移动机器人与非完整约束Mobile Robots & Nonholonomic Constraints

1为什么机器人需要它Why robots need this

2直觉：一个轮子只会往前滚Intuition: a wheel only knows how to roll forward

• 让左轮停住、右轮转动（左轮速度接近 0，右轮转速大于 3 rad/s）。ICR 在哪里？半径是多少？（可以点"绕一个轮子转"预设。）Stop the left wheel and turn the right one (left near 0, right above 3 rad/s). Where is the ICR, and what is the radius? (The "Pivot" preset does it.)
• 切换到"全向"，让机器人纯侧移（\(v_y>0.5\)，\(v_x\) 与 \(\omega\) 接近 0）。差速机器人为什么做不到？Switch to "Omni" and make the robot slide purely sideways (\(v_y>0.5\), \(v_x\) and \(\omega\) near 0). Why can't a differential-drive robot do this?
• 回到差速模式，打开"里程计"，把右轮半径误差设成 3% 左右，让机器人大致走直线：估计位姿与真实位姿相差超过 0.3 m 需要走多远？Back in differential mode, turn "Odometry" on, set the right-wheel radius error to about 3%, and drive roughly straight: how far must the robot go before the estimate is off by more than 0.3 m?

3动手玩：不能横着滑，却到得了Play: you cannot slide sideways, but you can get there

• 拖动小车，把鼠标向侧面拉出至少 0.5 m。小车的侧向位置变了吗？Drag the robot and pull the pointer at least 0.5 m to the side. Does the robot's sideways position change?
• 先用 \(\varepsilon\le 0.3\) 摆动一次，再用 \(\varepsilon\ge 0.6\) 摆动一次。第二次的横移是第一次的几倍？和 \(\varepsilon^2\) 的比值一致吗？Wiggle once with \(\varepsilon\le 0.3\), then once with \(\varepsilon\ge 0.6\). How many times larger is the second shift, and does it match the ratio of \(\varepsilon^2\)?
• 用摆动（和前进/后退微调）把小车停进车位：轮轴中点进入车位中心附近，朝向偏差小于 7°。Use wiggles (plus forward/back nudges) to park the robot in the slot: axle midpoint near the slot center, heading within 7°.

4第二、第三视角：底盘家族、可达集与最短路径Second and third views: the base zoo, reachable sets and shortest paths

四种常见底盘一张表Four common bases in one table

能到哪里：可达集Where can you get to: reachable sets

• 把目标放在车身正侧面（\(|x|<0.6\)，\(|y|\) 在 1 到 3 之间），让汽车的最短路径长度超过直线距离的 3 倍。减小 \(R\) 会怎样？Put the target right at the car's side (\(|x|<0.6\), \(|y|\) between 1 and 3) so that the car's shortest path is more than 3 times the straight-line distance. What happens as you reduce \(R\)?

指定终点朝向：Dubins 最短路径With a prescribed final heading: Dubins shortest paths

• 把终点拖到离起点很近的地方（比如侧前方不到 \(2R\)），让最短路径变成 RLR 或 LRL（三段全是圆弧）。直线段去哪了？Drag the goal close to the start (say less than \(2R\) ahead and to the side) until the shortest path becomes RLR or LRL (three arcs). Where did the straight segment go?

如果允许倒车，最短路径会换成 Reeds–Shepp 曲线（Reeds & Shepp，1990），有倒车尖点（换向点）的形式；它的组合更多，但结构相同：直线和最大曲率圆弧的拼接。

If reversing is allowed, the shortest paths become Reeds–Shepp curves (Reeds & Shepp, 1990), which contain cusps where the car changes direction; there are more words, but the structure is the same: straight segments and maximum-curvature arcs glued together.

5再看数学The math

提示：把鼠标悬停在公式中带底色的符号上，FIG 1、FIG 2、FIG 5、FIG 6 里对应的元素会被高亮。Tip: hover the highlighted symbols in the formulas; the matching elements in FIG 1, 2, 5 and 6 light up.

运动学模型Kinematic models

独轮车是最基本的模型：位姿 \(q=(x,y,\theta)\)，控制 \(v\)（沿车头方向的速度）和 \(\omega\)。它是 SE(2) 上的"左不变"系统（5.4–5.5 节）：

The unicycle is the basic model: pose \(q=(x,y,\theta)\), controls \(v\) (speed along the heading) and \(\omega\). It is a left-invariant system on SE(2) (Lessons 5.4–5.5):

差速驱动只是它的另一套坐标：轮半径 \(r\)，轮距 \(W\)，左右轮角速度 \(\omega_L,\omega_R\)，则

Differential drive is the same model in other coordinates: wheel radius \(r\), track \(W\), wheel rates \(\omega_L,\omega_R\), then

\(R\) 是从机器人中心沿车轴量到 ICR 的有向距离（\(R>0\) 在左边）。ICR 在世界坐标系中的位置是 \(c=p+\tfrac1\omega(-\dot y,\dot x)\)，即把速度向量逆时针转 90° 再除以 \(\omega\)。

\(R\) is the signed distance from the robot center to the ICR along the axle (\(R>0\) on the left). In the world frame the ICR is at \(c=p+\tfrac1\omega(-\dot y,\dot x)\): rotate the velocity vector by 90° counter-clockwise and divide by \(\omega\).

自行车 / 阿克曼模型：把前后两对轮各合并成一个轮子，轴距 \(L\)，参考点取后轴中点，前轮转角 \(\delta\)：

Bicycle / Ackermann model: merge each pair of wheels into one, wheelbase \(L\), reference point at the rear-axle midpoint, front steering angle \(\delta\):

真实汽车内外轮转角不同（内轮转得更多，\(\tan\delta_{\text{in}}=L/(R-w/2)\)，\(w\) 为轮距），使四个轮子的垂线交于同一个 ICR，这就是阿克曼转向几何。全向（麦克纳姆）底盘则没有约束：体坐标系下的速度 \((v_x,v_y,\omega)\) 可以任意给定；四个麦克纳姆轮（常见的 X 形滚子布局，\(\ell\) 为前后半轴距与左右半轮距之和）需要的转速为

Real cars steer inner and outer wheels differently (the inner one more, \(\tan\delta_{\text{in}}=L/(R-w/2)\) with \(w\) the track) so that the perpendiculars of all four wheels meet at one ICR: Ackermann geometry. An omnidirectional (mecanum) base has no constraint: body-frame velocity \((v_x,v_y,\omega)\) can be chosen freely; four mecanum wheels (the common X-roller layout, \(\ell\) the sum of the half-wheelbase and half-track) need the wheel rates

（符号取决于滚子方向和轮子编号，换了安装方式矩阵里的正负号会变；列向量两两正交，所以正解是 \((J^\top J)^{-1}J^\top=r\,\operatorname{diag}(\tfrac14,\tfrac14,\tfrac1{4\ell^2})J^\top\)，\(J\) 为上面的 4×3 矩阵。）三个控制对三个自由度，\(J\) 列满秩（有左逆），所以是完整系统。

(Signs depend on roller orientation and wheel numbering and change with the mounting; the columns are mutually orthogonal, so the forward map is \((J^\top J)^{-1}J^\top=r\,\operatorname{diag}(\tfrac14,\tfrac14,\tfrac1{4\ell^2})J^\top\), with \(J\) the 4×3 matrix above.) Three controls for three DoF and \(J\) has full column rank (a left inverse): a holonomic system.

约束：Pfaffian 形式、可积性与 FrobeniusThe constraint: Pfaffian form, integrability and Frobenius

"轮子不侧滑"写成对速度的线性约束（Pfaffian 约束）：

"The wheel does not slip sideways" is a linear constraint on velocities (a Pfaffian constraint):

\(\Delta(q)\) 是每一点处"允许的速度方向"，是 \(\mathbb R^3\) 里的一个二维平面（"分布"）。如果存在函数 \(\varphi(q)\) 使得 \(\mathrm d\varphi\) 与 \(A\) 成比例，那么 \(\varphi(q)=\text{const}\) 是一个曲面，机器人永远待在上面，约束是完整的，构型空间降维。Frobenius 定理说：这当且仅当分布 \(\Delta\) 对合（involutive）：\(\Delta\) 中任意两个向量场的李括号仍然在 \(\Delta\) 里。

\(\Delta(q)\) is the set of allowed velocity directions at each point: a two-dimensional plane (a "distribution") inside \(\mathbb R^3\). If a function \(\varphi(q)\) exists whose \(\mathrm d\varphi\) is proportional to \(A\), then \(\varphi(q)=\text{const}\) is a surface the robot can never leave, the constraint is holonomic and the configuration space loses a dimension. Frobenius' theorem says this happens iff the distribution \(\Delta\) is involutive: the Lie bracket of any two vector fields in \(\Delta\) stays in \(\Delta\).

李括号、摆动与 ε²The Lie bracket, the wiggle and ε²

两个向量场的李括号（3.6 节的流的对易子）定义为 \([f,g]=\tfrac{\partial g}{\partial q}f-\tfrac{\partial f}{\partial q}g\)。独轮车的两个场：

The Lie bracket of two vector fields (the commutator of flows from Lesson 3.6) is \([f,g]=\tfrac{\partial g}{\partial q}f-\tfrac{\partial f}{\partial q}g\). For the unicycle's two fields:

它指向车的右侧，不在 \(\Delta\) 里，所以对合性失败，约束是非完整的。它的几何意义是流的对易子：先沿 \(g_1\) 流 \(\varepsilon\)，再沿 \(g_2\) 流 \(\varepsilon\)，再沿 \(-g_1\)，最后沿 \(-g_2\)，则

It points to the car's right and is not in \(\Delta\), so involutivity fails and the constraint is nonholonomic. Its geometric meaning is the commutator of flows: flow along \(g_1\) for \(\varepsilon\), then \(g_2\) for \(\varepsilon\), then \(-g_1\), finally \(-g_2\), and

独轮车的精确结果可以直接算：在车体坐标系下，净位移是 \((\varepsilon(1-\cos\varepsilon),\,-\varepsilon\sin\varepsilon)\approx(\tfrac{\varepsilon^3}{2},\,-\varepsilon^2)\)，朝向变化严格为零。FIG 2 右图用数值积分验证了这一点：横移 \(/\varepsilon^2\) 在 \(\varepsilon=0.0125\) 时等于 0.99997。一次摆动要前后共开 \(2\varepsilon\)，再原地来回转 \(2\varepsilon\) 弧度，取 \(\varepsilon=\sqrt d\) 时行驶距离 \(\approx2\sqrt d\)：想横移 1 cm 就要来回开约 20 cm，外加来回转约 0.2 rad（约 11°），小位移的"汇率"很差。

For the unicycle the exact result can be computed directly: in the body frame the net displacement is \((\varepsilon(1-\cos\varepsilon),\,-\varepsilon\sin\varepsilon)\approx(\tfrac{\varepsilon^3}{2},\,-\varepsilon^2)\), with exactly zero change in heading. The right panel of FIG 2 checks this numerically: sideways shift \(/\varepsilon^2\) equals 0.99997 at \(\varepsilon=0.0125\). One wiggle drives \(2\varepsilon\) forward and back and turns \(2\varepsilon\) rad in place, so with \(\varepsilon=\sqrt d\) the distance driven is \(\approx2\sqrt d\): to shift 1 cm sideways you drive about 20 cm back and forth, plus about 0.2 rad (11°) of turning, a poor exchange rate for small shifts.

可控性：Chow–RashevskiiControllability: Chow–Rashevskii

把 \(g_1,g_2\) 以及它们反复取李括号得到的所有向量场放在一起，如果在每一点都张成整个 \(\mathbb R^n\)（"李代数秩条件"），则系统在连通的构型空间里可控：任意两个位姿都能用允许的运动连起来（Chow 1939，Rashevskii 1938）。独轮车：\(\operatorname{span}\{g_1,g_2,[g_1,g_2]\}=\mathbb R^3\) 处处成立（\(\det[g_1,g_2,[g_1,g_2]]=1\)），所以可以停到任何位姿，只是代价是每个"缺失方向"上平方根的路程（亚黎曼几何里的"非完整度"）。对比：一个不可积约束（冰刀）的可达集是整个构型空间，一个可积约束（珠子在铁丝上）的可达集只是一条曲线。

Collect \(g_1,g_2\) and all vector fields obtained from them by repeated Lie brackets. If these span all of \(\mathbb R^n\) at every point (the "Lie algebra rank condition"), the system is controllable on a connected configuration space: any two poses can be joined by allowed motions (Chow 1939, Rashevskii 1938). For the unicycle \(\operatorname{span}\{g_1,g_2,[g_1,g_2]\}=\mathbb R^3\) everywhere (\(\det[g_1,g_2,[g_1,g_2]]=1\)), so every pose is reachable, at the price of a square-root-sized detour for each "missing direction" (the "degree of nonholonomy" of sub-Riemannian geometry). Compare: with a non-integrable constraint (the skate) the reachable set is the whole configuration space; with an integrable one (the bead on a wire) it is just a curve.

Dubins 路径的闭式解Closed-form Dubins paths

把起点平移旋转到原点、朝向 +x，取 \(R=1\)，终点距离 \(d\)，起终点朝向相对连线方向的夹角为 \(\alpha,\beta\)。每个字都有闭式解，比如 LSL：

Move the start to the origin with heading +x, set \(R=1\), let the goal be at distance \(d\) and let \(\alpha,\beta\) be the start and goal headings relative to the line joining them. Every word has a closed form; for example LSL:

（\(t,q\) 是两段圆弧的转角，\(p\) 是直线段长度，总长 \(R(t+p+q)\)；这是 Shkel–Lumelsky 的归一化。）六个字里有些在几何上不存在（\(p^2<0\) 或余弦超出 \([-1,1]\)），取存在的里面最短的即可。我们对 5000 对随机的起终点位姿做了测试：把路径沿弧积分到终点，最大位置误差 \(1.8\times10^{-14}\)，最大朝向误差 \(7.5\times10^{-15}\)；另外对 150 对随机位姿，用 Newton 迭代独立地解"三段式"方程，没有找到比这六个字更短的解。

(\(t,q\) are the turning angles of the two arcs, \(p\) is the straight length, and the total length is \(R(t+p+q)\); this is the Shkel–Lumelsky normalization.) Some of the six words do not exist geometrically (\(p^2<0\) or a cosine outside \([-1,1]\)); take the shortest of those that do. We tested 5000 random start/goal pose pairs: integrating the path along its arcs to the end gives a maximum position error of \(1.8\times10^{-14}\) and heading error of \(7.5\times10^{-15}\); and for 150 random pairs an independent Newton solver for three-segment words found nothing shorter than these six words.

里程计：航位推算Odometry: dead reckoning

一个采样周期里左右轮编码器给出行驶弧长 \(\Delta s_L,\Delta s_R\)，则 \(\Delta s=\tfrac12(\Delta s_R+\Delta s_L)\)，\(\Delta\theta=(\Delta s_R-\Delta s_L)/W\)。若周期内 \(v,\omega\) 近似不变，机器人走的是一段圆弧，精确更新是（见 2.5 节，对常值输入的 ODE 精确积分）：

In one sampling period the wheel encoders give the arc lengths \(\Delta s_L,\Delta s_R\); then \(\Delta s=\tfrac12(\Delta s_R+\Delta s_L)\) and \(\Delta\theta=(\Delta s_R-\Delta s_L)/W\). If \(v,\omega\) are roughly constant within the period the robot follows an arc, and the exact update (the exact integral of the ODE with constant input, Lesson 2.5) is

这是弦长（\(2\tfrac{\Delta s}{\Delta\theta}\sin\tfrac{\Delta\theta}2\)）沿弦的方向走，比"用起始朝向走 \(\Delta s\)"的欧拉法准确得多。误差：朝向误差 \(\delta\theta\) 一旦出现就不会自己消失，之后每走一米，横向误差增加约 \(\delta\theta\) 米，位置误差随路程线性甚至更快增长；轮径、轮距的系统误差还会给出一个"稳定的"圆弧漂移（FIG 1 的任务 3）。编码器噪声积分后成为随机游走，协方差随路程增长（4.5 节的不确定性传播）。所以里程计只在短时间内可靠，长期定位要靠地标、激光或视觉的修正（Part IV：SLAM、定位）。

This moves along the chord (length \(2\tfrac{\Delta s}{\Delta\theta}\sin\tfrac{\Delta\theta}2\)) in the chord's direction, much more accurate than Euler's "advance \(\Delta s\) along the starting heading". Error: a heading error \(\delta\theta\), once present, never goes away, and every meter driven afterwards adds about \(\delta\theta\) meters of cross error, so the position error grows linearly or faster with distance; systematic errors in wheel radius or track add a "steady" arc drift (Task 3 of FIG 1). Integrated encoder noise becomes a random walk whose covariance grows with distance (uncertainty propagation, Lesson 4.5). So odometry is reliable only over short spans; long-term localization needs corrections from landmarks, lidar or vision (Part IV: SLAM, localization).

纯追踪：一个几何控制律Pure pursuit: a geometric control law

在路径上取离后轴中点距离为 \(\hl{look}{L_d}\) 的"前视点"，让车沿着一段过后轴中点、与车头相切、并通过前视点的圆弧走。设前视点在车体系下为 \((l_x,l_y)\)，与车头夹角为 \(\alpha\)，圆弧曲率由弦几何得到：圆心在 \((0,1/\kappa)\)，经过 \((l_x,l_y)\) 即 \(l_x^2+(l_y-1/\kappa)^2=1/\kappa^2\)，化简得

Pick a "lookahead point" on the path at distance \(\hl{look}{L_d}\) from the rear-axle midpoint and drive along the circular arc that passes through the rear-axle midpoint, is tangent to the heading and goes through the lookahead point. With the point at \((l_x,l_y)\) in the body frame and angle \(\alpha\) to the heading, chord geometry gives the curvature: the center is at \((0,1/\kappa)\) and the circle passes through \((l_x,l_y)\), i.e. \(l_x^2+(l_y-1/\kappa)^2=1/\kappa^2\), which simplifies to

\(L_d\) 小，车贴近路径但转向激烈，对转向滞后敏感，容易振荡；\(L_d\) 大，运动平顺，却会"切弯"。工程上 \(L_d\) 通常随速度增大（例如 \(L_d=k\,v\)）。

A small \(L_d\) hugs the path but steers hard and is sensitive to steering lag, so it oscillates easily; a large \(L_d\) is smooth but "cuts corners". In practice \(L_d\) grows with speed (e.g. \(L_d=k\,v\)).

位姿镇定：极坐标律与 Brockett 障碍Pose regulation: the polar-coordinate law and Brockett's obstruction

要把独轮车开到目标位姿 \((x_g,y_g,\theta_g)\)，Siegwart 等的教材（沿用 Astolfi 的极坐标镇定思路）改用极坐标：\(\rho\) 是到目标的距离，\(\hl{alpha}{\alpha}\) 是车头到视线方向的夹角，\(\hl{beta}{\beta}\) 是视线方向到目标朝向的夹角（都取 \((-\pi,\pi]\)）：

To drive the unicycle to a goal pose \((x_g,y_g,\theta_g)\), Siegwart et al.'s textbook, following Astolfi, switches to polar coordinates: \(\rho\) is the distance to the goal, \(\hl{alpha}{\alpha}\) is the angle from the heading to the line of sight, and \(\hl{beta}{\beta}\) is the angle from the line of sight to the goal heading (all wrapped to \((-\pi,\pi]\)):

代入后 \(v/\rho=k_\rho\) 是常数，\((\alpha,\beta)\) 在原点附近的线性化为 \(\dot{[\alpha,\beta]}^\top=\begin{bmatrix}-(k_\alpha-k_\rho)&-k_\beta\\-k_\rho&0\end{bmatrix}[\alpha,\beta]^\top\)，特征多项式 \(\lambda^2+(k_\alpha-k_\rho)\lambda-k_\rho k_\beta\)，所以稳定条件为

Substituting, \(v/\rho=k_\rho\) is a constant, and the linearization of \((\alpha,\beta)\) near the origin is \(\dot{[\alpha,\beta]}^\top=\begin{bmatrix}-(k_\alpha-k_\rho)&-k_\beta\\-k_\rho&0\end{bmatrix}[\alpha,\beta]^\top\), with characteristic polynomial \(\lambda^2+(k_\alpha-k_\rho)\lambda-k_\rho k_\beta\); hence stability requires

经典取值 \(k_\rho=3,\ k_\alpha=8,\ k_\beta=-1.5\)（Siegwart 等教材）。当目标在车身后方时（\(|\alpha|>\pi/2\)），把"车头"换成车尾，倒车驶向目标。我们在节点脚本里对多组不同的起终点位姿（包括目标在车后的倒车情形，也包括速度、角速度限幅）做了数值验证，终点位置误差趋于 0，朝向误差收敛到 \(10^{-7}\) 以下；FIG 6 就用这个律。

The textbook values are \(k_\rho=3,\ k_\alpha=8,\ k_\beta=-1.5\) (Siegwart et al.). When the goal lies behind the car (\(|\alpha|>\pi/2\)) swap the roles of front and rear and drive to it in reverse. We checked the law numerically from many different start/goal poses (including reverse driving when the goal is behind, and with speed and rate saturation): position error goes to 0 and heading error converges below \(10^{-7}\). FIG 6 uses this law.

注意：这个律在目标处 \(\rho=0\) 无定义，\(\omega\) 在 \(\rho\to0\) 时并不趋于 0，所以在原来的 \((x,y,\theta)\) 坐标下它是不连续的。这不是缺陷，而是必然：Brockett 定理（1983）说，系统 \(\dot q=f(q,u)\) 要想被某个连续的、与时间无关的反馈 \(u=k(q)\) 渐近镇定到 \(q=0\)，映射 \((q,u)\mapsto f(q,u)\) 的像必须包含 \(0\) 的一个邻域。独轮车每一点的可用速度只有二维平面 \(\Delta(q)\)，永远盖不住"侧向"一小点速度（比如 \((0,\epsilon,0)\)），所以任何光滑、定常的反馈都镇定不了它，这和它是否可控无关（它是可控的）。出路只有三条：不连续反馈（上面的极坐标律、滑模）、时变反馈、或者干脆先规划一条满足约束的路径（Dubins、MPC 等）再去跟踪：跟踪一条可行轨迹并没有障碍。

Note that the law is undefined at the goal, where \(\rho=0\), and \(\omega\) does not tend to 0 as \(\rho\to0\), so in the original \((x,y,\theta)\) coordinates it is discontinuous. That is not a flaw but a necessity: Brockett's theorem (1983) says that for \(\dot q=f(q,u)\) to be asymptotically stabilized to \(q=0\) by a continuous time-invariant feedback \(u=k(q)\), the image of the map \((q,u)\mapsto f(q,u)\) must contain a neighborhood of \(0\). At each point the unicycle's available velocities form only the plane \(\Delta(q)\) and can never cover a small "sideways" velocity such as \((0,\epsilon,0)\), so no smooth, time-invariant feedback stabilizes it, even though it is controllable. There are three ways out: discontinuous feedback (the polar law above, sliding mode), time-varying feedback, or simply to plan a feasible path first (Dubins, MPC, ...) and then track it: tracking a feasible trajectory has no such obstruction.

6机器人中的它：跟踪路径，停到位姿In robotics: tracking a path and stopping at a pose

• 让车振荡：转向滞后 \(\tau_s\ge0.25\) s，前视距离 \(L_d\le0.55\) m，横向误差反复穿过 0 且幅度超过 0.2 m。Make the car oscillate: steering lag \(\tau_s\ge0.25\) s and lookahead \(L_d\le0.55\) m, with the cross-track error repeatedly crossing 0 at an amplitude above 0.2 m.
• 选"直角弯"，取 \(L_d\ge1.5\) m，让车切弯：走完全程后最大横向误差超过 0.3 m。再把 \(L_d\) 调到 0.8 m 左右比较。Choose "Corner" with \(L_d\ge1.5\) m so the car cuts the corner: after the full run the maximum cross-track error exceeds 0.3 m. Then compare with \(L_d\approx0.8\) m.

停到位姿：极坐标律Stopping at a pose: the polar-coordinate law

• 运行并让机器人停到目标位姿：位置误差小于 1 cm，朝向误差小于 1°。试一个目标在车后方的起点。Run it and have the robot settle at the goal pose: position error below 1 cm, heading error below 1°. Try a start with the goal behind the robot.
• 违反稳定条件（\(k_\beta>0\) 或 \(k_\alpha<k_\rho\)），运行到结束：朝向误差还剩多少？Violate the stability condition (\(k_\beta>0\) or \(k_\alpha<k_\rho\)) and run to the end: how much heading error is left?

7常见误区Common pitfalls

8小练习Check yourself

9延伸阅读Further reading

• R. Siegwart, I. Nourbakhsh & D. Scaramuzza, Introduction to Autonomous Mobile Robots, 2nd ed., MIT Press, 2011, Ch. 3 — 移动机器人运动学：差速、阿克曼、全向轮模型，运动学约束，极坐标位姿控制律（3.6 节）Mobile robot kinematics: differential, Ackermann and omni-wheel models, kinematic constraints, and the polar pose-control law (Sec. 3.6)
• K. Lynch & F. Park, Modern Robotics, Ch. 13 — 轮式移动机器人：全向轮、非完整约束、里程计、运动规划与反馈控制Wheeled mobile robots: omni wheels, nonholonomic constraints, odometry, motion planning and feedback control
• S. M. LaValle, Planning Algorithms, Cambridge University Press, 2006, Ch. 13 — 微分约束：Pfaffian 约束、可积性、可控性、Dubins 车与 Reeds–Shepp 车（在线免费）Differential models: Pfaffian constraints, integrability, controllability, the Dubins and Reeds–Shepp cars (free online)
• R. M. Murray, Z. Li & S. S. Sastry, A Mathematical Introduction to Robotic Manipulation, CRC Press, 1994, Ch. 7 — 机器人系统中的非完整行为：李括号、Chow 定理、路径规划Nonholonomic behavior in robotic systems: Lie brackets, Chow's theorem, path planning
• L. E. Dubins, On curves of minimal length with a constraint on average curvature, and with prescribed initial and terminal positions and tangents, American Journal of Mathematics 79(3), 1957 — Dubins 最短路径的原始论文The original Dubins shortest-path paper
• J. A. Reeds & L. A. Shepp, Optimal paths for a car that goes both forwards and backwards, Pacific Journal of Mathematics 145(2), 1990 — 允许倒车的最短路径Shortest paths when reversing is allowed
• R. W. Brockett, Asymptotic stability and feedback stabilization, in Differential Geometric Control Theory, Birkhäuser, 1983 — Brockett 必要条件Brockett's necessary condition
• M. Aicardi, G. Casalino, A. Bicchi & A. Balestrino, Closed loop steering of unicycle-like vehicles via Lyapunov techniques, IEEE Robotics & Automation Magazine 2(1), 1995 — 用 Lyapunov 方法设计独轮车类车辆的闭环镇定律（与正文的线性极坐标律不同的另一种设计）Closed-loop steering of unicycle-like vehicles designed with Lyapunov techniques (a different design from the linear polar law in the text)
• R. C. Coulter, Implementation of the Pure Pursuit Path Tracking Algorithm, CMU-RI-TR-92-01, 1992; and B. Paden et al., A Survey of Motion Planning and Control Techniques for Self-Driving Urban Vehicles, IEEE Trans. Intelligent Vehicles 1(1), 2016 — 纯追踪与自动驾驶规划、控制综述Pure pursuit, and a survey of planning and control for self-driving vehicles