1. Clipping Clipping Sutherland-Hodgman Clipping
Cohen-Sutherland Clipping
Liang-Barsky Clipping

2. The Story So Far for 3D Graphics
We now understand:
how to model objects as a set of polygons and create a 3D world
how to view the 3D world with a camera model, projecting the polygons to a 2D image plane
What should come next:
determine visible polygons in a scene (clipping and hidden surface removal)
rasterise objects – turn 2D objects represented by edge coordinates to pixels (filling polygons)
calculate lighting/texture values for pixels

3. Clipping
Clipping removes parts of the geometry that are outside the view/clipping region
Clipping edge/plane: an infinite line or plane and we want to output only the geometry on one side of it
Clip region
Result

4. Inside-Outside Testing
Suppose lines/planes have a normal vector pointing toward the outside of the clip region (or towards inside)
Dot products give inside/outside information
x is any point on the clip line/plane
Outside
Inside x f n i s

5. Clipping Points Against a View Volume
A point is inside the view volume if it is on the inside of all the clipping planes
Now we see why clipping is done in canonical view space. For instance, to check against the left plane: x coordinate must be > -1
The normals to the clip planes can be considered to point inward, toward the visible region (user defined)
In general, if a point p’(x’,y’,z’) is inside, then for a clipping plane nxx + nyy +nzz + d = 0, with (nx,ny,nz) pointing inward, nxx’ + nyy’ +nzz’ + d > 0

6. Remember the Plane Equation?
Given a point A on the plane, and the plane’s normal vector n, the plane equation can be obtained using the fact that every vector on the plane should be perpendicular to n so dot product is 0 P r
nxx+nyy+nzz+d=0
(d = -n.a), n= (nx,ny,nz) So

7. Sutherland-Hodgman p2 p3 p1 p4 p0 p0 p1 p2 p3 p4
Consider the polygon as a list of vertices
Clip the polygon against each edge of the clip region in turn
Rewrite the polygon one vertex at a time – the rewritten polygon will be the clipped polygon p2 p3 p1 p4 p0 p0 p1 p2 p3 p4
Clip Top
Vertices in
Clip Right
Clip Bottom
Clip Left
Clipped vertices out
Clip Far
Clip Near

8. Sutherland-Hodgman Inside Outside Inside Outside Inside Outside Insidep i p s p s i p s
Output p
Output i
No output
Output i and p
Case 1: Wholly inside visible region - save endpoint
Case 2: Leave (from inside to outside) visible region - save the intersection
Case 3: Wholly outside visible region - save nothing
Case 4: Enter (from outside to inside) visible region - save intersection and endpoint

9. Cohen-Sutherland Clip Rectangle
We extend the edges of the clip rectangle to divide the plane of the clip rectangle into nine regions
Each region is assigned a 4-bit code (outcode) determined by where the region lies with respect to the clip edges
Each bit in the outcode is set to either 1 (true) or 0 (false), depending on the following conditions:
Bit 1: above top edge Y > Ymax
Bit 2: below bottom edge Y < Ymin
Bit 3: right of right edge X > Xmax
Bit 4: left of left edge X < Xmin
1001
1000
1010
0001
0000
0010
0101
0100
0110
Clip Rectangle

10. Cohen-Sutherland
Use outcode to record end point in/out against each clipping line/plane
An outcode has 4 bits in 2D, 6 bits in 3D, one bit per clipping line
1st and 2nd bits for Y (>Ymax, <Ymin) clipping lines, 3rd and 4th bits for X (>Xmax,<Xmin) clipping lines, 5th and 6th for Z (>Zmin,<Zmax) clipping lines
Bit is 1 if point is outside the corresponding clipping line, 0 if inside

11. Cohen-Sutherland Say a=outcode (P1), b=outcode(P2)
If a=b=0 then both ends inside so line inside – trivial accept
If (a | b) = 0, then one inside one outside – inconclusive - compute intersection point and check outcode for the intersection point
If a & b != 0 (bitwise AND) then both ends on the same side of the window – trivial reject
If a & b = 0 both ends are outside, but on the outside of different edges of the window - inconclusive - compute intersection point and check outcode for the intersection point

12. Cohen-Sutherland E a=b=0 B D A (a | b) = 0 C a & b != 0 H a & b = 0 JF G H I J
a=b=0
(a | b) = 0
a & b != 0
a & b = 0
a & b = 0
End points pairs are check for trivial acceptance or trivial rejection
If not accepted or rejected, divided into two segments at a clip edge
Repeat the process on resulting line segments until
completely inside or rejected

13. Example P1 P1’ P2 Clip against left, right, bottom, top
boundaries in turn.
P1: 1001
P2: 0100
First clip to left
edge, giving
P1’P2
P1’: 1000
P1’ P2
x=xmin

14. Example P1’ P1’’ P2’ P2 x=xmin P1’: 1000 P2 : 0100 No need to clip
against right edge
Clip against
bottom gives P1’P2’
Clip against top
gives P1’’P2’
P1’
P1’’
P2’
P2’=0000, P1’=0000 P2
x=xmin

15. Calculating the Intersection
To calculate intersection of P1P2 with, say left edge:
Left edge: x = xmin
Line : y - y2 = m (x - x2)
where m = (y2 - y1) / (x2 - x1)
Thus intersection is (xmin, y*)
where y* = y2 + m (xmin - x2) P2 P1

16. Liang-Barsky Clipping
This uses the parametric equations for a line and solves four inequalities to find the range of the parameter for which the line is visible (within the viewport)
The parametric equation of the line segment gives x values and y values for every point in terms of a parameter t that ranges from 0 to 1. The equations are
We can see that when t = 0, the point computed is P(x1,y1); and when t = 1, the point computed is P(x2,y2)

17. Liang-Barsky Clipping
We want parameter values that are inside all the clipping planes
Last parameter value to enter is the start of the visible segment
First parameter value to leave is the end of the visible segment
If we leave some clip plane before we enter another, we cannot see any part of the line
Liang-Barsky is more efficient than Cohen-Sutherland - computing intersection vertices is most expensive part of clipping

18. Parametric Intersection
Find parametric intersections
tright
ttop
tbottom
tleft
tb<tl< tt<tr
tright
tleft
ttop
tb<tt< tl<tr
tbottom

19. Liang-Barsky Clipping
Set tmin=0 and tmax=1
If t< tmin or t > tmax then ignore it and go to the next edge. Otherwise classify the t value as entering or leaving value (using inner product to classify)
If t is entering value set tmin =t; if t is leaving value set tmax =t
If tmin < tmax then draw a line from
(x1 + Dx tmin, y1 + Dy tmin) to (x1 + Dx tmax, y1 + Dy tmax)
If the line crosses over the window, you will see the two points are intersections between the line and clipping edges

20. Entering or Leaving
We can classify if entering or leaving value by using inner product
Let P(x1,y1), Q(x2,y2) be the line and n be the normal vector (outward)
If the parameter t is entering
If the parameter t is leaving
Let P = (1,3), Q = (-4,2) while the edge equation is x+2y-4 = 0.
Determine if the vector from P to Q is entering or leaving the edge.
We can determine n from the equation ax + by + c = o, where n = (a,b).
Therefore, n = (1,2).
Q - P = (-4-1,2-3) = (-5,-1).
n. (Q - P) = (1,2) * (-5,-1) = -5-2 = -7 which is less than 0.
So, this line is entering the edge

21. T Values for Intersection with Clip Edges
T value intersection with left edge: x = L
With right edge: x = R
With top edge: y = T
With bottom edge: y = B
e.g.

22. Line and Clipping Edge Intersection
From the parametric equation and the t values, we can calculate the intersection between a line and an edge as follows
Intersection with left edge
Intersection with right edge
Intersection with top edge
Intersection with bottom edge

23. Example
Consider if t value is entering or exiting by using inner product.
(Q-P) = (15+5,9-3) = (20,6)
At left edge (Q-P).nL = (20,6)(-10,0) = -200 < 0 entering so we set tmin = 1/4
At right edge (Q-P)nR = (20,6)(10,0) = 200 > 0 exiting so we set tmax = 3/4
Because tmin < tmax then we draw a line from
(-5+(20)*(1/4), 3+(6)*(1/4)) to (-5+(20)*(3/4), 3+(6)*(3/4))

24. Example Consider if tvalue is entering
or leaving by using inner product.
(Q-P) = (2+8,14-2) = (10,12)
At top edge (Q-P).nT =
(10,12).(0,10) = 120 > 0 exiting
so we set tmax = 8/12
At left edge (Q-P).nL =
(10,12).(-10,0) = -100 < 0 entering
so we set tmin = 8/10
Because tmin > tmax then
we don't draw a line.