ragazzi ciao =) , una domanda. Ma i file che vengono nominati nel manuale, dove li posso trovare? Perchè quando ho scaricato il manuale mi è uscito solo il pdf. (es. pag 89)
Added by chiar molino at 11:10am on November 1, 2017
th 6 branches and 96 values per branch.
How can I find the weighted values of each branch in the tree?
For each branch, I need to take each item and subtract it from every other item in the branch. Then take the length of the branch - 1 (95) divided by the sum of each subtraction. http://upload.wikimedia.org/wikipedia/en/math/a/e/7/ae79a689e288c866e2046eef70fe030c.png
I am able to do this by grabbing 1 branch and doing the calculations, but this requires me to flatten the list at points, which obviously will not work when I try to do the computation on the entire tree and not an isolated branch.
Attached is an image of the definition I built to compute the weight values for a specific branch.
Thank you.…
closest point to the very first would be removed from the list, so the initial list reduces from 100 to 98. From the 98 i pick one and search the remaining 97 for the closest. From the remaining 96 i pick again one and search in the 95,...
(The product I want to result is:
having a number of random lines in 3D space, produced by an even number of points as discribed, this shall be the initial springs for a ("selfadjusting") tensegrity. Each one of these lines (later springs in kangaroo) get divided in three areas - that means four points. These four points again are the "attractor points" of neighbor springs, so the strut "knows" where to set the next elastic connection,...the rest I´ll have to figure out)
angelos…
exact formula is inside /lib/skybright.cal if this can help you to find the name.
{ RCSid: $Id$ } { Sky brightness function for sunny and cloudy skies.
Additional arguments required for calculation of skybright:
A1 - 1 for CIE clear, 2 for CIE overcast, 3 for uniform, 4 for CIE intermediate A2 - zenith brightness A3 - ground plane brightness A4 - normalization factor based on sun direction A5,A6,A7 - sun direction }
cosgamma = Dx*A5 + Dy*A6 + Dz*A7;
gamma = Acos(cosgamma); { angle from sun to this point in sky }
zt = Acos(A7); { angle from zenith to sun }
eta = Acos(Dz); { angle from zenith to this point in sky }
wmean(a, x, b, y) : (a*x + b*y) / (a + b);
skybr = wmean((Dz+1.01)^10, select(A1, sunnysky, cloudysky, unifsky, intersky), (Dz+1.01)^-10, A3);
sunnysky = A2 * (.91 + 10*exp(-3*gamma) + .45*cosgamma*cosgamma) * if( Dz - .01, 1.0 - exp(-.32/Dz), 1.0) / A4;
cloudysky = A2 * (1 + 2*Dz)/3;
unifsky = A2;
intersky = A2 * ( (1.35*sin(5.631-3.59*eta)+3.12)*sin(4.396-2.6*zt) + 6.37 - eta ) / 2.326 * exp(gamma*-.563*((2.629-eta)*(1.562-zt)+.812)) / A4;
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ng long in the x axis and three in the y and they don't all intersect each other... I wrote a script to Boolean difference them but its not working like i want it to . I included a rhino result that id like to achieve in the file. THX -ethan
heres the script:
import rhinoscriptsyntax as rs
b1 = []for i in range(b1L): b1.append (x)print b1bb= len(b1)print bbb2 = []for j in range(b2L): b2.append (y)print b2bc = len(b2)print bc
def bool ():....for i in range (bb):........for j in range(bc):............a = rs.BooleanDifference( b1,b2, False).....return (a) a = bool ()…