WeekNo.%04 Imaging%Geometry% (course:%Computer%Vision)• Imaging%Geometry% • Translaon% • Scaling% • Rotaon% Naeem%A.%Mahoto% Outline% Wednesday,%August5,%2015%

Naeem A. Mahoto e-‐mail: [email protected]

Department of So9ware Engineering, Mehran UET

Jamshoro, Sind, Pakistan

Wednesday, August 5, 2015

Week No. 04 Imaging Geometry

(course: Computer Vision)

•  Imaging Geometry

•  TranslaRon

•  Scaling

•  RotaRon

Naeem A. Mahoto

Outline


•  Any real world scene exists in a 3-‐dimensional space. •  Imaging Geometry deals with the techniques and

transformaRon for mapping Real world objects and their characterisRcs on Image plane

•  Any point in 3D co ordinate system is represented by BLOCK le]ers -‐ World Coordinate system -‐ W(X, Y, Z)

•  Any point in 2D co-‐ordinate system/Image plane is represented by small le]ers -‐ Camera coordinate system – f (x, y)

•  The process of acquiring an image of a 3-‐D scene involves projecRng the scene on a 2-‐D surface.

•  The projecRon from 3-‐D to 2-‐D is a perspecRve projecRon. Naeem A. Mahoto

Imaging Geometry


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TranslaRon


X

Z

P1(X1,Y1,Z1)

P2(X2,Y2,Z2)

(dx,dy,dz)

YConsider to translate a point P1, with coordinate values (x1 , y1 , z1 ) to new locaRon P2 by a displacement (dx, dy, dz)

The translaRon is easily accomplished by using the equaRon: I. X2 = X1 + dx II. Y2 = Y1 + dy III. Z2 = Z1 + dz

•  Where (X2, Y2, Z2) are new points and can be represented in Matrix form, as:

Naeem A. Mahoto

TranslaRon


1 0 0 X2 Y2 Z2

X1 Y1 Z1

dx dy dz

0 1 0 0 0 1

=

V* = T V

where, V* is the new point, T is the translaRon matrix, and

V is the original point.

•  The product of Matrices T & V seems Invalid, as Columns of T ≠ Rows of V

•  Generalize this transformaRon by inserRng a DUMMY row in all three matrices

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TranslaRon


1 0 0 X2 Y2 Z2

X1 Y1 Z1

dx dy dz

0 1 0 0 0 1

= V* = T V

1 1 1 0 0 0

1 0 0 dx dy dz

0 1 0 0 0 1

1 0 0 0

T = Transla'on Matrix

•  The IniRal Point P1 can be traced back by applying the Inverse TranslaRon, that is:

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Inverse TranslaRon


1 0 0 X2 Y2 Z2

X1 Y1 Z1

-‐dx -‐dy -‐dz

0 1 0 0 0 1

=

V = T-‐1 V*

1 1 1 0 0 0

1 0 0 -‐dx -‐dy -‐dz

0 1 0 0 0 1

1 0 0 0

T-‐1 = Inverse Transla'on Matrix

T . T-‐1 = I (IdenRty Matrix) IdenRty Matrix: Square matrix having diagonal elements equal to one

Naeem A. Mahoto

TranslaRon


Original Image Translated Image

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TranslaRon


Translated Image Translated Image

This property is used in adjusRng the Objects on Monitor screen

•  The scale operator performs a geometric transformaRon which can be used to shrink or zoom the size of an image

•  The shrink or Zoom, depends upon the vales of scaling factor

•  For values between 0 &1, the resultant Image is shrinked

•  For values greater than 1, the resultant Image is expanded by that factor

Naeem A. Mahoto

Scaling


•  The Scaling is easily accomplished by using the equaRons: I.  X2 = X1 * sx II.  Y2 = Y1 * sy III.  Z2 = Z1 * sz

•  Where (X2, Y2, Z2) are new points and expressed in Matrix form as:

Naeem A. Mahoto

Scaling


X2 Y2 Z2

X1 Y1 Z1

=

V* = S V

1 1

Where S represents the Scaling Matrix

sx 0 0 0 0 0

0 sy 0 0 0 sz

1 0 0 0

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Scaling


Original Image Scaled Image at 0.5*0.5

•  The IniRal Scale can be traced back by applying the Inverse Scaling, that is:

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Inverse Scaling


1/sx 0 0 X2 Y2 Z2

X1 Y1 Z1

0 0 0

0 1/sy 0 0 0 1/sz

=

V = S-‐1 V*

1 1 1 0 0 0

S-‐1 = Inverse Scaling Matrix

S . S-‐1 = I (IdenRty Matrix) IdenRty Matrix: Square matrix having diagonal elements equal to one

1/sx 0 0 0 0 0

0 1/sy 0 0 0 1/sz

1 0 0 0

•  The transformaRon used for 3-‐D rotaRon is inherently more complex than the TransformaRons (TranslaRon and Scaling)

•  The simplest from of these transformaRons is for RotaRon of a point about the coordinate axes

•  To rotate a point about another arbitrary Point in space, we require three transformaRon –  The first translate the arbitrary point to the Origin, –  The second performs the rotaRon, and –  The third translates the point back to its Original posiRon

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RotaRon


•  RotaRon of a point about the Z coordinate axis by an angle θ is achieved by using the transformaRon

•  The rotaRon angle θ is measured clockwise when looking at the region from a Point on the +Z axis. –  This transformaRon affects only the values of X and Y coordinates

•  The rotaRon operator performs a geometric transform which

maps the posiRon ( x1, y1 ) of a picture element in an input image onto a posiRon ( x2, y2 ) in an output image by rotaRng it through a user-‐specified angle θ about an origin (0,0)

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RotaRon


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RotaRon along Z-‐axis


P1(X1,Y1,Z1)

P2(X2,Y2,Z2)

0 X2

X1

Y 1 Y 2

φ

θ

X

Z

Y

R R

R is length of vector

φ is iniRal angle with X axis.

RotaRon is in clockwise direction.

•  Assume that the Vector (R) is making an angle of φ with X axis

•  IniRally, the components can be achieved as: –  X1 = R cos φ –  Y1 = R sin φ –  Z1 = Z1

•  If we rotate the vector R by θ around Z-‐axis in Counter clock

wise DirecRon, then now point formed will be given by: –  X2 = R cos (φ+θ) –  Y2 = R sin (φ+θ ) –  Z2 = Z1 (Un changed axis)

Naeem A. Mahoto



•  By expanding the terms of Sine and Cosine, we get: –  X2 = R cosθ cosφ -‐ R sinφ sinθ –  Y2 = R sinθ cosφ + R sinφ cosθ –  Z2 = Z1

•  But, we have assumed iniRally that: –  R cos φ = X1 –  R sin φ = Y1

•  So the new point (x2,Y2,z2) becomes:

–  X2 = X1 cosθ -‐ Y1 sinθ –  Y2 = X1 sinθ + Y1 cosθ –  Z2 = Z1

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•  For matrix representaRon, these equaRons can be interpreted as: –  X2 = X1 cosθ -‐ Y1 sinθ + 0 –  Y2 = X1 sinθ + Y1 cosθ + 0 –  Z2 = 0 + 0 + Z1

•  In form of matrices, the rotaRon operaRon can be given as:

Naeem A. Mahoto



cosθ -‐sinθ 0 X2 Y2 Z2

X1 Y1 Z1

sinθ cosθ 0 0 0 1

=

V* = RθZ V

•  In form of matrices, the Inverse rotaRon operaRon can be given as:

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Inverse RotaRon along Z-‐axis


cosθ sinθ 0 X1 Y1 Z1

X2 Y2 Z2

-‐sinθ cosθ 0 0 0 1

=

V = (RθZ)-1 V*

(RθZ)-‐1 represents the Inverse RotaRon Matrix

(RθZ). (RθZ)-‐1 = I (IdenRty Matrix)

•  Assume that the Vector (R) is making an angle of φ with Z-‐axis

•  IniRally, the components can be achieved as: –  X1 = R sin φ –  Y1 = Y1 –  Z1 = R cos φ

•  If we rotate the vector R by β around Y-‐axis in Counter clock wise DirecRon, then now point formed will be given by: –  X2 = R sin (φ+ β) –  Y2 = Y1 (Unchanged axis) –  Z2 = R cos (φ+ β)

Naeem A. Mahoto

RotaRon along Y-‐axis


•  By expanding the terms of Sine and Cosine, we get: –  X2 = R sinφ cosβ + R cosφ sinβ –  Y2 = Y1 –  Z2 = R cosφ cosβ -‐ R sinφ sinβ

•  But, we have assumed iniRally that: –  R sin φ = X1 –  R cos φ = Z1


–  X2 = X1 cosβ + Z1 sinβ –  Y2 = Y1 –  Z2 = Z1 cosβ -‐ X1 sinβ

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•  For matrix representaRon, these equaRons can be interpreted as: –  X2 = X1 cosβ + Z1 sinβ + 0 –  Y2 = 0 + Y1 + 0 –  Z2 = -‐ X1 sinβ + 0 + Z1 cosβ


Naeem A. Mahoto



cosβ 0 sinβ X2 Y2 Z2

X1 Y1 Z1

0 1 0 -‐sinβ 0 cosβ

=

V* = Rβy V


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Inverse RotaRon along Y-‐axis


V = (Rβy)-1 V*

(Rβy)-‐1 represents the Inverse RotaRon Matrix

(Rβy). (Rβ

y)-‐1 = I (IdenRty Matrix)

cosβ 0 -‐sinβ X1 Y1 Z1

X2 Y2 Z2

0 1 0 sinβ 0 cosβ

=

•  Assume that the Vector (R) is making an angle of φ with Y-‐axis

•  IniRally, the components can be achieved as: –  X1 = X1 –  Y1 = R cos φ –  Z1 = R sin φ

•  If we rotate the vector R by α around X-‐axis in Counter clock

wise DirecRon, then now point formed will be given by: –  X2 = X1 (Un changed axis) –  Y2 = R cos (φ+ α ) –  Z2 = R sin(φ+ α )

Naeem A. Mahoto

RotaRon along X-‐axis


•  By expanding the terms of Sine and Cosine, we get: –  X2 = X1 –  Y2 = R cosφ cos α -‐ R sinφ sin α –  Z2 = R sinφ cos α + R cosφ sin α

•  But, we have assumed iniRally that: –  R cos φ = Y1 –  R sin φ = Z1


–  X2 = X1 –  Y2 = Y1 cos α -‐ Z1 sin α –  Z2 = Z1 cos α + Y1 sin α

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•  For matrix representaRon, these equaRons can be interpreted as: –  X2 = X1 + 0 + 0 –  Y2 = 0 + Y1 cos α -‐ Z1 sin α –  Z2 = 0 + Y1 sin α + Z1 cos α


Naeem A. Mahoto



1 0 0 X2 Y2 Z2

X1 Y1 Z1

0 cosα -‐sinα

0 sinα cosα

=

V* = Rαx V


Naeem A. Mahoto

Inverse RotaRon along X-‐axis


V = (Rαx)-1 V*

(Rαx)-‐1 represents the Inverse RotaRon Matrix

(Rαx). (Rα

x)-‐1 = I (IdenRty Matrix)

X1 Y1 Z1 =

1 0 0 0 cosα sinα

0 -‐sinα cosα

X2 Y2 Z2

•  The Inverse rotaRon of an angle, say (θo), around an axis is, equivalent to the rotaRon by (–θo) around same axis

•  For example, Inverse RotaRon of θ around Z-‐axis is equivalent to the rotaRon of –θ around Z-‐axis

Naeem A. Mahoto

Inverse RotaRon


(RθZ)-‐1 = R-‐θ

Z

cosθ sinθ 0 -‐sinθ cosθ 0

0 0 1

cosθ sinθ 0 -‐sinθ cosθ 0

0 0 1

=

⎩⎨⎧ =

=⋅otherwise0

1 jiRR j

Ti

Remember that Rotation Matrices are Ortho normal!

(RθZ)T.(Rθ

Z) = (RθZ).(Rθ

Z)T = I (IdenRty Matrix)

Therefore, inverse of a rotation matrix is just its transpose.

Naeem A. Mahoto

RotaRon


Original Image

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RotaRon


Rotated @ 180o Rotated @ 270o

•  A class of linear 2D geometric transformaRon, which maps variables (i.e., pixel intensity values) located at (x1, y1) of input image into new variables (x2, y2) of output image.

•  The general affine transformaRon is commonly wri]en in homogeneous coordinates as shown:

•  Homogenous coordinates: A coordinate system used in projecRve geometry including points at infinity. It represents infinity using finite coordinates

Naeem A. Mahoto

Affine TransformaRon


•  Lets consider a, w two real numbers, then v = a/w tends to infinity if w tends to zero. If w != 0 then v = a/w, and if w = 0, infinite value is represented by v = (a,0). Thus concept of infinity is represented as number pair like (a,w)

•  Replace x-‐y coordinates in digital image with x/w, y/w. A funcRon f(x,y) becomes f(x/w, y/w)

•  •  By defining only the B matrix, this transformaRon can carry out

pure TranslaRon:

Naeem A. Mahoto

Affine TransformaRon: TranslaRon


•  Similarly, pure Scaling is:

Naeem A. Mahoto

Affine TransformaRon: Scaling


•  Pure RotaRon uses the A matrix and is defined as (for posiRve angles being clockwise rotaRons):

Naeem A. Mahoto

Affine TransformaRon: RotaRon


•  Some of the important characterisRcs of PerspecRve ProjecRon are: –  Objects which are of the same size but at different distances from the

image plane would appear to be of different sizes. –  Parallel lines in the scene would appear to converge at some finite

point. –  The expanse of an object along the line of sight appears much smaller

compared to the expanse along a direcRon perpendicular to the line of sight.

–  Consider a plane which is parallel to the image plane and passing through the center of projecRons. All the world points lying on this plane will not get projected on to the image plane. Naeem A. Mahoto

PerspecRve ProjecRon


1.  Show that following idenRRes are true: I.   R90

Y . R90X = R270

Z . R90Y

II.   R90Y . R180

X = R180Z . R90

Y III.   R180

Y . R90X = R180

Z . R270X

IV.  R180Y . R270

X = R180Z . R90

X

3.  Consider a vector (7,3,2) which is rotated around Z-‐axis by 900, and then rotated around Y-‐axis by 900 and finally translated by (4,-‐3,7). Find the new coordinates of the vector. All rotaRons are clockwise.

Naeem A. Mahoto

Homework


3.  What is rotaRon for an object rotated by 300 around z-‐axis, followed by 600 around x-‐axis and followed by 900 a rotaRon of around y-‐axis. All rotaRons are counter clockwise.

Naeem A. Mahoto

Homework


Naeem A. Mahoto Wednesday, August 5, 2015

Documents

WeekNo.%04 Imaging%Geometry% (course:%Computer%Vision)• Imaging%Geometry% • Translaon% • Scaling% • Rotaon% Naeem%A.%Mahoto% Outline% Wednesday,%August5,%2015%