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George wrote:

> Bill is worrying away at this question of the shape of the altitude curve of
> the Sun, around noon, and the effect on it of the speed and direction of the
> vessel.
>
> Let me take, first, his final question, which was-
>
> "What about east/west displacement if trying to determine longitude from
> LAN?", and will assume, for that purpose, that the ship has no motion in the
> North-South direction, but is travelling East-West only, and the Sun's
> decllination is unchanging, at a solstice..

A scholarly reply, and thought provoking as always. Of course the
declination change of my "sea monkey" example referred to equinoxes  where
declination changes relatively rapidly, as opposed to your solstice example
where change is too small to matter given the practical limits of cel nav.
It was also a tongue-in-cheek attempt at humor.

Much water has passed over the damn since my first draft of the following,
but before we get to the details and the dance, a summary:

1. Regarding my "sea monkey" post, I stand by my analysis. Let me be more
specific, as the post was glib. Given a fixed position at equinox where the
observer is at the location on earth where declination is 23.44d at LAN, the
Sun will not rise at exactly 90d, nor will it set at exactly 270d.  The day
length, (as measured from sunrise to sunset, upper limbs on horizon) will
not be exactly 12 hours solar time. I was proposing azimuth and day length.
Whether there is a point about which the elevation vs. time curve is
symmetrical; be it called LAN, noon, or peak altitude was and is superfluous
for that post.

2. I understand that given a true east or west COG and constant velocity,
the curve of Sun altitude vs. time will be altered. Vessel course west, the
curve will be "flatter;" vessel course east the curve will be steeper with a
more pronounced peak.  Maximum elevation will depend on declination at the
time and location.

3. I now see your point about the north or south motion of the vessel
shifting the point of symmetry away from LAN (LAN defined as the vessel and
Sun being on the same meridian), and that indeed the curve will be
symmetrical about a peak shifted away from LAN on the time axis.

From my simplistic vantage point, (vessel Lat is north of declination/same
therefore Sun south of vessel, and vessel moving south--or other variations
on the theme) there is a time when the Sun would "stall out" vertically at
LAN to an observer at a fixed position. The vessel's motion would keep
elevation climbing to an onboard observer until the rate of climb induced by
the vessel cancelled out the rate of fall after LAN (to an observer at a
fixed position).

That was a learning experience, as I had not read or considered that shift,
so thank you George and gentlemen for again slightly decreasing my sphere of
ignorance. (One might argue that at this stage of life, I am forgetting what
I knew faster than I am learning, so the sphere of ignorance is increasing.
That is probably a safe bet.)

4. Regarding the finer points that NavList gurus may wish to explore/debate:
If expertise of the subject were to be exchanged for weapons, I would be the
guy armed with a rubber knife trying to defend myself during an ICBM
exchange ;-)
================================================
To give you an overview of my initial frame of reference, the first time I
saw a sextant outside of print in Chapman was a television series on
sailing, one part on cel nav.  A woman on a coastline with a sextant doing
LAN observations to determine Lat and lon.  Calibrate the sextant (IC?). Ten
minutes (or so?) before predicted LAN take an elevation reading of the Sun
and record that angle and the time. Wait for the Sun to reach its peak where
it stands still (vertically) for a few moments and record that elevation and
time.  Starting a few minutes earlier than the difference between the first
observation and peak altitude, recalibrate the sextant and set it to the
initial pre-LAN elevation (with any adjustments due to re-calibration). Note
the time when the Sun returns to the initial elevation.  Divide the
difference between the first and last times of equal elevation and you have
the precise moment of LAN, from which you can derive longitude (as well as
latitude from the peak elevation). No mention of dip, refraction, temp/BP
corrections, NA etc. Just sweet and simple.  That segment is what made
learning cel nav a long-term goal for me.  I can do that!

As I later understood it, finding latitude was not a major problem, just a
matter of UT and elevation when the Sun hits its peak--time allowing the
observer to pick up declination from the NA. A bit less accurate before the
chronometer?

One of my first self-assignments for using my trusty cardboard sextant (on
the water) was creating equations to predict the time of LAN based on the
vessels course and speed--after calculating the time of LAN at my current
AP. I have two equations (one for the eastern component and one for the
western component of vessel motion).  The north and south components of the
course were only of interest as course was combined with distance for
polar-to-rectangular coordinate conversions to obtain the east/west change
in distance--which was then converted to longitudinal arc.

In a perfect world, I could predict the exact second of LAN on the run. The
reality is the equations seem to work, but are limited by input.  If the
starting AP/fix is off, or if COG and SOG are not spot on, the best I can
hope for is an estimated time of LAN.  Which is all I wanted. Now I learn of
a shift in time for the peak due to north-south motion is also a factor in
observed LAN. Oy vey.

Noting this was a cardboard sextant, I had no illusions (delusions) of
picking off longitude.  But now I am past the cardboard sextant days and you
are past the Ebco days.

Taking the east-west motion curve argument to the extreme, let's imagine the
following hypothetical. No change in declination, the earth is a perfect
sphere, no EQ of time, and we ignore little wobbles here and there etc.  We
are working only on Sun time, our chronometers or digital watches are in our
pockets.

First scenario:  We are on board an aircraft (Kant) moving west at the same
angular velocity as the Sun. We pass over the prime meridian exactly at LAN
and circumnavigation the earth at the same latitude and altitude. 24 hours
later we arrive back at the prime meridian. It has been local noon the
entire trip, so no "sun" time has passed as we (passengers or pilots)
perceive it.  We pull the digital watches out of our collective pockets and
the calendar indicates one day has passed (as we would have noted if we
watched UT change during the trip). Of note, our "curve" would be a straight
line.

Second scenario:  We are on board an aircraft (Kafka) moving east at the
same angular velocity as the Sun is moving west. We pass over the prime
meridian exactly at LAN and circle the earth at the same latitude and
altitude. 24 hours later we arrive back at the prime meridian. Two
"sun-time" days have passed as we (passengers or pilots) perceive it. Pull
the digital watches out of our pockets and the calendar indicates one day
has passed (as we would have noted if we watched UT change during the trip.)
Our "curve" would be bimodal.

 > ... which allows an
> observer to take either the altitude at LAN, or the maximum altitude,
> without bothering much about the difference.

At this point I was totally lost, perhaps semantics I thought.  We agree
that "The moment of LAN (Local Apparent Noon) is when the Sun and the ship
are on the same meridian." It *was* my understanding that instant is also
the moment when the Sun is directly north or south, and at its peak
elevation.  How can we use peak altitude *or* LAN?  I understand the
difference between peak and LAN with north or south motion now.

> But not so the difference in the timing of that central value, to anyone
> trying to use Frank's proposed method to find his longitude. The correction
> that has to be made for the ship's south-going speed of 10 knots is all of 5
> minutes of time, or 1.25 degrees of longitude, which has to accommodate the
> speed of the ship, any tidal current and any declination changes in the Sun
> position. With modern shipping commonly travelling at over 25 knots, it's
> obvious what an enormous correction this must be, and how precisely it would
> have to be made.
>
> And yet Frank is claiming that he can derive the central moment of that
> altitude curve, and then make that correction, to provide an overall error
> in the whole process of no more than 5 miles in longitude, which corresponds
> to 35 seconds of time. No wonder he is reluctant to disclose the details.
>
> The moment of LAN (Local Apparent Noon) is when the Sun and the ship are on
> the same meridian. The Sun will always be crossing meridians, going
> Westward, at almost exactly 15 degrees per hour. The ship will be crossing
> meridians travelling much more slowly, at a few tens of minutes in an hour,
> perhaps in the same direction as the Sun, perhaps the opposite direction.
> The moment when they cross the same meridian will depend on that speed, so
> to calculate that moment you must take the speed into account, but the
> maximum height of the Sun will occur at that moment and will not depend on
> the speed of the vessel.  The only effect you would see, if you examined the
> curve of altitude against time more closely, was that it would be very
> slightly peakier, changing slightly faster with time if the vessel was
> travelling Eastward than vice versa. However, it would still have the same
> maximum value, at the same moment, of LAN. There would be no surprises.
>
> North-South travel, around noon, has a very different effect, the details of
> which Bill questions. Judging by his words, he finds it hard to accept that
> it can result in a symmetrical curve of altitude against time: symmetrical,
> not about noon, but about a moment displaced, earlier or later, in time.
> It's because of the way two quite different changes combine together.

You misjudge the intent of my question(s), although I may well have stated
my question(s) very poorly.

> The speed, Northward or Southward, of a vessel, has a very similar but
> greater effect than the changing declination of the Sun (which can be up to
> 1 knot at the equinoxes) So just let's consider a vessel moving Southward at
> say 10 knots, at Winter solstice, at a position of say 56 deg North. And
> let's make an enormous supposition, that at the moment of LAN, we could stop
> the rotation of the Earth. The Sun would then really "hang " in the sky,
> absolutely stationary. But to our observer on the Ship, not quite
> stationary. He would still be travelling Southward, toward the Sun, at 10
> arcminutes per hour, so to him the Sun would be slowly increasing in
> altitude, at that rate.
>
> Now put that picture to one side, and set the Earth rotating again. And now,
> take the vessel to be stationary. Now our observer sees a familiar curve:
> the Sun climbs, reaches a peak, then falls, quite symmetrically about LAN.
> Actually, near the peak, that's as close as you like to a parabolic curve,
> where the drop, from the peak value, changes with the square of the time
> from the peak.
>
> And now, in real life, we have to combine those two motions together, the
> symmetrical parabola with the steadily changing altitude, increasing at 10
> minutes an hour. And I suspect that the only way to convince Bill will be to
> persuade him to take a bit of graph paper and combine them for himself.

I am very visual, so love graphing.   Below was where I had severe
problems comprehending.
>
> As an example, let's take the ship approaching the Clyde in Winter, that I
> tried to get Frank to consider as a noon-longitude exercise (but he ducked).
> The noon Sun (neglecting corrections) would be at 10 deg 30' altitude.
> Before and after that moment, if the ship is stationary, it will be somewhat
> less. Here are some numbers.
>
> First column, time in minutes before and after LAN.
> Second column, Sun altitude with stationary observer.
> Third column, change in altitude due to observer's motion.
> Fourth column, combination of the two.
>
> time (min)     alt1     shift  sum
> LAN-20      10d 23.1    -3.3   10d 19.7
> LAN-15      10d 26.2    -2.5   10d 23.7
> LAN-10      10d 28.3    -1.7   10d 26.6
> LAN-5       10d 29.6    -0.8   10d 28.8
> LAN         10d 30.0     0.0   10d 30.0
> LAN+5       10d 29.6    +0.8   10d 30.4
> LAN+10      10d 28.3    +1.7   10d 30.0
> LAN+15      10d 26.2    +2.5   10d 28.7
> LAN+20      10d 23.1    +3.3   10d 26.4

> Anyone who cares to plot out the fourth column will find it to be just
> another parabola, perfectly symmetrical in itself, but with its centre of
> symmetry displaced to be about 5 minutes later than LAN. It also has a
> slightly higher peak value but not very significantly so,...

Addressing your four-column table, I had trouble putting it in context and
creating a comprehensive mental image, especially as it seemed to refer to a
post(s) I do not recall reading. I had trouble grasping the argument you
were advancing.

While I appreciate the Clyde problem you posted both for the prose and word
pictures, it seemed to me that N 56 during the winter solstice would produce
a pretty flat curve.  For my edification I used June 20, 2008, 23:00:00 UT.
Lat N 40, W 164d 33!9 lon for the static position and a starting lon for the
vessel in motion,.  Vessel motion south at 12 kt.

I ran my scenario from 25 minutes before and after LAN at 5 minute
intervals. 5 minutes pre and post LAN I used 1 minute intervals.

At first blush it appears that calculated latitude (no instrument or
observer error) will only be shifted by 0d 00!2 towards north (plot away).
Time of peak altitude vs. LAN is late by *nominally* 1.5 minutes. That
equates to 23 nm, or 0d 29' lon at N 40.  That's a ton to me, especially as
it is a theoretical and does not include other potential errors.

George, NavList 5130

> But much of this is old ground, which has been well-trodden on this list
> before, and I doubt whether any minds are going to be changed.

References to the "old ground" please. Thread name plus month a help.

Thanks

Bill


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