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    Re: Help with Letcher's Method Example
    From: George Huxtable
    Date: 2002 Mar 4, 20:42 +0000

    Chuck Griffiths asked-
    >Can anyone help me work through the solutions for P and R in George Huxtable's
    >"About Lunars, Part 3" under the Letcher's Method? I'm having trouble with both
    >equations and it would be helpful if someone could provide the solutions worked
    >through with intermediate values. I imagine it's an order of operations problem
    >I'm having in both cases.
    >Chuck Griffiths
    George Huxtable replies-
    Fine, let's go through it, in some detail for Chuck and anyone else that
    has had problems.
    First, I will copy out all the stuff about Letcher's method that appeared
    in "About Lunars, part 3", so nobody needs to look that up again.
    In "Self-contained celestial navigation using H.O.208" (1977), John S
    Letcher describes such an approximate method for clearing the lunar
    distance. The way he proposes for its use would be suitable only for lunar
    distances up to 90 degrees, whereas any lunar observer would certainly
    require its application to angles up to 120 degrees. I will therefore give
    a modified calculation method that will cover the full range of useful
    sextant angles.
    I am not sure where this method comes from, as Letcher does not quote any
    reference, nor does he show how it is derived from first-principles. It
    might be his own invention, perhaps. To me, it seems very clever. I will
    christen it "Letcher's method" until we find out more about its provenance.
    If anyone recognises it from another publication, I would be interested to
    Let's investigate Letcher's Method.
    We will assume that all sextant observations have been corrected for index
    As before, it starts with the observed lunar distance d, after correction
    for semidiameters. This is to be corrected by adding, separately,-
    a) The correction P, for the combined parallaxes of the Moon and the Sun,
    which may be a positive or negative amount, never greater than about 60
    b) The correction R, for the refractions of Moon and Sun, always positive,
    never more than 11 min.
    Both P and R need to be calculated to within 0.1 min.
    These two corrections require knowledge of-
    d, the observed lunar distance between centres (i.e. corrected for
    m, the observed altitude of the Moon's centre above the true horizon (i.e.
    corrected for dip and semidiameter)
    s, the observed altitude of the centre of the Sun (or other body), above
    the true horizon (i.e. corrected for dip and, if necessary, semidiameter).
    HP, the Moon's horizontal parallax, in minutes of arc, at the hour of
    The HP of the Sun (or other body) is ignored, thus losing a bit of precision.
    It is no longer necessary (as it was in the other methods) to make parallax
    and refraction corrections to m and s individually, as M and S are no
    longer needed. These corrections are taken into account automatically, as
    part of the clearance procedure. That is partly why the method is so much
    Let's go through it.
    First, obtain B = (cos d sin m - sin s) / sin d  (B is just an intermediate
    angle, used in the computation).
    Then the parallax correction in minutes is-
    P = HP*B + (HP)^2*[(cos m)^2 - B^2)]/ (6900*tan d)
    P may be a positive or negative number, in minutes, to be added or
    subtracted from d.
    The second term of P is usually a very small quantity, but still needs to
    be evaluated.
    Now for the refraction term R. This, very cleverly, includes its own
    built-in calculation of the way refraction changes with altitude.
    R = .95* (sin s/sin m + sin m/sin s - 2*cos d) / sin d ,in minutes of arc.
    So the end result is D = d + P + R , giving the Corrected Lunar Distance.
    Let's compare this method with the others, using Steven Wepster's Atlantic
    observations of 2001 Apr 02 once again.
    The inputs we need are, as before-
    d = 107.382 deg    (observed lunar distance between centres)
    m = 049.875 deg   (observed moon-centre altitude above true horizontal)
    s = 021.173 deg   (obs Sun or body-centre altitude above true horizontal)
    HP = 59.4 min.
    By pocket calculator we get-
    B= -0.6178
    so P = -36.7 - 0.1 = -36.8 min,   and R = 3.2 min
    As d (in minutes) is 107deg 22.9, then for the Corrected Lunar Distance,
    D= d + P + R, or 107deg 22.9 - 36.8 + 3.2
    Therefore D = 106deg 49.3 by Letcher's method.
    This should be compared with 106 deg 49.2, by Borda's method using tables,
    and 106deg 49.3 using Young's method with a calculator. Of these, I would
    choose the last, 106deg 49.3, as being the more precise. Really, what
    better agreement could anyone wish for, between three different methods?
    However, don't expect to always obtain that same precision using Letcher's
    method. He concedes that by ignoring some of the contributions such as the
    Sun's parallax and the ellipsoidal figure-of-the-Earth, errors may combine
    up to a total of 0.3 minutes. Even so, when compared with the likely errors
    inherent in measuring the lunar distance, due to the motion of a small
    craft, a possible error of 0.3 minutes in the clearing of it may be very
    acceptable. My vote would go to Letcher. But it's your choice...
    My thanks to Bill Murdoch for alerting me to this useful Letcher
    publication. If you can find a secondhand copy, I think it would be a
    worthwhile buy. It was written in the days when scientific calculators
    existed but were uncommon; hence the emphasis on using tables.
    Now to work through that example, but this time in excruciating detail, to
    show each little step.
    we have, as before,-
    d = 107.382 deg    (observed lunar distance between centres)
    m = 049.875 deg   (observed moon-centre altitude above true horizontal)
    s = 021.173 deg   (obs Sun or body-centre altitude above true horizontal)
    HP = 59.4 min.
    so cos d = -.2987 (remember cos d = -cos (180-d) = -cos 72.618 = -.2987)
       sin d =  .9543 (remember sin d =  sin (180-d) =  sin 72.618 =  .9543
       tan d =-3.195  (remember tan d = -tan (180-d) = -tan 72.618 =-3.195
       sin m =  .7646
       sin s =  .3612
    First thing we need to work out is the parallax correction P, and the fist
    step in doing that is to work out the value of B. Above, I described B as
    an angle, but it's not an angle, really, just a number.
    Get B from B = (cos d sin m - sin s) / sin d
    so cos d sin m =  -.2284
    cosd sin m - sin s = -.2284 -.3612 = -.5896
    B = (cos d sin m- sin s)/ sin d  = -.5896 / .9543  = -.6178
    Now we need to calculate P from-
    P = HP*B + (HP)^2*[(cos m)^2 - B^2)]/ (6900*tan d)
    First work out the term in square brackets, [(cos m)^2 - B^2] where the
    symbol "^2" means "to the power of 2" or more simply "squared".
    cos m = cos 49.875 = .6445, so (cos m) squared = .4153
    and we have just calculated B as -.6178 so B squared is .3817 (it loses its
    negative sign on being squared, of course)
    So the term in the square brackets is .4153 - .3817 = .0336
    Multiply this by HP squared, which is 3528, to give 118.5
    Divide it by 6900 to give .01717
    Now we need to divide that by tan d, or -3.195
    so .01717 / -3.195 gives -.00537. This is the second part of P, which now
    needs to be added to HP*B, or 59.4 * -.6178, or -36.7 minutes
    (-36.70) + (-.00537) = -36.71 minutes of arc, which we can call 36.7.
    There's a difference of 0.1 minutes between this recalculation of P and the
    value of -36.8 that was obtained last time, which probably suffered from a
    rounding error in the second term.
    Now for R., for which we need
    R = .95* (sin s/sin m + sin m/sin s - 2*cos d) / sin d ,in minutes of arc
    sin s/sin m is .3612 / .7646, or .4604, and sin m/sin s is .7646/.3612, or
    summing these terms gives 2.5772. now subtract 2cosd or 2*(-.2987), or (-.5974)
    This gives 3.1746.
    Muliply by .95 to get 3.0159. Divide by sin d which is .9543.
    3.0159 / .9543 = 3.16 minutes which we can round to 3.2 minutes to give R
    So the result for D = d + P + R
    d in decimal degrees was 107.382 which in degrees and minutes we can express as
    d = 107 deg 22.9 min
    so d+P+R is 107 deg 22.9 min +  (-36.7 min) + 3.2 min, or 106 deg 49.4 min
    This is the end-result, within a whisker of what we found before.
    Clearly this has given some trouble to Chuck Griffiths and maybe others,
    and I would be interested to learn just where they came unstuck.
    George Huxtable.
    George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    Tel. 01865 820222 or (int.) +44 1865 820222.

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