Appendix A
The proof for the uniqueness of solution to linear system (8)
Proof. The linear system (8) can be simplified as MX = B. The coefficient matrix M with the size
(4N − 2) × (4N − 2) can be expressed
12 / 5 3 / 5 1

 3 / 5 12 / 5 2 1



 1

2
2
1


1
1




2 1 12 / 5 3 / 5 1


1
3 / 5 12 / 5 2 1




1
2


1




... ..



1 2
2
1 


1
1



2 1 12 / 5 3 / 5 


1
3 / 5 12 / 5

(A1)
and
B = (9/5P0+ 6/5P1, 6/5P0+ 9/5P1, 0, 2P1,..., 9/5PN−1+ 6/5PN, 6/5PN−1+ 9/5PN)T.
(A2)
To prove the linear system (8) has the unique solution, it is necessary to prove M is a nonsingular
matrix.
In fact, the matrix M can be written as M = S +A, where S is composed by 2×2 block matrices in
the diagonal of M. Hence S and A can be denoted as respectively
 S1




S





 0
  AT
 1

A




0
S2
0
...
0
A1
0
A2
 A2T
0
A3
...
0
 A2TN  2









S N 




.

A2 N  2 

0 
It is not difficult to find that S is a symmetric matrix and A is an anti-symmetric matrix. Assume
that X = (x1, x2 ... x4N−3, x4N−2)T is the solution to the equation MX = 0. To prove M is invertible,
it is only needed to prove x = 0. Since XTMX= XTMTX = 0,
M = S + A, MT= S − A, then we have XTSX = XTAX = 0. Denote X = (Y1, Y2 ... Y2N−1)T, where Yi=
N
12 / 5
3/5 
(x2i−1,x2i). Because XTSX = 0,  Y2Ti1 SiY2i 1  0 and in S the block-matrix Si  
 is
 3 / 5 12 / 5
i 1
positive-definite, Y2Ti 1 SiY2i 1  0, thus
Y2i-1=0 for i=1...N. Therefore X= (0, Y2, 0, Y4 ... Y2N−2, 0)T, SX = 0.
On the other hand, MX = 0, thus AX = 0 which is equivalent to be
 A1
  AT
 2






A3
 A4T
A5
...
...
 A2TN  4
  Y2T 
 T 
  Y4 
  Y6T 
0



A2 N 3  Y2TN  4 


 A2TN  2  Y2TN  2 
The coefficient matrix of linear equation above is a block-diagonal matrix and Ai is nonsingular
for any i, therefore Y2i = 0 for i = 1...N-1. Now it has been proved that X = 0. Therefore M is a
nonsingular matrix.
Appendix B The real solution to Eq.(10)
Eq.(10) can be simplified to be a quintic equation
F (u )  [( ri (u )  Pi )  ri(u )]( Pi 1  Pi ) 2
 [(ri (u )  Pi )  ( Pi 1  Pi )]
[ri(u )  ( Pi 1  Pi )]  0
(B1)
From eq.(9), since Di(0) = Di(1) = 0, equation F(u) = 0 must have at least one real root on the open
interval (0,1) and both sides of Eq.(B1) have two factors u and u
simplified to be a cubic equation about unknown u having the following form
au3+bu2+cu+d=0
(B2)
where
a  (3bi  Pi  3ai  Pi 1 )2 ( Pi 1  Pi )2
 [(3bi  Pi  3ai  Pi 1 )  ( Pi 1  Pi )]2 ;
b  [(3bi  Pi  3ai  Pi 1 )  (4 Pi 1  Pi  2bi  7ai )]
( Pi 1  Pi ) 2  [(3bi  Pi  3ai  Pi 1 )  ( Pi 1  Pi )]
[(4 Pi 1  Pi  2bi  7ai )  ( Pi 1  Pi )];
c  [(3bi  Pi  3ai  Pi 1 )  (2 Pi 1  Pi  ai )
 (2 Pi 1  Pi  3ai )  (2 Pi 1  2bi  4ai )]
( Pi 1  Pi ) 2  [(3bi  Pi  3ai  Pi 1 )
 ( Pi 1  Pi )][(2 Pi 1  Pi  ai )  ( Pi 1  Pi )]
 [(2 Pi 1  Pi  3ai )  ( Pi 1  Pi )]
[(2 Pi 1  2bi  4ai )  ( Pi 1  Pi )];
d  [(( Pi 1  Pi )  (2 Pi 1  2bi  4ai )
 (2 Pi 1  Pi  3ai )  (ai  Pi 1 ))]
( Pi 1  Pi ) 2  ( Pi 1  Pi ) 2 [(2 Pi 1  2bi  4ai )
(B3)
 ( Pi 1  Pi )]  [(2 Pi 1  Pi  3ai )  ( Pi 1  Pi )]
[(ai  Pi 1 )  ( Pi 1  Pi )].
If a = 0, Eq. (B2) is a quadratic equation, we can use the formula for the root of quadratic equation
ui 
c  c 2  4bd
.
2b
and find that ones in the interval (0,1). Take the larger D(ui*) among the values as the chord error
Erri.
If a≠0, then the method of finding the real roots is given below. Let A = b2 ac, B = bc ad, C =
c2 bd and ∆ = B2 AC.
If A = B = 0, then Eq.(B2) has three repeated real roots
ui1  ui 2  ui 3  
b
.
3a
If ∆ > 0, then equation has only one real root
ui 
b  ( 3 Y1  3 Y2 )
3a
,
where
  B  B 2  4 AC
Y1,2  Ab  3a 

2


.


If ∆ = 0 and A ≠ 0, then equation has three real roots
ui1  
b B
B
 , ui 2  ui 3  
.
a A
2A
If ∆ < 0, the equation has three different real roots
ui1 
b  2 A cos
3a
  arccos T , T 

3,u
i 2,i 3



b  A  cos  3 sin 
3
3


, where
3a
2 Ab  3aB
2 A3
( A  0, 1  T  1).
Find all the real roots lying on the open interval (0,1) and select ui* such that D(ui*) is the largest
one among all. Then at the segment Pi−1Pi, the chord error is Erri= D(ui*).
Appendix C The proof of Theorem 1
Proof. By the property of G2 continuity, then c1d1 and d2c2 are coplanar. In fact, presume that c1d1||
aibi. Since c1, d1, c2, d2 are on OSQ by the property of G2 and OSQ ∩ OSi = c2d2, then c2d2||aibi.
Besides, since the path is G2 at Pi−1 and Pi, it is known that c1d1  1 ai bi and c2d2  2 ai bi ,
where
1,
2>
0.
Let d1Q  t d1c2 . Hence from the Bezier segment ri1(u) between Pi−1 and Q,
OQ 
ri1 (1)  ri1 (1) d1c1  d1Q
 a b d c

  21 i i 1 32 .
3
3

|| ri1 (1) ||
|| 3d1Q ||
t || 3d1c2 ||
From Bezier segment ri2(u) between Q and Pi,
(C1)
OQ 
ri2 (0)  ri2 (0) c2 d 2  Qc2
2 ai bi  d1c2


.
3
3
|| ri2 (0) ||
|| 3Qc2 ||
(1  t ) 2 || 3d1c2 ||3
(C2)
Therefore for the general case (the control polygon Pi−1aibiPi is spatial), we have OQ  OQ for
any
1,
2,
t. That deducts the contradiction that the transitional curve at Q has no G2 continuity.
Hence it is proved that the new ridge line c1d1 must intersect the old ridge line aibi. Similarly we
can conclude that d2c2 must intersect aibi.
For the case that Pi−1ai and Pibi are skew lines. Claim that c1d1 ∦c2d2. In fact, since c1d1 and d2c2
intersect aibi, OSi−1 and OSi coincide if c1d1||c2d2. This contradicts that Pi−1ai and Pibi are skew
lines. Let e1 be the intersection point of c1d1 and aibi, and e2 be the intersection point of d2c2 and
aibi.
Assume e1≠e2. Since e1e2 in the plane OSQ, line aibi is on the plane OSQ. Thus ai, bi ∈ OSQ. On
the other hand, c1, d1, c2, d2∈OSQ, hence c1ai∈OSQ. That is Pi−1ai∈OSQ. Similarly we can know
that Pibi ∈OSQ. This contradicts Pi−1ai and Pibi are skew lines. Hence e1= e2.
Appendix D The proof of Theorem 2
Before proving Theorem 2, note there is an important equivalent condition for G2 continuity [16].
Theorem 3. [Equivalent condition of G2 continuity] The parametric curve r(u) is G2 if and only if
both the vectors T and T   N are continuous, where T is the unit tangent vector, N is the
unit normal vector and κ is curvature magnitude along the curve r(u).
The proof of Theorem 2:
Proof. For G2 continuity at Pi−1, since the curvature vector of the path at Pi−1 is κ( Pi 1 )  κi (0),
by the equivalent condition of G2 continuity,
c1d1  Pi 1c1
3
|| Pi 1c1 ||

ai bi  Pi 1ai
|| Pi 1ai ||3
.
Since c1d1  k4 c1e  k4 (1  k1 ) Pi 1ai  k3k4 ai bi , the sufficient and necessary condition for G2
continuity at Pi−1 is
k22=k3k4.
(D1)
Similarly for
continuity at Pi−1, it can be obtained that the sufficient and necessary condition
2
for G continuity at Pi is
k22= (1 − k3)k5.
(D2)
2
For G continuity at Q, then
G2
c2 d 2  Qc2
3
|| Qc2 ||

d1c1  d1Q
|| d1Q ||3
Since
Qc2  (1  k6 )d1c2  (1  k6 )[(1  k4 )c1e  (1  k5 )ed2 ]
.
and
d1Q  k6 d1c2  k6 [(1  k4 )c1e  (1  k5 )ed2 ],
then it is easy to obtain that the sufficient and necessary condition for G2 continuity at Q is
k5 (1  k4 )
(1  k6 )
2

k4 (1  k5 )
k62
.
(D3)