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Drawing elliptical orbit in Python (using numpy, matplotlib)

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轻奢々 2020-12-21 13:12

I wonder how can I draw elliptical orbit by using the equation ay² + bxy + cx + dy + e = x² ?

I have first determined the a,b,c,d,e constants a

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夕颜 (楼主)

2020-12-21 13:34
Intro

Easiest thing would be to parametrize this equation. As @Escualo suggests, you could introduce a variable t and parametrize x and y along that. Parametrizing means separating your equation into two separate equations for x and y individually in terms of t. So you would have x = f(t) and y = g(t) for some values of t. You could then plot the x, y pairs that result for each value of t.

The catch here is that your ellipse is rotated (the x*y coupling term is an indication of that). To separate the equations, you have to first transform the equation to get rid of the coupling term. This is the same as finding a set of axes that are rotated by the same angle as the ellipse, parametrzing along those axes, then rotating the result back. Check out this forum post for a general overview.

Math

You first need to find the angle of rotation of the ellipse's axes with respect to the x-y coordinate plane.

$\theta = \frac{1}{2} tan^{-1}(\frac{b}{-1 - a})$

Your equation then transforms to

$a'x'^2+b'y'^2+c'x+d'y'+e'=0$

where

$a'=-cos^2(\theta)+b*sin(\theta)cos(\theta)+a*sin^2(\theta)$

$b'=-sin^2(\theta)-b*sin(\theta)cos(\theta)+a*cos^2(\theta)$

$c'=c*cos(\theta)+d*sin(\theta)$

$d'=-c*sin(\theta)+d*cos(\theta)$

$e'=e$

To find the (nearly) standard form of the ellipse, you can complete the squares for the $x'$ and $y'$ portions and rearrange the equation slightly:

$\frac{(x'-h)^2}{a''^2}+\frac{(y'-k)^2}{b''^2}=c''$

where

$a''=\frac{1}{\sqrt{a'}}$

$b''=\frac{1}{\sqrt{b'}}$

$c''=\frac{c'^2}{4a'}+\frac{d'^2}{4b'}-e'$

$h=-\frac{c'}{2a'}$

$k=-\frac{d'}{2b'}$

Since you know $\theta$ , you can now parametrize the equations for $x'$ and $y'$ :

$x'=h+a''\sqrt{c''}sin(t)$

$y'=k+b''\sqrt{c''}cos(t)$

You would then rotate back into normal x-y space using the formulas

$x=x'cos(\theta)-y'sin(\theta)$

and

$y=x'sin(\theta)+y'cos(\theta)$

Code

The code to get the x- and y- arrays to pass to plt.plot is now relatively straightforward:
```
def computeEllipse(a, b, c, d, e):
    """
    Returns x-y arrays for ellipse coordinates.
    Equation is of the form a*y**2 + b*x*y + c*x + d*y + e = x**2
    """
    # Convert x**2 coordinate to +1
    a = -a
    b = -b
    c = -c
    d = -d
    e = -e
    # Rotation angle
    theta = 0.5 * math.atan(b / (1 - a))
    # Rotated equation
    sin = math.sin(theta)
    cos = math.cos(theta)
    aa = cos**2 + b * sin * cos + a * sin**2
    bb = sin**2 - b * cos * sin + a * cos**2
    cc = c * cos + d * sin
    dd = -c * sin + d * cos
    ee = e
    # Standard Form
    axMaj = 1 / math.sqrt(aa)
    axMin = 1 / math.sqrt(bb)
    scale = math.sqrt(cc**2 / (4 * aa) + dd**2 / (4 * bb) - ee)
    h = -cc / (2 * aa)
    k = -dd / (2 * bb)
    # Parametrized Equation
    t = np.linspace(0, 2 * math.pi, 1000)
    xx = h + axMaj * scale * np.sin(t)
    yy = k + axMin * scale * np.cos(t)
    # Un-rotated coordinates
    x = xx * cos - yy * sin
    y = xx * sin + yy * cos

    return x, y
```
To actually use the code:
```
from matplotlib import pyplot as plt

a = -4.10267300566
b = 1.10642410023
c = 0.39735696603
d = 3.05101004127
e = -0.370426134994

lines = plt.plot(*computeEllipse(a, b, c, d, e))
```
To overplot your original points on the ellipse:
```
x = [1.02, 0.95, 0.87, 0.77, 0.67, 0.56, 0.44, 0.30, 0.16, 0.01]
y = [0.39, 0.32, 0.27, 0.22, 0.18, 0.15, 0.13, 0.12, 0.12, 0.15]
ax = lines[0].axes
ax.plot(x, y, 'r.')
```
The result is the following image:

Note

Keep in mind that the forum post I linked to uses a different notation than the one you do. Their equation is Ax² + Bxy + Cy² + Dx + Ey + F = 0. This is a bit more standard than your form of ay² + bxy - x² + cx + dy + e = 0. All of the math is in terms of your notation.
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