Q&A forum: Catenary Calculator
(Point Load)

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Why does the graph appear to show beta more like -45 instead of -89?

What is the purpose of the two red points?

The line depicting the direction of the force is intended to be indicative only. If it is drawn vertically, it will obscure the point number (3).

ℓ₁ & ℓ₂ are points at the bottom of the loops (1 & 2).
It is true, their purpose is a little unclear. I shall clarify this on your website.
Please note the paragraph in the technical help page:
If your catenary is tight (i.e. does not generate a loop between ends '1', '2' and/or '3'), the position of the associated imaginary loop (ℓ₁ and/or ℓ₂) will appear outside the catenary in the image.

Hi tech help,
Some parameters that work just fine in the Catenary program DO NOT WORK in the Catenary+ program.

L = 592.5
W = 12.44464
X = 587.85
Y = 63.59

It doesn’t matter what values are used for F or the angle or placement of the load F. The above parameters cannot be solved in Catenary+. When I try to enter x = 587.85 it errors with " 0 < x < 586.575 "

Please help. I need to apply a point load and I need it to work!


Whilst the 'Catenary' calculation for x,y co-ordinates is iterative, it completes its calculation using only one iteration and is therefore able to resolve all unknowns very easily and accurately.

The calculation method used in 'Catenary+', however, uses numerous (6) nested iterations and would therefore find it more difficult to achieve accurate results in extreme situations. CalQlata has therefore incorporated operational limits in order to ensure acceptable error margins for all successful calculations.

Your calculation pull's a line 592.5 long between two points 591.2793846 apart which can be considered 'bar-tight' (see 3rd paragraph in the program's Technical Help) or very close to it. This is the reason why you have found an upper distance limit for 'x' of <586.575 for your calculation. This co-ordinate limit is artificially imposed by the program.

CalQlata could provide you with a modified version of the program for your personal use but you would need to heed the following:
1) CalQlata would have to make a charge for creating the program; and
2) In extreme calculation situations the results may fall outside the recommended error limits designated by CalQlata; and
3) It may be possible that some (iterative) calculations fail to resolve

I hope the above is of some help