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?
…
a seed, and instead creating a pattern where each color has a seed/control slider for each row? For example, row 1: brown 2, tan 6, yellow 7, purple 3, repeat. row 2: brown 6, tan 1, yellow 4, purple 10, repeat. row 3: yellow 5, purple 1, brown 3, tan 10, repeat. row 4: purple 2, brown 7, tan 3, yellow 4, repeat. Then repeat that sequence up the wall? For each color, the number in the sequence should be adjustable.
Thank you again for your help!…
say about revit 2017, I hope it will work for revit 2017.
1. Install Chameleon for 2012 following the install instructions.
2. Copy the files in this folder: C:Users*your user name*AppDataRoamingAutodeskRevitAddins2012
3. to this folder: C:Users*your user name*AppDataRoamingAutodeskRevitAddins2013
4. Open Revit…
s 8, 4, 2, 10, 1, 3, 8, 4, 2, 0. But then for the end result to maintain all numbers above 5 but replace all numbers below with a defined number..Let's say zero. So then the list would read...8, 0, 0, 10, 0, 0, 8, 0, 0.…
the paths.
Structure one (paths = 2)
{1;0;0;0;0}(N=10)
{2;0;0;0;0}(N=10)
Structure two (paths = 2)
{1;0;0;0}(N=10)
{2;0;0;0}(N=10)
If i merge the two lists i don't get a structure with 2 paths:
Structure result af merging (2 paths)
{1}(N=20)
{2}(N=20)
as i had expected, but instead a structure with 4 paths because of the difference in amount of zeroes:
Structure result af merging (4 paths)
{1;0;0;0;0}(N=10)
{2;0;0;0;0}(N=10)
{1;0;0;0}(N=10)
{2;0;0;0}(N=10)
The amount of zeroes changes all the time when working on the definition, so what im asking is if there some way to adress paths with * number of zeroes behind.…
W, X, Y, Z}
----------------------------------------
and if I set this
SetA = {U, V, W, X, Y, Z}
SetB = {1, 2, 3}
Imap = {2, 2, 2}
I will get this?
result = {U, V, 3, 2, 1, W, X, Y, Z}
----------------------------------
And what if I set this?:
SetA = {U, V, W, X, Y, Z}
SetB = {1, 2, 3}
Imap = {2, 2}…
Added by Frane Zilic at 3:26pm on September 10, 2010
branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…