternaries. The aim here is to elucidate whether mathematical training in this area might improve designs.

Recently, at least one mechanical igloo making device, the "Icebox" [see: http://www.grandshelters.com/icebox-design.html] claims to employ a catenary curve section for the snow shelter constructed, it is reported for the strength of this form, but probably also for its aesthetic. It would be interesting to examine the mechanical device involved (which I have not seen) in relation to the mathematics of the article. It actually constructs a shell structure representing a 360 degree rotation of the curve (of course with entrances, etc.-- technically I suppose it is a 180 degree rotation of the full curve, 360 degrees for the half profile). My impression is that this may be a different curve than that found in traditional, more hemispherical, arctic igloos which employ wind compacted and sintered snow, but possibly one more suitable for the hand packed snow used in this case, which would be found at lower latitudes in loose condition and becomes packed on stuffing into the device's form -- not that this tidbit is particularly germane to the article...comes from living in Canada and building snow forts and shelters as a child.

Is there room for strengthening the article in this direction a little, or perhaps starting a parallel article on the catenary curve in design and architecture? More on Gaudi's use of ropes to do his designing would be interesting.

--FurnaldHall (talk) 11:27, 12 December 2008 (UTC)

Any info about catenary domes?
I have seen the idea of building a dome using the same idea as when constructing arches, namely using a catenary form (though the equation would become different because a dome is 2-dimensional while an arch is 1-dimensional), see Hexagonal Geodesic Domes - Catenary Domes. I was wondering if this is a good idea and if it could improve the stability of a dome much? Or is a dome in general already stable as it is, thanks to it's mean curvature? --Kri (talk) 19:26, 27 January 2009 (UTC)


 * I've tried looking this up but can't find a lot of material. As near as I can tell, with a dome each element is subject not only to the forces of the elements immediately above and below it as in the arch, but the elements to either side. The net effect of the side elements is a horizontal force that can vary to balance out the other forces involved. The upshot is that, as long as certain inequalities are met to keep the horizontal forces pointing inward, there is a lot more possible variation the shape possible.--RDBury (talk) 22:06, 21 June 2009 (UTC)

Problems with Towed cables section, class promotion
My feeling is that this article is nearly ready to be promoted to B+ class but the Towed cable section is standing in the way of this. The problems with this section are outlined in a previous comment. I'm currently working on a more mathematical version of the section which uses more familiar terminology and idealizes the physics (ignoring gravity and tangential drag) so that it becomes possible to find a closed form equation for the curve. From what I can find, towed cables are a well-studied engineering topic which is sufficiently notable for a separate article. But such an article should be located on the Engineering portal and not here.--RDBury (talk) 18:35, 5 August 2009 (UTC)
 * The new version of the section is done. I also added Towed cable to the requests for new articles.--RDBury (talk) 15:14, 8 August 2009 (UTC)

Elastic Catenary
After looking at the equation relating mass density at a point to the unstretched mass density, $$\lambda = \frac{\lambda_0}{1+\epsilon T}$$ I believe it to be incorrect. Here is my argument: This formula says that the density can be given by knowing the unstretched mass density, the spring constant of the cable, and the tension in the cable. I will exibit two springs with equal parameters when unstretched, but with differing parameters when stretched.

Consider two springs, the first an inch long, the second a mile long. The small spring is made from a stretchy material compared to the longer spring, so that the difference in length makes their spring constants equal (since longer springs tend to have smaller spring constants.) Further, the mass density of these springs when unstretched can be made the same by distributing the mass equally (so the mile long spring is much more massive than the inch long one). Now suppose we apply an equal tension to both springs, enough to extend the small spring twice its unstretched length (1 additional inch). Since the larger spring also has the same spring constant, it will, too, stretch one inch. Notice that the small spring's mass density has halved (since the mass is the same but the spring is twice as long), but the density of the mile long spring has hardly changed at all, since 1 inch is negligible in comparison to a mile (by a factor of 1 in 63360). This shows that two springs with the same tension, unstretched density, and spring constant can have different stretched densities, which proves that the formula for stretched density cannot be a function of these parameters alone.

Where on earth did this formula come from? It isn't even referenced.

Danielkwalsh (talk) 08:14, 25 July 2010 (UTC)

However, If $$\epsilon$$ is not a spring constant but instead is defined by $$\epsilon=k L_0$$, then this formula holds. Perhaps this should be made more clear. I think I may make this correction to the page.

Danielkwalsh (talk) 09:25, 25 July 2010 (UTC)

Vandalism in Reference
Reference #11 is vandalized. I couldn't find the original reference in the recent diffs. Somebody needs to fix it. —Preceding unsigned comment added by Subh83 (talk • contribs) 17:20, 13 September 2010 (UTC)
 * The anagram was introduced in this revision and apart from being moved to a reference it seems to have been unchanged. I can't see any vandalism. I think its actually got too many e's I make it 5 e's and 8 i's giving
 * abcccddeeeeefggiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuvx
 * rather than
 * abcccddeeeeeefggiiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux
 * The original phrase was "Ut pendet continuum flexile, sic stabit contiguum rigidum inversum".--Salix (talk): 18:00, 13 September 2010 (UTC)

I discussed the error in the anagram on my blog in July 2010: http://newtonexcelbach.wordpress.com/2010/07/14/arches-anagrams-and-plagiarism/. At the time there were 121 Google hits for an exact search on the incorrect version, and just 3 to the correct version (one of which was a plagiarised version of my words). There is now one more site with the correct version, here. DougAJ4 (talk) 10:35, 18 January 2011 (UTC)

Ctesiphon Arch
According to the article, the Ctesiphon Arch is "roughly but not exactly a catenary". The problem is that it is also "roughly but not exactly" any number of other simple curves, including an ellipse and a parabola. I have never seen any convincing mathematical evidence that it is any closer to a catenary than to other candidate curves. Is there any? —Preceding unsigned comment added by 86.186.36.105 (talk) 01:15, 26 October 2010 (UTC)

I have examined the shape of the arch, compared with a catenary and parabola, here: http://newtonexcelbach.wordpress.com/2008/06/08/the-roof-of-the-taq-i-kisra/. The analysis is certainly a bit rough and ready (due to lack of detailed information about the actual shape of the arch), but it does suggest that it is closer to a catenary than a parabola. In the following post I have looked at the effect of different shapes on the stresses in the arch. DougAJ4 (talk) 10:44, 18 January 2011 (UTC)

Proposal of new figure and rewrite of "Derivation"
I have a new figure that I think is very explanatory and a new derivation that I think is shorter, clearer and more elementary then what presently is in the article. Any protest against a re-write (about!) like this:

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

The forces acting on a segment of catenary curve are shown in the figure below

The vector sum of the forces acting on the segment from the two extremities and from the gravitational force must be zero. As the gravitational force is directed downwards the horizontal components of the forces acting on the extremes must have the same magnitude. As this is true for any segment of the catenary this is a fixed constant for the whole of the catenary. Denoting this constant with $$f$$ one gets that the vertical component of the force at the left extreme $$ x_1$$ is $$ -f \ y^\prime(x_1)$$ and at the right extreme $$ x_2$$ is $$ f \ y^\prime(x_2)$$ The path length of the curve representing a function $$ y(x)$$ with $$ x $$ varying from $$ x_1 $$ to $$ x_2 $$ is

\int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx $$

If $$g$$ is the gravitational constant and $$\rho$$ is the mass per length unit of the chain the gravitational force acting on the arc from $$ x_1 $$ to $$ x_2 $$ is $$g \rho \int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx \,$$

This force must be compensated by the vertical components of the forces acting on the two extremes of the arc, i.e.

For the function
 * $$ y(x) = a \cosh \frac{x}{a}$$

one has that

and that

It follows that

From ($$), ($$) and ($$) follows that if $$ a = \frac {f}{ g \rho} $$ the vector sum of the vertical force components at the extremes $$ x_1 $$ and $$ x_2 $$ for any arc will take the value needed to compensate for the gravitational force acting on the arc.

From the relation
 * $$\cosh (x -\delta) - \cosh (-x -\delta ) = -2\ \sinh (x)\ \sinh (\delta)$$

follows that for any two points $$ (x_1\ ,\ y_1)$$ and $$ (x_2\ ,\ y_2)$$ in a vertical plane and any positive number $$ a$$ one can find values $$ x_0\ ,\ y_0$$ such that the function $$ y = y_0 + a \cosh \frac{x - x_0}{a}$$ passes through both points, i.e. such that:


 * $$ y_1 = y_0 + a \cosh \frac{x_1 - x_0}{a}$$


 * $$ y_2 = y_0 + a \cosh \frac{x_2 - x_0}{a}$$

By selecting the parameter $$ a$$ properly the length of the curve as given by ($$) can be given any desired value larger then $$ r$$ where $$ r$$ is the distance between the two points.

From the figure is clear that the tension of the chain at any point $$(x\ ,\ y)$$ is $$f \sqrt{1+{y^\prime}^2} = f \cosh \frac{x-x_0}{a}=f\ \frac{y-y_0}{a}$$

Stamcose (talk) 20:32, 30 November 2010 (UTC)

I would like to replace the sections "Derivation", "Alternative1" and "Alternative2" to (essentially!) the text above next week! Any objections? Please word these objections (if any!) the next few days!

Stamcose (talk) 20:21, 5 December 2010 (UTC)


 * I think that your derivation is good but I don't think it should replace the other ones. You should add it as another method. The method you present is "guess the solution and try it", which is completely valid but kind of relies on the fact that you know the answer. I know it strictly doesn't require previous knowledge but I think you'll know what I mean. Also I don't see any reason to reduce the amount of information on Wikipedia.


 * Phancy Physicist (talk) 04:33, 6 December 2010 (UTC)

Why does so much space appear above the figure? I don't remember enough about SVG at the moment to look at it, but could that be cleaned up? / ninly ( talk ) 00:14, 6 December 2010 (UTC)


 * The figure above is good but I would like to see some labels on the force vectors.


 * Phancy Physicist (talk) 04:37, 6 December 2010 (UTC)

It is true that it always is problematic to remove stuff. Although I do not think the present derivations are good or even rigorous (correct!) I do not insist on removing them. If other people want to keep them (and certainly the author!) it is OK with me.It is certainly an inevitable effect of the Wikipedia system that the articles get (too?) voluminous and poorly structured!

The space on the top of the figure annoyed me too! It comes from the conversion tool PostScript => SVG that is recommended by Wikipedia (in the Help pages). Now I have made myself (without any third party tool!) a properly sized SVG: The analytical solution of differential equations is really always to guess/find a class of solutions from which the uniquely defined solution satisfying the boundary conditions can be constructed. A typical example is the elementary differential equation

$$\ddot{y}+A^2 y=0$$ where one guesses and verifies that $$\cos Ax$$ and $$\sin Ax$$ and any linear combination of these are solutions from what follows that any solution is a linear combination of these functions as the boundary values completely define the solution

Stamcose (talk) 23:08, 6 December 2010 (UTC)


 * What part of the old proofs do you question? I am not the original author so don't worry about offending me. I just don't see anything wrong with them.


 * Phancy Physicist (talk) 05:19, 8 December 2010 (UTC)

The problem with the present derivation is that author absolutely wants to use the "natural parametrization" $$s$$ which is the arc length. But in this very case this is a detour that brings nothing but confusion!

The present
 * $$\mathbf{T}=\mathbf{T}(s)$$

is with my notations (without s)
 * $$ (f\ ,\ f \ y^\prime )$$

where $$f$$ is the constant horizontal component of the force what in the present text is denoted $$c$$.

As $$\tan \varphi $$ simply is $$y^\prime$$ and $$ s=\int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx$$ the expression
 * $$\tan \varphi = \frac{\lambda g s}{c}$$

is simply
 * $$y^\prime = \frac{\lambda g }{c} \int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx$$

With my notations this is

as by definition $$x_1$$ has been selected such that $$y^\prime(x_1)=0$$

This relation is clear from my figure and my very short argument saving the confusing round about in the present text.

I simply say that the two parameter family of functions
 * $$ y(x) = a \cosh \frac{x-x_0}{a}+y_0$$

are the only solutions as by selecting $$x_0$$ and $$y_0$$ properly any value for $$ y(x_1)$$ and $$y^\prime(x_1)$$ can be obtained.

One can also go ahead as follows what clearly is the intention of the present text:

Denoting $$y^\prime$$ with $$z$$ and taking the derivative of both sides one gets the equation
 * $$z^\prime = \frac{dz}{dx}= \frac{1 }{a} \sqrt{1+z^2}$$

or
 * $$\frac{dx}{dz}= \frac{a }{\sqrt{1+z^2}}$$

which is integrate to


 * $$x= a \ \sinh^{-1}(z) + x_0$$

where $$x_0$$ is the constant of integration

The other constant of integration $$y_0$$ is then obtained integrating $$y^\prime = z$$

I do not think it is much difference between "guessing and finding" a primitive function to $$\frac{a }{\sqrt{1+z^2}}$$ and to "guessing and finding" a two parameter family of solutions to a second order differential equation!

I do not myself understand the very last arguments of the present text but the arguments above are the (clear!) arguments that should be given!

Stamcose (talk) 20:44, 8 December 2010 (UTC) And this should be the final figure with the right size for Wikipedia and with subtitle thanks to "frame" in the tag!

Stamcose (talk) 20:44, 8 December 2010 (UTC)

I think I am more or less ready with a new version of "Derivation"! Any protests? Any additional suggestions?

++++++++++++++++

The forces acting on a segment of catenary curve are shown in the figure below

The vector sum of the forces acting on the segment from the two extremities and from the gravitational force must be zero. As the gravitational force is directed downwards the horizontal components of the forces acting on the extremes must have the same magnitude. As this is true for any segment of the catenary this is a fixed constant for the whole of the catenary. Denoting this constant with $$f$$ one gets that the vertical component of the force at the left extreme $$ x_1$$ is $$ -f \ y^\prime(x_1)$$ and at the right extreme $$ x_2$$ is $$ f \ y^\prime(x_2)$$ The path length of the curve representing a function $$ y(x)$$ with $$ x $$ varying from $$ x_1 $$ to $$ x_2 $$ is

\int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx $$

If $$g$$ is the gravitational constant and $$\rho$$ is the mass per length unit of the chain the gravitational force acting on the arc from $$ x_1 $$ to $$ x_2 $$ is $$g \rho \int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx \,$$

This force must be compensated by the vertical components of the forces acting on the two extremes of the arc, i.e.

Denoting the constant ratio $$\frac {f}{ g \rho}$$ with $$a$$ and taking the derivative of equation ($$) with respect to the upper limit of the integral, i.e. with respect to $$x_2$$, one gets


 * $$y^{\prime\prime} = \frac{1 }{a} \sqrt{1+{y^\prime}^2}$$

Denoting $$y^\prime$$ with $$z$$ this equation takes the form
 * $$z^\prime = \frac{dz}{dx}= \frac{1 }{a} \sqrt{1+z^2}$$

what means that for the inverse function $$x(z)$$ one has


 * $$\frac{dx}{dz}= \frac{a }{\sqrt{1+z^2}}$$

which is integrate to


 * $$x= a \ \sinh^{-1}(z) + x_0$$

where $$x_0$$ is the constant of integration or equivalently
 * $$z=y^\prime=\sinh(\frac{x-x_0 }{a})$$

Again integrating with respect to $$x$$ one gets where $$y_0$$ is the second constant of integration

The length of the curve given by ($$) from $$x=x_1$$ to $$x=x_2$$ is {NumBlk|:|$$l=\int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx = \left [a \sinh \frac{x-x_0}{a} \right ]_{x_1}^{x_2}\,$$|$$}}

This family of solutions is parametrized with the 3 parameters $$a,\ x_0,\ y_0 $$. For any concrete case these 3 parameters must be computed to fit the boundary value conditions. In a typical case the form of a chain having a given length l and being attached in two fixed point with the coordinates $$x_1,\ y_1$$ and $$x_2,\ y_2$$ relative a vertical coordinate system should be computed.

This means that $$a,\ x_0,\ y_0 $$ have to be determined such that

Setting
 * $$x_m=\frac{x_2+x_1 }{2}$$
 * $$\Delta x=\frac{x_2-x_1 }{2}$$

subtracting ($$) from ($$) and then dividing with $$a$$ one gets

For any given values $$x_1,\ y_1,\ x_2,\ y_2,\ a$$ one can determine $$\sinh \frac{x_m-x_0 }{a}$$ from ($$)

When $$\sinh \frac{x_m-x_0 }{a}$$ has been determined $$\frac{x_m-x_0 }{a}$$ is computed by solving a quadratic equation.

With $$x_0$$ known ($$) or ($$) can subsequently be used to determine $$y_0$$

Having determined $$x_0$$ with the algorithm just described the curve length $$l$$ corresponding to the selected $$a$$ value can be computed from ($$). With an iterative algorithm the $$a$$ value that corresponds to a certain curve length $$l$$ can finally be derived.

From the figure it is further clear that the tension of the chain at any point $$(x\ ,\ y)$$ is $$f \sqrt{1+{y^\prime}^2} = f \cosh \frac{x-x_0}{a}=f\ \frac{y-y_0}{a}$$ where $$f=a\ g\ \rho$$ is the magnitude of the constant horizontal force component

If the mass density $$\rho$$ is not constant but varies depending on some law the resulting differential equation will not have an exact analytic solution anymore. But the resulting curve can be determined with arbitrary accuracy also in this case using an algorithm for the numerical integration of ordinary differential equations. Taking the derivative of equation ($$) with respect to the upper limit of the integral, i.e. with respect to $$x_2$$, without assuming $$\rho$$ to be constant one gets


 * $$y^{\prime\prime} = \frac{g}{f}\ \ \rho \sqrt{1+{y^\prime}^2}$$

Denoting $$y^\prime$$ with $$z$$ one gets the following first order differential equations
 * $$y^\prime = z$$
 * $$z^\prime = \frac{g}{f}\ \ \rho \sqrt{1+{y^\prime}^2}$$

Given any initial values for $$y(x1)$$ and $$z(x1)$$ and any value for the parameter $$f$$ these differential equations can be propagated to $$x=x_2$$ with $$\rho$$ specified as any function of the state variables $$x,y,z$$. The free parameters to be iteratively adjusted to fit the boundary constraints are now $$z(x1)$$ and $$f$$. They can for example be adjusted iteratively such that $$y(x2)=y_2$$ where $$(x_2\ ,\ y_2)$$ is the second attachment point. This leaves an additional degree of freedom corresponding to for example the length of the curve.

For the "elastic catenary" one would for example have that $$\rho = \frac{\rho_0}{1+\epsilon\ f\sqrt{1+{y^\prime}^2}}= \frac{\rho_0}{1+\epsilon\ f\sqrt{1+{z}^2}}$$ where $$\rho_0$$ is the mass density without any stress.

Stamcose (talk) 17:15, 9 December 2010 (UTC)

To check the case "elastic catenory" with a mass density (mass per unit lenth) of $$\rho=\frac {\rho_0}{ 1+\epsilon\ F}$$ where $$F$$ where is the "stress force" I integrate numerically the differential equations


 * $$y^\prime = z$$
 * $$z^\prime = \frac{1}{a\ (1+\epsilon\ f\sqrt{1+z^2})}\ \ \sqrt{1+z^2}$$

Where $$a=\frac {f}{ g \rho_0}$$

But the resulting curve had only small deviations to a normal catenary! Even the physically completely un-realistic case of an $$\epsilon\ f$$ product with the value 10 the deviation was un-spectacular, just about visible for the bare eye on a plot.

As the theoretical analysis of this case also is "sick" I would propose to remove these items completly! It was an interesting idea to include these generalisations (in the beginning of 2008) but a closer analysis has now shown that it was not all that fruitful after all!

Stamcose (talk) 16:16, 10 December 2010 (UTC)

Proposal of new figure and rewrite of "Towed cables"
The present derivation is absolutely correct but it refers to the formula
 * $$\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a}$$

that is stated in this article but not has been proven or explained in detail here. The proof I give here is in fact essentially a proof of this relation!

In this talk page it is also stated (above) that "clarification is needed"!

I think this more straightforward version illustrated by a figure is more digestible for the reader

++++++++++++++++++++

The following figure illustrates a segment of a cable that is fixed in both ends and exposed to drag.

The velocity relative to the cable is assumed to be constant and the coordinate system is selected such that this velocity is in the -y direction, i.e. $$\mathbf{v} = (0, -v)$$. To compute the force due to drag, write $$\mathbf{v} = \mathbf{v}_u + \mathbf{v}_n$$

where $$\mathbf{v}_u$$ and $$\mathbf{v}_n$$ respectively are the components parallel to and orthogonal to the cable. The cable is assumed to be smooth so the force on the cable due to $$\mathbf{v}_u$$ is taken to be negligible. The force acting on the cable, per unit length, following the Drag equation is therefore
 * $$\mathbf{G} = f(x) \mathbf{n}$$

with

where $$c$$ is a constant depending on the density of the fluid, the diameter of the cable, and the Drag coefficient and $$\mathbf{n} = (n_x\ ,\ n_y)$$ denotes the unit normal vector.

For any curve $$y(x)$$ the tangent (unit vector) is

and the normal (unit vector) is

From ($$) and ($$) follows that

From ($$) and ($$) follows that the x-component of the total force on the segment of the curve from $$x= x_1$$ to $$x= x_2$$ is

and the component in the y-direction is

If now
 * $$y^'=\sinh(\frac{x-x_0}{a})$$

one has that


 * $$1+{y^'}^2=\cosh^2(\frac{x-x_0}{a})$$

and from ($$),($$) and ($$) that

If the now the force in the cable is
 * $$F\ =c\ v^2\ a\ $$

the force at the right extreme of the cable segment is
 * $$ c\ v^2 \ a\ ( t_x(x_2)\ ,\ t_y(x_2) ) $$

and at the left extreme
 * $$ -c\ v^2 \ a\ ( t_x(x_1)\ ,\ t_y(x_1) ) $$

From ($$) and ($$) follows that the vector sum of these forces is precisely the force needed to counter act the forces on the segment caused by the drag

Stamcose (talk) 13:28, 17 December 2010 (UTC)

Citations and sources are needed
Please be sure that all additions to the Catenary article are verifiable. Any new items added to the article should have inline citations for each claim made. As a courtesy to editors who may have added claims previously, before Wikipedia citation policy is what it is today, many of the existing unsourced claims have been tagged citation needed to allow some time for sources to be added.

More specifically, to all of the recent discussion on derivations, and a new derivation that may soon be added to the article, Wikipedia is an encyclopedia for verifiable that can be sourced from reliable secondary sources with a citation for substantial assertions. It is specifically not for original reseach. While I personally do not plan to challenge any of the unsourced proofs at this time, I think it is fair to say that a lot of unsourced information won't stand the test of time, and will be winnowed out over the longer term as Wikipedia emerges from myriad individual actions. Net: I think you should source the new derivation you are planning to add to the article. N2e (talk) 16:57, 18 December 2010 (UTC)


 * The only place in the mathematical part where "Citations and sources" is lacking is in the "Equation" part. But
 * $$y = a \, \cosh \left ({x \over a} \right ) = {a \over 2} \, \left (e^{x/a} + e^{-x/a} \right )$$,


 * is just the definition so the tag "citation needed" is irrelevant here. On the contrary a reference to the derivation of
 * $$\tan \varphi = \frac{s}{a}$$.
 * $$\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a}$$
 * $$\kappa=\frac{a}{s^2+a^2}$$.


 * could be relevant!


 * Stamcose (talk) 10:34, 22 December 2010 (UTC)


 * Actually, even a simple definition, which is an absolute "truth" to you and I who understand mathematics, is not something that is "known" to all readers of this encyclopedia. That is why WP has the core policy of verifiability, let's get it cited and improve the article. N2e (talk) 15:59, 22 December 2010 (UTC)


 * Ref (a current calculus textbook) added for the simple definition. / ninly ( talk ) 17:27, 2 January 2011 (UTC)

Modifications by 165.246.99.227
In my opinion the contribution now called "5.3 Analysis" should be fully replaced with what presently is "7. Alternative analysis". But by preference (as a courtecy!) the main author of this section should agree to this first and do this himself. The mathematical imperfections of "5.3 Analysis" has already been discussed above. If "7. Alternative analysis" appears longer it is because it goes much further, it explains how to solve the boundary value problem! And "Towed cables" is also in the pipeline!

Who else has an opinion about this matter?

Stamcose (talk) 16:58, 21 December 2010 (UTC)
 * I think the 5.3 section start a lot better than the 7 section, it introduces all the different forces explaining what they are rather than just referring to a picture with unlabelled force vectors. Both are much to long, is not wikipedia's job to give a detailed derivation more indicate the key points and provide references where these can be checked. Both currently fail WP:OR. --Salix (talk): 17:41, 21 December 2010 (UTC)

Most readers that not are mathematicians will certainly just glance over this type of text without the intention of follow it in detail. Then it should simply be as short as possible. But for the benefit of those who really make the effort to go into the details the mathematics should be correct and logical. This is not really the case with the present "5.3 Analysis". For example:

The present
 * $$\mathbf{T}=\mathbf{T}(s)$$

is with my notations (without s)
 * $$ (f\ ,\ f \ y^\prime )$$

where $$f$$ is the constant horizontal component of the force what in the present text is denoted $$c$$.

As $$\tan \varphi $$ simply is $$y^\prime$$ and $$ s=\int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx$$ the expression
 * $$\tan \varphi = \frac{\lambda g s}{c}$$

is simply
 * $$y^\prime = \frac{\lambda g }{c} \int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx$$

With my notations this is

as by definition $$x_1$$ has been selected such that $$y^\prime(x_1)=0$$

This relation is clear from my figure and my very short argument saving the confusing round about in the present text.

Stamcose (talk) 10:16, 22 December 2010 (UTC)

Too many analyses and "alternative analyses" without reference
This article contains too many different analyses and alternative analyses with lots of mathematical derivations but not referencing a single reliable source as to address why such analyses are useful or let readers verify whether they are correct. If proper references are given, there may not even be the need of documenting all the detailed derivations in Wikipedia. However I think much of these are original research and are candidates for removal (see WP:NOR). - Subh83 (talk) 06:04, 14 March 2011 (UTC)


 * I concur. The maths could be better organized, and much reduced given that a lot of it is unsourced.  Don't have time for much more analysis tonight, but will be glad to help weigh in on improving the article.  Ping me if you want any other looks.  N2e (talk) 04:14, 15 March 2011 (UTC)


 * I agree even though I contributed much of the math. The material does seem to be hard to organize though. There are several ways to analyze the problem and many ways to generalize it (non-uniform density, add elasticity, more general force) which are covered in the literature on the subject. One method of analysis may be best for some generalization and not for others, so I think presenting several methods gives a better understanding of subject.


 * As I see it, the analysis can be divided into three parts: I) Derive a differential equation(s) for the curve. II) Integrate the differential equation(s) to obtain a Cartesian equation. III) Apply parts I and II to derive other properties. Part I can proceed by a) analyzing a small section of the curve, b) analyzing a large section of the curve and applying the principle that the curve at equilibrium may be treated as rigid, c) using calculus of variations. From my recent reading there is actually more material that could be added, but I am planning to go through the existing material with an eye toward more rigorous referencing and to remove what seems to be OR. Much of the existing derivation comes from and related pages. I'm also thinking that it would be better to put a very simple derivation for the basic catenary first, rather than derive general formulas which is the current approach.--RDBury (talk) 15:25, 11 September 2011 (UTC)
 * I did some reorganization and rewriting of the analysis sections so everything either has cites or is marked with an unreferenced tag. I used Routh as a source for most of it since it seems to be a definitive text, though Minchin seems nearly as complete. I added some new material as well but there is still more that could be added.--RDBury (talk) 01:07, 16 September 2011 (UTC)
 * I think that the reorganization and rewriting has made the article much better, and the overall sourcing and cites are quite good. Thanks for doing that RDBury!  N2e (talk) 00:22, 13 November 2011 (UTC)

Towed cables
Most towed cables are fixed at only one end, e.g., at the towing ship. See, for example: A. P. Dowling, “The dynamics of towed flexible cylinders. Part 2. Negatively buoyant elements.” Journal of Fluid Mechanics 187, 533-571, 1988. Figure 1 in that reference depicts a typical geometry for towed cable catenary calculations.

The assumption of negligible drag parallel to the cable is not typically made for these types of calculations. Many such cables are a mile long or longer, and the tangential drag becomes a major contributor to the cable tension, which can have a significant effect on the catenary. Most such solutions must be performed numerically. Psalm 119:105 (talk) 02:06, 20 May 2011 (UTC)
 * The Dowling ref. was in the article several years ago but was removed. In the old version of the section it looks like the formulas were copied from the article with little explanation of how they were derived or what the variables meant. In any case, the towed cable problem seems to be a variation of Bernoulli's sail problem, and there is much more material available on that, so I'm planning to replace the section with a fully referenced one on sails.--RDBury (talk) 13:16, 21 September 2011 (UTC)

Simple suspension bridges, citation needed
I restored the statement that Simple suspension bridges follow a catenary curve though I don't have a source at the moment. It is relevant, in fact it's kind of the point of the section, and rather obvious since in this case the bridge is the chain, not to mention that this type of bridge is also called a catenary bridge. This is mentioned in the other article but I haven't been able to verify the sources there since they always seem to be in "not included in preview" pages.--RDBury (talk) 15:20, 6 September 2011 (UTC)
 * Since there is an editor who keeps deleting the statement, and since it's kind of the point of the section, I've commented out the section until the issue can be resolved. Hopefully someone can find a cite so the material can be restored.
 * It seems blindingly obvious. -- 202.124.74.120 (talk) 04:58, 13 November 2011 (UTC)

Spider web
Re change of web photo text from "... multiple elastic catenaries..." to "...multiple (approximate) catenaries...". Change comment is "Source needed to be specific". As far as I can tell the article describes elastic catenaries and so is self referential. Is this a valid reference? Mtpaley (talk) 18:11, 7 September 2011 (UTC)
 * I'm trying to strike a balance between having verifiable captions and being so careful that real world images can't be used. I added several of these images in an effort to make the subject more interesting for non-mathematical readers. I think at some point an appeal to WP:Likely to be challenged must be made. That the spider web threads follow an elastic catenary could be challenged since there are several assumptions involved, for example that the threads follow Hooke's law. But I don't think there can be much doubt that the threads follow the basic assumptions of catenary, that they are flexible threads where the forces of tension and its own weight are at equilibrium, and therefore follow the curve at least approximately.--RDBury (talk) 23:51, 8 September 2011 (UTC)

Huygens vs. Jeffeson
The sentence "Thomas Jefferson is usually credited with the English word catenary." seems to be contradicted in MacTutor, which has "Huygens was the first to use the term catenary in a letter to Leibniz in 1690... ." This is repeated in the MathWorld article. The passage in question seems to be Appendix II of the letter dated 9 October 1690 which can be found here. The passage is in Latin (though the main body of the letter is in French) and the word Huygens uses is catena (or a conjugate such as catenae, never catenaria as claimed in the article). The title of the appendix has the phrase curvae catenae which would translate to "chain curve". The appendix does seem interesting from a historical point of view, it's basically a summary of what was known at the time on the subject, but it seems doubtful that the claim in MacTutor can be substantiated. I'm therefore removing our mention of the Huygens letter from the article, at least until there is a more clear explanation of how it contributed to the etymology of the word.--RDBury (talk) 14:27, 12 September 2011 (UTC)
 * PS. MacTutor was apparently using Lockwood as a source, "The name was the first used by Huygens in a letter to Leibniz in 1690." It's not clear from the context what "name" is meant, indeed it was apparently misinterpreted by MacTutor. It's possible that what Lockwood meant is that Huygens first used the phrase curvae catenae, but I think it's more likely that Lockwood was quoting (and possibly misinterpreting) yet another source.--RDBury (talk) 15:13, 12 September 2011 (UTC)

"Catenary" bridges follow a catenary curve?
We've gone back and and forth a couple times on whether the statement that a catenary bridge follows a catenary curve should be included in the catenary article. It seems to be that there is insufficient reason to delete the statement since WP:V only specifies that statements which are "challenged or likely to be challenged must be attributed". To me the statement is obvious and is not likely to challenged, and so should not be removed since it is encyclopedic and relevant to the subject. I have not removed the cite needed tag though since a reference would be nice to have to bring the article to GA standard. Do you have a reason to challenge the truth of the statement? If so then I should point out that it is repeated several times in Simple suspension bridge. In any case, without the statement the section is only about parabolas and bridges and has little to do with the catenary, so I've commented it out until the issue can be resolved.--RDBury (talk) 11:18, 12 November 2011 (UTC)


 * Well, as I understood the statics in a mechanical design class I took once, these bridges only approximate a catenary shape, for a variety of practical reasons like differential unit mass, non-uniformity of attachment points to the support cables, non-ideal atmospheric reality, etc. But as usual in Wikipedia, we should not make such claims, either way, out of our heads and based on what we recall from former studies, we should make claims that are verifiable, or leave them out until someone can support the claim with a reliable source.  That is why I removed the claim, which had not been supported with a citation despite one having been asked for over a year ago.  Cheers.  N2e (talk) 13:20, 12 November 2011 (UTC)
 * Your objection seems to be that the mathematical model of the bridge can never be exactly equal to an actual bridge, and this is true if you go into fine enough detail. But this is like saying that a billiard ball is not a sphere because if you examine it under a microscope you can find imperfections in the surface, or that a laser beam does not follow a straight line because of changes in air density. Any mathematical statement about the real world is really about an idealized representation of an object rather than the actual object, this is understood implicitly and without this kind of assumption mathematics is only an intellectual exercise with no application. If you still have the text from your design class then perhaps you can add what it has to say about the bridges to the article. I'll be happy to have some kind of statement, even if it's an approximate one, rather than not covering the material at all.--RDBury (talk) 18:46, 12 November 2011 (UTC)


 * Well, a couple of comments. First off, the article is excellent overall now and I will be happy with what you decide, either way.  To your specific comments, however, I should respond directly.  I think that your comment about the catenary not following catenary shape (like a billiard ball to a perfect sphere) is a valid point, but is most applicable to, say, an anchor chain rode, of identical and equal segments, each connected at a single point, having identical mass both above and below the centerline of the connecting points, identical segments in all parts of the bridge, etc.  Most "catenary" bridges are much rougher approximations due to a rather extensive set of nonlinearties in the bridge mass, and particularly because the center of mass of the bridge structure is typically far from the centerline of the tensioned structural member.  So that is my thought about your specific comment.


 * Having said all that, you MIGHT want to consider the term "approximates a catenary shape" or "closely approximates a catenary shape" — as well as continue to look for the source that you, quite fairly, have noted in the article we are still looking for — but I will support whichever decision you come to. You have done an awesome job on improving this article!


 * Thanks for handling the discussion with such scholarly aplomb. Cheers.  N2e (talk) 00:22, 13 November 2011 (UTC)


 * I'm not sure where this business about the centreline of the connecting points being far from the cables comes from. I haven't seen anything like that happening, you're not thinking of a normal suspension bridge are you? It would be possible even so to have something that deviated significantly from a catenary by using cables that were thinner in the middle, one doesn't need quite the same strength in the centre, but in actual practice they just use a cable or chain with a constant cross section, the essence of catenary bridges is simplicity. They follow a catenary very exactly except of course when somebody is walking over them. Dmcq (talk) 01:11, 13 November 2011 (UTC)


 * The simple suspension bridge is essentially just a thick chain. It's as close to a catenary as a chain is (except when somebody is walking over it, as Dmcq says). I have added a reference to stressed ribbon bridges, for which clear sources do exist. -- 202.124.74.120 (talk) 05:19, 13 November 2011 (UTC)
 * Thanks for finding the references and the other improvements, well done. Imo the article the ready for a GA review now, though I'll let the dust settle a bit before nominating it. @N2e, You're welcome though I can't say "aplomb" is something I always manage to maintain. Actually I owe you a thanks as well for giving the issue enough of a kick so that it got fixed. Thanks also for recognizing my contributions but it's a collaborative effort as any good WP article is, and my efforts are minor compared to what others have done.--RDBury (talk) 22:14, 13 November 2011 (UTC)

Catenary bridge?
I would suggest a lot of caution in this section of the article. First, there is not a clear definition for a "catenary bridge," which you have named a section in the current article. Because bridge engineering is a well-researched science, use of the term "catenary bridge" could be original research. I recommend "catenary-shaped bridges." Second, there is not a clear engineering definition of simple suspension bridge that matches the definition used in this article (see discussions at Talk:Simple_suspension_bridge, and Talk:Suspension_bridge) Also, this edit uses a non-bridge engineering reference to make the major point of this section, that simple suspension bridge is a defined term and engineering research shows that when the deck follows the cables, the curve is a catenary. I think that this is a poor source for such the technical, well-researched subject of bridge engineering. - ¢Spender1983 (talk) 03:49, 14 November 2011 (UTC)


 * Also, this reliable mathematical reference used in the article says that a suspension bridge will be either a parabola or something between a catenary and parabola. So where is the reference that definitively says a simple suspension bridge that has a deck following the curve of the cable is a catenary? - ¢Spender1983 (talk) 04:18, 14 November 2011 (UTC)

In response to both of those, the technical references (such as Lockwood, which you cite, but which you seem to have misunderstood) all say clearly that: To explicitly state the obvious fact that (2) applies to simple suspension bridges, a non-bridge engineering reference (but nevertheless a reliable reference) was used. The phrase "catenary bridge" for a simple suspension bridge (and sometimes for a stressed ribbon bridge) is quite common in the bridge engineering literature. -- 202.124.73.186 (talk) 11:50, 14 November 2011 (UTC)
 * 1) if the load is distributed horizontally, as in a bridge with a suspended horizontal deck, the curve is a parabola,
 * 2) if the load follows the chain, it is a catenary, and
 * 3) if some load follows the chain and some load is distributed horizontally, intermediate curves result.


 * Thanks for the addition of the Troyano reference. It shows that there was some error in my original statement. Troyano gives a technical source for the term Catenary bridge and this source also meets the highest level of what counts as a reliable source: it is an academic publication from the field of bridge engineering. I still question the use of the Trinks reference in the article. Is a book about furnaces, written by a mechanical engineering professor, considered a good source for the academic subjects of either catenary curves or bridges? Especially now that a more reliable source has been added. - ¢Spender1983 (talk) 19:18, 19 November 2011 (UTC)


 * There seems to be some disagreement over terminology, which is common problem in writing an encyclopedia because in many cases different authors use different terms for the same thing, or same term for different things, or even talk about a subject without giving it a name. The name "simple suspension bridge" is used here to mean bridges such as the Capilano Suspension Bridge where the roadway follows the suspending cables. If there is a more commonly used term for this type of bridge, as opposed to the usual kind of suspension bridge, then I have no objection to changing the terminology, but for now this term appears to be the most recognized. When Lockwood used the term "suspension bridge" he is clearly not including this type of bridge in the discussion so his conclusions don't apply here. In fact, though in general bridge design is well researched and covered in the literature, there seems to be very little on simple suspension bridges. This isn't surprising since they aren't practical except as foot bridges and they are comparatively low-budget, often hand crafted affairs. So I don't see a problem with using a non-bridge engineering source; if there is a bridge engineering source which says that the curve is not a catenary then I might give it more weight, but these sources don't seem to cover the subject so it's better to use a another source since the material is clearly encyclopedic.--RDBury (talk) 06:38, 16 November 2011 (UTC)

Review of new sections
addedd these new sections to the catenary article. I deleted these new sections because: The author has contested this deletion on my talk page. Other editors' views and opinions are requested. Gandalf61 (talk) 17:49, 17 November 2012 (UTC)
 * 1) They were unsourced.
 * 2) They were too long, adding over 12kb to a 39kb article.
 * 3) They mostlty derive results and properties given elsewhere in the article.
 * 4) Lengthy derivations and proofs do not belong in Wikipedia - Wikipedia is not a text book.
 * I agree with the above. I would add that it could be considered WP:OR without any sources to backup this treatment.--Salix (talk): 18:04, 17 November 2012 (UTC)

Deletion by User Gandalf61
1. A basic principle of Wikipedia is that you do not unilaterally delete text of others. Your personal opinion is not necessarily shared by others. That text is added but not deleted tend to make articles somewhat longish but this is not really a problem. People can read the part of the article that appeals to them.

I am strongly of the opinion that this "alternative analysis" is by far superior to the text it "duplicates". In addition interesting plots are included!

Stamcose (talk) 03:10, 19 November 2012 (UTC)


 * What basic principle? See WP:BRD. —Tamfang (talk) 07:12, 10 December 2012 (UTC)

And in addition "unsourced" is clearly irrelevant here. This is a mathematical analysis that by its very nature is "selfsourcing".

Stamcose (talk) 03:18, 19 November 2012 (UTC)


 * That's a definite problem in mathematical articles, but completely opposed by Wikipedia policies. WP:CALC specifies the degree you can claim "self-sourcing"; anything beyond that requires reliable sources.  Proofs can be placed in Wikibooks or Wikiversity (if it's still up) without worrying about Wikipedia 's rules on original research.  — Arthur Rubin  (talk) 04:35, 19 November 2012 (UTC)

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Asymmetric Catenary
I have found very little information on the subject of asymmetric catenaries. I would like to see the equation of a cord catenary, where the ends are supported at different heights. The horizontal distance between the supports is known as are the properties of the cord (tension, weight/unit length etc) — Preceding unsigned comment added by Coli.white (talk • contribs) 22:20, 3 April 2018 (UTC)

Electrified railways
A long, detailed article without a single mention of railway overhead electrification, of which catenary-suspended wires are an essential component! -- Picapica (talk) 21:50, 24 April 2018 (UTC)

Analysis Section is confusing
The equations under analysis are not straightforward at all. s is being used for the length of the chain, but then the variable s is also being used for an equation describing length of the curve. It's really confusing and I can't make any sense of it. Breebree214 (talk) 18:25, 11 March 2019 (UTC)

In the description of a catenary and parabola, below the picture it says: The catenary and parabola equations are respectively, y = cosh(x) and y =x2 I think this should be y = cosh(x) and y =1 + ½ x2 87.254.72.195 (talk) 14:56, 27 November 2020 (UTC)

Inconsistent formulae comparing catenary with parabola
In the illustration comparing a catenary with parabola, below the picture it says: "The catenary and parabola equations are respectively, y = cosh(x) and y = x^2" I think this should be "y = cosh(x) and y = 1 + ½ x^2" 87.254.72.195 (talk) 15:02, 27 November 2020 (UTC)

Alternate derivation
New editor has recently added another derivation. (It is included below, collapsed, for reader convenience.) It is unsourced, and has been reverted as such. I think it is worthwhile to discuss (1) is it possible that it could be reliably sourced? and (2) if it were reliably sourced, would it be worth including in the article? --JBL (talk) 22:21, 2 February 2021 (UTC)

Derivation via Variational Calculus
This derivation avoids the complicated integrals encountered in other methods.

Given 2 points in 2D space, $$(-d,y_d)$$ and $$(d,y_d)$$, between which a rope will freely hang, find its shape and length $$2l_0$$.

This in essence is an optimization problem; the solution is a path between the 2 points for which the potential energy is at a minimum. Calculus of Variations deals with that exactly: Given a path-integral dependent quantity, the Euler-Lagrange equation produces the path for which the quantity is extremized. In our case the quantity is the potential energy and the extremum is a minima.

The potential energy,$$U = \int dU = \int dmgh$$where $$dm$$ is the differential mass element, $$g$$ is the gravitational acceleration and $$h$$ is the height of the differential mass-element.

Denote the rope density as $$\rho$$, so $dm = \rho dl$. The differential length element rewrite as $dl = \sqrt{dx^2 +dy^2} = dx \sqrt{1+y'^2}$, and now identify $$h$$ as $$y(x)$$.

With this we rewrite,$$U = \rho g\int y \sqrt{1+y'^2} dx$$Solving the optimization problem means we found the $$y(x)$$ such that the value of this integral is at minimum.

The Euler-Lagrange equation,$$\frac{d}{dx} \left( \frac{\partial U}{\partial y'} \right) = \frac{\partial U}{\partial y}$$treats $$y$$ and $$y'$$ as independent variables, so the above differentiation has to be carried out accordingly,$$\frac{d}{dx} \left( \frac{yy'}{\sqrt{1+y'^2}} \right) = \sqrt{1+y'^2}$$Manufacture the differential on the LHS,$$\frac{yy'}{\sqrt{1+y'^2}} d \left( \frac{yy'}{\sqrt{1+y'^2}} \right)= ydy$$this is now simply integrated into,$$\frac{1}{2} \left( \frac{yy'}{\sqrt{1+y'^2}} \right)^2 = \frac{1}{2} y^2 + C$$where $$C$$ is some constant of integration.

rearrange,$$\frac{1}{2}y^2 \left( \frac{y'^2} -1\right) =C \Rightarrow \frac{1}{2}y^2 \left( \frac{-1}\right) =C$$$$C$$ is 0 iff $$y$$ is a constant 0, which is not out solution. Since $$C$$ is yet to be determined, we can rewrite this equation more compactly as,$$Cy^2 = 1 +y'^2 \quad (*)$$and considering this form, we see this $$C$$ has to be positive.

We have now arrived at a 1st-order non-linear ODE which one can solve by either guessing $$e^{rx}$$ and solving the characteristic equation, or by differentiating to get,$$Cyy' = y'y \Rightarrow Cy=y$$which is a well-known 2nd-order linear ODE which we know is solved by exponents of $$\pm \sqrt{C}x$$. Any linear combination of these exponents is a solution to the ODE, but we are looking for the one that produces a symmetric curve (rope is horizontal at the vertex). One can verify that the only combination to satisfy this is the $$\cosh$$ function, so our solution is of the form,$$y(x) = y_0 \cosh(\sqrt{C}x)$$Now sub this into $$(*)$$ to determine $$C$$,$$Cy_0^2\cosh^2(\sqrt{C}x)=1+y_0^2C\sinh^2(\sqrt{C}x)$$considering the hyperbolic identity $\cosh^2\varphi-\sinh^2\varphi =1$, this equation holds only if,$$y_0^2C=1 \Rightarrow \sqrt{C}= \pm\frac{1}{y_0} $$choose either since $$\cosh$$ is symmetric, and the solution is,$$y(x) = y_0 \cosh \left({\frac{x}{y_0}}\right)$$To get the $$y_0$$, fix the equation at the end of the rope,$$y_d \equiv y(d) = y_0 \cosh \left( \frac{d}{y_0} \right)$$solve this numerically for $$y_0$$ and the shape is fully determined.

The length can now be computed by directly integrating over the rope's length,

$$2l_0 = \int dl = \int_{-d}^{d} \sqrt{dx^2 +dy^2} = \int_{-d}^{d} dx \sqrt{1+y'^2} = \int_{-d}^{d} dx \cosh \left( \frac{x}{y_0} \right) = y_0 \sinh \left( \frac{x}{y_0} \right) \bigg|_{-d}^{+d} = 2y_0 \sinh \left( \frac{d}{y_0} \right) $$$$\Rightarrow {l_0} = y_0 \sinh \left( \frac{d}{y_0} \right)$$using again the hyperbolic identity, one can rewrite this as,$$l_0^2 = y_d^2 -y_0^2$$from which the length is immediately determined.

Path of a moving charge in a uniform electric field
The claim in the "Science" section that the path of a moving charge in a uniform electric field follows a catenary, is incorrect. The path described in the cited book is actually a "squashed" catenary $$y = b \cosh \left(\frac{x}{a}\right)$$ with $$b > a$$, even though the book says "along a catenary curve". Note that the cited book uses x and y differently; I here use the sense used in the Wikipedia article. Specifically, $$\frac{b}{a} = \frac{\sqrt{m_0^2c^4+p_x^2c^2}}{p_xc}$$ where $$m_0$$ is the invariant mass of the moving charge, and $$p_x$$ is its (constant) momentum along the x axis. Only in the limit where the moving charge travels at the speed of light does this become a true catenary.

I am uncertain how to best correct the claim (maybe just delete it?), as I am not a frequent contributor. For anyone who wants to verify this issue, note that equation (20.5) in the cited book has a typo: the right-hand numerator should be $$eEy$$ instead of $$eE_y$$. 81.191.117.202 (talk) 15:26, 10 December 2022 (UTC)