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11-06-2013
02:39 AM

11-06-2013
02:39 AM

One catenary problem - continued (again!)

For some reason when I try to post a reply to Fred here: http://communities.ptc.com/message/225402#225402 it seems to wipe out all the other posts in that thread. Hence I’ve started this continuation.

Fred said:

I agree. It would only be correct if Fx were the tension in the cable, but then some other equations would be wrong (though your equation above still has mismatched units Fred!).

Alan

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13 REPLIES 13

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11-06-2013
02:44 AM

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11-06-2013
03:16 AM

11-06-2013
03:16 AM

Re: One catenary problem - continued (again!)

Sorry

g*G

And second!

Where ara the sheet and the Animation

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11-06-2013
03:36 AM

11-06-2013
03:36 AM

Re: One catenary problem - continued (again!)

Valery Ochkov wrote:

Where ara the sheet and the Animation

See attached.

Alan

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11-06-2013
07:23 AM

11-06-2013
07:23 AM

Re: One catenary problem - continued (again!)

Thanks, Alan and Fred!

Len's we begin with a common problem - H.1 not equal H.2, x.0 not equal L/2, etc

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11-06-2013
08:25 AM

11-06-2013
08:25 AM

Re: One catenary problem - continued (again!)

Not tension, only the horizontal component:

Sorry, my image didn't come thru well. alpha is the angle of the tangent to the chain at the weight,

tangent of aplha is equal to the slope (dy/dx) AND the ratio of vertical tension to horizontal tension. If the weight G is shared equally (only at the midpoint) then the math follows.

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11-06-2013
02:03 PM

11-06-2013
02:03 PM

Re: One catenary problem - continued (again!)

I'll leave the animation to others; but I think this is the full solution.

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11-07-2013
03:39 AM

11-07-2013
03:39 AM

Re: One catenary problem - continued (again!)

Fred Kohlhepp wrote:

I'll leave the animation to others; but I think this is the full solution.

Thanks, Fred!

But see

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11-07-2013
08:22 AM

11-07-2013
08:22 AM

Re: One catenary problem - continued (again!)

Really?

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11-07-2013
08:57 AM

11-07-2013
08:57 AM

Re: One catenary problem - continued (again!)

Sorry one more:

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11-07-2013
10:39 AM

11-07-2013
10:39 AM

Re: One catenary problem - continued (again!)

Now I know you're joking! 😉

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11-07-2013
02:35 PM

11-07-2013
02:35 PM

Re: One catenary problem - continued (again!)

Great job fellows!

Do we have any novel way to calculate Guess values - or are they just guestimates? At least I am having issue to make this "...2b.xmcd" to find solution in some real world cases. Solution sheet also fails with Fwft:= 0N.

"I think this is the full solution." For point load part maby, but for orginal probem set with elastic catenary no.

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11-07-2013
05:03 PM

11-07-2013
05:03 PM

Re: One catenary problem - continued (again!)

The sheet fails at no concentrated weight (Fwgt = 0N)because of the requirement thst the slopes be positive. If you disable those [sinh(x/a)>0], it will find a solution.

It was not my intent to provide a verified, cast iron, bullet proof tool; merely to show a general solution approach. (Borrowed heavily from Tom Gutman's earlier work.) The fact that the graph must be tweaked to make the absolute end heights plot correctly (the solution engine constrains the delta height) then tweaked again to plot for whatever height is desired and not cut the top off the graph does not detract from the solution.

Learn. Modify. Impress us!

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11-07-2013
11:06 PM

11-07-2013
11:06 PM

Re: One catenary problem - continued (again!)

Yes, it is a grait Job!

But sorry - H.1 must be H.1 and H.2 must be H.2 - in fitst line! As in my jobs

Or I do not understand some one?