TNM046: Datorgrafik Perspective Projection, Clipping, Hidden surface removal

TNM046: Datorgrafik Perspective Projection, Clipping, Hidden surface

removal

Sasan Gooran

VT 2014

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

z y

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

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

look like the original scene.

Perspective Projection

View plane, z=z

z y

P : x y z

!

"

##

#

$

%

&

&

&

P' : x ' y' z'

!

"

##

#

$

%

&

&

&

x '

x = z

y' z

y = z

z

x ' = x z

z y' = y z

z

x ' = x d

−z y' = y d

−z

à à

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

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

y

z

h

!

"

# #

# #

#

$

%

&

&

&

&

&

=

d 0 0 0

0 d 0 0

0 0 d 0

0 0 −1 0

!

"

# #

# #

$

%

&

&

&

&

x y z 1

!

"

#

# #

#

$

%

&

&

&

&

Perspective Projection

x

= d ⋅ x y

= d ⋅ y z

= 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 = d ⋅ x

−z y' = y

h = d ⋅ y

−z z' = z

= −d

Exercise 34 a, b

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

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

is transformed to z=1 and z=-z

is transformed to z=-1.

Notice that z is independent of x and y and therefore only

9

Perspective Projection – Clipping OpenGL

x

y

z

h

!

"

# #

#

# #

$

%

&

&

&

&

&

=

d 0 0 0

0 d 0 0

0 0 A B

0 0 −1 0

!

"

#

# #

#

$

%

&

&

&

&

x y z 1

!

"

# #

# #

$

%

&

&

&

&

z' = z

h = Az + B

−z

−1 = −Az

+ B

z

→ −Az

+ B = −z

1 = −Az

+ B

z

→ −Az

+ B = z

A = − z

+ z

z

− z

B = − 2z

z

z

− z

Which gives:

and

Perspective Projection – Clipping OpenGL

x

y

z

h

!

"

# #

#

# #

$

%

&

&

&

&

&

=

d ? 0 0 0

0 d ? 0 0

0 0 − z

+ z

z

− z

− 2z

z

z

− z

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

11

Perspective Projection – Clipping OpenGL

x_h y_h z_h h

!

"

##

##

#

$

%

&

&

&

&

&

=

d / s 0 0 0

0 d / s 0 0

0 0 −z_far + z_near

z_far − z_near − 2z_nearz_far z_far − z_near

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.

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

clip plane P

C

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

≤ x ≤ xv

yv

≤ y ≤ yv

"

# $

%$

Notice that here we have xv_min = yv_min = 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

yv_max P₃

P₂ P₄

P₁

P₇

P₈ P₅

P₆

Clipping

x y

View plane xv_max xv_min

yv_min 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

+ ux

= x

+ u(x

− x

) y = (1− u)y

+ uy

= y

+ u(y

− y

)

"

# $

%$

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.

P₇

P₈ P₅

P₆

15

Clipping

x y

View plane xv_max xv_min

yv_min 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

P

:

Clipping

x y

View plane xv_max xv_min

yv_min 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

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

, P

, P

(triangle list) and the vertex list.

We calculate the intersection points I

and I

between the triangle edges and the view plane boundary.

A temporary vertex list and triangle list take points P

, I

and I

.

P

and P

are discarded. The clipped triangle is plotted.

Hidden Surface Removal - Culling

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

View direction

The normal of triangle j is N

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

   

V 

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

and N

are both ‘visible’

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

V 

N₁ N₂

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.

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

T 9

Sorting à

Triangle Distance

T₅ 14

T₁ 11

T₄ 9

T 6

23

Painters Algorithm

Triangle Distance

T₅ 14

T₁ 11

T₄ 9

T₂ 6

T₃ 3

x

y T

T

T

T

T

The painters algorithm works if the typical distances are large compared to

the triangle sizes, then the ordering is simple.

Painters Algorithm

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

x y

z

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