9

I am trying to 3D plot the magnification factor in vibrations for multiple types of damping. To simplify it for those who have no idea what it is, basically, you have 3 variables:

  • beta, which varies between 0 and infinite, but I would like to visualize it from 0 to 3, in 0.2 intervals.
  • damping ratio, d, which varies between 0 and infinite, but I would like to plot it from 0 to 1, in 0.1 intervals.
  • finally, nu, which is a function that varies according to the two variables before.

2D function, y is nu, x is beta

My intuition says that I should plot this with (X,Y,Z) = (beta, d, nu), but I am just starting to use this library and I am kind of new to python, I just use it when I need to visualize or calculate problems in class. I tried creating 2 arrays for beta and d, but I don't know I should create the array for nu, since it depends on both.

This is the piece of code I have until now:

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


nu = []
b = [0.1 + i / 100 for i in range(0, 510)]
damp = [0.1 + i/10 for i in range(0,510)]

for d in damp:
    nu_new = []
    nu.append(nu_new)
    for beta in b:
        nu_new.append( math.sqrt(1+(2*d*beta)**2)/ math.sqrt((1-beta**2)**2+(2*d*beta)**2))

fig = plt.figure()
ax = Axes3D(fig)
ax.plot(b, d, nu)
plt.show()

I am kind of stuck trying to plot this, so if you have any suggestion I would be glad.

Improve this question

4 Answers 4

Reset to default
14

If you're using numpy, then don't use the math module. Numpy as all of the math functions built in but they work on numpy arrays far better. We can calculate nu at all of our b, d values with the aid of a meshgrid.

A meshgrid can take 2 1D arrays, and return 2 2D arrays such that every index in the arrays corresponds to a unique pair of elements from the original 1D arrays.

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

b = np.arange(0.2, 3.2, 0.2)
d = np.arange(0.1, 1.0, 0.1)

B, D = np.meshgrid(b, d)
nu = np.sqrt( 1 + (2*D*B)**2 ) / np.sqrt( (1-B**2)**2 + (2*D*B)**2)

fig = plt.figure()
ax = Axes3D(fig)
ax.plot_surface(B, D, nu)
plt.xlabel('b')
plt.ylabel('d')
plt.show()

This produces: enter image description here

Additionally, 3D plots tend to block seeing all of the data (because a spike hides things behind it). I would recommend a pcolormesh or a contourf plot. In the later case the last 6 lines become:

plt.contourf(B, D, nu)
plt.colorbar()
plt.xlabel('b')
plt.ylabel('d')
plt.show()

which produces: enter image description here

Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

3

This should work: I'm not a Python expert and especially the two for loops might be very unpythonic, but it gets the job done.

import math
import matplotlib.pyplot as plt
import numpy as np

b = np.arange(0.2, 3.2, 0.2)
d = np.arange(0.1, 1.0, 0.1)
nu = np.zeros( (b.size, d.size) )
counter_y = 0

for deta in d:
    counter_x = 0
    for beta in b:
        nu[counter_x, counter_y] = math.sqrt( 1 + (2*deta*beta)**2 ) / math.sqrt( (1-beta**2)**2 + (2*deta*beta)**2)
        counter_x += 1
    counter_y += 1

X, Y = np.meshgrid(d, b)

fig = plt.figure()
ax = fig.add_subplot(111, projection = '3d')
ax.plot_surface(X, Y, nu)
Improve this answer

Comments

1

what you need is to first create a matplotlib figure. 

fig = plt.figure()
ax = Axes3D(fig)
ax.plot(b, d, nu)
plt.show()

Further, all your variable should be of same size. So, your variable d should be an array of same length as the others.

If you turn your variable d into array of 0.1 of length 510 as others, you get following.

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


nu = []
b = [0.1 + i / 100 for i in range(0, 510)]
d = 0.1


for beta in b:
    nu.append( math.sqrt(1+(2*d*beta)**2)/ math.sqrt((1-beta**2)**2+(2*d*beta)**2))

#turned d into array of length 510 with 0.1 for each value
d = np.ones(510)*0.1


fig = plt.figure()
ax = Axes3D(fig)
ax.plot(b, d, nu)
plt.show()

you get:

enter image description here

Improve this answer

2 Comments

thank you! that makes sense according to the documentation. But I am also having trouble populating nu. How can I populate it if it varies with 2 variables?
its exactly that I want but for multiple d's. The problem is I don't know how to create multiple nu's for the multiple d's
0

Here is the answer of MalteHerrmann but without for loops.

import math
import matplotlib.pyplot as plt
import numpy as np

b = np.arange(0.2, 3.2, 0.2)
d = np.arange(0.1, 1.6, 0.1) # changed a bit the interval
nu = np.zeros( (b.size, d.size) )

for i in range (len(d)):
    for j in range (len(b)):
        nu[i,j] = math.sqrt( 1 + (2*d[i]*b[j])**2 ) / math.sqrt( (1-b[j]**2)**2 + (2*d[i]*b[j])**2)

X, Y = np.meshgrid(d, b)

fig = plt.figure()
ax = fig.add_subplot(111, projection = '3d')
ax.plot_surface(X, Y, nu)
Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.