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A question about polygons

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A question about polygons

Devising a computational method for determining if a point lies in a plane convex polygon

- Recall our ‘globe.cpp’ ray-tracing demo
- It depicted a scene showing two objects
- One of the objects was a square tabletop
- Its edges ran parallel to x- and z-axes
- That fact made it easy to determine if the spot where a ray hit the tabletop fell inside or outside of the square’s boundary edges
- We were lucky!

- What if we wanted to depict our globe as resting on a tabletop that wasn’t a nicely aligned square? For example, could we show a tabletop that was a hexagon?
- The only change is that the edges of this six-sided tabletop can’t all be lined up with the coordinate system axes
- We need a new approach to determining if a ray hits a spot that’s on our tabletop

- How can we tell if a point lies in a triangle?

c

q

p

b

a

Point p lies inside triangle Δabc

Point q lies outside triangle Δabc

- Draw vectors from a to b and from a to c
- We can regard these vectors as the axes for a “skewed” coordinate system
- Then every point in the triangle’s plane would have a unique pair of coordinates
- We can compute those coordinates using Cramer’s Rule (from linear algebra)

c

ap = c1*ab + c2*ac

p

b

a

c1 = det( ap, ac )/det( ab, ac )

c2 = det( ab, ap )/det( ab, ac )

y-axis

p = (c1,c2)

x-axis

x + y = 1

p lies outside the triangle if c1 < 0 or c2 < 0 or c1+c2 > 1

typedef float scalar_t;

typedef struct { scalar_t x, y; } vector_t;

scalar_t det( vector_t a, vector_t b )

{

returna.x * b.y – b.x * a.y;

}

theta = 2*PI / 6

( cos(2*theta), sin(2*theta) )

( cos(1*theta), sin(1*theta) )

( cos(3*theta), sin(3*theta) )

( cos(0*theta), sin(0*theta) )

( cos(5*theta), sin(5*theta) )

( cos(4*theta), sin(4*theta) )

A point p lies outside the hexagon -- unless

it lies inside one of these four sub-triangles

- Demo program ‘hexagon.cpp’ illustrates the use of our just-described algorithm
- Every convex polygon can be subdivided into triangles, so the same ideas can be applied to any regular n-sided polygon
- Exercise: modify the demo-program so it draws an octagon, a pentagon, a septagon

- A tetrahedron is a 3D analog of a triangle
- It has 4 vertices, located in space (but not all vertices can lie in the same plane)
- Each face of a tetrahedron is a triangle

c

b

o

a

typedef float scalar_t;

typedef struct { scalar_t x, y, z; } vector_t;

scalar_t det( vector_t a, vector_t b, vector_t c )

{

scalar_tsum = 0.0;

sum += a.x * b.y * c.z – a.x * b.z * c.y;

sum += a.y * b.z * c.x – a.y * b.x * c.z;

sum += a.z * b.x * c.y – a.z * b.y * c.x;

returnsum;

}

- Let o, a, b, c be vertices of a tetrahedron
- Form the three vectors oa, ob, oc and regard them as the coordinate axes in a “skewed” 3D coordinate system
- Then any point p in space has a unique triple of coordinates:op = c1*oa + c2*ob + c3*oc
- These three coordinates can be computed using Cramer’s Rule

c1 = det( op, ob, oc )/det( oa, ob, oc )

c2 = det( oa, op, oc )/det( oa, ob, oc )

c3 = det( oa, ob, op )/det( oa, ob, oc )

Point p lies inside the tetrahedron – unless

c1 < 0 or c2 < 0 or c3 < 0

orc1 + c2 + c3 > 1

- Just as a convex polygon can be divided into subtriangles, any convex polyhedron can be divided into several tetrahedrons
- We can tell if a point lies in the polyhedron by testing to see it lies in one of the solid’s tetrahedral parts
- An example: the regular dodecahedron can be raytraced by using these ideas!