de Casteljau's algorithm General formula ∑ Bernstein polynomials

Curves
• So far we’ve seen polylines
• GL_LINE_STRIP, etc.
• Smooth curves would be better for
• building models
• animation
• Possible representations
• Explicit y=f (x)
• can use only if the curve is a function
• Implicit f (x,y,z) = 0
• difficult to work with
• Parametric (f (u), g(u))
Function Not a function!
• Our choice: parametric curves where the functions
n
are all polynomials in the parameter x(
u)
=
• easy (and efficient) to compute
∑ak
k=
0
• infinitely differentiable
• we’ll look at ones defined by control points
n
k
y(
u)
=
and an algorithm operating on them
∑bku
52438S Computer Graphics Winter 2004 1
k=
0
Kari Pulli
V ′ =
0
V ′ =
1
V ′
=
0
=
=
B(
u)
=
( 1−
u)
( 1−
u)
K
General formula
( 1−
u)
V ′ 0 + uV
′ 1
( 1−
u)
( ( 1−
u)
V0
+ uV1
) + u(
( 1−
u)
V1
+ uV2
)
2 ( 1−
u)
V + 2(
1−
u)
uV
2
+ u V
K
n
∑
k
V
⎛n
⎞
⎜ ⎟
⎝k
⎠
0
+ uV
V + uV
1
0
1
2
( n−k
) k ( 1−
u)
u Vk
1
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Kari Pulli
2
binomial coefficient
1 3 3 1
1 4
1
11
1 2 1
1 51010
5
u
6 4 1
1
k
Pascal's
triangle
⎛n
⎞ n!
⎜ ⎟ =
⎝k
⎠ k!
( n − k)!
V0
Bézier curve: de Casteljau’s
algorithm
• Evaluate point B(u) by recursive linear interpolation
V1 V’1
V”1
V2
V’2
V’0 = (1-u) V0 + u V1
V’1 = (1-u) V1 + u V2
V’2 = (1-u) V2 + u V3
V”0 = (1-u) V’0 + u V’1
= (1-u) ((1-u)V0+uV1) + u((1-u)V1+uV2)
= (1-u)
V’0
V”0
V3
• What if u=0? u=1?
• What’s the relationship between the number of control points
and the degree of the polynomials?
2 V0 + 2(1-u)uV1 + u2 V2
V”1 = (1-u) V’1 + u V’2
= (1-u) ((1-u)V1+uV2) + u((1-u)V2+uV3)
= (1-u) 2 V1 + 2(1-u)uV2 + u2 V”’0
V3
V”’0 = (1-u) V”0 + u V”1
= ...
= (1-u) 3V0 + 3(1-u) 2uV1 + 3(1-u)u2V2 + u3 How much is u?
V'0 divides V0-V1 1:3
so 3 parts V0 and
1 part V1 so
0.75 V0 + 0.25 V1
V3
u=0 => V0, u=1 => V3, so interpolates endpoints
Degree of polynomial one less than # controls
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Kari Pulli
Bernstein polynomials
• The coefficients of the control
points are functions called
the Bernstein polynomials
3
B0
( u)
= ( 1−
u)
2
B1
( u)
= 3u(
1−
u)
2
B2
( u)
= 3u
( 1−
u)
3
B ( u)
= u
• Useful properties (when 0

Bezier Java applet
• Try this online at
http://www.gris.uni-tuebingen.de/projects/ilo/repository/bezier.jar
• Move the
• interpolation point, see how the others (and the point on
curve) move
• control points (can even make loops)
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Kari Pulli
V0
V0 '
V1
V0 "
Subdivide and conquer
V1 '
Q(u)
L = [V0,V’0,V”0,V”’0]
R = [V”’0,V”1,V’2,V3]
evaluated at u=0.5
note: V”’0 = Q(0.5)
V1 "
V2
V2 '
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Kari Pulli
V3
def DisplayBezier(V):
if FlatEnough(V):
Line(V[0],V[3])
else:
L,R = Subdivide(V)
DisplayBezier(L)
DisplayBezier(R)
Displaying Bézier curves
• How could we draw
one of these things?
• It would be nice if we
had an adaptive
algorithm, that would
take into account
curvature / flatness
V0
def DisplayBezier(V):
assert(len(V) == 4)
if FlatEnough(V):
Line(V[0],V[3])
else:
something
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Kari Pulli
Testing for flatness
V0
• Compare total length of control polygon to length of
line connecting endpoints:
0
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Kari Pulli
V1
3
V1
V2
V0
−V1
+ V1
−V2
+ V2
−V3
< 1+
ε
V −V
V2
V3
V3

More complex curves
• Suppose we want to draw a more complex curve
Why not use a high-order Bézier?
High order polynomials are difficult to control
• Instead, we’ll splice together a curve from individual
segments that are cubic Béziers
Why cubic?
Lowest dimension with control for the second derivative
Lowest dimension for non-planar polynomial curves
• There are three properties we’d like to have in our
newly constructed splines…
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Kari Pulli
Interpolation
• Bézier curves are
approximating
• The curve does not
(necessarily) pass through all
the control points
• Each point pulls the curve
toward it, but other points are
pulling as well
• Instead, we may prefer a
spline that is interpolating
• That is, that always passes
through every control point
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Kari Pulli
• Every control point affects
every point on the curve
(except the endpoints)
• Moving a single control
point affects the whole
curve!
Local control
• We’d like our spline to have local control
• that is, have each control point affect some well-defined
neighborhood around that point
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Kari Pulli
C 1 only
Continuity
• We want our curve to have continuity
• There shouldn’t be an abrupt change when we move from
one segment to the next.
• There are nested degrees of continuity:
Not C0 : C0 :
C 1 ,C 2 : C 3 , C 4 , …:
Superscript tells how many derivatives
are continuous
C
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Kari Pulli
2

Ensuring continuity
• C2 continuous curves would be nice
• Since the functions defining a Bézier curve are
polynomial
• all their derivatives exist and are continuous
• therefore, we only need to worry about the derivatives at the
endpoints of the curve
3 ⎛ 3⎞
( 3−k
) k
B(
u)
= ∑⎜
⎟(
1−
u)
u Vk
k ⎝k
⎠
3
= 1− u V
2
2
+ 3 1−
u uV + 3 1−
u u V
3
+ u V
=
( ) 0 ( ) 1 ( ) 2 3
2 3
2 3
2 3
3
( 1− 3u
+ 3u
− u ) V + 3(
u − 2u
+ u ) V1
+ 3(
u − u ) V2
+ u V3
3 ⎡−
u
⎢ 2
⎢ 3u
=
⎢−
3u
⎢
⎣ 1
3u
− 6u
3
2
3u
0
0
3
− 3u
3u
2
0
0
3
u ⎤ ⎡V0
⎤
⎥ ⎢ ⎥
0⎥
⎢
V1
⎥ = u
0⎥
⎢V
⎥ 2
⎥ ⎢ ⎥
0⎥⎦
⎣V
⎦
3 2 [ u u 1]
⎡ −1
⎢
⎢
3
⎢−
3
⎢
⎣ 1
1⎤
⎡V0
⎤
0
⎥ ⎢ ⎥
⎢
V
⎥ 1 ⎥
0⎥
⎢V
⎥ 2
⎥ ⎢ ⎥
0⎦
⎣V
⎦
3
3
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Kari Pulli
3
− 6
Derivatives at the endpoints
B′
( 0)
= 3(
V −V
B′
( 1)
= 3(
V −V
B′
′ ( 0)
= 6(
V − 2V
+ V
1
3
0
B′
′ ( 1)
= 6(
V − 2V
+ V )
1
0
2
)
)
1
2
2
3
)
V0
• In general, the nth derivative at an endpoint depends
only on the n+1 points nearest that endpoint
• Geometrical interpretation of derivatives?
1 st derivative at start 3 times the vector V1 –V0 (V3 –V2 at the end)
2 nd derivative 6 times the vector sum of V0 –V1 and V2 –V1
V1
V2
3
0
− 3
3
0
0
V3
• B''(1)?
Evaluating derivatives
3
B(
u)
= [ u
2
u u
⎡ −1
⎢
] ⎢
3
1
⎢−
3
⎢
⎣ 1
• How do we get derivatives
w.r.t u?
• What is B'(0)? ⎡ −1
3 − 3 1⎤
⎢
3 − 6
⎢−
3
⎢
⎣ 1
3
0
3 0
⎥
0 0⎥
⎥
0 0⎦
− 6
[ 0 0 1 0]
⎢
⎥ = [ − 3 3 0 0]
− 3
1⎤
⎡V0
⎤
0
⎥ ⎢ ⎥
⎥ ⎢
V1
⎥
0⎥
⎢V
⎥ 2
⎥ ⎢ ⎥
0⎦
⎣V3
⎦
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Kari Pulli
3
3
0
3
0
0
Just differentiate [u 3 u 2 u 1]
[3u2 2u 1 0](u=0) =[0 0 1 0]
=> -3V0 + 3V1
[6u 2 0 0](u=1) =[6 2 0 0]
=> 6V1 - 12V2 + 6V3
Ensuring C 2 continuity
• Given a cubic Bézier segment (V0,V1,V2,V3)
• add another curve (W0,W1,W2,W3) to it
• in such a way that the joint is C 2
• But first, if a and b are points, what is (2a-b)?
2a-b
2a-b = a + (a-b)
”mirror b w.r.t. a”
a b
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• C 0 constraint:
• C 1 constraint:
• C 2 constraint:
B”(0) = B”(1)
⇒ W0-2W1+W2 = V1-2V2+V3
⇒ W2 = 2W1 + V1-2V2
= 2W1-(2V2-V1)
Ensuring C 2 continuity
”W2 = First mirror V1 w.r.t V2,
then that w.r.t W1”
B(0) = B(1)
⇒ W0 = V3
”W0 = Attach end points”
B’(0) = B’(1)
⇒ W1-W0 = V3-V2
⇒ W1 = 2V3-V2
”W1 = Mirror V2 w.r.t V3”
V1
V0
Only W3 remains free!
B′
( 0)
= 3(
V −V
)
B′
( 1)
= 3(
V −V
B′
′ ( 1)
= 6(
V − 2V
+ V )
W2 W3
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Kari Pulli
V2
W0
V3
Beziers in Blender
• There are four handle
types
• SHIFT-h ”auto” (yellow)
• tries to keep curve smooth
not really C 2 though
• h toggles ”aligned” (pink)
and ”free” (black)
• aligned is tangent continuous
but not really C 1
• free is only C 0
• v ”vertex” (green)
• like free except tangents aim
directly at the other ends of
segments
1
3
B′
′ ( 0)
= 6(
V − 2V
+ V )
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Kari Pulli
W1
0
1
0
2
)
1
2
2
3
Beziers in Blender
• Add 2 curves
• SPACE -> Add -> Curve -> Bezier Curve
• C 0 continuous
• Connect the segments
• select ends with Rclick and SHIFT+Rclick
• hit f (”make a face”)
• The Bezier controls for a segment are
• two consecutive vertices on curve (V0, V3)
• the handles of V0 and V3 on segment’s side
• Add vertices
• select one vertex with right button
• CTRL + left click adds them
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Kari Pulli
Building a complex spline
• Constraining a Bezier curve made of many segments
to be C 2 continuous is a lot of work
• for each new segment we have to add 3 new control points
• only one of the control points is really free
• B-splines are easier (and C2 )
• First specify 4 vertices (de Boor points), then one per
segment
B1
B2
B0
B3
B4
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Kari Pulli
B5
V0
V3
V1
V2
How many segments here?
3, four de Boor points
for the first segment,
then one per segment

Building a complex spline
• Where are the Bézier control points?
B0
B1
V0
V1 V2
V3
W0
• Express them in terms of the de Boor points
• V1 = (2B1 + B2) / 3
• V0 =
B2
W2
[(B0 + 2B1)/3 + (2B1 + B2)/3] / 2 = (B0 + 4B1 + B2) / 6
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Kari Pulli
W1
B3
W3
Example
• How to get Bezier controls from B-spline controls?
• split B1 –B2 segment into three parts, put V1 and V2 there
• split B0 –B1 segment into three parts, put V0 in the middle
between V1 and the segment closest to B1
• repeat for V3
52438S Computer Graphics Winter 2004 23
Kari Pulli
B5
B4
B0
Building a complex spline
• How do we get de Boor’s points from the Bézier control
points?
• B1 =
• B0 =
B1
2V1 –V2
V0
V1 V2
W0
V3
B2
W2
52438S Computer Graphics Winter 2004 22
Kari Pulli
W1
B3
W3
3[ (2V0 –V1) – B1] + B1 =6V0 –3V1 –2B1
= 6V0 –3V1 –2(2V1 –V2) = 2V2-V1 + 6(V0 –V1)
= B2 + 6(V0 –V1)
Example
B0
• How to get Bezier controls from B-spline controls?
• extend V1 –V2 so it’s three times longer, get B1 and B2
• reflect V1 w.r.t. V0, get a helper point
• extend the helper point three-fold, get B0
• repeat for B3
52438S Computer Graphics Winter 2004 24
Kari Pulli
B1
B5
B4
B2
B3

Endpoints of B-splines
• We can see that B-splines don’t interpolate the de
Boor points
• It would be nice if we could at least control the
endpoints of the splines explicitly
• There’s a hack to make the spline begin and end at
control points by repeating them
• How many times?
3
V0 = (B0 + 4B1 + B2) / 6
three control points affect
a beginning or ending
Bezier point
B
0
B
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Kari Pulli
Finding the derivatives
• Now what we need to do is solve for the derivatives
• To do this we’ll use the C2 continuity requirement
• end points match
• tangents match
• second derivatives match
V
V
V
V
0
1
2
3
= C
= C
0
0
= C −
1
= C
1
+
1
3
1
3
D
D
1
0
52438S Computer Graphics Winter 2004 27
Kari Pulli
1
B
2
W
W
W
B
0
1
2
3
3
B
4
= C
1
1
= C
= C
2
2
B
W = C +
6( V 1 − 2V2
+ V3
) = 6(
W0
− 2W1
+ W2
)
=> 3V1 -6V2 = -6W1 + 3W2
=> 3C0 + D0 -6C1 + 2D1 = -6C1 -2D1 + 3C2 -D2
=> D0 + 4D1 + D2 = 3(C2 -C0)
−
5
1
3
1
3
D
D
1
2
C 2 interpolating splines
• Of the 3 nice things (continuity, local control, interpolation) we
don’t have the last one
• Here’s the idea behind C
D1
2 interpolating splines
• suppose we had cubic Béziers connecting control points C0, C1, C2, …
• and that we somehow knew the first derivative at each point
• (which we don't…)
C0
D0
• Find the V and W control points in terms of Cs and Ds and use
continuity to solve for Ds
52438S Computer Graphics Winter 2004 26
Kari Pulli
Finding the derivatives, cont.
• Here’s what we’ve got so far:
D
m−2
0
1
+ 4D
2
m−1
= 3(
C
• How many equations is this?
• How many unknowns are we solving for?
1
+ D
2
3
m
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Kari Pulli
2
3
m
C1
D + 4D
+ D = 3(
C − C )
D + 4D
+ D = 3(
C − C )
M
0
1
− C
m−2
)
C2
D2
m-1
m+1
D3
C3

Not quite done yet
• We have two additional degrees of freedom, which
we can nail down by imposing more conditions on the
curve
• There are various ways to do this. We’ll use the
variant called natural C2 interpolating splines,
which requires the second derivative to be zero at the
endpoints
• This condition gives us the two additional equations
we need. At the starting point, it is:
( V − 2V
+ V ) = 0
6 0 1 2
=> C0 –2(C0 + D0/3) + C1 –D1/3 = 0
=> – 3C0 -2D0 + 3C1 –D1 = 0
=> 2D0+D1 = 3(C1-C0)
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Kari Pulli
C 2 interpolating spline
• Once we’ve solved for the real Dis, we can plug them
in to find our Bézier control points and draw the final
spline:
C0
• Have we lost anything?
D0
C1
D1
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Kari Pulli
C2
D2
Yes, local control.
Change one C somewhere, all D’s are going to change,
so the curve will be different everywhere (except at controls).
C3
D3
Solving for the derivatives
• Let’s collect our m+1 equations into a single linear
system:
⎡2
⎢
⎢
1
⎢
⎢
⎢
⎢
⎢
⎣
1
4
1
1
4
1
O
1
4
1
⎤ ⎡ D0
⎤ ⎡ 3(
C1
− C0
) ⎤
⎥ ⎢ ⎥ ⎢
⎥
⎥ ⎢
D1
⎥ ⎢
3(
C2
− C0
)
⎥
⎥ ⎢ D ⎥ ⎢ − ⎥
2 3(
C3
C1)
⎥ ⎢ ⎥ = ⎢
⎥
⎥ ⎢ M ⎥ ⎢ M ⎥
1⎥
⎢D
⎥ ⎢ − ⎥
m−1
3(
Cm
Cm−
2 )
⎥ ⎢ ⎥ ⎢
⎥
2⎥⎦
⎢⎣
Dm
⎥⎦
⎢⎣
3(
Cm
− Cm−1
) ⎥⎦
• It’s easier to solve than it looks
• We can use forward elimination to zero out
everything below the diagonal, then back
substitution to compute each D value
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Kari Pulli
A third option
• If we’re willing to sacrifice C2 continuity, we can get
interpolation and local control
• Instead of finding the derivatives by solving a system
of continuity equations, just pick something arbitrary
but local
C4
• If we set each derivative to
C1
be a constant multiple of
the vector between the
previous and next controls, C0
we get a
Catmull-Rom spline
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Kari Pulli
C2
C3

Catmull-Rom splines
• The math for Catmull-Rom splines is pretty simple:
V
V
V
0
1
2
3
= C
1
V = C +
1
= C
= C
2
2
+
t
3
( C
t
3
2
( C
3
− C
0
− C )
1
)
C0
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D2
Kari Pulli
Constructing surfaces of
revolution
• Given: A curve C(u) in the yz-plane:
• Let Rz (v) be a rotation about the z-axis
• Find: A surface S(u,v) which is C(u) rotated about
the z-axis
⎡sin(
2πv)
c y ( u)
⎤
⎢
cos(
2πv)
c ( u)
⎥
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Kari Pulli
D0
⎡ 0 ⎤
⎢
c
⎥
⎢ y(
u)
C(
u)
= ⎥
⎢cz
( u)
⎥
⎢ ⎥
⎣ 1 ⎦
S(
u,
v)
= ⎢
⎢
⎢
⎣
y
cz
( u)
1
⎥
⎥
⎥
⎦
C1
C2
D1
D4
C4
C3
D3
Surfaces of revolution
• Idea: rotate a 2D profile curve around
an axis
How to do it in Blender:
• clear (x)
• switch to front view (Numpad 1)
• add curve (SPC, Add Curve Bezier)
• select & move (Rclick, g)
• add vertices (CTRL-Lclick)
• exit editmode (TAB)
• convert to mesh (ALT-c)
• back to editmode, select all (TAB, a)
• go to top view (Numpad 7)
the rotation axis is perpendicular to
the view direction, around 3d cursor
• in edit buttons (F9)
select spin angle & steps, hit spin button
• cursor turns to question mark, click a
window that has top view selected
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Kari Pulli
General sweep surfaces
• The surface of revolution is a special case of a
swept surface
• Idea: Trace out surface
S(u,v) by moving
a profile curve C(u)
along
a trajectory curve T(v)
• More specifically:
• suppose that C(u) lies in an (x c,y c) coordinate system with
origin O c
• for every point along T(v), lay C(u) so that O c coincides with
T(v)
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Kari Pulli

Orientation
• The big issue:
• How to orient C(u) as it moves along T(v)?
• Here are two options:
• 1. Fixed (or static)
• just translate Oc along T(v)
• 2. Moving
• use the Frenet frame of T(v)
• allows smoothly varying orientation
• permits surfaces of revolution, for example
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Kari Pulli
Frenet swept surfaces
• Orient the profile curve C(u) using the Frenet frame of
the trajectory T(v):
• put C(u) in the normal plane nb
• place O c on T(v)
• align x c for C(u) with -n
• align y c for C(u) with b
• If T(v) is a circle, you get a surface of revolution
exactly!
• Several variations are possible:
• scale C(u) as it moves, possibly using length of T(v) as a
scale factor
• morph C(u) into some other curve as it moves along T(v)
• …
52438S Computer Graphics Winter 2004 39
Kari Pulli
Frenet frames
• Motivation: Given a curve T(v), we want to attach a smoothly
varying coordinate system
n
• To get a 3D coordinate system, we need 3 independent
direction vectors
tˆ(
v)
= normalize(
T ′ ( v))
bˆ
( v)
= normalize(
T ′ ( v)
× T ′
( v))
nˆ
( v)
= bˆ
( v)
× tˆ(
v)
• As we move along T(v), the Frenet frame (t,b,n) varies
smoothly
b
t
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Kari Pulli
Tensor product Bézier surfaces
V 30
V 31
V 20
V 32
v
V 21
V 10
V 33
V 22
V 00
• Given a grid of control points V ij , forming a control net,
construct a surface S(u,v) by:
• treating rows of V as control points for curves V 0(u),…, V n(u)
• treating V 0(u),…, V n(u) as control points for a curve parameterized by v
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Kari Pulli
V 11
V 01
V 23
V 12 V13
V 02
u
V 03

Tensor product surfaces, cont.
• Which control points are interpolated by the surface?
Corners
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Kari Pulli
Tensor product B-spline surfaces
• As with spline curves, we can
piece together a sequence of
Bézier surfaces to make a
B31 spline surface
• If we enforce C2 continuity
B and local control,
30
we get B-spline surfaces:
• treat rows of B as control points to generate control points
in u
• treat those as control points in v
• generate surface points from control points in v
52438S Computer Graphics Winter 2004 43
Kari Pulli
B 32
B 20
B 21
B 10
B 33
B 22
B 00
B 11
B 01
B 23
B 12
B 02
B 13
B 03
Matrix form
• Tensor product surfaces can be written out explicitly:
n n
S(
u,
v)
= ∑∑Vij
Bi
( u)
B j ( v)
=
n
n
i=
0 j=
0
where M Bézier
3 2 [ v v v 1]
⎡ −1
⎢
⎢
3
=
⎢−
3
⎢
⎣ 1
V M
3 ⎡u
⎤
⎢ 2 ⎥
⎢u
⎥
⎢ u ⎥
⎢ ⎥
⎢⎣
1 ⎥⎦
52438S Computer Graphics Winter 2004 42
Kari Pulli
M
3
− 6
3
0
Bézier
− 3
3
0
0
1⎤
0
⎥
⎥
0⎥
⎥
0⎦
T
Bézier
Tensor product B-splines, cont.
• Which B-spline control points are interpolated by the surface?
In general, none
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Kari Pulli

Trimmed NURBS surfaces
• Uniform B-spline surfaces are a special case of
NURBS surfaces
• Sometimes, we want to have control over which parts
of a NURBS surface get drawn
• We can do this by trimming the u-v domain
• define a closed curve in the u-v domain (a trim curve)
• do not draw the surface points inside of this curve
• It’s really hard to maintain continuity in these regions,
especially while animating
52438S Computer Graphics Winter 2004 45
Kari Pulli
Subdivision surfaces
• We defined Bezier curves through subdivision
• let’s generalize that idea for surfaces
• Iteratively refine a control polyhedron (or control mesh) to
produce the limit surface using splitting and averaging steps
• splitting creates a denser mesh (usually doesn’t change shape)
• averaging moves control points (usually makes the mesh smoother)
• There are two types of splitting steps:
• vertex schemes
• face schemes
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Kari Pulli
Building more complex models
• Bezier and
Bspline surfaces
require a regular
control point
network
• Wouldn’t it be
nice to use
arbitrary meshes
as a control
mesh?
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Kari Pulli
Vertex schemes
• A vertex surrounded by n faces is split into n
subvertices, one for each face:
• Doo-Sabin subdivision:
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Kari Pulli

Face schemes
• Each quadrilateral face is split into four subfaces:
• Catmull-Clark subdivision:
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Kari Pulli
Averaging step
• We can use masks for the averaging step:
• (these masks for Loop’s scheme)
3
even/old
vertex
odd/new
vertex
2
n(
1−
a(
n))
5 ( 3 + 2cos(
2π
/ n))
c ( n)
= a(
n)
= −
ci
= 1
• where 0
a(
n)
4 32
• new vertex location is a weighted sum: v
• if you average only new vertices, surface
interpolates the control points
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Kari Pulli
1
0 '
=
3
1
n
n
∑ c
i=
iv
0 i ∑i=
0
c
i
Face schemes, cont.
• Each triangular face is split into four subfaces:
• Loop subdivision:
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Kari Pulli
Adding creases (sharp edges)
• We can tag the mesh
• smooth/sharp edge
• smooth/cusp/corner/dart vertex
• the subdivision mask depends on the edge/vertex types
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Kari Pulli

Trick for semi-sharp creases
• Here’s an example using
Catmull-Clark surfaces of
the kind found in Geri’s
Game
• Notice that the creases
are not infinitely sharp
• The trick:
• perform a few subdivisions
using the sharp masks
• change the masks back to
smooth
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Kari Pulli
Editing the surfaces
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Kari Pulli
Rendering subdivision surfaces
• The real surface is defined as the limit of the
subdivision process
• Don't need to go quite that deep…
• At any level can calculate the limit
• location
• tangents => normal
• Typically a few levels of subdivision are enough
52438S Computer Graphics Winter 2004 54
Kari Pulli