Convert plot to a surface plot, matplotlib?

Question

Win 10 x64 Anaconda Python 2.7

I'm plotting an involute spiral onto a Gaussian surface with the following code..

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# Spiral parameters
samNum = 1000
spConst = 10.0
t = np.linspace(0, 6*np.pi, samNum)

# Coordinates of involute spiral on xy-plane
coords = np.zeros([samNum, 3])
coords[:,0] = spConst * (np.cos(t) + t * np.sin(t)) # x coord
coords[:,1] = spConst * (np.sin(t) - t * np.cos(t)) # y coord

# Paramters for 2D Gaussian surface 
amp = 200
sigma_x = 75.0
sigma_y = 75.0
theta = np.pi
a = np.cos(theta)**2 / (2 * sigma_x**2) + np.sin(theta)**2 / (2 * sigma_y**2)
b = -np.sin(2 * theta) / (4 * sigma_x**2) + np.sin(2 * theta) / (4 * sigma_y**2)
c = np.sin(theta)**2 / (2 * sigma_x**2) + np.cos(theta)**2 / (2 * sigma_y**2)

# z coords of spiral projected onto Gaussian surface
coords[:,2] = amp * np.exp(-(a * coords[:,0]**2 - 2 * b * coords[:,0]*coords[:,1] + c * coords[:,1]**2)) # z coord

# plot 3D spiral
ax.scatter(coords[:,0], coords[:,1], coords[:,2],  s=1, c='k')

# plot lines projecting 3D spiral on to the xy-plane
for p in range(samNum):  
    ax.plot([coords[p,0], coords[p,0]], [coords[p,1], coords[p,1]], [0, coords[p,2]], color='g', linewidth=0.1)

ax.set_xlabel('X axis')
ax.set_ylabel('Y axis')
ax.set_zlabel('Z axis')

This gives the following output...

I would like to convert the green ribbon into a continuous surface. I have had a look at parametric surfaces in matplotlib but cant get my head around how to covert this into a surface.

So is this possible? Any pointers appreciated.

ImportanceOfBeingErnest · Accepted Answer · 2017-05-11 22:57:20Z

3

In principle you have everything you need already there,

t = np.linspace(0, 6*np.pi, samNum)

T, Z = np.meshgrid(t, [0,1])
X = spConst * (np.cos(T) + T* np.sin(T))
Y = spConst * (np.sin(T) - T * np.cos(T))

gives you the X and Y coordinates, and the upper Z coordinate is obtained via Z[1,:] = coords[:,2].

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# Spiral parameters
samNum = 1000
spConst = 10.0
t = np.linspace(0, 6*np.pi, samNum)

T, Z = np.meshgrid(t, [0,1])
X = spConst * (np.cos(T) + T* np.sin(T))
Y = spConst * (np.sin(T) - T * np.cos(T))

# Coordinates of involute spiral on xy-plane
coords = np.zeros([samNum, 3])
coords[:,0] = spConst * (np.cos(t) + t * np.sin(t)) # x coord
coords[:,1] = spConst * (np.sin(t) - t * np.cos(t)) # y coord

# Paramters for 2D Gaussian surface 
amp = 200
sigma_x = 75.0
sigma_y = 75.0
theta = np.pi
a = np.cos(theta)**2 / (2 * sigma_x**2) + np.sin(theta)**2 / (2 * sigma_y**2)
b = -np.sin(2 * theta) / (4 * sigma_x**2) + np.sin(2 * theta) / (4 * sigma_y**2)
c = np.sin(theta)**2 / (2 * sigma_x**2) + np.cos(theta)**2 / (2 * sigma_y**2)

# z coords of spiral projected onto Gaussian surface
coords[:,2] = amp * np.exp(-(a * coords[:,0]**2 - 2 * b * coords[:,0]*coords[:,1] + c * coords[:,1]**2)) # z coord

Z[1,:] = coords[:,2] 
ax.plot_surface(X,Y,Z)

# plot 3D spiral
ax.scatter(coords[:,0], coords[:,1], coords[:,2],  s=1, c='k')


ax.set_xlabel('X axis')
ax.set_ylabel('Y axis')
ax.set_zlabel('Z axis')

plt.show()

answered May 11, 2017 at 22:57

ImportanceOfBeingErnest

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3 Comments

DrBwts Over a year ago

I think its meshgrid I'm struggling with. What am I looking at when I'm looking at X, Y & Z? I can see that X[0,:]=X[1,:] (same with Y) but cant get my head around why do that? What are these meshgrid objects (no I cant make head nor tail our of the meshgrid docs). Am I making sense?

ImportanceOfBeingErnest Over a year ago

Is this a problem of understanding how meshgrid works or a problem of why the return of meshgrid is ideally suited to produce the kind of plot we're having here? (In any case I feel like I cannot give a good answer to that in the comments section of a different question.)

DrBwts Over a year ago

its both to be honest.

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