Calculating parabola of drooping chain

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    Chain Parabola
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Discussion Overview

The discussion revolves around the mathematical modeling of a chain suspended between two posts that droops in the middle, forming a curve. Participants explore the variables involved in determining the shape of the curve, which some identify as a catenary rather than a parabola. The conversation includes both theoretical considerations and practical problem-solving approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests that the relationship determining the curve involves chain tension, weight, and angle, and seeks additional variables.
  • Another participant corrects the initial assumption that the shape is a parabola, stating it is actually a catenary.
  • Several participants discuss a specific problem involving two posts and a chain, with one participant noting it can be solved without knowledge of the catenary.
  • Hints are provided about visualizing the problem, including drawing to scale and approximating the chain's path with a triangle.
  • Some participants express frustration over the problem's difficulty, while others find humor in the challenge.
  • There are mentions of integrating the length of the chain into smaller segments to analyze the curve, but no consensus on the best approach is reached.
  • One participant reflects on how those less familiar with mathematical concepts might solve the problem more easily than those who are knowledgeable.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the problem or the nature of the curve, with multiple competing views and approaches remaining throughout the discussion.

Contextual Notes

The discussion includes various assumptions about the problem's parameters and the mathematical relationships involved, but these assumptions are not universally agreed upon. The complexity of the problem and the differing levels of mathematical understanding among participants contribute to the ongoing debate.

Molydood
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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?

looking forward to hearing your comments

thanks,
Martin
 
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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.
 
This site says it all:
http://teacher.sduhsd.k12.ca.us/abrown/Activities/Matching/answers/Catenary.htm
 
Last edited by a moderator:
thanks guys, that's great!
 
Molydood
which I managed to solve without the use of formulae/mathematics
How so?
 
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?
 
Molydood said:
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?

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)
 
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. :smile:
 
Yeah that brain teaser's pretty dumb... hehe.

- Warren
 
  • #10
I can't believe that I did not see the solution instantly, had to think about it for a few minutes.
 
  • #11
Nice brain teaser. I'll challenge my friends at school. I bet it'll drive them crazy!
 
  • #12
So how is it solved without using the caternary formula?
 
  • #13
Here's a hint: Draw it to scale.

cookiemonster
 
  • #14
It's a tad hard to draw something that fits this equation

[tex]y = \frac{e^{ax} + e^{-ax}}{2a}[/tex]

Besides, it aparently can be done without using that formula.
 
  • #15
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.

cookiemonster
 
  • #16
LOL, come on... Think man.

Consider the problem statement.

You are going to hate yourself when the solution pops into your head.
 
  • #17
The only thing I can think of doing is breaking the length of the chain into little triangles then integrating it.
 
  • #18
Let it all hang out!
 
  • #19
You're working way too hard here, Shawn.

cookiemonster
 
  • #20
Well what's the answer? I've asked a 3rd year chem honors student as well as a 2nd year mechanical engineer and neither of them can figure it out. I'm stumped as well.
 
  • #21
cookiemonster said:
Calculate a distance or two, like how far the chain is drooping and how long the chain is.

That there is probably too big a hint...

cookiemonster
 
  • #22
The chain is drooping 3 and it's 6 long. Those are both given in the question.
 
  • #23
Yeah, now look at those two numbers for just one second.

cookiemonster
 
  • #24
Think degenerate solution. :wink:
 
  • #25
Yes I've seen that relationship but it still has to satisfy that equation.

[tex]y = \frac{e^0 + \frac{1}{e^0}}{2a}[/tex]

[tex]y = \frac{1 + \frac{1}{1}}{2a}[/tex]

[tex]y = \frac{2}{2a}[/tex]

[tex]y = \frac{1}{a}[/tex]

*
Nevermind, the 0 probably screws up the equation. 0's tend to do that.
 
Last edited:
  • #26
ShawnD said:
The only thing I can think of doing is breaking the length of the chain into little triangles then integrating it.

Draw it as one triangle.
 
  • #27
Shawn, did you ever get the answer? I'm interested to hear your reaction when you get it.
Martin
 
  • #28
I'm more interested to see the look on his face when he gets it. :biggrin:
 
  • #29
ShawnD said:
The chain is drooping 3 and it's 6 long. Those are both given in the question.
Consider approximating the chain's path with a triangle.
 
  • #30
NateTG said:
Consider approximating the chain's path with a triangle.

lol! Got it.
 

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