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108 lines
2.8 KiB
108 lines
2.8 KiB
4 years ago
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# Tetrahedron
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# Tetrahedron bounding box size:
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# Figure 1: Side View
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#
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# *_ -
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# / \ ^
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# / | \_
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# / | \ depth
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# / | \_ v
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# *----0-------* -
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#
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# Figure 2: Top View
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#
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# *_B
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# | \_
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# |\ \_
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# | \ \_
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# | \ \_
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# | \A \__
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# |<a> *-<b>-__*D
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# | / _/
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# | / _/
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# | / _/
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# |/ _/
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# |_/
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# *C
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# From these diagrams, you can tell that the width is the height of an
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# equilateral triangle. If we assume the edge of the tetrahedron is s, then:
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define @ns s - 4
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define @half_height s / 2
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define @nhalf_height @ns / 2
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# For the depth, a few calculations are necessary:
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#
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# a + b = s * sin(⅓π)
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# a² + (s/2)² = b²
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define @a (s/2)*tan(⅙π)
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define @na (@ns/2)*tan(⅙π)
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define @b s * (sin(⅓π) - tan(⅙π)/2)
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define @nb @ns * (sin(⅓π) - tan(⅙π)/2)
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# a² + d² = (s*sin(⅓π))²
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define @depth s * sqrt((sin(⅓π))^2 - ((1/2)*tan(⅙π))^2)
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define @ndepth @ns * sqrt((sin(⅓π))^2 - ((1/2)*tan(⅙π))^2)
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# Considering that the origin is in the center of the base, the min x is -a,
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# the min y is -s/2, and the min z is 0
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-@a < x < @b
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-@half_height < y < @half_height
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0 < z < @depth
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# A can be given by:
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#
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# (0,0,depth)
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# B and C can be given by:
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#
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# (-a,±half_height,0)
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# D can be given by:
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#
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# (b,0,0)
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# The tetrahedron can be given by the intersection of 4 semi-spaces.
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# However, one of the semi-spaces is z >= 0, which is implicit in the
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# declaration of the z boundaries, so it will not require a condition.
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# Cross Product Cheatsheet
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#
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# a ⨯ b = | i j k |
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# | ax ay az |
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# | bx by bz |
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#
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# = (ay bz - az by, az bx - ax bz, ax by - ay bx)
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# The normal vector for the plane ABC is the cross product of AB and AC
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#
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# AB = (-a, half_height, -depth)
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# AC = (-a, -half_height, -depth)
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#
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# vABC = (half_height * (-depth) - (-depth) * (-half_height),
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# (-depth) * (-a) - (-a) * (-depth),
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# (-a) * (-half_height) - half_height * (-a)) =
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# (- s * depth, 0, s * a)
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# The normal vector for the plane ADB is the cross product of AD and AB
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#
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# AD = (b,0,-depth)
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# AB = (-a, half_height, -depth)
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#
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# vABD = (depth * half_height, depth * a + b * depth, b * half_height)
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# The normal vector for the plane ACD is equal to the previous one but
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# reflected on the y axis. Thus, we only have to flip a plus to a minus.
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{x * (- s * @depth) + (z - @depth) * (s * @a) ≤ 0 ∧ \
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x * @depth * @half_height + y * @depth * (@a + @b) + (z - @depth) * @b * @half_height ≤ 0 ∧ \
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x * @depth * @half_height - y * @depth * (@a + @b) + (z - @depth) * @b * @half_height ≤ 0} \
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⊻ \
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{x * (- @ns * @ndepth) + (z - @ndepth) * (@ns * @na) ≤ 0 ∧ \
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x * @ndepth * @nhalf_height + y * @ndepth * (@na + @nb) + (z - @ndepth) * @nb * @nhalf_height ≤ 0 ∧ \
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x * @ndepth * @nhalf_height - y * @ndepth * (@na + @nb) + (z - @ndepth) * @nb * @nhalf_height ≤ 0 ∧ \
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z > 1}
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