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CreeD wrote:
> It's shaped like an equilateral triangle with the 3 points clipped
> off. In other words, viewed from above it's a hexagon with 3 long
> sides and 3 short ones. The short sides should equal exactly 1/3rd
> of the long ones. Can someone give me the gist of how to come up
> with some exact coordinates for such a shape (or, ideally, post the
> code?)
The corners are all on a regular triangular lattice ...
b = sqrt(3/4); // sin(pi/3)
P0 = < 0 , 0 >
P1 = < 3 , 0 >
P2 = < 3.5, b >
P3 = < 2 ,4b >
P4 = < 1 ,4b >
P5 = <-0.5, b >
What you need most is the second coordinate of the centroid, which is
5b/3, or 5/sqrt(12) -- that's its distance from the line {P0,P1}.
Subtract it from 4b to get 7b/3 = 7/sqrt(12), the distance from the
centroid to the line {P3,P4}. Thus:
intersection {
plane { z, 5/sqrt(12) }
plane { z, 5/sqrt(12) rotate 120*y }
plane { z, 5/sqrt(12) rotate -120*y }
plane { -z, 7/sqrt(12) }
plane { -z, 7/sqrt(12) rotate 120*y }
plane { -z, 7/sqrt(12) rotate -120*y }
}
--
Anton Sherwood -- br0### [at] p0b0x com -- http://ogre.nu/
(not an advanced user)
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