2D TRANSFORMATIONS
COMPUTER GRAPHICS
2D Transformations
“Transformations are the operations applied to
geometrical description of an object to change its
position, orientation, or size are called geometric
transformations”.
Translation
❖ Translation is a process of changing the position
of an object in a straight-line path from one co-
ordinate location to another.
❖ We can translate a two dimensional point by
adding translation distances, tx and ty.
❖ Suppose the original position is (x ,y) then new
position is (x’, y’).
❖ Here x’=x + tx and y’=y + ty.
Click icon to add picture
Translation
❖ Matrix form of the equations:
X’ = X + tx and Y’ = Y + ty is
P = x
y
P’ = x’
y’
T= tx
ty
❖ we can write it,
P’= P + T
❖ Translate a polygon with co-ordinates A(2,5) B(7,10) and C(10,2) by 3
units in X direction and 4 units in Y direction.
❖ A’ = A +T
= 2 +
5
3 =
4
5
9
❖ B’ = B + T
= 7
10
C’ = C + T
+ 3 =
4
10
14
❖
= 10 + 3 =
2 4
13
7
Rotation
❖ A two dimensional rotation is applied to an object by
repositioning it along a circular path in the xy plane.
❖ Using standard trigonometric equations , we can express
the transformed co-ordinates in terms of
x’ = r cos( coscosr sinsin
y’ = r sin( cosr sin
❖ The original co-ordinates of the point is
x = r cos
y = r sin
Click icon to add picture
After substituting equation 2 in equation 1 we get
x’=x cos
Y’=x sin + y cos
Rotation
❖ That equation can be represented in matrix form
x’ y’ = x y cos
-sin cos
❖ we can write this equation as,
P’ = P . R
❖ Where R is a rotation matrix and it is given as
R = cos
-sin cos
❖ A point (4,3) is rotated counterclockwise by angle of 45.
find the rotation matrix and the resultant point.
❖ R = cos =
- cos
cos45
-sin45
= 1/√2
sin45
cos45
1/√2
- 1/√2 1/√2
P’ = 4 3 1/√2 1/√2
- 1/√2 1/√2
= 4/√2 – 3/√2 4/√2 + 3/√2
= 1/√2 7/√2
Scaling
❖ A scaling transformation changes the size of an object.
❖ This operation can be carried out for polygons by
multiplying the co-ordinates values (x , y) of each vertex
by scaling factors Sx and Sy to produce the transformed
co-ordinates (x’ , y’).
x’ = x . Sx
y’ = y . Sy
❖ In the matrix form
x’ y’ = x y Sx 0
0 Sy
= P . S
Scaling
• Uniform Scaling Un-uniform Scaling
Homogeneous co-ordinates for Translation
❖ The homogeneous co-ordinates for translation are given as
T = 1 0 0
0 1 0
tx ty 1
❖ Therefore , we have
x’ y’ 1 = x y 1 1 0 0
0 1 0
tx ty 1
= x + tx y + ty 1
Homogeneous co-ordinates for rotation
❖ The homogeneous co-ordinates for rotation are given as
R = cos sin
-sin cos
0 0 1
❖ Therefore , we have
x’ y’ 1 = x y 1 cos sin
-sin cos
0 0 1
= x cos - y sin x sin + y cos 1
Homogeneous co-ordinates for scaling
❖ The homogeneous co-ordinate for scaling are given as
S = Sx 0 0
0 Sy 0
0 0 1
= x . Sx y . Sy
❖ Therefore , we have
x’ y’ 1 = x y 1 Sx 0 0
0 Sy 0
0 0 1
1
CONCLUSION
T o manipulate the initially created
object and to display the
modified object without having to
redraw it, we use
Transformations.
THANK YOU

2Dand3D transformationppt.pptx

  • 1.
  • 2.
    2D Transformations “Transformations arethe operations applied to geometrical description of an object to change its position, orientation, or size are called geometric transformations”.
  • 3.
    Translation ❖ Translation isa process of changing the position of an object in a straight-line path from one co- ordinate location to another. ❖ We can translate a two dimensional point by adding translation distances, tx and ty. ❖ Suppose the original position is (x ,y) then new position is (x’, y’). ❖ Here x’=x + tx and y’=y + ty.
  • 4.
    Click icon toadd picture
  • 5.
  • 6.
    ❖ Matrix formof the equations: X’ = X + tx and Y’ = Y + ty is P = x y P’ = x’ y’ T= tx ty ❖ we can write it, P’= P + T
  • 7.
    ❖ Translate apolygon with co-ordinates A(2,5) B(7,10) and C(10,2) by 3 units in X direction and 4 units in Y direction. ❖ A’ = A +T = 2 + 5 3 = 4 5 9 ❖ B’ = B + T = 7 10 C’ = C + T + 3 = 4 10 14 ❖ = 10 + 3 = 2 4 13 7
  • 8.
    Rotation ❖ A twodimensional rotation is applied to an object by repositioning it along a circular path in the xy plane. ❖ Using standard trigonometric equations , we can express the transformed co-ordinates in terms of x’ = r cos( coscosr sinsin y’ = r sin( cosr sin ❖ The original co-ordinates of the point is x = r cos y = r sin
  • 9.
    Click icon toadd picture After substituting equation 2 in equation 1 we get x’=x cos Y’=x sin + y cos
  • 10.
  • 11.
    ❖ That equationcan be represented in matrix form x’ y’ = x y cos -sin cos ❖ we can write this equation as, P’ = P . R ❖ Where R is a rotation matrix and it is given as R = cos -sin cos
  • 12.
    ❖ A point(4,3) is rotated counterclockwise by angle of 45. find the rotation matrix and the resultant point. ❖ R = cos = - cos cos45 -sin45 = 1/√2 sin45 cos45 1/√2 - 1/√2 1/√2 P’ = 4 3 1/√2 1/√2 - 1/√2 1/√2 = 4/√2 – 3/√2 4/√2 + 3/√2 = 1/√2 7/√2
  • 13.
    Scaling ❖ A scalingtransformation changes the size of an object. ❖ This operation can be carried out for polygons by multiplying the co-ordinates values (x , y) of each vertex by scaling factors Sx and Sy to produce the transformed co-ordinates (x’ , y’). x’ = x . Sx y’ = y . Sy ❖ In the matrix form x’ y’ = x y Sx 0 0 Sy = P . S
  • 14.
    Scaling • Uniform ScalingUn-uniform Scaling
  • 15.
    Homogeneous co-ordinates forTranslation ❖ The homogeneous co-ordinates for translation are given as T = 1 0 0 0 1 0 tx ty 1 ❖ Therefore , we have x’ y’ 1 = x y 1 1 0 0 0 1 0 tx ty 1 = x + tx y + ty 1
  • 16.
    Homogeneous co-ordinates forrotation ❖ The homogeneous co-ordinates for rotation are given as R = cos sin -sin cos 0 0 1 ❖ Therefore , we have x’ y’ 1 = x y 1 cos sin -sin cos 0 0 1 = x cos - y sin x sin + y cos 1
  • 17.
    Homogeneous co-ordinates forscaling ❖ The homogeneous co-ordinate for scaling are given as S = Sx 0 0 0 Sy 0 0 0 1 = x . Sx y . Sy ❖ Therefore , we have x’ y’ 1 = x y 1 Sx 0 0 0 Sy 0 0 0 1 1
  • 28.
    CONCLUSION T o manipulatethe initially created object and to display the modified object without having to redraw it, we use Transformations.
  • 29.