Catenaries for the numerically impaired

[jeff4136]

a = 4.29728
x = -4*cos(t*180)
y = a*cosh(x/a)-a

...gets me a (desired) catenary
starting at -4,2
thru 0,0
ending 4,2

The value chosen for "a" was trial and error (look-up and bracketing). Does anyone know a simple (algebraic) solution for "a" for given x & y spans?

Many thanks,
Jeff

 

[dlongmi]

Jeff,


I searched through my memory and evenmy Statics bookand all I can come up with is atrial and error approach to find the bottom of the parabola too. Although my memory is fuzzy i do recall: Fsub(h), the horizontal component of tensile force at "any" point along the cable is used along with constants and is integrated to yield the curve's equation. Since the tensile force in the cable changes continuously in both magnitude and direction a simple algebraic equation is out, I think. However....I did find this equation written in my book. Can't remember writing it but there it is. I have no idea what it is for but it may get you something.


y = 4h/L(squared) * x(squared)


h is the vertical height of the cable from the support.


L is total span


x is distance from 0,0


Hope I have not done more damage

Let me know if I was full of it.

 

[jeff4136]

Thank you for that. It does appear to be for a parabola (y = x^2) vs a catenary. Think I had it on a crib sheet at one time, long since lost. It's good to have it again and I intend to scribble it in the margins of a couple of texts.