Catenary tangent to plane

Hi everybody,

I want to create an arc between two points, using the catenary curve.
The condition:
The catenary curve should touch a sloping plane above in one point. in this point, the tangent of the catenary will be a line on the sloping plane.

Since the catenary command in Grasshopper possesses a curve length input, i was trying to approximate the length using Galapagos. Unfortunately I couldn't figure out the conditions properly.

One could be something like "find the circle between two curves with the smallest possible radius".. but that would also apply on intersections, which I need to avoid.

Does anybody have an idea, for clearly defining conditions??

Thanks,
Benjamin

Replies to This Discussion

Permalink Reply by David Rutten on July 18, 2012 at 2:46pm

Here's my approach. Basic idea is to solve the two intersection points between the catenary and the plane, then try and minimize the distance between those points. Problem of course is that not all catenary lengths result in two intersection points but the solver seems to be able to deal with that.

David Rutten

david@mcneel.com

Poprad, Slovakia

Attachments:

CatenarySolver.gh, 3 KB

Permalink Reply by Benjamin Felbrich on July 19, 2012 at 3:04am

Thanks a lot David, great idea

I'll try it later today.

Permalink Reply by Benjamin Felbrich on July 19, 2012 at 8:15am

Okay, I just tried it.

works so far. I assume, it is due to the way grasshopper creates the catenary, by approximating it with a certain amount of polyline segments, that my two points won't get closer than in the picture.
I guess there is no way to increase the accuracy of the catenary, without going into the script, right??
Anyways, the result is satisfying.

Thanks you very much
Benjamin

Permalink Reply by Benjamin Felbrich on July 19, 2012 at 8:35am

Hmm this is funny,

when I use the annealing solver, I get this error after some time:

The good thing is: the two intersection points seem to have become one point, which is exactly what i wanted. perfect

Permalink Reply by David Rutten on July 19, 2012 at 10:59am

You could create a nurbs curve from the catenary, then you will no longer be plagued by the linear segment problem. Also, increase the number of decimals on the slider to get a more accurate solution.

David Rutten

david@mcneel.com

Poprad, Slovakia

Permalink Reply by Benjamin Felbrich on July 20, 2012 at 12:22am

I did that. That is why I wondered about the two intersection points. It seems that the catenary is created in incremental steps and the NURBS-curve adopts this characteristic

Anyways the result is close enough

But what about this:
A catenary with three ends. connected only in the peak point, touching a sloped plane. The rest points are of different heights.
I wonder how a catenary with three ends behaves.

Here is what I figured out so far:
The three catenaries meet in one point, if every catenary connects a point Pi (imagine the i as an index-i) and its coherent mirrored through the center C of the connecting circle.

Now it would be interesting to know about the relation between position of the peak point H to the other points, or the plane of our circle K. Or any relations in this setting.

Any information is greatly appreciated, i will try to update this post, if i get to know something....

Benjamin

Permalink Reply by Benjamin Felbrich on July 20, 2012 at 12:25am

And most important!!!:

Is this set stable? or will it crash? I could imagine, that every catenary segment, since it is connected to two others (which have double the weight as in the case of a normal catenary) behaves different.

Permalink Reply by David Rutten on July 20, 2012 at 3:47am

A catenary curve only simulates the ideal shape for a single curve. When you start to create a network the shapes will change. The image you posted is almost certainly not stable.

You can probably solve a network of cables using Kangaroo, I'm pretty sure I saw a video of that once.

David Rutten

david@mcneel.com

Poprad, Slovakia

RSS