1. CS 325 Introduction to Computer Graphics 02 / 26 / 2010 Instructor: Michael Eckmann

2. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Today’s Topics Questions/comments? View volumes and specifying arbitrary views Arbitrary view examples

3. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Specifying an arbitrary view To specify an arbitrary view, we should be able to place the view plane anywhere in 3d. –we'll need to specify the direction of the plane and where it lives within the world reference coordinate system (WRC)‏ –we'll also need to know which direction is up (to know what is displayed at the top of the image when we transform to viewport. A common way to specify an arbitrary view is to specify the following: a View Reference Point (VRP) which is a point on the plane a View Plane Normal (VPN) which is the normal vector to the plane a View Up Vector (VUP) which is a vector from which we determine which way is up See diagram. Is that enough info to specify an arbitrary view in your opinion? One note, the VUP vector is allowed to be specified as not perpendicular to VPN. The up direction (determined by a relation of the directions of VPN and VUP) though is perpendicular to VPN.

4. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Specifying an arbitrary view The VRP, VPN and VUP create another reference coordinate system. We call this the view reference coordinate system (VRC). We name the principle axes u, v and n. Within VRC we –specify a window on the view plane with Center of Window (CW) and min and max u and v values –a Projection Reference Point (PRP) which is the CoP for perspective views –Front and Back clipping planes specified as distances F and B, from VRP along the VPN –see next diagram. In the World Reference Coordinate (WRC) system we –define VRP, VPN and VUP In the View Reference Coordinate (VRC) system we –define CW, PRP, F and B

5. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 View Volumes To be able to clip against the canonical view volume and still allow any desired arbitrary view volume we'll need to normalize the desired view volume to the canonical view volume. So, the procedure to project from 3d to 2d given a finite view volume, will be as follows: –apply a normalizing transform to get to the canonical view volume –clip against the canonical view volume –project onto the view plane –transform into viewport

6. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Normalizing to CVV Now we're ready to develop the normalizing transformation for perspective projections. This will transform world coordinate positions so that the view volume is transformed into the canonical view volume. After this transform is applied, we would clip against the CVV and then project onto the view plane (via a perspective projection matrix).

7. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Normalizing to CVV The steps to do this are as follows: –Given the following: VRP, VPN, VUP, PRP, u and v min and max, F and B 1. Translate VRP (view reference point) to origin 2. Rotate the VRC (view reference coordinate system) so that VPN (n-axis) lies on the z-axis, the u-axis lies on the x-axis and the v-axis lies on the y-axis 3. Translate PRP (the Projection Reference Point which is CoP) to the origin 4. Shear so the center line of the view volume lies on the z-axis 5. Scale so that the view volume becomes the canonical view volume Take a look at the pictures

8. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Translate VRP to origin 1. T(-VRP) = [ 1 0 0 -VRP x ] [ 0 1 0 -VRP y ] [ 0 0 1 -VRP z ] [ 0 0 0 1 ]

9. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Rotate VRC 2. Rotate the VRC in the following way: we want u to go to (1, 0, 0)‏ in x,y,z coordinates we want v to go to (0, 1, 0)‏ in x,y,z coordinates we want n to go to (0, 0, 1)‏ in x,y,z coordinates make them have the correct directions and magnitude 1 n = VPN / | VPN | u = (VUP x n) / | VUP x n | v = n x u

10. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Rotate VRC 2. we want u to go to (1, 0, 0)‏ we want v to go to (0, 1, 0)‏ we want n to go to (0, 0, 1)‏ R = [ u x u y u z 0 ] [ v x v y v z 0 ] [ n x n y n z 0 ] [ 0 0 0 1 ] example: This matrix transforms the v vector to (0, 1, 0)‏

11. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Rotate VRC 2. To check, show this matrix transforms the v vector to (0, 1, 0)‏ [ u x u y u z 0 ] [ v x ] [ u. v ] [ 0 ] [ v x v y v z 0 ] [ v y ] = [ v. v ] = [ 1 ] [ n x n y n z 0 ] [ v z ] [ n. v ] [ 0 ] [ 0 0 0 1 ] [ 1 ] [ 1 ] [ 1 ] dot product of perpendicular vectors is 0 (cos 90 = 0)‏ and dot product of a vector with itself is its magnitude squared 1*1 = 1

12. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Translate PRP to origin 3. T(-PRP) = [ 1 0 0 -PRP u ] [ 0 1 0 -PRP v ] [ 0 0 1 -PRP n ] [ 0 0 0 1 ]

13. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Shear 4. Now we want to shear so the center line is on z-axis. (To see why we don't simply want to rotate look at the diagram on the handout to see the cross- section of the view volume after the first 3 steps are performed.)‏ Notice the CW is on that line and so is the origin (which PRP got translated to.)‏ So, to get that center line on the z-axis, we want the direction of the vector CW – PRP to be in the (DoP) direction of projection [0,0,z].

14. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Shear 4. [ ( u min + u max ) /2 ] [ PRP u ] CW = [ ( v min + v max ) /2 ] PRP = [ PRP v ] [ 0 ] [ PRP n ] [ ( u min + u max ) /2 – PRP u ] CW – PRP = [ ( v min + v max ) /2 – PRP v ] [ 0 – PRP n ]

15. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Shear 4. SH per = [ 1 0 SH x 0 ] [ 0 1 SH y 0 ] [ 0 0 1 0 ] [ 0 0 0 1 ] [ 1 0 SH x 0 ] [ ( u min + u max ) /2 – PRP u ] [ 0 ] [ 0 1 SH y 0 ] [ ( v min + v max ) /2 – PRP v ] = [ 0 ] [ 0 0 1 0 ] [ 0 – PRP n ] [ DoP z ] [ 0 0 0 1 ] [ 1 ] [ 1 ] So, DoP z = – PRP n Solve for SH x and SH y and get SH x = ( ( u min + u max ) /2 – PRP u ) / PRP n SH y = ( ( v min + v max ) /2 – PRP v ) / PRP n

16. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Scale 5. A few notes about the diagram that shows the scaling There is a mistake where y= (v max – v min )/2 and y= -(v max – v min )/2 are pointing to the top of the back clipping plane. Instead they should be pointing to the top of the viewing window which is the middle vertical line. Second, the diagram shows a value vrp' z which is equal to -PRP n

17. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Scale 5. scaling done in 2 steps first scale in x and y (to make the sloped planes be unit slopes)‏ second scale uniformly (in x,y,z) so that back clipping plane is at z = -1, and the unit slopes remain unit slopes To scale in x and y we have a matrix of the form: [ s x1 0 0 0 ] [ 0 s y1 0 0 ] [ 0 0 1 0 ] [ 0 0 0 1 ] From the diagram (a) y= -(v max – v min )/2 is the y value of the bottom of the window. We want that bottom side of the view volume to lie on the y=z plane which is a unit slope, so we want y= -(v max – v min )/2 = z. z for the viewing window is -PRP n. So, we need to figure out what scale factor will make -(v max – v min )/2 equal to -PRP n. s y1 = 2 PRP n / (v max – v min )‏

18. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Scale 5. (see diagram)‏ This is similar in the x direction, but the viewing window range in x direction is (u max to u min ) so, to scale in x and y so that the 4 sloped planes are unit slope, we set the scales to be: s x1 = 2 PRP n / (u max – u min )‏ s y1 = 2 PRP n / (v max – v min )‏

19. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Scale 5. (see diagram)‏ to then scale so that back clipping plane is at z = -1 (do the scaling uniformly (in x y and z) so that the 4 sloped planes remain unit slope.) We have a matrix of the form [ s x2 0 0 0 ] [ 0 s y2 0 0 ] [ 0 0 s z2 0 ] [ 0 0 0 1 ] we want the z = -PRP n + B plane to be the z = -1 plane. So, we need to figure out what scale factor will make -PRP n + B be -1. s x2 = -1 / (-PRP n + B)‏ s y2 = -1 / (-PRP n + B)‏ s z2 = -1 / (-PRP n + B)‏ where B is the distance to the back clipping plane from -PRP n

20. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Scale 5. final scale matrix S per = [ s x2 0 0 0 ] [ s x1 0 0 0 ] [ 0 s y2 0 0 ] [ 0 s y1 0 0 ] [ 0 0 s z2 0 ] [ 0 0 1 0 ] [ 0 0 0 1 ]

21. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Perspective Normalization Composite matrix transformation to do Normalization of arbitrary perspective projection view volume to canonical view volume N per = S per SH per T(-PRP) R T(-VRP)‏