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
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…
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
ements between mid axes of elements ( being perpendicular to both). On the attached image, the links are the small elements connectiong nodes 4 to 5, 6 to 7 and 8 to 9. All nodes (1 to 9 including 1', 2', 3') are defined in Karamba as fixed supports but nodes 1,1', 2, 2', 3, and 3' have hinges added with the beam joint component. The freed rotations are shown on the figure.
I wondered if that was the correct way of defining such a structure in Karamba bearing in mind that nodes 1', 2' and3' are free nodes in the reality.
Thanks again for your help !
Yousef…
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013
plit each edge into 3 and join the dots to make 9 triangle on each face.
3. project the points of these triangles out onto a sphere with the same radius as the icosohedron.
maybe you can take the same approach but project out onto your olive shape? I don't think you will end up with only 4 tessellating shapes though....
If the base icosohedron of a geodesic dome is split into 9 triangles on each face you get a geodesic dome with 3 different corner geometries, 2 different triangular panels and 3 different length triangle edges. If the dome was elongated carefully you could probably restrict the resulting components of an olive shaped dome.
This will be a slightly different dome to your Rhombic Enneacontahedron too!…
onsecutive points at the same height then your 'Break at discontinuities' component eliminates the middle point completely and then the 'Interpolate Curve' component gives a much bigger bump in the wrong direction. This was enough to get curves to meet from opposite sides.
I fixed this by changing the heights to 1.1 or 2.9, rather than 1.0 and 3.0, but it took a little while to work it out! Sigh.
I attach a new version. But I actually preferred it as it was before. See what you think!
Bob
p.s. in the first list, elements 11, 12, 23 and 24 go from 1 to 3; elements 17 and 18 go from 3 to 1. In the second list, elements 6, 17, 18 and 29 go from 1 to 3; elements 12 and 23 go from 3 to 1. Given the above fix, these can be easily seen.…
Added by Bob Mackay at 10:40pm on November 24, 2015