the first area and the first number, the second area and the second number and the third are and the third number. For example, let's assume we have the following areas {65, 15, 20}. The absolute difference between these two sets equals {abs(44-65), abs(39-15), abs(17-20)} == {21, 24, 3}. The sum-total of all these absolute differences is your fitness, i.e. 21+24+3 = 48. This number has to go to zero.
If we enter the results you just got, then the absolute differences look like this: {abs(44-44), abs(39-17), abs(17-39)} == {0, 22, 22}, which results in a fitness of 44. Only an exact match will result in a fitness of zero.
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David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 12:44pm on November 13, 2010
sis by applying the loads on the deformed shape. This is done by steps or, in other words, in each step karamba applies a portion of the loads, computes the deformation and on this new configuration applies another portion of the loads and so on.
Now focus on what “portion” means.
One way of “portioning” the loads is to divide them in n parts (whit n= number of steps) and then applying in each step one part. For example, if you set 10 increments, in the first step the system will be loaded with the 10% of the loads, in the second with the 20% and so on. (I really don’t know if this is a linear proportion or not, but the concept is the same: a numerical division of the total loads).
Another way is to take that part of the loads that produces a fixed displacement. You know that F=ku: k is a property of the system, so it is fixed; setting the MaxDisp input you can compute u=Maxdisp/inc; so you can get F.
As I said before I am not sure of what I have written.
Tell me what you think!
Francesco
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original list 40% and 60% and then the latter again by 20% and 80% The three percentages I get would be:
40%
12%
48%
You can't really directly manipulate the resulting percentages, can you? And it'll be more problematic if I want more than 3 percentages?
Thank you though! I'm using your method for the time being!…
tern. The question raised to solve was the space problem of those who pray together in mosques under any weather conditions. While stepping up for a solution, learning Grasshopper was the target to reach.
The first definition is downloaded from Jennifer Coppin's weblog[1]. Then it was regenerated in Grasshopper to learn how it works. Having influenced by Daniel Piker's Kangaroo video, I was in the pursuit of a new definition[2]. The whole definition was finally completed after a few days of work.
I then looked for a tangible result which might be a stimulus to produce a canopy which is adaptable. The result, as it is rendered, was not satisfactory but a bright idea happened to me having seen the examples of Islamic calligraphy[3]. It was then a question if there is a correlation between Islamic calligraphy and star patterns in terms of geometrical base or as a part of a holistic and unconscious approach.
All comments appreciated.Thank you.
[1] jennifercoppin.blogspot.co.uk/2011/09/islamic-deformations.html[2] vimeo.com/8842130[3] kalemguzeli.org/hatteserleriayrinti.php?KNO=695&HKNO=20…
esn't do just that. I have 3 inputs which contain curves, one of the inputs usually has <null> items in it, so I am trying to filter out those nulls. The 3 inputs have different amounts of curves in them, normally one of the inputs is grafted as well.
So one would assume it just gets rid of the nulls and thats it. Well it doesn't. In my case it simply takes the curve from the first input (1 curve) and copies it 20 times (the second input contains 20 curves). The nulls are gone, but so are 19 other curves!? Maybe I don't understand the component properly, but its not doing what I am expecting.
Using a Merge component works as expected.
I am aware of checking for Null Items and using Dispatch, but I thought Combine Data is exactly for that purpose!?
Flattening the inputs makes no difference.
…
Added by Armin Seltz at 4:56am on January 25, 2016
0;3} (N = 2)
{0;0;0;4} (N = 2)
{0;0;1;0} (N = 2)
{0;0;1;1} (N = 2)
{0;0;1;2} (N = 2)
{0;0;1;3} (N = 2)
{0;0;1;4} (N = 2)
Flattening this structure using the Flatten component would result in:
{0} (N = 20)
However, using a Path Mapper with the following masks will flatten is somewhat more intelligently:
{A;B;C;D} -> {A;B;C}
Now, you get:
{0;0;0} (N = 10)
{0;0;1} (N = 10)
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:19am on December 14, 2009
I tell you what I had to do and how I did it.
I have the following situation. A urban context with a square plot 40m x 40m surrounded by buildings.
If I extrude the plot I get 4 surfaces and I need to calculate the minimum daily quantity of direct sunlight hours each test point receives in the period from 22nd of April to 22nd of August. For example for the test point at index 21 of surface with index 1 (I am just creating these numbers in my mind) the minimum is on 27th of April and the test point receive 8 hours (this is also invented for the sake of the example) of direct sunlight. All the other days it receives more. So the values I have to found are these minimums for all the test points. Now how to calculate these minimum quantities is a different issue of the topic of this post and actually I manage it.
Continuing with the explanation of what I had to... so I have only the initial plot that generate 4 surfaces, then I want to test smaller plots generated by an offset of 4 m of the original one, and the relative 4 surfaces for each smaller plot.
So in this case I think I cannot use your suggestion because the object don't exist yet.
I managed creating a loop with Anemone, the loop generate an offset starting from the original at 0 until 4 (then I multiply it by 4 to obtain the offset at 0, 4, 8, 12 and 16. Then I did like you also suggest I record every time the result with the DataRecorder and I create for each result a different branch with the index coming from the loop (0, 1, 2, 3 and 4) with the Flatten component.
In this image you can see all the surfaces saved in the same way as described above and in white the test points that receive minmum or equal than 2.5 hours per day of direct sunlight in the period from from 22nd of April to 22nd of August and in dark gray the test points that receive less.
The main point of this discussion is just the fact that instead use this tricky way I used, or the one you suggest, to analyze separately (because they shade each other) 20 geometries (in this case 20 they could be many more) it would be good if it would be possible just to input all the geometries at the same time and they would not shade each other so to get directly all the results with one run and in a more simple way.
Francesco
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t works? when i use the AddInterpCurve command with the knot style option set to chord:rs.AddInterpCurve(pointList, degree=3, knotstyle=1)i get the following message:Runtime error (TypeErrorException): Cannot convert numeric value 1 to CurveKnotStyle. The value must be zero.
rs.AddInterpCurve(pointList, degree=3, knotstyle=0)
works fine. Hope anybody can help...…