Homogeneous Transformation
Ref: Richard Paul Chap. 1
1
Notation
Vector: v
Plane: P
Discussion is in 3-space
Frame: I, A
Point in space: p
Point p as a vector v in frame E: Ev
Same point as a vector w in frame H: Hw
2
Vectors
Homogeneous coordinate w
Inner (dot) product
Outer (cross) product
[a,b,c,0]T: point at infinity
3
Planes
Compared with ax+by+cz+d = 0 …
Point v on a plane:
[x/w,y/w,z/w]
[a/m,b/m,c/m]
-d/m
4
Transformation
v = Hu
With homogeneous coordinates, translate and
rotation become linear transformations in R4
5
Plane Equation after Transformation
Given the transform:
Q (the plane P after transformation H):
Proof:
We require:
(=0)
6
Example
The plane P defined by these points:
(0,0,2), (1,0,2), (0,1,2) is [0,0,1,-2]
The transform H:
Transformed points are
(6,-3,7), (6,-2,7), (6,-3,8)
The plane after transformation:
0
0
1
Q  PH  0 0 1  2
1

0
1
0
0
0
0 3
1  7
 1 0 0  6
0  4

0 1
How to compute H-1 (see next page) …
7
Inverse Transformation
Assumption: [n o a] is orthogonal
Verify:
n  n n  o n  a n  p  p  n 
o  n o  o o  a o  p  p  o 
I
T 1T  
a  n a  o a  a a  p  p  a 


0
0
1
 0

8
Recall Normal Matrix
In OpenGL, normal vectors are
transformed by normal matrix into eye
space
Normal matrix is the inverse transpose
of modelview matrix (M-T)
ne  M T no  neT  noT M 1
Normal vector and plane equation are
related!
9
Point rotation is closely
related to coordinate
transformation
(next page)
Rotating a Point
x  2uˆ1  uˆ 2
uˆi  Ruˆi , i  1,2
x
A
B
x’
x  2uˆ1  uˆ 2
 2 Ruˆ1  Ruˆ 2
(same coordinates
in new bases)
 R 2uˆ1  uˆ 2   Rx
cos(90)  sin(90)  0 1
R



sin(

90
)
cos(

90
)

1
0

 

 0 1 2   1 
    
x  

  1 0  1    2 
10
Coordinate Transformation
 2  B R : Rotation that takes
x    A frame B to frame A
1
cos(90)  sin(90)  0 1
B

AR  


sin(

90
)
cos(

90
)

1
0

 

A
B
B
A
 0 1 2   1 
    
x R x  

  1 0  1    2 
B
A
A
x
11
Ex: Coordinate Transform
x
1
0
A
x  1, B x   1
1
 1 
B
A
A
x ABT Bx
A
B
T : T takingframeA toframeB
1  0 1 2  0 
1   1 0 1   1
 
 
1  0 0 1   1 
T  Transl(2,1) Rot(90)
A
B
 0 1 2 1 0 2   0 1 0
  1 0 1   0 1 1    1 0 0 

 


 0 0 1 0 0 1  0 0 1
12
Ex: Coordinate Transform
x
A
B
 2
 2
A
x  1 , B x   0 
1 
 1 
B
AT  Transl( 1,2) Rot(90)
1 0  1  0
0 1  2 1


0 0 1  0
  2  0  1
 0   1 0
  
 1  0 0
B
B
x 
AT
 1 0  0  1  1 
0 0  1 0  2
0 1 0 0
1 
 1   2
 2 1 
1  1 
A
x
13
Coordinate Transform
glTranslatef (2,1,0);
glRotatef (30,0,0,1);
drawtank();
• Use the transformation of the tank
(and its local coordinates) to find the
world coordinates of specific points.
GiveAx, Find W x
xWAT Ax
W
W
A
T : T takingframeW toframeA
A
(3,0)
W
1
0
W

AT 
0

0
0
1
0
0
0
0
1
0
2 cos30  sin 30
1  sin 30 cos30
0  0
0

1  0
0
0
0
1
0
0 cos30  sin 30
0  sin 30 cos30

0  0
0
 
1  0
0
0
0
1
0
tank
• Implemented by SVL (ex: tip of tank)
Vec3 X = proj (HTrans4(vec3(2,1,0))*HRot4(Vec3(0,0,1),30*3.14/180)*vec4(3,0,0,1);
14
2
1
0

1
Extra
15
Relative Transform & Frames
Rot(z,90)
Rot(y,90)
Trans(4,-3,7)
16
Reference Frame
17
Reference Frame (cont)
The transformed vector is the same vector
described w.r.t. the reference frame
18
Transform Equation
omit
superscript
s
19
Transform Equation
OA AB = OB
The Problem
cos30  sin 30 4  23

O
A   sin 30 cos30 6   12
 0
0
1  0
 23

O 1
A   12
0

A
0
3
2
0


 6 12 

 6 23 

1

3
2
1
2
cos(30)  sin(30) 12  23

O
B   sin(30) cos(30) 2    12

0
0
1   0
AB
A
O

 4
 4
1
2
3
2
4

6
1 
 12
B
B A
 23
 1
 2
0

O


0
 34  14

  23
 0

3
2
3
4
 14
0
0
12

2
1 
1 O
B


 4 23  6 12   23

 4 21  6 23   12
 0
1

1
2
3
2
1
2
3
2
1
2
3
2
0
12

2
1 
6 3  1  2 3  3   12
 
 6  3  2  3 3    23
 0
1
 
3
2
1
2
0
20
4 32 

 4  2 3

1
