lconcepts in parametric design and exercises using Rhino, Grasshopper, andPython. Each of the 3 workshops corresponds to learning different software skillsthe softwares and applying those skills to a creative design challenge.…
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But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
ifically: I have a 100' vertical plane lofted between curved top and bottom profiles. I contour it every 8' (normal direction is Z, giving me 13 horizontal curves). I use Divide Curve to divide each contour into 10 segments. The "Points" output of Divide Curve now yields 13 branches with 11 items each, corresponding to 13 contours with 11 points from the left end of the curve to its right.
I now want to string "vertical" lines, and connect all the 2nd items in each branch together, all the 3rd items, etc... in order to make a polyline that travels between each 2nd point or 3rd point. i don't want to use Cull Pattern/Nth/Index because the number of subdivisions could change (11 could become 20, etc).
How do I connect the Nth item of each branch in this tree? Moreover, how do I connect all values in a branch with their corresponding values in all other branches?
Thanks for any replies,
Richman Neumann
Solomon Cordwell Buenz Architects
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output will show a tree with 3 branches of 4 integers each that I can pass on to other components. What is the best way to do it?
I have tried creating a tree and using a for loop to do so, but it didn't work.
Thank you for your help.
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7, 9, 12 and 13 to be able to rotate freely around the y axis at nodes 2, 3, 6, 7, 10 and 11 respectively. The last 2 conditions, for elements 12 and 13, doesn't give any problems, but the first 4 does.
Any help?
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e now contains (40x3x11=)1320 points with a branch structure of {0-39,0-2} i.e
{0;0}
{0;1}
{0;2}
{1;0}
...
{39;2}
with each branch containing 11 points.
I now want to create lines from the points on line {x;0} to {x;1}, {x;1} to {x;2}, and {x;2} to {x;0}, which will give me a triangular grid on each triangle. How do i do this? I think I need to split the tree 40 times to give me 40 single layered trees but I imagine there's a cleverer way of doing this!
This feel like it should be simple, but I;m having trouble working with the double-layered branch structure. Any help would be greatly appreciated!
Thanks,
Matt…
ample.
1.) Generate a series of 11 values.
2.) Multiply series by 30 degrees.
3.) Convert to Radians
4.) Rotate Point A about Point B 11 x 30
PI is available from the Maths Tab Utility Section or alternatively double click on the canvas and type PI hit enter.
If you haven't seen the Icons view with Fancy Wires before then these are available from the View Menu and are optional depending on your personal preference. Fancy Wires display the contents of a data stream in a graphical way. single wire = single data, double wire = multiple data on a single branch, double dashed wire = multiple data on multiple branches.
In the example below there are now two pivot points each generating a different 11 points in a circle.
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