Finding point of sag on catenary curve with only catenary segment

I have a project where we have multiple catenary curves between two points which do not produce the lowest point (sag point). However, in order to calculate the reactions of the forces on each catenary, we need this point known.

Extend curve unfortunately does not have a catenary option and so I have been working to calculate this sag point but have been met with some resistance. I have scoured the net and various civil engineering and geometry books but many do not have a formula for finding the sag.

I have reviewed all the related grasshopper forum questions to develop my own catenary script (for my own understanding but entending this to produce a completed “idealized” catenary for purposes of the reactions has evaded me. If anyone has any suggestions. Let me know.

Much appreciated

Catenary (14.9 KB)

I’m not sure this could be of some help to you, anyway, you can do curvature test with many sample points on your catenary curve and find the point with the smallest radius of curvature circle.

Catenary (21.2 KB)


That certainly is helpful and right on with one of the conversations I had this morning with a colleague at my office. Unfortunately, it doesn’t look like it can project there the sag point is if the right/ left point is not beyond the y-axis.

Where as the condition I have is that I need to find a way of projecting the catenary to develop the sag location out beyond that of the catenary segment.

I really do appreciate the assistance though and the speedy reply!

  • Alex

Does this help?

Catenary (14.0 KB)

Unfortunately, I am looking for a condition where the segment does not hit the apex (sag) of the catenary and project the theoretical sag point beyond that of the catenary segment.

If I’m understanding your question right, you have a segment of a catenary which does not include the point at the bottom which is tangent to the horizontal, and you want to find a continuation of this same catenary curve that does include that point.
You can find this geometrically by taking a full catenary which does include the low point, finding the point on it which shares the same tangent as the top of your partial curve, move it so these points coincide, then scale2d your full catenary so it matches the partial one.



Thank you very much. I really appreciate you taking the time to write and develop the step by step. I have been so focused on trying to mathmatically resolve this but it makes perfect sense that you can brute force it by using another idealized catenary. Superb!

I am not sure if it is possible but It might be a very cool component to add to kangraroo, a catenary curve extend and sag point. I plan on using this logic for my study and if I can get a cluster to work Ill post it on here for everyone to use.

Many thanks again!