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    Re: LOP by Sextant Box Shadow
    From: Frank Reed
    Date: 2008 Jul 19, 18:14 -0400

    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.
    PS: works with the Moon, too.
    Navigation List archive: www.fer3.com/arc
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