it is about 3D transformation-Rotation also explained numerical it is part of Computer Graphics Unit-3

1. 3D TRANSFORMATION
(Rotation)
By:
Arvind Kumar
Assistant Professor
(Vidya College of Engineering)

2. 3D Rotation
By: Arvind Kumar
3D
About the Coordinate-Axes
–X-axis rotation
–Y-axis rotation
–Z-axis rotation
Composite 3D Rotations
–Rotation about an axis that is parallel to one
of the coordinate axes
–Rotation about an arbitrary axis
3D

3. 3D Rotation Coordinate axis
By: Arvind Kumar
3D
About X-Axis
3D






































11000
0cossin0
0sincos0
0001
1
'
'
'
z
y
x
z
y
x


z
y
x

4. 3D Rotation Coordinate axis
By: Arvind Kumar
3D
About Y-Axis
3D
z
y
x






































11000
0cos0sin
0010
0sin0cos
1
'
'
'
z
y
x
z
y
x



5. 3D Rotation Coordinate axis
By: Arvind Kumar
3D
About Z-Axis
3D
z
y
x























 













11000
0100
00cossin
00sincos
1
'
'
'
z
y
x
z
y
x



6. 3D Rotation about an parallel axis
By: Arvind Kumar
3D
Steps to be follows:
–Translate the object so that the rotation axis
coincides with the parallel coordinate axis
–Rotation about that axis.
–Translate Back so that the rotation axis is moved
back to its original position
x

7. Rotation about an arbitrary Axis
By: Arvind Kumar
3D
Steps to be Follows
1. Translate (x1, y1, z1) to the
origin
2. Rotate (x’2, y’2, z’2) on to the z
axis
3. Rotate the object around the z-
axis
4. Rotate the axis to the original
orientation
5. Translate the rotation axis to
the original position
Rxy-1
T-1
Rxy
T
           TRRRRRTR  xyzyx
111 

Rz
(x2,y2,z2)
(x1,y1,z1)
x
z
y

8. Rotation about an arbitrary Axis
By: Arvind Kumar
3D
Step 1. Translation
















1000
100
010
001
1
1
1
tz
ty
tx
T
(x2,y2,z2)
(x1,y1,z1)
x
z
y

9. Rotation about an arbitrary Axis
By: Arvind Kumar
3D
Step 2. Rotation about x-axis
 




























1000
0//0
0//0
0001
1000
0cossin0
0sincos0
0001
dcdb
dbdc
x


R
(0,b,c)

d
c
cb
c
d
b
cb
b






22
22
cos
sin
y
z
(a,b,c)
x

10. Rotation about an arbitrary Axis
By: Arvind Kumar
3D
• Step 3. Rotation about y axis
22
222222
cos,sin
cbd
dacbal
l
d
l
a


 
 











 












 

1000
0/0/
0010
0/0/
1000
0cos0sin
0010
0sin0cos
ldla
lald
y


R
(a,b,c)

l
d
x
y
z
(a,0,d)

11. Rotation about an arbitrary Axis
By: Arvind Kumar
3D
Step 4. Rotatation about z axis
 











 

1000
0100
00cossin
00sincos


zR

l
y
x
z

12. Rotation about an arbitrary Axis
By: Arvind Kumar
3D
Step 5. Apply the reverse transformation to
place the axis back in its initial position
   










































1000
0cos0sin
0010
0sin0cos
1000
0cossin0
0sincos0
0001
1000
100
010
001
1
1
1
111





z
y
x
yx RRT
           TRRRRRTR  xyzyx
111 

y
l
l
z
x

13. Numerical
By: Arvind Kumar
3D
Q1. Find the new coordinates of a unit cube 90ᵒ rotated about
an axis defined by its endpoints A(2,1,0) and B(3,3,1).

14. Solution:
By: Arvind Kumar
3D
Step1. Translate point A to the origin
x
z
y
B’(1,2,1)















1000
0100
1010
2001
T
A’(0,0,0)

15. Solution:
By: Arvind Kumar
3D
Step2. Rotate axis A’B’ about the x axis by and
angle , until it lies on the xz plane
 


















1000
0
5
5
5
52
0
0
5
52
5
5
0
0001
xR
6121
5
5
5
1
cos
5
52
5
2
12
2
sin
222
22





l
x
z
y
l
B’(1,2,1)

Projected point
(0,2,1)
B”(1,0,5)
A’

16. Solution:
By: Arvind Kumar
3D
Step3. Rotate axis A’B’’ about the y axis, until it
coincides with the z axis.
 


















1000
0
6
30
0
6
6
0010
0
6
6
0
6
30
yR
6
30
6
5
cos
6
6
6
1
sin




x
z
y
l

B”(1,0,  5)
A’(0,0,5)

17. Solution:
By: Arvind Kumar
3D
Step4. Rotate the cube 90° about the z axis
 











 

1000
0100
0001
0010
90zR
           TRRRRRTR  xyzyx  
90111
Finally, Matrix about the arbitrary axis

18. Solution:
By: Arvind Kumar
3D
 











































































 














































1000
560.0167.0741.0650.0
151.1075.0667.0742.0
742.1983.0075.0166.0
1000
0100
1010
2001
1000
0
5
5
5
52
0
0
5
52
5
5
0
0001
1000
0
6
30
0
6
6
0010
0
6
6
0
6
30
1000
0100
0001
0010
1000
0
6
30
0
6
6
0010
0
6
6
0
6
30
1000
0
5
5
5
52
0
0
5
52
5
5
0
0001
1000
0100
1010
2001
R

19. Solution:
By: Arvind Kumar
3D
Step5. Multiplying R(θ) by the point matrix of the
original cube
     PRP  
 











































11111111
076.0091.0560.0726.0817.0650.0301.1467.1
483.0409.0151.1225.1184.0258.0484.0558.0
891.2909.1742.1725.2816.2834.1667.1650.2
11111111
10011001
00001111
11001100
1000
560.0167.0741.0650.0
151.1075.0667.0742.0
742.1983.0075.0166.0
P