How does Equatorial System work?

morls (Stephen) · #1 05-01-2011, 02:52 PM

Hi,

For my holiday reading I'm checking out "The Night Sky" by Steve Massey. I've only made it in a few pages before getting myself all confused - the equatorial system is tying my brain in knots!

Here's my problem: (paraphrasing from book..)

- "earth rotates through 360º in 24 hours, hence one hour of arc = 15º"
I get it. 15x24=360. Simple mathematics.

- "each hour is further subdivided into minutes and seconds of arc"
I get it. Sounds perfectly reasonable.

Now comes the bit where my brain gets jammed:

- "For example, 1º of sky is equal to 60 arc minutes and 1 minute of sky is equal to 60 arc seconds"

I don't get it.

I thought 1 hour of arc = 15º.

given that, shouldn't 60 arc minutes = 1 arc hour = 15º?

I don't know where my thinking is going wrong here, can anyone help?

higginsdj · #2 05-01-2011, 03:06 PM

1 hr of RA = 15 degrees, not 1 hr of arc. RA is a measure based on time, ie the time it takes to rotate 360 degrees. 2 different units of measure. So 1 min of RA is not equal to 1 arc minute but 15 arc minutes and 1 sec in RA is not 1 arc second but 15 arc seconds. Declination on the other hand is measured in degrees.

Cheers

morls (Stephen) · #3 05-01-2011, 03:24 PM

Thanks David,

I'm not sure I understand your answer yet....I'll go off and give it some thought.

I think my decision to start with a dobsonian is looking better all the time!

Cheers

Stephen

RickS (Rick) · #4 05-01-2011, 03:32 PM

The terminology is confusing. Arc minutes and arc seconds are just subdivisions of a degree. They don't have any direct relationship to minutes and seconds of time.

morls (Stephen) · #5 05-01-2011, 03:55 PM

Ah, that helps. Thanks Rick.

barx1963 (Malcolm) · #6 06-01-2011, 12:24 AM

Remember that co-ordinates in the sky are measued by 2 references RA (right ascension) and Dec (declination)
Dec uses degrees, minutes and seconds and measures how far from the celestial equator the object is, so something directly above the equator of the earth is at 0deg in dec, something halfway between the equator and the South Celestial pole is at -45deg in dec and the South Celestial Pole is at -90deg Dec. Each degree in dec is divided into 60 minutes and each minute is divided into 60 seconds.
RA is measured in TIME and is in Hours, Minutes, and seconds. Think of its as the distance an object moves in the sky in that time period. Obviously an object over the equator moves Faster than one closer to the pole.
The one hour equaling 15deg only applies at the equator. Logically a star sitting directly on one of the poles won't move no matter how long you wait!
Malcolm

morls (Stephen) · #7 06-01-2011, 03:16 PM

Thanks Malcolm.

It's a very gradual process for me, and I'm trying to learn new terminology and think about what I'm seeing, and trying to visualise, in a different way...so please forgive me if I'm paraphrasing like mad here..

So am I right to think in RA we have hours, minutes and seconds, up to 23 hours, 59 minutes, 59 Seconds, and these do relate to time? Is Greenwich 0hr 0' 0"?

In DEC, is it between +90º (N) and -90º (S)? And each degree is also divided into arc minutes and arc seconds, but in this case they have nothing to do with time, but are just subdivisions of a degree of angle?

Thanks for all your replies, I really appreciate the help

bojan · #8 06-01-2011, 03:30 PM

Quote:

Originally Posted by morls

Thanks Malcolm.
So am I right to think in RA we have hours, minutes and seconds, up to 23 hours, 59 minutes, 59 Seconds, and these do relate to time? Is Greenwich 0hr 0' 0"?

Not quite

RA (or more correctly AR - Ascension Recta - see here: http://books.google.com.au/books?id=...0recta&f=false) is a (sideral) time, elapsed between culmination moment of Ascending node (intersection point of Ecliptic and Celestial equator) and the culmination moment of the object in question.
Or, if we divide the celestial equatorial circle by 24, and each section we call "hour", then if we divide each "hour" by 60 "minutes" and so on, then AR is distance between celestial meridians of Ascending node and the meridian of a said object.

Quote:

Originally Posted by morls

In DEC, is it between +90º (N) and -90º (S)? And each degree is also divided into arc minutes and arc seconds, but in this case they have nothing to do with time, but are just subdivisions of a degree of angle?

Quite right

!
That is why we call those units "arc minutes" and arc seconds", sometimes even omitting word "arc"

morls (Stephen) · #9 06-01-2011, 04:40 PM

Thanks Bojan, although I'm still confused....

Quote:

Originally Posted by bojan

RA (or more correctly AR - Ascension Recta - see here: http://books.google.com.au/books?id=...0recta&f=false) is a (sideral) time, elapsed between culmination moment of Ascending node (intersection point of Ecliptic and Celestial equator) and the culmination moment of the object in question.

Does this mean the point where the ecliptic and celestial equator intersect is 0hr 0' 0"? What do you mean by culmination moment?

Quote:

Originally Posted by bojan

Or, if we divide the celestial equatorial circle by 24, and each section we call "hour", then if we divide each "hour" by 60 "minutes" and so on, then AR is distance between celestial meridians of Ascending node and the meridian of a said object.

I'm a bit confused by this...I thought the ascending node was the point at which an object's plane intersects the ecliptic as it moves from south to north?

bojan · #10 06-01-2011, 05:32 PM

Quote:

Originally Posted by morls

Does this mean the point where the ecliptic and celestial equator intersect is 0hr 0' 0"? What do you mean by culmination moment?

Exactly right

Culmination is a moment when particular object (or point on celestial sphere) is highest above horizon.

Quote:

Originally Posted by morls

I'm a bit confused by this...I thought the ascending node was the point at which an object's plane intersects the ecliptic as it moves from south to north?

Sorry, I made a mistake here

What I meant was Vernal Equinox point (http://en.wikipedia.org/wiki/Ecliptic)

morls (Stephen) · #11 06-01-2011, 07:00 PM

Thanks again Bojan

with both your explanations and the links you gave I'm getting a lot closer to understanding RA....seems to make sense if I think of the Vernal Equinox point being the celestial equivalent of Greenwich...

bojan · #12 06-01-2011, 07:32 PM

Yes, vernal equinox point is the starting point, the origin of the celestial coordinate system.

It is also worth mentioning, that the local sideral time is actually the time elapsed since the vernal equinox point crossed the local meridian.

BTW, this is the historical reason why RA is expressed in time units and not in angle: in the past RA of the object was measured with the help of a "Meridian instrument" (http://en.wikipedia.org/wiki/Transit_instrument).
This was a telescope mounted such that it could move only in altitude. The Azimuth was fixed and oriented in South-North direction.
Astronomer would measure the exact time of passing of object over the centre of the field of view (with the help of cross hairs in the eyepiece).
The local sideral time of that passage was actually RA of the object

NB: 24h of sideral time is the time Earth takes for one rotation against stars - and this time is 3m 56s shorter than solar day.

morls (Stephen) · #13 07-01-2011, 02:57 PM

Thanks Bojan, and everyone else who has helped me with this...it's great of you all to share your knowledge

As I move a bit beyond just pointing my dob at this object or that, and actually start to think about how people have been describing not only what is out there, but where it is in relation to us and everything else I'm more and more amazed....