1. Clip against view volume Project onto projection plane
Viewing in 3D
Conceptual model of the 3D viewing process
2D device coordinates
3D world-coordinate output primitives
Clipped world coordinates
Clip against view volume
Project onto projection plane
Transform into viewport in 2D device coordinates for display
2-Dec-18
Computer Graphics : Viewing in 3D

2. (for computer graphics)
3D Coordinate Systems
y axis
y axis
z axis -1 +1
x axis
x axis +1 -1
z axis
Right-handed
(for mathematics)
Left-handed
(for computer graphics)
2-Dec-18
Computer Graphics : Viewing in 3D

3. 3D Primitives: Point & Line
Any point in 3D coordinate is defined by (x,y,z)
Any line in 3D coordinate is defined with 2 points:
(x1, y1, z1) and (x2, y2, z2)
Line equations: or
2-Dec-18
Computer Graphics : Viewing in 3D

4. 240-422 Computer Graphics : Viewing in 3D
3D Primitives: Planes
A plane is specified as
Ax + By + Cz + D = (A, B, C, D are constants)
or x + B1y + C1z + D1 = 0
(where B1 = B/A, C1 = C/A and D1 = D/A)
If we know 3 points (not all in a line), we can determine the plain equation from
x1 + B1y1 + C1z1 + D1 = 0
x2 + B1y2 + C1z2 + D1 = 0
x3 + B1y3 + C1z3 + D1 = 0
Solving for B1, C1, D1 yields the plain equation
2-Dec-18
Computer Graphics : Viewing in 3D

5. 3D Primitives: Planes (cont’d)
A plane can also be specified with a single point in the plane (xp, yp, zp) and the normal vector (direction perpendicular to the plane) [Nx, Ny, Nz]
[Nx, Ny, Nz]
(xp, yp, zp)
Plane equation:
Nx(x-xp) + Ny(y-yp) +Nz(z-zp) = 0
2-Dec-18
Computer Graphics : Viewing in 3D

6. Geometrical Projections
Parallel: determined by Direction of Projection (DOP) (projectors are parallel—do not converge to an “eye” or COP)
Perspective: determined by Center of Projection
(COP) (in our diagrams the “eye”)
2-Dec-18
Computer Graphics : Viewing in 3D

7. 240-422 Computer Graphics : Viewing in 3D
Parallel Projection
Suppose the direction of projection is given by the vector [xp, yp, zp] and the image is to be projected onto the xy plane
If we have a point on the object at (x1, y1, z1), we wish to determine where the projected point (x2, y2) will lie.
Writing the equations for a line passing the point (x,y,z) and in the direction of projection:
x = x1 + xpu
y = y1 + ypu
z = z1 + zpu
2-Dec-18
Computer Graphics : Viewing in 3D

8. 240-422 Computer Graphics : Viewing in 3D
Parallel Projection
Where does this line intersect the xy plane (z=0)?
Since u = -z1/zp
substituting this into the first two equations gives
x2 = x1 - z1 (xp/zp)
y2 = y1 - z1 (yp/zp)
This can be written in matrix form
2-Dec-18
Computer Graphics : Viewing in 3D

9. Perspective Projection
The further an object is from the viewer, the smaller it appears
If the center of projection is at (xc, yc, zc) and the point on the object is (x1, y1, z1) then the project ray will be the line containing these points and will be given by
x = xc + (x1-xc)u
y = yc + (y1-yc)u
z = zc + (z1-zc)u
2-Dec-18
Computer Graphics : Viewing in 3D

10. Perspective Projection
The projected point (x2, y2) will be the point where this line intersects the xy plane
For this intersection point (z=0), we can compute u
u = -zc / (z1-zc)
Substituting into the first two equations gives or
2-Dec-18
Computer Graphics : Viewing in 3D

11. Perspective Projection
In matrix form
2-Dec-18
Computer Graphics : Viewing in 3D