<p>Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit.
Earth's north pole vector in
<a href="https://en.wikipedia.org/wiki/Equatorial_coordinate_system" rel="nofollow">equatorial coordinates</a> is
$$\vec N_{\oplus,eq} = (0, 0, 1)$$
To transform this to
<a href="https://en.wikipedia.org/wiki/Ecliptic_coordinate_system" rel="nofollow">ecliptic coordinates</a>, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get
$$\vec N_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$
In a spherical coordinate system, two angles define a unique direction.
In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$.
This <a href="http://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2009reprint.pdf" rel="nofollow">IAU report</a>, table 1,
gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years.
To convert these to rectangular form
$$(x_{eq}, y_{eq}, z_{eq}) =
(\cos \alpha \cos \delta,
\sin \alpha \cos \delta,
\sin \delta)$$
Then if you want J2000 ecliptic coordinates
$$(x_{ecl}, y_{ecl}, z_{ecl}) =
(x_{eq},
y_{eq} \cos \varepsilon + z_{eq} \sin \varepsilon,
z_{eq} \cos \varepsilon - y_{eq} \sin \varepsilon)$$
But if you want another planet's orbital plane to be the $xy$-plane,
then you also need the orbit's inclination and longitude of ascending node;
I leave that transformation as an exercise for the motivated reader.</p>

CC BY-SA 3.0
2016-09-02T10:06:07.810Z
Mike, thank's for the answer and to correct my english (which is not my mother langage). If i understand well, the norm is : rotation of obliquity counter clockwise around  the vector defining the ascending node (x in the case of the eath). I already done the work to translate planet position thank to their orbit using kepler formula, what i missed was the norm used to define the rotation axis for the title... maybe a misunderstanding of the concept of ascending node. I start studying astronomy to no more than a week.. in all case many thanks.
Guillaume Pelletier
CC BY-SA 3.0
2016-09-02T13:22:09.743Z
Take care not to confuse orbital elements with rotational elements. Another planet's ascending node relates its orbital plane to Earth's orbital plane, independent of either planet's rotation.
Mike G
CC BY-SA 3.0
2016-09-03T08:44:23.050Z
Yes i do not confuse, but for a planet the axis tilt is defined thanks to its orbital plane by a rotation of obliquity using X vector ( X defined into the orbit plane, toward the point of ascending node ).  Is that right ??
Guillaume Pelletier
CC BY-SA 3.0
2016-09-03T14:13:17.250Z
No, that is only valid for Earth. In general there is no such shortcut; a single angle is insufficient to define a unique direction. Rotating the ecliptic pole about another planet's ascending node (not the x axis) by that planet's orbital inclination gives you the pole of the other planet's orbit. The obliquity of the planet's rotational axis gives you an angle away from that orbital axis but does not determine direction. You really need both $\alpha$ and $\delta$ as above.
Mike G
CC BY-SA 3.0
2016-09-03T14:53:36.033Z
You express my concern in FAR better way.. it' s mean this direction has NO rule and Has to be acquired by observation ??
Guillaume Pelletier
CC BY-SA 3.0
2016-09-03T15:19:25.950Z
Yes. Fortunately astronomers have made, analyzed, and published those observations.
Mike G
CC BY-SA 4.0
2019-12-09T03:30:42.283Z
@uhoh Table 1 of the IAU WGCCRE report gives most planets' polar $\alpha_0$ and $\delta_0$ [ICRF] as functions of $T$ [Julian centuries; T(J2000.0) = 0].
Mike G
CC BY-SA 4.0
2019-12-09T04:36:04.933Z
For some reason I totally missed that you'd already explained that, sorry for the erroneous comment.  fyi I've linked to this answer [here](https://space.stackexchange.com/q/40325/12102)
uhoh
CC BY-SA 4.0
2019-12-09T23:12:30.023Z
If i only had read this post 1 year ago when i did my bachelor thesis on coordinate transformations... i was so confused about why the time dependent Longitude of the ascending node of the earth was Long(t) = 0°, because after the Aquinoktium has been defined at J2000.0, why shouldnt earths orbit shift? And the answer is somehow still unclear to me...
Andres Huster
CC BY-SA 4.0
2019-12-10T01:04:46.427Z
@AndresHuster That sounds like it could be a new question.
Mike G
CC BY-SA 3.0
2016-09-02T00:22:13.210Z
2016-09-02T00:22:13.210Z
Mike G
<p>According to the orbital plane XYZ, I understand that the axial tilt is the angle between the planet axis rotation and any normal of the plane (actually Z in right handed system). BUT, this is not sufficient to define a vector direction, I need another angle into the XY plane. Where do I find this information?  Actually I believe it's relative to the direction of the periapsis. Am I right?</p>

<p>Following all the nice conversation on the answers, I maybe need to ask </p>

<p>How to get the axial tilt vector(x,y,z) relative to ecliptic or orbital plane, given ONLY ONE angle, i.e. 23.439281 deg for the Earth, 25.19 for Mars, etc.</p>

CC BY-SA 3.0
2016-09-01T16:48:14.947Z
Does [this post](http://astronomy.stackexchange.com/questions/8776/how-precisely-is-planetary-tilt-defined-the-tilt-direction-not-just-the-angle) answer your question?
called2voyage
CC BY-SA 3.0
2016-09-01T17:19:02.280Z
nope, still discussion from specialist and no one give simple definition.. one is speaking about the velocity vector, the other to another referential.. nothing clear for a newbiz like me. BTW many thank to taking the time to answer.
Guillaume Pelletier
CC BY-SA 3.0
2016-09-01T17:20:58.457Z
This is the answer: http://astronomy.stackexchange.com/a/8800/6 Unfortunately I'm not sure how to simply it from there.
called2voyage
CC BY-SA 3.0
2016-09-01T17:45:45.553Z
When you say "pole" do you mean "How do they define north and south?"
called2voyage
CC BY-SA 3.0
2016-09-01T17:58:11.707Z
no, i just want to obtain the vector of the axial tile relative to the ecliptic plane (z to the eccliptic pole) or relativ to the planet orbital plane, which must have the form of (alpha, theta) or (x,y,z). Actually, for the earth, it's seem a rotation X is doing the job. but for others planet with orbital plane different than the eccliptic i do not find any clear explanation anywhere..
Guillaume Pelletier
CC BY-SA 3.0
2016-09-01T17:58:41.917Z
To answer your question "How is the pole of rotation of an astronomical body defined", it is simply the axis of rotation, where the vector direction is defined by a right hand rule in relation to the rotation direction.
zephyr
CC BY-SA 3.0
2016-09-01T17:59:05.480Z
@GuillaumePelletier That is what is answered in the link I gave you above.
called2voyage
CC BY-SA 3.0
2016-09-01T18:08:55.893Z
nope, still no one give me how to get the axial tile vector(x,y,z) relative to eliptic or orbital plane, given ONLY ONE angle.. ie: 23.439281 deg for the earth, 25.19 for mars.. etc
Guillaume Pelletier
CC BY-SA 3.0
2016-09-01T19:34:30.553Z
@GuillaumePelletier If you only have a single angle, you cannot possibly determine the axial tilt vector. With only one angle (and knowing the planet rotation direction) at best you can define the vector to exist somewhere on a cone with opening angle equal to your given angle and direction defined by the rotation direction. You need more information to get the full, unique vector.
zephyr
CC BY-SA 4.0
2019-12-08T23:48:39.107Z
Try NASA JPL Horizons Ephemeris. This gives planetary axial tilt. I am trying to figure out how to determine the javians magnetic tilt when given a daily axial tilt
Gordon
<planet><orbit><rotation><coordinate><ecliptic>
How to get the axial tilt vector(x,y,z) relative to ecliptic
Guillaume Pelletier
# How to get the axial tilt vector(x,y,z) relative to ecliptic

(Asked by: username_0 on 2016-09-01)

According to the orbital plane XYZ, I understand that the axial tilt is the angle between the planet axis rotation and any normal of the plane (actually Z in right handed system). BUT, this is not sufficient to define a vector direction, I need another angle into the XY plane. Where do I find this information? Actually I believe it's relative to the direction of the periapsis. Am I right?

Following all the nice conversation on the answers, I maybe need to ask

How to get the axial tilt vector(x,y,z) relative to ecliptic or orbital plane, given ONLY ONE angle, i.e. 23.439281 deg for the Earth, 25.19 for Mars, etc.

**Tags:** planet, orbit, rotation, coordinate, ecliptic
> **Comments:**
> **username_1**: Does [this post](http://astronomy.stackexchange.com/questions/8776/how-precisely-is-planetary-tilt-defined-the-tilt-direction-not-just-the-angle) answer your question?  
> **username_0**: nope, still discussion from specialist and no one give simple definition.. one is speaking about the velocity vector, the other to another referential.. nothing clear for a newbiz like me. BTW many thank to taking the time to answer.  
> **username_1**: This is the answer: http://astronomy.stackexchange.com/a/8800/6 Unfortunately I'm not sure how to simply it from there.  
> **username_1**: When you say "pole" do you mean "How do they define north and south?"  
> **username_0**: no, i just want to obtain the vector of the axial tile relative to the ecliptic plane (z to the eccliptic pole) or relativ to the planet orbital plane, which must have the form of (alpha, theta) or (x,y,z). Actually, for the earth, it's seem a rotation X is doing the job. but for others planet with orbital plane different than the eccliptic i do not find any clear explanation anywhere..  
> **username_2**: To answer your question "How is the pole of rotation of an astronomical body defined", it is simply the axis of rotation, where the vector direction is defined by a right hand rule in relation to the rotation direction.  
> **username_1**: @GuillaumePelletier That is what is answered in the link I gave you above.  
> **username_0**: nope, still no one give me how to get the axial tile vector(x,y,z) relative to eliptic or orbital plane, given ONLY ONE angle.. ie: 23.439281 deg for the earth, 25.19 for mars.. etc  
> **username_2**: @GuillaumePelletier If you only have a single angle, you cannot possibly determine the axial tilt vector. With only one angle (and knowing the planet rotation direction) at best you can define the vector to exist somewhere on a cone with opening angle equal to your given angle and direction defined by the rotation direction. You need more information to get the full, unique vector.  
> **username_3**: Try NASA JPL Horizons Ephemeris. This gives planetary axial tilt. I am trying to figure out how to determine the javians magnetic tilt when given a daily axial tilt  

---

## 1 answers

### Answer by username_4 on 2016-09-02. Score: 4 | Accepted answer

Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit.
Earth's north pole vector in
[equatorial coordinates](https://en.wikipedia.org/wiki/Equatorial_coordinate_system) is
$$\vec N\_{\oplus,eq} = (0, 0, 1)$$
To transform this to
[ecliptic coordinates](https://en.wikipedia.org/wiki/Ecliptic_coordinate_system), we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get
$$\vec N\_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$
In a spherical coordinate system, two angles define a unique direction.
In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$.
This [IAU report](http://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2009reprint.pdf), table 1,
gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years.
To convert these to rectangular form
$$(x\_{eq}, y\_{eq}, z\_{eq}) =
(\cos \alpha \cos \delta,
\sin \alpha \cos \delta,
\sin \delta)$$
Then if you want J2000 ecliptic coordinates
$$(x\_{ecl}, y\_{ecl}, z\_{ecl}) =
(x\_{eq},
y\_{eq} \cos \varepsilon + z\_{eq} \sin \varepsilon,
z\_{eq} \cos \varepsilon - y\_{eq} \sin \varepsilon)$$
But if you want another planet's orbital plane to be the $xy$-plane,
then you also need the orbit's inclination and longitude of ascending node;
I leave that transformation as an exercise for the motivated reader.

> **Comments:**
> **username_0**: Mike, thank's for the answer and to correct my english (which is not my mother langage). If i understand well, the norm is : rotation of obliquity counter clockwise around  the vector defining the ascending node (x in the case of the eath). I already done the work to translate planet position thank to their orbit using kepler formula, what i missed was the norm used to define the rotation axis for the title... maybe a misunderstanding of the concept of ascending node. I start studying astronomy to no more than a week.. in all case many thanks.  
> **username_4**: Take care not to confuse orbital elements with rotational elements. Another planet's ascending node relates its orbital plane to Earth's orbital plane, independent of either planet's rotation.  
> **username_0**: Yes i do not confuse, but for a planet the axis tilt is defined thanks to its orbital plane by a rotation of obliquity using X vector ( X defined into the orbit plane, toward the point of ascending node ).  Is that right ??  
> **username_4**: No, that is only valid for Earth. In general there is no such shortcut; a single angle is insufficient to define a unique direction. Rotating the ecliptic pole about another planet's ascending node (not the x axis) by that planet's orbital inclination gives you the pole of the other planet's orbit. The obliquity of the planet's rotational axis gives you an angle away from that orbital axis but does not determine direction. You really need both $\alpha$ and $\delta$ as above.  
> **username_0**: You express my concern in FAR better way.. it' s mean this direction has NO rule and Has to be acquired by observation ??  
> **username_4**: Yes. Fortunately astronomers have made, analyzed, and published those observations.  
> **username_4**: @uhoh Table 1 of the IAU WGCCRE report gives most planets' polar $\alpha_0$ and $\delta_0$ [ICRF] as functions of $T$ [Julian centuries; T(J2000.0) = 0].  
> **username_5**: For some reason I totally missed that you'd already explained that, sorry for the erroneous comment.  fyi I've linked to this answer [here](https://space.stackexchange.com/q/40325/12102)  
> **username_6**: If i only had read this post 1 year ago when i did my bachelor thesis on coordinate transformations... i was so confused about why the time dependent Longitude of the ascending node of the earth was Long(t) = 0°, because after the Aquinoktium has been defined at J2000.0, why shouldnt earths orbit shift? And the answer is somehow still unclear to me...  
> **username_4**: @AndresHuster That sounds like it could be a new question.  

---

