ts, a value for each point.This script is wrinting in C#, not C++. And of course, can be done with standard components, perhaps someone else help you with that.…
cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014
tName_FinalProject_PartD.pdf
Below is the desk crit list, please sign up for a spot in the comments below:
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See you Monday!…
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015
Hi,
I want to divide curve with distance between points so it will be like this:
1--2---3----4-----5------6-------7-----, ...
with values in range 1 to 50, must be simple but im stuck..
tnx
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…
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
you want each "element" to be a single Item or a single item for ALL elements. See Below
0. 20
1. 30
2. 59
3. 60
4. {9,45,29}
5. 0.0
6. 3.0
7. 6.0
Or
0. 20 30 59 60 {9,45,29} 0.0 3.0 6.0
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Added by Danny Boyes at 3:13am on October 29, 2013