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    Re: LOP by Sextant Box Shadow
    From: Lu Abel
    Date: 2008 Jul 19, 15:38 -0700

    In a similar vein, once on a very clear, calm night I looked aft of my
    boat and saw a crystal-clear image of the moon on the water.   I
    wondered if I could use the image as one would in an artificial
    horizon.  I grabbed my sextant and tried it.  The result was pretty
    reasonable -- my sight was off by about 10 miles, which I attributed
    more to night vision challenges and/or my own shaky hand than
    difficulties with the concept.
    
    Lu Abel
    
    frankreed@HistoricalAtlas.net wrote:
    > Greg, you wrote:
    > "My sextant box was sitting before me on the dock casting a shadow.  Could
    > this shadow provide me with a LOP? The answer is yes. The inverse tangent of
    > the sextant box height divided by the length of the cast shadow generates an
    > hs. "
    >
    > The limiting factor in these sights is determining whether the surface is
    > level. It would be interesting to test some built surfaces (houses, docks,
    > parking lots, etc.) and find out what sort of typical scatter there is in
    > leveling. When I was growing up, the scatter for dock surfaces was five
    > degrees at least with some outliers at 20 degrees tilt, but they build 'em
    > better now. :-)
    >
    > And you wrote:
    > "I wasn't sure whether to treat this hs as an upper or lower limb so I just
    > did the reduction as an upper limb to see what would happen."
    >
    > You can figure out which limb is associated with the different parts of the
    > shadow by imagining what an ant would see. Imagine an ant behind the box
    > where the Sun is completely concealed. As it crawls out (in a direction away
    > from the Sun), it encounters a little "penumbral band" where the shadow
    > tansitions from fully dark to fully light. If the ant looks over his
    > shoulder just as he enters the penumbral band, he will see the Sun's upper
    > limb just appearing. When he is dead center in the penumbral band, he will
    > see the center of the Sun just clearing the top of the box. And as he
    > finally exists the penumbra, he sees the lower limb of the Sun just clearing
    > the top of the box. You can try this yourself with the shadow of a building.
    > The order is reversed, of course, when the shadow is being cast by an
    > overhanging eave.
    >
    > Shadow fringes and spots of light on the ground under trees contain some
    > interesting information. Since the Sun's angular diameter is nearly
    > constant, the width of the shadow fringe, or penumbral band, is related in a
    > simple way to the distance between the object casting the shadow and the
    > shadow itself. An angle of 32 minutes of arc is a ratio of 107:1. So if I
    > see a tall building casting a shadow with a penumbral band that is five feet
    > wide, then the portion of the building creating that portion of the shadow
    > would be about 535 feet away. Note that you have to measure the shadow width
    > in a direction that is perpendicular to the light rays from the Sun. If the
    > shadows are faint or confused, you can do this also by walking back and
    > forth. Find the spot where the Sun's limb first appears. Then walk until the
    > whole disk of the Sun is clearly visible. The distance between those two
    > places, multiplied by 107, gives the distance to the object in question.
    >
    > Similarly, if you're walking down a shaded sidewalk and you see circles of
    > light on the ground along your way, each of those circles is a simple image
    > of the Sun created by small gaps in the foliage above you. The gaps are not
    > circular. It's the Sun's circular disk that makes the circular patches of
    > light (and during a partial solar eclipse, you would find that the images
    > match the partially obscured Sun). As with the penumbral shadow fringes, the
    > distance to the gaps in the foliage creating the patches of light can be
    > determined by multiplying by 107. This is an easy way to get the height of a
    > tree. You find the end of the tree's shadow and then you look for the first
    > few circular sun images inside the main shadow --they're created by gaps in
    > the foliage near the very top of the tree. A little geometry then converts
    > that slant distance to height. I figure it's accurate to about +/-10%
    > without detailed measurements.
    >
    >  -FER
    > PS: works with the Moon, too.
    >
    >
    >
    > >
    >
    
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