CS 376 Introduction to Computer Graphics 02 / 26 / 2007 Instructor: Michael Eckmann

CS 376Introduction to Computer Graphics

02 / 26 / 2007

Instructor: Michael Eckmann

Michael Eckmann - Skidmore College - CS 376 - Spring 2007

Today’s Topics• Questions?

• Specifying an arbitrary perspective view

• Developing the composite transform for an arbitrary perspective view

• Arbitrary view examples

• 3D view volume clipping


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.


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


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


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).


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


Translate VRP to origin1. T(-VRP) =

[ 1 0 0 -VRPx ]

[ 0 1 0 -VRPy ]

[ 0 0 1 -VRPz ]

[ 0 0 0 1 ]


Rotate VRC2. Rotate the VRC in the following way:

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)

make them have the correct directions and magnitude 1

n = VPN / | VPN |

u = (VUP x n) / | VUP x n |

v = n x u


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)


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


Translate PRP to origin3. T(-PRP) =

[ 1 0 0 -PRPu ]

[ 0 1 0 -PRPv ]

[ 0 0 1 -PRPn ]

[ 0 0 0 1 ]


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].


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 ]


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


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


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 )


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 )


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


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 ]


Perspective NormalizationComposit matrix transformation to do Normalization of arbitrary perspective

projection view volume to canonical view volume

Nper

= Sper

SHper

T(-PRP) R T(-VRP)


Arbitrary 3d View examples• Let's take a look at a couple of examples in the handout.

• 1st let's reacquaint ourselves with the world coordinates of

the object (house)

• 2nd look at figure 6.29 and then its results in 6.28

• 3rd look at figure 6.30 which gives the same result with

different parameters (change VRP, causes a required

change in PRP and the Window Coordinates too to give

the same result.)• Because VRP changes, the origin of the u-v-n VRC

system changes. Therefore, since PRP and CW and the u

min, u

max, v

min and v

max are all within VRC, they need to

change if we want to have the same view.


Arbitrary 3d View examples• Let's look at the two point perspective view given in 6.22

• We need to orient the view plane so that it cuts both the x and z

coordinate axes. See figure 6.32 and that VPN is (1,0,1) which

is a vector from the origin on the x-z plane at 45 degrees

between the positive x and positive z axes. (I'll draw on board.)

• So, that VPN makes the view plane oriented such that it cuts

both x and z axes at the same angle. What can you say about

the view plane in relation to y?

• The parameters as stated in the handout have the VRP at the

bottom right corner of object, so that makes the view plane

oriented via the VPN and positioned so that VRP is on it.


Arbitrary 3d View examples• Also, VRP being where it is has an effect on where you place

PRP and the window parameters because these are in the VRC

system of which VRP (in WCS) is at the origin (in VRC).


Arbitrary 3d View examples• In the handout description of the viewing parameters for figure

6.34 it says all parameters are the same except view up vector

is 10 degrees away from y axis.

• Notice the house is tilted slightly to the right (compared to the

figure 6.22) so the up direction is tilted towards the left.

• To figure out a vector that's 10 degrees away from (0,1,0) we

can keep 0 and 1 constant and solve for z of the new vector.


Arbitrary 3d View examples• So, the dot product of v1=(0,1,0) with v2=(0,1,z) is 1.

• Then use other equation for dot product: |v1||v2|cosA = 1

• |v1| = 1

• |v2| = sqrt(1+z*z)

• So, sqrt(1+z*z) cos A = 1

• A=10 degrees. so, cosA is about .985

• So, sqrt(1+z*z) = 1/.985, then square both sides to get

• 1+z*z = 1.0307, z*z = 0.0307, then sqrt of both to get • z = 0.175

• A Vup vector of (0, 1, 0.175) could have approximately that

effect.


View volume


3d Clipping• The view volume on the last slide has 6 faces

• For a canonical view volume the faces are the on the

following planes:

• x = z, x = -z, y = z, y = -z, z = -zmin

, z = -1

• the 2 vertical faces (the front and back clipping planes) are on z = -z

min, z = -1

• the two side faces are on x = z, x = -z• the top and bottom faces are on y = z, y = -z

• the view volume lives totally in -z, so which plane is the top face on? y = z or y = -z


3d Clipping• Cohen Sutherland's extension into 3d from 2d.

• A six bit (as opposed to 4 in 2d) outcode is used here.• 1 = true, 0 = false

– Bit 1 = above view volume (y > -z)– Bit 2 = below view volume (y < z)– Bit 3 = right of view volume (x > -z)– Bit 4 = left of view volume (x < z)– Bit 5 = behind view volume (z < -1)– Bit 6 = in front of view volume (z > z

min)

This leads to 27 different outcode volumes

See drawing on board.


3d Clipping• Trivially accept if both endpoints have outcodes of

000000.

• Trivially reject if logical AND of the outcodes of the

endpoints is 000000.

• When would the logical AND of the outcodes be 000000?

• Calculate intersection with borders in order if can't

trivially accept or reject.


3d Clipping• the parametric equation of a line is:

• x = x0 + t (x

1 – x

0)

• y = y0 + t (y

1 – y

0)

• z = z0 + t (z

1 – z

0), 0 <= t <= 1

• To calculate the intersections of lines with the unit slope planes

of the cvv, is easy.

• For y = z, y0 + t (y

1 – y

0) = z

0 + t (z

1 – z

0) and we can solve

for t. Then use t to find the x and y coordinate and z = y so we

already know z.


3d Clipping• Then, once we have the x y and z coordinates of the

intersection we can determine if• the intersection is actually on a face of the view volume• or the intersection is not on a face of the view volume

– if the intersection is not on a face, then we exclude the portion of the line outside that plane (away from the view volume) and recalculate the region code of the intersection

– if the intersection is on a face, we keep that intersection as a final endpoint and exclude the portion of the line outside that plane, recalculate region code too.

• Another nice feature of this algorithm is that when given two

endpoints of the line segment and their region codes, even if we

can't trivially accept or reject them, we can tell which planes

they intersect (when corresponding bits are different.)

Therefore, we only have to clip that line against those planes.


3d Clipping• Calculating the intersections at unit slopes is easier than at

arbitrary slopes, hence the decision to normalize to a canonical

view volume.• x = x

0 + t (x

1 – x

0)

• y = y0 + t (y

1 – y

0)

• z = z0 + t (z

1 – z

0), 0 <= t <= 1

• Recall an arbitrary plane is Ax + By + Cz + D = 0 (the equation

of a plane.)

• Now, it should be obvious that more calculations are involved

in finding the intersection of a line with an arbitrary plane than

with simple planes like y = z, z = -1, etc.

Documents

CS 376 Introduction to Computer Graphics 02 / 26 / 2007 Instructor: Michael Eckmann