基本术语与标架约定
一条参数曲线称为正则曲线 ，是指它在参数区间内的一阶导数始终不为零。这一条件排除了参数化速度在某个参数值处变为零的情形，并保证参数区间中的每个参数值都对应确定的切向方向。即使曲线图像存在自交，同一空间点所对应的不同参数值仍分别具有各自的切向方向。曲线属于 C 2 C^2 C 2
本文直接采用弧长参数 s s s
r : I ⟶ R 2 , s ⟼ r ( s ) . \mathbf r:I\longrightarrow\mathbb R^2,
\qquad s\longmapsto\mathbf r(s). r : I ⟶ R 2 , s ⟼ r ( s ) . 其中 I I I r \mathbf r r C 2 C^2 C 2
∥ d r d s ∥ = 1 , \left\|\frac{d\mathbf r}{ds}\right\|=1, d s d r = 1 , 所以它自动是正则曲线。单位切向量定义为
T ( s ) : = d r d s , ∥ T ( s ) ∥ = 1. \mathbf T(s):=\frac{d\mathbf r}{ds},
\qquad
\|\mathbf T(s)\|=1. T ( s ) := d s d r , ∥ T ( s ) ∥ = 1. 本文采用直角坐标平面的标准定向 ：从 x x x y y y
J : = [ 0 − 1 1 0 ] , J:=
\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}, J := [ 0 1 − 1 0 ] , 即逆时针旋转 90 ∘ 90^\circ 9 0 ∘
N ( s ) : = J T ( s ) . \mathbf N(s):=J\mathbf T(s). N ( s ) := J T ( s ) . 对每个弧长位置 s s s r ( s ) \mathbf r(s) r ( s ) ( T ( s ) , N ( s ) ) (\mathbf T(s),\mathbf N(s)) ( T ( s ) , N ( s )) s s s r ( s ) \mathbf r(s) r ( s ) s s s T \mathbf T T N \mathbf N N T \mathbf T T 90 ∘ 90^\circ 9 0 ∘ N \mathbf N N ( T , N ) (\mathbf T,\mathbf N) ( T , N ) 正向标准正交标架 。本文将这个由单位切向量和左单位法向量组成的移动标架称为有符号 Frenet 标架 。
标架 （frame）是微分几何中的标准术语，但它与坐标系并不等同。坐标系为一个区域内的点赋予坐标值；标架则是在指定位置选取一组有序基向量，用来表示该位置处的向量。本文的 ( T , N ) (\mathbf T,\mathbf N) ( T , N ) T \mathbf T T N \mathbf N N
有符号曲率及其几何意义
曲率向量定义为单位切向量关于弧长的导数：
K ( s ) : = d T d s . \mathbf K(s):=\frac{d\mathbf T}{ds}. K ( s ) := d s d T . 曲率向量的范数定义为无符号曲率：
κ ( s ) : = ∥ K ( s ) ∥ . \kappa(s):=\|\mathbf K(s)\|. κ ( s ) := ∥ K ( s ) ∥. 有符号曲率定义为曲率向量在左单位法向量方向上的分量：
κ r ( s ) : = K ( s ) ⋅ N ( s ) = d T d s ⋅ N ( s ) . \begin{aligned}
\kappa_r(s)
&:=\mathbf K(s)\cdot\mathbf N(s)\\[0.45em]
&=\frac{d\mathbf T}{ds}\cdot\mathbf N(s).
\end{aligned} κ r ( s ) := K ( s ) ⋅ N ( s ) = d s d T ⋅ N ( s ) . 无符号曲率 κ \kappa κ κ r \kappa_r κ r
κ = ∣ κ r ∣ . \kappa=|\kappa_r|. κ = ∣ κ r ∣. 图 1 在固定左法向约定下对比有符号曲率为正、零、负的三种情形，并显示零曲率处仍然存在的正向标准正交标架。
图 1. 固定左法向约定下的平面有符号 Frenet
标架与曲率符号。沿曲线前进时，左转对应 K \mathbf K K N \mathbf N N κ r = 0 \kappa_r=0 κ r = 0 K = 0 \mathbf K=\mathbf 0 K = 0 N = J T \mathbf N=J\mathbf T N = J T
平面有符号 Frenet 公式
正交标架推导
切向量导数
由 ∥ T ∥ = 1 \|\mathbf T\|=1 ∥ T ∥ = 1
⟨ T , T ⟩ = 1. \langle\mathbf T,\mathbf T\rangle=1. ⟨ T , T ⟩ = 1. 对该恒等式关于弧长 s s s
⟨ d T d s , T ⟩ = 0. \left\langle
\frac{d\mathbf T}{ds},
\mathbf T
\right\rangle
=0. ⟨ d s d T , T ⟩ = 0. 结合 K = d T / d s \mathbf K=d\mathbf T/ds K = d T / d s
K ⊥ T . \mathbf K\perp\mathbf T. K ⊥ T . 由于 ( T , N ) (\mathbf T,\mathbf N) ( T , N ) K \mathbf K K
K = ( K ⋅ T ) T + ( K ⋅ N ) N = κ r N . \mathbf K =
(\mathbf K\cdot\mathbf T)\mathbf T
+
(\mathbf K\cdot\mathbf N)\mathbf N =
\kappa_r\mathbf N. K = ( K ⋅ T ) T + ( K ⋅ N ) N = κ r N . 最后一个等号同时使用了 K ⋅ T = 0 \mathbf K\cdot\mathbf T=0 K ⋅ T = 0 K ⋅ N = κ r \mathbf K\cdot\mathbf N=\kappa_r K ⋅ N = κ r K = d T / d s \mathbf K=d\mathbf T/ds K = d T / d s
d T d s = κ r N . \frac{d\mathbf T}{ds} =
\kappa_r\mathbf N. d s d T = κ r N . 该关系也给出无符号曲率与有符号曲率之间的联系：
κ = ∥ d T d s ∥ = ∥ κ r N ∥ = ∣ κ r ∣ . \kappa =
\left\|
\frac{d\mathbf T}{ds}
\right\| =
\|\kappa_r\mathbf N\| =
|\kappa_r|. κ = d s d T = ∥ κ r N ∥ = ∣ κ r ∣. 法向量导数
为了确定 d N / d s d\mathbf N/ds d N / d s
⟨ N , N ⟩ = 1 , ⟨ T , N ⟩ = 0. \langle\mathbf N,\mathbf N\rangle=1,
\qquad
\langle\mathbf T,\mathbf N\rangle=0. ⟨ N , N ⟩ = 1 , ⟨ T , N ⟩ = 0. 分别求导后得到
⟨ d N d s , N ⟩ = 0 , ⟨ d T d s , N ⟩ + ⟨ T , d N d s ⟩ = 0. \left\langle
\frac{d\mathbf N}{ds},
\mathbf N
\right\rangle
=0,
\qquad
\left\langle
\frac{d\mathbf T}{ds},
\mathbf N
\right\rangle
+
\left\langle
\mathbf T,
\frac{d\mathbf N}{ds}
\right\rangle
=0. ⟨ d s d N , N ⟩ = 0 , ⟨ d s d T , N ⟩ + ⟨ T , d s d N ⟩ = 0. 第一式表明 d N / d s d\mathbf N/ds d N / d s N \mathbf N N d T / d s = κ r N d\mathbf T/ds=\kappa_r\mathbf N d T / d s = κ r N ∥ N ∥ = 1 \|\mathbf N\|=1 ∥ N ∥ = 1
⟨ T , d N d s ⟩ = − κ r . \left\langle
\mathbf T,
\frac{d\mathbf N}{ds}
\right\rangle =
-\kappa_r. ⟨ T , d s d N ⟩ = − κ r . 因此，d N / d s d\mathbf N/ds d N / d s T \mathbf T T − κ r -\kappa_r − κ r N \mathbf N N
d N d s = − κ r T . \frac{d\mathbf N}{ds} =
-\kappa_r\mathbf T. d s d N = − κ r T . 至此，切向量与左单位法向量对弧长的导数都已由有符号曲率确定。上述正交性推导的标准形式见 [1, §6.A, pp. 54–55] 。
旋转算子的直接验证
第二条关系也可以直接由 N = J T \mathbf N=J\mathbf T N = J T J J J s s s 90 ∘ 90^\circ 9 0 ∘ 180 ∘ 180^\circ 18 0 ∘
J 2 = − I . J^2=-I. J 2 = − I . 对 N = J T \mathbf N=J\mathbf T N = J T d T / d s = κ r N d\mathbf T/ds=\kappa_r\mathbf N d T / d s = κ r N
d N d s = J d T d s = κ r J N = κ r J 2 T = − κ r T . \begin{aligned}
\frac{d\mathbf N}{ds}
&=
J\frac{d\mathbf T}{ds}\\[0.55em]
&=
\kappa_rJ\mathbf N\\[0.55em]
&=
\kappa_rJ^2\mathbf T\\[0.55em]
&=
-\kappa_r\mathbf T.
\end{aligned} d s d N = J d s d T = κ r J N = κ r J 2 T = − κ r T . 该计算只提供 d N / d s d\mathbf N/ds d N / d s d T / d s = κ r N d\mathbf T/ds=\kappa_r\mathbf N d T / d s = κ r N
航向角坐标推导
同一组关系可以在直角坐标中显式表示。设
r ( s ) = [ x r ( s ) y r ( s ) ] , \mathbf r(s) =
\begin{bmatrix}
x_r(s)\\
y_r(s)
\end{bmatrix}, r ( s ) = [ x r ( s ) y r ( s ) ] , 并在所讨论的参数区间上选择可微且连续展开的参考路径航向角
θ r ( s ) \theta_r(s) θ r ( s ) 2 π 2\pi 2 π 2 k π 2k\pi 2 k π 连续展开 是指在沿曲线前进时选择连续变化的角度值，避免角度经过 π \pi π − π -\pi − π
T ( s ) = [ cos θ r ( s ) sin θ r ( s ) ] , N ( s ) = [ − sin θ r ( s ) cos θ r ( s ) ] . \mathbf T(s) =
\begin{bmatrix}
\cos\theta_r(s)\\
\sin\theta_r(s)
\end{bmatrix},
\qquad
\mathbf N(s) =
\begin{bmatrix}
-\sin\theta_r(s)\\
\cos\theta_r(s)
\end{bmatrix}. T ( s ) = [ cos θ r ( s ) sin θ r ( s ) ] , N ( s ) = [ − sin θ r ( s ) cos θ r ( s ) ] . 在这一表示下，有符号曲率与切向角满足
κ r ( s ) = d θ r d s . \kappa_r(s) =
\frac{d\theta_r}{ds}. κ r ( s ) = d s d θ r . 图 2(a) 给出切向量与左单位法向量的直角坐标分量；图 2(b) 则以相邻切向量的转角说明 κ r = d θ r / d s \kappa_r=d\theta_r/ds κ r = d θ r / d s
图 2. 航向角的坐标表示及其沿弧长的变化。（a）T \mathbf T T N = J T \mathbf N=J\mathbf
T N = J T Δ θ r \Delta\theta_r Δ θ r Δ s \Delta s Δ s
为得到该关系以及标架导数的显式形式，分别对 T \mathbf T T N \mathbf N N
d T d s = d θ r d s [ − sin θ r cos θ r ] = κ r N , d N d s = d θ r d s [ − cos θ r − sin θ r ] = − κ r T . \begin{aligned}
\frac{d\mathbf T}{ds}
&=
\frac{d\theta_r}{ds}
\begin{bmatrix}
-\sin\theta_r\\
\cos\theta_r
\end{bmatrix} =
\kappa_r\mathbf N,\\[0.8em]
\frac{d\mathbf N}{ds}
&=
\frac{d\theta_r}{ds}
\begin{bmatrix}
-\cos\theta_r\\
-\sin\theta_r
\end{bmatrix} =
-\kappa_r\mathbf T.
\end{aligned} d s d T d s d N = d s d θ r [ − sin θ r cos θ r ] = κ r N , = d s d θ r [ − cos θ r − sin θ r ] = − κ r T . 第一行表明，切向角相对于弧长的变化率正是曲率向量在左单位法向量方向上的有符号分量；第二行则与正交性推导所得的法向量导数一致。切向角表示及关系
κ r = d θ r / d s \kappa_r=d\theta_r/ds κ r = d θ r / d s [2, Theorem 1.17, p. 8] 。
Frenet 公式的矩阵形式
综合正交标架推导与航向角坐标表示，得到平面有符号 Frenet 公式
[ d T d s d N d s ] = [ 0 κ r − κ r 0 ] [ T N ] . \begin{bmatrix}
\dfrac{d\mathbf T}{ds}\\[0.8em]
\dfrac{d\mathbf N}{ds}
\end{bmatrix} =
\begin{bmatrix}
0 & \kappa_r\\
-\kappa_r & 0
\end{bmatrix}
\begin{bmatrix}
\mathbf T\\
\mathbf N
\end{bmatrix}. d s d T d s d N = [ 0 − κ r κ r 0 ] [ T N ] . 正交标架推导不需要选取 x , y x,y x , y κ r = d θ r / d s \kappa_r=d\theta_r/ds κ r = d θ r / d s κ r = 0 \kappa_r=0 κ r = 0
与经典主法向量的关系
本文的 N = J T \mathbf N=J\mathbf T N = J T 左单位法向量 ，需要与经典 Frenet 标架中的主法向量 区分。经典主法向量按
N p : = T ′ ∥ T ′ ∥ \mathbf N_{\mathrm p}:=
\frac{\mathbf T'}{\|\mathbf T'\|} N p := ∥ T ′ ∥ T ′ 定义，因此只在无符号曲率 κ = ∥ T ′ ∥ > 0 \kappa=\|\mathbf T'\|>0 κ = ∥ T ′ ∥ > 0 κ = 0 \kappa=0 κ = 0 T ′ = 0 \mathbf T'=\mathbf 0 T ′ = 0 0 / 0 \mathbf 0/0 0 /0 N = J T \mathbf N=J\mathbf T N = J T J J J T \mathbf T T ( T , N ) (\mathbf T,\mathbf N) ( T , N ) [1, §§2.D, 3.A–B, 6.A] ；经典 Frenet 标架在零曲率处失效的问题见 [3, §2.1, p. 98] 。
这不是经典 Frenet 标架定义内部的矛盾，而是它明确的适用条件：经典构造只在
κ > 0 \kappa>0 κ > 0 { s ∈ I : κ ( s ) > 0 } \{s\in I:\kappa(s)>0\} { s ∈ I : κ ( s ) > 0 }
由前文已经得到的 d T / d s = κ r N d\mathbf T/ds=\kappa_r\mathbf N d T / d s = κ r N κ > 0 \kappa>0 κ > 0
N p = T ′ ∥ T ′ ∥ = κ r ∣ κ r ∣ N = sgn ( κ r ) N . \begin{aligned}
\mathbf N_{\mathrm p}
&=
\frac{\mathbf T'}{\|\mathbf T'\|}\\[0.55em]
&=
\frac{\kappa_r}{|\kappa_r|}\mathbf N\\[0.55em]
&=
\operatorname{sgn}(\kappa_r)\mathbf N.
\end{aligned} N p = ∥ T ′ ∥ T ′ = ∣ κ r ∣ κ r N = sgn ( κ r ) N . 其中 sgn \operatorname{sgn} sgn
sgn ( x ) = { 1 , x > 0 , 0 , x = 0 , − 1 , x < 0. \operatorname{sgn}(x)=
\begin{cases}
1, & x>0,\\
0, & x=0,\\
-1, & x<0.
\end{cases} sgn ( x ) = ⎩ ⎨ ⎧ 1 , 0 , − 1 , x > 0 , x = 0 , x < 0. 上式只讨论 κ > 0 \kappa>0 κ > 0 sgn ( κ r ) = ± 1 \operatorname{sgn}(\kappa_r)=\pm1 sgn ( κ r ) = ± 1 κ r = 0 \kappa_r=0 κ r = 0 N p \mathbf N_{\mathrm p} N p
这说明二者只在左转的 κ r > 0 \kappa_r>0 κ r > 0 κ r < 0 \kappa_r<0 κ r < 0 − N -\mathbf N − N κ r = 0 \kappa_r=0 κ r = 0 N p \mathbf N_{\mathrm p} N p N = J T \mathbf N=J\mathbf T N = J T
曲率变号示例
零曲率造成的困难并不总能通过取极限消除。下面给出一个主法向量必然发生翻转的例子。设路径的弧长参数满足 s ∈ [ 0 , L ] s\in[0,L] s ∈ [ 0 , L ] s ∗ ∈ ( 0 , L ) s_{\ast}\in(0,L) s ∗ ∈ ( 0 , L )
T ( s ) = [ cos ( ( s − s ∗ ) 2 / 2 ) sin ( ( s − s ∗ ) 2 / 2 ) ] \mathbf T(s)=
\begin{bmatrix}
\cos\bigl((s-s_{\ast})^2/2\bigr)\\
\sin\bigl((s-s_{\ast})^2/2\bigr)
\end{bmatrix} T ( s ) = [ cos ( ( s − s ∗ ) 2 /2 ) sin ( ( s − s ∗ ) 2 /2 ) ] 它的长度恒为 1 1 1
r ( s ) : = r ( 0 ) + ∫ 0 s T ( u ) d u , \mathbf r(s)
:=
\mathbf r(0)+
\int_0^s \mathbf T(u)\,du, r ( s ) := r ( 0 ) + ∫ 0 s T ( u ) d u , 便有 r ′ ( s ) = T ( s ) \mathbf r'(s)=\mathbf T(s) r ′ ( s ) = T ( s ) ∥ r ′ ( s ) ∥ = 1 \|\mathbf r'(s)\|=1 ∥ r ′ ( s ) ∥ = 1
θ r ( s ) = ( s − s ∗ ) 2 2 . \theta_r(s)=\frac{(s-s_{\ast})^2}{2}. θ r ( s ) = 2 ( s − s ∗ ) 2 . 因此
N ( s ) = J T ( s ) = [ − sin ( ( s − s ∗ ) 2 / 2 ) cos ( ( s − s ∗ ) 2 / 2 ) ] , \mathbf N(s)=J\mathbf T(s)=
\begin{bmatrix}
-\sin\bigl((s-s_{\ast})^2/2\bigr)\\
\cos\bigl((s-s_{\ast})^2/2\bigr)
\end{bmatrix}, N ( s ) = J T ( s ) = [ − sin ( ( s − s ∗ ) 2 /2 ) cos ( ( s − s ∗ ) 2 /2 ) ] , 并且
d T d s = ( s − s ∗ ) [ − sin ( ( s − s ∗ ) 2 / 2 ) cos ( ( s − s ∗ ) 2 / 2 ) ] = ( s − s ∗ ) N ( s ) . \frac{d\mathbf T}{ds}
=
(s-s_{\ast})
\begin{bmatrix}
-\sin\bigl((s-s_{\ast})^2/2\bigr)\\
\cos\bigl((s-s_{\ast})^2/2\bigr)
\end{bmatrix}
=
(s-s_{\ast})\mathbf N(s). d s d T = ( s − s ∗ ) [ − sin ( ( s − s ∗ ) 2 /2 ) cos ( ( s − s ∗ ) 2 /2 ) ] = ( s − s ∗ ) N ( s ) . 与 d T / d s = κ r N d\mathbf T/ds=\kappa_r\mathbf N d T / d s = κ r N
κ r ( s ) = s − s ∗ , κ ( s ) = ∣ s − s ∗ ∣ . \kappa_r(s)=s-s_{\ast},
\qquad
\kappa(s)=|s-s_{\ast}|. κ r ( s ) = s − s ∗ , κ ( s ) = ∣ s − s ∗ ∣. 这意味着曲线在 s < s ∗ s<s_{\ast} s < s ∗ s > s ∗ s>s_{\ast} s > s ∗ s = s ∗ s=s_{\ast} s = s ∗ s ≠ s ∗ s\ne s_{\ast} s = s ∗
N p ( s ) = d T / d s ∥ d T / d s ∥ = s − s ∗ ∣ s − s ∗ ∣ N ( s ) = { − N ( s ) , s < s ∗ , N ( s ) , s > s ∗ . \mathbf N_{\mathrm p}(s)
=
\frac{d\mathbf T/ds}{\|d\mathbf T/ds\|}
=
\frac{s-s_{\ast}}{|s-s_{\ast}|}\mathbf N(s)
=
\begin{cases}
-\mathbf N(s), & s<s_{\ast},\\
\mathbf N(s), & s>s_{\ast}.
\end{cases} N p ( s ) = ∥ d T / d s ∥ d T / d s = ∣ s − s ∗ ∣ s − s ∗ N ( s ) = { − N ( s ) , N ( s ) , s < s ∗ , s > s ∗ . 由于 N ( s ∗ ) = ( 0 , 1 ) T \mathbf N(s_{\ast})=(0,1)^{\mathsf T} N ( s ∗ ) = ( 0 , 1 ) T
lim s → s ∗ − N p ( s ) = [ 0 − 1 ] , lim s → s ∗ + N p ( s ) = [ 0 1 ] . \lim_{s\to s_{\ast}^-}\mathbf N_{\mathrm p}(s)
=
\begin{bmatrix}0\\-1\end{bmatrix},
\qquad
\lim_{s\to s_{\ast}^+}\mathbf N_{\mathrm p}(s)
=
\begin{bmatrix}0\\1\end{bmatrix}. s → s ∗ − lim N p ( s ) = [ 0 − 1 ] , s → s ∗ + lim N p ( s ) = [ 0 1 ] . 两个极限方向相反，所以不可能在 s = s ∗ s=s_{\ast} s = s ∗ N p \mathbf N_{\mathrm p} N p N = J T \mathbf N=J\mathbf T N = J T
这个例子说明的不是“每个孤立零曲率点都会导致翻转”，而是经典定义并不保证主法向量能够跨过零曲率点连续延拓。如果 κ r \kappa_r κ r T ′ / ∥ T ′ ∥ \mathbf T'/\|\mathbf T'\| T ′ /∥ T ′ ∥ T ′ = 0 \mathbf T'=0 T ′ = 0
后续模型使用的几何关系
后续 Frenet 坐标下的车辆运动学将直接使用以下参考路径几何关系：
d r d s = T , N = J T , d T d s = κ r N , d N d s = − κ r T , d θ r d s = κ r . \begin{aligned}
\frac{d\mathbf r}{ds} &= \mathbf T,\\[0.5em]
\mathbf N &= J\mathbf T,\\[0.5em]
\frac{d\mathbf T}{ds} &= \kappa_r\mathbf N,\\[0.5em]
\frac{d\mathbf N}{ds} &= -\kappa_r\mathbf T,\\[0.5em]
\frac{d\theta_r}{ds} &= \kappa_r.
\end{aligned} d s d r N d s d T d s d N d s d θ r = T , = J T , = κ r N , = − κ r T , = κ r . 这些关系统一了参考路径的位置、切向、左法向、航向角和有符号曲率。后续将在该标架上定义车辆相对于参考路径的纵向位置、横向偏差与航向误差，并推导 Frenet 坐标下的车辆运动学。
参考文献
Anton Petrunin and Sergio Zamora Barrera,
What Is Differential Geometry? Curves and Surfaces
, arXiv:2012.11814v7, 2025.
Neil Donaldson,
Math 162A: Introduction to Differential Geometry
, course notes, University of California, Irvine, Winter 2024.
Daniel Carroll, Emek Köse, and Ivan Sterling,
Improving Frenet’s Frame Using Bishop’s Frame
, Journal of Mathematics Research , vol. 5, no. 4, 2013, pp. 97–106.