# Calculating parabola of drooping chain

1. Mar 25, 2004

### Molydood

Hi,

I was giving a pub quiz type quesstion the other day, which I managed to solve without the use of formulae/mathematics, but it got me thinking about how to solve it mathematically too.

Imagine a chain stretched between two posts, that then droops in the middle, and forms a natural parabola. What are the variables involved and what is the relationship that determines the curve?

I guess that it would mean calculating for all positions on the chain (using calculus?), in order to give a final curve, but this would require a basic starting formulae stating the relationship.

So variable would be, for any specific point on the chain:
Chain tension
Chain weight
Chain angle (tangent to curve)

Any others?
Any idea on the relationship?

thanks,
Martin

2. Mar 25, 2004

### Integral

Staff Emeritus
You are on the right track. Trouble is the shape is not a parabola, it is a catenary.

I cannot help you off the top of my head, but this type of analysis can be found in books on Partial Differential equations.

3. Mar 25, 2004

### Chi Meson

4. Mar 25, 2004

### Molydood

thanks guys, that's great!

5. Mar 25, 2004

### Loren Booda

Molydood
How so?

6. Mar 26, 2004

### Molydood

Hi,

'Solve' is perhaps the wrong word. I'll give you the problem then post the answer a little later if required:
two posts of 4 metre height support a chain of length 6 metres. The chain hangs 1 metre from the floor at its lowest point. How far apart are the posts?

7. Mar 26, 2004

### Chi Meson

AHA! THis is a brain teaser that does not require any knowledge of the catenary! (I've heard this one before, and I never got the chance to discover it for myself, so I won't give it away)

8. Mar 26, 2004

### Chen

As Chi Meson said this has nothing to do with physics... but if the said chain hung 2 meters from the floor you couldn't solve it.

9. Mar 26, 2004

### chroot

Staff Emeritus
Yeah that brain teaser's pretty dumb... hehe.

- Warren

10. Mar 26, 2004

### Integral

Staff Emeritus
I can't believe that I did not see the solution instantly, had to think about it for a few minutes.

11. Mar 28, 2004

### Meithan

Nice brain teaser. I'll challenge my friends at school. I bet it'll drive them crazy!

12. Mar 28, 2004

### ShawnD

So how is it solved without using the caternary formula?

13. Mar 28, 2004

Here's a hint: Draw it to scale.

14. Mar 28, 2004

### ShawnD

It's a tad hard to draw something that fits this equation

$$y = \frac{e^{ax} + e^{-ax}}{2a}$$

Besides, it aparently can be done without using that formula.

15. Mar 28, 2004

Heh, start with the posts. Then add in the minimum height off the ground. Calculate a distance or two, like how far the chain is drooping and how long the chain is. The rest comes pretty naturally.

16. Mar 28, 2004

### Integral

Staff Emeritus
LOL, come on.... Think man.

Consider the problem statement.

You are going to hate yourself when the solution pops into your head.

17. Mar 28, 2004

### ShawnD

The only thing I can think of doing is breaking the length of the chain into little triangles then integrating it.

18. Mar 28, 2004

### Integral

Staff Emeritus
Let it all hang out!

19. Mar 28, 2004