vernal equinox

[pallas15]

as a noob ive been trying to get my head around the different ways of finding stars ,planets etc.I pretty much get the azimuthal grid but the part thats confusing me is the j2000 equatorial grid referencing. For example declination is measured either +(x) degrees or -(x) degrees.Thats fair enough but that is measured from the equatorial line, but I cant for the life of me figure out where this should be to actually measure the coordinates.Is it a specific number of degrees from the zenith or a certain angle depending on your direction and/ or elevation???..Next is right ascension.I totally get that each 15degree angle of direction is an hour RA, i.e. each turn to N/S/E/W is 6 hours in time and the whole grid is 360degrees in 24 hours AND that right ascension coordinates travel left to right..for example Hamal is(at time of posting) around 2hours 10 from the vernal equinox which is (if ive got this correct) the point where the ecliptic line crosses the equatorial line that is in pisces.NOW.. this is where it gets confusing.Sitting in front of my laptop with stellarium and star calc i can just count from the grids to get my info.Outside is a different matter.Yes I recognize some stars and constellations but id like to know how I can use these coordinates without a reference point in the sky as the pointers on the charts are constantly changing as time rolls on.

At the time of writing the vernal equinox is at about 120 degress...it was 97 earlier. The equator was 36 degrees above my facing position of 155 degrees but as i moved the equators angle moved (obviously),..There must be a method to get a solid reference point but im unsure as to the simplest way.If I turn all the compass markers and grids off stellarium I would like to be able to work out coordinates but the only constant I can think of is polaris....My head hurts,never been too good at mathematics ....hope some one can help....

[Mr Q]

The RA/DEC coordinates system is used to find objects on a star atlas and thus, in the night (and day) sky.

DEC is the equivalent to our terristial maps lattitude but labled in degrees, minuets and seconds. Zero degrees is the "ecliptic", which is the path our solar system objects take as they pass through the sky. Negative numbers are below (to the south) and positive numbers are above (to the north celestial pole at Polaris) this line.

RA is the equivalent to our Earth maps' longitude and is measured in hours, minuets and seconds.

Each direction is measured in a 360 degree circle with RA hours broken down into hours (24 for the complete celestial circle), minuets (60 minuets = 1 hour) and then seconds (60 seconds = 1 minuet) and DEC is measured in degrees (360 for the full circle), with 60 minuets = 1 degree and 60 seconds = 1 degree.

It sounds complicated at first but once you get used to it, it's a simple plotting system that is necessary in locating objects using a star atlas and plotting an object's position in the night sky.

Think of the celestial coordinates system as marking the inside of a sphere (the whole night sky) that encompasses the Earth (our viewing point).

[pallas15]

The RA/DEC coordinates system is used to find objects on a star atlas and thus, in the night (and day) sky.

DEC is the equivalent to our terristial maps lattitude but labled in degrees, minuets and seconds. Zero degrees is the "ecliptic", which is the path our solar system objects take as they pass through the sky. Negative numbers are below (to the south) and positive numbers are above (to the north celestial pole at Polaris) this line.

RA is the equivalent to our Earth maps' longitude and is measured in hours, minuets and seconds.

Each direction is measured in a 360 degree circle with RA hours broken down into hours (24 for the complete celestial circle), minuets (60 minuets = 1 hour) and then seconds (60 seconds = 1 minuet) and DEC is measured in degrees (360 for the full circle), with 60 minuets = 1 degree and 60 seconds = 1 degree.

It sounds complicated at first but once you get used to it, it's a simple plotting system that is necessary in locating objects using a star atlas and plotting an object's position in the night sky.

Think of the celestial coordinates system as marking the inside of a sphere (the whole night sky) that encompasses the Earth (our viewing point).

thanks for the reply but I kinda said that in my post.I want to know if I go outside with a set of coordinates such as RA/DE (J2000): 23H 37M 6.2s/ +2 03' 31.8" (the moon)where I get my reference point from to start measuring.I know from my location the meridian is at 0 degrees and 180 degrees.So facing SOUTH at around 5.44 am the above coordinates say that the moon is 2degrees 3minutes and 31 seconds ABOVE the equatorial line and as roughly every 15 degrees R/A is an hour.and the equatorial grid + hours to the left and - hours to the right that the moon is roughly just over half way between 0degrees SOUTH and the next 15degrees /1hour increment west of the meridian.These coordinates are only correct at this day and time obviously as we revolve in our orbit.So if I wanted to find the moon at a different time and it was 0 visibility (for example) what could I measure from to find it.I know 0 degrees is NORTH and POLARIS would be around 52/53 degrees alt which is my latitude,and pretty constant, visible or not.Polaris' coordinates say that it is 89 degrees above the equator but HOW DO I KNOW WHERE THE EQUATOR IS ?? .It also says polaris is RA 2H 31m..so using the same grids(eq j2000) would make its position around 45 degrees from N 0 degrees.Is this because the RA of polaris IS a constant..so from my location if I measured 52degrees up and 45 degrees (2h30m) from 90 degrees(E) then polaris will always be there?.However 45 degrees from 90 IS 45 degrees but 2H 30M from 90 degrees (E) is around 37 degrees,then youve got the hour angle(2H 47m) which seems closer.That gives a compass direction of around 47 degrees NE all RA coordinates are measured from a none constant that is the vernal equinox in as much as its position changes but its relationship to polaris does not.The discrepancies must come from the various measuring methods-j2000-az/alt-date-hour angle.So going by the simplest way use North 0 degrees as a direction...51 degrees (uk)DE (=89(j2000) up ,then that should be polaris but then what constant can I use to find or identify objects that im unfamiliar with.If RA is measured from0H 00M00S from the vernal equinox ,but that moves how can I use an RA coordinate without visual clues..az/alt is simple its angles both ways ,but the J2000 method uses a method calculated from a moving point..how can I find that point...???

[themos]

All good questions, it is very confusing at first.

When you go out, the first thing you want to know is where is North and South. If you can spot Polaris in the sky (I do it by following the Ursa Major stars that point to it) then you know where North is, roughly. So, South is opposite that.

Where is the equator? Point at Polaris and then imagine a plane perpendicular to your arm. Use your other arm to point at points on that plane. As you swing that other arm, you are pointing at the celestial equator. It's highest in the South and dips towards the horizon East and West. It's below the horizon in the North. If you are at Latitude 51 degrees North, then the equator gets as high as 90-51=39 degrees in the South. Declination is measured relatively to that plane.

For RA, you can open Stellarium, set the right time and place and click on a star very near the South. Read off its RA coordinate. That is known as the Local Sidereal Time, by the way. Say you get 18h 21m. Then your 23h 37m object will cross the meridian in 5 hours 16 minutes. You can convert that to degrees by multiplying by 15 which will be 79 degrees. Put one arm pointing South to the Equator and make an angle of 79 degrees (to the East) with the other arm, staying on the equatorial plane (keep both arms perpendicular to Polaris!). Now that you have the RA direction, move that arm towards Polaris (up, that is) for positive declination or away from Polaris (down) for negative declination. I hope that makes sense.