TNM046: Datorgrafik Perspective Projection, Clipping, Hidden surface removal

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TNM046: Datorgrafik Perspective Projection, Clipping, Hidden surface

removal

Sasan Gooran VT 2014

2

Pinhole camera

Film plane Pinhole

z y

The data on the film plane need to be flipped later to obtain the desired photo. Mathematically we can avoid this step if we instead model the pinhole camera with the film plane in front of the pinhole

X-axis is pointing towards us

3

Pinhole camera

Film plane at z=z

_vp

z y

Once the picture is taken if we hold up the photograph at z=z

_vp

plane and observe it with our own eye placed at the origin it will

look like the original scene.

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4

Perspective Projection

View plane, z=z

_vp

z y

P : x y z

!

"

#

$

%

&

& P' :

x ' y' z'

!

"

#

$

%

&

x ' x = z

vp

y' z y = z

_vp

z' = z

vp

z

x ' = x z

vp

z y' = y z

_vp

z' = z

vp

z

x ' = x d

−z y' = y d z' = −d −z

à à

Assume the distance from the eye to the view plane is d, (d=-zvp)

5

Perspective Projection

x ' = x d

−z y' = y d

−z z' = −d

How can we write this as a matrix multiplication using homogenous coordinate representation?

Notice the division by –z! So multiply everything with –z and write it as a matrix multiplication and try to get –z from the last row.

x

_h

y

_h

z

_h

h

!

"

#

# #

$

%

&

=

d 0 0 0

0 d 0 0

0 0 d 0

0 0 −1 0

!

"

#

# #

#

$

%

&

x y z 1

!

"

#

$

%

&

6

Perspective Projection

x

_h

= d ⋅ x y

_h

= d ⋅ y z

_h

= d ⋅ z h = −z

This is what we get from the matrix multiplication on previous page.

So far we always had the last element, h, equal to 1. From now on whenever it is not 1, then divide the three xyz coordinates by h. If we do so we get what we want, which is:

x ' = x

h

h = d ⋅ x

−z y' = y

_h

h = d ⋅ y

−z z' = z

_h

h = −d

Exercise 34 a, b

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7

Clipping

No objects closer than the near clip and farther than the far clip should be shown.

Clipping must separate the visible from the invisible objects (or parts of objects).

z y

View plane Near clip plane

Far clip plane View volume

8

Perspective Projection - Clipping OpenGL

z y

View plane Near clip plane

Far clip plane View volume

-Z_near

-Z_far

In OpenGL we should arrange the matrix such that z=-z

_far

is transformed to z=1 and z=-z

_near

is transformed to z=-1.

Notice that z is independent of x and y and therefore only column 3 and 4 are affected. See next page:

9

Perspective Projection – Clipping OpenGL

x

_h

y

_h

z

_h

h

!

"

#

# #

$

%

&

=

d 0 0 0

0 d 0 0

0 0 A B

0 0 −1 0

!

"

#

# #

#

$

%

&

x y z 1

!

"

#

# #

$

%

&

z' = z

_h

h = Az + B

−z

−1 = −Az

_near

+ B z

near

→ −Az

near

+ B = −z

near

1 = −Az

far

+ B

z

_far

→ −Az

far

+ B = z

far

A = − z

_far

+ z

_near

z

_far

− z

_near

B = − 2z

_near

z

_far

z

_far

− z

_near

Which gives:

and

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10

Perspective Projection – Clipping OpenGL

x

h

y

_h

z

_h

h

!

"

#

# #

$

%

&

=

d ? 0 0 0

0 d ? 0 0

0 0 − z

_far

+ z

near

z

far

− z

near

− 2z

_near

z

_far

z

far

− z

near

0 0 −1 0

!

"

#

# #

$

%

&

x y z 1

!

"

# #

#

$

%

&

In OpenGL we also want the x and y coordinates to be between -1 and 1. Assume that the view plane is limited to (-s, s) in both x and y directions. Thus we want (-s, s) to be transformed to (-1, 1), which means we just need to divide the x and y values with s.

11

Perspective Projection – Clipping OpenGL

x_h yh

z_h h

!

"

##

#

##

$

%

&

=

d / s 0 0 0

0 d / s 0 0

0 0 −zfar+ znear

zfar− znear

−2znearzfar

zfar− znear

0 0 −1 0

!

"

#

##

#

##

$

%

&

x y z 1

!

"

#

###

$

%

&

&&

z y

View plane s

d

This angle is called vertical field of view, vfov, and we have:

cot( vfov

2 ) = d

s

This is why the perspective matrix in the lab material (Section 2.4.3) looks like as it does.

12

Clipping

z y

clip plane

Objects can consist of millions of triangles, testing each triangle against the clip planes is expensive.

We use instead bounding boxes or spheres. The bounding object, rather than the individual triangles, can now be tested against the clip planes.

Example: We wrap around the object with a sphere with radius R and the center C. If the distance from C to the upper clip plane (CP) is bigger than R then the upper clip plane does not intersect the bounding sphere, thus, any object triangle doesn’t either.

clip plane P

C

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13

Clipping- View Coordinate-system

Point clipping: Assume that the point P has the coordinate (x,y) in the view system. It is saved for display if:

xv

min

≤ x ≤ xv

max

yv

_min

≤ y ≤ yv

max

"

# $

%$

Notice that here we have xvmin= yvmin= 0

Line clipping: First, test whether it lies completely inside the clipping window (a line with both endpoints inside all clipping boundaries lies inside the window, ex: P₁P₂).

If not, determine whether it lies completely outside the window (a line with both endpoints outside any one of the clipping boundaries lies outside the window, ex:

P₃P₄). Notice that for line P₇P₈, P₇ is outside the bottom boundary but P₈ is not outside that boundary. On the other hand, P8 is outside the right boundary but P7 is not. Therefore we cannot say that this line is outside the window.

Finally if we cannot determine either, we must perform intersection calculations with one or more clipping boundaries.

x y

View plane xv_max xv_min

yv_min yvmax

P₃ P₂

P₄

P1

P₇ P₈

P5 P6

14

Clipping

x y

View plane xv_max xvmin

yvmin

yv_max

For a line segment with endpoints (x1, y1) and (x₂, y₂) and one or both endpoints outside the clipping rectangle we can use the parametric representation of the line segment.

x = (1− u)x

1

+ ux

2

= x

1

+ u(x

2

− x

1

) y = (1− u)y

1

+ uy

2

= y

1

+ u(y

2

− y

1

)

"

# $

%$

with 0 ≤ u ≤ 1

If the value of u for an intersection with a boundary edge is outside the range 0 to 1, the line does not enter the window at that boundary.

If the value of u is within this range, the line segment crosses into the clipping area.

This method can be applied to each clipping boundary edge P₇

P8

P₅ P₆

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Clipping

x y

View plane xv_max xvmin

yvmin

yv_max

Example: Assume that

xv_max=4, yv_max=3, P₇:(3, -1) and P₈: (5, 2)

P₇ P₈

Since 5>xv_max then P₈ is outside the right boundary but 3<xv_max which means P₇ is not outside that boundary. P₇ is outside the bottom boundary (-1<0) but P8 is not (2>0). Then we must perform intersection calculations.

x = 3+ u(5 − 3) = 3+ 2u y = −1+ u(2 − (−1)) = −1+ 3u

"

# $

%$

with 0 ≤ u ≤ 1

The parametric representation of P

₇

P

₈

:

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16

Clipping

x y

View plane xvmax

xvmin

yvmin

yv_max

P₇ P8

x = 3+ u(5 − 3) = 3+ 2u y = −1+ u(2 − (−1)) = −1+ 3u

"

# $

%$

with 0 ≤ u ≤ 1

Left boundary (x=0):

0 = 3+ 2u → u = −3 / 2 < 0, does not enter at this boundary

Top boundary (y=3):

3 = −1+ 3u → u = 4 / 3 > 1, does not enter at this boundary

Bottom boundary (y=0):

0 = −1+ 3u → 0< u = 1 / 3 < 1, does enter at this boundary

Intersection point is (11/3, 0) and this point is inside the boundary

Right boundary (x=4):

Intersection point is (4, 1/2) and this point is inside the boundary

4 = 3+ 2u → 0< u = 1 / 2 < 1, does enter at this boundary

Exercise 36 b, c

17

Clipping

x y

View plane

P₃

P₂ P₁

I₂ I₁

Example: Clipping of a triangle against the view plane.

Assumption: The triangle has been projected into the view plane.

Given: The points P

₁

, P

₂

, P

₃

(triangle list) and the vertex list.

We calculate the intersection points I

₁

and I

₂

between the triangle edges and the view plane boundary.

A temporary vertex list and triangle list take points P

₁

, I

₁

and I

₂

. P

₂

and P

₃

are discarded. The clipped triangle is plotted.

18

Hidden Surface Removal - Culling

How do we determine whether or not a triangle is visible?

View direction

The normal of triangle j is N

_j

The triangle is visible if its normal points towards us (the eye):

Triangle j is visible if,  V • 

N

_j

≤ 0 and not visible if  V • 

N

_j

> 0

V 

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19

Culling

Culling works fine if we have one convex object, but for two objects it may fail.

View direction Invisible Visible

Invisible Visible

Culling will remove the backside of both objects.

The triangles with normals N

₁

and N

₂

are both ‘visible’

à Culling cannot remove the front side of the left object.

V 

N₁ N₂

20

Culling

Culling can fail even if we only have one object. (concave)

This object (torus) from the side can appear as two objects:

Culling fails.

If we look it from above, it is a ring (one object):

Culling works.

21

Hidden Surface Removal – Painters Algorithm

Assume XYZ is the view coordinate system and the view direction is Z and we use parallel projection. We make use of the distances.

x y

z

P₁ P₂

T₁ T₂

We have two triangles, T1 and T2. Select points P1 and P2. Take the Z-values of these points, Z1 and Z2. Since the projection is parallel then Z₁ and Z₂ are distances from triangle T₁ and T₂ to the view plane, respectively.

Generate a list and store the z-values of the points.

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22

Painters Algorithm

Assume that we have 5 triangles, which are ordered randomly according to their distance to the view plane (Left table).

We sort the triangles according to their distance (right table).

Draw the triangles from the top to the bottom in the list (right table).

If there is an overlap between triangles, the closer one over-plots the farther one.

Triangle Distance

T1 11

T2 6

T₃ 3

T₄ 9

T₅ 14

Sorting à

T5 14

T1 11

T₄ 9

T₂ 6

T₃ 3

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Painters Algorithm

T5 14

T₁ 11

T₄ 9

T2 6

T3 3

x

y T

₅

T

₁

T

₄

T

₂

T

₃

The painters algorithm works if the typical distances are large compared to the triangle sizes, then the ordering is simple.

24

Painters Algorithm

Problem: Consider triangles that are close compared to their size.

x y

z

^P2

P₁ T₁ T₂

The point P1 determines the distance of T1.

If we use the point P₂ to determine the distance of T2, then T2 is farther away.

If we use the point P3 to determine the distance of T₂, then T₂ is closer.

The problem gets worse if the triangles intersect.

P₃