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Visible Surface Determination CS32310 H Holstein Oct 2012 1

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Visible Surface Determination. CS32310 H Holstein Oct 2012. Visible Surface Determination. Hidden surface removal Why? Occlusion from view point General approach Depth sort. Sorting can be carried out in - PowerPoint PPT Presentation

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Page 1: Visible Surface Determination

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Visible Surface Determination

CS32310H HolsteinOct 2012

Page 2: Visible Surface Determination

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Visible Surface Determination

• Hidden surface removal• Why?• Occlusion from view point• General approach– Depth sort

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• Sorting can be carried out in – image precision - that is, a decision is made at

each pixel, at screen resolution, asto which surface the pixel represents, and thus, how it is to be rendered.

– object precision - that is, decisions are based on the geometry of the objects, irrespective of screen resolution, but dependent (in practice) of the available floating point precision.

– a mixture of object precision (eliminate certain surfaces from consideration) and image precision (resolve the remaining conflicts at the pixel level).

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• If variable resolution is required (zooming),– then image precision methods have to be

reworked entirely, – while object precision methods (using sorting

independently of image resolution) can use their previous results, and need only render on the finer (or coarser) pixel grid.

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• Complexity– If a scene has n surfaces, we might expect an object

precision algorithm to take O(n2) time, since every surfaces may have to be tested against every other surface for visibility.

– On the other hand, if the are N pixels, we might expect an image precision algorithm to take O(nN) time, since every pixel may have to be tested for the visibility in n surfaces.

– Since the number of pixels N usually greatly exceeds the number of surfaces, the number of decisions to be made is much fewer in the object precision case.

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• Complexity continued– Different algorithms try to reduce these basic

counts. – Thus, one can consider bounding volumes (or

“extents”) to determine roughly whether objects cannot overlap – this reduces the sorting time.

– With a good sorting algorithm, O(n2) may be reducible to a more manageable O(n log n).

– Concepts such as depth coherence (the depth of a point on a surface may be predicable from the depth known at a nearby point) can cut down the number of arithmetic steps to be performed.

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z-Buffer approach

• Image precision algorithms may benefit from hardware acceleration.

• The z-buffer algorithm requires a depth buffer (z-buffer) d[x][y] to record the nearest the rendering of the nearest point encountered so far, to by placed at pixel (x,y).

• The algorithm relies on the fact that if a nearer object occupying (x,y) is found, then the depth buffer is overwritten with the rendering information from this nearer surface.

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z-Buffer algorithm

• Image precision algorithms may benefit from hardware acceleration.

• The z-buffer algorithm requires a depth buffer (z-buffer) d[x][y] to record the nearest the rendering of the nearest point encountered so far, to by placed at pixel (x,y).

• The algorithm relies on the fact that if a nearer object occupying (x,y) is found, then the depth buffer is overwritten with the rendering information from this nearer surface.

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for (every pixel) { // initialize the colour to the background // initialize the depth buffer to “far” for every pixel [x][y] for (each facet F){ for (each pixel (x,y) on the facet) if (depth(x,y) < buffer[x][y]){ // F is closest so far

set pixel(x,y) to colour of F; d[x][y] = depth(x,y);}}}

}

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z-Buffer: pros & cons

• Advantages– Very general - deals with intersecting surfaces of

arbitrary shape – can benefit from depth coherence – allows incremental building up of picture in any order – can be implemented in hardware

• Disadvantages– Large amount of memory needed (for depths)– Brute force

• Same pixel may be rendered many times

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Warnock’s algorithm

• Based on quadrant subdivision – a divide and conquer strategy.– If a region is “simple”, render it, otherwise

subdivide into four parts. – Decision strategies as to whether a region is

“simple” are crucial to efficiency. – Subdivisions can proceed to pixel level.– trade-off between the complexity of testing each

facet and doing further subdivision.

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colour = proc(F)if (size < 1) return ; // nothing to draw if (size == 1) draw closest pixel; else if isSimple(F) colour(F); else {colour(NW); //recursivecalls colour(NE);colour(SW); colour(SE);}

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Warnock’s alg.: pros & cons

Of historical interest!• Advantage– Simple to implement

• Disadvantages– Enforces a seemingly haphazard rendering order

(unimportant if results are assembled in a buffer)– Scanline coherence difficult to apply– Subdivisions may be forced to pixel level

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Only some subdivisions are shown.

The intersection between the pyramid and block can be solved • either geometrically, • or by subdivision

until a solution becomes feasible.

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Painter’s algorithm

• As each facet is painted into the frame buffer, its colour paints over whatever was drawn there before, just as a painter covers old layers of paint with new ones.

• Thus, the colour shown at a pixel is the most recently drawn colour.

(i) depth-sort the facets(ii) paint entire facets, in order, from the farthest to

the closest.

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Painter’s algorithm

• The algorithm works in this simple form only if the surfaces have disjoint depth extents.

• When the extents overlap, it may be possible to partition the surfaces into disjoint extents, e.g. as is done in a BSP tree (see below).

Painter’s algorithm fails:Neither surface is fully in front or behind the other.

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BSP trees – binary space partition

• Binary tree sort• Recursive• Build the tree– Insert nodes, in some order, in the left or right sub-

tree, according to potential occlusion by root• Traverse tree – according to view

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BSP trees – binary space partition

• This algorithm uses the concept of “facet plane”– associated with each facet of every solid object in

the scene.• A facet plane is an infinite plane containing a

planar facet of a solid object, – It partitions the space into regions “in front of” and

“to the back of” the facet plane.• “In front of” is determined by the outward facet

normal.

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BSP trees – append new node

A

B

Normal (outward)

C

D

Append on the right

ABAB

CD

Insert a node for facet CD, when CD is in the front half-space of the root facet AB.

Sub-tree root

Back half-spaceInside of AB

Front half-spaceOutside of AB

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A

B

Normal (outward)

C

D

AB

CD

Append on the left

ABABInsert a node for facet CD, when CD is in the back half-space of the root facet AB.

Sub-tree root

BSP trees – append new node

Back half-spaceInside of AB

Front half-spaceOutside of AB

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A

B

Normal (outward)

C

D

AB

ED

Append sub-facets on the left and right

ABABInsert nodes for facet CD, when CD is cut by the plane of the root facet.

Sub-tree root

BSP trees – append new node

E

CE

Back half-spaceInside of AB

Front half-spaceOutside of AB

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AB

C

Plane B of facet A cuts facet C.Need to calculate the partitions of C.

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BSP trees – Build Tree Example

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BSP trees – Build Tree Example

• Facet 2 lies behind 1, insert 2 to the left of 1.• Facet 3 lies behind 1, (go left); behind 2, insert left of 2.• Facet 4 lies bh 1, (left); bh 2, (left); bh 3, insert left of 3• Facet 5 lies bh 1; bh 2; in front 3, insert right of 3• Facet 6 lies bh 1; bh 2; in fr 3; bh 5, insert left of 5

• Tree construction• Start with an arbitrary facet (1), and

put into the root. – Facet 1 divides the space into “front” and

“back”, according to the outward normal.

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BSP trees – Build Tree Example

• Facet 7 lies bh 1; bh 2; to the left of 1.• Facet 7 lies both behind 3 and in front of 3, so split 7 into 7a In

front of 3) and 7b (behind 3). Insert both portions, starting at node 3.

• Facet 7a lies in fr 3 (right); bh 5, (left); bh 6, insert left of 6• Facet 7b lies bh 3; bh 4; insert left of 4

• Insertion of facet 7

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BSP trees – Build Tree Example

• Facet 8 lies bh 1; bh 2; to the left of 1.• Facet 8 lies both behind 3 and in front of 3, so split 8 into 8a (In

front of 3) and 8b (behind 3). Insert both portions, starting at node 3.

• Facet 8a lies in fr 3; bh 5, (left); bh 6; bh 7a, insert left of 7a• Facet 8b lies bh 3; bh 4; bh 7b, insert left of 7b

• Insertion of facet 8

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Tree build - Comments

• Build the BSP tree in this way– Possibly for several objects

• The tree is view-independent, – Depends only on the way each facet partitions the

space into two halves• Shape of tree affected by order of facet node

insertion– This may affect facet sub-partitioning– Does not affect the semantics

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Tree Traversal

• View dependent– For each facet, there is a viewing vector• Possibly constant, as in parallel projection• Possibly position vector dependent, as in perspective

projection– For each facet, the outward normal is required– View direction and normal determine whether the

facet is forward or back facing to the viewer• Traversal influenced at each facet node by

forward/back facing property to viewer

Page 29: Visible Surface Determination

BSP trees – traversal: root front facing

29

CD could obscure something on the plane through AB or behind it:

X

A

B

Normal (outward)

Viewing point

C

D Root

first render AB (root)and then CD

AB

CD

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E

F

BSP trees – traversal: root front facing

X

A

B

Normal (outward)

Viewing point

Root

Something on the plane through AB could obscure EF:

first render EFand then AB (root)

EF

AB

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E

F

BSP trees – traversal: root front facing

X

A

B

Normal (outward)

Viewing point

1. First render EF (behind root) 2. then AB (root)3. then CD (front of root)

RootA

C

D

1

2

3

Left sub-tree

Right sub-tree

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BSP trees – traversal: root front facing

• When the root is seen as forward facing(+), carry out a Left-Root-Right recursive traversal

• Render the left sub-tree, then the root, then the right sub-tree

1

2+

3behind root

In front of root

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BSP trees – traversal: root front facing

• When the root is seen as forward facing(+), carry out a Left-Root-Right recursive traversal

• Render the left sub-tree, then the root, then the right sub-tree

1

2+

3behind root

In front of root

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E

F

BSP trees – traversal: root back facing

X

A

B

Normal (back facing)

Viewing point

1. First render EF (front of root) 2. then AB (root)3. then CD

RootA

C

D

1

2

3

Right sub-tree

Left sub-tree

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BSP trees – traversal: back facing

• When the root is seen as back facing(-), carry out a Right-Root-Left recursive traversal

• Traversal is ‘in-order’, so take action (render, as required) on second encountering with the root.

3

2-

1behind root

In front of root

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BSP trees – traversal: back facing

• When the root is seen as back facing(-), carry out a Right-Root-Left recursive traversal

• Traversal is ‘in-order’, so take action (render, as required) on second encountering with the root.

3

2-

1behind root

In front of root

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A

B

Normal (outward)

C

D

AB

ED

Previously append sub-facets on the left and right

ABABTraversal will not encounter cases in which a facet is neither totally in front or behind the root facet, because of prior node splitting.

Sub-tree root

BSP trees – node splitting

E

CE

Front half-space

Back half-space

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BSP trees – traversal

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BSP tree example – traversalTraversal convention

L: Left link, R: Right link

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BSP tree example – traversalTraversal convention

L: Left link, R: Right link

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BSP tree example – traversalTraversal convention

Traversal sequence (by rows): 1R -1 1L 2L 3L 4R

-4 4L 7bL 8bR -8b 8bL+7b 7bR +3 3R -5R -56L 7aL 8aR -8a 8aL +7a7aR +6 6R +2 2R exit

L: Left link, R: Right link

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BSP tree example – traversal

Traversal sequence (by rows):

The facets in this sequence are:-1, -4, -8b, +7b, +3, -5, -8a, +7a, +6, +2

Of these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

1R -1 1L 2L 3L 4R-4 4L 7bL 8bR -8b 8bL+7b 7bR +3 3R -5R -56L 7aL 8aR -8a 8aL +7a7aR +6 6R +2 2R exit

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BSP tree example – traversalOf these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

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BSP tree example – traversalOf these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

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BSP tree example – traversalOf these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

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BSP tree example – traversalOf these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

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BSP tree example – traversalOf these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

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BSP tree example – traversalOf these, only the ‘+’ facets are forward facing, so the final rendering sequence is+7b, +3, +7a, +6, +2

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Backface culling

• This was considered in the lecture notes on projections

• Easy to apply• Object precision method – can sometimes be

used with other algorithms to reduce the facet candidates that need to be considered for visible surface determination (e.g. z-buffer).

• An integral part of the BSP-tree method.

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Backface culling – a comment

• In perspective view, different parts of a plane would have different viewing directions

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

n

v1

v2

v

XViewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

n

v1

v2X

Viewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

• Consider

n

v1

v2

v2.n - v1.n = (v2 – v1).n

XViewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

• Consider

n

v1

v2

(v2 – v

1)

v2.n - v1.n = (v2 – v1).n

XViewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

• Consider

n

v1

v2

(v2 – v

1), perpendicular to n

v2.n - v1.n = (v2 – v1).n

XViewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

• Consider

n

v1

v2

(v2 – v

1), perpendicular to n

v2.n - v1.n = (v2 – v1).n = 0

XViewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n?

• Consider

n

v1

v2

(v2 – v

1), perpendicular to n

v2.n - v1.n = (v2 – v1).n = 0

So v2.n = v1.nIngore this

XViewing point

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Backface culling – a comment

• Does it matter which viewing direction v is used in computing the sign of v.n? NO!

• A position vector v to an arbitrary point on the plane will do (e.g. any vertex position vector).

n

v

XViewing point

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END