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    Re: Latitude by Noon Sun for Beginners
    From: Jean-Philippe Planas
    Date: 2009 Apr 29, 02:15 -0700
    It does work for me and I have no subscription.

    --- On Wed, 4/29/09, Gary LaPook <glapook@pacbell.net> wrote:
    From: Gary LaPook <glapook@pacbell.net>
    Subject: [NavList 8097] Re: Latitude by Noon Sun for Beginners
    To: NavList@fer3.com
    Date: Wednesday, April 29, 2009, 3:30 AM

    I posted a link to the problem in Ocean Navigator. I subscribe to the magazine but i am just wondering if the link works for everybody or only for those with subscriptions.


    Gary LaPook wrote:
    Here is a link to the problem:


    The data for the problem is DR 59º 25' S 68º 25' W

    January 25, 2008 16:46:20 Z

    Hs 59º 09.3'

    Height of eye 8 feet

    Lower limb of the sun

    No index error or watch error.

    For a LAN latitude shot your DR plays no part, LAN is when it is, when
    the sun is highest in the sky. The GHA of the sun at the time given is
    68º 31.2' W so your DR longitude is off by 6.2'. If you were actually at
    your DR, LAN should have occurred only 25 seconds earlier so I don't
    know why you mention a 12 minute difference, are you looking at the 2008

    The time of the sight is not critical for determining latitude (within
    reason) since you only need it to determine the declination of the sun
    which changes slowly and would have been the same 25 seconds earlier as
    at the stated time which was 18º 31.2' S. (Depending on the accuracy
    needed for latitude you might be able to get by with just a calendar for
    time as they did for centuries, before the invention of chronometers and
    the development of the lunar distance procedure.)

    First you correct the sextant altitude, subtract dip and refraction and
    add semi diameter for the lower limb if using the Air Almanac, or
    subtract dip and use the combined correction table in the Nautical
    Almanac which combines refraction, parallax and semi diameter for the
    lower limb and you should get the answer given in the magazine, Ho= 53º
    22'. (Or, doing it in your head, the square root of 8 is almost 3,
    refraction between 33º and 63º is 1' and SD is 16' for a total
    correction of plus 12' giving an Ho of 53º 21'.)

    Subtract Ho from 90º to get zenith distance, the distance you are either
    north or south of the sun's GP. Instead of writing 90º is is easier to
    write 89º 60' (which is the same thing.) Subtracting Ho from 89º 60' you
    get a zenith distance of 36º 38' which is your distance south of the
    sun's GP since you are looking north at LAN. Since you must be closer to
    the pole than the sun you no you add the zenith distance to the
    declination of the sun and get a noon latitude of 55º 37.8' S, the same
    as in the magazine.

    The magazine goes on to plot an EP which does take into account the DR.
    To method used to find an EP is to plot a perpendicular from the DR to
    the LOP. Since in this case the LOP is a latitude line the perpendicular
    goes straight south from the DR to intersect the latitude LOP at the DR

    So which longitude is correct, the DR longitude or the longitude
    corresponding to the sun's GHA at the time of LAN given as 68º 31.2' W
    at 16:46:20 Z?

    If you have been following the discussions you should have noticed that
    due to the difficulty of accurately determining the exact second when
    the sun "hangs" that there is some uncertainty in that time which then
    causes uncertainty in a longitude based on the time of observed LAN so
    it is taken that the DR longitude is more accurate which is why the EP
    is plotted on the latitude line at the DR longitude.

    Another way to do these types of sights is to treat them exactly like
    any other sight, use a sight reduction table with the LHA of zero. Doing
    it this way you see why the exact time is not so important when
    determining latitude since LHA equals zero for a four minute period, you
    simply move the AP east or west but the Az stays the same, 360º in this

    Doing it this way you find an Hc of 53º 59.8' for an AP of 55º 00' S,
    68º 31.2' W, resulting in an intercept of 37.8 NM Away from an AZ of
    360º resulting is an LOP running east and west at latitude 55º 37.8' S,
    the same answer as before.

    Theoretically the LAN sight would produce a slightly more accurate
    latitude due to the slight change in altitude during the plus and minus
    two minute period when LHA = zero although nobody worries about this
    when doing normal sights on other bodies. To see how big a difference
    this might make I added two minutes to the time of the sight and used
    the same AP and the Hc came out to be 53º 59.6' a difference of only
    0.2', a difference small enough to be lost in the "noise" of celestial


    JKP@obec.com wrote:
    I've been following with great interest, if with less than complete comprehension, the recent threads on finding position at and around noon.

    I'd like to divert a tiny new rivulet from that stream-- a back-to-basics thread on noon sights for latitude. I don't like those titles for how-to books that end in "...for Dummies," so I'm using the kinder term "...for Beginners."

    I've been working the latest Nav Problem in Ocean Navigator magazine, and I'm finding I can't get the latitude the editor does from the noon sight data presented. My main stumbling block seems to be that the time given for the shot is not the time I consider Local Apparent Noon for the DR longitude used. Wouldn't that be essential? If the sun is observed at 12 minutes or more after LAN, how does one account for that difference?

    My main source of instruction is Dutton's (12th edition), which I find usually gives excellent and thorough explanations of procedures. The article on latitude by noon sun focuses on how to determine LAN and what to do with the Ho obtained then. I've found nothing so far about accounting for a time difference and the fact that the sun is not at its zenith when observed before or after LAN. Much is made of combining the vessel's change of position east or west with that of the sun to pinpoint LAN, but the magazine problem provides no data about course or speed.

    I'll bet one you can tell me somnething that will make me slap my forehead and exclaim "Of course!" Fire away!

    Thanks in advance,
    John P.


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