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Trying to calculate Orbit figures with C# and 3D space

  1. Aug 30, 2014 #1
    Hello smart people!

    I am trying to make a simulation of an orbit with C#. I have come so far that I more or less only need a way to convert the apoapsis, periapsis and ascending node to 3D space vector3's. If it is possible, I want to avoid longitude and latitude?

    Anyone have any suggestions?

    my code: http://pastebin.com/5mbUVm99 [Broken]
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 30, 2014 #2
    Not sure what you mean by this, but at some point you will need spherical coordinates. From your code it looks like you have a good understanding of trigonometry so this shouldn't be a problem.

    One possible cause of confusion is reference frames. Have you defined one? If so, which? If not, do you need help with that? Have you heard of geocentric-ecliptic, geocentric-equatorial and heliocentric-ecliptic frames?

    The point is that latitude and longitude depend on Earth's rotation so it's not inertial. Totally understand you want to avoid that.
  4. Aug 30, 2014 #3
    Thank you for the quick reply :)

    - Im a noob at Orbital Mechanics and stuff like this, but I can learn quickly if a subject interests me :D

    I was hoping that a Quaternion was enough for the rotation of the primary, to calculate positions and degrees from. (even if I only know that a Quaternion is a rotation in 3d space)

    Im not even sure what a reference frame is, On some wikipages I saw a reference plane, that was a circle around equator. - I have not defined one :/

    I have not heard of any of these: geocentric-ecliptic, geocentric-equatorial and heliocentric-ecliptic frames - but maybe the equatorial is the equator plane I saw on the wiki page? :D
  5. Aug 30, 2014 #4
    Ok, it's much simpler than it seems, but extremely important if you want to work with x,y,z and define angles.

    In the celestial sphere astronomers define the celestial equator (a projection of Earth's equator) and the ecliptic, which is the apparent motion of the Sun through the sphere during the year. Of course, the ecliptic is basically a projection of Earth's orbit into the apparent sphere, so when we say "the ecliptic plane" we mean the plane of our orbit.

    The point where the ecliptic and the celestial equator intersect is the vernal equinox (Northern Hemisphere seasons). On March 21st this is the apparent position of the Sun in the sky. The line between this point and Earth's center is also the intersection between the equatorial plane and the ecliptic plane.

    The geocentric-equatorial reference frame has the origin on Earth's center, the Z axis through the North pole, the X axis pointing to the vernal equinox, and the Y axis follows the right hand rule. Both X and Y will be on the plane of the equator. X will also happen to be on the ecliptic plane, but not Y.

    The heliocentric-ecliptic reference frame has the origin on the Sun's center, the X axis pointing to the vernal equinox, the Y axis 90º ahead in the direction of Earth's orbit, and the Z axis pointing North perpendicular to the equatorial plane (it's not coincident with the Sun's North pole).

    The perifocal coordinate system makes the XY plane coincident with the satellite's orbit. The X axis points to the periapsis, the Y axis is 90º ahead in the direction of its motion and the Z axis completes the right hand rule.

    Source for the 3 above: Fundamentals of Astrodynamics by Bate, Mueller and White; ISBN 0-486-60061-0.

    I haven't found a geocentric-ecliptic frame in bibliography, but I have used it and I'm sure it does exist. Basically it's centered on Earth, X axis to the vernal equinox, Z axis pointing North perpendicular to the ecliptic plane and the Y axis completes the right hand rule.

    If you use geocentric-ecliptic and heliocentric-ecliptic there's a nice simplification as the XY plane is coincident, so you just have to add angles to convert from one to the other. The perifocal frame doesn't have this property so you need a rotation to convert, but it's nice because you only need one angle to describe your satellite's position.

    Are you familiar with Spherical coordinate systems? I assume yes because you have used trigonometry in your code, if not you'll be able to understand it.

    So when you use 3d vectors and x,y,z variables you must know what reference frame they refer to. I think that's all you need to do the conversions, though you may need a pen and a paper to draw some arrows and angles before you understand everything.
  6. Aug 30, 2014 #5
    actually nope.. the most of the calculations in my code I have done with using equations from: http://www.braeunig.us/space/orbmech.htm - I use Unity3D to show me the graphics, in my test case I placed the sun at xyz 0,0,0 - and the earth at around 147098291000 meters from it on X axis, so I had something that looked like something I could refer to. and gave earth and velocity of 30288.01431 m/s on the Z axis, so that it would have an orbit period of 365 days.

    After reading your reply, I think I need to study some more in basic trigonometry, I'll just continue to try figure the whole thing out, and some day it should probably work like I want it to :D

    Thank you for helping :)
  7. Aug 30, 2014 #6
    Don't worry, it's not complicated if you already know what sine and cosine are. It'll take little time to grasp the concept and apply it to your code.
  8. Sep 3, 2014 #7


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    I made an online calculator that does this: http://orbitsimulator.com/formulas/OrbitalElements.html
    It converts between vectors and orbital elements. It's a JavaScript, so you can view the code. The code is pretty much using straight-forward methods found in many celestial mechanic textbooks, and even Wikipedia.
  9. Dec 14, 2015 #8
    hello , I was asked to develop what you already did, could you please share the code with me since the link you provided is not working, and if you know other way to do it instead of using OpenGL cause i never used it
    Last edited by a moderator: May 7, 2017
  10. Dec 19, 2015 #9
    Several books here would probably help you.
    None of it is terribly difficult to understand, but it takes a while to get your head into that mode of thinking.
    Last edited by a moderator: May 7, 2017
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