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116 lines
3.7 KiB
116 lines
3.7 KiB
# Cube bounding box size: |
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# Figure 1: |
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# _*_ - |
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# _/ | \_ ^ |
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# _/ | \_ | |
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# * | * h |
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# | | | e |
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# | _*_ | i |
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# | _/ \_ | g |
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# |_/ \_| h |
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# *_ _* t |
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# \_ _/ | |
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# \_ _/ v |
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# * - |
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# |
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# |<--width-->| |
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# From this diagram, you can tell that the width is the diagonal of the square. |
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# We know as a corollary of Pythagoras's theorem that the diagonal of a square |
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# of side s is s * √2 |
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define @width s * √2 |
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# Figure 2: |
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# _*__ - |
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# / \__ ^ |
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# _/ \__ d |
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# _*__ _* e |
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# _/ \__ / p |
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# *__ \__ _/ t |
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# \__ _* h |
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# \__ _/ v |
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# * - |
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# |
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# |<----height----->| |
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# From this diagram, you can tell that the depth is the diagonal of the cube. |
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# From Pythagoras's theorem we know that the diagonal of a cube of side s is |
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# √(((√2)^2)s^2 + s^2) = s * √3 |
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define @depth s * √3 |
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# If we consider the x = 0 section of the image above, we get the following: |
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# Figure 3: |
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# _*__ - |
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# / \__ ^ |
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# _/ | \__ d |
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# _/ b _* e |
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# _/ | / p |
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# *__ -a- -* _/ t |
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# \__ | _/ h |
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# \__c_/ v |
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# * - |
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# |
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# |<----height----->| |
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# We know that: |
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# a^2 + b^2 = (s * √2)^2 |
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# a^2 + c^2 = s |
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# b + c = s * √3 |
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# From that, we get that a = s * √(2/3), b = (2*s)/(√3), c = s/(√3) |
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# And we know that height = 2 * a |
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define @half_height s * √(2/3) |
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# Cube bounding box position: |
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# Considering that the cube's center is the origin, we just need to negate and |
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# half the bounding box sizes to get the minimum boundaries for each variable. |
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-@width / 2 < x < @width / 2 |
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-@half_height < y < @half_height |
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-@depth / 2 < z < @depth / 2 |
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# The cube is the intersection of 6 semi-spaces. |
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# Each semi-space is defined by an inequality based on the plane that defines |
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# its boundary. |
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# The plane equations are achieved through a point belonging to the plane, as |
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# well as a vector normal to the plane. |
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# The easiest plane to construct is the one that is invariant in the x |
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# dimension, that is, parallel to the x axis. |
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# For ease of calculation, the point used is the point (0,0,@half_height). |
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# The vector used will have the coordinates (0,-a,c) |
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# This means the plane equation starts out as: |
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# -a*y + c*(z-((s * √3)/2)) <= 0 |
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# Which can be simplified to: |
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# -a*y + c*z <= (s^2)/2 |
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# Given that a and c are both expressed as functions of s, we can factor out |
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# the s and get the following: |
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# y * (-√(2/3)) + z * √(1/3) <= s / 2 |
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# By using the same point and rotating the vector by +-120 degrees or +-2*pi/3 |
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# radians, we get the following two plane inequalities. |
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# (-(√2)/2) * x + y * √(1/6) + z * √(1/3) <= s / 2 |
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# ((√2)/2) * x + y * √(1/6) + z * √(1/3) <= s / 2 |
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# By replacing the point with (0,0,-@half_height) and scaling each of the |
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# three vectors by (-1), we get the remaining three plane inequalities. |
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{y * (-√(2/3)) + z * √(1/3) <= s / 2 ∧ \ |
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(-(√2)/2) * x + y * √(1/6) + z * √(1/3) <= s / 2 ∧ \ |
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((√2)/2) * x + y * √(1/6) + z * √(1/3) <= s / 2 ∧ \ |
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y * √(2/3) - z * √(1/3) <= s / 2 ∧ \ |
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(-(√2)/2) * x - y * √(1/6) - z * √(1/3) <= s / 2 ∧ \ |
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((√2)/2) * x - y * √(1/6) - z * √(1/3) <= s / 2} \ |
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⊻ \ |
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{y * (-√(2/3)) + z * √(1/3) <= (s - 2) / 2 ∧ \ |
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(-(√2)/2) * x + y * √(1/6) + z * √(1/3) <= (s - 2) / 2 ∧ \ |
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((√2)/2) * x + y * √(1/6) + z * √(1/3) <= (s - 2) / 2 ∧ \ |
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y * √(2/3) - z * √(1/3) <= (s - 2) / 2 ∧ \ |
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(-(√2)/2) * x - y * √(1/6) - z * √(1/3) <= (s - 2) / 2 ∧ \ |
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((√2)/2) * x - y * √(1/6) - z * √(1/3) <= (s - 2) / 2}
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