e whole "surface" 10 times along y axis.
I expected that the numbering of the cubes would follow the creation, but no. So, I want element number 1 to be the initial cube, number 10 the last element of the column, 11 the 1st element of the second column, 100 the last element of the last front column etc....
Any ideas??
Thanks in advance …
otate 9°, the second module 18°, the third 27°, etc. I can only seem to get all 11 modules to rotate the same... I know it probably has something to do with series but I can't figure it out. the screen shot shows the 11 modules rotating to the same degree and the drawing is a plan of what I am trying to accomplish. Thank you so much in advance!…
Added by Daniel Lamm at 6:54pm on September 4, 2014
elated with the Topology outputs:
So let's try to do (via components) the face reconstruction stage (the missing 4 as above):
Alias crenelatedEdgesTree as polylineTree.
Imagine a Lst that samples all the edges per Face ("changed" and "unchanged") as Curves.
1. Let's take face 3: this is surrounded by edges 10,11,12,13 and 37.
2. Has edge 10 "changed" (to polyline) ? No because in the polylineTree there's no branch {10} ... thus sample edge 10 from the EList (Note: apparently that's a boundary edge). Has edge 11 "changed" ? No ... blah, blah.
3. Has edge 12 "changed" ? Yes because in the polylineTree there's a branch {12} ... thus sample the item from that branch. Same for 13 ... etc etc.
4. Thus we have sampled all the surrounding edges as Curves and the next step is to join them > yielding a closed Curve.
5. Then we must "planarize" that Curve (by projecting it into the corresponding Brep Face plane) ... and the rest are history.
So ... try it and report any issue encountered.…
are on their own paths, but the first branch contains 3 curves and the second one 2 curves. If you want the same result for all pairs of curves you'd need to split up the first and second branches, so that all curves are on their own branch.…
Added by Lars Renklint at 4:33am on September 6, 2009
d Y it will create points by matching each index of each list in order. For example index[0] of list X with index[0] of list Y , etc...
If the lists are are of uneven length the component will match the last index of the shortest list with the remaining indices of the longest list. For example list X has 10 items and list Y has 12 items. The matching logic would be list X [9] list Y [10] and list X [9] list Y [11]
Cheers
Nicholas…
hat position you would like in order:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
You still must show up on time and you still must stay and work on your project during the duration of class.
See you all next week, I hope I won't still have this cough!
-Joseph Iwaskiw…