une 7 at Madison Square Garden. Promoted by Miguel Cotto Promotions, Top Rank®, DiBella Entertainment and Sampson Boxing, in association with Maravilla Box, Tecate and Madison Square Garden.Fight Tickets and PPV InfoCotto vs Martinez tickets are now available for purchase online at Ticketnetwork. To charge by phone with a major credit card, call Buy Tickets Center at (855) 859-4045. The fight will be produced and distributed live by HBO Pay-Per-View®, beginning at 9:00 p.m. ET/6:00 p.m. PT.Hall of Fame trainer Freddie Roach is devising a cunning and effective strategy to, badger, outfox, and hound WBC middleweight champion Sergio “Maravilla” Martinez (51-2-2, 28KOs) into a brush with defeat, on June 7th at Madison Square Garden. Roach is training former three division world champion Miguel Cotto (38-4, 31KOs) for the match.“We have seen Martinez fights. I observed him closely when I was in Julio Cesar Chavez Jr’s corner, and he’s very fast. But…like all opponents he can be outboxed. If Miguel focuses on his body, Martinez will loose his speed and that’s when we will knock him out,” Roach mused.Cotto (38-4, 31 KO) and Martinez (51-2-2, 28 KO) will meet for Martinez’s WBC middleweight title and The Ring Magazine championship at Madison Square Garden, promoted by Top Rank, DiBella Entertainment, and Miguel Cotto Promotions.“Face Off” will premiere on Saturday, May 24 at Midnight EDT (so technically, the first minutes of Sunday, May 25, technically), after the May 24 HBO Boxing After Dark card, featuring Bryant Jennings vs Mike Perez, Daniel Geale vs Matthew Macklin, and Edwin Rodriguez vs Marcus Johnson.Other air times:HBO air times include: May 24 (12:00 a.m.), 25 (10:00 a.m.), 27 (3:15 p.m. & 2:30 a.m.), 28 (12:30 a.m.), 29 (5:30 p.m.), 30 (10:30 a.m.), 31 (7:00 p.m.) and June 3 (12:30 p.m. & 3:30 a.m.), 5 (1:15 a.m.), 6 (6:45 p.m.) and 7 (10:45 a.m.).HBO2 air times include: May 25 (5:30 p.m.), 26 (10:45 a.m. & 3:30 a.m.) 30 (4:00 p.m.) and 31 (11:30 p.m.).…
une 7 at Madison Square Garden. Promoted by Miguel Cotto Promotions, Top Rank®, DiBella Entertainment and Sampson Boxing, in association with Maravilla Box, Tecate and Madison Square Garden.Fight Tickets and PPV InfoCotto vs Martinez tickets are now available for purchase online at Ticketnetwork. To charge by phone with a major credit card, call Buy Tickets Center at (855) 859-4045. The fight will be produced and distributed live by HBO Pay-Per-View®, beginning at 9:00 p.m. ET/6:00 p.m. PT.Hall of Fame trainer Freddie Roach is devising a cunning and effective strategy to, badger, outfox, and hound WBC middleweight champion Sergio “Maravilla” Martinez (51-2-2, 28KOs) into a brush with defeat, on June 7th at Madison Square Garden. Roach is training former three division world champion Miguel Cotto (38-4, 31KOs) for the match.“We have seen Martinez fights. I observed him closely when I was in Julio Cesar Chavez Jr’s corner, and he’s very fast. But…like all opponents he can be outboxed. If Miguel focuses on his body, Martinez will loose his speed and that’s when we will knock him out,” Roach mused.Cotto (38-4, 31 KO) and Martinez (51-2-2, 28 KO) will meet for Martinez’s WBC middleweight title and The Ring Magazine championship at Madison Square Garden, promoted by Top Rank, DiBella Entertainment, and Miguel Cotto Promotions.“Face Off” will premiere on Saturday, May 24 at Midnight EDT (so technically, the first minutes of Sunday, May 25, technically), after the May 24 HBO Boxing After Dark card, featuring Bryant Jennings vs Mike Perez, Daniel Geale vs Matthew Macklin, and Edwin Rodriguez vs Marcus Johnson.Other air times:HBO air times include: May 24 (12:00 a.m.), 25 (10:00 a.m.), 27 (3:15 p.m. & 2:30 a.m.), 28 (12:30 a.m.), 29 (5:30 p.m.), 30 (10:30 a.m.), 31 (7:00 p.m.) and June 3 (12:30 p.m. & 3:30 a.m.), 5 (1:15 a.m.), 6 (6:45 p.m.) and 7 (10:45 a.m.).HBO2 air times include: May 25 (5:30 p.m.), 26 (10:45 a.m. & 3:30 a.m.) 30 (4:00 p.m.) and 31 (11:30 p.m.).…
une 7 at Madison Square Garden. Promoted by Miguel Cotto Promotions, Top Rank®, DiBella Entertainment and Sampson Boxing, in association with Maravilla Box, Tecate and Madison Square Garden.Fight Tickets and PPV InfoCotto vs Martinez tickets are now available for purchase online at Ticketnetwork. To charge by phone with a major credit card, call Buy Tickets Center at (855) 859-4045. The fight will be produced and distributed live by HBO Pay-Per-View®, beginning at 9:00 p.m. ET/6:00 p.m. PT.Hall of Fame trainer Freddie Roach is devising a cunning and effective strategy to, badger, outfox, and hound WBC middleweight champion Sergio “Maravilla” Martinez (51-2-2, 28KOs) into a brush with defeat, on June 7th at Madison Square Garden. Roach is training former three division world champion Miguel Cotto (38-4, 31KOs) for the match.“We have seen Martinez fights. I observed him closely when I was in Julio Cesar Chavez Jr’s corner, and he’s very fast. But…like all opponents he can be outboxed. If Miguel focuses on his body, Martinez will loose his speed and that’s when we will knock him out,” Roach mused.Cotto (38-4, 31 KO) and Martinez (51-2-2, 28 KO) will meet for Martinez’s WBC middleweight title and The Ring Magazine championship at Madison Square Garden, promoted by Top Rank, DiBella Entertainment, and Miguel Cotto Promotions.“Face Off” will premiere on Saturday, May 24 at Midnight EDT (so technically, the first minutes of Sunday, May 25, technically), after the May 24 HBO Boxing After Dark card, featuring Bryant Jennings vs Mike Perez, Daniel Geale vs Matthew Macklin, and Edwin Rodriguez vs Marcus Johnson.Other air times:HBO air times include: May 24 (12:00 a.m.), 25 (10:00 a.m.), 27 (3:15 p.m. & 2:30 a.m.), 28 (12:30 a.m.), 29 (5:30 p.m.), 30 (10:30 a.m.), 31 (7:00 p.m.) and June 3 (12:30 p.m. & 3:30 a.m.), 5 (1:15 a.m.), 6 (6:45 p.m.) and 7 (10:45 a.m.).HBO2 air times include: May 25 (5:30 p.m.), 26 (10:45 a.m. & 3:30 a.m.) 30 (4:00 p.m.) and 31 (11:30 p.m.).…
9)
http://www.grasshopper3d.com/forum/topics/circle-packing-1
In any case, i need a posibility to define diameters of circles (0.1 mm resolution) and, for each diameter, how many circles i want to be used for calculations.
For example. 3 circles from 4mm, 5 circles from 2.5 mm, 7 circles from 2.1mm and rest of the surface to be filled with 1.3mm circles...
In some cases i will use maybe 3 or 4 different diameters, in some cases maybe more than 10.
But mostly will be from 3 to 10 different diameters.
Also, it would be great if i could predefine positions for some of the circles. Mostly for greatest diameters.
Thank you and sorry for my bad english...
…
pe and its surface.
However, I don't have that much knowledge about both grasshopper and Mathematica.. I mean I can only make assumptions and think about relations of certain functions but that's all.
If you can help me on this, I would appreciate it so much.
You can see a screenshot of the code and model of the demonstration from mathematica in attachment.
And here is the mathematica code;
Manipulate[ Module[{\[CurlyEpsilon] = 10^-6, c1 = Tan[a1], c2 = Tan[a2], c3 = Tan[a3], c4 = Tan[a4], c5 = Tan[a5], c6 = Tan[a6]}, ContourPlot3D[ Evaluate[ c6 Sin[3 x] Sin[2 y] Sin[z] + c4 Sin[2 x] Sin[3 y] Sin[z] + c5 Sin[3 x] Sin[y] Sin[2 z] + c2 Sin[x] Sin[3 y] Sin[2 z] + c3 Sin[2 x] Sin[y] Sin[3 z] + c1 Sin[x] Sin[2 y] Sin[3 z] == 0], {x, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, {y, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, {z, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, Mesh -> False, ImageSize -> {400, 400}, Boxed -> False, Axes -> False, NormalsFunction -> "Average", PlotPoints -> ControlActive[10, 30], PerformanceGoal -> "Speed"]], {{a1, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(1\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a2, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(2\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a3, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(3\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a4, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(4\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a5, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(5\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a6, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(6\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, AutorunSequencing -> {1, 3, 5}, ControlPlacement -> Left]…
number of divisions on that curve as in the defintion (i.e. by 4). The offset in the def is slightly different and should cull two or three more curves as in the lists that show my aim below.
Basically I want to look into each branch of the groups of points from each closed curve . Marking in a list whether it contains a one or a zero (0= outside 1 = coincidents).
{0;0}0. 21. 22. 23. 2 {0;1} 0. 01. 22. 03. 2 {0;2}0. 01. 02. 03. 0 {0;3}0. 21. 22. 23. 2 {0;4}0. 21. 22. 23. 2 {0;5}0. 21. 22. 23. 2 {0;6}0. 01. 22. 23. 1 {0;7}0. 21. 22. 03. 0 {0;8}0. 21. 22. 23. 2 {0;9}0. 21. 22. 23. 2 {0;10}0. 21. 22. 23. 2 {0;11}0. 21. 22. 23. 2 {0;12}0. 21. 22. 23. 2 {0;13}0. 01. 22. 23. 0 {0;14}0. 21. 22. 23. 2
I want to create a list from these points. That marks each curve that pokes out, in a cull pattern as such:
20022210222202
Using a 1 where there are co-incidents in the curve points and the boundary. A 2 for true (outside points) and a 0 for containment. So I might be able to use the 1 in future developments - however if a true false list is easiest I can live with that.
So could I use F(x) function? - to look for 0 or 1's in each bunch of points and thus list as such for a cull pattern? or will Path mapper help me here? Or can I rely on simply grafting and splitting??
I am usure of the neatest solution and would love to learn. Hope you can direct me.rgrds
J.…
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.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 12:06pm on March 15, 2013
}
0,3
{7}
1,2
{8}
1,3
{9}
2,3
{10}
0,1,2
{11}
0,1,3
{12}
0,2,3
{13}
1,2,3
{14}
0,1,2,3
I want it to be parametric so I can have it for any set of indexes
I have tried to solve it with points and then removing duplicate points but it's not helping much
any help is truly appreciated
…
the join component and then iterates these curves through a move component(10 vectors) all is in order. Yet when I connect more than a single path the order of the generated curves as they are iterated through the move componenet seems to become arbitrary. All the curves connect, meaning Im pretty sure the first curve in path a is connected to the first curve in path b and c, yet this new curve is now curve 5 for instance.
As suggested I have attached a graf tree component which now seperates these curves in different paths before attaching them to the join slider. The result though, as I connect the latter to the move component, is that the ten resultant curves are iteraed 10 times. So 100 curves altogether instead of 10...
Any help would be awesome... Why does the join command destroy the order of the curves even when it seems to be fulfilling its (joning) function properly?
…