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
This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010
ggle A
7. Toggle A
8. Toggle 2
9. Toggle 3
10. Toggle A
11. Toggle A
12. Toggle 3
I was thinking to use somehow slider and animate option....but without luck
Any idea would be appreciated…
not sure about my method used to obtain the result.I think is really complex; do you know a smart method to simplify all my equation?
thanks guys!! :D…
ere a way to set this up in a way that lets say when there are 28 it will go for one of 4, 7 or 14 and if it is a primer number it will exclude enough number of boats to obtain the closest suitable number - lets say there are 29 boats but it will only show 28 and go back to selecting 4,7 or 14.…
Added by Levent Ozruh at 10:40am on October 14, 2014
ave 63 branches in list B which shell be sorted by the indices from flattened list A. first indice is 3 so Branch 3 should be first in the new list, branch 45 second and so on, right?…
was stopped and the solver returned 3 options with the highest fitness score.
(Details are here: http://dimak1999.blogspot.com/2011/02/i-have-not-had-chance-to-use-galapagos.html)…
etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …
} (N=11) {0;1} (N=11) {0;2}(N = 11) {0;3}(N = 11) {0;4}(N = 11)
2. I run the Points that are coming out from the Divide Curve Components through the Path Mapper components with this definition:
{A;B} (i) > {A} (i)
3. I run data coming out from Path Mapper component through:
a) Parameter Viewer component and the result is:
{0} N=11 (data with 1 branches)
b) Point > Panel and the result is:
collection of 11 point (N=11) which is the exactly the same as the collection of point belonging to {0;4} (N = 11).
So, here is the question:
why the collection of points coming out from the Path Mapper {A;B} (i) > {A} (i) component is the same as the collection of points belonging to the curve {0;4}(N = 11) ?
Anyway ... It 's the first time I ask a question here... so I would like to thank you for what you do with your work! Thank you! You are really great!…
byte-accuracy red, green, blue channels) = 27 bytes. More likely 28 bytes as colours are probably stored as 32-bit integers, allowing for an unused alpha channel.
28 * 800,000 equals roughly 22 megabytes, which is way down from 9 gigabytes. That's a 400 fold memory overhead, which is pretty hefty.
Grasshopper stores points as instances of classes, so on 64-bit systems it actually takes 64+64+3*8 = 152 bytes per point*, which adds up to 122MB, still way less than 9GB. It would be interesting to know where all the memory goes...
* Grasshopper points also store reference data, in case they come from the Rhino document. This data will not exist, but even so it will require 64-bits of storage.…
Added by David Rutten at 4:13pm on December 11, 2014