如何从 x、y 点列表和偏移距离中获取偏移样条曲线的 x、y 坐标

Question

我需要制作翼型轮廓曲线的偏移平行封闭，但我无法弄清楚如何使所有点与主轮廓曲线上的点在所需距离处等距。

这是我的示例机翼剖面

这是我最好但不是很好的方法

EDIT @Patrick 距离 0.2 的解决方案

Answer 1

你得用infinity/zero的特例斜率，但基本方法是用插值法计算一个点的斜率，然后求垂直斜率，然后计算点处的斜率那个距离。

我修改了 here to add a second graph. It works with the data file you provided 中的示例，但您可能需要更改不同信封的符号计算。

编辑根据你关于希望信封是连续的评论，我在最后添加了一个俗气的半圆，非常接近 为你做这件事。本质上，在创建信封时，您可以将其做得越圆越凸，效果就越好。另外，开头和结尾要重叠，不然会有空隙。

此外，几乎可以肯定它可以变得更有效率——无论如何我都不是 numpy 专家，所以这只是纯粹的 Python。

def offset(coordinates, distance):
    coordinates = iter(coordinates)
    x1, y1 = coordinates.next()
    z = distance
    points = []
    for x2, y2 in coordinates:
        # tangential slope approximation
        try:
            slope = (y2 - y1) / (x2 - x1)
            # perpendicular slope
            pslope = -1/slope  # (might be 1/slope depending on direction of travel)
        except ZeroDivisionError:
            continue
        mid_x = (x1 + x2) / 2
        mid_y = (y1 + y2) / 2

        sign = ((pslope > 0) == (x1 > x2)) * 2 - 1

        # if z is the distance to your parallel curve,
        # then your delta-x and delta-y calculations are:
        #   z**2 = x**2 + y**2
        #   y = pslope * x
        #   z**2 = x**2 + (pslope * x)**2
        #   z**2 = x**2 + pslope**2 * x**2
        #   z**2 = (1 + pslope**2) * x**2
        #   z**2 / (1 + pslope**2) = x**2
        #   z / (1 + pslope**2)**0.5 = x

        delta_x = sign * z / ((1 + pslope**2)**0.5)
        delta_y = pslope * delta_x

        points.append((mid_x + delta_x, mid_y + delta_y))
        x1, y1 = x2, y2
    return points

def add_semicircle(x_origin, y_origin, radius, num_x = 50):
    points = []
    for index in range(num_x):
        x = radius * index / num_x
        y = (radius ** 2 - x ** 2) ** 0.5
        points.append((x, -y))
    points += [(x, -y) for x, y in reversed(points)]
    return [(x + x_origin, y + y_origin) for x, y in points]

def round_data(data):
    # Add infinitesimal rounding of the envelope
    assert data[-1] == data[0]
    x0, y0 = data[0]
    x1, y1 = data[1]
    xe, ye = data[-2]

    x = x0 - (x0 - x1) * .01
    y = y0 - (y0 - y1) * .01
    yn = (x - xe) / (x0 - xe) * (y0 - ye) + ye
    data[0] = x, y
    data[-1] = x, yn
    data.extend(add_semicircle(x, (y + yn) / 2, abs((y - yn) / 2)))
    del data[-18:]

from pylab import *

with open('ah79100c.dat', 'rb') as f:
    f.next()
    data = [[float(x) for x in line.split()] for line in f if line.strip()]

t = [x[0] for x in data]
s = [x[1] for x in data]


round_data(data)

parallel = offset(data, 0.1)
t2 = [x[0] for x in parallel]
s2 = [x[1] for x in parallel]

plot(t, s, 'g', t2, s2, 'b', lw=1)

title('Wing with envelope')
grid(True)

axes().set_aspect('equal', 'datalim')

savefig("test.png")
show()

Answer 2

如果您愿意（并且能够）安装第三方工具，我强烈推荐 Shapely 模块。这是一个向内和向外偏移的小样本：

from StringIO import StringIO
import matplotlib.pyplot as plt
import numpy as np
import requests
import shapely.geometry as shp

# Read the points    
AFURL = 'http://m-selig.ae.illinois.edu/ads/coord_seligFmt/ah79100c.dat'
afpts = np.loadtxt(StringIO(requests.get(AFURL).content), skiprows=1)

# Create a Polygon from the nx2 array in `afpts`
afpoly = shp.Polygon(afpts)

# Create offset airfoils, both inward and outward
poffafpoly = afpoly.buffer(0.03)  # Outward offset
noffafpoly = afpoly.buffer(-0.03)  # Inward offset

# Turn polygon points into numpy arrays for plotting
afpolypts = np.array(afpoly.exterior)
poffafpolypts = np.array(poffafpoly.exterior)
noffafpolypts = np.array(noffafpoly.exterior)

# Plot points
plt.plot(*afpolypts.T, color='black')
plt.plot(*poffafpolypts.T, color='red')
plt.plot(*noffafpolypts.T, color='green')
plt.axis('equal')
plt.show()

这是输出；注意向内偏移量上的 'bowties'（自交点）是如何自动删除的：

如何从 x、y 点列表和偏移距离中获取偏移样条曲线的 x、y 坐标

How to get the x,y coordinates of a offset spline from a x,y list of points and offset distance

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