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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
fully

On Tue, 7 Jun 2005 15:09:29 +0100 George Huxtable
 writes:
> 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@huxtable.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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