# Plotting implicit equations in 3d

### Question

I'd like to plot implicit equation F(x,y,z) = 0 in 3D. Is it possible in Matplotlib?

1
32
1/21/2011 8:15:49 AM

You can trick matplotlib into plotting implicit equations in 3D. Just make a one-level contour plot of the equation for each z value within the desired limits. You can repeat the process along the y and z axes as well for a more solid-looking shape.

``````from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
import numpy as np

def plot_implicit(fn, bbox=(-2.5,2.5)):
''' create a plot of an implicit function
fn  ...implicit function (plot where fn==0)
bbox ..the x,y,and z limits of plotted interval'''
xmin, xmax, ymin, ymax, zmin, zmax = bbox*3
fig = plt.figure()
A = np.linspace(xmin, xmax, 100) # resolution of the contour
B = np.linspace(xmin, xmax, 15) # number of slices
A1,A2 = np.meshgrid(A,A) # grid on which the contour is plotted

for z in B: # plot contours in the XY plane
X,Y = A1,A2
Z = fn(X,Y,z)
cset = ax.contour(X, Y, Z+z, [z], zdir='z')
# [z] defines the only level to plot for this contour for this value of z

for y in B: # plot contours in the XZ plane
X,Z = A1,A2
Y = fn(X,y,Z)
cset = ax.contour(X, Y+y, Z, [y], zdir='y')

for x in B: # plot contours in the YZ plane
Y,Z = A1,A2
X = fn(x,Y,Z)
cset = ax.contour(X+x, Y, Z, [x], zdir='x')

# must set plot limits because the contour will likely extend
# way beyond the displayed level.  Otherwise matplotlib extends the plot limits
# to encompass all values in the contour.
ax.set_zlim3d(zmin,zmax)
ax.set_xlim3d(xmin,xmax)
ax.set_ylim3d(ymin,ymax)

plt.show()
``````

Here's the plot of the Goursat Tangle:

``````def goursat_tangle(x,y,z):
a,b,c = 0.0,-5.0,11.8
return x**4+y**4+z**4+a*(x**2+y**2+z**2)**2+b*(x**2+y**2+z**2)+c

plot_implicit(goursat_tangle)
`````` You can make it easier to visualize by adding depth cues with creative colormapping: Here's how the OP's plot looks:

``````def hyp_part1(x,y,z):
return -(x**2) - (y**2) + (z**2) - 1

plot_implicit(hyp_part1, bbox=(-100.,100.))
`````` Bonus: You can use python to functionally combine these implicit functions:

``````def sphere(x,y,z):
return x**2 + y**2 + z**2 - 2.0**2

def translate(fn,x,y,z):
return lambda a,b,c: fn(x-a,y-b,z-c)

def union(*fns):
return lambda x,y,z: np.min(
[fn(x,y,z) for fn in fns], 0)

def intersect(*fns):
return lambda x,y,z: np.max(
[fn(x,y,z) for fn in fns], 0)

def subtract(fn1, fn2):
return intersect(fn1, lambda *args:-fn2(*args))

plot_implicit(union(sphere,translate(sphere, 1.,1.,1.)), (-2.,3.))
`````` 48
1/19/2011 10:35:32 PM

As far as I know, it is not possible. You have to solve this equation numerically by yourself. Using scipy.optimize is a good idea. The simplest case is that you know the range of the surface that you want to plot, and just make a regular grid in x and y, and try to solve equation F(xi,yi,z)=0 for z, giving a starting point of z. Following is a very dirty code that might help you

``````from scipy import *
from scipy import optimize

xrange = (0,1)
yrange = (0,1)
density = 100
startz = 1

def F(x,y,z):
return x**2+y**2+z**2-10

x = linspace(xrange,xrange,density)
y = linspace(yrange,yrange,density)

points = []
for xi in x:
for yi in y:
g = lambda z:F(xi,yi,z)
res = optimize.fsolve(g, startz, full_output=1)
if res == 1:
zi = res
points.append([xi,yi,zi])

points = array(points)
``````