Computer Graphic - Transformations in 2D

 2 x 1 matrix:
 General Problem: [B] = [T] [A]
[T] represents a generic operator to be applied to the
points in A. T is the geometric transformation matrix.
If A & T are known, the transformed points are obtained
by calculating B.

 Solid body transformations – the above equation is
valid for all set of points and lines of the object being
transformed.

2x3 3x4
 Since A is a 2x3 matrix and B is a 3x4 matrix, the
product AB is a 2x4 matrix

 T= identity matrix:
a=d=1, b=c=0 => x’=x, y’=y.
 Scaling & Reflections:
b=0, c=0 => x' = a.x, y' = d.y; This is scaling by a in x, d in
y.
If, a = d > 1, we have enlargement (scale up) & uniform
scaling.
If, a ≠ d > 1, we have enlargement (scale up) & non
uniform scaling.
If, 0 < a = d < 1, we have compression (scale down) &
uniform scaling.
If, 0 < a ≠ d < 1, we have compression (scale down) & non
uniform scaling.
1 0
0 1
Sx
0
0

 Let Sx= 3, Sy=2:
 What if Sx and/or Sy are negative?
Get reflections through an axis or plane.
 Only diagonal elements are involved in scaling and
reflections. (off diagonal = 0)

 EX: Let Sx= 5, Sy=5:
=
X’=5X, Y’=5Y so its scale up & uniform scale.
 EX: Let Sx= 1/5, Sy=1/5:
=
X’=5/X, Y’=5/Y so its scale down & uniform scale.
5 0
0 5
X’
Y’
X
Y
X’
Y’
1/5 0
0 1/5
X
Y

Reflection about Matrix T
Y = X X’ = Y
Y’ = X
EX: (3,4) => (4,3)
Y = -X X’ = -Y
Y’ = -X
EX: (3,4) => (-4,-3)
Y = 0 Axis (or X Axis) X’ = X
Y’ = -Y
EX: (3,4) => (3,-4)
X = 0 Axis (or Y Axis) X’ = -X
Y’ = Y
EX: (3,4) => (-3,4)
X = 0 and Y = 0 (Y axis & X axis) X’ = -X
Y’ = -Y
EX: (3,4) => (-3,-4)

 Off diagonal terms are involved in Shearing.
 y' depends linearly on x ; This effect is called shear.

 Positive Rotations: counter clockwise about the origin.

θ (in degrees) Matrix T
90
180
270 or -90
360 or 0

 Rotate the following shape by θ=270 anticlockwise

 B = A + Td , where Td = [tx ty]T.
 x’= x + tx , y’= y + ty.
 Where else are translations introduced?
1. Rotations - when objects are not centered at the
origin.
2. Scaling - when objects/lines are not centered at the
origin. if line intersects the origin then no
translation.
 Note: we cannot directly represent translations as
matrix multiplication, as we know so far.
 We can represent translations in our general
transformation by using homogeneous coordinates.

 Use a 3 x 3 matrix:
 x' = ax + cy + tx
 y' = bx + cy + ty
 w’ = w
 Each point is now represented by a triplet: (x, y, w).
 (x/w, y/w) are called the Cartesian coordinates of the
homogeneous points.

 EX: Given 4 homogeneous points
 P0=(3,4,2,.5) P1=(24,32,16,4) P2=(9,12,6,1) P3=(18,24,12,3)
 Which of the homogeneous points represent a different 3d
point?
 P0= (3/.5,4/.5,2/.5)= (6,8,4)
 P1=(24/4,32/4,16/4)= (6,8,4)
 P2=(9/1,12/1,6/1)= (9,12,6)
 P3=(18/3,24/3,12/3)= (6,8,4)
 So p2 is different

 General Purpose 2D transformations in homogeneous
coordinate representation:
 Parameters involved in scaling, rotation, reflection
and shear are: a, b, c, d.
 For translation:
If [B]=[T][A] then p,q are
translation Parameter.
if [B]=[A][T] then m,n are
translation parameter.

 There are three steps for translation about an
arbitrary point in space:
1. Translate by (-Tx, -Ty).
2. Scale by Sx, Sy, where Sx = new coordinate for x/ old
Sy = new coordinate for y/ old
3. Translate back by (Tx, Ty)

 EX: Apply translation to the matrix according to the
shape below.
1 0 16
0 1 4
0 0 1
1 0 -2
0 1 -2
0 0 1
8 0 0
0 2 0
0 0 1
X
Y
Z

 There are three steps for scaling about an arbitrary
point in space:
1. Translate by (-Tx, -Ty)
2. Scale by Sx, Sy
3. Translate back by (Tx, Ty)

 EX: Apply Scaling up to the matrix such that Tx= 1,
Ty= 2, Sx= 4, Sy= 12.
=
= =
 x’= x+7w y’=y+16w w’=w
1 0 -1
0 1 -2
0 0 1
X’
Y’
Z’
4 0 0
0 12 0
0 0 1
1 0 4
0 1 12
0 0 1
X
Y
Z
1 0 3
0 1 4
0 0 1
1 0 4
0 1 12
0 0 1
X
Y
Z
1 0 7
0 1 16
0 0 1
X
Y
Z

 EX: Apply Scaling down to the matrix such that Tx= 1,
Ty= 2, Sx= 4, Sy= 12.
=
1 0 -4
0 1 -12
0 0 1
X’
Y’
Z’
1/4 0 0
0 1/6 0
0 0 1
1 0 1
0 1 2
0 0 1
X
Y
Z

 There are three steps for rotation about an arbitrary
point in space:
1. Translate by (-Px, -Py).
2. Rotate.
3. Translate back by (Px, Py)

 EX: Apply rotation to the matrix such that Px= 1, Py= 4
by θ = 90 .
=
1 0 -1
0 1 -4
0 0 1
X’
Y’
Z’
0 -1 0
1 0 0
0 0 1
1 0 1
0 1 4
0 0 1
X
Y
Z

 There are five steps for reflection about an arbitrary
point in space:
1. Translate line to the origin.
2. Rotation about the origin.
3. Reflection matrix.
4. Reverse the rotation.
5. Translate line back.
 If the line pass the origin we need only 3 steps
(remove step 1 and step 5).

 If we want to apply a series of transformations T1, T2,
T3 to a set of points, we can do it in two ways:
1. We can calculate p'=T1*p, p''= T2*p', p'''=T3*p''
2. Calculate T=T1*T2*T3 then p'''=T*p.
 Translations: Translate the points by
tx1, ty1, then by tx2, ty2.
 Scaling: Similar to translations.
 Rotations: Rotate by θ1, then by θ2 and its done by 2
ways:
1. Replace θ1, θ2 by θ=θ1+θ2.
2. calculate T1 for θ1, then T2 for θ2 then multiply
them.

 EX: Apply two scaling to the matrix such that Sx1= 2,
Sy1= 3, Sx2=4, Sy2=2.
 Way #1
 P’= =
 P’’= =
2 0 0
0 3 0
0 0 1
1 0 0
0 2 0
0 0 1
1 0 0
0 2 0
0 0 1
2 0 0
0 6 0
0 0 1
4 0 0
0 2 0
0 0 1
2 0 0
0 6 0
0 0 1
8 0 0
0 12 0
0 0 1

 Way #2
 T = =
 P’’= =
2 0 0
0 3 0
0 0 1
4 0 0
0 2 0
0 0 1
8 0 0
0 6 0
0 0 1
8 0 0
0 6 0
0 0 1
1 0 0
0 2 0
0 0 1
8 0 0
0 12 0
0 0 1

 EX: Find the new coordinate for the point (3,7,1) if you
rotate the point by 180 degree then by 90 degree with
clockwise.
 Composite rotation : 180 + 90 = 270 degree
 Since w=1 the Cartesian coordinate for the point is
(3,7)
1 0 -3
0 1 -7
0 0 1
Cos270 Sin270 0
-Sin270 Cos270 0
0 0 1
1 0 3
0 1 7
0 0 1

 EX: Suppose you have the point (2,4),rotate the point
by 90 degree then by 180 degree the by 45 degree then
by 45 degree then scale it by Sx=2 and Sy=4.
 Composite rotation : 90+180+45+45 = 360 degree
2 0 0
0 4 0
0 0 1
Cos360 Sin360 0
-Sin360 Cos360 0
0 0 1
2
4
1

 EX: Transform the point from left to right.
 Sx1= 4/2= 2, Sy1= 5/2= 2.5, Sx2= 7/4, Sy2= 8/5
 Sx= 2*7/4= 3.5, Sy= 2.5*8/5= 4
1 0 -2
0 1 -2
0 0 1
3.5 0 0
0 4 0
0 0 1
1 0 7
0 1 8
0 0 1

 EX: Transform the point from left to right.
 Sx= 3/1= 3, Sy= 3/1= 3
1 0 -1
0 1 -1
0 0 1
3 0 0
0 3 0
0 0 1
1 0 3
0 1 3
0 0 1

 Cases where T1 * T2 = T2 * T1:
T1 T2
Translation Translation
Scaling Scaling
Rotation Rotation
Uniform Scaling Rotation

 EX: If we scale, translate to origin then translate back
is it equivalent to translate to origin, scale then
translate back? (Sx=2, Sy=3, tx=3, tx=2)
 Answer: No, the order of matrix is important.
=
=
2 0 0
0 3 0
0 0 1
1 0 -3
0 1 -2
0 0 1
1 0 3
0 1 2
0 0 1
2 0 0
0 3 0
0 0 1
2 0 0
0 3 0
0 0 1
1 0 -3
0 1 -2
0 0 1
1 0 3
0 1 2
0 0 1
2 0 3
0 3 4
0 0 1

 EX: If we rotate the object by 90 degree then scale it
,is it equivalent to scale it then rotate it by 90 degree ?
(Sx=2, Sy=3)
=
=
0 -1 0
1 0 0
0 0 1
2 0 0
0 3 0
0 0 1
0 -3 0
2 0 0
0 0 1
0 -1 0
1 0 0
0 0 1
2 0 0
0 3 0
0 0 1
0 -2 0
3 0 0
0 0 1

 EX: If we rotate the object by 90 degree then scale it
,is it equivalent to scale it then rotate it by 90 degree ?
(Sx=4, Sy=4)
=
=
0 -1 0
1 0 0
0 0 1
4 0 0
0 4 0
0 0 1
0 -4 0
4 0 0
0 0 1
0 -1 0
1 0 0
0 0 1
4 0 0
0 4 0
0 0 1
0 -4 0
4 0 0
0 0 1

 EX: Prove that rotation then translation not equal to
translation then rotation?
=
= not equal to the
above
Cosθ -Sinθ
0
Sinθ Cosθ
0
0 0
1
1 0 -
tx
0 1 -
ty
0 0
1
Cosθ –Sinθ -txcosθ+tysinθ
Sinθ Cosθ -txsinθ-tycosθ
0 0 1
1 0 -
tx
0 1 -
ty
0 0
1
Cosθ -Sinθ
0
Sinθ Cosθ
0
0 0
1

 EX: Given a solid object, if we apply uniform scale
then translation to the body, is it the same as we apply
translation then uniform scale?
=
=
Sx 0
0
0 Sy
0
0 0
1
1 0
tx
0 1
ty
0 0
1
Sx Sy Sxtx
0 Sy Syty
0 0 11 0
tx
0 1
ty
0 0
1
Sx 0
0
0 Sy
0
0 0
1
Sx 0 tx
0 Sy ty
0 0 1

 Screen Coordinates: The coordinate system used to
address the screen (device coordinates).
 World Coordinates: A user-defined application specific
coordinate system having its own units of measure,
axis, origin, etc.
 Window: The rectangular region of the world that is
visible.
 Viewport: The rectangular region of the screen space
that is used to display the window.

 The Purpose is to find the transformation matrix that
maps the window in world coordinates to the viewport
in screen coordinates.
 Window: (x, y space) denoted by: xmin, ymin, xmax, ymax.
 Viewport: (u, v space) denoted by: umin, vmin, umax, vmax.

 The overall transformation:
Translate the window to the origin.
Scale it to the size of the viewport.
Translate it to the viewport location.

 EX: Given (xmin=3, ymin=12, xmax=8, ymax=16)
(umin=1, vmin=2, umax=2, vmax=4)
Sx=1/5,Sy=1/2
Map from window to view port.
0 0 -3
0 1 -12
0 0
1
1/5 0 0
0 1/2 0
0 0
1
1 0 1
0 1 2
0 0 1

 Map from window to view port.
 (xmin=20, ymin=10, xmax=40, ymax=50)
 (umin=5, vmin=4, umax=10, vmax=12)
 Sx=(10-5)/(40-20)= 0.25, Sy=(12-4)/(50-10)=0.2
1 0 -20
0 1 -10
0 0
1
.25 0 0
0 .2 0
0 0 1
1 0 5
0 1 4
0 0 1

Computer Graphic - Transformations in 2D

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