(Easy) Removing Outside Row of Curves

Hello,

I am working with drawing a series of circles onto a surface and than removing the curves from the surface. I have successfully drawn the curves and can easily subtract them from the surface, but I do not want to cut out from the edge of the surface itself. I would like to remove the outside rows of circles so that the edge stays intact. I believe that this is done with a simple (x-1) expression added, I am just not sure which part of the definition to add the expression to.

Any help would be greatly appreciated.

Thanks,

Surge

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divided surface with attractor points.gh, 7 KB
JPEG.png, 240 KB

Replies to This Discussion

Permalink Reply by Diego G on April 7, 2013 at 6:05pm

Hello Surge.

Can you post a picture or a sketch of what you want to do. It would make it far easier to give you advise.

Permalink Reply by Surge Troost on April 7, 2013 at 10:22pm

Permalink Reply by Erick Vásquez on April 7, 2013 at 10:51pm

Contaiment relationship (0=outside, 1=coincident, 2=inside)

why you search the outside? just test equality to 1 or test with 2 and don't invert cull pattern.

Permalink Reply by Surge Troost on April 8, 2013 at 11:34am

thanks just had to change the 0 to 1

Permalink Reply by Michael Pryor on April 7, 2013 at 11:21pm

Or use isotrim. Then evaluate each sub surfaces center. Then put the frames at those points.

Permalink Reply by Diego G on April 8, 2013 at 8:04am

Hi Surge

It took a bit of time to understand what you wanted to do. If I understood you correctly it is quite simple to achieve what you want to do. You want to preserve the edge of the surface by removing any circle that could be "on top" of it. If that is correct then it can be achieved with the next logic>

First extract the boundary curve of the surface (You already did that).

Intersect each inner circle with the boundary curve.

If the intersection is successful then it would produce one or more points.

Test what curves have 0 intersection points. Cull the rest.

Here is your definition with that part already added in a green group and a red group of what I believe is a redundant part of your definition.

Hope it helps

Diego

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