CS 325 Introduction to Computer Graphics 03 / 08 / 2010 Instructor: Michael Eckmann.

CS 325Introduction to Computer Graphics

03 / 08 / 2010

Instructor: Michael Eckmann

Michael Eckmann - Skidmore College - CS 325 - Spring 2010

Today’s Topics• Questions/comments?

• 3d view volume clipping

• Visible Surface determination– back face culling– depth buffer (aka z-buffer)


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 NOT 000000.

• Calculate intersection with borders of clipping planes 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

• Calculating the intersections of lines with the unit slope planes

of the cvv is easy.

• For example, for the y = z plane, y0 + t (y

1 – y

0) = z

0 + t (z

1 –

z0) and we can solve for t. We'll then know the y and z

coordinates of the intersection (z = y), so just use t to find the x

coordinate.


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.)

• 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.


Visible Surface Determination• Types

• Image precisionfor (each pixel in display){

Find closest object that is pierced by the projector through that pixel

Draw the pixel at the appropriate color in frame buffer}

• Object precisionfor (each object in world){

Determine parts of object whose view is unobstructed by other parts of it or by other objects

Save the visible parts}Go through all the visible parts and display at proper

resolution in frame buffer• Some combination of image and object precision


Visible Surface Determination• The visible surface determination algorithms we'll discuss are:

• Back Face Culling (removal)• Depth Buffer method (aka z-buffer)

– A-Buffer method• Depth Sort• BSP Tree• Scanline Algorithm


Back Face Culling (removal)• This is a technique to remove unseen faces (polygons) from

convex, solid (completely closed) polyhedra

• If one polyhedron occludes another or if a polyhedron is

concave then back face culling isn't enough to remove all the

invisible surfaces.

• Example drawings on board.– A convex polyhedron– A concave polyhedron– A polyhedron that occludes another polyhedron


Back Face Culling (removal)• If we are working in a Right Handed Coordinate system and we

specify a polygon by listing its vertices in counter-clockwise order,

we can determine whether or not it is back facing or front facing by

doing the following:

• Compute the plane equation for a face with the surface normal

pointing out of the front. End up with Ax+By+Cz+D with specific

values for A B C and D.

• Recall that the plane equation will equal 0 if the x,y,z is on the plane,

non-zero otherwise. For points behind the plane, the plane equation

will be < 0. For points in front, plane equation is > 0.

• Let's see how the normal vector and the plane equation can be

calculated now.


Back Face Culling (removal)• Given counter-clockwise order of vertices allows us to compute the

normal to the plane defined by those vertices. The direction of the

normal is important --- we want it to point out of the front face.

• Example: Given a rectangle with vertices P1, P2, P3, P4 in

counterclockwise order to compute the normal to the plane that that

rectangle is on, in the direction pointing out of its front face we can

do the following:

• [P2 – P1] x [P4 – P1]

• Let's see a drawing on the board.

• Note well: To determine the direction of a vector when subtracting

two points (a-b) the direction is from b to a (not a to b).


Back Face Culling (removal)• So a general way to generate the normal vector with direction out of

the front face, given counter-clockwise order of vertices, the two

vectors V1 and V2 are:

• V1 = 2nd vertex minus the 1st vertex

• V2 = last vertex minus the 1st vertex

• Then take cross product: V1 x V2 = N

• N is a normal vector to the plane containing the vertices and is in the

direction pointing out of the front face.


Back Face Culling (removal)• Now that we have N, the next step is to find the plane equation. How

can we do that now?


Back Face Culling (removal)• Now that we have N, the next step is to find the plane equation. How

can we do that now?

• N = (A, B, C) so A, B and C are known.

• The plane equation is: Ax + By + Cz + D = 0.

• D is the only unknown. To find D, just plug in the coordinates of one

of the vertices (x1,y

1,z

1) and solve for D:

• D = - ( Ax1 + By

1 + C z

1)

• Now we have the equation of the plane on which the polygon lies and

the normal to that plane is pointing out of the front face.


Back Face Culling (removal)• How do we use this plane equation now to determine if the polygon

is front facing or backfacing (in relation to the PRP (or CoP))?


Back Face Culling (removal)• How do we use this plane equation now to determine if the polygon

is front facing or backfacing (in relation to the PRP (or CoP))?

• We plug in the PRP (Projection Reference Point) which is the Center

of Projection (CoP) when dealing with perspective projections to the

plane equation and if A*PRPx+B*PRP

y+C*PRP

z+D < 0 it is a

backface and we remove it from processing.


Back Face Culling (removal)• The text shows that we can simplify this idea by just taking the dot

product of a View vector (in direction from PRP) and the Surface

Normal to the plane containing the polygon and it's a backface if this

dot product is > 0. That is, a polygon is a backface if Vview

● N > 0.

• Why is that true?


Back Face Culling (removal)• A polygon is a backface if V

view ● N > 0.

• Why is that true?

• Recall: V1 . V2 = |V1| |V2|cosA

• The only thing affecting the sign of the dot product is the cosA (note:

the magnitudes are always positive)

• When A = [0, 90), then cosA > 0

• When A = (90, 180], then cosA < 0

• When A = 90, cosA = 0

• Let's see this on the drawing on the board.

Back Face Culling (removal)• Further simplification is when the View vector is parallel to (and in

the direction of) the -z axis. That is, the negative z axis, and pointing

away from the origin. THIS ONLY WORKS IF YOU

TRANSFORM THE PERSPECTIVE TO PARALLEL!!!

• This simplification can be used if after we transformed our arbitrary

view volume to the perspective canonical view volume• we then transform the perspective canonical view volume to a

parallel-projection canonical view volume. See handout (p275).

• If we do this, our projectors all have a direction of projection of

[0,0,-1]. And so, if the z coordinate of the plane normal is < 0, then it

is back facing. The z-coordinate of the plane normal in the next

diagram is named C.

CS 325 Introduction to Computer Graphics 03 / 08 / 2010 Instructor: Michael Eckmann.

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