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    Re: Sundog
    From: Brad Morris
    Date: 2013 Feb 16, 15:41 -0500

    Hi Greg

    To answer my own question, yes, of course we can use our sextant as a sundial to find the APPARENT time.  I will use Martelli's method, my copy is published in 1923.

    Since the mechanics of Martelli's method have been well investigated on list, I will just jump to the solution(s). 
    If we assume that 5 deg is the true, corrected altitude, then the apparent time is 4hrs 44min 25sec PM. 
    If the 5 deg is observed, I accept Bill's corrected altitude of 4 deg 47 min, yielding an apparent time of 4hrs 42min 30sec PM.

    Skipping to HO71 Azimuths of the Sun, 1934, I will interpolate between the declination entries of 14 deg and 15 deg to obtain time entries for 14 deg 4 min. 
    4:40 112d 2.933m
    4:50  112d 44.866m
    Interpolating between those values for the apparent time(s) found above yields
    4:44:25PM  112d 21.45m
    4:42:30PM   112d 13.41m

    Subtract from 360 to find the true azimuth(s) of the sun
    4:44:25PM  247d 38.55m
    4:42:30PM  247d 46.19m

    Add the static 22 deg sundog correction to find the direction(s) you are facing. 
    4:44:25PM, true alt 5d, facing 269 d 38.55 m
    4:42:30PM, true alt 4d 47m, facing 269 d 46.19 m

    Naturally, all of the fine points can be debated.  The corrections for refraction can change the result.  What was the pressure and temperature?  Are we experiencing anomalous dip? The same can be argued for height of eye and, since the picture shows dip short waters, the dip correction.   These will have modest affect on the result.

    Regards
    Brad


    On Feb 16, 2013 2:55 PM, "John H" <apacherunner@gmail.com> wrote:

    Two "nudgey-pedantic" details.   22 degrees is the minimum angle of bending of light from a hexagonal ice crystal.   It can certainly be larger, much larger.   It turns out that for random orientations of an ice crystal, you get most of the "action" around 22 degrees.   That's because, as you vary the angle of incidence of the sunlight (or moonlight), there's a very broad minimum for refraction, and 22 degrees is the minimum value, so you get most of the light concentrated there, and a sharp cut-off.

    Second - the horizontal formation is due to the "plate-like" shape of the ice crystals.   As they settle out of the atmosphere, they tend to 'flatten' - try dropping a piece of paper and you'll see what I'm talking about - the long flat side aligns horizontally.   Sundogs and moondogs formed when the sun or moon are close to the horizon have this effect, BUT, when they're high in the sky, the circles around the sun or moon are pretty much circular and symmetric.

    In any case, the 22 degree number and assuming it's flat for a setting sun is pretty good - I couldn't resist being the pedant. 

    On Sat, Feb 16, 2013 at 2:11 PM, Bill B <billyrem42---net> wrote:

    On 2/16/2013 12:54 PM, Greg Rudzinski wrote:
    > After closer inspection of the image ......large snip....
    > Is there a UT time for the image ?
    
    Excuse my OCD. To solve a problem we first need to understand the
    problem. Frank stated:
    Lat 41d (Franks approx. location 41.5d N, 71.4d W)
    Hilly country ("Hilly" is relative. A hill in KS... ;-)
    YOU are facing the sundog
    YOUR sundog approx 5d high (using what as a horizon?)
    Sun blocked by hill to left
    Date February 10, (2013 assumed but not stated)
    
    Of importance Frank did not specify this is the sundog he saw, but
    rather the sundog YOU are facing. I think Frank's sundog was closer to
    41.5d N.
    
    To me this means MY sundog could be at any longitude I specify near
    sunset. Where I place myself determines time and therefore declination.
    For simplicity of calculations MY sundog was at 41d N and 75d W. (a bit
    north of Philadelphia, PA.) 41d N and not too far from Frank, and hilly.
    (A look at a topo map of Frank's immediate area does not suggest it is a
    natural for development of a downhill ski resort.)
    
    Once you set YOUR longitude and determine approx time (and declination)
    when observed CENTER of sun would be 5d above the horizon and adjust for
    dip, refraction and temp/pressure just plug 41d, adjusted HO/HC and
    declination into the sine formula for azimuth. Add 22d and voila.
    
    That's my take on the king's new cloths ;-)
    
    Bill B
    
    
    
    On 2/16/2013 12:54 PM, Greg Rudzinski wrote:
    > After closer inspection of the image ......large snip....
    > Is there a UT time for the image ?
    
    
    
    
    
    

    View and reply to this message: http://fer3.com/arc/m2.aspx?i=122386


    View and reply to this message: http://fer3.com/arc/m2.aspx?i=122387

       
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