Computer Graphics II

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Thomasine Lloyd
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1 Computer Graphics II Autumn

2 Outline 1 Transformations

3 Outline Transformations 1 Transformations

4 Orthogonal Projections Suppose we want to project a point x onto a plane given by its unit-length normal, N N n N p x y The point y is the projection of x onto the plane; the point p is some given point on the plane From the vector y p, we have N (y p) = 0

5 Orthogonal Projections Suppose we want to project a point x onto a plane given by its unit-length normal, N N n N p x y The point y is the projection of x onto the plane; the point p is some given point on the plane From the vector y p, we have N (y p) = 0

6 Orthogonal Projections (contd.) Straight away we can see that or x = y + n N y = x n N But what is n? The two components of the vector x p are its component in the plane, y p, and an orthogonal component which we can write as n N x p = (y p) + n N Since y p and N are perpendicular, taking the dot product on both sides with N yields N (x p) = n N N = n

7 Orthogonal Projections (contd.) Then y = x Nn = x N( N (x p)) = x N( N t (x p)) = x N N t (x p) = (I N N t )x + N N t p Notice that this is of the form y = Mx + b, for points y, x, b This is of the form y is a multiple of x plus some constant, b This is what we call an affine transformation Our previous example projecting onto a line was also an affine transformation (See also first answer to this question.)

8 Outline Transformations 1 Transformations

9 Oblique Projections Transformations Given a point x how do we project it obliquely onto a plane given by its unit-length normal, N N p x y D Let D be the direction in which we project points; w.l.o.g. we can assume that D is unit-length D cannot be parallel to plane! D N = N D 0

10 Oblique Projections Transformations Given a point x how do we project it obliquely onto a plane given by its unit-length normal, N N p x y D Let D be the direction in which we project points; w.l.o.g. we can assume that D is unit-length D cannot be parallel to plane! D N = N D 0

11 Oblique Projections (contd.) From previous picture y = x + d D, and so Dotting both sides with N, and y p = (x p) + d D 0 = N (x p) + d N D N (x p) d = N D, N D 0 (x p) D N (x p) y p = (x p) N D D ( D ) N t = I D t N (x p)

12 Oblique Projections (contd.) Finally, ( D ) N t y = I D t N D N t x + D t N p Note: N D = D N = N t D = D t N D N t Since this is of the form y = Mx + b again, we can say that oblique projections are also in the class of affine transformations

13 Oblique Projections (contd.) Finally, ( D ) N t y = I D t N D N t x + D t N p Since this is of the form y = Mx + b again, we can say that oblique projections are also in the class of affine transformations

14 Outline Transformations 1 Transformations

15 Perspective Projections Graphics look much more realistic when drawn in the perspective style In this style points are projected onto a view plane not in parallel (as with orthogonal and oblique) but according to rays that originate at a single point This requires a special type of projection

16 Perspective Projections (contd.) N e p x y D With source, e, direction of projection is not uniform here but depends instead on the vector x e and y = e + t(x e), for some scalar t

17 Perspective Projections (contd.) Then since y p is orthogonal to N and, solving for t we get y p = (e p) + t(x e) 0 = N (e p) + t N (x e) N (e p) t = N (x e) > 0 This makes sense since the vectors (e p) and (x e) have opposite directions w.r.t. N so one will contribute a positive DP and the other negative.

18 Perspective Projections (contd.) So N (e p) y = e (x e) N (x e) N (x e) = e N (x e) N (e p) (x e) N (x e) = (e N t N (e p)i)(x e) N (x e) (The last line is because N (x e) = N t (x e).) This cannot be expressed as an affine transformation due to x appearing in the denominator.

19 Beginnings of a Camera System In the coordinate system of viewing direction D, up vector U, how does a point x = e + d D + u U + r R get projected onto a viewing plane situated at position d min units on the D axis in front of us? The viewing plane will be square to the viewing direction. Straight away we can say that d = D (x e), u = U (x e) and r = R (x e)

20 Beginnings of a Camera System (contd.) We will say that the plane normal N is D, that is, pointing back towards us. Then the closest point p on that plane is the one straight ahead (along the viewing direction) p = e + d min D, e p = d min D for dmin > 0 and Putting this back in to the previous (2 slides ago), we get N (e p) y = e (x e) N (x e) = e D ( d min D) D (x e) (x e) = e + d min d (x e) = e + d D + u U + r R m, m = d/d min

21 Beginnings of a Camera System (contd.) So a world point x described as x = e + d D + u U + r R which is the point (d, u, r) in the D, U, R co-ordinate system (that is centered at e ) is associated with (gets projected to) (d, u, r)/(d/d min ) on the viewing plane situated at d min on the D axis Exercise: check where the point (d, u, r) = (d min, 20, 75) gets projected to Note the scaling factor, d/d min, determines all three components; we have been able to deal with scaling factors from much earlier but the problem here is that the scaling factor is different for every point

Chapter 1. Linear Equations and Straight Lines. 2 of 71. Copyright 2014, 2010, 2007 Pearson Education, Inc.

Chapter 1. Linear Equations and Straight Lines. 2 of 71. Copyright 2014, 2010, 2007 Pearson Education, Inc. Chapter 1 Linear Equations and Straight Lines 2 of 71 Outline 1.1 Coordinate Systems and Graphs 1.4 The Slope of a Straight Line 1.3 The Intersection Point of a Pair of Lines 1.2 Linear Inequalities 1.5