Grasshopper

algorithmic modeling for Rhino

hey guys!

im dealing with a catenary roof. it should be paneled with planar quads.

i try to use kangaroo have as much planar panels as possible.

but i got stuck at 4...

i tried to generate the cateneary at first and have a second run for the planar panels.

but no success...

could you maybe helb me out?

thanks

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Pretty sure those aren't planar.

you need to give edge freedom to move, otherwise you will never get them planar.. and Dave is right those panels in your image isnt planar, they look like ETFE cushions, which is a plastic, and is therefore not as problematic to use as glass... 

Specifically if they were planar, then the diagonals of each panel would intersect at their midpoints, since all panels are symmetric as the bottom-up photo shows:

hey guys!

i know that these etfe panels are not planar.

but is it possible to make them planar, to make a cladding out of glas???

these picture are just to show the expexted result, as inspiration.

how to give them edge freedome?

No, it is definitely not possible with this grid and boundary.
To see why, consider one of the panels in the corner - it has 3 points fixed in the plane of the boundary, 3 points define a plane, and for the panel to be planar, the 4th point must lie in this same plane. Therefore the 2 panels adjacent to this corner panel also have 3 points fixed in the boundary plane. Thus also their neighbours, and so on by induction the only way for all panels to be planar is if the entire mesh is flat and in the plane of the boundary.

To design a grid which can be planarized, try following the curvature directions of the surface.

try to allow the points to move freely in z, but then constrain them in x & y there is a component in Kangaroo that does this... 

Just to clarify, I meant the edge points :) 

it is possible, but not watertight;)

Hi Bea,

Please refer to the attached. I have played around with some of Daniel's K2 examples a few weeks ago. I had an example with the K1 solver but I can't seem to find it. It's  a proof of concept of what Daniel is mentioning. You need to have your boundaries to follow the curvature of the catenary, otherwise this is not possible.

Best,

M.

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