## Catenary Problem

All help appreciated!

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### Catenary Problem

I've been trying to solve a little problem involving catenary curves, and it's driving me nuts.

According to the standard derivation of the catenary curve, formed when a uniform heavy chain is supported above ground, the chain forms a curve of the form:

y = c cosh(x/c)

where c is a constant equal to T0/σg - here, T0 is the value of the tension in the chain at the bottom (where the tangent angle θ is zero), σ is the mass per unit length, and g is the acceleration due to gravity.

I've been considering a problem, illustrated by this image: Catenary Problem Image.jpg (23.61 KiB) Viewed 1250 times

Here, a heavy chain of constant mass per unit length, is suspended from two supports, both of height h, each support being a distance d from the y-axis. I'm interested in knowing what happens when L, the length of the chain, is increased, whilst h and d are kept constant.

Now in the original derivation of the catenary curve, the coordinate system used assumes that at x=0, y=c. In the above setup, this no longer holds, and instead, the curve is more properly considered to be:

y = k cosh(x/k) + Q

where Q is a vertical scaling constant that moves the curve so that when x=0, y=c, regardless of the value of k (k now being the value that is genuinely equal to T0/σg). Since cosh(0) = 1, we have:

c = k + Q, therefore Q = c - k 

Therefore our actual equation for the curve in the above setup is:

y = k cosh(x/k) + c - k 

We also take note of the fact that the arc length of the curve is given by:

s = k sinh(x,k) 

and we therefore have the following results:

L/2 = k sinh(d/k) 

h = k cosh(d/k) + c - k 

I am interested to find out what happens when L is increased, while keeping d and h constant, until the point where the curve touches the ground (this limiting value of L I shall refer to as Lmax).

Now, from an elementary geometrical standpoint, in a Euclidean space, any curved arc connecting two points has an arc length that is longer than the straight ling segment connecting those two points. Therefore, in the limiting case for Lmax, we expect ½Lmax to be greater than the length of the line joining (0,0) to (d,h), which, of course, is given by the elementary Pythagoras relation as (d2+h2)½. Therefore (d2+h2)½ constitutes a lower bound on ½Lmax. Let's refer to this lower bound condition as Lemma 1.

So, I now try to find a relationship between L and c, so that I can determine what happens to L as c vanishes in the limit - as L increases, so c diminishes. This is where matters become annoying in the extreme.

We now take advantage of the hyperbolic identity:

cosh2z - sinh2z = 1 

Reworking  to give:

(h - c + k) = k cosh(d/k) 

we now take  and  together, and derive:

(h - c + k)2 - (L/2)2 = k2 

Expanding the term (h + c - k)2, we have:

(h + c - k)2 = h2 + 2ch - 2hk +c2 - 2ck + k2 

Substituting  into , eliminating terms and rearranging gives:

h2 + (2h - c)(k - c) - ck = (L/2)2 

Letting c->0 in the limit gives us that Lmax/2 = h, which is in clear violation of Lemma 1, and therefore untenable as a result.

So just what have I missed above, in order to arrive at this contradictory result?
Signature temporarily on hold until I can find a reliable image host ... Calilasseia
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### Re: Catenary Problem

Letting c->0 in equation  yields:
h2 + 2hk = (Lmax/2)2
I am, somehow, less interested in the weight and convolutions of Einstein’s brain than in the near certainty that people of equal talent have lived and died in cotton fields and sweatshops. - Stephen J. Gould newolder

Name: Albert Ross
Posts: 7308 Country: Feudal Estate number 9
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### Re: Catenary Problem

Bloody obvious now I see it! Thanks!

EDIT: what you're seeing above is what happens when I try doing mathematics at 2am while drifting off to sleep in bed Signature temporarily on hold until I can find a reliable image host ... Calilasseia
RS Donator

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Age: 58 Country: England Print view this post

### Re: Catenary Problem

I wouldn't worry too much. I'm wide awake and I thought you had a Fringillidae issue.
A most evolved electron. Animavore

Name: The Scribbler
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### Re: Catenary Problem

Happy to be of help. My own brain fades are more frequent than just before bedtimes and can last for extended periods these days. Heigh ho.
I am, somehow, less interested in the weight and convolutions of Einstein’s brain than in the near certainty that people of equal talent have lived and died in cotton fields and sweatshops. - Stephen J. Gould newolder

Name: Albert Ross
Posts: 7308 Country: Feudal Estate number 9
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### Re: Catenary Problem

hmmmmph....almost glad I didn't go for astrophysics... Travel photos > https://500px.com/macdoc/galleries
EO Wilson in On Human Nature wrote:
We are not compelled to believe in biological uniformity in order to affirm human freedom and dignity. Macdoc

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