Catenary
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In physics and geometry, the catenary is the curve that an idealised hanging chain or cable assumes when supported at its ends and acted on only by its own weight. The curve is the graph of the hyperbolic cosine function, and has a U-like shape, superficially similar in appearance to a parabola (though mathematically quite different). Its surface of revolution, the catenoid, is a minimal surface and is the shape assumed by a soap film bounded by two parallel circular rings.
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[edit] History
The word catenary is derived from the Latin word catena, which means "chain". Huygens first used the term catenaria in a letter to Leibniz in 1690. However, Thomas Jefferson is usually credited with the English word catenary.<ref>""Catenary" at Math Words". Pballew.net. 1995-11-21. http://www.pballew.net/arithme8.html#catenary. Retrieved 2010-11-17.</ref> The curve is also called the "alysoid", "chainette",<ref name="MathWorld">MathWorld</ref> or, particularly in the material sciences, "funicular".<ref>e.g.: Shodek, Daniel L. (2004). Structures (5th ed.). Prentice Hall. p. 22. ISBN 9780130488794. OCLC 148137330.</ref>
It is often stated<ref>For example Lockwood p. 124</ref> that Galileo thought that the curve followed by a hanging chain is a parabola. A careful reading of his book Two new sciences<ref>Galileo Galilei (1914). Dialogues concerning two new sciences. Trans. Henry Crew & Alfonso de Salvio. Macmillan. pp. 149, 290. http://books.google.com/books?id=SPhnaiERbWcC. </ref> shows this to be an oversimplification. Galileo discusses the catenary in two places; in the dialog of the Second Day he states that a hanging chain resembles a parabola. But later, in the dialog of the Fourth Day, he gives more details, and states that a hanging cord is approximated by a parabola, correctly observing that this approximation improves as the curvature gets smaller and is almost exact when the elevation is less than 45o. That the curve followed by a chain is not a parabola was proven by Joachim Jungius (1587–1657) and published posthumously in 1669.<ref>Swetz, Faauvel, Bekken, "Learn from the Masters," 1997, MAA ISBN 0-88385-703-0, pp.128-9</ref><ref name="Lockwood p. 124">Lockwood p. 124</ref>
The application of the catenary to the construction of arches is due to Robert Hooke, who discovered it in the context of the rebuilding of St Paul's Cathedral,<ref>"Monuments and Microscopes: Scientific Thinking on a Grand Scale in the Early Royal Society" by Lisa Jardine</ref> possibly having seen Huygens' work on the catenary. (Some much older arches are also approximate catenaries.)
In 1671, Hooke announced to the Royal Society that he had solved the problem of the optimal shape of an arch, and in 1675 published an encrypted solution as a Latin anagram<ref>cf. the anagram for Hooke's law, which appeared in the next paragraph.</ref> in an appendix to his Description of Helioscopes,<ref>"Arch Design". Lindahall.org. 2002-10-28. http://www.lindahall.org/events_exhib/exhibit/exhibits/civil/design.shtml. Retrieved 2010-11-17.</ref> where he wrote that he had found "a true mathematical and mechanical form of all manner of Arches for Building." He did not publish the solution of this anagram<ref>The original anagram was "abcccddeeeeefggiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux": the letters of the Latin phrase, alphabetized.</ref> in his lifetime, but in 1705 his executor provided it as Ut pendet continuum flexile, sic stabit contiguum rigidum inversum, meaning "As hangs a flexible cable so, inverted, stand the touching pieces of an arch."
In 1691 Gottfried Leibniz, Christiaan Huygens, and Johann Bernoulli derived the equation in response to a challenge by Jakob Bernoulli. David Gregory wrote a treatise on the catenary in 1697.<ref name="Lockwood p. 124"/>
Euler proved in 1744 that the catenary is the curve which, when rotated about the x-axis, gives the surface of minimum surface area (the catenoid) for the given bounding circle.<ref name="MathWorld"/>
[edit] The inverted catenary arch
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Hooke discovered that the catenary is the ideal curve for an arch of uniform density and thickness which supports only its own weight. When the centerline of an arch is made to follow the curve of an up-side-down (i.e. inverted) catenary, the arch endures almost pure compression, in which no significant bending moment occurs inside the material.[citation needed]
Catenary arches are often used in the construction of kilns. In this construction technique, the shape of a hanging chain of the desired dimensions is transferred to a form which is then used as a guide for the placement of bricks or other building material.<ref>Minogue, Coll; Sanderson, Robert (2000). Wood-fired Ceramics: Contemporary Practices. University of Pennsylvania. p. 42. ISBN 0812235142.</ref><ref> Peterson, Susan; Peterson, Jan (2003). The Craft and Art of Clay: A Complete Potter's Handbook. Laurence King. p. 224. ISBN 1856693546.</ref>
However the conditions for a catenary to be the ideal arch are almost never fulfilled: arches usually support more than their own weight, and on the rare occasions when they are freestanding they are sometimes not of uniform thickness.[citation needed]
The Gateway Arch in St. Louis, Missouri, United States is sometimes said to be an (inverted) catenary, but this is incorrect.<ref>Osserman, Robert (2010), "Mathematics of the Gateway Arch", Notices of the American Mathematical Society 57 (2): 220–229, ISSN 0002-9920, http://www.ams.org/notices/201002/index.html</ref> It is close to a more general curve called a flattened catenary, with equation y=Acosh(Bx). (A catenary would have AB=1.) While a catenary is the ideal shape for a freestanding arch of constant thickness, the Gateway Arch is narrower near the top. According to the U.S. National Historic Landmark nomination for the arch, it is a "weighted catenary" instead. Its shape corresponds to the shape that a weighted chain, having lighter links in the middle, would form.<ref name="nrhpinv2">Laura Soullière Harrison (1985) (PDF), National Register of Historic Places Inventory-Nomination: Jefferson National Expansion Memorial Gateway Arch / Gateway Arch; or "The Arch", National Park Service, http://pdfhost.focus.nps.gov/docs/NHLS/Text/87001423.pdf, retrieved 2009-06-21 and Accompanying one photo, aerial, from 1975PDF (578 KB)</ref>
[edit] Simple suspension bridges
Free-hanging chains follow the catenary curve, but suspension bridge chains or cables do not hang freely since they support the weight of the bridge. In most cases the weight of the cable is negligible compared with the weight being supported. When the force exerted is uniform with respect to the length of the chain, as in a simple suspension bridge, the result is a catenary.[citation needed]
When the force exerted is uniform with respect to horizontal distance, as in a suspension bridge, the result is a parabola.<ref> Paul Kunkel (June 30, 2006). "Hanging With Galileo". Whistler Alley Mathematics. http://whistleralley.com/hanging/hanging.htm. Retrieved March 27, 2009.</ref>
When suspension bridges are constructed, the suspension cables initially sag as the catenary curve, before being tied to the deck below, and then gradually assume a parabolic curve as additional connecting cables are tied to connect the main suspension cables with the bridge deck below.[citation needed]
[edit] Anchoring of marine objects
The catenary form given by gravity is taken advantage of in its presence in heavy anchor rodes. An anchor rode (or anchor line) usually consists of chain and/or cable. Anchor rodes are used by ships, oilrigs, docks, wind turbines and other marine assets which must be anchored to the seabed.
Particularly with larger vessels, the catenary curve given by the weight of the rode presents a lower angle of pull on the anchor or mooring device. This assists the performance of the anchor and raises the level of force it will resist before dragging. With smaller vessels and in shallow water it is less effective.<ref>"Chain, Rope, and Catenary - Anchor Systems For Small Boats". Petersmith.net.nz. http://www.petersmith.net.nz/boat-anchors/catenary.php. Retrieved 2010-11-17.</ref>
The catenary curve in this context is only fully present in the anchoring system when the rode has been lifted clear of the seabed by the vessel's pull, as the seabed obviously affects its shape while it supports the chain or cable. There is also typically a section of rode above the water and thus unaffected by buoyancy, creating a slightly more complicated curve.
[edit] Mathematical description
[edit] Equation
The equation of a catenary in Cartesian coordinates has the form<ref>Larson, Ron; Edwards, Bruce H. (2010). Calculus. Belmont, California: Brooks/Cole, Cengage Learning. p. 393. ISBN 0-547-16702-4.</ref>
- <math>y = a \, \cosh \left ({x \over a} \right ) = {a \over 2} \, \left (e^{x/a} + e^{-x/a} \right )</math>,
where <math>\cosh</math> is the hyperbolic cosine function.
The Whewell equation for the catenary is
- <math>\tan \varphi = \frac{s}{a}</math>.
Differentiating gives
- <math>\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a}</math>
and eliminating <math>\varphi</math> gives the Cesàro equation:
- <math>\kappa=\frac{a}{s^2+a^2}</math>.
[edit] Other properties
All catenary curves are similar to each other. Changing the parameter a is equivalent to a uniform scaling of the curve.<ref>"Catenary". Xahlee.org. 2003-05-28. http://xahlee.org/SpecialPlaneCurves_dir/Catenary_dir/catenary.html. Retrieved 2010-11-17.</ref>
A parabola rolled along a straight line traces out a catenary (see roulette) with its focus.<ref name="MathWorld"/>
Square wheels can roll perfectly smoothly if the road has evenly spaced bumps in the shape of a series of inverted catenary curves. The wheels can be any regular polygon except a triangle, but the catenary must have parameters corresponding to the shape and dimensions of the wheels.<ref>"Roulette: A Comfortable Ride on an n-gon Bicycle" by Borut Levart, Wolfram Demonstrations Project, 2007.</ref>
A charge in a uniform electric field moves along a catenary (which tends to a parabola if the charge velocity is much less than the speed of light c).[citation needed]
The surface of revolution with fixed radii at either end that has minimum surface area is a catenary revolved about the x-axis.
Over any horizontal interval [a,b], the ratio of the area under the caternary to its length equals a, independent of the interval selected. The catenary is the only plane curve other than a horizontal line with this property. Also, the geometric centroid of the area under a stretch of catenary is the midpoint of the perpendicular segment connecting the centroid of the curve itself and the x-axis.<ref>Parker, Edward (2010), "A Property Characterizing the Catenary", Mathematics Magazine 83: 63-64</ref>
[edit] Analysis
We assume that the path followed by the chain is given parametrically by <math>\mathbf{r} = (x, y) = (x(s),\ y(s))</math> where <math>s</math> represents arc length and <math>\mathbf{r}</math> is the position vector. This is the natural parameterization and has the property that <math>\frac{d\mathbf{r}}{ds}</math> is the unit tangent vector, <math>\mathbf{u}</math>. The derivation of the curve for an optimal arch is similar except that the forces of tension become forces of compression and everything is inverted.
It is now possible to derive two equations which together define the shape of the curve and the tension of the chain at each point. This is done by a careful inspection of the various forces acting on a small segment of the chain and using the fact that these forces must be in balance if the chain is in static equilibrium.
First, let <math>\mathbf{T}=\mathbf{T}(s)</math> be the force of tension as a function of <math>s</math>. The chain is flexible so it can only exert a force parallel to itself. Since tension is defined as the force that the chain exerts on itself, <math>\mathbf{T}</math> must be parallel to the chain. In other words,
- <math>(1)\qquad\mathbf{T} = T \mathbf{u}</math>
where <math>T</math> is the magnitude of <math>\mathbf{T}</math>, a positive scalar function of <math>s</math>.
Second, let <math>\mathbf{G}=\mathbf{G}(s)</math> be the external force per unit length acting on a small segment of a chain as a function of <math>s</math>. The forces acting on the segment of the chain between <math>s</math> and <math>s+\Delta s</math> are the force of tension <math>\mathbf{T}(s+\Delta s)</math> at one end of the segment, the nearly opposite force <math>-\mathbf{T}(s)</math> at the other end, and the external force acting on the segment which is approximately <math>\mathbf{G}\Delta s</math>. These forces must balance so
- <math>\mathbf{T}(s+\Delta s)-\mathbf{T}(s)+\mathbf{G}\Delta s \approx \mathbf{0}</math>.
Divide by <math>\Delta s</math> and take the limit as <math>\Delta s \to 0</math> to obtain
- <math>(2)\qquad\frac{d\mathbf{T}}{ds} + \mathbf{G} = \mathbf{0}</math>.
Note that, up till now, no assumptions have been made regarding the force <math>\mathbf{G}</math>, so equations (1) and (2) can be used as the starting point in the analysis of a flexible chain acting under any external force. The next step is to put in the specific expression for <math>\mathbf{G}</math> and solve the resulting equations.
In this case, <math>\mathbf{G} = (0, -\lambda g)</math> where the chain has constant mass per unit length <math>\lambda</math> and the only external force acting on the chain is that of a uniform gravitational field <math>\mathbf{g} = (0, -g)</math>. So we have
- <math>\frac{d\mathbf{T}}{ds} = (0, \lambda g)</math>.
Integrating we get,
- <math>\mathbf{T} = (c, \lambda g s + d)</math>.
Note that at the minimum the curve is horizontal and <math>\mathbf{T} = T \mathbf{u} = T (1, 0) = (T, 0) = (c, \lambda g s + d)</math>. So <math>c</math> is the tension of the chain at its lowest point this point occurs at <math>s = -d/\lambda g</math>. The point from which <math>s</math> is measured is arbitrary, so pick this point to be the minimum, giving <math>d = 0</math>. The equation becomes
- <math>\mathbf{T} = (c, \lambda g s)</math>.
Note that the horizontal component of the tension is a constant.
From here, we can continue the derivation in two ways.
[edit] Alternative 1
If <math>\varphi</math> is the tangential angle of the curve then <math>\mathbf{T}</math> is parallel to <math>(1, \tan \varphi)</math> so
- <math>\tan \varphi = \frac{\lambda g s}{c}</math>.
Write <math>a = \frac{c}{\lambda g}</math> to combine constants and obtain the Whewell equation for the curve,
- <math>\tan \varphi = \frac{s}{a}</math>.
In general, parametric equations can be obtained from a Whewell equation by integrating:
- <math>x = \int \cos \varphi \, ds</math>
- <math>y = \int \sin \varphi \, ds</math>
To find these integrals, make the substitution <math>\tan{\varphi} = \sinh u</math> (or <math>\varphi = \mbox{gd} u</math> where <math>\mbox{gd}</math> is the Gudermannian function).
Then <math>s = a \ \sinh u</math> and
- <math>x = \int \cos {\varphi} ds = \int \mbox{sech} u \, a \cosh u du = a \int du = au + \alpha</math>
- <math>y = \int \sin {\varphi} ds = \int \tanh u \, a \cosh u du = a \int \sinh u du = a \cosh u + \beta</math>.
We can eliminate u to obtain
- <math>y = a \cosh \frac{x-\alpha}{a} + \beta</math>
where <math>\alpha</math> and <math>\beta</math> are constants to be determined, along with <math>a</math>, by the boundary conditions of the problem. Usually these conditions include two points from which the chain is being suspended and the length of the chain.
[edit] Alternative 2
From
- <math>\mathbf{T} = T \mathbf{u} = T \left(\frac{dx}{ds}, \frac{dy}{ds}\right) = (c, \lambda g s)</math>,
- <math>\frac{dy}{dx} = \frac{dy}{ds}\left/\frac{dx}{ds}\right. = \frac{\lambda g s}{c} = \frac{s}{a}</math>,
where <math>a = \frac{c}{\lambda g},</math> same as before. Then
- <math>\frac{dx}{ds} =1\left/\frac{ds}{dx}\right. = \frac{1}{\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}} = \frac{a}{\sqrt{a^2+s^2}}</math>
and
- <math>\frac{dy}{ds} = \frac{dy}{dx}\cdot\frac{dx}{ds} = \frac{s}{\sqrt{a^2+s^2}}</math>.
The integrals of the right hand sides of these equations can be found using standard techniques giving
- <math>x = a\ \operatorname{arcsinh}(s/a) + \alpha,\ y = \sqrt{a^2+s^2} + \beta</math>.
Isolating <math>s</math> in the first equation and using the result to substitute <math>s</math> in the second equation gives
- <math>y = a \cosh \frac{x-\alpha}{a} + \beta</math>
as before, <math>\alpha</math> and <math>\beta</math> are constants to be determined, along with <math>a</math>, by the boundary conditions of the problem, which is exact the same result as that obtained with Alternative 1.
[edit] Variations
[edit] Elastic catenary
In an elastic catenary, the cable replaced by a spring and is no longer assumed to be of fixed density, but is allowed to stretch in accordance with Hooke's Law. In this case, the mass per unit length is no longer constant but can be given as
- <math>\lambda = \frac{\lambda_0}{1+\epsilon T}</math>
where <math>\lambda_0</math> is the mass per unit length for the chain in its relaxed state and <math>\epsilon</math> is the spring constant. As in the earlier derivation,
- <math>\frac{d\mathbf{T}}{ds} = (0, \lambda g)</math>.
So the horizontal component of <math>\mathbf{T}</math>, <math>T \cos \varphi</math> is a constant c. Putting this into the equation for density produces
- <math>\lambda = \frac{\lambda_0}{1+ c\epsilon \sec \varphi}</math>.
Then the equation for the vertical component of <math>\mathbf{T}</math> is
- <math>\frac{d}{ds}(T \sin \varphi) = c\frac{d}{ds} (\tan \varphi) = \frac{g\lambda_0}{1+ c\epsilon \sec \varphi}</math>,
or, combining constants,
- <math>\frac{d}{ds} (\tan \varphi) = \frac{1}{a+ b\sec \varphi}</math>.
Using the substitution <math>\tan{\varphi} = \sinh u</math> gives
- <math>\frac{d}{ds} (\sinh u) = \cosh u \frac{du}{ds} = \frac{1}{a+ b\cosh u}</math>
or
- <math> \cosh u (a+ b\cosh u) = \frac{ds}{du}</math>.
Parametric equations can be obtained by integrating:
- <math>x = \int \cos \varphi \, ds = \int \mbox{sech} u \cosh u (a+ b\cosh u)du = \int (a+ b\cosh u)du= au + b\sinh u + \alpha </math>,
- <math>y = \int \sin \varphi \, ds = \int \tanh u \cosh u (a+ b\cosh u)du = \int \sinh u(a+ b\cosh u)du = a\cosh u + \tfrac{b}{2}\sinh^2 u + \beta </math>.
When b = 0, corresponding to a completely inelastic cable, this is simply the catenary. When a = 0, corresponding to the case there the cable essentially has length 0 in its relaxed state, similar to a Slinky, this is a parabola. When a and b are both >0 then the curve is intermediate between a catenary and a parabola.
[edit] Equal resistance catenary
In an equal resistance catenary, cable is strengthened according to the magnitude of the tension at each point, so its resistance to breaking is constant along its length. Assuming that the strength of the cable is proportional to its density, the mass per unit length can be given as
- <math>\lambda = \lambda_r T</math>
where <math>\lambda_r</math> is the mass per unit length per unit of tension force required for the chain to resist breaking. As in the earlier derivation,
- <math>\frac{d\mathbf{T}}{ds} = (0, \lambda g)</math>.
So the horizontal component of <math>\mathbf{T}</math>, <math>T \cos \varphi</math> is a constant c. Putting this into the equation for density produces
- <math>\lambda = \lambda_r c \sec \varphi</math>.
Then the equation for the vertical component of <math>\mathbf{T}</math> is
- <math>\frac{d}{ds}(T \sin \varphi) = c\frac{d}{ds} (\tan \varphi) = g \lambda_r c \sec \varphi</math>,
or, combining constants,
- <math>\frac{d}{ds} (\tan \varphi) = \frac{1}{a} \sec \varphi</math>
or
- <math>\frac{d}{ds}\left(\frac{dy}{dx}\right) = \frac{1}{a} \frac{ds}{dx}</math>.
Multiplying both sides by <math>ds/dx</math> gives
- <math>\frac{d^2y}{dx^2} = \frac{1}{a} \left(\frac{ds}{dx}\right)^2 = \frac{1}{a} \left[1+\left(\frac{dy}{dx}\right)^2\right]</math>.
This can be reduced to a differential equation of degree one using separation of variables to obtain
- <math>\arctan \frac{dy}{dx} = \frac{x}{a}-\alpha</math>
or
- <math>\frac{dy}{dx} = \tan\left(\frac{x}{a}-\alpha\right)</math>.
Another integration produces
- <math>y = -a\ln \cos\left(\frac{x}{a}-\alpha\right) + \beta</math>.
[edit] Towed cables
Instead of gravity, we assume we have a cylindrical cable that is acted on by drag forces due to the movement of some surrounding fluid (e.g. air or water). The velocity relative to the cable is assumed to be a constant <math>\mathbf{v} = (0, -v)</math>. (Velocity is assumed to be vertical here to preserve similarities with the gravitational case.) To compute the force due to drag, write <math>\mathbf{v} = \mathbf{v}_u + \mathbf{v}_n</math> where <math>\mathbf{v}_u</math> and <math>\mathbf{v}_n</math> 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 <math>\mathbf{v}_u</math> is taken to be negligible. The force acting on the cable, following the Drag equation is
- <math>\mathbf{G} = -c |\mathbf{v}_n|^2 \mathbf{n}</math>
where <math>c</math> is a constant depending on the density of the fluid, the diameter of the cable, and the Drag coefficient. If <math>\mathbf{n} = (-\sin\varphi,\ \cos\varphi)</math> denotes the unit normal vector, then
- <math>\mathbf{v}_n = \mathbf{v}\cdot\mathbf{n}\ \mathbf{n} = -\cos\varphi\ \mathbf{n}</math>.
So
- <math>\mathbf{G} = -c \cos^2\varphi\ \mathbf{n}</math>.
From equations (1) and (2) above,
- <math>\frac{d\mathbf{T}}{ds} = \frac{dT}{ds} \mathbf{u} + T\frac{d\mathbf{u}}{ds}
= \frac{dT}{ds} \mathbf{u} + T\frac{d\varphi}{ds}\mathbf{n} = c \cos^2\varphi\ \mathbf{n}</math>. Setting the coefficients of <math>\mathbf{u}</math> and <math>\mathbf{n}</math> equal produces
- <math>\frac{dT}{ds} = 0,\,T\frac{d\varphi}{ds} = c \cos^2\varphi</math>.
So T is a constant in this case and combining constants in the second equation gives
- <math>\frac{d\varphi}{ds} = \frac{\cos^2\varphi}{a}</math>
which is one of the equations for the catenary given above. This is a case where a different expression for the force acting on the chain/cable produce the same curve but a different expression for tension.
In applications, the force of gravity and additional terms in the force due to drag may be added to the expression for force, yielding equations that must be solved numerically.
[edit] Alternative analysis
The forces acting on a segment of catenary curve are shown in the figure at right.
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 <math>f</math> one gets that the vertical component of the force at the left extreme <math> x_1</math> is <math> -f \ y^\prime(x_1)</math> and at the right extreme <math> x_2</math> is <math> f \ y^\prime(x_2)</math> The path length of the curve representing a function <math> y(x)</math> with <math> x </math> varying from <math> x_1 </math> to <math> x_2 </math> is
- <math>
\int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx </math>
If <math>g</math> is the gravitational constant and <math>\rho</math> is the mass per length unit of the chain the gravitational force acting on the arc from <math> x_1 </math> to <math> x_2 </math> is <math>g \rho \int\limits_{x_1}^{x_2}\sqrt{1+{y^\prime}^2} dx \,</math>
This force must be compensated by the vertical components of the forces acting on the two extremes of the arc, i.e.
|
</p> | ({{{3}}}) |
Denoting the constant ratio <math>\frac {f}{ g \rho}</math> with <math>a</math> and taking the derivative of equation (1) with respect to the upper limit of the integral, i.e. with respect to <math>x_2</math>, one gets
- <math>y^{\prime\prime} = \frac{1 }{a} \sqrt{1+{y^\prime}^2}</math>
Denoting <math>y^\prime</math> with <math>z</math> this equation takes the form
- <math>z^\prime = \frac{dz}{dx}= \frac{1 }{a} \sqrt{1+z^2}</math>
what means that for the inverse function <math>x(z)</math> one has
- <math>\frac{dx}{dz}= \frac{a }{\sqrt{1+z^2}}</math>
which is integrate to
- <math>x= a \ \sinh^{-1}(z) + x_0</math>
where <math>x_0</math> is the constant of integration or equivalently
- <math>z=y^\prime=\sinh(\frac{x-x_0 }{a})</math>
Again integrating with respect to <math>x</math> one gets
|
</p> | ({{{3}}}) |
where <math>y_0</math> is the second constant of integration
The lowest point of this curve has the coordinates <math>(\ x_0\ ,\ y_0 + a)</math>
The length of the curve given by (2) from <math>x=x_1</math> to <math>x=x_2</math> is
|
</p> | ({{{3}}}) |
This family of solutions is parametrized with the 3 parameters <math>a,\ x_0,\ y_0 </math>. 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 <math>x_1,\ y_1</math> and <math>x_2,\ y_2</math> relative a vertical coordinate system should be computed.
This means that <math>a,\ x_0,\ y_0 </math> have to be determined such that
|
</p> | ({{{3}}}) |
|
</p> | ({{{3}}}) |
|
</p> | ({{{3}}}) |
Setting
- <math>x_m=\frac{x_2+x_1 }{2}</math>
- <math>\Delta x=\frac{x_2-x_1 }{2}</math>
subtracting (4) from (5) and then dividing with <math>a</math> one gets
|
</p> | ({{{3}}}) |
For any given values <math>x_1,\ y_1,\ x_2,\ y_2,\ a</math> one can determine <math>\sinh \frac{x_m-x_0 }{a}</math> from (7)
When <math>\sinh \frac{x_m-x_0 }{a}</math> has been determined <math>\frac{x_m-x_0 }{a}</math> is computed by solving a quadratic equation.
In case <math>y_1=y_2</math>, i.e. in the case that the two attachment points are at the same heigth, one has that <math>x_0=x_m</math> and that the length is <math>l= 2\ a \left (\sinh \frac{\Delta x}{a} \right )</math>
With <math>x_0</math> known (4) or (5) can subsequently be used to determine <math>y_0</math>.
Having determined <math>x_0</math> with the algorithm just described the curve length <math>l</math> corresponding to the selected <math>a</math> value can be computed from (6). With an iterative algorithm the <math>a</math> value that corresponds to a certain curve length <math>l</math> can finally be derived.
From figure 1 it is further clear that the tension of the chain at any point <math>(x\ ,\ y)</math> is <math>f \sqrt{1+{y^\prime}^2} = f \cosh \frac{x-x_0}{a}=f\ \frac{y-y_0}{a}</math> where <math>f=a\ g\ \rho</math> is the magnitude of the constant horizontal force component
If the mass density <math>\rho</math> is not constant but varies depending on some law the resulting differential equation will in most cases not have a closed form analytic solution. But the resulting curve can still be determined with arbitrary accuracy by the numerical integration of the differential equations
- <math>y^\prime = z</math>
- <math>z^\prime = \frac{g}{f}\ \ \rho(z) \sqrt{1+z^2}</math>
Given any initial values for <math>y(x1)</math> and <math>z(x1)</math> and any value for the parameter <math>f</math> these differential equations can be propagated to <math>x=x_2</math> with <math>\rho</math> specified as any function of the state variable <math>z</math>. The free parameters to be iteratively adjusted to fit the boundary constraints are now <math>z(x1)</math> and <math>f</math>. They can for example be adjusted iteratively such that <math>y(x2)=y_2</math> where <math>(x_2\ ,\ y_2)</math> is the second attachment point. This leaves an additional degree of freedom for the two parameters that can be used to get the correct length of the curve.
An example is the "elastic catenary" for which the force
- <math>f \sqrt{1+{y^\prime}^2} = f \sqrt{1+z^2}</math>
stretches the material with a factor
- <math>1\ +\ \epsilon f \ \sqrt{1+z^2}</math>
where <math>\epsilon</math> is an elasticity coefficient and that therefore the mass density (mass per unit length) is
- <math>\rho = \frac {\rho_0} {1\ +\ \epsilon f\ \sqrt{1+z^2}}</math>
where <math>\rho_0 </math> is the mass density of the material in the absence of stress.
A case where a closed form mathematical solution is possible is the case of "the equal resistance catenary" where the mass density (mass per unit length) is proportional to the force <math>f \sqrt{1+{y^\prime}^2} = f \sqrt{1+z^2}</math>, i.e.
- <math>\rho = \rho_0 \sqrt{1+z^2}</math>
where <math>\rho_0 </math> is the density at the lowest point
Setting <math>a=\frac {f}{ g \rho_0}</math> the differential equations now take the form
- <math>y^\prime = z</math>
- <math>z^\prime = \frac{1}{a}\ \left( 1+z^2 \right )</math>
what means that for the inverse function <math>x(z)</math> one has
- <math>\frac{dx}{dz}= \frac{a }{1+z^2}</math>
which is integrate to
- <math>x= a \ \arctan(z) + x_0</math>
where <math>x_0</math> is the constant of integration or equivalently
- <math>z=y^\prime=\tan(\frac{x-x_0 }{a})</math>
where <math>x</math> is constraint to an interval <math>x_0 -a \ \frac{\pi}{2} < x < x_0 +a \ \frac{\pi}{2}</math>
Again integrating with respect to <math>x</math> one gets
- <math>y=y_0 \ -a\ln \left (\cos \left (\frac{x-x_0 }{a}\right )\right )</math>
where <math>y_0</math> is the second constant of integration.
As <math>a \ \sinh \frac{C}{a} \to \infty</math> when <math>a \ \to 0</math> for any constant <math>C</math> it follows from (6) that by making a catenary that is fixed at two points sufficiently long the constant horizontal force component <math>f</math> can be made arbitrarily small. For this generalized "catenary of equal resistance" this is no more true, as <math> a </math> must be larger then <math>\frac{|x-x_0|}{\frac{\pi}{2}}</math> for any <math> x </math> between <math> x_1 </math> and <math> x_2 </math> the positions of the two attachment points and the density <math>\rho_0</math> at the lowest point impose a lower limit for the fixed horizontal force component <math>f</math>
[edit] Alternative analysis "towed cables"
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. <math>\mathbf{v} = (0, -v)</math>. To compute the force due to drag, write <math>\mathbf{v} = \mathbf{v}_u + \mathbf{v}_n</math>
where <math>\mathbf{v}_u</math> and <math>\mathbf{v}_n</math> 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 <math>\mathbf{v}_u</math> is taken to be negligible. The force acting on the cable, per unit length, following the Drag equation is therefore
- <math>\mathbf{G} = f(x) \mathbf{n}</math>
with
|
</p> | ({{{3}}}) |
where <math>c</math> is a constant depending on the density of the fluid, the diameter of the cable, and the Drag coefficient and <math>\mathbf{n} = (n_x\ ,\ n_y)</math> denotes the unit normal vector.
For any curve <math>y(x)</math> the tangent (unit vector) is
{{{2}}} |
</p> | ({{{3}}}) |
and the normal (unit vector) is
{{{2}}} |
</p> | ({{{3}}}) |
|
</p> | ({{{3}}}) |
From (3) and (4) follows that the x-component of the total force on the segment of the curve from <math>x= x_1</math> to <math>x= x_2</math> is
|
</p> | ({{{3}}}) |
and the component in the y-direction is
|
</p> | ({{{3}}}) |
{{{2}}} |
</p> | ({{{3}}}) |
\end{align} </math>|}}
If the now the force in the cable is
- <math>F\ =c\ v^2\ a\ </math>
the force at the right extreme of the cable segment is
- <math> c\ v^2 \ a\ ( t_x(x_2)\ ,\ t_y(x_2) ) </math>
and at the left extreme
- <math> -c\ v^2 \ a\ ( t_x(x_1)\ ,\ t_y(x_1) ) </math>
From (7) and (8) 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
[edit] See also
- Overhead lines
- Roulette (curve) - an elliptic/hyperbolic catenary
- Troposkein - the shape of a spun rope
[edit] References
[edit] Bibliography
- Lockwood, E.H. (1961). "Chapter 13: The Tractrix and Catenary". A Book of Curves. Cambridge. http://www.archive.org/details/bookofcurves006299mbp.
- Weisstein, Eric W., "Catenary" from MathWorld.
- O'Connor, John J.; Robertson, Edmund F., "Catenary", MacTutor History of Mathematics archive, University of St Andrews, http://www-history.mcs.st-andrews.ac.uk/Curves/Catenary.html.
- "Chaînette" at Encyclopédie des Formes Mathématiques Remarquables
- "Chaînette élastique" at Encyclopédie des Formes Mathématiques Remarquables
- "Courbe de la corde à sauter" at Encyclopédie des Formes Mathématiques Remarquables
[edit] External links
| 40x40px | Wikimedia Commons has media related to: Catenary |
| 40x40px | Wikisource has the text of the 1911 Encyclopædia Britannica article Catenary. |
- "Catenary of equal resistance" at Encyclopédie des Formes Mathématiques Remarquables
- "Catenary" at Visual Dictionary of Special Plane Curves
- Hanging With Galileo - mathematical derivation of formula for suspended and free-hanging chains; interactive graphical demo of parabolic vs. hyperbolic suspensions.
- Catenary Demonstration Experiment - An easy way to demonstrate the Mathematical properties of a cosh using the hanging cable effect. Devised by Jonathan Lansey
- Horizontal Conveyor Arrangement - Diagrams of different horizontal conveyor layouts showing options for the catenary section both supported and unsupported
- Catenary curve derived - The shape of a catenary is derived, plus examples of a chain hanging between 2 points of unequal height, including C program to calculate the curve.
- Cable Sag Error Calculator - Calculates the deviation from a straight line of a catenary curve and provides derivation of the calculator and references.
- Hexagonal Geodesic Domes - Catenary Domes, an article about creating catenary domes
- Dynamic as well as static cetenary curve equations derived - The equations governing the shape (static case) as well as dynamics (dynamic case) of a centenary is derived. Solution to the equations discussed.af:Kettinglyn
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