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    Re: Lunar trouble, need help
    From: George Huxtable
    Date: 2008 Jun 24, 10:01 +0100

    Kent Nordstrom's lunar calculation is now starting to converge with my own.
    However, I find much to disagree with in the following paragraph-
    "My refraction values are calculated as Tan app. Altitude x 60,53� for
    standard atmospheric conditions (760 mm Hg and +10 degr. C). Then a
    correction for actual T and B is made for altitudes from 20 degrees and
    upwards . For lower altitudes another way of correction for B and T is done.
    You could always argue about the relevance of this approach, we all know
    that refraction is difficult to model, in particular on low levels. Anyway,
    with the approach taken my results coincide with refraction values taken
    from tables quite well (Tables from 1845, based on Janet Taylor, Henry Raper
    I assume)."
    Where on Earth does "Tan app. Altitude x 60,53� " come from? For one thing,
    it's quite the wrong way up, showing refraction increasing with altitude
    rather than the other way round, and I suspect that must be the result of an
    accidental slip in his email text, rather than in his calculation.. It would
    be more correct (but for high altitudes only) if the tan were changed to
    cot, or (which amounts to the same thing) altitude were changed to
    zenith-angle, and if, at the same time, the constant was changed from 60.53"
    to 58.293" (see Meeus, also Smart).
    Then that amended formula would provide reasonable predictions, for
    altitudes greater than 20�. But it breaks, down, and badly, at lower
    altitudes, and then a more complicated expression is required. These errors
    have NOTHING to do with any corrections for temperature and barometer, for
    which refraction can ALWAYS be modelled, very accurately, as being simply
    proportional to air density at the observer.
    No easy analytic function exists for correcting refraction over the whole
    range of altitudes, but a good empirical approach by Bennett (see Meeus)
    gives, for refraction in arc-minutes at apparent altitude h0 degrees -
    R = cot (h0 + 7.31 / (h0 + 4.4))
    which I like to tinker with by changing it to-
    R = cot ( h0 -.000861 h0 + 7.31 / (h0 +4.4)), which makes refraction zero at
    90�, as it has to be.
    I've no doubt that the high-altitude refraction figures that Kent actually
    used, in correcting that lunar, were a reasonable shot at reality, and would
    not much upset his lunar distance. It's his text-explanation that I find
    He invokes the names of Taylor and Raper, both of whom provide tables of
    refraction, but neither explains the details of their basis, and I am pretty
    sure they are NOT calculated according to Kent's precepts.
    Kent wrote-
    "I agree with George that I should have used 06-20-13 instead of 06-20-44
    referring the moon�s altitude to the mean time of LD�s. This change has a
    very small effect on my LD."
    Yes, we agree about that, but now, does his recalcalculated Moon mean
    observation-time agree with mine?
    George wrote �But next, the bit that really puzzles me about Kent's
    treatment is how he
    manages to invoke sidereal time. That would come in only if he was getting
    his astronomical positions from an Astronomer's almanac that gave
    sky-positions in terms of right-ascension, rather than hour-angle�.
    And Kent replied-
    "My approach for calculating the MT is as I understand the very same as used
    in the old days.
    This is of course for finding the difference between the GMT and the MT,
    i.e. the longitude. It is a separate calculation. So let me outline what I
      1.. Find the LHA at the predicted GMT (we know (roughly) the latitide,
    from the generation of true distances at say time 3, time 6 etc. we can
    predict the declination at the GMT. True altitude is from the altitude
    reduction. Aries GHA to the GMT can be predicted as well).
      2.. Find the Aries GHA for upper meridian passage in Greenwich, (normally
    one day before).
      3.. Convert this GHA to time.
      4.. Correct for assumed longitude and then you get the time for Aries UMP
    on your location.
      5.. Find the difference between Aries GHA and the body�s GHA at GMT. This
    is the RA.
      6.. Find the sum of the LHA and the RA, which is the sideral time.
      7.. Subtract the value for Aries UMP at the location. Now you have the
    difference in sideral time.
      8.. Convert this difference in sideral time and you get the MT at
    And now you can easliy find the difference between the GMT and the MT. I
    hope this explaination clarifies. I should also add that one must be very
    observant when doing this calculation, it is easy to do wrong."
    Well, I have struggled a bit with that, withouf fully understanding what
    Kent is doing, which may well be my problem rather than his. It seems a very
    roundabout procedure. It might help me if Kent would kindly provide the
    actual numbers he has worked with in this case, including the almanac
    predictions for Sun and Moon (and Aries?) positions, and from which almanac 
    they were
    contact George Huxtable at george@huxtable.u-net.com
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    Navigation List archive: www.fer3.com/arc
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