it,
[3] the upper limit, [4] the slider position.
What do you think?
Matt
Radlab, Incorporated
25 Drydock Ave, Floor 6
Boston, MA 02210
W| radlabinc dot com
E| matt@radlabinc
P| 617.440.3588…
Added by Matt Trimble at 7:54pm on November 17, 2009
l at each point intersection, less 14. align holes to common angle between each 2 points of intersection (so ovals align with curve)5. copy 4. 360/60 about center circle (creates 6 curves rotated thru 360)6. it appears there a 3 more sets of curves that need to be taken care in the same way as 1 thru 4 (see colander pic)6. project the oval patterns onto, 1/2 a sphere somewhat larger that the surface circle, to avoid extreme oval distortion.7. needs some Boolean subtraction of holes from sphere surface
Does this simple road map have some merit?
…
rk perfectly. line always connect branch wgich is shifted by 6 ( 0 to 6, 1 to 7) but second loft connecting wrong ( 0 to 3 and then 1 to 30)
Please advise what I am doing wrong?
David…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
Int32 = 0 To y.count - 1
Dim c As Char = Convert.ToChar(y(i).substring(7, 1))
Dim d As Char = Convert.ToChar(y(i).substring(8, 1))
Dim p As Integer
If c = ";" Then
p = convert.ToInt32(y(i).substring(6, 1))
Else If d = ";" Then
p = convert.ToInt32(y(i).substring(6, 2))
Else
p = convert.ToInt32(y(i).substring(6, 3))
End If
Dim path As New EH_Path(p)
For j As int32 = sum To sum + z(i) - 1
tree.Add(x(j), path)
Next
sum = sum + z(i)
Next
A = tree…