Grasshopper

algorithmic modeling for Rhino

How to get 4 usable pts from a grid of pts generated with Fibonacci Sequence

Hi Guys,

I have been generating a grid of points from a fibonacci sequence.
I couldn't manage to get 4 pts out of it to place a surface.
Any help would be appreciated.

Many thanks,

Tushar

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Try this definition. It can probably be done in some much simpler way, but this works. Kind of old-school feeling, pre-paths approach.

Also, I made so that the cell sizes follow the fib series, as opposed to the grid lines following it.

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Hey,

Thanks a lot for that.. I had done something similar as to what u have done, in ur project the cell size change, in my project the rib size change according to the sequence.. I am stuck realtime..Do u want me to send the Rhino File and the Model Images, if u can give me a simplier idea to proceed..

Regards,

Tushar
Here. I changed to the same grid you had.
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My question is what is the easiest way to take any generic grid of points and get the 4 subsets of corner points to make grid cells from the points.
hey,
somehow the file u ghx file u sent me isnt opening..
Not sure if it is same for you, but when I download Grasshopper definitions from this site, it saves to my machine as an XML file. I have to go back and change the file extension to .ghx to make it open.
Chris,

I think it depends on how you generate your "generic" points. Here would be a tree example from a subdivided surface:

its not giving me the save option ..when i click on it, i can see a long text script.. can we not zip it into a folder and send..
Tushar,
right click on the link and choose "save target as" and then make sure it downloads as Ghx and not Xml...
Taz what if you don't start from a surface but from a grid of point?
Hey

Thank you Arthur.. managed to save it now..
My question that resulted from Tushars question was if you have any arbitrary grid of points and want create cells from them. Here's my version, but I'm wondering if anyone has a superclean version that uses like 3 components instead of 15.

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