The Cartesian equation is: y = a cosh(x/a)

  The difination of: cosh(z) = (1/2)*(e^z + e^-z)

therefore:  y = (a/2)*(e^(x/a) + e^(-x/a))


so both equations that you came up with are really the same.



some addition info can be found here

Catenary Curve Equation  from
ARCH HISTORY AND ARCHITECTURAL INFORMATION website:
http://www.nps.gov/jeff/equation.htm

Thanks, Rich!  Now how about this one - Is there an equation for finding the slope at any point on the catenary without using high order Calculus or Diff Eq?!

Boy, did I hate Multivariable Calculus....

Fair go, pylko.

It doesn't really need any 'high order' calculus, beyond remebering that the slope = dy/dx.

For y=cosh(x), dy/dx=sinh(x) (In my ancient text that is shown only three pages after hyperbolic functions are introduced).

Thus, for y=a*cosh(x/a),
slope =  dy/dx = a* sinh(x/a)*(1/a) = sinh(x/a)

The equation for finding the slope at any point on the catenary:

slope(x) = (1/2)* sinh(x/a)

Rich

Hi, Rich2001.

Are you really sure of your catenary slope equation?  

If we make a=1 to make life simple, we get a catenary equation of y=cosh(x), for which dy/dx=sinh(x).  ie, I don't believe that your factor of (1/2) should be there.

If we go back to your other form of the catenary (and still keep a=1),
y=(1/2)*(e^x+e^(-x)),  
dy/dx=(1/2)*(e^x-e^(-x))=sinh(x), again no factor of (1/2).

Or try a simple numeric example (still at a=1), around x=1
for y= 0.9999 to 1.0001, the values of y=cosh(x) are  
x=0.999,   y=1.541906
x=1.000,   y=1.543081
x=1.001,   y=1.544257

At x=1.000 that gives me an approximate slope of
(1.544257-1.541906)/(1.001-0.999)=1.1755.  (working with higher accuracy on my calculator gives 1.1752014).  That result is to be compared with sinh(1.000)=1.1752011.  Again, clearly no factor of (1/2).










 

You got me. I should do Calculus in my head as I drive.

Thanks.  You guys minor in math?!

No, I just enjoy math. I does help that I taught Statics and Dynamics for several years and my son is taking calculus now so I have been using it to show him how to solve his homework.

My apologies for rather overdoing the extent of the verification of my earlier posting (sledgehammer to crack a nut, etc).

I have little doubt that my maths capability was considerably better 50 years ago. (That 'ancient' text was originally written in 1897, third edition in 1919, and reprinted six times up to 1942 when my eldest brother used it at school.  I inherited it in 1949 or so).

It's all just a matter of trying to keep hold of the basics (which in this particular case have not changed over the past 100 years!).

Keeping with basics.... An architect wants a catenary cable walkway.  He hasn't decided on the exact geometry of the catenary (ie: 1 rise, 6 run, etc...) but the direction of the reaction should be the slope at the end of the cable (whereever the end winds up being).

Thanks for your help