Inventor General Discussion

## Inventor General Discussion

New Member
Posts: 2
Registered: ‎01-28-2013
Message 1 of 3 (986 Views)

# Equation for catenary curve

986 Views, 2 Replies
01-28-2013 05:57 AM

I am designing a part that has the shape of a catenary. In order to make it right the first time, I want to shape the catenary af the exact equation and not by making a spline between points.

I am not a math genius nor an Inventor genius, so I someone could tell me the input (see attachment) I would be pleased!

The equation I need to draw (metric) is this: f(x) = -(2.0272/2)*(e^(x/2.0272)+e^(-x/2.0272))

The part is 3800mm wide and 2200mm high (see attachment 2)

I am looking forward to your suggestions!

*Expert Elite*
Posts: 4,198
Registered: ‎04-27-2005
Message 2 of 3 (932 Views)

# Re: Equation for catenary curve

01-28-2013 12:03 PM in reply to: ehaubjerg

I could not get this to look exactly like your picture. I think my equation is correct.  Here is what I used for the equations.  Set your units to Meters.

x(t) = t

yt) = -1.0136 ul * ( pow(2.71828 ul;( t / 2.0272 ul )) + pow(2.71828 ul;( -t / 2.0272 ul )))

tmin = -1.9   tMax = 1.9

Here is what this looks like.  Maybe you can tweak it and get it to work. If I get some more time, I will also look at this again.

New Member
Posts: 2
Registered: ‎01-28-2013
Message 3 of 3 (881 Views)

# Re: Equation for catenary curve

01-29-2013 02:39 AM in reply to: karthur1

Hi again.

Thank you very much for your reply. I worked out perfect. I am sorry, however, to tell you that I provided you the wrong equation - no wonder that you had trouble getting it right. The correct equation is f(x)= -(1063.35/2)*(e^(x/1063.35)+e^(-x/1063.35))

Sorry for that. However, I simply change the 2.0272 with 1063.35 and there it was!

You saved my day!

BR,

Esben

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