Andries van Dam September 18, 2003 3D Viewing III

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Andries van Dam September 18,
2003 3D Viewing III
CIS 736 Computer Graphics Review of Basics 5 of
5 Viewing III Friday, 06 February 2004 Reading
Transformations Adapted with Permission W. H.
Hsu http//www.kddresearch.org
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Andries van Dam September 18,
2003 3D Viewing III 1/42
Viewing in Three Dimensions
Mathematics of Projections

• Review from last time
• specifying a 3D viewsetting up the synthetic
  camera
• the shape of the 3D view volume which is the
  clipping region
• The mathematics of planar geometric projections
• how do we get from the view specification to the
  2D picture?
• Lecture roadmap
• deriving 2D picture from 3D view parameters is a
  hard problem
• our method relies on the fact that its easier to
  make a picture from the canonical view volume (3D
  parallel projection cuboid) at the canonical view
  position (camera at the origin, looking down the
  negative z-axis) than it is from the users
  specified arbitrary position. Well break the
  process into three easier steps
• get parameters for view specification (covered in
  last lecture)
• transform from specified view volume into
  canonical view volume
• using canonical view, clip and project scene to
  make 2D picture

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Stage One Specifying a View
Volume

• Reduce degrees of freedom five steps to
  specifying view volume
• position the camera (and therefore its view/film
  plane)
• point it at what you want to see, with the camera
  in the orientation you want
• define the field of view (for a perspective view
  volume, aspect ratio of film and angle of view
  somewhere between wide angle, normal, and zoom
  for a parallel view volume, width and height)
• choose perspective or parallel projection
• determine the focal distance

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Examples of a View Volume (1/2)

• Perspective Projection Truncated Pyramid
  Frustum
• Look vector is the center line of the pyramid,
  the vector that lines up with the barrel of the
  lens
• Note For ease of specification , up vector need
  not to be perpendicular to Look vector, but they
  cannot be collinear

Height angle Aspect ratio
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Examples of a View Volume (2/2)

• Orthographic Parallel Projection Truncated View
  Volume Cuboid
• Orthographic parallel projection has no view
  angle parameter

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Specifying Arbitrary 3D Views

• Placement of view volume (visible part of world)
  specified by cameras position and orientation
• Position (a point)
• Look and Up vectors
• Shape of view volume specified by
• horizontal and vertical view angles
• front and back clipping planes
• Perspective projection projectors intersect at
  Position
• Parallel projection projectors parallel to Look
  vector, but never intersect (or intersect at
  infinity)
• Coordinate Systems
• world coordinates standard right-handed xyz
  3-space
• viewing reference coordinates camera-space
  right handed coordinate system (u, v, n) origin
  at Position and axes rotated by orientation used
  for transforming arbitrary view into canonical
  view

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Andries van Dam September 18,
2003 3D Viewing III 6/42
Arbitrary View Volume Too
Complex

• We have now specified an arbitrary view using our
  viewing parameters
• Problem map arbitrary view specification to 2D
  picture of scene. This is hard, both for
  clipping and for projection
• Solution reduce to a simpler problem and solve
• Note Look vector along negative, not positive,
  z-axis is arbitrary but makes math easier

• there is a view specification from which it is
  easy to take a picture. Well call it the
  canonical view from the origin, looking down the
  negative z-axis

think of the scene as lying behind a window and
were looking through the window

• parallel projection
• sits at origin Position (0, 0, 0)
• looks along negative z-axis Look vector (0,
  0, 1)
• oriented upright Up vector (0, 1, 0)
• film plane extending from 1 to 1 in x and y

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Normalizing to the Canonical
View Volume

• Our goal is to transform our arbitrary view and
  the world to the canonical view volume,
  maintaining the relationship between view volume
  and world, then take picture
• for parallel view volume, transformation is
  affine made up of translations, rotations, and
  scales
• in the case of a perspective view volume, it also
  contains a non-affine perspective transformation
  that frustum into a parallel view volume, a
  cuboid
• the composite transformation that will transform
  the arbitrary view volume to the canonical view
  volume, named the normalizing transformation, is
  still a 4x4 homogeneous coordinate matrix that
  typically has an inverse
• easy to clip against this canonical view volume
  clipping planes are axis-aligned!
• projection using the canonical view volume is
  even easier just omit the z-coordinate
• for oblique parallel projection, a shearing
  transform is part of the composite transform, to
  de-oblique the view volume

Affine transformations preserve parallelism but
not lengths and angles. The perspective
transformation is a type of non-affine
transformation known as a projective
transformation, which does not preserve
parallelism
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Viewing Transformation
Normalizing Transformation

• Problem of taking a picture has now been reduced
  to problem of finding correct normalizing
  transformation
• It is a bit tricky to find the rotation component
  of the normalizing transformation. But it is
  easier to find the inverse of this rotational
  component (trust us)
• So well digress for a moment and focus our
  attention on the inverse of the normalizing
  transformation, which is called the viewing
  transformation. The viewing transformation turns
  the canonical view into the arbitrary view, or
  (x, y, z) to (u, v, n)

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Building Viewing Transformation
from View Specification

• We know the view specification Position, Look
  vector, and Up vector
• We need to derive an affine transformation from
  these parameters that will translate and rotate
  the canonical view into our arbitrary view
• the scaling of the film (i.e. the cross-section
  of the view volume) to make a square
  cross-section will happen at a later stage, as
  will clipping
• Translation is easy to find we want to translate
  the origin to the point Position therefore, the
  translation matrix is
• Rotation is harder how do we generate a rotation
  matrix from the viewing specifications that will
  turn x, y, z, into u, v, n?
• a digression on rotation will help answer this

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2003 3D Viewing III 10/42
Rotation (1/5)

• 3 x 3 rotation matrices
• We learned about 3 x 3 matrices that rotate the
  world (were leaving out the homogeneous
  coordinate for simplicity)
• When they do, the three unit vectors that used to
  point along the x, y, and z axes are moved to new
  positions
• Because it is a rigid-body rotation
• the new vectors are still unit vectors
• the new vectors are still perpendicular to each
  other
• the new vectors still satisfy the right hand
  rule
• Any matrix transformation that has these three
  properties is a rotation about some axis by some
  amount!
• Lets call the three x-axis, y-axis, and
  z-axis-aligned unit vectors e1, e2, e3
• Writing out

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Rotation (2/5)

• Lets call our rotation matrix M and suppose that
  it has columns v1, v2, and v3
• When we multiply M by e1, what do we get?
• Similarly for e2 and e3

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Rotation (3/5)

• Thus, for any matrix M, we know that Me1 is the
  first column of M
• If M is a rotation matrix, we know that Me1
  (i.e., where e1 got rotated to) must be a
  unit-length vector (because rotations preserve
  length)
• Since Me1 v1, the first column of any rotation
  matrix M must be a unit vector
• Also, the vectors e1 and e2 are perpendicular
• So if M is a rotation matrix, the vectors Me1 and
  Me2 are perpendicular (if you start with
  perpendicular vectors and rotate them, theyre
  still perpendicular)
• But these are the first and second columns of M
  Ditto for the other two pairs
• As we noted in the slide on rotation matrices,
  for a rotation matrix with columns vi
• columns must be unit vectors vi 1
• columns are perpendicular vi vj 0 (i ? j)

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Rotation (4/5)

• Therefore (for rotation matrices)
• We can write this matrix of vivj dot products as
• where MT is a matrix whose rows are v1, v2, and
  v3
• Also, for matrices in general, M-1M I,
  (actually, M-1 exists only for well-behaved
  matrices)
• Therefore, for rotation matrices only we have
  just shown that M-1 is simply MT
• MT is trivial to compute, M-1 takes considerable
  work big win!

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Rotation (5/5)

• Summary
• If M is a rotation matrix, then its columns are
  pairwise perpendicular and have unit length
• Inversely, if the columns of a matrix are
  pairwise perpendicular and have unit length and
  satisfy the right-hand rule, then the matrix is a
  rotation
• For such a matrix,

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Building the Orientation Matrix

• Now we know that a rotation matrix is a matrix
  that will turn the perpendicular unit vectors e1,
  e2, and e3 into the new perpendicular unit
  vectors v1, v2, and v3
• Additionally, we know that the columns of this
  matrix will be the vectors v1, v2, and v3
• Lets call the vectors e1, e2, and e3 the
  world-space axes x, y, and z, and lets call the
  vectors v1, v2, and v3 the camera-space unit
  vector axes u, v, and n
• We want to build a rotation matrix that will
  normalize (u, v, n) into (x, y, z). We just found
  a matrix, M, that will turn (x, y, z) into (u, v,
  n). Conversely, M-1MT turns (u, v, n) into (x,
  y, z). The rows of MT will just be u, v, and n
• Reduces the problem of finding the correct
  rotation matrix into finding the correct
  perpendicular unit vectors u, v, and n
• Restatement of rotation problem find u, v, and n
  using Position, and the Look vector and Up
  vector, then applying MT to get the x, y, z unit
  vectors e1, e2, e3

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2003 3D Viewing III 16/42
Finding u, v, and n from Position, Look, and Up
(1/6)

• We know that the (u, v, n) axes we want will have
  the properties that
• our arbitrary Look Vector will lie along the
  negative n-axis
• a projection of the Up Vector into the plane
  perpendicular to the n-axis will lie along the
  v-axis
• The u-axis will be mutually perpendicular to the
  v and n-axes, and will form a right-handed
  coordinate system
• Plan of attack first find n from Look, then find
  u from n and Up, then find v from n and u

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Finding u, v, and n (2/6)

• Finding n
• Finding n is easy. Look vector in canonical
  volume lies on z. Since z maps to n, n is a
  normalized vector pointing in the opposite
  direction from our arbitrary Look vector

Look
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Finding u, v, and n (3/6)

• Finding u
• Problem the orientation of the u-axis is
  constrained by the v and n-axes . . . But we
  dont have a v-axis yet. How can we proceed?
• Solution we will use the Up vector instead of
  the v-axis. The only problem with this is that
  (u, Up, n) will not necessarily form an
  orthogonal basis, i.e., mutually perpendicular
  and unit vectors (because Up is at an arbitrary
  angle to n). Well correct this later.
• The constraints on u are
• 1. must be perpendicular to plane spanned by
  n-axis and Up vector
• 2. (u, Up, n) must form a right-handed (though
  not mutually perpendicular) coordinate system
• To satisfy (1), we can use cross-product, but
  which one should we use?
• n X Up and Up X n are both perpendicular to the
  plane, but in different directions . . .
• Answer cross-products are right-handed, so use
  Up X n to satisfy (2) as well
• Use the cross-product applet to convince yourself
  that order matters in cross-product
  Applets-gtLinear Algebra-gtCross Product

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Finding u, v, and n (4/6)

• Finding u, contd.
• So, the formula is
• Note n is unit length, but Up vector might not
  be unit length or perpendicular to n, so we had
  to normalize u
• Problem what happens if Up and n are parallel?
• Answer the length Up x n is zero, which
  means we would divide by 0 in the above equation.
  Bad stuff, dont let it happen!
• try it out on the applet
• As a reminder, the cross product of two vectors a
  and b is

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2003 3D Viewing III 20/42
Finding u, v, and n (5/6)

• Finding v
• From previous slide, we currently have (u, Up, n)
  but this coordinate system is not mutually
  perpendicular
• Finding v is just as easy as finding u v must be
  perpendicular to u and n, and (u, v, n) must be a
  right-handed coordinate system
• Again using right hand rule
• Do not need to normalize v . . . why? Since n
  and u are unit length and mutually perpendicular,
  then their cross product is also unit length (see
  the applet!).

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Finding u, v, and n (6/6)

• To summarize
• The viewing transformation is now fully specified
• knowing u, v, and n, we can rotate the canonical
  view into the user-specified orientation
• we already know how to translate the view
• Important Note that we dont actually apply the
  forward viewing transformation. Instead, the
  inverse viewing transformation, namely the
  normalizing transformation, will be used to map
  the arbitrary view into the canonical view

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Transforming to the Canonical
View

• The Viewing Problem for Parallel Projection
• Given a parallel view specification and vertices
  of a bunch of objects, we use the normalizing
  transformation, i.e., the inverse viewing
  transformation, to normalize the view volume to a
  cuboid at the origin, then clip, and then project
  by ignoring z
• The Viewing Problem for Perspective Projection
• Normalize the perspective view specification to a
  unit frustum at the origin looking down the z
  axis then transform the perspective view volume
  into a parallel (cuboid) view volume, simplifying
  both clipping and projection

• Note its a cuboid, not a cube (makes
  transformation arithmetic and clipping easy)

(1, 1, 0)
Back clip plane transforms to z -1
Front clip plane transforms to z 0
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Steps for Normalizing View
Volume

• The Parallel Case
• Decomposes into multiple steps
• Each step defined by a matrix transformation
• The product of these matrices defines the whole
  transformation in one, more complex, composite
  matrix. The steps are
• move the eye/camera to the origin
• transform the view so that (u, v, n) is aligned
  with (x, y, z)
• adjust the scales so that the view volume fits
  between 1 and 1 in x and y, the back clip plane
  lies at z 1, the front plane at z 0
• The Perspective Case
• Same as parallel, but add one more step
• distort pyramid to cuboid to achieve perspective
  distortion and aligns the front clip plane with z
  0

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2003 3D Viewing III 24/42
Perspective (1/4)

• Get the eye to the origin
• We want a matrix that will take
• (Posx, Posy, Posz) to (0, 0, 0)
• Solution its just the inverse of the
  translation we found for producing the viewing
  transform translation
• (tx, ty, tz) (Posx, Posy, Posz)
• We will take the matrix
• and we will multiply all vertices explicitly
  (and the camera implicitly) to preserve
  relationship between camera and scene, i.e., for
  all vertices p
• This will move Position to (0, 0, 0)

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2003 3D Viewing III 25/42
The Current Situation

• Position now at origin
• But were hardly done! Still need to
• align orientation with x,y,z world coordinate
  system
• normalize proportions of the view volume

y
Look
z
x
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Perspective (2/4)

• Rotate the view and align with the world
  coordinate
• system
• We found out that the view transformation matrix
  M with columns u, v, and n would rotate the x, y,
  z axes into the u, v, and n axes
• We now apply inverse (transpose) of that
  rotation, MT, to scene. That is, a matrix with
  rows u, v, and n will rotate the axes u, v, and n
  into the axes x, y, and z
• Now every vertex in the scene (and the camera
  implicitly) get multiplied by the composite
  matrix
• Weve translated and rotated, so that the
  Position is at the origin, and the (u, v, n)
  coordinates and the (x, y, z) coordinates are
  aligned

current situation
Look
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2003 3D Viewing III 27/42
Perspective (3/4)

• Weve gotten things more or less to the right
  place, but the proportions of the view volume
  need to be normalized. Last affine
  transformation scaling
• Adjust so that the corners of far clipping plane
  eventually lie at (1, 1, 1)
• One mathematical operation works for both
  parallel and perspective view volumes
• Imagine vectors emanating from origin passing
  through corners of far clipping plane. For
  perspective view volume, these are edges of
  volume. For parallel, these lie inside view
  volume
• First step force vectors into 45-degree angles
  with x and y axes
• Well do this by scaling in x and y

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2003 3D Viewing III 28/42
Perspective (4/4)

• Scaling Clipping Planes
• Looking down from above, we see this
• Want to scale in x to make angle 90 degrees
• Need to scale in x by
• Similarly in y

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The xy scaling matrix

• The scale matrix we need looks like this
• So our current composite transformation looks
  like this

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One more scaling matrix

• Relative proportions of view volume planes now
  correct, but back clipping plane probably lying
  at some z ? 1
• We need to shrink the back plane to be at z 1
• The z distance from the eye to that point has not
  changed its still far (distance to far clipping
  plane)
• If we scale in z only, proportions of volume will
  change instead well scale uniformly

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The Current Situation

• Far plane at z 1.
• Near clip plane now at z k (note k gt 0)

y
z
z -1
x
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The Results

• Our near-final composite normalizing
  transformation for the canonical perspective view
  volume is
• Exactly the same tricks suffice to get the
  parallel view volume to the canonical one

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The Perspective Transformation
(1/5)

• Weve put the perspective view volume into
  canonical position, orientation and size
• Lets look at a particular point on the original
  near clipping plane lying on the Look vector
• It gets moved to a location
• on the negative z-axis, say

Look
Position
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The Perspective Transformation
(2/5)

• What is the value of k? Trace through the steps.
• p first gets moved to just (near)Look
• This point is the rotated to (near)(e3)
• The xy scaling has no effect, and the xyz
• scaling changes this to , so

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The Perspective Transformation
(3/5)

• We now transform the points in the standard
  perspective view volume between k and 1 into
  the standard parallel view volume
• The z-buffer, used for visible surface
  calculation, needs the z values to be 0 1, not
  1 0. So the perspective transformation must
  also transform the scene to the range z 0 to z
  1
• The matrix
• that does this is just
• (Remember that
• 0lt k lt1 )

Flip the z-axis and unhinge
Note not 1!
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The Perspective Transformation
(4/5)

• Take a typical point and rewrite it
  as
• where p is parametrized by
  distance along frustum
• when d 0, p is on near clip plane when d
  (1-k), p is on far clip plane
• depending on x, y, and z, p may or may not
  actually fall within frustum
• Apply D (from previous slide) to p to get new
  point p
• Note must divide through by !!!
• this causes x and y to be perspectivized, with
  points closest to near clip plane being scaled up
  most
• understanding homogenous coordinates is essential
  for correct viewing

Try values of d 0, 1-k, ½(1-k), -1, 1
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The Perspective Transformation
(5/5)

• Again consider
• What happens to x and y when d gets very large?
• note when we let d increase beyond (1-k), p is
  now beyond the frustum (such lines will be
  clipped)
• This result provides perspective foreshortening
  parallel lines converge to a vanishing point
• What happens when d is negative?
• here p is now in front of the near clip plane,
  possibly behind the eye point (COP)
• result when (kd) becomes negative, the sign of
  x and y will be flipped text would be
  upside-down and backwards
• you wont see these points because they are
  clipped
• What happens after perspective transformation?
• Answer parallel projection is applied to
  determine location of points onto the film plane
• projection is easy drop z!
• however, we still need to keep the z ordering
  intact for visible surface determination

Andries van Dam September 18,
2003 3D Viewing III 37/42
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2003 3D Viewing III 38/42
The End Result

• Final transformation
• Note that once the viewing parameters (Position,
  Up vector, Look vector, Height angle, Aspect
  ratio, Near, and Far) are known, the matrices
• can all be computed and multiplied together to
  get a single 4x4 matrix that is applied to all
  points of all objects to get them from world
  space to the standard parallel view volume. This
  is a huge win for homogeneous coordinates

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Clipping

• We said that taking the picture from the
  canonical view would be easy final steps are
  clipping and projecting onto film plane
• We need to clip scene against sides of view
  volume
• However, weve normalized our view volume into an
  axis-aligned cuboid that extends from 1 to 1 in
  x and y and from 0 to 1 in z
• Note This is the flipped (in z) version of the
  canonical view volume
• Clipping is easy! Test x and y components of
  vertices against /-1. Test z components against
  0 and 1

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Clipping (cont.)

• Vertices falling within these values are saved,
  and vertices falling outside get clipped edges
  get clipped by knowing x,y or z value at an
  intersection plane. Substitute x, y, or z 1 in
  the corresponding parametric line equations to
  solve for t
• In 2D

(x1, y1, z1 )
t1
t0
(x0, y0, z0 )
y
(1, 1)
(x1, y1 )
x1
x
(x0, y0 )
(-1, -1)
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Projection

• We can make a picture by the simple strategy of
  taking each point and ignoring z to project it
  onto the xy-plane
• A point (x,y,z) where
• turns into the point (x, y) in screen space
  (assuming viewport is the entire screen) with
• by
• - ignoring z
• -
• -
• If viewport is inside a Window Managers window,
  then we need to scale down and translate to
  produce window coordinates
• Note because its a parallel projection we could
  have projected onto the front plane, the back
  plane, or any intermediate plane the final
  pixmap would have been the same

1. Note that these functions are not exactly
correct since if x or y is ever 1, then we will
get x or y to be 1024 which is out of our
range, so we need to make sure that we handle
these cases gracefully. In most cases, making
sure that we get the floor of 512(x1) will
address this problem since the desired x will be
less than 1.
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Summary

• In summary, the entire problem can be reduced to
  a composite matrix multiplication of vertices,
  clipping, and a final matrix multiplication to
  produce screen coordinates. The final composite
  matrix (CTM) is the composite of all modeling
  (instance) transformations (CMTM) accumulated
  during scene graph traversal from root to leaf,
  composited with the final composite normalizing
  transformation N applied to the root/world
  coordinate system
• Thats all there is to it, folks!