CS 325Introduction to Computer Graphics
02 / 26 / 2010
Instructor: Michael Eckmann
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Today’s Topics• Questions/comments?
• View volumes and specifying arbitrary views
• Arbitrary view examples
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 planea View Plane Normal (VPN) which is the normal vector to the planea 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.
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
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
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).
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 origin2. 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-axis5. Scale so that the view volume becomes the canonical view volume
Take a look at the pictures
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Translate VRP to origin1. T(-VRP) =
[ 1 0 0 -VRPx ]
[ 0 1 0 -VRPy ]
[ 0 0 1 -VRPz ]
[ 0 0 0 1 ]
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Rotate VRC2. Rotate the VRC in the following way:
we want u to go to (1, 0, 0) in x,y,z coordinateswe want v to go to (0, 1, 0) in x,y,z coordinateswe 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
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Rotate VRC2.
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 =
[ ux u
y u
z 0 ]
[ vx v
y v
z 0 ]
[ nx n
y n
z 0 ]
[ 0 0 0 1 ]
example: This matrix transforms the v vector to (0, 1, 0)
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Rotate VRC2. To check, show this matrix transforms the v vector to (0, 1, 0)
[ ux u
y u
z 0 ] [ v
x ] [ u . v ] [ 0 ]
[ vx v
y v
z 0 ] [ v
y ] = [ v . v ] = [ 1 ]
[ nx 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
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Translate PRP to origin3. T(-PRP) =
[ 1 0 0 -PRPu ]
[ 0 1 0 -PRPv ]
[ 0 0 1 -PRPn ]
[ 0 0 0 1 ]
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Shear4. 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].
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Shear4.
[ ( umin
+ umax
) /2 ] [ PRPu ]
CW = [ ( vmin
+ vmax
) /2 ] PRP = [ PRPv ]
[ 0 ] [ PRPn ]
[ ( umin
+ umax
) /2 – PRPu ]
CW – PRP = [ ( vmin
+ vmax
) /2 – PRPv ]
[ 0 – PRPn ]
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Shear4. SH
per =
[ 1 0 SHx 0 ]
[ 0 1 SHy 0 ]
[ 0 0 1 0 ][ 0 0 0 1 ]
[ 1 0 SHx 0 ] [ ( u
min + u
max) /2 – PRP
u ] [ 0 ]
[ 0 1 SHy 0 ] [ ( v
min + v
max) /2 – PRP
v ] = [ 0 ]
[ 0 0 1 0 ] [ 0 – PRPn ] [ DoP
z ]
[ 0 0 0 1 ] [ 1 ] [ 1 ]
So, DoPz = – PRP
n
Solve for SHx and SH
y and get
SHx = ( ( u
min + u
max) /2 – PRP
u ) / PRP
n
SHy = ( ( v
min + v
max) /2 – PRP
v ) / PRP
n
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Scale5. A few notes about the diagram that shows the scaling
There is a mistake where y= (vmax
– vmin
)/2 and y= -(vmax
– vmin
)/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
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Scale5. 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 sy1
0 0 ][ 0 0 1 0 ][ 0 0 0 1 ]
From the diagram (a) y= -(vmax
– vmin
)/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 -PRPn . So, we need to figure out what scale
factor will make -(vmax
– vmin
)/2 equal to -PRPn .
sy1
= 2 PRPn / (v
max – v
min )
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Scale5. (see diagram)
This is similar in the x direction, but the viewing window range in x direction
is (umax
to umin
) so, to scale in x and y so that the 4 sloped planes are unit
slope, we set the scales to be:
sx1
= 2 PRPn / (u
max – u
min )
sy1
= 2 PRPn / (v
max – v
min )
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Scale5. (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
[ sx2
0 0 0 ]
[ 0 sy2
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 -PRPn + B be -1.
sx2
= -1 / (-PRPn + B)
sy2
= -1 / (-PRPn + B)
sz2
= -1 / (-PRPn + B)
where B is the distance to the back clipping plane from -PRPn
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Scale5. final scale matrix S
per =
[ sx2
0 0 0 ] [ sx1
0 0 0 ]
[ 0 sy2
0 0 ] [ 0 sy1
0 0 ][ 0 0 s
z2 0 ] [ 0 0 1 0 ]
[ 0 0 0 1 ] [ 0 0 0 1 ]
Michael Eckmann - Skidmore College - CS 325 - Spring 2010
Perspective NormalizationComposite matrix transformation to do Normalization of arbitrary perspective
projection view volume to canonical view volume
Nper
= Sper
SHper
T(-PRP) R T(-VRP)