more than 2 Point Attractors for scaling circles

Replies to This Discussion

Permalink Reply by Vicente Soler on May 17, 2009 at 12:14pm

Probably yes. If you want to add an arbitrary number of vectors, try decomposing them to its x,y,z numbers then add each list of numbers individually using the mass addition component and then reconstruct the vectors again with the vectorXYZ component. Depending how the definition works you may have to use paths somewhere.

Permalink Reply by John Dolci on May 17, 2009 at 1:06pm

I am not familiar with the definition you're talking about, but if it is a distance logic you're using to manipulate the radius of the circles, this is what I would do: take the distance from each attractor point to the centers of all of the circles, then filter the results through MIN components to acquire the smallest distance from each circle center to its closest attractor.

Permalink Reply by Maria Vas on May 17, 2009 at 1:53pm

Hi John, thanks for replying.
This is exactly what I did, and then I add them together.
The problem is that I cannot add more than two in the 'addition' operator.
So I guess I have to use another operator like mass addition or sth, but it didn't seem to work still.
I'm sending you the file if you want and have time to take a look at it.
thanks!
maria

Attachments:

• circles.ghx, 118 KB

Permalink Reply by corneel cannaerts on May 17, 2009 at 2:12pm

if it is just one or two extra attractor points, you could just use an extra addition opperator for each point.

just chain them after eachother

Permalink Reply by Vicente Soler on May 17, 2009 at 2:27pm

Or you can make the definition independent on the number of points. Well, almost. If you want different influence values for every point, add a slider per point in the merge component. If you want all points to have the same influence value, just use one slider and delete the rest.
The following example is set up for 5 points:

Permalink Reply by Maria Vas on May 17, 2009 at 3:41pm

thanks a lot vicente for the interest to make this for me,
this approach looks also interesting!
but I guess you have a newer version and it shows the components differently.
I have a question of what is the component with the 2 little leaves (after the grid).
Is it possible for you to send me the ghx. file?
thanks for the help so far anyway!
maria

Permalink Reply by Vicente Soler on May 17, 2009 at 4:02pm

That component creates a new path structure for every item, it's available only in the latest beta. The official version (0.5) works only with one dimensional lists. The latest beta (available here:
http://www.grasshopper3d.com/page/next-build) allows you to create lists within lists. This is a huge step forward, I'd recommend to install it (even though it has a few bugs) unless you are in the middle of a big project since compatibility issues may arise.

Permalink Reply by John Dolci on May 17, 2009 at 2:39pm

corneel said it. duplicate another copy of everything between your rectangular grid and the MIN component, set your third attractor, then use another ADD component to add the result of the first ADD component and the result of the third MIN component. This is going to leave you with larger values going into the radius field of the circle component, giving you very different results. you will either have to mess with the influence factors, or you could multiply the list from the final addition by 2/x, x being the total number of attractor points, and put that into the radius field. That will give you circles close to the same size as when you were using two attractors. See the attached file to see how I did it.

Attachments:

• circles_3attractor.ghx, 194 KB

Permalink Reply by Maria Vas on May 17, 2009 at 3:38pm

thanks john+corneel, it worked, so simple at the end..!
but it's because I'm still a beginner ;)
@John ps. I could not open the file, it gives an error, but no worries

Permalink Reply by Maria Vas on May 18, 2009 at 4:23am

Hi John,
there is something that I didn't do from what you said..
The strange result of the circle radius that I get, and how to manipulate it,
with the multiplication of 2/x.
i tried to do it, but I don't know if it's correct, because it doesn't do anything

If you have time is it possible to send the file again, or a printscreen of it?
thanks!