Talk:Catenary

message: Old posts are archived here.
Phancy Physicist (talk) 05:05, 4 August 2010 (UTC)
thread_title:  [[/Archive_1|Archive 1]] 

message: 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)
thread_title:  Towed cables - clarification needed 

message: 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)
thread_title:  Physics of the Catenary 

message: 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)
thread_title:  Explanation 

message: 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)
thread_title:  Who discovered the equation? 

message: 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)
thread_title:  Artistic and Practical Applications of Approximate Catenary Curves

message: 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)
replies: text: 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)
thread_title:  Any info about catenary domes? 

message: 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)
replies: text: 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)
thread_title:  Problems with Towed cables section, class promotion 

message: 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)
thread_title:  Elastic Catenary 

message: 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)
thread_title:  Ctesiphon Arch 

message: 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)
thread_title:  Proposal of new figure and rewrite of "Towed cables" 

message: 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)
replies: text: 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)
thread_title: Towed cables

message: 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)
replies: text: 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)
thread_title: Spider web

message: 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)
replies: text: 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)
thread_title:  Huygens vs. Jeffeson 

message: 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." </br>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#Verifiable_reference_needed, and Talk:Suspension_bridge#7th_Century_Maya_bridge) </br>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)
replies: text: 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)
replies: text: 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)

text: 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)
thread_title:  Catenary bridge? 

message: addedd these new sections to the catenary article. I deleted these new sections because:
#They were unsourced.
#They were too long, adding over 12kb to a 39kb article.
#They mostlty derive results and properties given elsewhere in the article.
#Lengthy derivations and proofs do not belong in Wikipedia - Wikipedia is not a text book.
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)
replies: text: 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)
thread_title:  Review of new sections 

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

text: 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)
thread_title:  Deletion by User Gandalf61 

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thread_title:  External links modified 

message: 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)
thread_title:  Asymmetric Catenary 

message: 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)
thread_title:  Electrified railways 

message: 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)
thread_title:  Analysis Section is confusing 

message: 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)
thread_title:  Inconsistent formulae comparing catenary with parabola 

message: 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)
thread_title:  Alternate derivation 

message: 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)
thread_title:  Path of a moving charge in a uniform electric field