自动驾驶规划 - 5次多项式拟合

简介

自动驾驶运动规划中会用到各种曲线，主要用于生成车辆的轨迹，常见的轨迹生成算法，如贝塞尔曲线，样条曲线，以及apollo EM Planner的五次多项式曲线，城市场景中使用的是分段多项式曲线，在EM Planner和Lattice Planner中思路是，都是先通过动态规划生成点，再用5次多项式生成曲线连接两个点（虽然后面的版本改动很大，至少lattice planner目前还是这个方法）。

在这里可以看出5次多项式的作用，就是生成轨迹，这里的轨迹不一定是车行驶的轨迹，比如S—T图中的线，是用来做速度规划的。 如下图：

在apollo里面用到了，3-5次多项式，

• cubic_polynomial_curve1d，三次多项式曲线
• quartic_polynomial_curve1d，四次多项式曲线-
• quintic_polynomial_curve1d，五次多项式曲线

3次多项式

$$f(x)=C_0+C_{1}x+C_2{x}^2+C_3{x}^3----(1)$$
对x求导====>
$$f^{\prime }(x)=C_1+2C_{2}x+3C_3{x}^2$$
$$f(x)=2C_2+6C_{3}x$$
因为是连接两个已知点，所以两个轨迹点的信息是已知的。
如已知，当x=0，代如得：
$$f(0)=C_0；f^{\prime }(0)=C_1；f^{\prime \prime }(0)=2C_2$$

所以
$$C_0=f(0)；C_1=f^{\prime }(0)；C_2=f^{\prime \prime }(0)/2$$
将终点 (X_end，f(X_end))带入得到C_3。

$$C_3=(f(x_p)-C_0-C_1{x}_p-C_2{x}_p^2)/x_p^3$$

（1）式的所有未知参数都求出，代入x可以得到两点间的轨迹。

apollo中3次多项式的计算

// cubic_polynomial_curve1d.cc
CubicPolynomialCurve1d::CubicPolynomialCurve1d(const double x0,
                                               const double dx0,
                                               const double ddx0,
                                               const double x1,
                                               const double param) {
  ComputeCoefficients(x0, dx0, ddx0, x1, param);
  param_ = param;
  start_condition_[0] = x0;
  start_condition_[1] = dx0;
  start_condition_[2] = ddx0;
  end_condition_ = x1;
}

void CubicPolynomialCurve1d::ComputeCoefficients(const double x0,
                                                 const double dx0,
                                                 const double ddx0,
                                                 const double x1,
                                                 const double param) {
  DCHECK(param > 0.0);
  const double p2 = param * param;
  const double p3 = param * p2;
  coef_[0] = x0;
  coef_[1] = dx0;
  coef_[2] = 0.5 * ddx0;
  coef_[3] = (x1 - x0 - dx0 * param - coef_[2] * p2) / p3;
}

5次多项式

$$f(x)=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4+C_5{x}^5----(3-1)$$
对x求导====>
$$f^{\prime }(x)=C_1+2C_{2}x+3C_3{x}^2+4C_4{x}^3+5C_5{x}^4----(3-2)$$

$$f^{\prime \prime }(x)=2C_2+6C_{3}x+12C_4{x}^2+20C_5{x}^3----(3-3)$$
因为是连接两个已知点，所以两个轨迹点的信息是已知的。
如已知，当x=0，代如得：
$$f(0)=C_0；f^{\prime }(0)=C_1；f^{\prime \prime }(0)=2C_2$$

所以
$$C_0=f(0)；C_1=f^{\prime }(0)；C_2=f^{\prime \prime }(0)/2$$
将终点 (x_e，f(x_e))带入 (3-1) (3-2) (3-3)。

得到
$$f(x_e)=C_0+C_1{x}_e+C_2{x}_e^2+C_3{x}_e^3+C_4{x}_e^4+C_5{x}_e^5----(3-4)$$
对x求导====>
$$f^{\prime }(x)=C_1+2C_2{x}_e+3C_3{x}_e^2+4C_4{x}_e^3+5C_5{x}_e^4----(3-5)$$

$$f^{\prime \prime }(x)=2C_2+6C_3{x}_e+12C_4{x}_e^2+20C_5{x}_e^3----(3-6)$$

联立可得
$$C_3=\frac{10(f(x_p)-1/2\ast f^{\prime \prime }(0)x_p^2-f^{\prime }(0)x_p-f(0))-4(f^{\prime }(x_p)-f^{\prime \prime }(0)x_p-f^{\prime }(0))x_p+1/2\ast (f^{\prime \prime }(x_p)-f^{\prime \prime }(0))x_p^2}{x_p^3}$$

$$C_4=\frac{-15(f(x_p)-1/2\ast f^{\prime \prime }(0)x_p^2-f^{\prime }(0)x_p-f(0))+7(f^{\prime }(x_p)-f^{\prime \prime }(0)x_p-f^{\prime }(0))x_p-1/2\ast (f^{\prime \prime }(x_p)-f^{\prime \prime }(0))x_p^2}{x_p^4}$$

$$C_5=\frac{6(f(x_p)-1/2\ast f^{\prime \prime }(0)x_p^2-f^{\prime }(0)x_p-f(0))-3(f^{\prime }(x_p)-f^{\prime \prime }(0)x_p-f^{\prime }(0))x_p+1/2\ast (f^{\prime \prime }(x_p)-f^{\prime \prime }(0))x_p^2}{x_p^5}$$

apollo中5次多项式的计算

void QuinticPolynomialCurve1d::ComputeCoefficients(
    const double x0, const double dx0, const double ddx0, const double x1,
    const double dx1, const double ddx1, const double p) {
  CHECK_GT(p, 0.0);

  coef_[0] = x0;
  coef_[1] = dx0;
  coef_[2] = ddx0 / 2.0;

  const double p2 = p * p;
  const double p3 = p * p2;

  // the direct analytical method is at least 6 times faster than using matrix
  // inversion.
  const double c0 = (x1 - 0.5 * p2 * ddx0 - dx0 * p - x0) / p3;
  const double c1 = (dx1 - ddx0 * p - dx0) / p2;
  const double c2 = (ddx1 - ddx0) / p;

  coef_[3] = 0.5 * (20.0 * c0 - 8.0 * c1 + c2);
  coef_[4] = (-15.0 * c0 + 7.0 * c1 - c2) / p;
  coef_[5] = (6.0 * c0 - 3.0 * c1 + 0.5 * c2) / p2;
}

5次多项式拟合

import math

import matplotlib.pyplot as plt
import numpy as np

# parameter
MAX_T = 100.0  # maximum time to the goal [s]
MIN_T = 5.0  # minimum time to the goal[s]

show_animation = True

class QuinticPolynomial:

    def __init__(self, xs, vxs, axs, xe, vxe, axe, time):
        # calc coefficient of quintic polynomial
        # See jupyter notebook document for derivation of this equation.
        # 根据初始状态c0 c1 c2
        self.a0 = xs # x(t)
        self.a1 = vxs # v(t)
        self.a2 = axs / 2.0 # a(t)

        A = np.array([[time ** 3, time ** 4, time ** 5],
                      [3 * time ** 2, 4 * time ** 3, 5 * time ** 4],
                      [6 * time, 12 * time ** 2, 20 * time ** 3]])
        b = np.array([xe - self.a0 - self.a1 * time - self.a2 * time ** 2,
                      vxe - self.a1 - 2 * self.a2 * time,
                      axe - 2 * self.a2])
        # Ax=b 求解x
        x = np.linalg.solve(A, b)
        # 计算c3 c4 c5
        self.a3 = x[0]
        self.a4 = x[1]
        self.a5 = x[2]
    # 根据时间计算点x(t)
    def calc_point(self, t):
        xt = self.a0 + self.a1 * t + self.a2 * t ** 2 + \
             self.a3 * t ** 3 + self.a4 * t ** 4 + self.a5 * t ** 5
        return xt
    # 计算v(t)
    def calc_first_derivative(self, t):
        xt = self.a1 + 2 * self.a2 * t + \
             3 * self.a3 * t ** 2 + 4 * self.a4 * t ** 3 + 5 * self.a5 * t ** 4

        return xt
    # 计算a(t)  
    def calc_second_derivative(self, t):
        xt = 2 * self.a2 + 6 * self.a3 * t + 12 * self.a4 * t ** 2 + 20 * self.a5 * t ** 3

        return xt
    # 计算jerk(t)  
    def calc_third_derivative(self, t):
        xt = 6 * self.a3 + 24 * self.a4 * t + 60 * self.a5 * t ** 2

        return xt


def quintic_polynomials_planner(sx, sy, syaw, sv, sa, gx, gy, gyaw, gv, ga, max_accel, max_jerk, dt):
    """
    quintic polynomial planner

    input
        s_x: start x position [m]
        s_y: start y position [m]
        s_yaw: start yaw angle [rad]
        sa: start accel [m/ss]
        gx: goal x position [m]
        gy: goal y position [m]
        gyaw: goal yaw angle [rad]
        ga: goal accel [m/ss]
        max_accel: maximum accel [m/ss]
        max_jerk: maximum jerk [m/sss]
        dt: time tick [s]

    return
        time: time result
        rx: x position result list
        ry: y position result list
        ryaw: yaw angle result list
        rv: velocity result list
        ra: accel result list

    """

    vxs = sv * math.cos(syaw)# 起点速度在x方向分量
    vys = sv * math.sin(syaw)# 起点速度在y方向分量
    vxg = gv * math.cos(gyaw)# 终点速度在x方向分量
    vyg = gv * math.sin(gyaw)# 终点速度在y方向分量

    axs = sa * math.cos(syaw)# 起点加速度在x方向分量
    ays = sa * math.sin(syaw)# 终点加速度在y方向分量
    axg = ga * math.cos(gyaw)# 起点加速度在x方向分量
    ayg = ga * math.sin(gyaw)# 终点加速度在y方向分量

    time, rx, ry, ryaw, rv, ra, rj = [], [], [], [], [], [], []

    for T in np.arange(MIN_T, MAX_T, MIN_T):
        xqp = QuinticPolynomial(sx, vxs, axs, gx, vxg, axg, T)
        yqp = QuinticPolynomial(sy, vys, ays, gy, vyg, ayg, T)

        time, rx, ry, ryaw, rv, ra, rj = [], [], [], [], [], [], []

        for t in np.arange(0.0, T + dt, dt):
            time.append(t)
            rx.append(xqp.calc_point(t))
            ry.append(yqp.calc_point(t))

            vx = xqp.calc_first_derivative(t)
            vy = yqp.calc_first_derivative(t)
            v = np.hypot(vx, vy)
            yaw = math.atan2(vy, vx)
            rv.append(v)
            ryaw.append(yaw)

            ax = xqp.calc_second_derivative(t)
            ay = yqp.calc_second_derivative(t)
            a = np.hypot(ax, ay)
            if len(rv) >= 2 and rv[-1] - rv[-2] < 0.0:
                a *= -1
            ra.append(a)

            jx = xqp.calc_third_derivative(t)
            jy = yqp.calc_third_derivative(t)
            j = np.hypot(jx, jy)
            if len(ra) >= 2 and ra[-1] - ra[-2] < 0.0:
                j *= -1
            rj.append(j)

        if max([abs(i) for i in ra]) <= max_accel and max([abs(i) for i in rj]) <= max_jerk:
            print("find path!!")
            break

    if show_animation:  # pragma: no cover
        for i, _ in enumerate(time):
            plt.cla()
            # for stopping simulation with the esc key.
            plt.gcf().canvas.mpl_connect('key_release_event',
                                         lambda event: [exit(0) if event.key == 'escape' else None])
            plt.grid(True)
            plt.axis("equal")
            plot_arrow(sx, sy, syaw)
            plot_arrow(gx, gy, gyaw)
            plot_arrow(rx[i], ry[i], ryaw[i])
            plt.title("Time[s]:" + str(time[i])[0:4] +
                      " v[m/s]:" + str(rv[i])[0:4] +
                      " a[m/ss]:" + str(ra[i])[0:4] +
                      " jerk[m/sss]:" + str(rj[i])[0:4],
                      )
            plt.pause(0.001)

    return time, rx, ry, ryaw, rv, ra, rj


def plot_arrow(x, y, yaw, length=1.0, width=0.5, fc="r", ec="k"):  # pragma: no cover
    """
    Plot arrow
    """

    if not isinstance(x, float):
        for (ix, iy, iyaw) in zip(x, y, yaw):
            plot_arrow(ix, iy, iyaw)
    else:
        plt.arrow(x, y, length * math.cos(yaw), length * math.sin(yaw),
                  fc=fc, ec=ec, head_width=width, head_length=width)
        plt.plot(x, y)


def main():

    sx = 10.0  # start x position [m]
    sy = 10.0  # start y position [m]
    syaw = np.deg2rad(10.0)  # start yaw angle [rad]
    sv = 1.0  # start speed [m/s]
    sa = 0.1  # start accel [m/ss]
    gx = 30.0  # goal x position [m]
    gy = -10.0  # goal y position [m]
    gyaw = np.deg2rad(20.0)  # goal yaw angle [rad]
    gv = 1.0  # goal speed [m/s]
    ga = 0.1  # goal accel [m/ss]
    max_accel = 1.0  # max accel [m/ss]
    max_jerk = 0.5  # max jerk [m/sss]
    dt = 0.1  # time tick [s]

    time, x, y, yaw, v, a, j = quintic_polynomials_planner(
        sx, sy, syaw, sv, sa, gx, gy, gyaw, gv, ga, max_accel, max_jerk, dt)

    if show_animation:  # pragma: no cover
        plt.plot(x, y, "-r")

        plt.subplots()
        plt.plot(time, [np.rad2deg(i) for i in yaw], "-r")
        plt.xlabel("Time[s]")
        plt.ylabel("Yaw[deg]")
        plt.grid(True)

        plt.subplots()
        plt.plot(time, v, "-r")
        plt.xlabel("Time[s]")
        plt.ylabel("Speed[m/s]")
        plt.grid(True)

        plt.subplots()
        plt.plot(time, a, "-r")
        plt.xlabel("Time[s]")
        plt.ylabel("accel[m/ss]")
        plt.grid(True)

        plt.subplots()
        plt.plot(time, j, "-r")
        plt.xlabel("Time[s]")
        plt.ylabel("jerk[m/sss]")
        plt.grid(True)

        plt.show()


if __name__ == '__main__':
    main()

上图模拟了车辆曲线的拟合过程的轨迹，箭头代表方向。

过程的速度

感觉5次多项式和贝塞尔曲线还是挺像的，有时间写一篇五次多项式、贝塞尔曲线、样条曲线的对比。

还有一个问题就是为啥要用5次多项式，而不是4次，6次更高阶或者低阶呢？下回复习一下老王的讲解。
参考：文章来源地址https://www.toymoban.com/news/detail-436810.html

• https://blog.csdn.net/qq_36458461/article/details/110007656
• https://zhuanlan.zhihu.com/p/567375631

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