explaining right ascension

[fistfullofnails]

i have been trying to figure this out for two days now to no avail.  from what i understand, the coordinates for right ascension and declination do not change for an object.  im apparently thinking about it all wrong because i cannot understand how something that moves throughout the sky over time can keep the same coordinates.  apparently the 0hr point moves throughout the sky as well maybe?  if so,  how do i keep up with that point?  im having my first telescope delivered friday and im trying to learn all i can ahead of time so i might be half prepared to jump right into it.  

[triton1]

It's us that is moving not the sky.

[ronin]

You are OK so far, it is a fixed point that is moving. :BangHead:

Where is the problem in that?

Bit like being on a boat and training binoculars on a lighthouse you are sailing past.

You have to move the binoculars but the object (lighthouse) is stationary.

Look at a tree or a building from a train, same thing.

[Gazabone]

Yep, exactly what Triton1 said. The stars etc don't move but you're on a planet that is spinning round (once every 23 hours, 59 minutes and 4 seconds). It might help to think of it like this; suppose you line up an object against 2 fixed points on earth (eg 2 telegraph poles) and say for example that RA is 15 hours RA then in 1 hours time the point you've lined up will be 14 hours RA because 15hours RA has moved on 1 more hour.

I've mentioned the time it takes for the earth to spin exactly once; if you have an object at exactly RA 15 hours at exactly 22:00 tonight then it will be in exactly the same place in the sky RELATIVE TO YOU at exactly 21:57:04 tomorrow night. (Sorry about shouty words, couldn't underline on my iPhone but wanted to emphasise).

Once you've got your head around it, it's easy but, as I've just found out, trying to explain it is another matter!

[Cornelius Varley]

Image that the stars are fixed points on a sphere with the earth at the centre of the sphere (the earth rotates within the larger sphere). The stars have celestial coordinates on that sphere (RA / DEC) just as locations on the earth have terrestrial coordinates (longitude / latitude)  . From a point on the earth's surface it appears that  the stars move across the sky.

[fistfullofnails]

so, how do I go about finding 0 RA at any given time?  Im thinking everything else is useless without knowing that.

[fistfullofnails]

Yep, exactly what Triton1 said. The stars etc don't move but you're on a planet that is spinning round (once every 23 hours, 59 minutes and 4 seconds). It might help to think of it like this; suppose you line up an object against 2 fixed points on earth (eg 2 telegraph poles) and say for example that RA is 15 hours RA then in 1 hours time the point you've lined up will be 14 hours RA because 15hours RA has moved on 1 more hour.

I've mentioned the time it takes for the earth to spin exactly once; if you have an object at exactly RA 15 hours at exactly 22:00 tonight then it will be in exactly the same place in the sky RELATIVE TO YOU at exactly 21:57:04 tomorrow night. (Sorry about shouty words, couldn't underline on my iPhone but wanted to emphasise).

Once you've got your head around it, it's easy but, as I've just found out, trying to explain it is another matter!

so then I take it 0 RA will move across the sky just the same.  ive been thinking if theoretically i woke up on march 21 and noted where the sun crossed the equatorial plane that day and marked that spot on my window, then at all other times and all other days that spot would still mark 0 RA.

[stu640]

Hopefully this will sound simple enough! If your telescope is going to have RA and Dec setting circles which spin freely, you may use this method to find an object. Find an object you know close enough to the one you're looking for. Set its coordinates on the setting circles. Your telescope mount now has the sky's 'grid' set for that particular moment in time. You won't have long to now slew the scope to the coordinates of your target on you setting circles, before the sky moves on. Now that's the theory I've picked up - I've not tried it yet!

[Merlin66]

By far the easiest way of using the RA/Dec circles on your mount is to know the position of a bright star in your night sky......

Then use that to re-set the index on your circles - point your scope to the known star (after aligning the polar axis to the pole!!!)

and move the circles until they match the known coordinates. If your mount has the usual RA drive then the RA circle on your mount will continue to rotate with the sky, maintaining the known coordinates while you observe the known star....

Ok, when you're ready to find an object based on it's RA/Dec just move the mount until the index pointers show the required RA/Dec - you're object should be close to the centre of the field of view.

Having said all that, you should also be aware that the circles fitted to the average mounts are pretty average to be kind(!!).

Give it a go and see what you think.

BTW what mount are you going to be using??

[ronin]

I think part of the problem is that RA is more a time difference as a position.

People say it is like Latitude but that sort of leads you off down a bit of a false path, as Latitude is fixed.

An RA of 4 hours means that the object is 4 hours away from being on the Greenwich Meridian line.

The Greenwich Meridian Line being the accepted datum from which RA is measuered, the Datum Time at which the measurement is made is I assume 00:00.

[fistfullofnails]

By far the easiest way of using the RA/Dec circles on your mount is to know the position of a bright star in your night sky......

Then use that to re-set the index on your circles - point your scope to the known star (after aligning the polar axis to the pole!!!)

and move the circles until they match the known coordinates. If your mount has the usual RA drive then the RA circle on your mount will continue to rotate with the sky, maintaining the known coordinates while you observe the known star....

Ok, when you're ready to find an object based on it's RA/Dec just move the mount until the index pointers show the required RA/Dec - you're object should be close to the centre of the field of view.

Having said all that, you should also be aware that the circles fitted to the average mounts are pretty average to be kind(!!).

Give it a go and see what you think.

BTW what mount are you going to be using??

this is the telescope that is being delivered.  

Orion 8945 SkyQuest XT8 Classic Dobsonian Telescope

[Cornelius Varley]

Wikipedia explanation of right ascension http://en.wikipedia.org/wiki/Right_ascension

[Merlin66]

Hmmmmm

The Dobbie moves in Alt/Az rather than the RA/Dec that the Equatorial will use......

You'll have difficulty see the RA/Dec circles on a Dobbie !!!!