s: one is curve degree 3 , other side is more cells with degree of 1.
CNC router used for cutting 1/2" birch plywood. Air chamber made using two 4' x 8' long sheets of latex.
When air is pumped into air chamber, system inflates.
LED Lights inside the assembly can change color and strobe, flash, etc.
http://thatsnotarchitecture.tumblr.com/…
{8} become {1) etc
(see image attached)
I've played around with series components and feeding the data into a tree branch in stages (0-4 5-9 etc), however the amount of branches i need to merge (5 in this case) is a variable that plan on changing in the future so I want a more sophisticated solution than manually copying groups of series components and feeding them into separate tree branches.Thanks in advance for any help…
he first problem:
1. Define 7 as the start point.
2. Iterate through the remaining list of points, find the closest point
3. Now, what do you do: find the next closest point from the previous one? Or stick testing from point 7? Note that if you stick with 7, how do you then deal with NOT creating a line between 6? (Or put another way, what will tell the computer its found all the shortest paths and to begin testing from the next set?) If you test each path as the alternative, you will end up with a diagram completely different to what you have drawn in numerous solutions, in fact, they would just be incorrect, not without an exhaustive and overly complex set of conditionals.
You could define zones - circular extents - but that again will produce results differently to what you have drawn, namely connections such as 2 & 3, 2 & 8, 3 & 4, and of course 1 & 2 would be problematic with this approach. Do you see why this isn't as simple to solve using a computer?
Mario, here's your moment....(i hope youre still young, it might take some time, and it needs to work for any possible combination, not just the one above! I'll pass)…
Int32 = 0 To y.count - 1
Dim c As Char = Convert.ToChar(y(i).substring(7, 1))
Dim d As Char = Convert.ToChar(y(i).substring(8, 1))
Dim p As Integer
If c = ";" Then
p = convert.ToInt32(y(i).substring(6, 1))
Else If d = ";" Then
p = convert.ToInt32(y(i).substring(6, 2))
Else
p = convert.ToInt32(y(i).substring(6, 3))
End If
Dim path As New EH_Path(p)
For j As int32 = sum To sum + z(i) - 1
tree.Add(x(j), path)
Next
sum = sum + z(i)
Next
A = tree…
byte-accuracy red, green, blue channels) = 27 bytes. More likely 28 bytes as colours are probably stored as 32-bit integers, allowing for an unused alpha channel.
28 * 800,000 equals roughly 22 megabytes, which is way down from 9 gigabytes. That's a 400 fold memory overhead, which is pretty hefty.
Grasshopper stores points as instances of classes, so on 64-bit systems it actually takes 64+64+3*8 = 152 bytes per point*, which adds up to 122MB, still way less than 9GB. It would be interesting to know where all the memory goes...
* Grasshopper points also store reference data, in case they come from the Rhino document. This data will not exist, but even so it will require 64-bits of storage.…
Added by David Rutten at 4:13pm on December 11, 2014