1. 2D Geometric Transformations
Graphics Systems /
Computer Graphics and Interfaces
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

2. 2D Geometric Transformations
The geometric transformations in computer graphics are essential to position, change the orientation and scale objects in the scene created. The movement is also implemented by the processing parameters vary over time.
Transformations:
Translation
Scaling
Rotation
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

3. Translation xT= X + Tx yT= Y + Ty
Vertices: (4,5) and (7,5)
TxT = 3y= -4
The pair of translation is called for translation vector. Each vertex is assigned a displacement T:
In the form of matrix product:
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

4. Scaling xT= X * Sx yT* Y = Sy Regarding the origin. Sx = 2 Sy = 2
100
xT= X * Sx
yT* Y = Sy
Regarding the origin.
120 40 80
150 20 40 60 80 40
120
200
Sx = 2
Sy = 2
Sx = 2
Sy = -1.5
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

5. Scaling In matrix form: Scale Factor: > Increase the subject 1
<1 reduces the object
sx= Sy Uniform scaling factor  does not distort the object
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

6. Rotation x = R.cos (a) y = R.sen (a) Around the origin.
100
xRYR
x = R.cos (a) R b
x, y
y = R.sen (a)
Around the origin.
the
xRR.cos = (a + b) = R.cos (a). Cos (b) - R.sen (a). Sin (b) = x.cos (b) - y.sen (b)
yRR.sen = (a + b) = R. Sin (b).cos (a) + R.sen (a). Cos (b) = x.sen (b) + y.cos (b)
Obtains a new position on the previous and the relative angle of rotation.
Rotation of -45
Vertices: (20.0) (60.0) (40,100)
Vertices: (14:14, ), (42.43, ), (98.99, 42.43)
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

7. Rotation + In matrix form:
Note: b positive movement in the opposite clockwise direction. +
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

8. Composition of TransformationsTc
The application of a sequence of operations =
1 single transformation
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

9. Composition of Transformations
Initial situation
(20.0)
(60.0)
(40,100)
After rotation
(0, -20)
(0, -60)
(100, -40)
a = - 90 °
After translation
(-80, -20)
(-80, -60)
(20, -40)
Tx = -80, Ty = 0
-80
0.0
100 20 60
Replacing the transformations:
After translation
(-60.0)
(-20.0)
( )
Tx = -80, Ty = 0
After rotation
(0.60)
(0.20)
(100.40)
a = - 90 °
100 20 60
Initial situation
(20.0)
(60.0)
(40,100)
100 20 60
Conclusion: The application of the transformations is not commutative
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

10. Homogeneous coordinates
The previous sequence, rotation, translation then applied to each vertex can be written as: 1
If the matrices representing the transformations were of the same size would be able to combine. 2
However, the above transformations can also be written as (homogeneously)
We can then write:
The matrix product is:
Associative
In general Noncommutative
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

11. Homogeneous coordinates - summary
Rotation Matrix
Matrix Translation
Escalation Matrix
In homogeneous coordinates of an object n dimensions is represented in space n +1 dimensions.
(X, y)  (x.h, y.h h)
2D 3D
We consider h = 1
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

12. Transformations - Examples
P '= T (dx1, Dy1) * P
T (dx2, Dy2)
T (dx1, Dy1)
R'' = T (dx2, Dy2) * P '
P'' P
T (?,?)
R'' = T (dx2, Dy2) * T (dx1, Dy1) * P
T (dx2, Dy2) * T (dx1, Dy1) = T (dx1+ Dx2, Dy1+ Dy2)
Check that:
O (sx2, Sy2) S * (sx1, Sy1) S = (sx1* Sx2, Sy1* Sy2)
R (the2) * R (the1) = R (a2+ A1)
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

13. Transformations on a arbitrary point (pivot)
Rotation
The rotation moves objects around the origin.
Rotation of -45
Solution:
Traverse the object so that the point Anchorman coincides with the origin
Rotating the object around the origin
Making the translation of the object so that the point Anchorman return to starting position (inverse of first)
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

14. Transformations on a arbitrary point (pivot)
Transformation matrix
Scaling
Traverse the object so that the point Anchorman coincides with the origin
Climb the object
Making the translation of the object so that the point Anchorman return to starting position (inverse of first)
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

15. Exercise Determine the transformation matrix for: T (-x1,-y1)
If P1= (1,2) and P2= (3.3) determine the matrix of equivalent transformation.
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

16. Other transformations
Reflection
In relation to axis x corresponds to a 180 ° rotation in 3D space around the axis of reflection, which results in a scaling S (1, -1) y x x y
In relation to axis y:
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

17. Other transformations
Reflection relative to the line y = x x y x y
Reflection relative to the line y =-x
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

18. Inverse Transformations
If a transformation possibly composite is given by a matrix Mdimensions of 3x3, then the inverse transformation that puts the object in its initial position (ie no transformation) is given by M-1.
Once M represents one or more transformations, the inverse matrix should be.
M.M-1 = I
For some transformations is easy to find the inverse matrix:
Translation:
Scaling:
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004

19. Exercise (Question of the May 23, 2002 test)
T(1,7)S(1,-3)T(0,-7). Nota: é necessário calcula-la [1 0 1] [ ] [0 0 1]
Sim, porque o escalamento é feito sobre Y. Se fosse também sobre X a distância entre as letras aumentava.
COLLEGE OF ENGINEERING UNIVERSITY OF PORTO
COMPUTER GRAPHICS AND INTERFACES /
GRAPHICS SYSTEMS JGB / AAS 2004