tName_FinalProject_PartD.pdf
Below is the desk crit list, please sign up for a spot in the comments below:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
See you Monday!…
where each branch contains all the points generated by dividing each curve, so if you divide into 10 segments, you'll get:
{0;0}(N = 11)
{0;1}(N = 11)
{0;2}(N = 11)
{0;3}(N = 11)
{0;4}(N = 11)
Where the second integer in the curly brackets refers back to the index of the curve in the original list.
Another way to look at this data is to see it as a table. It's got 5 rows (one for each original curve) and 11 columns, where every column contains a specific division point.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Hi Rassul,
I think you need to subtract 1 from the input to the SDivide component's UV inputs as you are getting 10*10 = 100 subsurfaces and 11*11 = 121 image samples
First use a series component with start=1, step=42, count=3
Use the output to create a new series component with start=existingseries, step=1, count=11
onsecutive points at the same height then your 'Break at discontinuities' component eliminates the middle point completely and then the 'Interpolate Curve' component gives a much bigger bump in the wrong direction. This was enough to get curves to meet from opposite sides.
I fixed this by changing the heights to 1.1 or 2.9, rather than 1.0 and 3.0, but it took a little while to work it out! Sigh.
I attach a new version. But I actually preferred it as it was before. See what you think!
Bob
p.s. in the first list, elements 11, 12, 23 and 24 go from 1 to 3; elements 17 and 18 go from 3 to 1. In the second list, elements 6, 17, 18 and 29 go from 1 to 3; elements 12 and 23 go from 3 to 1. Given the above fix, these can be easily seen.…
Added by Bob Mackay at 10:40pm on November 24, 2015
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
7 -18.2
8 5.02
9 12.4
10 18.1
11 7.01
12 5.11
13 2.35
this data is waveform data.
i want to pick out only climax(for example in upper data, number 1 and 10) from this data.
i have this problem...that is, i do not know to do what.
i think my English is strange...so you are difficult to understanding my question.…