Unfinished and uncertain)):
Since I am astrographer, I wanted to understand these two planes rotation, since in galactic plane are most deepsky objects and ecliptic plane (zodiacal signs) are essential for positioning. 

My guess is that really since north ecliptic is l=97 degrees (Lacerta), b=+30 degrees, from earth with 23,5° earth axis tilt it will appear to be Cassiopeia 120° (hence 23,5° offset, which goes statically in this direction - away from center galaxy (celestial sphere is pointing just "only" 3° lower(than ecliptic) toward galaxy disk, ***b*** = 27°).


I ve add **same pic in bellow** as author did, but with stars, which can anyone observe and so can be more familiar with (in appendix it is explained). Milky way is faint there, but you can compare it with the **second bellow pic.**, that shows the constellations in galactic plane surrounding our solar system. **Third pic bellow** shows Earth position in equinoxes and solstices on its orbit around the sun.

You may use something like **Stellarium** and check positions of these planes in solstices and equinoxes (as I did) and it may help you to get feeling for their orbits. Some pictures bellow also may help you to visualize it:


 False following **toward Cassiopeia** (in 60° tilt down): If you on the equator watch 60° up to north (to compensate solar system downward angle) at **vernal equinox** day moment, when sun is at highest point, which is now in Pieces constellation (RA – 0h), than you would see Cassiopeia (you can check it in Stellarium when sun light is turned off). At this time/ place 23,5° Earth axis does not offset you on equator, but as you have been going up, you went up 23,5° off (hence Lacerta - 97°).

If you watch (on equator) at **Autumn equinox** 120° up thru ground (or went 90° north, than 30° south, than watch directly overhead)) you would again see Cassiopeia.

You may also wait to night on 60° long/ 0° lat. and wait for align galactic and azimuthal sphere (which will be in 2020 in 23. 9 at 02: 43 minutes) and you will see Cassiopeia on zenith - you can than measure galactic long., which will be 123° (+/1 1°). So you can also see that at around midnight on Spring eq. you see (logically) stars/ signs of mid-day in Autumn eq.

It may help that 60° angle is down in direction away from galaxy center: like you are on top of the roof representing solar system, behind you back is Milky way center and solar system plane is falling down (at 60°) toward Cassiopeia.

In **Summer solstice** at night you can clearly see that on -23,5° you are seeing Sagittarius at your zenith which is in 0° galactic longitude. In mid EU – 50° this is -23 – 50 → 73° south from zenith, so only about 20° above the ground. Ecliptic is highest (hence summer), but zodiacal signs at night are lowest.

In **Winter solstice** you can clearly see that galactic plane (signs in galactic plane) and ecliptic (zodiacal signs) are again crossing. You can see in night on 23,5° (to compensate Earth axis tilt) galactic sing Auriga (180° gal. long.) above your head. Ecliptic (zodiacal signs) reaches in north at night highest point – you can see top of Gemini reaches less than 20° from zenith (in day ecl. is in its lowest – hence winter).

**Double check** -if you compare Galactic (Polaris Galacticum Borealis) and celestial north (Polaris) you will find out, that link continues to Cassiopeia. But key is here Galactic and Ecliptic north - which goes from gal. n. to 97° (Lacerta).

*Fourth picture* shows galactic, ecliptic plane with celestial equator – shows that ecliptic is 60° (author claimed 63°, but corrected it - anyone can get wrong heh?))


**Appendix**
**Right Ascension** define coordinates of object on Celestial sphere in analogy of longitude on Earth, but since Earth is rotating, it set 0 longitude like **0 hour** at the moment of vernal equinox – when sun is at highest point on 0° lat, which is now in Pieces (Greek had it in Aries and Babylonians in Taurus – it shifted due to precession). Than the hours/ min/ sec define how long behind this point object is. To get RA working in your time/ place, you have subtract RA value from local sidereal times – it so called Hour angle. If you set value to you telescope with equatorial mount (its RA value) you will find the object.

**Six hours** define same way highest point of sun in *summer solstice*, **12 h** in *Fall equinox*, **18 h** *winter solstice*.

**Declination** defines position on Celestial sphere: to get you local Dec, you just subtract from Dec your latitude - means how far from your zenith object is, minus value means south direction/ plus - north.

[![enter image description here][1]][1]

[![enter image description here][2]][2]

[![enter image description here][3]][3]

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  [1]: https://i.sstatic.net/XFlHa.jpg
  [2]: https://i.sstatic.net/UnGeP.jpg
  [3]: https://i.sstatic.net/RgCif.jpg
  [4]: https://i.sstatic.net/0FgN9.jpg