R.A and Dec question

Could someone please point me to a good tutorial for working out R.A and
Dec? For example I saw in a magazine article that M4 can be found at R.A
16h24m Dec -26. 32 How do I go outside and point my telescope at R.A 16h24m
Dec -26. 32 ?

Any help very much appreciated!!

Steve

"Steven Gill" wrote in message
...
Could someone please point me to a good tutorial for working out R.A and
Dec? For example I saw in a magazine article that M4 can be found at R.A
16h24m Dec -26. 32 How do I go outside and point my telescope at R.A
16h24m Dec -26. 32 ?

Any help very much appreciated!!

Steve
Without knowing what your scope is, the answer will be based on the
mathematics of how to calculate alt/az, from ra/dec. However if your scope
is mounted on an 'equatorial' mount (has a tilted axis, which points to
the celestial pole), then many of these have a simple calculator 'built
in' to the calibrations on the axes, which allow this to be done
relatively easily.
The Dec, is the angle above/below the celestial equator. From the UK,
'-26', is going to be very low in the sky. The Ra, is an 'angle', but done
just like a clock (in hours/minutes etc.), round the sky from the 'vernal
equinox' (the point where the Sun appears to be on approx Mar 21). The
easiest way will be a quick browse online, where there are dozens of
coordinate converter programs, which will give you an Ra/Dec, from Alt/Az.
A quick 'guess' calculation, by me, gives this as being south, around
10PM, and only just over 10 degrees above the southern horizon. Probably
impossible to see at this time.

Best Wishes

Steven Gill wrote:
Could someone please point me to a good tutorial for working out R.A
and Dec? For example I saw in a magazine article that M4 can be found
at R.A 16h24m Dec -26. 32 How do I go outside and point my telescope
at R.A 16h24m Dec -26. 32 ?

Any help very much appreciated!!

Steve

Hi,

http://liftoff.msfc.nasa.gov/academy...rse/RADEC.HTML
http://en.wikipedia.org/wiki/Right_ascension

R.A stands for "right ascension" and Dec stands for "declination". It is a
coordinate system which for all practical purposes stays fixed relative to
the stars, and therefore appears to be rotating around the earth once per
(sidereal) day. Of course, this is because the earth rotates around its own
pole-to-pole axis in the opposite direction. So if you want to point your
telescope to a particular point measured in RA and DEC, you must also take
the observation time into consideration in some way.

There are several ways this can be done in practice, depending on your
equipment. But often it means you must

a) Assuming you have an equatorial mount, you should polar align your mount
http://arnholm.org/astro/polar_alignment/ .

b) Find a bright star, near the relevant position, on a star chart. Find
that star in the sky and point the telescope to it. Adjust the setting
circles on the mount to show the coordinates of that star.

c) Move the telescope until the setting circles show the positions you want,
and M4 should be in your field of view.

Of course, this can be done using GOTO or other methods. I use a method
similar to the above, but instead of using setting circles, I have written a
PC program to do the GOTO job.

Clear skies
Carsten A. Arnholm
http://arnholm.org/
N59.776 E10.457

Many thanks for your help lads.
I am using a Meade ETX125 by the way


"Steven Gill" wrote in message
...
Could someone please point me to a good tutorial for working out R.A and
Dec? For example I saw in a magazine article that M4 can be found at R.A
16h24m Dec -26. 32 How do I go outside and point my telescope at R.A
16h24m Dec -26. 32 ?

Any help very much appreciated!!

Steve

Thank you for that wonderful explanation Carsten.

Now here is the justification in graphic terms -

http://www.pfm.howard.edu/astronomy/...S/AACHCIR0.JPG

Can you please explain to the original poster why you believe in a
system where the Earth covers a greater distance in its orbital
circumference the further from the Sun it is.Just put the sidereal
justification in an elliptical framework to affirm this unsightly
spectacle.

The actual relationship between clocks,terrestial geometry and
astronomy is complicated and goes back to remote antiquity.The most
important and the most historically visible complimentary addition
was the application of the Equation of Time principles of the 24 hour
day to axial rotation at 15 degrees per hour.The point being that to
justify an Ra/Dec system astronomically is fatal to astronomical
principles and indeed all else.






Carsten A. Arnholm wrote:
Steven Gill wrote:
Could someone please point me to a good tutorial for working out R.A
and Dec? For example I saw in a magazine article that M4 can be found
at R.A 16h24m Dec -26. 32 How do I go outside and point my telescope
at R.A 16h24m Dec -26. 32 ?

Any help very much appreciated!!

Steve

Hi,

http://liftoff.msfc.nasa.gov/academy...rse/RADEC.HTML
http://en.wikipedia.org/wiki/Right_ascension

R.A stands for "right ascension" and Dec stands for "declination". It is a
coordinate system which for all practical purposes stays fixed relative to
the stars, and therefore appears to be rotating around the earth once per
(sidereal) day. Of course, this is because the earth rotates around its own
pole-to-pole axis in the opposite direction. So if you want to point your
telescope to a particular point measured in RA and DEC, you must also take
the observation time into consideration in some way.

There are several ways this can be done in practice, depending on your
equipment. But often it means you must

a) Assuming you have an equatorial mount, you should polar align your mount
http://arnholm.org/astro/polar_alignment/ .

b) Find a bright star, near the relevant position, on a star chart. Find
that star in the sky and point the telescope to it. Adjust the setting
circles on the mount to show the coordinates of that star.

c) Move the telescope until the setting circles show the positions you want,
and M4 should be in your field of view.

Of course, this can be done using GOTO or other methods. I use a method
similar to the above, but instead of using setting circles, I have written a
PC program to do the GOTO job.

Clear skies
Carsten A. Arnholm
http://arnholm.org/
N59.776 E10.457

On or about 2006-06-29,
Steven Gill illuminated us with:
Many thanks for your help lads.
I am using a Meade ETX125 by the way

Then you can get it to display the current RA & Dec position on the
controller. I can't recall the key to press at the mo', but it is in the
manual. Hold the cancel key down for 2 seconds or somesuch? Once you've
aligned it properly of course. You may even be able to tell it to go to a
specific RA/Dec location, but I've not tried that with my ETX105.

--
Mark
Real email address |
is mark at | Common Sense is very Uncommon.
ayliffe dot org |

Cheers Mark


"Mark Ayliffe" wrote in message
...
On or about 2006-06-29,
Steven Gill illuminated us with:
Many thanks for your help lads.
I am using a Meade ETX125 by the way

Then you can get it to display the current RA & Dec position on the
controller. I can't recall the key to press at the mo', but it is in the
manual. Hold the cancel key down for 2 seconds or somesuch? Once you've
aligned it properly of course. You may even be able to tell it to go to a
specific RA/Dec location, but I've not tried that with my ETX105.

--
Mark
Real email address |
is mark at | Common Sense is very
Uncommon.
ayliffe dot org |

JRS: In article , dated Wed, 28 Jun
2006 16:05:05 remote, seen in news:uk.sci.astronomy, Steven Gill
posted :
Could someone please point me to a good tutorial for working out R.A and
Dec? For example I saw in a magazine article that M4 can be found at R.A
16h24m Dec -26. 32 How do I go outside and point my telescope at R.A 16h24m
Dec -26. 32 ?


Through the door? IMHO the first move should be to look it up in an
atlas to find out the approximate location - and for example if it turns
out to be in Orion I'd not bother looking tonight.

Actually, it's in Scorpius, a degree or two away from Antares and
currently low in the SSW after dark, so you'll need a clear horizon. It
won't get much, if any, higher tonight, and will set about 1 am. But if
you're in Scotland ...

--
© John Stockton, Surrey, UK. Turnpike v4.00 MIME. ©
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.