=1), {1,1}(N=1)...{1,15}(N=1). While Y is made of {0}(N=6), {1}(N=15).
What I need to do, is to cycle through all the branched items of the X lists with every single item of the corresponding list of Y.
So:
{0,0}(N=1) with 1st item of Y{0},
{0,1}(N=1) with 1st item of Y{0}
...
{0,15}(N=1) with 1st item of Y{0} (and this is the first cycle).
Then (cycle 2)
{0,0}(N=1) with 2nd item of Y{0},
{0,1}(N=1) with 2nd item of Y{0}
...
{0,15}(N=1) with 2nd item of Y{0}
Then do the same thing with the X list {1,0} .. {1,15} with the corresponding Y{1}
As a result I need the following list structure:
{0,0}N=6, {0,1}N=6, {0,2}N=6 ... {0,15}N=6,
{1,0}N=15, {1,1}N=15, {1,2}N=15 ... {0,15}N=15.
If I use the component you suggested "shift path" it will match the items per item-order instead of cycling through all the items of the first sublist...
Hope this makes sense.
Any suggestion?…
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…
9 8 7 6
5 4 3 2 1 0
I am triangulating this surface. I want to select just the red vertices. As you can note, I just need the inner vertices of this surface. I could do it mannually, but if I want to change the mesh density later, I will have to pick all of them manually again later.
Can someone help me?
Tks
…
out the difference between the target and 400 and subtract this from the numbers in the series that are between 6 and 26 but you need to think about how exactly you will do this... you can't just divide the difference by 23 and take this from each number as it would make some numbers less than 6.
…
Added by martyn hogg at 3:40pm on November 19, 2015