找到一个点和一个曲线python之间的距离 [英] find the distance between a point and a curve python
问题描述
如何找到我的轨迹与（384400,0,0）之间的壁橱距离？
How can I find the closet distance between my trajectory and (384400,0,0)?
另外，如何从（384400,0,0）到 t = 197465 时的路径？
Also, how can I the distance from (384400,0,0) to the path at time t = 197465?
我知道数组有数据，但有办法让它检查所有点的距离（x，y，z ）并返回我想要的内容？
I understand that the arrays have the data but is there a way to have it check the distance of all the points (x,y,z) and return what I am looking for?
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
me = 5.974 * 10 ** (24) # mass of the earth
mm = 7.348 * 10 ** (22) # mass of the moon
G = 6.67259 * 10 ** (-20) # gravitational parameter
re = 6378.0 # radius of the earth in km
rm = 1737.0 # radius of the moon in km
r12 = 384400.0 # distance between the CoM of the earth and moon
M = me + mm
pi1 = me / M
pi2 = mm / M
mue = 398600.0 # gravitational parameter of earth km^3/sec^2
mum = G * mm # grav param of the moon
mu = mue + mum
omega = np.sqrt(mu / r12 ** 3)
nu = -129.21 * np.pi / 180 # true anomaly angle in radian
x = 327156.0 - 4671
# x location where the moon's SOI effects the spacecraft with the offset of the
# Earth not being at (0,0) in the Earth-Moon system
y = 33050.0 # y location
vbo = 10.85 # velocity at burnout
gamma = 0 * np.pi / 180 # angle in radians of the flight path
vx = vbo * (np.sin(gamma) * np.cos(nu) - np.cos(gamma) * np.sin(nu))
# velocity of the bo in the x direction
vy = vbo * (np.sin(gamma) * np.sin(nu) + np.cos(gamma) * np.cos(nu))
# velocity of the bo in the y direction
xrel = (re + 300.0) * np.cos(nu) - pi2 * r12
# spacecraft x location relative to the earth
yrel = (re + 300.0) * np.sin(nu)
# r0 = [xrel, yrel, 0]
# v0 = [vx, vy, 0]
u0 = [xrel, yrel, 0, vx, vy, 0]
def deriv(u, dt):
n1 = -((mue * (u[0] + pi2 * r12) / np.sqrt((u[0] + pi2 * r12) ** 2
+ u[1] ** 2) ** 3)
- (mum * (u[0] - pi1 * r12) / np.sqrt((u[0] - pi1 * r12) ** 2
+ u[1] ** 2) ** 3))
n2 = -((mue * u[1] / np.sqrt((u[0] + pi2 * r12) ** 2 + u[1] ** 2) ** 3)
- (mum * u[1] / np.sqrt((u[0] - pi1 * r12) ** 2 + u[1] ** 2) ** 3))
return [u[3], # dotu[0] = u[3]
u[4], # dotu[1] = u[4]
u[5], # dotu[2] = u[5]
2 * omega * u[5] + omega ** 2 * u[0] + n1, # dotu[3] = that
omega ** 2 * u[1] - 2 * omega * u[4] + n2, # dotu[4] = that
0] # dotu[5] = 0
dt = np.arange(0.0, 320000.0, 1) # 200000 secs to run the simulation
u = odeint(deriv, u0, dt)
x, y, z, x2, y2, z2 = u.T
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z)
plt.show()
推荐答案
要查找轨迹中每个点的距离：
To find the distance along each point in the trajectory:
my_x, my_y, my_z = (384400,0,0)
delta_x = x - my_x
delta_y = y - my_y
delta_z = z - my_z
distance = np.sqrt(np.power(delta_x, 2) +
np.power(delta_y, 2) +
np.power(delta_z, 2))
然后找到最小值和特定距离：
And then to find the min and that specific distance:
i = np.argmin(distance)
t_min = i / UNITS_SCALE # I can't tell how many units to a second. Is it 1.6?
d_197465 = distance[int(197465 * UNITS_SCALE)]
PS，我不是火箭科学家，因此实际上没有检查上面的东西 x，y，z，x2，y2，z2 = uT 。我假设那些是你的轨迹的位置和速度？
PS, I'm no rocket scientist, and thus haven't actually checked the stuff above x, y, z, x2, y2, z2 = u.T. I'm assuming that those are your trajectory's position and velocity?
这篇关于找到一个点和一个曲线python之间的距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
