INFOGR Computer Graphics. Jacco Bikker - April-July Lecture 3: Geometry. Welcome!

Transcription

1 INFOGR Computer Graphics Jacco Bikker - April-July Lecture 3: Geometry Welcome!

2 Today s Agenda: 2D Primitives 3D Primitives Textures

3 2D Primitives Recap Last time: vectors and their properties: Magnitude, direction Scalar product Null vector, normal Parallel, linear (in)depence Commutative addition & subtraction Dot product, cross product Concepts: 4 1 R d spaces (orthonormal) 2D basis, Cartesian Left handed, right handed

4 2D Primitives Implicit representation Implicit curve: Example: circle x 2 + y 2 r 2 = 0 p c f x, y = 0 Function f maps two-dimensional points to a real value, i.e. x, y f(x, y) The points for which this value is 0 are on the curve. If p = (x, y) is a point on the circle, and p is a vector from the origin to p, it s length must be r, so p = r. Example: circle with center c and radius r: (x-c x ) 2 + y c y 2 r 2 = 0 p

5 2D Primitives Implicit representation Implicit curve: f x, y = 0 Example: line Slope-intersect form: y = ax + c y = 2 3 x + 1 Function f maps two-dimensional points to a real value, i.e. x, y f(x, y) The points for which this value is 0 are on the curve. Implicit form: ax + y c = 0 In general: Ax + By + C = 0 x c y a = y x

6 2D Primitives Implicit line representation Ax + By + C = 0 In this case: y = 2 3 x + 1 Slope-intersect form: y = ax + c Implicit form: ax + y c = 0 A = 2 3, B = 1, C = 1 c General form: The vector (A,B) is a normal of the line. x y Ax + By + C = 0 p(y, x) p(x, y) p( x, y) p( y, x)

7 2D Primitives Implicit line representation Ax + By + C = 0 In this case: y = 2 3 x + 1 We can use the normal to calculate the distance of a point to the line: d = N p + C A = 2 3, B = 1, C = 1 c For p = 3,3 : The vector (A,B) is a normal of the line. x y d = = = 0 For p = 0,0 : d = = 1 (? )

8 2D Primitives Implicit line representation Ax + By + C = 0 Given point p 1 and p 2, we determine A, B and C as follows: l = p 2 p 1 N = l y, l x A = N x, B = N y, C = (N p 1 ) p C p 2 Test with the line from the previous slides: p 1 = 3, 1 p 2 = 3,3 l = 6,4 N = 4,6 A = 4, B = 6 C = ( ) = 6 4x + 6y 6 = 0 It is convenient to normalize the normal: Only when ǁN ǁ = 1, C is the distance of the line to the origin. 2 x + y 1 = 0 3

9 2D Primitives Parametric representation Parametric curve: p 1 x y = g(t) h(t) Example: line p 0 = x p0, y p0, p1 = (x p1, y p1 ) x y = x p0 y p0 + t x p1 x p0 y p1 y p0 Or p t = p 0 + t p 1 p 0, t R. In this example: p 1 is the support vector; p 1 p 0 is the direction vector. p 0

10 2D Primitives Slope-intercept: y = ax + c Implicit representation: ax + y c = 0 Ax + By + C = 0 x p 1 y Parametric representation: p 0 p t = p 0 + t p 1 p 0

11 opposite INFOGR Lecture 3 Geometry 2D Primitives Circle - parametric x y = x c + r cos φ y c + r sin φ cos φ = x r c φ r sin φ = y r y tan φ = y x φ adjacent x SOH CAH TOA

12 Today s Agenda: 2D Primitives 3D Primitives Textures

13 3D Primitives Circle sphere (implicit) Recall: the implicit representation for a circle with radius r and center c is: x xc 2 + y yc 2 r 2 = 0 or: p c 2 r 2 = 0 In R 3, we get: p c = r x c x 2 + y c y 2 + z cz 2 r 2 = 0 or: p c = r p c

14 3D Primitives Line plane (implicit) Recall: the implicit representation for a line is: p 2 Ax + By + C = 0 In R 3, we get a plane: Ax + By + Cz + D = 0 p 1

15 3D Primitives Parametric surfaces A parametric surface needs two parameters: x = f(u, v), y = g(u, v), z = h(u, v). θ For example, a sphere: x = r cos φ sin θ, y = r sin φ sin θ, z = r cos θ. Doesn t look very convenient (compared to the implicit form), but it will prove useful for texture mapping.

16 3D Primitives Parametric planes Recall the parametric line definition: p 0 v p t = p 0 + t p 1 p 0 For a plane, we need to parameters: y w p s, t = p 0 + s p 1 p 0 + t(p 2 p 0 ) or: p s, t = p 0 + s v + tw where: p 0 is a point on the plane; v and w are two linearly independent vectors on the plane; s, t R. z x

17 Today s Agenda: 2D Primitives 3D Primitives Textures

18 Textures

19 Textures Back to the world of graphics Given a plane: y = 0 (i.e., with a normal vector (0,1,0) ). Two vectors on the plane define a basis: u = (1,0,0) and v = (0,0,1). Using these vectors, any point on the plane can be reached: P = λ 1 u + λ 2 v. We can now use λ 1, λ 2 to define a color at P: F(λ 1, λ 2 ) =. u v

20 Textures Example: F(λ 1, λ 2 ) = sin(λ 1 ) Another example: F(λ 1, λ 2 ) = ( int (2 λ 1 ) + (int)λ 2 ) & 1

21 Textures Other examples (not explained here): Perlin noise Details: Voronoi / Worley noise Details: A cellular texture basis function, S. Worley, 1996.

22 Textures Obviously, not all textures can be generated procedurally. For the generic case, we lookup the color value in a pixel buffer x y = P u P v T width T height Note that we find the pixel to read based on P; we don t find a P for every pixel of the texture. 255 v P u

23 Textures Retrieving a pixel from a texture: x y = P u P v T width T height We don t want to read outside the texture. To prevent this, we have two options: 1. Clamping x y = clamp(p u, 0, 1) clamp(p v, 0,1) T width T height Tiling x y = frac(p u) frac(p v) T width T height Tiling is efficiently achieved using a bitmask. This requires texture dimensions that are a power of 2.

24 Textures Texture mapping: oversampling

25 Textures Texture mapping: undersampling

26 Textures Fixing oversampling Oversampling: reading the same pixel from a texture multiple times. Symptoms: blocky textures. Remedy: bilinear interpolation: Instead of clamping the pixel location to the nearest pixel, we read from four pixels. w p1 : (1 frac(x)) (1 frac(y)) w p2 : frac x (1 frac y ) w p3 : 1 frac x frac(y) w p4 : 1 (wp 1 + wp 2 + wp 3)

27 Textures Fixing oversampling

28 Textures Fixing undersampling Undersampling: skipping pixels while reading from a texture. Symptoms: Moiré, flickering, noise. Remedy: MIP-mapping. The texture is reduced to 25% by averaging 2x2 pixels. This is repeated until a 1x1 image remains. When undersampling occurs, we switch to the next MIP level.

29 Textures

30 Textures

31 Textures Trilinear interpolation: blending between MIP levels.

32 Today s Agenda: 2D Primitives 3D Primitives Textures

33 INFOGR Computer Graphics Jacco Bikker - April-July Lecture 3: Geometry END of Geometry next lecture: 3D Engine Fundamentals