Lecture 4: 3D viewing and projections

Transcription

1 Lecture 4: 3D viewing and projections

2 Today s lecture Rotations around an arbitrary axis (Continuation from the last lecture) The view coordinate system Change of coordinate system (same origin) Change of frame (change of coordinate system + translation) How to position the camera? Projections: Orthogonal and perspective

3 Pipeline Todays lecture in the pipeline

4 3D transformation (From last lecture) Translation x y z 1 = t x t y t z x y z 1 Scaling

5 X Rotation 3D transformation (From last lecture) Y Rotation Z Rotation

6 Problem from last lecture: Rotation about an arbitrary axis Rotation about an arbitrary axis v: Find a matrix M that aligns e.g. (1,0,0) with v. Apply M -1 to object. Rotate object around (1,0,0) Apply M to object. M is a change of coordinate system Given v, how do we find M?

7 Change of coordinate system 1 We wish to express the same point with two different sets of basis vectors. P=xe 1 +ye 2 = x f 1 +y f 2

8 Change of coordinate system 2 Assume that we know the basis vectors of the new coordinate system f in terms of the old coordinate system e. f 1 =ae 1 +be 2 f 2 =ce 1 +de 2

9 Change of coordinate system 3 P=xe 1 +ye 2 = x f 1 +y f 2 f 1 =ae 1 +be 2 f 2 =ce 1 +de 2 f 1 f 2 P=x (ae 1 +be 2 ) +y (ce 1 +de 2 ) =(x a+y c)e 1 +(x b+y d)e 2 x=x a+y c y=x b+y d (Coordinates in f transferred to e) but we want x and y! (Coordinates in e to f)

10 Matrix notations (Coordinates in f transferred to e) (Coordinates in e transferred to f)

11 Change of coordinate system, final x y = a b x y c d = M 1 x y x y = M x y M is a pure rotation matrix if (and only if) we are dealing with Ortho-Normal (ON)-bases. M is then orthogonal and thus M 1 = M T

12 Interlude: Orthogonal matrices An orthogonal matrix is a matrix where the rows and columns are mutually orthogonal unit-length vectors. Each of the simple axis rotation matrices is an example of an orthogonal matrix. Example: X Rotation

13 Properties of orthogonal matrices The product of two orthogonal matrices is an orthogonal matrix Always invertible Skriv en ekvation här. The inverse of an orthogonal matrix is equal to the transpose of the matrix a b c d T a = c b d x y = M 1 x y = MT x y a c = b d T x y = a c b d x y

14 Back on track: How do we construct an ON basis? (If we only have one vector v) v 1 = v v Find a vector v' that is not parallel to v 1. v 2 = v 1 v v 1 v v 3 = v 1 v 2 v 3 v 2 v 1 =v ON basis: orthogonal vectors with length 1.

15 How do we find v' in practice? 1. Pick any random vector v. 2. If v X v 1 = 0 repeat 1.

16 How do we find v' in practice? 1. Pick any random vector v. 2. If v X v 1 = 0 repeat 1. No! If v X v 0, we risk numerical errors.

17 A better procedure Pick two orthogonal vector u 1 and u 2, e.g., u 1 =(1,0,0), u 2 =(0,1,0). If u 1 X v 1 > u 2 X v 1, set v' = u 1. Otherwise, set v'=u 2.

18 Rotation around an arbitrary axis, final Rotation about an arbitrary axis v: Construct an ON-basis where v 1 =v is the first basis vector. (v 2 and v 3 are the other bases) Construct a corresponding change-of-coordinates matrix M, which will align (1, 0, 0) with v. M = v = v 1 = v 1x v 2x v 3x v 1y v 2y v 3y v 1z v 2z v 3z v 1x v 2x v 3x v 1y v 2y v 3y v 1z v 2z v 3z 1 0 0

19 Rotation around an arbitrary axis, final Apply M -1 to object (transfers coordinates to a coordinate system where v is aligned with (1,0,0) ) Rotate around (1,0,0) Apply M. (transfers coordinates back to the original coordinate system) With homogeneous coordinates: M = v 1x v 2x v 3x 0 v 1y v 2y v 3y 0 v 1z v 2z v 3z

20 Some coordinate systems...in the pipeline

21 The view transform

22 The view transform Put the observer at the origin. Align the x and y axes with the ones of the screen. Use a right hand coordinate system where the z- axis points backwards. This simplifies light, clip, HSR, and perspective calculations.

23 Change of frame The View transformation is a change of frame. A change of frame is a change of origin + a change of coordinate system. We can split the operation into a translation and a change of coordinate system.

24 Change of frame

25 Change of frame

26 Change of frame, final x y = M 1 T( P 0 ) x y Change of Coordinates Translation And, if M is orthonormal: x y = M T T( P 0 ) x y

27 The LookAt(eye, at, up) transform A convenient way of positioning the camera

28 Look-At Transform using GLM glm::lookat creates a viewing matrix derived from an eye point, a reference point indicating the center of the scene, and an UP vector. glm:: lookat(eye, at, up) eye, at, up are glm::vec3 datatypes. Builds the 4x4 transformation matrix for you, but now you know what goes on behind the scenes.

29 Projection

30 Projection and Perspective Duccio di Buoninsegna, The Last Supper, 1308

31 Projection and Perspective Leonardo da Vinci, the last Supper

32 Projection and Perspective Filippo Brunelleschi , Florence, Italy Credited with formulating a mathematical system for linear perspective which revolutionized painting and allowed for naturalistic styles to develop. Brunelleschi Albrecht Dürer ( ) Nürnberg,Germany Described both mathematical and mechanical methods for drawing perspective. Dürer

33 Projection and Perspective Drawing a lute. Woodcut by Albrecht Dürer, 1525

34 Projection p=(x,y,z) Viewplane Object

35 Projection p=(x,y,z) Centre of projection (COP) Viewplane, z=d Object

36 Orthographic projection COP at infinity Project points onto the view-plane along Direction of projection (DOP). Easy to implement in view-coordinates. Just set z=0. DOP p=(p x, p y, p z ) Viewplane, z=0 Object

37 Orthographic projection In homogenous coordinates: Projection Matrix

38 Orthographic projection

39 Orthographic projection

40 Perspective projection Image: globalcitizenblog.com

41 Perspective projection Center of projection (COP) p=(p x, p y, p z ) p'=(p x ', p y ', p z ') c=(c x, c y, c z ) Viewplane, z=d Object

42 Perspective projection Center of projection (COP) p=(p x, p y, p z ) p'=(p x ', p y ', p z ') c=(c x, c y, c z ) Viewplane, z=d Linear equation: (1) p'=tp+(1-t)c Object Find p by solving (1) and (2) (2) d=tp z +(1-t)c z

43 Perspective projection Center of projection (COP) p=(p x, p y, p z ) p'=(p' x, p' y, p' z ) c=(0, 0, 0) Viewplane, z=d Linear equation: (1) p'=tp+(1-t)c (2) d=tp z +(1-t)c z Object Find p by solving (1) and (2) We can remove these terms since c=(0, 0, 0) in viewing coordinates.

44 Perspective projection Linear equation: (1) p'=tp (2) d=tp z Solve (2) for t => t=d/p z We get: p'=(d/p z )p

45 Perspective projection p=(x, y, z) (Similar triangles) p'=(x, y, z ) y y y = z z = d z d Viewplane, z=d x COP z y = d z y Note that z < 0 and d < 0

46 Homogeneous coordinates Use W to fit the perspective transformation into a matrix multiplication. Divide by homogeneous coordinate to obtain final projected point (perspective divide). (! transformation (Note that this is not a linear

47 Perspective using GLM glm::mat4 glm::perspective(fovy, aspect, near, far) aspect = width/height Far plane Top plane Projection Viewport plane View frustum COP fovy Near plane Bottom plane

48 Perspective Image: globalcitizenblog.com

49 Orthographic vs. Perspective projection Image: db-in.com

50 Perspective projection Image:

51 Summary Change of coordinate system Change of frame (change of coordinate system + translation) The view transform is a change of frame Projections: Orthogonal and perspective

52 Assignment 2 4 Apr You will work with: Transformations: Translation Scaling Rotation Homogenuous coordinates Good luck!