The three planes of our astronomical neighborhood.

[Philippe Lemay]

Lately I've been attempting to better visualize my place in the universe, from Terra out to the Milky Way galaxy. Because they are (I find) easy to understand, compared to the rest of the universe anyway.

From what I can see... there are three primary "planes" that we use as baselines to measure the locations of other astronomical bodies lying in space. The celestial sphere, which is grounded on the celestial equator, this is dictated by the Earth's equator. The ecliptic plane, which is grounded on the Earth's orbit, or as I like to call it the Sun's default equator. And finally the galactic plane, which is kinda self explanatory.

I'm sure many of you all know this, I'm not being condescending, I'm just running through this so I can establish just what I do or do not understand.

Now... I wanted to use Sketchup to make a 3D starmap of local space. But I got stuck on the angles that seperate the three planes. I know that the equator is 23.45 degrees off of the ecliptic, and that the galactic is supposedly 60 degrees off of the ecliptic. The problem is, I don't know... in which direction the tilt is when considering all three.

Here's a picture to better show my confusion.
http://i215.photobucket.com/albums/c...sEquator-1.png

The blue lines represent the possible alignment of the galactic plane. Each one is 60 degrees off of the equator... but... they can't both be right, can they?

[01101001]

Astronomer Fiona Vincent: Positional Astronomy: Galactic coordinates has a picture that relates the three. I can't vouch for it. I haven't had my morning coffee.

Exercise:

The North Galactic Pole is at Right Ascension 12h49m, declination +27°24'.
What is the tilt of the galactic plane to the celestial equator?

Click here for the answer.

Doing a Google image search on the terms plane galactic celestial ecliptic looks to provide many illustrations of the directions. Some of them may even agree with each other.

[Cougar]

The blue lines represent the possible alignment of the galactic plane. Each one is 60 degrees off of the equator... but... they can't both be right, can they?

No, can't both be right. It looks like the correct one is the one going from bottom left to upper right, but check me on that.

[Jeff Root]

There is a coincidence that the galactic center happens to be
very close to the ecliptic.

Grant Hutchison did a diagram or two on this subject. Hello, Grant??

-- Jeff, in Minneapolis

[Philippe Lemay]

With the help of the very useful site http://www.atlasoftheuniverse.com/ I was able to build up a model using solely galactic coordinates. I got this, that zero is pointed towards the galactic center.

http://i215.photobucket.com/albums/c.../Starmap04.png

Now... I kind of realize this is exactly what I was initially looking for. A 3D map of some of the local stars and their orientation relative to our own star and the rest of the galaxy (I guess all I really needed was their galactic coordinates), but I am still curious about the relative placement of the ecliptic and celestial planes. Now that I think about it though... the angle of tilt isn't enough. I would need to know like... the orientation.

[Philippe Lemay]

I think I got it! I'm looking for the exact place where the galactic plane crosses the ecliptic plane. So in other words... I'm looking for the galactic-ecliptic equinoxes. If I find that I can tilt the plane in the right direction on my Sketchup map.

[astromark]

The first two plains of axis are easy enough to 'see'. Earths tilted over at 23.45 deg from the Solar Plain.
Any school room globe shows that.. its what our seasons are about... but even those two are further complicated.
Earths Moon does not orbit above the equatorial path. Its orbital path wanders north and south a surprising amount.
The constant east to west track of most celestial objects does help create a useful image. The planets are conforming.
Next on your list is the angle of the galactic plain. Because of the great distances and obscuring interstellar stuff...
We do not get a great view of the Milky Way. At different times of the year my view becomes as good as it gets.. 40deg South.
Looking straight up there is the hart of the Milky Way. When the sky is clear and away from city lighting.
So if I at about midnight on the 21st of June I measure the angle from the ecliptic to the central cloud of our galaxy...
yep. I can actually claim to see it. The other method I could suggest is to use the options panel on 'Celestia' and it will show you these ecliptic lines and you will 'see'. All will be clear. If its not raining which it is and is thus why I am here not there... at the observatory.

[Sophic]

Hmmm, you might want to try the program Celestia from this website, http://www.shatters.net/celestia/, its a full 3d astronomy program, it will let you float around all the different stars and zoom in a out, plus you can show multiple grids such as the celestial and galactic.

[CaptainToonces]

I think I got it! I'm looking for the exact place where the galactic plane crosses the ecliptic plane. So in other words... I'm looking for the galactic-ecliptic equinoxes. If I find that I can tilt the plane in the right direction on my Sketchup map.

two planes intersect in a line. When you're doing multiple 3d transformations, such as earth's rotational axis to earth's solar position, to the solar system's galactic position, and including orientations, it is a little complicated and one of the standard ways to do this is with matrix multiplication.

For example, you could give your body's position and orientation in Earth coordinates using 4 vectors. The first 3 vectors are all perpendicular to each other in 3d space and they specify the directions that are "up", to your "right", and the direction you are "looking". The 4th vector is your location translated from the origin of your coordinate system in this case the Earth's coordinate system. So you take those 4 vectors and make each of them a column in a matrix. You then create the same kind of matrix for the Earth's position and orientation in the Sun's coordinate system. You then multiply these two matrices by each other according to the rules of matrix multiplication (which can be found on wikiP) and oula you have your body's position in the Sun's coordinate system. You can then do this again according to the sun's galactic position to get your body's position in galactic coordinates.

Sorry if this doesn't help I have background in 3d graphics software engineering.

[astromark]

You might be able to do these things with your coordinated matrix of 4 vectors column's I have no idea what you are on about.
and before you jump on my lac of understanding just consider that in a graphics studio not too far from here was little
'Gollum' created and 'Avatar' brought to life... Mine is a 3d world. I need more help to understand your view.
I can only advise a search through Celestia...

[CaptainToonces]

I'm not equipped to go into it tonight but the point is if the problem is converting between coordinate frames such as my head to the earth to the sun to the galaxy, it is not that easy, and matrix multiplication with orthogonal vectors won't let you down if you pursue that angle of solution.

[astromark]

What might be for you a perfectly clear understanding of the terrestrial, solar and galictic plains. is not at issue for me. I did not ask the question...
I have no want to study your 'matrix multiplication with orthogonal vectors'... I am content with my understanding of the relationships of axial tilts in relation to a clear understanding of ecliptics and the three plains of the near buy objects of interest. This is not in any way to suggest that what you are saying is wrong. It most likely is correct. The fact that I do not understand you has been made...
Poor old Philippe Lemay... I think hes got it right regardless of us.

[EDG]

There are some excel co-ordinate converstion sheets that could help you here, Phillippe (this is from a project I did myself a while back):
http://www.evildrganymede.net/rpg/world/mapping.htm