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?
…
Like you've done. Use List Item and input an integer which represents the curve you want to select. Your slider will need to be able to handle integers 0, 1, 2, 3, 4, and 5 since you have 6 curves.
ersect (2, 3, 4, 5, 6) with the line and the ones which do not intersect (0, 1, 7). Intersect is done! But how to get the non intersecting vectors (0, 1, 7)?
So I e. g. could deselect vectors 2, 3, 4, 5, 6 so I would display/use only vectors 0, 1, 7 and the bounced ones.
Appreciate your help!
Rudi…
i mean, i want a slider that can do 3 sides, 4 sides, 5, 6, 7, 8, 9 and 10. for the grids because I dont want to use a fixed grid shape such as square grid (4 sides only).
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…
t item (see the image), is it possible to do this in another way (quickly) ?
Is it possible to divide that curve into 2 separate curves using a point that i've used for the division?
Thanks…
Added by luca.pavarin at 4:08pm on January 7, 2010