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  1. #1
    Tony's Avatar
    Tony Guest

    Default Right ascension & declination



    I'm a geriatric who just enjoys looking at the night sky - when it
    isn't raining. I'm familiar with the major constellations and
    understand altitude and azimuth but cannot understand how the subject
    of this post can help me locate what I'm searching for. I understand
    what they are but how do I, at 47*N, 7*W, at 1945hrs on 07/01/04, (or
    any other time and location) locate (say) Pegasus at RA 23'22 and Dec
    25.1*?
    I can find it without any trouble using guide stars but would like,
    before I pop my clogs, to be able to do it more scientifically. I
    imagine that declination involves right angled triangles, my latitude
    and the declination of the star, but HOW? And as for RA I'm clueless.
    Given a few more years, some warm clothing and a lot of clear nights I
    might, using HPplanet planetarium, be able to arrive at the answer
    inductively, but the only thing I can guarantee is the clothes.
    So can some plain speaking astronomical wizard help out please?

  2. #2
    Martin Frey's Avatar
    Martin Frey Guest

    Default Right ascension & declination

    anything@tonybo.evesham.net (Tony) wrote:


    Not a wizard but...

    lean on a deck chair so that your entire body is in line with the pole
    star. Then treat RA and dec angles just as you would alt az angles.
    You'll need an RA reference point, just as you need North when using
    az. Pick a star you know, find its RA, and then sweep round from
    there, remembering that the RA horizon is at right angles to your
    body!

    seriously, I wouldn't bother unless you have an equatorial mount.

    --
    Martin Frey
    http://www.hadastro.org.uk
    N 51 02 E 0 47

  3. #3
    Mike Dworetsky's Avatar
    Mike Dworetsky Guest

    Default Right ascension & declination



    "Tony" <anything@tonybo.evesham.net> wrote in message
    news:e53886a7.0401071202.3f4efdb4@posting.google.c om...

    Rather than get into a complicated calculation for which you seem
    ill-prepared, the best way to proceed is to use a simple device called a
    planisphere. These can be purchased from various planetariums and/or their
    on-line shops. Basically, they are devices with a map of the constellations
    on a disk that pivots around the celestial pole, with a cut-out window that
    represents the horizon. By turning the disk to the correct date and time of
    night, you should see a map of the current constellations on view, with
    compass directions and angles of elevation reasonably well represented.
    (The time and date settings should be explained on the planisphere
    instructions.)

    Doing the calculations would involve use of geometrical formulae of
    spherical trigonometry. But the designer of the planisphere has already
    done this for you.

    Be sure you buy a planisphere designed for your latitude (+/- 5 degrees),
    e.g., buy a British one for the UK, a 35 degree one for the southern USA,
    etc. Otherwise there will be serious distortions.

    Chances are you can buy one over the internet.

    --
    Mike Dworetsky

    (Remove "pants" spamblock to send e-mail)



  4. #4
    Richard @'s Avatar
    Richard @ Guest

    Default Right ascension & declination

    Mike,

    You make mention of using a compass! Is a compass more accurate than using
    Polaris, i.e. magnetic north versus Polaris' north?

    Thanks,

    Richard


    "Mike Dworetsky" <platinum198@pants.btinternet.com> wrote in message
    news:btj519$atr$1@hercules.btinternet.com...
    their
    constellations
    that
    of



  5. #5
    Dr John Stockton's Avatar
    Dr John Stockton Guest

    Default Right ascension & declination

    JRS: In article <e53886a7.0401071202.3f4efdb4@posting.google.com >, seen
    in news:uk.sci.astronomy, Tony <anything@tonybo.evesham.net> posted at
    Wed, 7 Jan 2004 12:02:40 :-

    At 47 N 7 W you should be swimming, not skygazing!



    You understand, no doubt, latitude and longitude on the spherical Earth,
    measuring from the Equator and a perpendicular line through Greenwich.

    Now consider a large concentric starry sphere, with its own celestial
    equator, tilted 23* with respect to ours; on it RA 0000h..2400h
    correspond to longitude 0*..360* and Dec corresponds to latitude. Now,
    holding that fixed, rotate the Earth once in 23h56m. Now do the
    geometry.

    If you do not know how, it's probably not worth trying unaided; it's
    tricky but of no particular interest. For aid, seek a book such as
    mentioned in the newsgroup FAQ or by Peter Duffett-Smith.

    Planetarium / sky map software often includes, IIRC, a converter; I have
    sometimes thought of writing one in Javascript.

    Probably one really ought to allow for the Earth being ellipsoidal, if
    the answer is to be good to a minute of arc.

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

  6. #6
    Mike Dworetsky's Avatar
    Mike Dworetsky Guest

    Default Right ascension & declination



    "Richard @" <c@virgin.com> wrote in message
    news:btk5mp$7hls0$1@ID-105234.news.uni-berlin.de...

    No, I only said that the planisphere will give a reasonable representation
    of the night sky and that stars on the map will be more or less at the right
    compass bearings. You have to orientate the planisphere to the directions
    in order to use it. I'm assuming that the poster is able to determine N, S,
    E, W either by use of a magnetic compass, a road map, or by finding Polaris.
    Or by knowing which way the Sun is at noon GMT.

    time
    degrees),
    USA,

    --
    Mike Dworetsky

    (Remove "pants" spamblock to send e-mail)




  7. #7
    Dr John Stockton's Avatar
    Dr John Stockton Guest

    Default Right ascension & declination

    JRS: In article <btk5mp$7hls0$1@ID-105234.news.uni-berlin.de>, seen in
    news:uk.sci.astronomy, Richard @ <c@virgin.com> posted at Thu, 8 Jan
    2004 17:58:15 :-

    Please put responses after trimmed quotes, as the considerate regulars
    here do.

    A compass is undoubtedly more accurate when Polaris is behind cloud.

    Otherwise, it depends on the compass, the locality, and the user. If
    you can borrow the gyro- or fibre-optic- compass off a submarine,
    results should be good; but you probably mean a magnetic compass.

    One should be able to read a sufficiently large magnetic compass to a
    fraction of a degree, reproducibly. But one must then allow for
    regional magnetic variation (geography), local variation (ores,
    buildings, etc.), and immediate variation (steel belt buckle; nails in
    table; etc.). With a good grade hiker's, or a small boat's, magnetic
    compass, you might get to a degree.

    With Polaris, one should allow for it not being exactly at the current
    true pole; the Pole is apparently about 5% of the way from Polaris to
    Beta Ursa(e?) Minoris, after checking that figure, you should easily get
    an error no more than about a tenth of a degree, at a guess.

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

  8. #8
    Tony's Avatar
    Tony Guest

    Default Right ascension & declination

    Dr John Stockton <spam@merlyn.demon.co.uk> wrote in message news:<ciKb0YJ86Z$$EwtB@merlyn.demon.co.uk>...
    Equatorial mounts, planisperes, books, computer programmes? Why do so
    many people assume that being old equates with being helpless or
    stupid?

    I'm simply trying to puzzle out how to convert az/ra to dec/alt. I
    think I've cracked the latter:-Alt = 90 -(Dec - Lat) {I hope}.If the
    result is greater than 90, then subtract from 180 which gives a
    southerly altitude. But I can't check until there are some cloud free
    skies and, of course, that is only at transit - which I don't
    understand (yet)- so application will have to wait until I manage to
    clarify RA.
    When I solve it I shall do nothing with it except check that I'm
    right, in much the same way as I attempt to understand the workings of
    cars, washing machines, computers etc - and then employ someone else
    to fix them. It's called curiosity.
    Anyway, thank you all for your comments and suggestions.









  9. #9
    Russell Healey's Avatar
    Russell Healey Guest

    Default Right ascension & declination

    > I'm simply trying to puzzle out how to convert az/ra to dec/alt. I

    Tony

    Your formula above is correct for the maximum altitude of
    an object whose declination is "Dec" viewed from latitude "Lat"
    at transit i.e. when it is due south. Another way of writing this is

    Max Alt = 90 - |Dec-Lat|

    where |x| means the magnitude of x i.e. disregard the sign, merely look
    at the difference (a postitive number) between Dec and Lat

    The corresponding formula for minimum altitude (when due Norths) is
    Min Alt = Lat + Dec - 90


    These sorts of problems are very interesting. Thanks for asking the
    question.
    I realise I've only confirmed one aspect of it!

    --
    Russell Healey



  10. #10
    Stephen Tonkin's Avatar
    Stephen Tonkin Guest

    Default Right ascension & declination

    Tony <anything@tonybo.evesham.net> wrote:

    See equation near bottom of:
    http://astunit.com/tutorials/positional.htm

    Also, a few decades ago, before I got into computers, I knocked up
    "modern" astrolabes that would do the job for me, using card and
    transparent plastic. I could dig out the designs if you're interested.

    Best,
    Stephen

    Remove footfrommouth to reply

    --
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    + Stephen Tonkin | ATM Resources; Astro-Tutorials; Astro Books +
    + (N51.162 E0.995) | <http://astunit.com> +
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