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60 lines
1.7 KiB
60 lines
1.7 KiB
# 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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# 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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# 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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{{ρ ≤ @a * sec(φ - ⅓π) * (1 - z/@depth) ∧ φ ≥ 0 ∧ φ ≤ 2 * ⅓π} ∨ \ |
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{ρ ≤ @a * sec(φ - π) * (1 - z/@depth) ∧ {φ ≥ 2 * ⅓π ∨ φ ≤ -2 * ⅓π}} ∨ \ |
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{ρ ≤ @a * sec(φ + ⅓π) * (1 - z/@depth) ∧ φ ≤ 0 ∧ φ ≥ -2 * ⅓π} ∨ ρ = 0} \ |
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⊻ \ |
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{z ≥ 1 ∧ z ≤ @ndepth ∧ \ |
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{{ρ ≤ @na * sec(φ - ⅓π) * (1 - z/@ndepth) ∧ φ ≥ 0 ∧ φ ≤ 2 * ⅓π} ∨ \ |
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{ρ ≤ @na * sec(φ - π) * (1 - z/@ndepth) ∧ {φ ≥ 2 * ⅓π ∨ φ ≤ -2 * ⅓π}} ∨ \ |
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{ρ ≤ @na * sec(φ + ⅓π) * (1 - z/@ndepth) ∧ φ ≤ 0 ∧ φ ≥ -2 * ⅓π} ∨ ρ = 0}}
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