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Frank Reed wrote-

>Next you need to correct for your speed towards or away from the Sun. For
>example, if we're sailing south and the Sun is to the south of us, then each
>altitude that we have measured will be a little higher as we get closer to the
>latitude where the Sun is straight up. We need to 'back out' this effect so
>that  the data can be used to get a fix at a specific point and time. This
>isn't
>hard.  First, we need the fraction of our speed that is in the north-south
>direction.  If I'm sailing SW at 10 knots, then the portion southbound (in the
>Sun's  direction) is about 7.1 knots. You can get this fraction by simple
>plotting or  an easy calculation. Next we need the Sun's speed. The
>position where
>the Sun is  straight overhead is moving north in  spring, stops around June
>21,  then heads south in fall, bottoming out around December 21 (season names
>are  northern hemisphere biased here). It is sufficient for the purposes
>of this
> method to say that the Sun's speed is 1 knot northbound in late winter
>through  mid spring, 1 knot southbound from late summer through mid
>autumn, and 0
>for a month or two around both solstices (it's easy to prepare a monthly table
> if you want a little more accuracy). Add these speeds up to find out how
>much  you're moving towards or away from the Sun. 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. Reverse the rules if
>you're moving  away from the Sun. Spelled out verbally like this, this speed
>correction can  sound tedious but the concept is really very simple and
>it's very
>easy to do.  [Incidentally, George Huxtable deserves credit for emphasizing
>the importance of  dealing with this issue (although I don't think he ever
>spelled out how to do  it)]

===========================

To start with the last bit. This question has come up before. Yes, I have
posted an explanation of how that job should be done. Not though, by the
tedious method that Frank proposes. The only advantage of Frank's method (a
significant advantage none the less) is that it's clear what's going on, as
the observer corrects each point on his altitude graph so that it becomes
the same as an observer at constant latitude would observe.

I can refer to a reply sent in response to a posting forwarded by Doug
Royer, on 8 Jan 04, threadname "Re: [NAV-L] Fw: Lat. and Lon at LAN."

This went as follows (I have since corrected an error which appeared in the
original posting and was put right soon after)

"2. Time difference between moment of maximum altitude and LAN on a moving ship.

This second problem affects Doug's big-ships more than it does the small
vessels that most of us sail. I don't know the speed of Doug's ships, but
let's choose 20 knots as a round number. Let's say that he is steaming
South at 20 knots, toward the Sun, from 51deg N, at the exact LAN of 11 h
57m 53s GMT. A stationary observer with a sextant, in a raft that he
passes, will indeed see the Sun "hanging" with a constant maximum altitude
at that time. On his bridge, however, Doug WON'T see the Sun's altitude as
stationary, he will see it as increasing at 20' per hour, because he is
steaming toward the Sun at 20 knots, or 20' per hour. To him, the Sun won't
appear to "hang" in the sky until somewhat later, when it's real altitude
(to the man in the raft) is falling by 20' per hour, and just matches the
rise that's caused by his own speed. So his LAN will appear to be too late.
By how much?

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. If lat = +51deg and dec = -
23.5deg, this works out at 511 sec, or all of 8.5 minutes late. Unless
corrected for, this will give rise to an error (not an uncertainty this
time, but an actual error) of 127.5' in the longitude!"

=================

The method described above is an alternative to Frank's proposed method of
correcting each altitude after it's been plotted on the graph. Instead, the
points are plotted uncorrected, exactly as observed, and the centre of
symmetry of the pattern is found, to determine the time of maximum
amplitude. Then a correction is made, of the time difference between
maximum altitude and meridian passage. If the vessel is travelling toward
the Sun, as in the example above, then that delays the moment of maximum
altitude, so in the example the time of meridian passage (for longitude) is
obtained by subtracting from the moment-of-symmetry an amount of 511
seconds of time. Correspondingly, if the vessel was moving away from the
Sun, (if the Sun's direction was aft of abeam) that same amount would have
to be added.

It's worthwhile estimating in advance what this correction is going to be,
and in which direction, because what the observer is looking to measure is
the centre-of-symmetry of his plot of observed altitude. His observation
period ought to be 20 to 30 minutes before and after that moment of maximum
altitude.

In the note above I didn't mention the effect of the changing declination
of the Sun. It's Northward motion, in Spring, or Southward, in Autumn, of
up to a knot should really be accounted for when assessing how fast the
vessel is approaching the Sun, or receding from it.

Note that the correct moment to determine latitude by Noon Sun altitude is
NOT at the moment of maximum altitude, but at the moment of meridian
passage, so the altitude at that moment (and not at the peak altitude)
should be read off from the altitude graph at the correct instant. Luckily,
at the speeds our small craft mostly make, the difference in latitude is
small. Not so, when a vessel is doing 20 knots, as that error goeas as the
square of the speed!

=================

Going back to Frank's procedure, a navigator will meet a bit of a problem
when he comes to-

"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."

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?

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. 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.

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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