x1c 0d1x的数据如图所示，需要计算光滑曲线的公式或方法。红线是求解后将看到终点的线。我将感谢你的帮助/讨论。我需要数学公式，可能是如何解决它使用scipy python包。
g0czyy6m1#
实际上，如果使用三次贝塞尔曲线，您将获得更大的灵活性。这是第一个问题：
import mathimport numpy as npimport matplotlib.pyplot as plt#-----------------------------def length( vector ): return np.linalg.norm( vector ) # length of a vectordef normalise( vector ): return vector / length(vector) # normalise a vector (numpy array)#-----------------------------class Bezier: '''class for Cubic Bezier curve''' def __init__( self, q0, q1, q2, q3 ): self.p0 = np.array( q0 ) self.p1 = np.array( q1 ) self.p2 = np.array( q2 ) self.p3 = np.array( q3 ) def pt( self, t ): return ( 1.0 - t ) ** 3 * self.p0 + 3.0 * t * ( 1.0 - t ) ** 2 * self.p1 + 3.0 * t ** 2 * ( 1.0 - t ) * self.p2 + t ** 3 * self.p3 def curve( self, n ): crv = [] for t in np.linspace( 0.0, 1.0, n ): crv.append( self.pt( t ) ) return crv#-----------------------------def drawBezier( x1, y1, heading1, x2, y2, heading2, n ): '''Turns two points plus headings into cubic Bezier control points [ P0, P1, P2, P3 ] by adding intermediate control points''' result = [] rad1, rad2 = heading1 * math.pi / 180, heading2 * math.pi / 180 d1 = np.array( [ np.sin( rad1 ), np.cos( rad1 ) ] ) # direction vector at point 1 d2 = np.array( [ np.sin( rad2 ), np.cos( rad2 ) ] ) p0 = np.array( ( x1, y1 ) ) # start point of Bezier curve p3 = np.array( ( x2, y2 ) ) # end point of Bezier curve p1 = p0 + d1 * length( p3 - p0 ) / 2.0 # use direction at P0 to get P1 p2 = p3 - d2 * length( p3 - p0 ) / 2.0 # use direction at p3 to get P2 arc = Bezier( p0, p1, p2, p3 ) # define a cubic Bezier object x = []; y = [] for pt in arc.curve( n ): x.append( pt[0] ) y.append( pt[1] ) plt.plot( x, y ) #-----------------------------x1, y1, heading1 = 0.0, 0.0, 0.0x2, y2, heading2 = 1.0, 1.0, 90.0drawBezier( x1, y1, heading1, x2, y2, heading2, 100 )plt.plot( [ x1, x2 ], [ y1, y2 ], 'o' ) # plot original support pointsplt.show()
import math
import numpy as np
import matplotlib.pyplot as plt
#-----------------------------
def length( vector ): return np.linalg.norm( vector ) # length of a vector
def normalise( vector ): return vector / length(vector) # normalise a vector (numpy array)
class Bezier:
'''class for Cubic Bezier curve'''
def __init__( self, q0, q1, q2, q3 ):
self.p0 = np.array( q0 )
self.p1 = np.array( q1 )
self.p2 = np.array( q2 )
self.p3 = np.array( q3 )
def pt( self, t ):
return ( 1.0 - t ) ** 3 * self.p0 + 3.0 * t * ( 1.0 - t ) ** 2 * self.p1 + 3.0 * t ** 2 * ( 1.0 - t ) * self.p2 + t ** 3 * self.p3
def curve( self, n ):
crv = []
for t in np.linspace( 0.0, 1.0, n ):
crv.append( self.pt( t ) )
return crv
def drawBezier( x1, y1, heading1, x2, y2, heading2, n ):
'''Turns two points plus headings into cubic Bezier control points [ P0, P1, P2, P3 ] by adding intermediate control points'''
result = []
rad1, rad2 = heading1 * math.pi / 180, heading2 * math.pi / 180
d1 = np.array( [ np.sin( rad1 ), np.cos( rad1 ) ] ) # direction vector at point 1
d2 = np.array( [ np.sin( rad2 ), np.cos( rad2 ) ] )
p0 = np.array( ( x1, y1 ) ) # start point of Bezier curve
p3 = np.array( ( x2, y2 ) ) # end point of Bezier curve
p1 = p0 + d1 * length( p3 - p0 ) / 2.0 # use direction at P0 to get P1
p2 = p3 - d2 * length( p3 - p0 ) / 2.0 # use direction at p3 to get P2
arc = Bezier( p0, p1, p2, p3 ) # define a cubic Bezier object
x = []; y = []
for pt in arc.curve( n ):
x.append( pt[0] )
y.append( pt[1] )
plt.plot( x, y )
x1, y1, heading1 = 0.0, 0.0, 0.0
x2, y2, heading2 = 1.0, 1.0, 90.0
drawBezier( x1, y1, heading1, x2, y2, heading2, 100 )
plt.plot( [ x1, x2 ], [ y1, y2 ], 'o' ) # plot original support points
plt.show()
字符串
的数据三次Bezier的优点是，您还可以无缝处理两条方向线不相交的情况。例如
x2, y2, heading2 = 1.0, 1.0, 0.0
型与输出
的
1条答案
按热度按时间g0czyy6m1#
实际上，如果使用三次贝塞尔曲线，您将获得更大的灵活性。
这是第一个问题：
字符串
的数据
三次Bezier的优点是，您还可以无缝处理两条方向线不相交的情况。例如
型
与输出
的