CODEWITHSUNDEEP
pythonnumpymatplotlib
dustin
dustin

Reputation: 4406

Using arctan / arctan2 to plot a from 0 to 2π

I am trying to replicate a plot in Orbital Mechanics by Curtis, but I just can't quite get it. However, I have made head way by switching to np.arctan2 from np.arctan.

Maybe I am implementing arctan2 incorrectly?

import pylab
import numpy as np

e = np.arange(0.0, 1.0, 0.15).reshape(-1, 1)

nu = np.linspace(0.001, 2 * np.pi - 0.001, 50000)
M2evals = (2 * np.arctan2(1, 1 / (((1 - e) / (1 + e)) ** 0.5 * np.tan(nu / 2) -
           e * (1 - e ** 2) ** 0.5 * np.sin(nu) / (1 + e * np.cos(nu)))))

fig2 = pylab.figure()
ax2 = fig2.add_subplot(111)

for Me2, _e in zip(M2evals, e.ravel()):
    ax2.plot(nu.ravel(), Me2, label = str(_e))

pylab.legend()
pylab.xlim((0, 7.75))
pylab.ylim((0, 2 * np.pi))
pylab.show()

In the image below, there are discontinuities popping up. The function is supposed to be smooth and connect at 0 and 2 pi in the y range of (0, 2pi) not touching 0 and 2pi.

Enter image description here

Textbook plot and equation:

Enter image description here

Enter image description here

At the request of Saullo Castro, I was told that:

The problem may lie in the arctan function which gives "principle values" as output.

Thus, arctan(tan(x)) does not yield x if x is an angle in the second or third quadrant. If you plot arctan(tan(x)) from x = 0 to x = Pi, you will find that it has a discontinuous jump at x = Pi/2.

For your case, instead of writing arctan(arg), I believe you would write arctan2(1, 1/arg) where arg is the argument of your arctan function. That way, when arg becomes negative, arctan2 will yield an angle in the second quadrant rather than the fourth."

Upvotes: 9

Views: 24076

Answers (2)

Tandeitnik
Tandeitnik

Reputation: 11

Given x and y coordinates, the following code will evaluate correctly the angle (assuming the angle is measured from the x-axis and counter-clockwise) between the range 0 to 2pi. It just sums 2pi if the output of arctan2 is negative in a simple and effective manner.

import numpy as np
np.arctan2(y,x) + -1*(np.sign(np.arctan2(y,x))-1)*np.pi

Upvotes: 1

Saullo G. P. Castro
Saullo G. P. Castro

Reputation: 58895

The common practice is to sum 2pi in the negative results of arctan(), which can be done efficiently. The OP's suggestion to replace arctan(x) by arctan2(1, 1/x), also suggested by Maple 15's documentation as pointed out by Yay295, produces the same results without the need to sum 2pi. Both are shown below:

import pylab
import numpy as np
e = np.arange(0.0, 1.0, 0.15).reshape(-1, 1)
nu = np.linspace(0, 2*np.pi, 50000)
x =  ((1-e)/(1+e))**0.5 * np.tan(nu/2.)
x2 = e*(1-e**2)**0.5 * np.sin(nu)/(1 + e*np.cos(nu))
using_arctan = True
using_OP_arctan2 = False

if using_arctan:
    M2evals = 2*np.arctan(x) - x2
    M2evals[M2evals<0] += 2*np.pi
elif using_OP_arctan2:
    M2evals = 2 * np.arctan2(1,1/x) - x2

fig2 = pylab.figure()
ax2 = fig2.add_subplot(111)
for M2e, _e in zip(M2evals, e.ravel()):
    ax2.plot(nu.ravel(), M2e, label = str(_e))
pylab.legend(loc='upper left')
pylab.show()

Enter image description here

Upvotes: 12

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