Some estimates of the height of rational
Bernstein-Bézier triangular surfaces∗
Renjiang Zhang
1,2,3
and Guojin Wang
1,2
1. Institute of Computer Images and Graphics, Zhejiang University, Hangzhou, 310027, PR.China
2. State Key Lab of CAD and CG, Zhejiang University, Hangzhou, 310027, PR.China
3. China Institute of Metrology, Zhejiang, Hangzhou, 310034, PR.China
Email:renjiang@mail.hz.zj.cn
Abstract
Subdivision, intersection and triangulation of the surface are the basic and common
operations in Computer Aided Design and computer graphics. A complicated surface
can be approximated by some triangular patches. The key technique of this problem
is the estimation of approximate error. This work has been done in some papers
by estimating second derivatives of the surface, thus very complicated computation
must be executed and hence time is overly consumed. Avoiding this estimation and
exerting the influence of the control points of the surfaces, this paper gives a formula
for computing the exact distance between a degree 2 Bernstein-Bézier triangular surface and the corresponding interpolating plane triangular patch. Also, this paper
gives estimates of maximal distance between a degree 2 or 3 rational Bernstein-Bézier
triangular surface and the corresponding interpolating plane triangular patch with a
high precision. These results are easy for programming, simple for computation, and
valuable for improving the subdivision/triangulation algorithms of common rational
Bézier surfaces in the design system.
Key Words: Bézier triangular surfaces, Rational Bézier triangular surfaces, Subdivision, Approximation, Error estimate.
§1
Introduction
Subdivision, intersection and triangulation of the surface are very common and basic operations in Computer Aided Design and computer graphics. A complicated surface may be subdivided as some triangular
patches, that is, taking a series points on the surface uniformly, then the surface can be approximated by a
triangular mesh created by using all above points. This technique has many advantages in applications. For
example, to find the intersection between two complicated surfaces is a difficult non-linear problem. After they
are subdivided as many triangular patches, the problem to find the intersections between these two surfaces may
be transformed into a linear problem to find the intersections between these two series of the plane triangular
patches. It is easy to be solved. This intersections is obviously an approximate solution. Then to reply the
∗ supported
by the National Natural Science Foundation of China (No.60173034), the National Natural Science Foundation for
Innovative Research Groups (No.60021201) and the Foundation of State Key Basic Research 973 Item (No. 2002CB312101).
1
problem, i.e., whether or not it can meet the requirement of the precision, we need make the researches on
computing the distances between these plane triangular patches and the given surfaces[1,2]. If the maximal
distance between that plane triangular patch and the original triangular surface interpolated by this plane triangular patch is less than the given tolerance , then the plane triangular patch can be instead of the original
surface. Some papers [1,2,3,4,5,6] have investigated the problem on estimating the maximal distance between
the interpolating plane triangular patch and the original triangular surface. A similar problem on the curve is
estimating the maximal distance between the straight line segment and the original curve segment interpolated
by this straight line segment [7,8,9,10,11,12,13]. In 1998, Nairn[11] et al obtained some sharp and quantitative
bounds on the distance between a polynomial piece and its Bernstein-Bézier control polygon.
On the aspect of linear approximation of scalar-valued function surface, in 2000, Rief [4] obtained the best
bounds on the deviation between the Bernstein-Bézier triangular surface and the corresponding control mesh.
This can be expressed as
|T n (u, v) − H n (u, v)| ≤
n ω3 (n)
−
3
3n
k∆2 Ti,j,k k, (u ≥ 0, v ≥ 0, u + v ≤ 1).
(1)
Where T n (u, v) is degree n Beinstein-Bézier function surface, H n (u, v) is the corresponding control polyhedron
mesh, ∆2 Ti,j,k is the mixed second difference of the inner Bézier ordinates, and ω3 (n) = 0 when n is a multiple
of 3, otherwise ω3 (n) = 1.
On the aspect of linear approximation of parametric function surface, in 1992, Shang and Hirsch[2] obtained
the bounds on the maximal distance between the plane triangular patch and the corresponding parametric
surface. In 2001, Sun et al [5] improved Shang’s result to give the following better sharp bounds: Let S(u, v) ∈ C 2
be a parameter surface and let l(u, v) be a plane triangular patch which interpolate S(u, v), i.e., there are three
points A, B, C such that
S(A) = l(A), S(B) = l(B), S(C) = l(C),
then
sup kS(u, v) − l(u, v)k ≤
(u,v)∈T
2 2
L (M1 + 2M2 + M3 ).
9
(2)
Where T is a triangle
in the parameter
2
space, L is the maximum
length of the edges of ∆ABC and M1 =
∂ 2 S(u,v) ∂ S(u,v) ∂ 2 S(u,v) sup ∂u2 ; M2 = sup ∂u∂v ; M3 = sup ∂v2 . Recently , Zhang and Wang [6] improved
(u,v)∈T
(u,v)∈T
(u,v)∈T
the above result again and obtained the best bounds about the approximation problem.
Applying the method which is similar to the above estimate (2) to investigate the approximation of rational
Bézier triangular surfaces which are commonly used in CAGD requires to estimate second derivative of the
surfaces. Thus, its calculation will be very complicated and hence will take long time, also the estimate
will be too rough (see latter Example 1 and Example 2), so it is’t an effective method. To overcome this
limitation, this paper give a method for computation and estimate, of the maximal distances between the
triangular surfaces and the corresponding plane triangular patches interpolating the original surfaces. We
obtain a formula for computing the exact distance between the degree 2 Bézier surfaces and the interpolating
plane triangular patches, also we show some estimates of maximal distance between the degree 2 or 3 rational
Bézier triangular surfaces and the corresponding interpolating plane triangular patches. These results are very
easy to be programmed and need less calculation. It is very useful in improving the efficiency of the algorithms
for subdivision, intersection and triangulation, of lower degree rational Bézier surfaces.
2
§2
Preliminaries
First we give the form of Bernstein-Bézier triangular surfaces. Let Ti,j,k ∈ R3 (i, j, k ≥ 0,
i + j + k = n) be
(n + 1)(n + 2)/2 point vectors and Ω be a coordinate triangle and let P = (u, v, w) be a point in the parameter
domain Ω in which (u, v, w) are the barycentric coordinates of the point P , then
T n (u, v, w) =
X
n
Bi,j,k
(u, v, w)Ti,j,k ,
(u, v, w) ∈ Ω,
u, v, w ≥ 0,
u+v+w =1
i+j+k=n
is called Bernstein-Bézier surface of degree n on the triangle Ω, where
n
Bi,j,k
(u, v, w) =
n! i j k
uv w ,
i!j!k!
i + j + k = n,
u, v, w ≥ 0,
u + v + w = 1,
which is called degree n Bernstein basis, Ti,j,k is called control point of the surface. Correspondingly, if wi,j,k > 0,
then
P
T n (u, v, w) =
i+j+k=n
n
Bi,j,k
(u, v, w)Ti,j,k wi,j,k
P
i+j+k=n
n (u, v, w)w
Bi,j,k
i,j,k
is called rational Bernstein-Bézier surface of degree n on the triangle Ω.
We discover that rational Bernstein-Bézier triangular surface of degree n has a good property which is
called shape-invariability under some linear fractional parameter transformation. That is, let
u=
αu0
;
αu0 + βv 0 + γw0
where
s
α=
v=
1
n
wn,0,0
βv 0
;
αu0 + βv 0 + γw0
s
;
β=
1
n
w0,n,0
w=
s
;
γ=
n
γw0
,
αu0 + βv 0 + γw0
1
w0,0,n
,
then the surface can be transformed into the normalized form as follows:
P
n
∗
Bi,j,k
(u, v, w)Ti,j,k wi,j,k
i+j+k=n
P
T n (u, v, w) =
n (u, v, w)w ∗
Bi,j,k
i,j,k .
i+j+k=n
For convenience, we still write u0 , v 0 , w0 as u, v, w respectively and set
∗
∗
∗
wn,0,0
= w0,n,0
= w0,0,n
= 1;
s
∗
wi,j,k
=
n
1
wn,0,0
i 1
w0,n,0
j 1
w0,0,n
k
wi,j,k , ((i, j, k) 6= (n, 0, 0), (0, n, 0), (0, 0, n)).
We need to point out that there are two ways to define the distance between a surface and its interpolating
plane triangular patch. One way is called the parametric distance, which is defined as the maximal distance
from the surface to its interpolating plane triangular patch at the same parameter. The other way is called
the geometric distance, which is defined as the maximal vertical distance from any point on the surface to its
interpolating plane triangular patch. In the general cases they are different, the latter is always less than or
equal to the former. In this paper we define the distance as the geometric distance and always write hn (u, v, w)
as the distance between the point T n (u, v, w) and the plane determined by the three points Tn,0,0 , T0,n,0 and
T0,0,n , namely, hn (u, v, w) is the height of the surface T n (u, v, w). Without loss of generality, this paper just
investigates the height of the normalize rational Bernstein-Bézier triangular surfaces, since any non-normalize
rational Bernstein-Bézier triangular surfaces can be transformed into the normalize rational Bernstein-Bézier
∗
triangular surfaces. Furthermore, for convenience we still write the weight factor wi,j,k
as wi,j,k .
3
§3
Estimate for degree 2
Theorem 1 For a degree 2 Bernstein-Bézier surface on the triangle ∆T2,0,0 T0,2,0 T0,0,2 , suppose a, b and
c denote the directed distances from its three central control points T0,1,1 , T1,0,1 and T1,1,0 to ∆T2,0,0 T0,2,0 T0,0,2
respectively, then the height of the surface has the following exact results:
(1)
If the three points T1,1,0 , T1,0,1 and T0,1,1 lie in the same side of ∆T2,0,0 T0,2,0 T0,0,2 and by using
| a |, | b | and | c | as three edges a triangle can be formed, then we have
max
(u,v,w)∈Ω
(2)
h2 (u, v, w) = |2abc/(2ab + 2bc + 2ca − a2 − b2 − c2 )|;
Otherwise, we have
max
(u,v,w)∈Ω
h2 (u, v, w) =
1
max(| a |, | b |, | c |).
2
where Ω = {(u, v, w)|u + v + w = 1, u, v, w ≥ 0}.
Proof: Without loss of generality, choose the plane which determined by ∆T2,0,0 T0,2,0 T0,0,2 as the xy
plane, and set up a space coordinate system, then the third component z(u, v, w) of the surface T 2 (u, v, w) at
the parameters (u, v, w) is just the directed distance between the point P = (u, v, w) and the xy plane. It’s easy
to know
h2 (u, v, w) = |z(u, v, w)|,
z(u, v, w) = 2avw + 2buw + 2cuv.
In the following we will find the maximum value of the function z(u, v, w) on Ω = {(u, v, w)|u + v + w =
1, u, v, w ≥ 0} by Lagrange’s multipliers method. Write
F (u, v, w) = z(u, v, w) − 2λ(u + v + w − 1),
then three partial derivatives of F (u, v, w) for u, v and w directions are
1/2Fu = bw + cv − λ = 0;
1/2Fv = aw + cu − λ = 0;
1/2Fw = av + bu − λ = 0
respectively. Combining with
u + v + w = 1,
we obtain a system of linear equation with four variables u, v, w and λ.
In case (1), without loss of generality, assume a, b and c be all positive. If regard a, b and c as three edge
lengthes a triangle can be formed, we have
a + b − c > 0,
c + a − b > 0,
b + c − a > 0.
It is easy to compute the coefficient determent of the above system of linear equations as follows:
0 c b −1 c 0 a −1 = 2ab + 2bc + 2ca − a2 − b2 − c2 .
D := b a 0 −1 1 1 1
0 Applying cosine law, we have
a2 + b2 + c2 = 2ab cos C + 2bc cos A + 2ca cos B,
4
where A, B and C are the three interior angles of the triangle which edge lengthes are a, b and c, respectively.
Thus we have
D = 2ab(1 − cos C) + 2bc(1 − cos A) + 2ca(1 − cos B) > 0.
Using Cram’s law, we have
u0 =
D1
,
D
v0 =
D2
,
D
w0 =
D3
,
D
where
D1 := a(b + c − a),
D2 := b(c + a − b),
D3 := c(a + b − c).
Then we have
(u0 , v0 , w0 ) ∈ Ω = {(u, v, w)|u + v + w = 1, u, v, w ≥ 0}.
Therefore in this case (u0 , v0 , w0 ) is a unique extreme point of the function z(u, v, w) in Ω, hence it is easy to
obtain
h2 (u0 , v0 , w0 ) = 2abc/(2ab + 2bc + 2ca − a2 − b2 − c2 ).
Now we consider the height of the surface in the three boundaries Ω1 = {v + w = 1, u = 0, v, w ≥ 0}, Ω2 =
{w + u = 1, v = 0, w, u ≥ 0}, Ω3 = {u + v = 1, w = 0, u, v ≥ 0}, of Ω. It’s easy to get
max
z(u, v, w) = a/2;
(u,v,w)∈Ω1
max
z(u, v, w) = b/2;
(u,v,w)∈Ω2
max
z(u, v, w) = c/2.
(u,v,w)∈Ω3
Next we compare the value 2abc/(2ab+2bc+2ca−a2 −b2 −c2 ) with a/2, b/2 and c/2. Without loss of generality,
assume a be the maximum value among a, b and c. Therefore
2a(b + c) ≤ a2 + (b + c)2 ,
then
2ab + 2bc + 2ca − a2 − b2 − c2 ≤ 4bc,
or
a/2 ≤ 2abc/(2ab + 2bc + 2ca − a2 − b2 − c2 ).
Thus, we have
h2 (u, v, w) = 2abc/(2ab + 2bc + 2ca − a2 − b2 − c2 ).
In case (2), we distinguish two cases as follows:
A) If the three points T1,1,0 , T1,0,1 and T0,1,1 lie in the same side of ∆T2,0,0 T0,2,0 T0,0,2 , but regarding | a |, | b |
and | c | as three edge lengthes any triangle can not be formed, thus it’s easy to prove the function z(u, v, w) has
no extreme point in Ω, so h2 (u, v, w) takes its maximum value on the boundary of Ω. In this case the theorem’s
result holds.
B) If the three points T1,1,0 , T1,0,1 and T0,1,1 do not lie in the same side of ∆T2,0,0 T0,2,0 T0,0,2 , without loss
of generality, assume that a > 0, b > 0, c < 0 and a ≥ b. In this case we distinguish two cases again:
1. D ≥ 0. It’s easy to prove that the function z(u, v, w) has no extreme point in Ω, so h2 (u, v, w) takes its
maximum value on the boundary of Ω. In this case the theorem’s result holds.
2. D < 0. From
[(b − a) + c]2 = a2 + b2 − 2ab + c2 + 2c(b − a) ≥ 0,
5
or
4bc ≥ D,
we have
a
1
2abc
≤ ≤ max{a, b, | c |}.
D
2
2
In this case the theorem’s result holds either. Thus completes the proof.
Now we put up two examples about Beinstain-Bézier function surface to compare the result of Theorem 1
with the other two results above.
Example 1 Suppose a Beinstain-Bézier triangular function surface of degree 2 with its controls T2,0,0 =
T0,2,0 = T0,0,2 = 0,T1,1,0 = T1,0,1 = 1,T0,1,1 = −1, then
T 2 (u, v) = 2uv + 2u(1 − u − v) − 2v(1 − u − v) = −2u2 + 2v 2 + 2uv + 2u − 2v.
Since k42 Ti,j,k k = 3, the estimate (1) gives the estimate value 32 . Since M1 = 4,M2 = 2,M3 = 4 and L =
the estimate (2) gives the estimate value 3. But Theorem 1 gives the exact value
1
2,
√
2,
which is much less than
the other two estimates.
Example 2 Suppose a Beinstain-Bézier triangular function surface of degree 2 with its controls T2,0,0 =
T0,2,0 = T0,0,2 = 0,T1,1,0 = 1,T1,0,1 = 2,T0,1,1 = 3, then
T 2 (u, v) = 2uv + 4u(1 − u − v) + 6v(1 − u − v) = −4u2 − 6v 2 − 8uv + 4u + 6v.
Since k42 Ti,j,k k = 4, the estimate (1) gives the estimate value 2. Since M1 = 8,M2 = 8,M3 = 12 and L =
the estimate (2) gives the estimate value 10. But Theorem 1 gives the exact value
3
2,
√
2,
which is much less than
the other two estimates.
Example 3 Suppose a Beinstain-Bézier triangular function surface of degree 2 with its controls T2,0,0 =
T0,2,0 = T0,0,2 = 0,T1,1,0 = T1,0,1 = T0,1,1 = 1, then
T 2 (u, v) = 2uv + 2u(1 − u − v) + 2v(1 − u − v) = −2u2 − 2v 2 − 2uv + 2u + 2v
Since M1 = 4,M2 = 2,M3 = 4 and L =
√
the estimate (1) gives the estimate value
2, the estimate (2) gives the estimate value 3. Since k42 Ti,j,k k = 1,
1
2.
And, Theorem 1 gives the exact value
2
3.
Although the estimate
value given by Theorem 1 is larger than that given by the estimate (1), but we must point out that in this
example Theorem 1 uses only one plane triangular patch to approximate the given surface and the estimate (1)
uses an entire mesh formed by the four plane triangular patches to approximate the given surface, so the above
two results show different. Considering it the result of Theorem 1 is still satisfactory.
Theorem 2
For a degree 2 rational Bernstein-Bézier surface on the triangle ∆T2,0,0 T0,2,0 T0,0,2 , suppose
a, b and c denote directed distances from its three central control points T0,1,1 , T1,0,1 and T1,1,0 to ∆T2,0,0 T0,2,0 T0,0,2
respectively, then the height of the surface has the following estimate:
h2 (u, v, w) ≤
2ω
max{|a|, |b|, |c|},
1 + 2ω
where ω = max{w1,1,0 , w1,0,1 , w0,1,1 }.
Proof: Without loss of generality, choose the plane which determined by ∆T2,0,0 T0,2,0 T0,0,2 as the xy
plane, and set up a space coordinate system, then the third component z(u, v, w) of the surface T 2 (u, v, w) at
the parameters (u, v, w) is a directed distance from the point P = (u, v, w) to the xy plane. It’s easy to know
2aw1,1,0 uv + 2bw0,1,1 vw + 2cw1,0,1 wu
| z(u, v, w) |= 2
u + v 2 + w2 + 2w1,1,0 uv + 2w0,1,1 vw + 2w1,0,1 wu 6
≤
2w1,1,0 uv + 2w0,1,1 vw + 2w1,0,1 wu
max(| a |, | b |, | c |).
1 + 2(w1,1,0 − 1)uv + 2(w0,1,1 − 1)vw + 2(w1,0,1 − 1)wu
Therefore
2w1,1,0 uv + 2w0,1,1 vw + 2w1,0,1 wu
1 + 2(w1,1,0 − 1)uv + 2(w0,1,1 − 1)vw + 2(w1,0,1 − 1)wu
≤
2ωuv + 2w0,1,1 vw + 2w1,0,1 wu
1 + 2(ω − 1)uv + 2(w0,1,1 − 1)vw + 2(w1,0,1 − 1)wu
≤
2ω(uv + vw + wu)
,
1 + 2(ω − 1)(uv + vw + wu)
Analogizing the proof of Theorem 1, we know that 1/4 ≤ uv + vw + wu ≤ 1/3 and the maximum value of the
function f (t) =
2ωt
1+2(ω−1)t
(t = uv + vw + wu) is
§4
Theorem 3
2ω
1+2ω .
Thus completes the proof.
Estimate for degree 3
For a degree 3 Bernstein-Bézier surface on the triangle ∆T2,0,0 T0,2,0 T0,0,2 , let ai,j,k be the
distance from the point Ti,j,k to ∆T3,0,0 , T0,3,0 , T0,0,3 , and write a = max{ai,j,k }, then
i,j,k
h3 (u, v, w) ≤
24a − s
,
27
p
p
p
where s = 6( (a − a2,1,0 )(a − a0,1,2 ) + (a − a1,2,0 )(a − a1,0,2 ) + (a − a0,2,1 )(a − a2,0,1 ) + (a − a1,1,1 )).
Proof: Without loss of generality, choose the plane which determined by ∆T3,0,0 T0,3,0 T0,0,3 as the xy
plane, and set up a space coordinate system, then the third component z(u, v, w) of the surface T 3 (u, v, w) is
the directed distance from the point P = (u, v, w) to the xy plane. Write zi,j,k as the coordinate of the point
Ti,j,k in z-axis direction, then
h3 (u, v, w) = |z(u, v, w)|
≤ 3a2,1,0 u2 v + 3a1,2,0 v 2 u + 3a0,1,2 w2 v + 3a0,2,1 v 2 w + 3a1,0,2 uw2 + 3a2,0,1 u2 w + 6a1,1,1 uvw
= a − [a(u3 + v 3 + w3 ) + 3(a − a2,1,0 )u2 v + 3(a − a0,1,2 )vw2 + 3(a − a1,2,0 )uv 2 +
3(a − a1,0,2 )uw2 + 3(a − a0,2,1 )v 2 w + 3(a − a2,0,1 )u2 w + 6(a − a1,1,1 )uvw]
≤ a − [a(u3 + v 3 + w3 ) + suvw].
Noting that in the sum s there is at least one term which is zero, we know s ≤ 18a. By using Lagrange’s
multipliers method again, we can prove that f (u, v, w) = a(u3 + v 3 + w3 ) + suvw takes its minimum value at
the parameters u = 1/3, v = 1/3, w = 1/3 on Ω, then
a(u3 + v 3 + w3 ) + suvw ≥
3a + s
.
27
Thus completes the proof.
Theorem 4 For a degree 3 rational Bernstein-Bézier surface on a triangle, let ai,j,k be the distance from
the point Ti,j,k to ∆T3,0,0 T0,3,0 T0,0,3 , and write a = max{ai,j,k } and ω = max{wi,j,k }, then
i,j,k
h3 (u, v, w) ≤
where t = 6(
p
(ω − w2,1,0 )(ω − w0,1,2 ) +
i,j,k
24ω − t
a.
3 + 24ω − t
p
(ω − w1,2,0 )(ω − w1,0,2 ) +
p
(ω − w0,2,1 )(ω − w2,0,1 ) + (ω − w1,1,1 )).
Proof: Without loss of generality, choose the plane which determined by ∆T3,0,0 T0,3,0 T0,0,3 as the xy
plane, and set up a space coordinate system, then the third component z(u, v, w) of the surface T 3 (u, v, w) is
7
the directed distance from the point P to the xy plane. Write zi,j,k as the coordinate of the point Ti,j,k in z-axis
direction, then by the proof of Theorem 3 and the inequality u3 + v 3 + w3 ≥ 19 , we know
h3 (u, v, w) = |z(u, v, w)|
≤
u3
+
v3
3w2,1,0 u2 v + 3w1,2,0 v 2 u + 3w0,1,2 w2 v + 3w0,2,1 v 2 w + 3w1,0,2 uw2 + 3w2,0,1 u2 w + 6w1,1,1 uvw
a.
+ w3 + 3w2,1,0 u2 v + 3w1,2,0 v 2 u + 3w0,1,2 w2 v + 3w0,2,1 v 2 w + 3w1,0,2 uw2 + 3w2,0,1 u2 w + 6w1,1,1 uvw
≤
u3
+
(24ω − t)/27
a
+ w3 + (24ω − t)/27
v3
≤
24ω − t
a
3 + 24ω − t
Thus completes the proof.
Theorem 5 For a degree n rational Bernstein-Bézier surface on a triangle, let ai,j,k be the distance from
the point Ti,j,k to ∆Tn,0,0 T0,n,0 T0,0,n , and write a = max{ai,j,k } and ω = max{wi,j,k }, then
i,j,k
hn (u, v, w) ≤
i,j,k
(3n−1 − 1)ω
a.
1 + (3n−1 − 1)ω
Proof The proof is simple, we omit it.
§5
An algorithm
In this section we will give a method to calculate the height of degree 2 Bernstein-Bézier surface. Using
the above results, this algorithm can be described as follows:
Step 1. Input the coordinates of the three angle points and the three central control points which lie on
the control mesh of the surface.
Step 2. Write the plane equation determined by the three angle points.
Step 3. Calculate direction distances from the three central control points to the plane respectively, and
write them as a, b, c.
Step 4. Check that whether a, b and c are all larger than zero or all less than zero. If not, then go to Step
7.
Step 5. Check that whether or not the following three expressions all hold :
| a | + | b |>| c |; | a | + | c |>| b |; | b | + | c |>| a | .
If not, then go to Step 7.
Step 6. Output |2abc/(2ab + 2bc + 2ca − a2 − b2 − c2 )|, end.
Step 7. Output
1
2
max(| a |, | b |, | c |), end.
¿From the analysis in this paper, we know that the estimates obtained by this paper are better than the
other two estimates (1) and (2) for many applications such as subdivision, intersection, display etc., of degree
2 or 3 rational Bernstein-Bézier triangular surfaces. For the Bernstein-Bézier triangular surface, whose degree
is larger than 3, applying the estimate (2) needs more calculation, and the estimate value is usually larger than
that obtained by the paper. The estimate (1) is the best bounds on the deviation
between
Bernstein-Bézier
ω3 (n)
n
triangular surface and the corresponding control mesh, but the constant 3 − 3n
which is on the right
hand of (1) will infinitely increase along with increasing the degree n, also it needs more triangular patches to
approximate the surface. So we suggest applying Theorem 5 is a good choice.
8
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