11 - Projective Geometry

11 - Projective Geometry

Greatly simplifies the basic operations in comparison to euclidean geometry and allows for projective transformations.

In 2D-projective geometry we use 3D vectors to represent all objects uniformly and operate on. We also use matrices to represent transformations.

2D Objects

The two 2D Objects (Point and Line) are both represented by a 3D Vector originating from the origin and representing the 2D object as projection on the 2D plane at height z = 1.

Any 3D Vector except (0,0,0) can represent any of the two geometric objects.

Homogenization

2D point at (x,y) represented as (x,y,1) which is the same as any λ(x,y,1) with λ being non-zero.

Normalization

To retrieve the projected 2D point one just divides by z: (x,y,z) -> (x/z,y/z,1) -> (x/z,y/z)

Points at Infinity

When z = 0 in (x,y,z) this means that we have a point at infinity in the direction (x,y).

Lines

A Line on the 2D plane is represented by a 3D Vector which is the normal of a plane going through the origin with its intersection at the 2D projected plane being the line.

Lines at Infinity

The one line (0,0,1) represents the line at infinity containing all the points at infinity.

All parallel lines meet at a point at infinity.

Operations

Is Point on Line

Point p is on line l iff vectors are orthogonal (dot product equals 0).

Line through two Points

Line through points p,q has to be orthogonal to both vectors (cross product).

Intersection of two Lines

Intersection point of lines l,m has to be on l and m (orthogonal to both => cross product).

Construct Parallel through Point

Parallel to line l through point p. Get infinite point in direction of l: q = l x (0,0,1). Then line through both points p,q. (cross product).

Construct Perpendicular through Point

Perpendicular to line l through point p. Get infinite point in direction of l: q = l x (0,0,1) = (x,y,0). Then get orthogonal to q: q_o = (y,-x,0). Then line through both points p,q (cross product).

Project Point onto Line

Project point p on line l. Get Perpendicular m to l. Then intersection of m,l (cross product).

Area of Triangle

Triangle given by points p,q,r. Normalize p,q,r. Obtain matrix M = (p q r). Area is 1/2 |det M| = 1/2 |p*(q x r)|.

Transformations

Rotation

    (cos -sin 0)
M = (sin  cos 0)
    (  0    0 1)

Translation

    (1 0 x)
M = (0 1 y)
    (0 0 1)

Affine Transformation

Preserving parallelity.

    (a b c)
M = (d e f)
    (0 0 1)

Projective Transformation

Preserving collinearity and point-on-line property. Uniquely determined by four non-collinear points and their images.

    (a b c)
M = (d e f)
    (g h i)

Find Projective Transformation

Find projective transformation matrix based on four non-collinear points a,b,c,d and their images a',b',c',d'.

Find two transformations M1 and M2:

• M1 is transformation from points (1,0,0),(0,1,0),(0,0,1),(1,1,1) to a',b',c',d'
• M2 is transformation from points (1,0,0),(0,1,0),(0,0,1),(1,1,1) to a,b,c,d

=> Transformation from a,b,c,d to a',b',c',d' will be M1 * M2^(-1)

Find M1, M2 by solving system of linear equations and inverting 3x3 matrix.