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Offline OchkovVF  
#1 Posted : 13 October 2024 19:13:28(UTC)
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CatenaryG.sm (24kb) downloaded 20 time(s). CatenaryG.pdf (300kb) downloaded 10 time(s).
CatenaryG-L.png

Edited by user 13 October 2024 19:29:07(UTC)  | Reason: Not specified

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on 16/10/2024(UTC)

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Offline Razonar  
#2 Posted : 16 October 2024 18:28:20(UTC)
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Hi Valery. This file uses Lagrange multipliers method for minimizing.

CatenaryG.sm (46kb) downloaded 10 time(s).
CatenaryG.pdf (204kb) downloaded 9 time(s).

Best regards.
Alvaro.
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Offline Razonar  
#3 Posted : 16 October 2024 20:52:48(UTC)
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This version uses maple for the symbolic evaluation of the integrals, reducing the calculation time a lot.

CatenaryG-maple.sm (47kb) downloaded 12 time(s).
CatenaryG-maple.pdf (206kb) downloaded 13 time(s).

Best regards.
Alvaro.
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Offline ioan92  
#4 Posted : 19 October 2024 04:17:06(UTC)
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Hi Valery, Hi Alvaro,

I find useful some additional comments on the subject targeted in yours interesting posts.

A. The catenary cable problem, you are dealing with, seems as being physically speaking this:

• A cable defined by its 2 endpoints (the 4 coordinates can be reduced to 2 information (H; V) if the origin point of the cable is taken in the origin of the axis system, and by the developed length L of the cable.

• A load (mass) is applied on the cable between its 2 ends.

Note: When the 3-input data of a cable (H; V; L) are given, its geometrical equation (the catenary) is completely determined by using 3 nonlinear equations. The nonlinear 3 equation system can be transformed in 3 independent equations, as it is shown here:

https://en.smath.com/for...nary-Cable-Analysis.aspx #14 Posted

B. If I’m not wrong, your objective is to find the new position of the cable (the cable being axially rigid) that assures the equilibrium.

Globally, this problem, IMHO, has 3 distinct technical instances:

i) The intermediary load is fixed on the cable length on a given distance over the developed cable. And your target is to find the final equilibrium position; that means 2 unknowns.

ii) This load can roll (slide) over the cable and you look for the position where the load finishes in equilibrium. This means 3 unknowns (the position of the rolling load on the cable length and the 2 coordinates of this point).

iii) This load can move only in one direction (vertical, for instance); the second DOF (horizontal, for instance) is blocked. That’s only 1 unknown - case technically pertinent also.

I find nice the graphical representation as the intersection of 2 distinct cables; the real solution also includes a third information - the position on the cable's length.

If I understood well your development, it seems that it’s belonging to the ii) case?

What do you think about some other solutions?

For instance, a method useful for a general case (more than a force between the cable’s ends) can be based on the concept of FEM (?).

Best regards,
Ioan


Edited by user 20 October 2024 22:45:02(UTC)  | Reason: A little bit more reflexion...

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Offline Razonar  
#5 Posted : 19 October 2024 13:14:15(UTC)
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Originally Posted by: ioan92 Go to Quoted Post
... What do you think about some other solutions?


Hi. I think this may be the solution to the problem with a sliding mass.

CatenaryG-sliding mass-maple.sm (37kb) downloaded 14 time(s).

Best regards.
Alvaro
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on 20/10/2024(UTC)
Offline ioan92  
#6 Posted : 20 October 2024 01:14:16(UTC)
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Hi Alvaro,

Once again, thanks for this nice piece of high-flying mathematics, based on Lagrangian optimization, handled by Maple.

Unfortunately, I can’t run your application completely because I received this kind of message:"Lexemple est incorrect"

I’m not sure how to set the optimization option? All on “symbolic”?

If I may a suggestion, it’s to prepare a little bit in advance certain values, as for instance, the length o a cable segment can be obtained without integrating at each use.

I wish you a nice Week-End,
Ioan

Do to others as you would like them to do to you!
Knowledge is of no value unless you put it into practice - Chekhov
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