How to draw a curve in Cartesian coordinates in a semicircle tube in polar coordinates?

Question

import numpy as np
import matplotlib.pylab as plt

def tube():
    theta = np.linspace(0, np.pi/2, 30)

    x = np.cos(theta)
    y = np.sin(theta)
    z = x*0.8
    w = y*0.8

    plt.plot(z,w)
    plt.plot(x,y)
    plt.axis("equal")
    plt.show()

print plt.figure(1);tube()

def euler():
    A, B, a = 40, 10, 2

    t  = 10  # time
    dt = 1e-3 # interval

    nbpt = int(t/dt)

    n = 1
    s = 1. # sign of the derivative, initially chosen
    y = [0]*nbpt # result

    while n < nbpt:
        yp2 = B - A*y[n-1]**a
        if yp2 < 0:
            s = -s
            n -= 1 # recalculating the previous value
        else:
            y[n] = y[n-1] + dt*s*np.sqrt(yp2)
            n += 1

    plt.plot(np.linspace(0,t,nbpt),y)
    plt.show()

print plt.figure(2);euler()

I want to draw the curve made with euler() in the tube made with tube() . I guess I must go from cartesian coordinates to polar coordinates but is there anyway to make the process easier with Python ?

Answer 1

There are many ways to do it since the question does not fully pin down what transformation you are looking for. However, assuming any transformation will do so long as the resultant curve oscillates between the boundary lines of the tube, you could use:

def polarmap(x, y):
    # normalize x and y from 0 to 1
    x = (x-x.min())/(x.max()-x.min())
    y = (y-y.min())/(y.max()-y.min())

    # make theta go from 0 to pi/2
    theta = np.pi*x/2

    # make r go from 0.8 to 1.0 (the min and max tube radius)
    r = 0.2*y + 0.8

    # convert polar to cartesian
    x = r*np.cos(theta)
    y = r*np.sin(theta)
    plt.plot(x, y)

---

For example,

import numpy as np
import matplotlib.pylab as plt


def tube():
    theta = np.linspace(0, np.pi/2, 30)

    x = np.cos(theta)
    y = np.sin(theta)
    z = x*0.8
    w = y*0.8

    plt.plot(z,w)
    plt.plot(x,y)

def euler():
    A, B, a = 40, 10, 2

    t  = 10  # time
    dt = 1e-3 # interval

    nbpt = int(t/dt)

    n = 1
    s = 1. # sign of the derivative, initially chosen
    y = [0]*nbpt # result

    while n < nbpt:
        yp2 = B - A*y[n-1]**a
        if yp2 < 0:
            s = -s
            n -= 1 # recalculating the previous value
        else:
            y[n] = y[n-1] + dt*s*np.sqrt(yp2)
            n += 1

    x = np.linspace(0,t,nbpt)
    y = np.array(y)
    return x, y

def polarmap(x, y):
    # normalize x and y from 0 to 1
    x = (x-x.min())/(x.max()-x.min())
    y = (y-y.min())/(y.max()-y.min())

    # make theta go from 0 to pi/2
    theta = np.pi*x/2

    # make r go from 0.8 to 1.0 (the min and max tube radius)
    r = 0.2*y + 0.8

    # convert polar to cartesian
    x = r*np.cos(theta)
    y = r*np.sin(theta)
    plt.plot(x, y)

fig, ax = plt.subplots()
tube()
x, y = euler()
polarmap(x, y)
plt.axis("equal")
plt.show()

which yields

---

Notice that in polarmap the first step was to normalize both x and y so that they both ranged from 0 to 1. You can think of them as parameters on equal footing. If you swap the two parameters before passing them to polarmap , eg:

x, y = euler()
x, y = y, x    # swap x and y
polarmap(x, y)

then you get