<p>i have no idea if i understand the issue.. i found this while looking for the ra and dec of the ecliptic, based on the icrf frame, because i have some work to do in my own simulator <a href="https://cubetronic.github.io" rel="nofollow noreferrer">https://cubetronic.github.io</a> ...<br />
but it sounds like you just need to do a simple transform.. but you'll need to know the ra and dec between ecliptic and laplace... there is a reference in the 2009 iau pdf for invariable to icrf... but not ecliptic.. anyway.. from my projects &quot;tilter.js&quot; file, i wrote out my own transform to use:</p>
<pre><code>
////////////////////////////////////////////////////////////////////////////////
// tilting (rotate/transform) between icrf and body frame (eci)
/*
    // see iau report: WGCCRE2009reprint.pdf
    // for reference, this is how bodies are oriented:
    // setup (threejs creates objects y-up even if defaultUp is z)
    body[moon].mesh.rotation.set(Math.PI / 2, 0, 0);

    // apply dec, then ra, then W eastward
    body[moon].mesh.rotateOnWorldAxis(
        yAxis, (Math.PI / 2) - body[moon].declination);
    body[moon].mesh.rotateOnWorldAxis(zAxis, body[moon].rightAscension);
    body[moon].mesh.rotation.y += (Math.PI / 2) + body[moon].primeMeridian;

// in three.js, tilting could be done like this:

// transform vector from icrf to tilt (z-up)
rotateOnWorldAxis(z, -ra)
rotateOnWorldAxis(y, dec - Math.PI / 2)

// transform vector from tilt to icrf (z-up)
rotateOnWorldAxis(y, Math.PI / 2 - dec)
rotateOnWorldAxis(z, ra)

*/

// to convert orbital coordinates to a local body frame,
// don't rotate to MATCH the body frame,
// but rather rotate in the equal and opposite direction

// transform vectors from icrf to local body frame (all z-up)
function icrfToEci(icrf, ra, dec) {
    let {x, y, z, vx, vy, vz} = icrf;

    // rotate on z axis, -ra
    let x2 = x * Math.cos(-ra) - y * Math.sin(-ra);
    let vx2 = vx * Math.cos(-ra) - vy * Math.sin(-ra);
    let y2 = y * Math.cos(-ra) + x * Math.sin(-ra);
    let vy2 = vy * Math.cos(-ra) + vx * Math.sin(-ra);

    // rotate on y axis (dec - Math.PI / 2) must swap signs
    let angle = -dec + Math.PI / 2;
    let x3 = x2 * Math.cos(angle) - z * Math.sin(angle);
    let vx3 = vx2 * Math.cos(angle) - vz * Math.sin(angle);
    let z2 = z * Math.cos(angle) + x2 * Math.sin(angle);
    let vz2 = vz * Math.cos(angle) + vx2 * Math.sin(angle);

    x = x3;
    vx = vx3;
    y = y2;
    vy = vy2;
    z = z2;
    vz = vz2;

    return {x, y, z, vx, vy, vz};
}

// transform vectors from local body frame to icrf (all z-up)
function eciToIcrf(eci, ra, dec) {
    let {x, y, z, vx, vy, vz} = eci;

    // rotate on y axis (Math.PI / 2 - dec) must swap signs
    let angle = -Math.PI / 2 + dec;
    let x2 = x * Math.cos(angle) - z * Math.sin(angle);
    let vx2 = vx * Math.cos(angle) - vz * Math.sin(angle);
    let z2 = z * Math.cos(angle) + x * Math.sin(angle);
    let vz2 = vz * Math.cos(angle) + vx * Math.sin(angle);

    // rotate on z axis, ra
    let x3 = x2 * Math.cos(ra) - y * Math.sin(ra);
    let vx3 = vx2 * Math.cos(ra) - vy * Math.sin(ra);
    let y2 = y * Math.cos(ra) + x2 * Math.sin(ra);
    let vy2 = vy * Math.cos(ra) + vx2 * Math.sin(ra);

    x = x3;
    vx = vx3;
    y = y2;
    vy = vy2;
    z = z2;
    vz = vz2;

    return {x, y, z, vx, vy, vz};
}
</code></pre>
<p>idk if that's what you need.. but i can see from the image it looks like you're using y-up not z-up, so this may not work as expected. good luck</p>

CC BY-SA 4.0
2022-01-04T04:33:14.447Z
2022-01-04T04:33:14.447Z
cubetronic
<p>The <a href="https://ssd.jpl.nasa.gov/?sat_elem" rel="nofollow noreferrer">NASA JPL websize</a> gives mean orbital elements for e.g. Iapetus or Titan in relation to the local laplace plane.</p>
<pre><code>// Mean orbital elements for Iapetus referred to the local Laplace planes
a(km)    e       w(deg)   M(deg)   i(deg)  node(deg)  n(deg/day)  P(days)  Pw(yr)   Pnode(yr) 
3560854  0.0293  271.606  201.789  8.298   81.105     4.5379416   79.33    1676.69  3438.73
// Laplace plane parameters (are Pw and Pnode part of it?)
R.A.(deg)  Dec.(deg)  Tilt(deg)
284.715    78.749     15.210
</code></pre>
<p>Via the <a href="https://ssd.jpl.nasa.gov/horizons.cgi" rel="nofollow noreferrer">JPL HORIZONS system website</a> I can also get the relative position of these objects in Ecliptic of J2000.0 reference frame. I've chosen the specific date of JD2454886.9461875 for this question, since Titan should create an eclipse on Saturn at that date, but I'll use Iapetus as the example here, as it should be much further away from the regular ecliptic.</p>
<pre><code>*******************************************************************************
Ephemeris / WWW_USER Tue Aug 24 09:07:33 2021 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: Iapetus (608)                   {source: sat427l_merged_DE438}
Center body name: Saturn (699)                    {source: sat427l_merged_DE438}
*******************************************************************************
$$SOE
2454886.946187500 = A.D. 2009-Feb-24 10:42:30.6000 TDB 
 X =-3.667428122525245E-03 Y =-2.292450955142063E-02 Z = 6.128563674523897E-03
 VX= 1.818570055804101E-03 VY=-3.151270839204952E-04 VZ=-2.929182822870737E-04
 LT= 1.386776951611314E-04 RG= 2.401129858878437E-02 RR=-5.166357246498433E-05
*******************************************************************************
</code></pre>
<p>Converting the orbital elements to Cartesian coordinates (in the local laplacian reference plane) is fairly simple. I use <a href="https://github.com/mgreter/astro.js" rel="nofollow noreferrer">my own JS implementation</a> to do that (i'm quite sure the result is correct, but feel free to challenge).</p>
<pre><code>var lapIapetusOrbital = {
  e: 0.0293,
  a: 3560854 * KM2AU,
  w: 271.606 * DEG2RAD,
  M: 201.789 * DEG2RAD,
  i: 8.298 * DEG2RAD,
  O: 81.105 * DEG2RAD,
  // lap_ra: 284.715 * DEG2RAD,
  // lap_dec: 78.749 * DEG2RAD,
  // lap_tilt: 15.210 * DEG2RAD,
  G: GMJY.sat, // AU and JY
  epoch: 0,
};

var time = JD2J2K(2454886.9461875);
console.log(new Orbit(lapIapetusOrbital).r(time));
</code></pre>
<p>Now the question is how to rotate the resulting Cartesian position to convert it into J2000 ecliptic reference frame, given the laplace frame parameters from the <a href="https://ssd.jpl.nasa.gov/?sat_elem" rel="nofollow noreferrer">JPL Website</a>? I really seem not to be able to figure it out.</p>
<pre><code>in =&gt; { x: -0.013035389203744634, y: -0.020207481562296, z: 0.001422614609286161 }
out =&gt; { x: -3.667428122525245E-03 y: -2.292450955142063E-02, z: 6.128563674523897E-03 }
</code></pre>
<p>I <a href="http://www.ocbnet.ch/laplace-frame-transform/" rel="nofollow noreferrer">created a small ThreeJS example</a> to illustrate the problem. It shows Saturn with Titan and Iapetus once with Horizon positions and once with calculated Cartesian coordinates. Goal is to rotate the calculated positions close to the Horizon positions.</p>
<p><a href="https://i.sstatic.net/BfWhh.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BfWhh.png" alt="Positions without correction" /></a></p>
<p>I'm aware that this method will not yield good results if the calculated time moves further away from the orbital elements epoch. But I'm still really curious how to do it as I couldn't for the life of me figure it out until now. You may also checkout a failed implementation at <a href="http://www.ocbnet.ch/solsys-preview-19/three-app-build/demo/solsys.html?stars=2#inert=saturn&amp;fov=65&amp;pos.x=0.01358624325612839&amp;pos.y=-0.001701195901759227&amp;pos.z=0.003044825090501039&amp;tgt.x=0&amp;tgt.y=0&amp;tgt.z=0&amp;jd=2454886.9461875&amp;speed=0.0002777777777777778" rel="nofollow noreferrer">my latest solsys explorer preview</a> (titan is in the wrong place).</p>
<p>Any help is greatly appreciated!</p>

CC BY-SA 4.0
2022-01-04T23:46:51.313Z
I am not familiar with the problem, so I can't say for sure, but chapter 9 of the Explanatory Supplement to the Astronomical Almanac 3rd ed seems to answer your question in a couple of different ways.  The process is too involved to put here.
Greg Miller
<orbital-mechanics><orbital-elements><frame-of-reference>
How to translate from local laplace plane to ecliptic J2000 (or ICRF)
Marcel Greter
# How to translate from local laplace plane to ecliptic J2000 (or ICRF)

(Asked by: username_0 on 2021-08-24)

The NASA JPL websize gives mean orbital elements for e.g. Iapetus or Titan in relation to the local laplace plane.
// Mean orbital elements for Iapetus referred to the local Laplace planes
a(km)    e       w(deg)   M(deg)   i(deg)  node(deg)  n(deg/day)  P(days)  Pw(yr)   Pnode(yr) 
3560854  0.0293  271.606  201.789  8.298   81.105     4.5379416   79.33    1676.69  3438.73
// Laplace plane parameters (are Pw and Pnode part of it?)
R.A.(deg)  Dec.(deg)  Tilt(deg)
284.715    78.749     15.210

Via the JPL HORIZONS system website I can also get the relative position of these objects in Ecliptic of J2000.0 reference frame. I've chosen the specific date of JD2454886.9461875 for this question, since Titan should create an eclipse on Saturn at that date, but I'll use Iapetus as the example here, as it should be much further away from the regular ecliptic.
*******************************************************************************
Ephemeris / WWW_USER Tue Aug 24 09:07:33 2021 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: Iapetus (608)                   {source: sat427l_merged_DE438}
Center body name: Saturn (699)                    {source: sat427l_merged_DE438}
*******************************************************************************
$$SOE
2454886.946187500 = A.D. 2009-Feb-24 10:42:30.6000 TDB 
 X =-3.667428122525245E-03 Y =-2.292450955142063E-02 Z = 6.128563674523897E-03
 VX= 1.818570055804101E-03 VY=-3.151270839204952E-04 VZ=-2.929182822870737E-04
 LT= 1.386776951611314E-04 RG= 2.401129858878437E-02 RR=-5.166357246498433E-05
*******************************************************************************

Converting the orbital elements to Cartesian coordinates (in the local laplacian reference plane) is fairly simple. I use my own JS implementation to do that (i'm quite sure the result is correct, but feel free to challenge).
var lapIapetusOrbital = {
  e: 0.0293,
  a: 3560854 * KM2AU,
  w: 271.606 * DEG2RAD,
  M: 201.789 * DEG2RAD,
  i: 8.298 * DEG2RAD,
  O: 81.105 * DEG2RAD,
  // lap_ra: 284.715 * DEG2RAD,
  // lap_dec: 78.749 * DEG2RAD,
  // lap_tilt: 15.210 * DEG2RAD,
  G: GMJY.sat, // AU and JY
  epoch: 0,
};

var time = JD2J2K(2454886.9461875);
console.log(new Orbit(lapIapetusOrbital).r(time));

Now the question is how to rotate the resulting Cartesian position to convert it into J2000 ecliptic reference frame, given the laplace frame parameters from the JPL Website? I really seem not to be able to figure it out.
in => { x: -0.013035389203744634, y: -0.020207481562296, z: 0.001422614609286161 }
out => { x: -3.667428122525245E-03 y: -2.292450955142063E-02, z: 6.128563674523897E-03 }

I created a small ThreeJS example to illustrate the problem. It shows Saturn with Titan and Iapetus once with Horizon positions and once with calculated Cartesian coordinates. Goal is to rotate the calculated positions close to the Horizon positions.

I'm aware that this method will not yield good results if the calculated time moves further away from the orbital elements epoch. But I'm still really curious how to do it as I couldn't for the life of me figure it out until now. You may also checkout a failed implementation at my latest solsys explorer preview (titan is in the wrong place).
Any help is greatly appreciated!

**Tags:** orbital-mechanics, orbital-elements, frame-of-reference
> **Comments:**
> **username_1**: I am not familiar with the problem, so I can't say for sure, but chapter 9 of the Explanatory Supplement to the Astronomical Almanac 3rd ed seems to answer your question in a couple of different ways.  The process is too involved to put here.  

---

## 1 answers

### Answer by username_2 on 2022-01-04. Score: 1

i have no idea if i understand the issue.. i found this while looking for the ra and dec of the ecliptic, based on the icrf frame, because i have some work to do in my own simulator https://cubetronic.github.io ...
but it sounds like you just need to do a simple transform.. but you'll need to know the ra and dec between ecliptic and laplace... there is a reference in the 2009 iau pdf for invariable to icrf... but not ecliptic.. anyway.. from my projects "tilter.js" file, i wrote out my own transform to use:

////////////////////////////////////////////////////////////////////////////////
// tilting (rotate/transform) between icrf and body frame (eci)
/*
    // see iau report: WGCCRE2009reprint.pdf
    // for reference, this is how bodies are oriented:
    // setup (threejs creates objects y-up even if defaultUp is z)
    body[moon].mesh.rotation.set(Math.PI / 2, 0, 0);

    // apply dec, then ra, then W eastward
    body[moon].mesh.rotateOnWorldAxis(
        yAxis, (Math.PI / 2) - body[moon].declination);
    body[moon].mesh.rotateOnWorldAxis(zAxis, body[moon].rightAscension);
    body[moon].mesh.rotation.y += (Math.PI / 2) + body[moon].primeMeridian;

// in three.js, tilting could be done like this:

// transform vector from icrf to tilt (z-up)
rotateOnWorldAxis(z, -ra)
rotateOnWorldAxis(y, dec - Math.PI / 2)

// transform vector from tilt to icrf (z-up)
rotateOnWorldAxis(y, Math.PI / 2 - dec)
rotateOnWorldAxis(z, ra)

*/

// to convert orbital coordinates to a local body frame,
// don't rotate to MATCH the body frame,
// but rather rotate in the equal and opposite direction

// transform vectors from icrf to local body frame (all z-up)
function icrfToEci(icrf, ra, dec) {
    let {x, y, z, vx, vy, vz} = icrf;

    // rotate on z axis, -ra
    let x2 = x * Math.cos(-ra) - y * Math.sin(-ra);
    let vx2 = vx * Math.cos(-ra) - vy * Math.sin(-ra);
    let y2 = y * Math.cos(-ra) + x * Math.sin(-ra);
    let vy2 = vy * Math.cos(-ra) + vx * Math.sin(-ra);

    // rotate on y axis (dec - Math.PI / 2) must swap signs
    let angle = -dec + Math.PI / 2;
    let x3 = x2 * Math.cos(angle) - z * Math.sin(angle);
    let vx3 = vx2 * Math.cos(angle) - vz * Math.sin(angle);
    let z2 = z * Math.cos(angle) + x2 * Math.sin(angle);
    let vz2 = vz * Math.cos(angle) + vx2 * Math.sin(angle);

    x = x3;
    vx = vx3;
    y = y2;
    vy = vy2;
    z = z2;
    vz = vz2;

    return {x, y, z, vx, vy, vz};
}

// transform vectors from local body frame to icrf (all z-up)
function eciToIcrf(eci, ra, dec) {
    let {x, y, z, vx, vy, vz} = eci;

    // rotate on y axis (Math.PI / 2 - dec) must swap signs
    let angle = -Math.PI / 2 + dec;
    let x2 = x * Math.cos(angle) - z * Math.sin(angle);
    let vx2 = vx * Math.cos(angle) - vz * Math.sin(angle);
    let z2 = z * Math.cos(angle) + x * Math.sin(angle);
    let vz2 = vz * Math.cos(angle) + vx * Math.sin(angle);

    // rotate on z axis, ra
    let x3 = x2 * Math.cos(ra) - y * Math.sin(ra);
    let vx3 = vx2 * Math.cos(ra) - vy * Math.sin(ra);
    let y2 = y * Math.cos(ra) + x2 * Math.sin(ra);
    let vy2 = vy * Math.cos(ra) + vx2 * Math.sin(ra);

    x = x3;
    vx = vx3;
    y = y2;
    vy = vy2;
    z = z2;
    vz = vz2;

    return {x, y, z, vx, vy, vz};
}

idk if that's what you need.. but i can see from the image it looks like you're using y-up not z-up, so this may not work as expected. good luck


---

