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    Re: Bowditch Table 15
    From: Trevor Kenchington
    Date: 2005 Jan 24, 18:30 -0400

    Nels,
    
    Firstly, the web page isn't mine. That is Jim Thompson's response to my
    question. (I know ... too many Atlantic Canadians on one mailing list!)
    
    The idea that Table 15 gives distance from the object to the observer's
    horizon, so that the total distance from object to observer needs
    addition of the observer's horizon distance from Table 12, is certainly
    one possibility. As you have shown, the numbers in Table 15 vary in the
    appropriate directions. But are they the right magnitudes?
    
    You wrote:
    
    > Here's a thought experiment: for a fixed height of eye, as you stand
    > on the beach and watch a tall ship sail toward you (your height of eye
    > is 10 feet, the ship 100 feet tall, so H-h=90).  When the angle is 0
    > (ship just in sight), you know that the ship must be
    > 1.17(sqrt(10)+sqrt(100))=15.4 miles off, from Table 13.  Enter table
    > 15 with angle=0 and H-h=90 and read 11.1, add your Dh=1.17sqrt(10)=3.7
    > to get 14.8.  Close to Table 13, but maybe not close enough.
    
    
    Maybe not.
    
    But you are not quite using Table 15 properly. It requires that the
    observer corrects for dip. With a height of eye of 10ft, the almanac's
    standard dip table gives -3.1'. So the corrected angle used to enter
    Table 15 should be -3.1'. The table then gives a distance of 15.4M (when
    interpolated for the 0.1'). That, of course, is the ship's true distance
    (as found using either Table 12 or Table 13). Yet we know that Table 15
    doesn't work as a table of total distances.
    
    Add in the observer's horizon distance, 3.7M, and the total distance to
    the ship is 19.1M, which would clearly be far beyond the range at which
    it could be seen. So that doesn't work either.
    
    > Here's another thought experiment:  start at 30 feet up the mast so
    > that H-h=70 and measure, say, 0 10', indicating that the ship is 3.5
    > miles beyond your 6.4 NM horizon.   Total distance =9.9 NM.  Now you
    > climb to 70 feet, and H-h=30.  Your horizon is now 9.8NM, the ship is
    > still 9.9NM off, so the angle you measure must be larger, and in fact
    > must be greater than 1 50', since the ship is only 0.1 mile beyond
    > your horizon.
    
    
    Again, you need to make allowance for dip. You also have an error in
    reading the tables: For H-h=30, the minimum tabulated distance is 0.5M
    with an angle of 35', so I don't think you can say "greater than 1 50'".
    
    But, beyond that, it is fine to say that the measured angle increases
    from 10' to 110' as the observer climbs from 30 to 70 feet. That is
    clearly the correct direction of change. But it is also clearly _not_
    the right magnitude. By my calculation, on a Flat Earth, a ship with
    100ft masts 9.9 miles away would subtend an angle of only 5.7'. Even if
    I have got that wrong, there is no way that it could subtend 110', which
    is nearly 4 times the diameter of the Full Moon as seen by the same
    observer.
    
    Even the 10' angle when seen from 30ft height of eye, with 9.9M total
    distance, is greater than the angle would be at that range if the ship's
    waterline was visible.
    
    
    Sadly, I think we can conclude that Table 15 does NOT show the distance
    from the observed object to the observer's horizon
    
    
    
    > There are some holes left in my understanding.
    >
    > Here's one problem:  When I try computing the formula for a few
    > numbers, I get very different results than are shown in Table 15.
    >
    > Degrees   H-h   D from Formula D from Table 15
    > 0              30    40.8                 6.4
    > 0              70    95.25               9.7
    > 0.5           30    40.8                 2.8
    >
    > -0 4'         30     50.4                12.8
    >
    > Here is an interesting point about the formula:  as presented in the
    > 2002 Bowditch, the angle only matters if it is negative.  When \alpha
    > is positive, \tan(\alpha) is positive.  In the first term, squaring
    > and then taking the square root merely takes the absolute value (i.e.,
    > throws away the sign), so if \alpha is in the range 0-90 degrees
    > (where I think it will always be in practice), the first term cancels
    > with the third term, and we're left with the second term:
    > (H-h)/constant.  Either I've completely misunderstood things, or there
    > is a series of long standing blunders in Bowditch.  I'm betting on the
    > former.
    
    
    Jim's web page cites George Huxtable as saying that the second term of
    the Table 15 equation should be included in the square root bracket. I
    am not sure when George posted that information but it might explain why
    your results are so different.
    
    
    
    Maybe George can enlighten us.
    
    
    
    Trevor Kenchington
    
    
    
    --
    Trevor J. Kenchington PhD                         Gadus{at}iStar.ca
    Gadus Associates,                                 Office(902) 889-9250
    R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
    Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555
    
                         Science Serving the Fisheries
                          http://home.istar.ca/~gadus
    
    
    

       
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