# Analytical expression for the ground track of the International Space Station

For a maths task in school, I am investigating the orbit of the International Space Station around the Earth. I understand that when the 3D movement in space is represented on a 2D surface, the relationship is not sinusoidal, however I have created the following (simple) model, which I am unsure is the most accurate. Below you can also find my calculated values with the formula (in red), compared to the actual values from an online API. Any help improving on this be greatly appreciated!

$$y=51.64*\sin(x-304)$$

(This only applies for one of the curves (the one pictured below), as the wave is translated 22.5 degrees to the right after each cycle.)

• It's an interesting question! I've made an edit but I'm not sure exactly what you would like to ask. Are you looking for a better equation that fits the ground track? One think you might notice is that the ground track does not repeat. After every 92 minute orbit the Earth has rotated by about 15 degrees underneath it, so the period of your sinusoid should be closer to 360-22.5=337.5 degrees. While your question is not answered there, see Why does the ISS track appear to be sinusoidal?
– uhoh
Apr 13, 2020 at 12:18
• Also see ground track.
– uhoh
Apr 13, 2020 at 12:24
• You are new to Stack Exchange so you don't know that it's generally discouraged to cross-post the same question to multiple SE sites. (question in Math SE) It might speed up getting an answer but it leads to answer fragmentation which is a problem for future readers. This is frequently confusing to new users, but we don't have a way to easily link answers to the same question that are posted on multiple sites.
– uhoh
Apr 13, 2020 at 13:01

tl;dr: use a parametric equation.

If the Earth were not rotating, then we would have something like

\begin{align} x & = \cos \omega (t-t_0)\\ y & = \sin \omega (t-t_0) \ \cos i\\ z & = \sin \omega (t-t_0) \ \sin i\\ \end{align}

where the radius of the orbit is 1, $$\omega$$ is $$2 \pi/T$$ and $$T$$ is the orbital period, and $$i$$ is the inclination of the orbit.

Then we would have

\begin{align} lon & = \arctan2(y, x) + const\\ lat & = \arcsin(z)\\ \end{align}

If the Earth is rotating then

$$lon = \arctan2(y, x) - \omega_E (t-t_0) + const$$

where $$\omega_E$$ is $$2 \pi/T_D$$ and $$T_D$$ is a sidereal day (23h, 56m, 4s roughly).

Solving this for longitude as a function of latitude looks like some serious work and I am not sure there is an analytical solution.

Instead you can try the parametric equation approach where you first make a hidden table of times, and then solve for $$lon(t)$$ and $$lat(t)$$ and plot $$lat$$ vs $$lon$$

Here is a plot, I haven't adjusted $$t_0$$ or $$const$$ and just used rough values for $$\omega$$, $$\omega_E$$ and $$i$$ but it should be enough to get you stared.

$$t_0$$ and $$const$$ represent the known starting conditions of the spacecraft that you are trying to plot; $$t_0$$ is the time at which it crosses the equator going north, and $$const$$ is the longitude on the Earth below the spacecraft at that time.

Python script:

import numpy as np
import matplotlib.pyplot as plt

twopi = 2*np.pi

omega = twopi/(92*60)
omega_E = twopi/(23*3600 + 56*60 + 4)

time = 60 * np.arange(101.) # 100 minutes

t0 = 1000. # arbitrary, you can fit this later
inc = 51.
const = 1.0  # arbitrary, you can fit this later

x = np.cos(omega * (time-t0))
y = np.sin(omega * (time-t0)) * np.cos(to_rads*inc)
z = np.sin(omega * (time-t0)) * np.sin(to_rads*inc)

lon = np.arctan2(y, x) - omega_E * (time-t0) + const
lat = np.arcsin(z)

if True:
plt.figure()
plt.plot(to_degs*lon, to_degs*lat, '.k')
plt.xlim(-180, 180)
plt.ylim(-60, 60)
#plt.gca().set_aspect('equal')
plt.show()

• Thank you very much! This is greatly appreciated! Apr 14, 2020 at 8:06
• Thanks a lot :) Apr 14, 2020 at 10:36
• @uhoh i'm sorry for writing this on a old post, I'd love to open a new question if I should! What is $t0 = 1000.$, and what is the $const = 1.0$? And I assume $time = 60 * np.arange(101.)$ is just the period of the orbit? Jan 13, 2021 at 15:36
• @lawndownunder Thanks for pointing that out! I've made some edits, how does it look? Orbital period was already explained: "$\omega$ is $2 \pi/T$ and $T$ is the orbital period" so $time$ just specifies where to put the dots on the plot; there are 101 dots space at 60 second intervals. Remember that time is the single independent parameter in this parametric solution.
– uhoh
Jan 13, 2021 at 22:54
• Thank you for the info @uhoh! I actually made a choice to complete re-ask the whole question, so I made a bigger edit and a title change, with a little bit more info. It also involves the answer from this question. Jan 16, 2021 at 9:21