我认为您正在处理的是Traveling-Salesman-Problem(TSP) (here the slide of a Phd course which talk about how to try to solve it)，而将激光上的应力最小化的路径是最小化移动它所需的力和力的变化的路径，因此它是曲率最小的路径，所以我认为最好的方法是用圆弧对 3 对点之间的路径进行圆角处理。
关于如何计算通过3个点的圆的参数的示例可以找到here
我不熟悉 C#，所以我将添加一个 Python 实现，希望你也觉得它有用。
这个想法是，对于每个点 A、B、C 的三元组，我找到通过这 3 个点的圆弧，并且该弧将是连接 B 和 C 的路径。
我还没有时间测试这个，所以可能有一些错误的迹象。
# Initial points
points = [(1,1),(2,3),(5,3),(-4.1),(12,3)]
#List of point in the order find by the solution of the TSP
spl = tsp_solve(points) # generic function to solve the TSP
# Append the first two point of the list so that I can iterate over the list
# and parse every triplet of points in the same way.
spl = spl + spl[:2]
# The list where will be added every path that connect the points
paths = []
# For each tirplets of sequential points
for A,B,C in zip(spl[:-2],spl[1:-1],spl[2:]):
# Calculate the angular coefficent of the two line that pass on A,B and B,C
coeff_ab = (B[1] - A[1]) / (B[0] - A[0])
coeff_bc = (C[1] - B[1]) / (C[0] - B[0])
# If the two line have the same coeff then all the 3 point are on the same line
# and therfore the best path is that line.
if(coeff_ab == coeff_bc):
offset_y = A[1] - coeff_ab * A[0]
delta_x = C[0] - B[0]
paths.append({"type":"Line","coeff":coeff_ab,"offset_y":offset_y,"deta_x":delta_x})
continue
# Calculate the x of the center of the circle
center_x = coeff_ab *coeff_bc *(C[0]-A[0])
center_x += coeff_ab *(B[0]+C[0])
center_x -= coeff_bc *(A[0]+B[0])
center_x /= 2*(coeff_ab - coeff_bc)
# Calculate the y of the center of the circle
center_y = (A[1]+B[1)/2
center_y -= (center_x - (A[0] + B[0])/2)
center_y /= coeff_bc
radius = sqrt(center_x**2 + center_y**2)
paths.append({"type":"Circle","Radius":radius,"center_x":center_x,"center_y":center_y})
# Function To Calculate the X and Y of the lines and circles.
def calculate_circle_x(circle,time):
"""Function that return the x of a circle at a given time"""
time = time + circle["time_off"]
return circle["radius"] * cos(2*pi*time) + circle["center_x"]
def calculate_circle_y(circle,time):
"""Function that return the y of a circle at a given time"""
time = time + circle["time_off"]
return circle["radius"] * sin(2*pi*time) + circle["center_y"]
def calculate_line_x(line,time):
"""Function that return the x of a line at a given time"""
time = (line['delta_x']*time) + line["time_off"]
return time
def calculate_line_y(line,time):
"""Function that return the y of a line at a given time"""
time = (line['delta_x']*time) + line["time_off"]
return time * line["coeff"] + line['offset_y']
def calculate_x(obj,time):
"""Function that return the x of whatever it's passed"""
if(obj['type'] == 'Circle'):
return calculate_circle_x(obj,time)
else:
return calculate_line_x(obj,time)
def calculate_y(obj,time):
"""Function that return the y of whatever it's passed"""
if(obj['type'] == 'Circle'):
return calculate_circle_y(obj,time)
else:
return calculate_line_y(obj,time)
# Calculate some sample of the global path to plot it or do whatever with it.
number_of_sample = 100000
path_points = []
number_of_paths = len(paths)
# Calculate some time equidistant point's sample
for i in range(number_of_sample):
# Calculate the global time
global_time = i*number_of_paths/number_of_sample
# Calculate in which path the point it is
path_number = int(global_time)
# Calculate which time of the path it is
local_time = global_time - path_number
path = paths[path_number]
# Calculate the sampled point
new_point = (calculate_x(path,local_time),calculate_y(path,local_time))
# Add the sampled point to the path_points list
path_points.append(new_point)
# Print the result of the path point sampled.
print(path_points)
现在您有了积分，或者至少有关于如何计算积分的示例，您可以将其转换为 C#。我尝试了很多评论，以便即使您不了解 Python 也能理解它。