ut no luck so far...
and for some reason, the number of curves in the lists is double what I think it should be... for instance, if I create a series of 20 curves (20 stories of the building), the output at the end of the definition says "40 locally defined values" ... not sure if this is a problem...…
to process
When the timer and the counter is started it runs through incrementally.
Once the counter has gone past 50, the baking is switched off
There is a log to keep track of which index it got up to just incase it needs to be aborted... and there is a start at indices input in the counter.…
is passable give different distance in list of point to delete
ex: point 1 - 30
point 2 - 20
point 3 - 10
...........
the point all in one list
i try a lot but didn't found the way ...
d an object, which allows me to define a value as an input and which then outputs an adjustable number of random values which - if summed - equal the input value. For example I want to define 50 as an input and the output should be 20 + 30 or 20 + 20 + 10 or 25 + 25,...
So basically I need something like a flipped addition object (but with an adjustable number of random output values) Is there anything like that in Grasshopper?
Thanks a lot in advance!
regards
Peter…
Added by Peter Steiner at 3:35pm on November 5, 2012
ld go
Slider at 1 -> 1°, 2°, 3°.
Slider at 2 -> 2°, 4°, 6°
Slider at 3, etc.
And when the angle was too much that it surpassed 90 before connecting the 20 points, that line was not created. Maybe using a series instead of a range...mmm
Everything helps, thanks
…
square units. Then you have an integral number of fragments on each side. This means that if all fragments need to have the same surface area, you can only have the following possibilities for side A:
1 fragment = 100 square units
2 fragments = 50 square units each
3 fragments = 33⅓ square units each
4 fragments = 25 square units each
5 fragments = 20 " "
6 fragments = 16⅔ " "
etc.
For side B, the numbers are mostly different
1 fragment = 300 unit²
2 fragments = 150 unit²
3 fragments = 100 unit²
4 fragments = 75 unit²
For side C they are different still. Unless you join fragments across on both sides of the edges of the box, I very much doubt you'll be able to pull this off.
The solution I attached will create fragments as identical as possible, but it's a very boring outcome...
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Hi Mario,
UVInterval stan = new UVInterval();
stan.U = new Interval(20, 25);
stan.V0 = 50;
stan.V1 = 68.4;
does this not work?
--
David Rutten
david@mcneel.com
Poprad, Slovakia