Talk:Catenary

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

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

Assuming Aλ=(0,-1) we get

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

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

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

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

k T = secφ

(k arbitrary)

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

tanφ = C + k s

(C arbitrary)

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

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

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

cos 2 φ = 1 / (1 + tan 2 φ)

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

x = arcsinh s

sin 2 φ = tan 2 φ / (1 + tan 2 φ)

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

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

142.177.169.142 04:42, 18 Aug 2004 (UTC)

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

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

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

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

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

1 + y' 2 = y y''

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

S'pose one could do the usual guessing until

y = a e α x + b e β x

comes up. Then

α + β = 0

and

a b (α - β) 2 = 1

which gives

y = cosh(α x + ln(2 a α)) / α

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

Exponential form

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

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

cords

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