. If i divide an interval in 20 steps, i end up with 21 numbers, so i'm creating a grid of points that consists of 21 x 21 points. Since the surface component asks for the number of points in the U direction, i get the number "20" and add 1 to end up with 21.
If you want to create an isosurface, it's not as easy since the points can't be arranged into a grid. You can display the boundary points of the surface easily but to create actual geometry you need something like a marching cubes algorithm. I did a definition for it but it works really slow. An alternative can be to bake the points and use rhino's mesh from points command.…
Added by Vicente Soler at 1:22pm on November 30, 2009
ay how many valid permutations exist.
But allow me to guesstimate a number for 20 components (no more, no less). Here are my starting assumptions:
Let's say the average input and output parameter count of any component is 2. So we have 20 components, each with 2 inputs and 2 outputs.
There are roughly 35 types of parameter, so the odds of connecting two parameters at random that have the same type are roughly 3%. However there are many conversions defined and often you want a parameter of type A to seed a parameter of type B. So let's say that 10% of random connections are in fact valid. (This assumption ignores the obvious fact that certain parameters (number, point, vector) are far more common than others, so the odds of connecting identical types are actually much higher than 3%)
Now even when data can be shared between two parameters, that doesn't mean that hooking them up will result in a valid operation (let's ignore for the time being that the far majority of combinations that are valid are also bullshit). So let's say that even when we manage to pick two parameters that can communicate, the odds of us ending up with a valid component combo are still only 1 in 2.
We will limit ourselves to only single connections between parameters. At no point will a single parameter seed more than one recipient and at no point will any parameter have more than one source. We do allow for parameters which do not share or receive data.
So let's start by creating the total number of permutations that are possible simply by positioning all 20 components from left to right. This is important because we're not allowed to make wires go from right to left. The left most component can be any one of 20. So we have 20 possible permutations for the first one. Then for each of those we have 19 options to fill the second-left-most slot. 20×19×18×17×...×3×2×1 = 20! ~2.5×1018.
We can now start drawing wires from the output of component #1 to the inputs of any of the other components. We can choose to share no outputs, output #1, output #2 or both with any of the downstream components (19 of them, with two inputs each). That's 2×(19×2) + (19×2)×(19×2-1) ~ 1500 possible connections we can make for the outputs of the first component. The second component is very similar, but it only has 18 possible targets and some of the inputs will already have been used. So now we have 2×(18×2-1) + (18×2-1)×(18×2-1) ~1300. If we very roughly (not to mention very incorrectly, but I'm too tired to do the math properly) extrapolate to the other 18 components where the number of possible connections decreases in a similar fashion thoughout, we end up with a total number of 1500×1300×1140×1007×891×789×697×...×83×51×24×1 which is roughly 6.5×1050. However note that only 10% of these wires connect compatible parameters and only 50% of those will connect compatible components. So the number of valid connections we can make is roughly 3×1049.
All we have to do now is multiply the total number of valid connection per permutation with the total number of possible permutations; 20! × 3×1049 which comes to 7×1067 or 72 unvigintillion as Wolfram|Alpha tells me.
Impressive as these numbers sound, remember that by far the most of these permutations result in utter nonsense. Nonsense that produces a result, but not a meaningful one.
EDIT: This computation is way off, see this response for an improved estimate.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 12:06pm on March 15, 2013
rated by "<" symbols. Examples: "2<10", "2<4<10", "Pow(2, 1)<5*Sin(3)<10".
The entered text contains 2 or 3 segments separated by two or more consecutive dots. Examples "2..10", "2..4..10", "Pow(2, 1)....5*Sin(3)..10".
If only two segments are provided, then the initial value will be the same as the minimum value. If a bounds number or a default value is written as a simple number, then the number of decimal places will be harvested. I.e. "2..4..10" is not the same as "2..4..10.00" as the former will result in an integer slider and the latter in a slider with two decimal places.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:08am on February 15, 2013
the curves on surface issue it's solved seting flatten to the surface control point output. Still didnt know how to group points like:
1;1, 2;2, 3;3.....
1;2, 2;3, 3;4....
1;3, 2;4, 3;5...
....
t case point 3 should be able to move from 20 to 33
so in other word
pt 3 depends on pt 2
the problem is if i only have sliders
pt2 could be 20
and pt 3 could be 20, that is higher than 33
so the loft loops in itself
gracias por tu ayuda
salu2
m…
20 is my U+1. Anyway, I tried 19, 20, and 21-- I tried everything and all I see is red. I'll try to upload it again. Do I need to include some sort of vector conversion?