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    Re: Impossible lunar example
    From: George Huxtable
    Date: 2010 Sep 1, 22:22 +0100

    It may be worth clearing this matter up.
    
    If Douglas had been following this thread, he would realise that the test
    for impossibility, that nobody has so far questioned, and applies to both
    plane triangles and spherical triangles, apparent lunar triangles, and
    cleared lunar triangles, is as follows-
    
    No side of any such triangle can be longer than the sum of the other two.
    
    If Douglas tries sketching out a few triangles it will become obvious to
    him that this must be the case.
    
    And the question is not about whether the trig process  "works", when
    starting with an "impossible" apparent triangle, to produce some sort of
    result. It does. The question is whether that result, of another
    "impossible" triangle, has any physical meaning.
    
    George.
    
    contact George Huxtable, at  george@hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ----- Original Message -----
    From: "Douglas Denny" 
    To: 
    Sent: Wednesday, September 01, 2010 9:56 PM
    Subject: [NavList] Re: Impossible lunar example
    
    
    
    I admit to not following this thread so might be talking out of turn: but
    do not understand why you call this triangle impossible?  It is not an
    impossible spherical triangle at all.
    
    The limiting case for a spherical triangle is where one side is 180 degrees
    in which case it occupies a hemisphere.
    
    So:
    each side or angle of a spherical triangle is < 180
    the sum of the three sides is between 0 and 360
    the sum of the three angles is between 180 and 540
    the area of of any sph. triangle must be less than 2.pi.R^2
    
    By co-incidence I have just recently programmed my HP50 for "clearing"
    Lunar distances with the standard formula which I used as a basis from
    Cotter's 'History of Nautical Astronomy' p.209
    
    cos D = sin S sin M + (cos d - sin s sin m )(cos S cos M)/(cos s cos m)
    
    D=lunar distance; S = true altitude star;  M = true altitude of Moon; s =
    apparent alt star; m = apparent altitude Moon; d = apparent lunar distance.
    
    Putting your figures into my calculator for clearing a lunar distance (it
    includes refraction correction) gives:
    
    For the example of the "impossible" triangle
    LD = 103
    ZDmoon = 71
    ZDsun = 19
    
    Cleared distance = 103.061 degrees.
    
    Douglas Denny.
    Chichester. England.
    
    ======================
    
    Original Message:-
    
    One way to explain why clearing the lunar distance on the impossible
    triangle does not fail and proceeds as, George put it, "without meeting
    along the way a sin or cos greater than 1, or a square root of -1" is that
    it is operationally equivalent to clearing the lunar distance on a
    particular "possible" triangle.
    
    To see this, consider clearing the lunar distance using the cosine formula
    of spherical trigonometry in the form
    cos(LD) = cos(ZDmoon)*cos(ZDsun) + sin(ZDmoon)*sin(ZDsun)*cos(Z)
    LD = lunar distance
    ZDmoon = Moon's zenithal distance
    ZDsun = Sun's zenithal distance
    Z = Sun and Moon azimuth difference
    
    For the example of the impossible triangle
    LD = 103
    ZDmoon = 71
    ZDsun = 19
    
    Note however that
    
    cos(ZDmoon)*cos(ZDsun) = cos(180 - ZDmoon)*cos(180 - ZDsun)
    sin(ZDmoon)*sin(ZDsun) = sin(180 - ZDmoon)*sin(180 - ZDsun)
    
    And hence clearing the lunar distance on the impossible triangle is
    operationally equivalent to clearing the lunar distance on a triangle with
    sides
    LD = 103
    ZDmoon = 180 - 71 = 109
    ZDsun = 180 - 19 = 161
    which is an entirely "possible" triangle. Of course with ZD's > 90 this is
    not one that would arise in practice but the operations performed in the
    course of lunar distance clearing are all trigonometrically valid and
    geometrically meaningful even though a positive correction (refraction +
    dip etc.) applied to the ZD's of the impossible triangle corresponds to a
    negative correction applied to the sides of its "possible" cousin,
    
    Robin Stuart
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