7, 9, 12 and 13 to be able to rotate freely around the y axis at nodes 2, 3, 6, 7, 10 and 11 respectively. The last 2 conditions, for elements 12 and 13, doesn't give any problems, but the first 4 does.
Any help?
…
circles that can be populated (for each radius size) is set as an integer (or slider)
(ie. radius 1.5 = 10 , radius 3= 6, radius 6 = 6, radius 9=4)
Conditions are:
1) Each of the circle has a radius of influence,
Radius of influence = double the radius of the circle)
(3, 6, 12, 18)
2) Any overlapping circles in either: Radius of influence or the Circles are removed so that
No circles overlap.
3) There must also be 4 circles set at the corner points of the grid - These must be circles with a radius of 3 or 6
If you can do that I will be amazed as i've been trying for weeks! :(
Ive attached a sketch of what im looking for…
I'm trying to retrieve data from a tag heuer timing system through rs 232. Firefly ports available says com ports 1 and 3 are available. Is one of those two the rs 232 9 pin? Thanks for your help.
cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014