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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 |
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R.A and Dec question
"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 |
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R.A and Dec question
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 |
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R.A and Dec question
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 |
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R.A and Dec question
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 |
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R.A and Dec question
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 | |
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R.A and Dec question
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 | |
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R.A and Dec question
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. |
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