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我怎样才能找到我的轨迹和之间的最近距离(384400,0,0)？

另外，我怎么能从(384400,0,0)时间到路径的距离t = 197465？

我知道数组有数据，但是有没有办法让它检查所有点的距离(x,y,z)并返回我正在寻找的东西？

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()
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1 回答 1

2

要找到沿轨迹中每个点的距离：

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：

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，因此实际上并没有检查过上面的东西x, y, z, x2, y2, z2 = u.T。我假设这些是你的轨迹的位置和速度？

于 2013-04-23T00:17:36.923 回答