bers of point) index
and I called the last point as indexMax
that what I wrote I am sure that I made some mistakes- so if one of you can help me I will be more then glad
If abc(sin(3 * pi() * ptList / ptLast)) < 0.5 Then harmony = 3 = z, 2 = x, 1 = y
A = 0
Else
A = 1
n = 0
For n < ptLast
If A(n) = A(n + 1) Then
Zf(n) = Z(n) + 12 * A(n)
n = n + 1
End If
(n + 4) < ptLast Then
Zf(n) = Z(n) + 12 * A(n)
Zf(n + 4) = Z(n + 4) + 12 * A(n + 4)
Zf(n + 2) = Z(n + 2) + 6
Zf(n + 1) = Z(n + 1) + 6 - 3 * (A(n + 4) - A(n))
Zf(n + 3) = Z(n + 1) + 6 + 3 * (A(n + 4) - A(n))
n = n + 5
End If
Else M = ptLast - n
For n<ptLast
Zf(n) = (ptLast - n) / M * 12 * A(ptLast - M) + Z(n)
n = n + 1
End
Zf(ptLast) = Z(ptLast)
…
Thank you Marios,
but I want to put boxes on all the plane: if I divide U domain in 4 parts and V domain in 3 parts I'll have 12 boxes with 12 different heights in W
How can be done?
regards
Maurizio
he same order of the list. for example i have a list with 4 different lenght of curve like this:
0= 10
1= 12
2= 8 (minimum)
3= 17 (maximum)
and wont to make a ranking that the longest curve gets the value 4 and the smallest the value 1, like this
0= 2
1= 3
2= 1
3= 4
i tried the sort list function, but it dosn`t work
can anybody help me!
thx a lot…
g from a list of 12 items I would find all the combinations taking just 4 at time.
I'd use a Stream gate that takes the indexes of the items and pass them to a list item in order to select just the items of the combination. Doing so I can choose a single combination of index at time to pass to the list item.
In this moment all the data come out from the first gate, all the others are empty.
If I pass these index to the list item it gives me an error (probably because of the data structure).
*long version*
I start from a list of 12 segments, all of them with the starting point in common and the ending point distributed regularly in the space. It's a quite simple starting point.
What I'm trying to achieve is to find all the possible spatial configurations made of 2, 3, 4 segments. I started with 2 segments so I've 12^2=144 possible configurations but just 4 different configurations that can intuitivelly be recognized (60°, 90°, 120°, 180°).
Doing the same with 3 segments generates 12^3=1728 configurations and I don't know how many different ones. With 4 segments I've got 12^4=20736 possible configurations.
As you can imagine many configurations are identical but just with a different orientation so at the end I'll have to parse geometrically the output to delete duplicates (I'll address this later on).
Please could you help me to figure out how to mix these segments in different configurations?
Thank you in advance.…
掌握编程过程中遇到的思路方面和技术方面的问题. 内容包括以下几个方面:
反向逻辑思维能力的培养;
建立清晰的编程逻辑思维能力;
GH 的程序设计理念;
并行数据结构深入理解和控制.
Grasshopper course of McNeel Asia focus on the cultivation of students flexible use of programming techniques, the ability to solve practical problems. Our course deep into the whole process of programming, from programming thinking model, the components principle to usage details do detailed explanation, help students complete mastery programming encountered in the process of thinking and technical aspects, include the following content:
Ability of reverse logical thinking;
Establishment of clear programming logical thinking ability;
The program design concept of Grasshopper;
Understanding parallel data tree structure and how to control it.
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授课讲师 Instructor 课程由Grasshopper原厂McNeel公司在中国地区的两位 Rhino 原厂技术推广工程师 – Dixon、Jessesn联合授课。课程结束后对达到授课预定目标的学员颁发唯一由Grasshopper原厂认证的结业证书.
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课程日期 Schedule 7/15-7/20 Beijing 北京 7/26-7/31 Shanghai 上海 7/07-7/12 Shenzhen 深圳
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Note: pls follow below comments by Jessesn to see the samples…
wunderbar ^^* !
maybe you would be interested in Jun Mitani's work ?
www.flickr.com/photos/jun_mitani/mitani.cs.tsukuba.ac.jhe published two books (keyword : 三谷 純)立体ふしぎ折り紙
ふしぎな 球体・立体折り紙