Rasterization Pipeline

1Georgia Tech, IIC, GVU, 2006MAGIC MAGIC LabLab

http://www.gvu.gatech.edu/~jarekJarek Rossignac

Rasterization Pipeline

Rasterization Clipping



Rasterization costs

Per vertex– Lighting calculations– Perspective transformation– Slopes calculation

Per pixel– Interpolations of z, c (and leading trailing edges, amortized)– Read, compare, right to update the z-buffer and frame buffer



Acceleration techniques

Reduce per pixel cost– Reduce resolution– Back-face culling

Reduce per vertex cost– T-strips and fans– Frustum culling– Simplification– Texture map

Reduce both– Occlusion culling



Back-face culling

Done in screen space by hardware Test sign of (xB-xA) (yC-yA) - (xC-xA) (yB-yA)

AB

C



T-strips and fans

Reuse 2 vertices of last triangle

fan strip

Was important when hardware had only 3 registers

Less important in modern graphics boards

Saves nearly 2/3 of vertex processingA vertex is still processed twice on average



Frustum culling

Build a tight sphere around each object– Center it at C=((xmax+xmin)/2, (ymax+ymin)/2, (zmax+zmin)/2 )– Compute radius r as min||CV|| for all vertices V

Test the sphere against each half-space of the frustum– If N•(EC)<-r then sphere is outside

NE

C



Simplification

Collapse edges

NA B

Collapse A to B if |N•(AB)|<e Or use better error estimates



Vertex clustering (Rossignac-Borrel)

Subdivide box around object into grid of cells Coalesce vertices in each cell into one “attractor” Remove degenerate triangles

– More than one vertex in a cell– Not needed for dangling edge or vertex



Vertex clustering example



Hausdorff distance between T-meshes

Expensive to compute, – because it can occur away from vertices and edges!

AA BB

The point of B that The point of B that is furthest away is furthest away from A is closest to from A is closest to 3 points on A that 3 points on A that are inside facesare inside faces



What is the error produced by edge collapses?

Volume of the solid bounded by old and new triangles– requires computing their intersection

Exact Hausdorff distance between old and new triangles– expensive to compute precisely

Distance between v and planes of all triangles incident on collapsed vertices– one dot product per plane (expensive, but guaranteed error bound)

Quadratic distance between v and planes old triangles– evaluate one quadratic form representing all the planes (cheap but not a bound)– leads to closed form solution for computing the optimal V’ (in least square sense)

Estimating edge-collapse error

v’v’ vv



Measuring Error with Quadrics (Garland) Given plane P, and point v, we can define a quadric distance Q(v) = D(v,P)2

– P goes through point p and has normal N– D(v,P) = pvN– Q(p) = (pvN)2, which is a quadratic polynomial in the coordinatges of v– If v=(x,y,z), Q(p)=a11x2+a22y2+a33z2+2a23yz+2a13xz+2a12xy+2b1x+2b2y+2b3z+c

Q may be written represented by 3x3 matrix A,a vector b,and a scalar c Q(v) may be evaluated as v(Av)+2bv+c

– or in matrix form as:

[ ] [ ] czyx

bbbzyx

aaaaaaaaa

zyx

cQ

+⎥⎥⎦

⎤⎢⎢⎣

⎡+⎥

⎥⎦

⎤⎢⎢⎣

⎡⎥⎥⎦

⎤⎢⎢⎣

⎡

++=

321332313232212131211

2

2)( vbvAvv TT



Optimal position for v Pick v to minimize average distance to planes Q(v)

– set dQ(v)/dv = 0, which yields: v = -A-1 b



Q(v) measures flatness Isosurfaces of Q

– Are ellipsoids– Stretch in flat direction– Have eigenvalues proportional to the principal curvatures



Texture map

Replace distant details with their images, pasted on simplified shapes or even no flat bill-boards



Texture mapping

textureMode(NORMALIZED); PImage maya = loadImage(maya.jpg"); beginShape(QUADS); texture(maya); vertex(-1, 1, 1, 1, 1); vertex( 1, 1, 1, 0, 1); vertex( 1, -1, 1, 0, 0 ); vertex(-1, -1, 1, 1, 0); …endShape();



Occlusion culling

Is ball containing an object hidden behind the triangle?



Occlusion culling in urban walkthrough

From Peter Wonka Need only display a few facades--the rest is hidden



Shadows / visibility duality Point light source

– Shadow (volume): set of points that do not see the light From-point visibility

– Occluded space: set of points hidden from the viewpoint

Area light source (assume flat for simplicity)– Umbra: set of points that do not see any light– Penumbra: set of points that see a portion of the light source– Clearing: set of points that see the entire light source

From-cell visibility (cell C)– Hidden space: set of points hidden from all points in C– Detectable space: set of points seen from at least one point in C– Exposed space: set of points visible from all points in C



From-point visibility What is visible from a given viewpoint

– Not from a given area

Usually computed at run-time for each frame– Should be fast– Or done in background before the user starts looking around

May be pre-computed for a set of important viewpoints– Can be slower– User will be invited to jump to these (street crossings)– Or to navigate along pre-computed paths (videos)

May be useful for pre-computing from-cell visibility– Which identifies all objects B that are invisible when the

viewpoint is in a cell A– The cell may be reduced to a trajectory (taxi metaphor)



From-point visibility

Viewpoint/light = {a} (a single point)

HIDDEN DETECTED CLEAR



Zen break

The moon is a perfect ball It is full and perfectly clear You see it through your window You wonder about the meaning of life You trace its outline on the glass

Is is a circle?

Can you prove it?



Nothing is perfect

In general, the outline of the moon is an ellipse Proof:

– The pencil of rays from your eye to the moon is a cone– The intersection of that cone with the window if a conic– If the moon is clear and visible through the window: it is an ellipse– When the window is orthogonal to the axis of the cone: it is a

circle



Cherchez la lune

The moon is a ball B(c,r) of center c and radius r The occluder is a triangle with vertices d, e, and f Assume viewer is at point a Write the geometric formulae for testing whether the moon

is hidden behind the triangle

HIDDEN DETECTED CLEARde

f

c



When does a triangle appear clockwise?

When do the vertices d, e, f, of a triangle appear clockwise when seen from a viewpoint a?

clockwise de

f

counter-clockwiseed

f



A triangle appears clockwise when

The vertices d, e, f, of a triangle appear clockwise when seen from a viewpoint a when s(a,d,e,f) is true

Procedure s(a,d,e,f) {RETURN ad•(aeaf)>0}– ad•(aeaf) is a 3x3 determinant– It is called the mixed product (or the triple product)

clockwise

de

f

counter-clockwise

ed

f

a a



Testing a point against a half-space?

Consider the plane through points d, e, and f– It splits space in 2 half-spaces

Consider the half-space H of all points that see d, e, f clockwise

How to test whether point c is in H?

clockwise

de

f

c



Testing a point against a half-space

Consider the plane through points d, e, and f Consider the half-space H of all points that see d, e, f

clockwise A point c is in H when s(c,d,e,f)

s(c,d,e,f) {RETURN cd•(cecf)>0}

clockwise

de

f

c



Is a point hidden by a triangle?

Is point c hidden from viewpoint a by triangle (d, e, f)?– Write down all the geometric test you need to perform

de

f

a

c



Test whether a point is hidden by a triangle

To test whether c is hidden from a by triangle d, e, f do– IF NOT s(a,d,e,f) THEN swap d and e– IF NOT s(a,d,e,c) THEN RETURN FALSE– IF NOT s(a,e,f,c) THEN RETURN FALSE– IF NOT s(a,f,d,c) THEN RETURN FALSE– IF s(c,d,e,f) THEN RETURN FALSE– RETURN TRUE Clockwise from a

de

f

a

c



Testing a ball against a half-space?

Consider the half-space H of all points that see (d, e, f) clockwise

When is a ball of center c and radius r in H?

Clockwise from a

de

f

c



Testing a ball against a half-space

Consider the half-space H of all points that see (d, e, f) clockwise

A ball of center c and radius r is in H when:

ec • (edef) > r ||edef||

Clockwise from a

de

f

c



Is the moon hidden by a triangle?

Is the ball of center c and radius r hidden from a by triangle with vertices d, e, and f?

Clockwise from ade

f

a

c



Is the moon hidden by a triangle?

To test whether the ball of center c and radius r hidden from a by triangle with vertices d, e, and f do:– IF NOT s(a,d,e,f) THEN swap d and e– IF ac • (adae) < r || adae || THEN RETURN FALSE– IF ac • (aeaf) < r || aeaf || THEN RETURN FALSE– IF ac • (afad) < r || afad || THEN RETURN FALSE– IF ec • (edef) < r || edef || THEN RETURN FALSE– RETURN TRUE

Explain what each line tests Are these test correct? Are they sufficient?

Clockwise from ade

f

a

c



Is the moon hiding a triangle?

Triangle d, e, f is hidden from viewpoint a by a ball of center c and radius r when its 3 vertices are (hidden).– Proof?– How to test it?

Hidden moond

f

c

Hidden triangle

d

f

c

ee



When is a planar set A convex?

seg(p,q) is the line segment from p to qA is convex ( pA qA seg(p,q)A )

pp qq

ppqq

pp

qq



What is the convex hull of a set A?

The convex hull H(A) is the intersection of all convex sets that contain A.



Is a convex occluder hiding a polyhedron?

A polyhedron P is hidden by a convex occluder O if all of its vertices are– Proof?– Is a subdivision surface hidden if its control vertices are?

We do not need to test all of the vertices– Only those of a container (minimax box, bounding ball, convex

hull)• Proof?

– Only those on the silhouette of a convex container• Proof?

What if the occluder is not convex?– Can it still work?



Strategies for non-convex occluders?

What if O is not convex?– replace O by an interior ball

• Not easy to compute a (nearly) largest ball inside O• The average of the vertices of O may be outside of O

– replace O by an interior axis-aligned box• Not easy to compute a largest box inside

– replace O by a convex set it contains• O may be too thin

Andujar, Saona NavazoCAD, 32 (13), 2000



The Hoop principle

The occluder does not have to be convex, provided that a portion of its surface appears convex to the viewer and that the object is behind it.



Hoop from non-convex occluders

Pick a candidate occluder S– large triangle mesh – close to the viewer

Select a set D of triangles of S whose border C is a hoop:– single loop– that has a convex projection on the screen

Build a shield O– on plane behind D– bounded by projection of C

O is a convex occluder– anything hidden by O– is hidden by D– and hence by S– sure?

OCD

S



When does a 3D polyloop appear convex?

Each pair of consecutive edges defines a linear half-space from where their common vertex appears convex.

The portion of space from which a loop C of edges appears convex is the intersection of these half-spaces or of their complements.



Two strategies for building hoops

Grow a patch D on the surface S, or Grow a curve C on the boundary of interior octree cells

– simplify it, ensuring that the simplified border lies inside the solid



Testing against a hoop

Given a hoop C, we can easily test whether the viewpoint or a whole cell sees it as convex.– Test whether the viewpoint (or whether all the vertices of the

cell,) lie in the intersections of all half-spaces.

But this is not sufficient if we want to construct a shield from it.– Why not?We must ensure that the shield O is behind the portion D

of the original surface S that is bounded by C.



Ensuring that O is behind D

Store with the hoop the thickness

(depth of projection of D on the normal to C)

Compute normal using sum of cross-products

Use it to place the shield O sufficiently behind the hoop C so that it does not intersect D

C

D

n O

CD

n O



Ensuring that D occludes O

Store with the hoop the thickness

(range of projection of D on the normal n to C)

Ensure that the viewer is not in that range

… or shoot a ray from viewer to O and check parity of hits with D

CD

n O

CD

n O



Growing a hoop around the moon

Find the first intersection, i, between ac and the mesh of S Consider the triangle (d, e, f) of S that contains i If the triangle hides the moon, we are done (return hidden) Otherwise grow it by adding neighboring triangles

– Let C be the boundary of this patch D of triangles– Identify bad vertices of C that appear in front of the moon– Add triangles to make these vertices interior– Ensuring that C remains a simple loop– Stop when C has no bad vertices (return hidden)– or when you can’t grow (return not hidden)

Test the surface patch D bounded by C– Count intersections with ac– If odd then return hidden

Clockwise from ade

f

a

c



Cell-to-cell visibility

Now, let us look at the problem of pre-computing whether any portion of a cell A is visible from a viewpoint in cell B.

Why? (Applications?)



Notation and assumptions

Two sets, A and B Two points aA, bB O is the occluder A, B, and O are disjoint ab is the line segment from a to b a–b O ab = , a sees b (ignore self-occlusion) a | b O ab ≠ , a is occluded from by O

a b

BA

O

a b

BA

O



(Jarek’s) Terminology and definitions? HIDDEN: HO(A,B)

– B is entirely hidden from A by O– ?

DETECTED: DO(A,B)– B can be detected from any point of A– ?

GUARDED: GO(A,B)– B can be guarded from a single point of A– ?

CLEAR: CO(A,B)– B can be guarded from any point of A– ?

a bBA O

BA

BAO

a bBA O

O



HIDDEN: HO(A,B)– B is entirely hidden from A– HO(A,B) a, b: a | b

DETECTED: DO(A,B)– B can be detected from any point of A– DO(A,B) a, b: a–b

GUARDED: GO(A,B)– B can be guarded from some point of A– GO(A,B) a, b: a–b

CLEAR: CO(A,B)– B can be guarded from any point of A– CO(A,B) a, b: a–b

Definitions

a bBA O

BAO

BAO

a bBA O



Properties (and proofs?)

CO(A,B) CO(B,A) and HO(A,B) HO(B,A) CO(A,B) GO(A,B) and GO(B,A) GO(A,B) DO(B,A) proof ?

a bBA O

HO(A,B) a, b: a | b

BAO

DO(A,B) a, b: a–b

BAO

GO(A,B) a, b: a–b

a bBA O

CO(A,B) a, b: a–b



Properties and proofs

CO(A,B) CO(B,A) and HO(A,B) HO(B,A) CO(A,B) GO(A,B) and CO(A,B) GO(B,A) GO(A,B) DO(B,A) proof

GO(A,B) a, b: a–b b, a: b–a DO(B,A)

a bBA O

HO(A,B) a, b: a | b

BAO

DO(A,B) a, b: a–b

BAO

GO(A,B) a, b: a–b

a bBA O

CO(A,B) a, b: a–b



Counter-example?

Draw O such that none of the 8 relations hold

a bBA O

HO(A,B) a, b: a | b

BAO

DO(A,B) a, b: a–b

BAO

GO(A,B) a, b: a–b

a bBA O

CO(A,B) a, b: a–b

BAO?

HO(A,B), GO(A,B), DO(A,B), CO(A,B), HO(B,A), GO(B,A), DO(B,A), CO(B,A) are all false



Counter-example?

Draw O such that none of the 8 relations hold

a bBA O

HO(A,B) a, b: a | b

BAO

DO(A,B) a, b: a–b

BAO

GO(A,B) a, b: a–b

a bBA O

CO(A,B) a, b: a–b

HO(A,B), GO(A,B), DO(A,B), CO(A,B), HO(B,A), GO(B,A), DO(B,A), CO(B,A) are all false

a b BAO



Cell-to-cell visibility (C2CV)

Viewpoint is in cell A Small objects in cell B need not be rendered if cell A is occluded from cell B by some surface S:

HS(A,B)?



Terminology Assume that surface S, cell A, and cell B are disjoint? R = convex hull of AB (contains all rays from A to B) T = tube (boundary of R not on A or B) C connected component of ST



The Occlusion TheoremThe surface S occludes cell A from cell B, if and only if:

• There exists a connected component, C, of ST that is bounding a portion D of S and has no boundary in the interior of R.

2. Given an arbitrary (non-singular) ray, ab, from aA to bB, ab has an odd number of intersections with DC.



Proof



Various cases



ShieldTester for triangle meshes

Obtain an edge eAB of T from a vertex of A to a vertex of BIF eAB S = THEN RETURN (“visible”)FOR EACH triangle x of S that is stabbed by eAB DO { Start a WALK that traces a connected component C of ST IF the WALK did not run into a boundary of S THEN {

INVADE the portion of S to the left of the triangles crossed by the WALK along C

IF the invaded portion has no hole THEN IF the number of triangles that were stabbed by eAB

and visited by the WALK and the INVADE is odd THEN RETURN (“occluded”)}

RETURN (“undecided”)



Walking one step



Turning left on S



Invade process



Hot edges



Results



Is the moon hidden by a tree? Can’t grow a patch of edge-connected triangles

– Leaves/branches are small and disjoint Use approximate shadow fusion

– Render tree(s)– Mark area of rendered pixels– Construct big rectangle inside it (how?)– Let M= max(z-buffer inside rectangle)– Push back rectangle to z=M– Use it as occluder for other objects (behind the tree)

Accuracy depends on screen resolution– Can be improved by multiple passes (zoomed 1/4 of image each)

Use bounding spheres around objects to test them– Conservative part– Use a hierarchy of bounding sphere (hierarchical z-buffer)



Occluded volumes

Assume that we have an area light source or a cell containing the viewpoint that is a polyhedron

Assume that all occluders are polyhedra Can we compute the hidden volume (umbra)?

Difficult because it is bounded by curved surfaces



What you must remember

How to build a rasterizer How to accelerate it



Examples of questions for exams

When is it no-longer useful to simplify the model? What are the 7 acceleration techniques? Which ones reduce per-vertex cost? How much savings should you expect from each technique?

Documents

Rasterization Pipeline