B样条曲线与贝塞尔曲线学习笔记

贝塞尔曲线

$$基本公式：B(t)=\sum _{i=0}^n\left({}_i^n\right)P_i(1-t{)}^{n-i}{t}^i，t\in [0,1]$$
三次贝塞尔曲线：
$$B(t)=P_0(1-t{)}^3+3P_{1}t(1-t{)}^2+3P_2{t}^2(1-t)+P_3{t}^3，t\in [0,1]$$
由此可见其系数规律：
$$11$$$$121$$$$1331$$$$14641$$
分别为一阶到四阶的系数规律，变化规律为杨辉三角，并且$t$与$(t-1)$的规律是一个逐渐转变的一个过程。

一段贝塞尔曲线拟合程序：

import matplotlib.pyplot as plt

x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
y = [12, 2, 78, 12, 34, 23, 67, 87, 98, 10]
xnew = []
ynew = []


def Bazier_3(m1, m2):
    for i in range(101):
        t = i / 100
        xnew.append(m1[0] * (1 - t) ** 3 + 3 * m1[1] * t * (1 - t) ** 2 + 3 * m1[2] * t ** 2 * (1 - t) + m1[3] * t ** 3)
        ynew.append(m2[0] * (1 - t) ** 3 + 3 * m2[1] * t * (1 - t) ** 2 + 3 * m2[2] * t ** 2 * (1 - t) + m2[3] * t ** 3)

for i in range(len(x) // 3):
    Bazier_3(x[i * 3:(i + 1) * 3 + 1], y[i * 3:(i + 1) * 3 + 1])
   
plt.plot(xnew, ynew)
plt.plot(x, y)
plt.scatter(x, y)
plt.show()

可以看到 ，在3，6这两个点，并不满族c2连续。

B样条曲线

$$基本公式：P(t)=\sum _{i=0}^n{P}_i{B}_{i,n}(t)$$

其基本形式与贝塞尔曲线相似

$$其中T=[t_0,t_1,t_2,...,t_n]$$

$$N_{i,0}(t)={\{}_{1t>t_i或t\ge t_{i+1}}^{0t_i\le t\le t_{i+1}}$$

$$N_{i,k}(t)=\frac{t-t_i}{t_{i+k}-t_i}{N}_{i,k-1}(t)+\frac{t_{i+k+1}-t}{t_{i+k+1}-t_{i+n}}{N}_{i+1,k-1}(t)k\ge 1$$
$$其中，N_{i,k}(t)中的i是控制点，k是次数$$
用的最多的是三次B样条曲线。
其中：
$N_{0,3}(t)=\frac{1}{6}(-t^3+3t^2-3t+1)$

$N_{1,3}(t)=\frac{1}{6}(3t^3-6t^2+4)$

$N_{2,3}(t)=\frac{1}{6}(-3t^3-3t^2+3t+1)$

$N_{1,3}(t)=\frac{1}{6}{t}^3$

为了使其闭合，要取最后一个点与与第一个控制点相同，即$P_{m+1}=P_0$，$P_{m+2}=P_1$，$P_{m+3}=P_2$，这样的曲线满足$c_2$连续。

一个拼接示例：

import matplotlib.pyplot as plt

x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
y = [12, 2, 78, 12, 34, 23, 67, 87, 98, 10]
xnew = []
ynew = []

arg = [[-1, 3, -3, 1], [3, -6, 0, 4], [-3, 3, 3, 1], [1, 0, 0, 0]]


def Ba(t, coefficient):
    return (coefficient[0] * t ** 3 + coefficient[1] * t ** 2 + coefficient[2] * t + coefficient[3]) / 6


def creat(n):
    for i in range(101):
        t = i / 100
        xnew.append(
            x[n + 0] * Ba(t, arg[0]) + x[n + 1] * Ba(t, arg[1]) + x[n + 2] * Ba(t, arg[2]) + x[n + 3] * Ba(t, arg[3]))
        ynew.append(
            y[n + 0] * Ba(t, arg[0]) + y[n + 1] * Ba(t, arg[1]) + y[n + 2] * Ba(t, arg[2]) + y[n + 3] * Ba(t, arg[3]))


for i in range(7):
    creat(i)

plt.plot(xnew, ynew)
plt.plot(x, y)
plt.scatter(x, y)
plt.show()

拼接后的图片如下所示，满足$c_2$连续。

贝塞尔曲线与B样条曲线的结合：

import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D

arg = [[-1, 3, -3, 1], [3, -6, 0, 4], [-3, 3, 3, 1], [1, 0, 0, 0]]  # B样条曲线的拟合参数
mar = [[0, -93.70105743, -7.71050644, 11.30164337],
       [0, -81.51637268, 2.65858841, 21.85936928],
       [0, -100.43165588, -506.84307861, -277.47509766],
       [0, -413.89691162, -244.77906799, -228.4907074],
       [0, -74.61241913, 22.23334312, 28.80702591],
       [0, -57.65986252, 33.13808441, 40.01465988],
       [0, -23.39715576, -46.36453247, -68.40002441],
       [0, -427.99655151, -242.35075378, -246.75854492],
       [0, -93.70105743, -7.71050644, 11.30164337]]  # 原数据


def Bazier_3(m1, m2):  # 贝塞尔曲线拟合
    x = []
    y = []
    for i in range(101):
        t = i / 100
        x.append(m1[0] * (1 - t) ** 3 + 3 * m1[1] * t * (1 - t) ** 2 + 3 * m1[2] * t ** 2 * (1 - t) + m1[3] * t ** 3)
        y.append(m2[0] * (1 - t) ** 3 + 3 * m2[1] * t * (1 - t) ** 2 + 3 * m2[2] * t ** 2 * (1 - t) + m2[3] * t ** 3)
    return x, y


def Ba(t, coefficient):  # 参数合成
    return (coefficient[0] * t ** 3 + coefficient[1] * t ** 2 + coefficient[2] * t + coefficient[3]) / 6


def creat_mart(mart):  # 贝塞尔曲线生成
    re = []
    for i in range(len(mart)):
        temp_x, temp_y = Bazier_3([0, 1, 2, 3], mart[i])
        re.append(temp_y)
    return re, temp_x


def creat_mart_finnally(data):  # 最终生成
    out = []
    times = data.shape[0]
    for j in range(times - 1):
        for i in range(45):
            t = i / 45
            temp = 0
            for k in range(4):
                temp += data[(j + k) % times] * Ba(t, arg[k])
            out.append(temp)
    return out


def draw(mat1, mat2, mat3):
    x = np.linspace(0, 8, 9)
    y = np.linspace(0, 3, 4)
    x, y = np.meshgrid(x, y)
    x = x.T
    y = y.T
    xs = np.ravel(x)
    ys = np.ravel(y)
    zs = np.ravel(mat1)

    xnew = np.linspace(0, 8, 360)
    ynew = np.linspace(0, 3, 101)
    xn, yn = np.meshgrid(xnew, ynew)

    x_min = np.linspace(0, 8, 9)
    y_min = np.linspace(0, 3, 101)
    x_min, y_min = np.meshgrid(x_min, y_min)

    plt.figure("原数据")
    ax = plt.subplot(1, 1, 1, projection='3d')
    ax.plot_trisurf(xs, ys, zs, cmap='coolwarm')
    ax.set_xlabel('angle')
    ax.set_ylabel('stepsize')
    ax.set_zlabel('Z')
    plt.title('raw')

    plt.figure("最终数据")
    ax2 = plt.subplot(1, 1, 1, projection='3d')
    ax2.plot_surface(xn.T, yn.T, mat2, rstride=2, cstride=2, cmap='coolwarm', linewidth=0.5, antialiased=True)
    ax2.set_xlabel('angle')
    ax2.set_ylabel('stepsize')
    ax2.set_zlabel('Z')
    plt.title('processed')

    plt.figure("中间数据")
    ax3 = plt.subplot(1, 1, 1, projection='3d')
    ax3.plot_surface(x_min.T, y_min.T, mat3, rstride=2, cstride=2, cmap='coolwarm', linewidth=0.5, antialiased=True)
    ax3.set_xlabel('angle')
    ax3.set_ylabel('stepsize')
    ax3.set_zlabel('Z')
    plt.title('min')

    plt.show()


mat_new, _ = creat_mart(mar)

mat_new = np.array(mat_new)

mat_f = []
for i in range(mat_new.shape[1]):
    mat_f.append(creat_mart_finnally(mat_new[:, i]))

mat_f = np.array(mat_f).T

draw(np.array(mar), mat_f, mat_new)

生成的图片如下：
源数据：

在步长方向上进行了贝塞尔曲线插值之后的数据是这样的：

在角度方向上面进行了B样条曲线插值：