06 projection.slides (1)

The graphics pipeline: part IWindowing transformsPerspective transform

Camera transformationWrap-up

Perspective projection (outline)

1 The graphics pipeline: part I

2 Windowing transforms

3 Perspective transform

4 Camera transformation

5 Wrap-up

Graphics, 1st period 2006/2007 Lecture 6: Orthographic and perspective projection



Projecting from arbitrary camera positionsCamera transformationOrthographic projection and the canonical view volumeWindowing transform

The graphics pipeline: part I

Given an arbitrary cameraposition, we want to displaythe objects in the model in animage

The projection should beperspective projection.





Camera transformation

We simplify by moving thecamera viewpoint to theorigin, such that we look intothe direction of the negativez-axis.

−Z





Orthographic projection

Orthographic projection is alot simpler than perspectiveprojection, so we transform theclipped view frustum to anaxis-parallel box

−Z





The canonical view volume

To simplify our calculations,we transform to the canonicalview volume

(−1,−1)

(1, 1)





Windowing transform

We apply a windowingtransform to display the square[−1, 1]× [−1, 1] onto anm× n image.





The graphics pipeline: part I

Every step in the sequence canbe represented by a matrixoperation, so the whole processcan be applied by performing asingle matrix operation!

(Well: almost.)




The canonical view volumeThe orthographic view volumeThe orthographic projection matrix

The canonical view volume

The canonical view volume is a2× 2× 2 box, centered at theorigin.

The clipped view frustum istransformed to this box (andthe objects within the viewfrustum undergo the sametransformation).

Vertices in the canonical viewvolume are orthographicallyprojected onto an m× nimage.

−z

x

y

(−1,−1, 1)

(1, 1,−1)

(1,−1,−1)





Mapping the canonical view volume

We need to map the square[−1, 1]2 onto a rectangle[−1

2 , m− 12 ]× [−1

2 , n− 12 ].

The following matrix takescare of that:m

2 0 m2 − 1

20 n

2n2 − 1

20 0 1

(−1,−1)

(1, 1)

m

n





The orthographic view volume

The orthographic view volumeis an axis-aligned box[l, r]× [b, t]× [n, f ].

Transforming it to thecanonical view volume is doneby

2r−l 0 0 00 2

t−b 0 00 0 2

n−f 00 0 0 1

1 0 0 − l+r2

0 1 0 − b+t2

0 0 1 −n+f2

0 0 0 1

−z

x

y

(−1,−1, 1)

(1, 1,−1)

(1,−1,−1)

(l, b, n)

(r, t, f)





The orthographic projection matrix

We can combine the matrices into one:

Mo =m2 0 0 m

2 − 12

0 n2 0 n

2 − 12

0 0 1 00 0 0 1

2r−l 0 0 00 2

t−b 0 00 0 2

n−f 00 0 0 1

1 0 0 − l+r2

0 1 0 − b+t2

0 0 1 −n+f2

0 0 0 1

Given a point p in the orthographic view volume, we map it to apixel [i, j] as follows:

ij

zcanonical

1

= Mo

xp

yp

zp

1




Transforming the view frustumPerspective transform matrixHomogeneous coordinates

Transforming the view frustum

We have to transform the viewfrustum into the orthographic viewvolume. The transformation needs to

Map lines through the origin tolines parallel to the z axis

Map points on the viewingplane to themselves.

Map points on the far plane to(other) points on the far plane.

Preserve the near-to-far order ofpoints on a line.






(cf. book, fig. 7.12)





Perspective transform matrix

The following matrix does the trick:

Mp =

1 0 0 00 1 0 00 0 n+f

n −f0 0 1

n 0

We have Mp

xyz1

=

xy

z n+fn − f

zn

Hmmmm. . . is that what wewant. . . ?





Homogeneous coordinates

We have seen homogeneous coordinates before, but so far, thefourth coordinate of a 3D point has alway been 1.

In general, however, the homogeneous representation of a point(x, y, z) in 3D is (hx, hy, hz, h). Choosing h = 1 has just been aconvenience.

So we have:

Mp

xyz1

=

xy

z n+fn − f

zn

homogenize−−−−−−−−→

nxz

nyz

n + f − fnz

1





Homogeneous coordinates

Note: moving and scaling still work properly

For example: translation1 0 0 tx0 1 0 ty0 0 1 tz0 0 0 1

hxhyhzh

=

hx + htxhy + htyhz + htz

h

homogenize−−−−−−−−→

x + txy + tyz + tz

1




Aligning coordinate systemsTransformation matrix

Aligning coordinate systems

Given a camera coordinate systemwith origin e and orthonormal base(u, v, w), we need to tranformobjects in world space coordinatesinto objects in camera coordinates.

This can alternatively be consideredas aligning the canonical coordinatesystem with the camera coordinatesystem.




Aligning coordinate systemsTransformation matrix

Camera transformation

The required transformation is takencare of by Mv =

xu yu zu 0xv yv zv 0xw yw zw 00 0 0 1

1 0 0 −xe

0 1 0 −ye

0 0 1 −ze

0 0 0 1




Wrap-up

If we combine all steps, we get:

compute Mo

compute Mv

compute Mp

M = MoMpMv

for each line segment (ai,bi) dop = Mai

q = Mbi

drawline(xp/hp, yp/hp, xq/hq, yq/hq)


06 projection.slides (1)

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