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Mike_B

Investigation into Catenary Curve

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I am a new member have a problem, the subject of which has interested me for years. The solution for a in the hyperbolic function (for a catenary)

 

y = aCosh(x/a), sometimes with +c added as a 'curve shifter'.

 

I have managed in the past by iterative trial and error if y & x are known, but very time consuming, ineligant, sometimes unsuccessful. The Casio calculator I use has a "solver", which is useful when it can find an answer, but often can't. I have a new all singing & dancing HP 50G with several solvers, which is far to complicated for me to use at all!

 

The problem is me. I do not understand why sometimes a cannot be found? Surely every Cosh curve must have an a, in fact the thing always has y=1 when x=0, the a being a 'scaling factor' to define what part of the curve is to be considered for span, sag, height above ground etc for a symmetrical portion of it. When a is negative the curve inverts, having vanished while passing through y=0.

 

Is there a way of transposing for a - (I can't do it in the simple formula above)?

What sort of equation is it anyway, not linear, not polynomial, not entirely logarithmic? So what 'solver' should be looking at it?

 

Please do not ask me to rush into Calculus. Tried it in my youth, can't do it. Sorry for my inepitude. I'm sure this is the right place however to admit this, and to ask for any help, for which I would be most interested & grateful.

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a is the intersection with the y axis - ie when x is zero, cosh x/a is one, and y is equal to a

 

What information are you working from in order to reach a value for a?

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!

Moderator Note

thanks Boss.

 

Thread Locked as this is a duplicate

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