Need help locating stars using right ascension and declination

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  • Member since
    December, 2008
Need help locating stars using right ascension and declination
Posted by floaty on Monday, December 15, 2008 10:07 PM

I've read several places about how the 0h 0m line of RA is marked by the vernal equinox and stuff, but I can't seem to wrap my brain around this concept.

 I mean, if I were to step outside with the coordinates of a star, how would I go about locating that star?

How do you determine where in the sky is 0h 0m RA?  I know it's when the sun passes the vernal equinox or whatever, but where is that in the sky?

 And then how would you interpret the declination?  Again, I know what it is.  I know it's the celestial equivalent to lines of latitude on the earth.  But what I can't understand is how you would apply these concepts to actually locate a star in the sky if you had their right ascension and declination.

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Posted by chipdatajeffB on Monday, December 15, 2008 11:45 PM

If you Google the terms "vernal equinox" or "autumnal equinox" you will find Web pages that describe this far better than I can, and they'll include illustrations that will help.

But to:

floaty

I mean, if I were to step outside with the coordinates of a star, how would I go about locating that star?

let me offer a shortcut: Get yourself a starchart, or a book like Norton's Star Atlas. You can find these at reasonable prices online. Alternatively, you can download a free planetarium program like Celestia or Stellarium, which you can use to print or display your own charts.

Look for the charts in your book which show the ecliptic (path of the Sun, Moon, and planets through the sky) and the Celestial Equator. On those charts, the Celestial Equator will be marked with the hours, minutes, and seconds of Right Ascension along the borders of the charts. In fact, most charts show the RA markers along the horizontal borders of the charts and the DEC markers along the vertical edges.

Alternatively, if you have a planisphere, rotate it so that it lines up Noon on the inner circle with March 21 on the outer circle. You'll see a line bisecting the word Noon and running up through the star chart to the rivet on which the chart turns in the elliptical window. That line marks 0:00:00 RA very nearly. You can work out the positions using RA from that point.

Once you know the names of some of the stars in the sky (and a planisphere is a great tool to help with that), you can look up their RA and remember one bright one visible at night per season. Then, using only those four stars, you can find your way around the sky fairly easily.

It also just so happens that if you stretch out your arm to the sky and then spread your pinkie finger and forefinger as far apart as you can, the measure of the sky so marked will be about 15 degrees, or one hour of arc (360 degrees, divided by 24 hours in a day, = 15 degrees of sky per hour).

So, to estimate an RA of three hours west of one of the stars you've memorized, you just estimate three times the span of the distance between your pinkie and forefinger "projected against the sky".

I think you'll find it far easier just to look these positions up in a star chart.

Hope that helps!

The universe is not only stranger than we imagine, it's stranger than we CAN imagine. --- JBS Haldane

Come visit me at Comanche Springs Astronomy Campus (we're on Google Maps) in Texas.

www.3rf.org

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  • Member since
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  • From: Texas
Posted by chipdatajeffB on Monday, December 15, 2008 11:53 PM

Click here for a simplified explanation and illustrations of the Celestial Sphere and locating objects using equatorial coordinates.

Click here for some good illustrations that should help (page down to Celestial Sphere).

The universe is not only stranger than we imagine, it's stranger than we CAN imagine. --- JBS Haldane

Come visit me at Comanche Springs Astronomy Campus (we're on Google Maps) in Texas.

www.3rf.org

  • Member since
    November, 2006
  • From: illinois, usa
Posted by saberscorpx on Tuesday, December 16, 2008 2:33 AM

Welcome floaty-

Until it becomes innate by repetition (anymore it takes a conscious effort for me not to see the labeled radec grid superimposed on the sky), use a star atlas to find the coordinates' constellation and a planisphere to determine that constellation's visibility. Right ascension and declination are the sky's longitude and latitude, so think of the constellations as countries and stars as cities.               Another option is to pass on the right ascension/declination coordinates in favor of a target's altaz position (altitude: horizon = 0 deg, zenith = 90 deg. azimuth: compass point along the horizon; north = 0 deg, east = 90 deg, south = 180 deg, west = 270 deg). Astro software programs often provide this info automatically. Here's the conversion math http://www.stargazing.net/kepler/altaz.html and an online utility: http://measurementsconverter.com/celestial_horizon_coordinates_calculator.html  

Basic altaz (altitude/azimuth) navigation: Everyone is equipped with close to a 10 degree fist at arm's length. 4 fists stacked vertically from the horizon gets us close to 40 deg altitude. Facing north, 5 horizontal fists to the right along the horizon takes us to 50 deg azimuth (northeast). Stick out your thumb for another 5 deg. 

 

Saber Does The Stars at

http://www.saberdoesthestars.wordpress.com

 

  • Member since
    September, 2007
Posted by Oriontaage on Thursday, January 22, 2009 12:55 PM

Hi,

I just uploaded a webpage on celestial motion to my website that tries to explain all of this at a beginner level. Perhaps this will help.

 http://www.astronomyasylum.com/celestial.html

The other comments/suggestions you have received look pretty good and I agree-try to find yourself a simple star wheel or chart and use this to learn your way around the night sky.

Bill

Homebuilt 10 inch. Ultralight Newtonian and Homebuilt GoTo GEM 

Celestron C8 and C90 (Classic Orange Tube Models)

Celestar 8 Mount with GPUSB Interface

Homebuilt 3.3 m Diameter Domed Observatory

Canon T3 DSLR, SPC900NC Webcam, and DSI Pro ccd Camera 

Visit my Observatory and Telescope Building Webpages at http://www.astronomyasylum.com

  • Member since
    March, 2009
Posted by MARTIN JR on Saturday, March 21, 2009 6:00 AM

You can enter dates as far back as Julian day 0, January 1, −4712 and as far into the future as you wish. Note that astronomers and historians use different conventions for years before A.D. 1. In history books, the year that preceded A.D. 1 is called 1 B.C., zero not having come into use in European culture at the time. Astronomers denote the year before A.D. 1 as “year 0”. Thus when an astronomer talks about an eclipse having occurred in the year −412, that's the year historians refer to as “413 B.C.”. In converting historical dates to Julian days, Your Sky assumes the canonical date for the adoption of the Gregorian calendar, Friday, October 15th, 1582. Many countries shifted to the Gregorian calendar much later; in Great Britain, not until 1752. When investigating events in history, make sure you express all dates after October 15th, 1582 in the Gregorian calendar.

 

  • Member since
    April, 2010
Posted by Peezerole on Tuesday, April 20, 2010 4:10 PM

 So often when people think they're are beginning to understand RA and Declination they tend to have one question that they think the answer to will make everything clear (The answer certainly made it clear to me).

Instead of answering this question lots people tend to rant on about how they aren't very good at explaining and then they give people web addresses to sites where the person has probably already been.

 The question that they ask is "where is the point 0,0 in the equatorial system - where is the point of the vernal equinox . . .?"

The point 0,0 is currently( I say currently because it does change over long periods. ) just below the head of pisces and you can see it on your planispher. Alternatively highlight some of the stars in the head of pisces in planitarium software to see that they are near to 0,0 .

I then looked for what the point 6hrs,0 was close to and 12 and 18 and this gave me a feel or the system. This really helped me when I first read about the equatorial system and if it is wrong then I am living under a mishapprehension and I will be corrected.

Peter
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Posted by chipdatajeffB on Tuesday, April 20, 2010 5:09 PM

Peezerole

 So often ... they ask ... "where is the point 0,0 in the equatorial system - where is the point of the vernal equinox . . .?"

The point 0,0 is currently( I say currently because it does change over long periods. ) just below the head of pisces and you can see it on your planispher. Alternatively highlight some of the stars in the head of pisces in planitarium software to see that they are near to 0,0 .

Yes, but that's not what the OP here asked ...

Another problem we have here is that (especially for folks who are new to the Forum) we most often don't know what they already know, so offering additional Web sites as resources is generally helpful.

In your case, there's no misapprehension: 0 in RA and 0 in Dec equate to a point, whereas the "zero hour" is an arc through the sky overhead (a great circle that runs through all the positions of Declination).

The universe is not only stranger than we imagine, it's stranger than we CAN imagine. --- JBS Haldane

Come visit me at Comanche Springs Astronomy Campus (we're on Google Maps) in Texas.

www.3rf.org

  • Member since
    April, 2010
Posted by Peezerole on Wednesday, April 21, 2010 12:50 PM

 In the original question the member "Floaty" asks as part of their question . . . "How do you determine where in the sky is 0h 0m RA?  I know it's when the sun passes the vernal equinox or whatever, but where is that in the sky?"

As I say once you know this position the whole system becomes a walk-in-the-park.The ancients put the stars onto a huge sphere that rotated around the earth and we, with all our modern knowledge, do the same. The start point 0,0 on this sphere is currently just below the head of Pisces, go up from this point and we have positive numbers up to +90 degress, go down and we have negative numbers to -90 degrees. Go west of this point and the hours increase from 0hrs up, go east of this point and the hours decrease from 24hrs down.

6hrs,0 is currently near to Betelgeuse

12hrs,0 is currently near the left arm of Virgo

18hrs,0 is currently between Ophiuchus and Serpens Cauda.

Peter
  • Member since
    August, 2009
Posted by TinRinnie on Wednesday, October 27, 2010 3:57 PM

Peter, this is EXACTLY the info I need(ed). THANK YOU.

I can't tell you how many times I gave up trying to understand RA & DEC. Sure, there was no problem (after a while) understanding all the lines drawn on the celestial sphere, but it didn't help me when I walked out my front door at night and looked up. I really despaired of ever understanding this. I just wanted to be able to go outside and be reasonably sure that I good see the celestial sphere in my mind's eye. If I could see Polaris, why not the rest?

When I started doing trig to figure this out, I knew I'd gone astray. I had gotten to the point where I realized that I can see only certain stars during certain times of the year, so, I reasoned, I should only be able to see certain hours of RA during that same time. You filled in the missing piece!!!

Thanks again Smile

 

 

  • Member since
    April, 2010
Posted by Peezerole on Thursday, October 28, 2010 10:00 AM

Normal 0 false false false EN-GB X-NONE X-NONE

I bought a Celestron SkyScout and with this you have the power to point to the 0,0 point wherever it is, day or night. This reinforces the fact that just because you can’t see it doesn’t mean it ain’t there . . . the celestial sphere doesn’t disappear in the day time and at any given time the observable sky represents one quarter of it (I think that’s right . . . it sounds about right).

Peter

Peter

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