Lecture14b Clipping

7/31/2019 Lecture14b Clipping

1/16

Line Clipping

Point clipping easy: Just check the inequalities

xmin < x < xmax

ymin < y < ymax

Line clipping more tricky


2/16

Cohen-Sutherland Line Clipping

Divide 2D space into 3x3 regions.

Middle region is the clipping window.

Each region is assigned a 4-bit code. Bit 1 is set to 1 if the region is to the left of

the clipping window, 0 otherwise. Similarly

for bits 2, 3 and 4.

4 3 2 1

Top Bottom Right Left


3/16


0000

1001

0001

0101 0100 0110

0010

10101000


4/16


To clip a line, find out which regions itstwo endpoints lie in.

If they are both in region 0000, then its

completely in. If the two region numbers both have a 1 in

the same bit position, the line is

completely out. Otherwise, we have to do some more

calculations.


5/16


For those lines that we cannot immediatelydetermine, we successively clip against eachboundary.

Then check the new endpoint for its region code.

How to find boundary intersection: To find ycoordinate at vertical intersection, substitute xvalue at the boundary into the line equation ofthe line to be clipped.

Other algorithms: Faster: Cyrus-Beck Even faster: Liang-Barsky Even faster: Nichol-Lee-Nichol


6/16

Sutherland-Hodgman

Polygon Clipping Input each edge (vertex pair) successively.

Output is a new list of vertices.

Each edge goes through 4 clippers.

The rule for each edge for each clipper is: If first input vertex is outside, and second is inside,

output the intersection and the second vertex

If first both input vertices are inside, then just output

second vertex If first input vertex is inside, and second is outside,

output is the intersection

If both vertices are outside, output is nothing


7/16

Sutherland-Hodgman Polygon Clipping:

Four possible scenarios at each clipper

outside inside

v1v1

v2

outside inside

v1

v2

outside inside

v1

v1

v2

outside inside

v1

v2

Outside to inside:

Output: v1 and v2

Inside to inside:

Output: v2

Inside to outside:

Output: v1

Outside to outside:

Output: nothing


8/16

Sutherland-Hodgman Polygon Clipping

v2

v1

v3

Right

Clipper

Bottom

Clipper

Top

Clipper

Left

Clipper

v1v2

v2v3

v3v1

v1

v2

v2

v2

v3v1

v2v2

v2v3

v3v1

v1v2

v2

v3

v1

v2 v2v2

v2v3

v3v1

v1v2

v3

Figure 6-27, page 332

v2

v1v2

v2 v2v2

v2v1

v1v2

v2v2

v1

v2

v2

v2

Edges Output Edges Output Edges Output Edges Output

Final


9/16

Weiler-Atherton Polygon Clipping

Sutherland-Hodgman

Weiler-Atherton


10/16

Clipping in 3D

Suppose the view volume has been

normalized. Then the clipping boundaries

are just:

xwmin = -1

ywmin = -1

zwmin = -1

xwmax = 1

ywmax = 1

zwmax = 1


11/16

Clipping Homogeneous

Coordinates in 3D Coordinates expressed in homogeneous

coordinates After geometric, viewing and projection

transformations, each vertex is: (xh, yh, zh, h)

Therefore, assuming coordinates have beennormalized to a (-1,1) volume, a point (xh, yh, zh, h) isinside the view volume if:

-1 < < 1 and -1 < < 1 and -1 < < 1

xh

h

yh

h

zh

h

Suppose that h > 0, which is true in normal cases, then

-h < xh < h and -h < yh < h and -h < zh < h


12/16

Remember Cohen-Sutherland 2D

Line Clipping Region Codes?

Divide 2D space into 3x3 regions.

Middle region is the clipping window.

Each region is assigned a 4-bit code. Bit 1 is set to 1 if the region is to the left of

the clipping window, 0 otherwise. Similarly

for bits 2, 3 and 4.

4 3 2 1

Top Bottom Right Left


13/16


Region Codes in 2D

0000

1001

0001

0101 0100 0110

0010

10101000


14/16

3D Cohen-Sutherland Region

Codes

Simply use 6 bits instead of 4.

4 3 2 1

Far Near Top Bottom Right Left

6 5

Example: If h + xh < 0, then bit 1 is set to 1.

This is because if -h > xh, then the point is to the left of the viewing volume.


15/16

Clipping Polygons

First, perform trivial acceptance and

rejection using, for example, its coordinate

extents.

Polygons usually split into triangle strips.

Then each triangle is clipped using 3D

extension of the Sutherland-Hodgman

method


16/16

Arbitrary Clipping Planes

Can specify an arbitrary clipping plane: Ax + By

+ Cz + D = 0.

Therefore, for any point, if Ax + By + Cz + D < 0,

it is not shown. To clip a line against an arbitrary plane,

If both end-points are in, then the line is in

If both end-points are out, then the line is out

If one end-point is in, and one is out, then we need to

find the intersection of the line and the plane

Documents

Lecture14b Clipping