Question

我有两个地理坐标;经度和纬度，相隔几米。我需要绘制一个相同的长轴和短轴的椭圆体，以地理坐标为中心

我使用了matplot lib和numpy来绘制椭圆体。我正在使用下面给出的公式计算经度的笛卡尔坐标为了将椭圆体移动到lat和long co-rodinates，我将所有x，y，z数组元素与cateian坐标经度x1，y1，z1

相加

这是移动椭球的正确方法，因此如果我绘制另一个具有不同地理坐标的椭球，我将能够可视化/计算它们的重叠区域

radius_of_world = 6371000
# Radii corresponding to the coefficients:
# http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node42.html ?
rx, ry, rz = 1/np.sqrt(coefs) # Is this correct for coefs gives as a,b,v

# Set of all spherical angles:
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)

# Postion of the lattitude and logitude on surface of the earth// ellipsoidal earth
x1 = radius_of_world * cos(longitude) * cos(latitude)
y1 = radius_of_world * np.sin(longitude) * np.cos(latitude)
z1 = radius_of_world * np.sin(latitude)

print 'Point location  is', x1,y1,z1

# Cartesian coordinates that correspond to the spherical angles:
# (this is the equation of an ellipsoid):
x = rx * np.outer(np.cos(u), np.sin(v)) 
y = ry * np.outer(np.sin(u), np.sin(v)) 
z = rz * np.outer(np.ones_like(u), np.cos(v))


# Plot:
#ax.plot_wireframe(x +x1 , y + y1, z + z1,  rstride=4, cstride=4, color=color)
ax.scatter(x +x1, y + y1, z + z1,color=color )

我不确定rx，ry，rz的计算是否正确;为了在2D中看到它，我稍微改了一下，现在我可以看到a和b有适当的长度

    Testing Elliptical Drawing in Spherical Corodinates

import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure(figsize=(12,12))  # Square figure

def drawCoverage(coefs,latitude,longitude,color):


# Radii corresponding to the coefficients:
# http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node42.html ??
rx, ry = coefs # This is what I changed

# Set of all spherical angles:
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)

# u = np.pi/2
# v = np.pi/2
# Cartesian coordinates that correspond to the spherical angles:
# (this is the equation of an ellipsoid):
x = rx * np.outer(np.cos(u), np.sin(v)) 
y = ry * np.outer(np.sin(u), np.sin(v)) 

#ax.autoscale(enable=True)
# Plot:
#plt.axis([0,100,0,100])
plt.plot(x , y , 'bo' )


latitude,longitude= 13.04738626,77.61946793   
coefs = (373, 258)  # Coefficients in a0/c x**2 + a1/c y**2 + a2/c z**2 = 1 
drawCoverage(coefs,latitude,longitude,'b')

plt.show()

我更改了coef计算，直接在3D模型中使用a，b，c，现在轴是正确的

# http://stackoverflow.com/questions/7819498/plotting-ellipsoid-with-matplotlib
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure(figsize=(12,12))  # Square figure
ax = fig.add_subplot(111, projection='3d')

def drawCoverage(coefs,latitude,longitude,color):

radius_of_world =  6372.8
# Radii corresponding to the coefficients:
# http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node42.html ??
rx, ry, rz = coefs

# Set of all spherical angles:
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)

# Postion of the lattitude and logitude on surface of the earth// ellipsoidal earth
x1 = radius_of_world * cos(longitude) * cos(latitude)
y1 = radius_of_world * np.sin(longitude) * np.cos(latitude)
z1 = radius_of_world * np.sin(latitude)

print 'Point location  is', x1,y1,z1

# Cartesian coordinates that correspond to the spherical angles:
# (this is the equation of an ellipsoid):
x = rx * np.outer(np.cos(u), np.sin(v)) 
y = ry * np.outer(np.sin(u), np.sin(v)) 
z = rz * np.outer(np.ones_like(u), np.cos(v))

#ax.autoscale(enable=True)
# Plot:
ax.plot_wireframe(x , y , z ,  rstride=4, cstride=4, color='b' )
ax.plot_wireframe(x + 111 , y + 111 , z + 111 ,  rstride=4, cstride=4, color='g')

#ax.scatter(x , y , z , color='b' )
#ax.scatter(x +111 , y +111 , z +111 , color='g')

#ax.set_xlim3d(0, 500)
#ax.set_ylim3d(0, 500)
#ax.set_zlim3d(0, 500)

# Adjustment of the axes, so that they all have the same span:
#max_radius = .25
#for axis in 'xyz':
#    getattr(ax, 'set_{}lim'.format(axis))((-max_radius, max_radius))



latitude,longitude= 13.04738626,77.61946793   
coefs = (373, 258, 258)  # Coefficients in a0/c x**2 + a1/c y**2 + a2/c z**2 = 1 
drawCoverage(coefs,latitude,longitude,'b')

latitude,longitude = 13.04638626, 77.61951605

# Distance between two lat.long is  0.1113 km
# http://www.movable-type.co.uk/scripts/latlong.html
#coefs = (258, 258, 373)
#drawCoverage(coefs,latitude,longitude,'g')
plt.show()

所以我在第二个代码片段中所做的是使用Haversine公式计算两个地理位置之间的距离（其距离大约为111米。我将这个111添加到第二个球体中，用于所有x，y和z点，以便将它分开并看到影响范围的重叠＆＃39;。这种方法是否正确

Answer 1

我猜这是方式

# drawing an ellipsoid - Oblate spheriod to show overlap for two geo points
at two altitudes

# http://stackoverflow.com/questions/7819498/plotting-ellipsoid-with-matplotlib
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure(figsize=(12,12))  # Square figure
ax = fig.add_subplot(111, projection='3d')

def haversine(lat1, lon1, lat2, lon2):

  R = 6372.8 # Earth radius in kilometers

  dLat = radians(lat2 - lat1)
  dLon = radians(lon2 - lon1)
  lat1 = radians(lat1)
  lat2 = radians(lat2)

  a = np.sin(dLat/2)**2 + np.cos(lat1)*np.cos(lat2)*np.sin(dLon/2)**2
  c = 2*np.arcsin(np.sqrt(a))

  return R * c

def drawCoverage(coefs,horizontal_distance,vertical_distance):
    """Coverage Overalp of two cells sperated by horizotal distance between two geographic co-ordinates and
    vertical distance - height
    """

    # Radii corresponding to the coefficients:
    # http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node42.html ??
    rx, ry, rz = coefs

    # Set of all spherical angles:
    u = np.linspace(0, 2 * np.pi, 100)
    v = np.linspace(0, np.pi, 100)

    # Cartesian coordinates that correspond to the spherical angles:
    # (this is the equation of an ellipsoid):
    x = rx * np.outer(np.cos(u), np.sin(v)) 
    y = ry * np.outer(np.sin(u), np.sin(v)) 
    z = rz * np.outer(np.ones_like(u), np.cos(v))

    # Plot:
    ax.plot_wireframe(x, y, z ,  rstride=4, cstride=4, color='b' )
    ax.plot_wireframe(x +horizontal_distance, y, z + vertical_distance,rstride=4, cstride=4, color='g')

    latitude1,longitude1 = 13.04738626,77.61946793   
    latitude2,longitude2 = 13.04638626, 77.61951605
    dist = haversine(latitude2,longitude2,latitude1,longitude1)
    height_in_meters = 258*2                                 
    coefs = (373, 258, 258)  # 373 major axis, 258 miniir axis - oblate spheroid,roate along x
    drawCoverage(coefs,0,height_in_meters)# for visual validity using just height

plt.show()

球形到笛卡尔坐标椭球重叠

1 个答案: