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Astronomy Hacks: Hack 17 - Understanding Celestial Coordinate Systems

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Old 10-09-2008, 12:49 AM
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Default Astronomy Hacks: Hack 17 - Understanding Celestial Coordinate Systems

I will do my best to summarize this chapter, understanding only some of it.

Horizontal Coordinates: This uses alitude and azimuth, with the local horizon as the equator. Objects above the horizon are positive, below are negative (ranging from -90 to +90 degrees). Azimuth specifies the angular location of the object relative to north (and this is where they lose me. I can understand altitude as so many degrees above the horizon, but I can't seem to visualize "angular location relative to north").

Equitorial Coordinates: This uses Declination and Right Ascention. Declination (similar to Latitude) is the angular separation of an object north or south of the celestial equator. Right Ascention (similar to Longitude)is the angular position of an object proceeding east from the Vernal Equinox (and again this is where they lose me. I am not really sure what proceeding east from the Vernal Equinox means and how that relates to my location).

They discuss other coordinate systems, but since they are not as common, and I have enough trouble with these two I will not try to summarize them.

If anyone wants to take a shot at explaining this better, I would love to hear it (especially if you can "explain it to me like I was six"). Like I said I only half understand this.

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Old 10-09-2008, 03:37 AM
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Hi Keith.

Altitude is measured from the observers horizon (0 degrees) to Zenith (90 degrees)

Azimuth is measured from a start point North (0 Degrees) and proceeds in an Easterly direction from North to complete a full circle, hence East is 90 degree, South 180 degrees West 270 degrees.

The Celestial equator is the projection of the Earth's equator onto the celestial sphere. It is halfway between the Celestial poles and marks 0 Dec. Dec then follows in degrees either to Plus 90 degrees North of the Celestial Equator, or to Minus 90 degrees, South of the celestial equator. Hence the NCP is at + 90 Dec and the SCP at - 90 Dec

RA is measured in Hours and minutes and seconds from 0hr to 24hr from west to east along the equator. A start point 0hrs has been decided (by whom and when I don't know, but there has to be a standard point to work from for RA) to be the first point in Aries where the Sun crosses the Celestial Equator

RA therefore follows the same principal as terrestrial Longitude, (which has its reference or zero point dictated by Greenwich, again don't ask me why), and Dec is similar to terrestrial latitude

But don't try to relate this to terrestrial coordinates.

Now open Stellarium and superimpose the EQ grid, then speed the time up a few clicks and hopefully this will come together in your mind. What you now see is a cartwheel centred or pivoted on the NCP in your case, with the stars moving in circles around it. Those grid lines for DEC are, like latitude, fixed like rims on a wheel, but the lines of RA are now rotating like the spokes of a wheel.

Now hopefully this explains the most important point. If you are tracking an object using an EQ mount and setting circles, once you are on target you lock down the RA ring so that it doesn't change as you are tracking (and dec is always constant). Then when you want to move to another object, you release the RA ring so that the indicator once again reads off RA as the mount is moved.

Clear as mud, hey. but about the best I can do for you

Please let me know if you are still unclear on any or all and I'll have another crack at it.
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Scopes: Vixen: VMC 200, NA120, ED100. Skywatcher: 127 Mak, ED80. Mounts: EQ6, HEQ5. EQ3-2

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Last edited by Vinnie; 10-09-2008 at 03:51 AM.
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Old 10-09-2008, 10:14 PM
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Thanks Vin. I did what you said and it looks exactly like a bycicle wheel spinning. I think the reason I have a problem with it is that the RA coordinates mean nothing to me. If you say that Lyra is at a declination of 40 degrees (give or take) I know that is almost directly over head; but if you tell me that Orion is at RA 6 hrs that tells me nothing of where the object is.
It really did help watching that grid spin and seeing the constellations move from east to west as they do in the real world.
Thanks again Vin
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Old 10-09-2008, 10:31 PM
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Yeah, Keith,

Although at this time you are using a Dob, I think it is good to get this concept somehow or other into your head. It becomes an integral part of understanding an EQ mount should you progress to one in the future.
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Scopes: Vixen: VMC 200, NA120, ED100. Skywatcher: 127 Mak, ED80. Mounts: EQ6, HEQ5. EQ3-2

"If a straight line is the shortest distance between two points then a circle is the longest distance between the same point, provided the circle is big enough."- Sellar and Yeatman
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Old 10-09-2008, 11:57 PM
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Keith,

Also when you get a chance, would you open a new thread here and just call it "Astronomy Hacks". You will only need to write maybe one sentence of introduction, (like "here is my chapter by chapter synopsis of the book Astronomy Hacks" or something similar, then I can start moving all the older posts under the one topic so that we can include it in our helpful links etc sticky.

I could start a main heading, but then it won't come up in your username.

Thanks
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Scopes: Vixen: VMC 200, NA120, ED100. Skywatcher: 127 Mak, ED80. Mounts: EQ6, HEQ5. EQ3-2

"If a straight line is the shortest distance between two points then a circle is the longest distance between the same point, provided the circle is big enough."- Sellar and Yeatman
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Old 10-10-2008, 12:15 AM
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Hi Vin
said thread has been posted, although with a somewhat lengthy intro.
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