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    Re: Northing correction to Noon longitudes.
    From: Henry Halboth
    Date: 2005 Jun 7, 14:17 -0400

    It all seem unnecessarily complicated. Within the appropriate time range,
    why not calculate a series of ex-meridian Latitudes for specific times
    and advance each to the time wanted, as the hydrographical surveyors did,
    and have the thing over with. I believe that Frank said at one time that
    he had a copy of Wharton + Fields book on hydrographical surveying, in
    which, if my recollection is correct, the matter is dealth with quite
    On Tue, 7 Jun 2005 15:09:29 +0100 George Huxtable
    > On 4 June, in "Latitude and Longitude by "Noon Sun"", Frank Reed
    > explained
    > his method for correcting "around noon" observations for longitude
    > for the
    > North-South component of vessel's speed, after explaining how to
    > obtain
    > that component, including declination change.
    > "If you're moving towards the
    > Sun,  then for every six minutes away from noon, add 0.1 minutes of
    > arc for
    > every knot  of speed to the altitudes before noon and subtract 0.1
    > minutes
    > of arc
    > for every  knot of speed to the altitudes after noon."
    > So, if there were 13 observations plotted, each of these (perhaps
    > only 12
    > of them) must be individally adjusted, by taking the time interval
    > in
    > minutes between each point and some (arbitrary?) time-zero, dividing
    > by 6,
    > and multiplying by 0.1 x the speed in knots, adding or subtracting
    > the
    > result from the altitude, and replotting a new point. It doesn't
    > sound like
    > a trivial operation to do 12 times over, does it?
    > Instead, I suggested that the original altitude data points be left
    > uncorrected, to provide (using Frank's folding-paper method) the
    > moment of
    > maximum altitude, on which-
    > "The moment of LAN is  delayed by 15.3 (tan lat - tan dec) * v where
    > v is the
    > Southerly component of  the speed in knots."
    > To which Frank responded, on 6 June,
    > "That adds needless complication to an otherwise  extremely simple
    > procedure."
    > Well, does it indeed? It appears to be a great simplification, to my
    > mind.
    > Perhaps list members will judge for themselves.
    > This correction doesn't need to be made very precisely because it's
    > only a
    > small one, but it certainly must be made. For the purpose, both Sun
    > dec and
    > ship's lat will be changing rather slowly. The simple trig
    > expression 15.3
    > (tan lat - tan dec) can readily be precalculated (and changes only
    > slowly
    > from one day to the next). It just needs multiplying by Northing
    > speed to
    > provide a result which is the time-difference in seconds between
    > maximum
    > altitude and meridian passage. So: no fiddling with the original
    > graph,
    > just one simple multiplication, followed by one time correction.
    > Which is
    > simplest?
    > By the way, I failed to mention, in previous postings, what should
    > be
    > rather obvious; that in the above expression both lat and dec should
    > be
    > taken as positive North, negative South.
    > ============================
    > Then I asked-
    > "The whole object of the exercise is to discover the  moment of
    > noon. So how
    > does the observer know how many minutes each plotted  point is away
    > from
    > noon in order to calculate that adjustment?"
    > I went on to answer my own question, but that part wasn't quoted-
    > "The answer is, I think, that it just doesn't matter, as far as
    > finding the
    > new centre-of-symmetry is concerned. For the purpose of making those
    > corrections, any point could arbitrarily be presumed to be the
    > moment-of-noon, and then the new centre-of-symmetry would show true
    > noon,
    > when the Sun was on the meridian."
    > Frank's answer was-
    > >It  makes no difference. Whatever point in time is picked as the
    > "zero"
    > >point, where  no adjustment for northing/southing is made, will be
    > the
    > >time of the
    > >fix. Being  able to label the fix as "noon" is not terribly
    > important but it
    > >is nicely  traditional. The real time on it, of course, is a moment
    > of  GMT.
    > It may be my fault, but I don't understand what Frank is saying
    > here. The
    > time that results from the paper-folding operation of the corrected
    > graph
    > is the moment of noon, surely, when the Sun crosses the meridian,
    > and what
    > we need to know to get the long is the chronometer reading of GMT at
    > that
    > moment (after equation of time is chucked in). I don't understand
    > how some
    > arbitrarily chosen moment, at which the corrections to altitude are
    > taken
    > to be zero, can be the "time of the fix", whatever that means. So I
    > suggest
    > that my own answer, above, to my question is the correct one.
    > Nevertheless,
    > we seem to agree that choosing a different zero-point for the
    > corrections
    > will not shift the timing of the corrected peak, which depends on
    > the slope
    > of the corrections, but not their amount.
    > Then I went on to-
    > >"However, it looks to me as if an error in
    > >that initial  presumption of noon would give rise to an error in
    > the deduced
    > >maximum  altitude, and so in the latitude. Perhaps Frank will
    > comment."
    > Frank did, as follows-
    > "Nope. No  error. See above."
    > However, I urge Frank to rethink his flippant dismissal of the point
    > that I
    > have made. What's needed, to calculate latitude simply, is the Sun's
    > altitude AT MERIDIAN PASSAGE, and not at any other time. To obtain
    > that,
    > Frank tells us to take the altitude from the peak value of the
    > corrected
    > Sun-altitude curve, at his "folding" point, which will be at
    > meridian
    > passage. But that's not the observed altitude, it's the corrected
    > altitude,
    > at meridian passage. The correction that's been made to observed
    > altitude,
    > at that moment, depends on how far it is away in time from the
    > zero-point
    > of his corrections, and that zero-moment was chosen quite
    > arbitrarily. Only
    > if the zero-point of the corrections happened to be at the moment of
    > meridian passage, would the peak of the corrected-altitude curve
    > correspond
    > to the observed altitude at that moment.
    > So I suggest that Frank's proposed method should be somewhat
    > modified. Yes,
    > certainly, use the corrected-altitude curve to determine, from its
    > symmetry, the moment of meridian passage. But then, read off,
    > corresponding
    > to that moment of meridian passage, the UNCORRECTED value of
    > altitude,
    > which will NOT in general be its peak value.
    > ====================
    > Finally, there's a curious comment, as follows-
    > "By the way, perhaps George could consider addressing people in the
    > second
    > person. Thanks in advance."
    > Can any list member, perhaps Frank himself, kindly explain what he
    > is on
    > about here? Otherwise, that request is completely lost on me.
    > George.
    > ================================================================
    > contact George Huxtable by email at george---.u-net.com, by
    > phone at
    > 01865 820222 (from outside UK, +44 1865 820222), or by mail at 1
    > Sandy
    > Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    > ================================================================

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