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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: running fix + lunar
From: Paul Hirose
Date: 2010 Nov 22, 21:31 -0800

```Andres Ruiz wrote:
> The "practical circumstance", is: one is sailing with known course and
> speed and to obtain a fix uses two altitudes with a sextant and a
> lunar distance at different times ant the chronometer or on-board
> watch has an error. Please Paul, correct me if I have not

You understand me perfectly. The idea of celestial sight reduction with
a set of linear equations is nothing new. An especially elaborate
example is the Kaplan paper, referenced in your other message. The only
new idea I've introduced is that time is only approximately known, and
must be found by a lunar. In reality this is an improbable scenario, but
I thought a few readers might find it interesting.

Here is another example. This time we'll shoot an afternoon lunar with
the Sun, followed by three altitudes at evening twilight. Normally, Sun
and Moon altitudes would be observed with the lunar distance, but to
make the problem more difficult let's say the horizon is bad at that
time. So the lunar distance must be combined with three altitudes
several hours later.

Here are the observations on 2010-12-01 and -02, with UTC from the
chronometer, and estimated positions.
21:36:02 at N25°57.7' W142°00.5' Moon-Sun 49°53.1'
02:27:11 at N26°35.3' W140°05.7' Jupiter 51°28.1'
02:32:56 at N26°36.1' W140°03.5' Schedar 51°11.7'
02:38:24 at N26°36.8' W140°01.4' Altair 50°21.1'

Those times and positions have considerable error. As before, assume the
same error affects all observations. The observed angles are accurate.
The lunar is near limb to near limb. Once again I'll try to describe the
method with simple language.

Begin by computing altitude (or lunar distance) at the given times and
locations. Compare these to the observed angles.

Moon - Sun computed separation angle is 1.7' greater than observed. One
degree north changes computed angle -.7'. One degree east changes it
+.6'. One minute later changes it -.4'.

I don't know a way to gauge the effect of latitude and longitude on
lunar distance, except to recompute from scratch for a different
position. However, I have a program which outputs the rate of change per
minute.

The "partial derivatives", as these are called, are much easier to
compute for the altitude observations. The increase in altitude equals
the (north) increase in latitude, times the cosine of azimuth. It also
equals the (east) increase in longitude, times the sine of azimuth,
times the cosine of latitude.

And since the passage of time has the effect of moving east in
longitude, its partial derivative is simply some fraction of the
longitude partial derivative. E.g., one minute of time is equivalent to
1/4 degree of longitude, so if 1 degree of longitude changes the
computed altitude by 40', 1 minute of time changes it 10'.

Getting back to our observations, Jupiter's computed altitude is 1°41'
less than observed. Moving one degree north changes computed altitude
-42'. Moving one degree east changes it +38'. One minute later changes
it +9.5'.

Schedar computed altitude is 3°09' less than observed. Moving one degree
north changes computed altitude +50'. One degree east changes it +29'.
One minute later changes it +7.3'.

Altair computed altitude is 4°05' greater than observed. One degree
north changes computed altitude -24'. One degree east changes it -49'.
One minute later changes it -12.3'.

Set up the equations as described in my previous message:
-1.7 = -.7∆φ + .6∆λ - .4∆t
101 = -42∆φ + 38∆λ + 9.5∆t
189 = 50∆φ + 29∆λ + 7.3∆t
-245 = -24∆φ - 49∆λ - 12.3∆t

Unlike before, we have three unknowns and four (not three) equations. A
solution may be found by the method of least squares. Results:
∆φ = +1.38° = N 1°23'
∆λ = +2.75° = E 2°45'
∆t = +5.99 m = +5m 59s

Add those corrections to the times and positions. The observed angles
are the same as before.
21:42:01 at N27 20.7 W139 15.5 Moon 49 53.1
02:33:10 at N27 58.3 W137 20.7 Jupiter 51 28.1
02:38:55 at N27 59.1 W137 18.5 Schedar 51 11.7
02:44:23 at N27 59.8 W137 16.4 Altair 50 21.1

Now repeat the process of determining (computed - observed), and the
partial derivatives. Set up new equations:
.6 = -.7∆φ + .4∆λ - .4∆t
11.4 = -46.6∆φ + 33.3∆λ + 8.3∆t
-2.8 = 50.9∆φ + 28.1∆λ + 7.0∆t
-4.3 = -22.4∆φ - 49.1∆λ - 12.3∆t

Note the values on the left side are much smaller than before. It shows
the first set of corrections were a big improvement. New solution:
∆φ = -.138° = S 8.2'
∆λ = +.369° = E 22.2'
∆t = -.878 m = -53 s

In the upper line of each pair below I show the corrected time and
position, and the corresponding computed lunar distance or altitude. The
lower line has the true values, which are unknown to the navigator
(except for the observed angles).

21:41:08 at N27 12.5 W138 53.3 Moon-Sun 49 53.1
21:41:18 at N27 11.8 W138 53.5 Moon-Sun 49 53.1

02:32:17 at N27 50.1 W136 58.5 Jupiter 51 28.1
02:32:27 at N27 50.0 W137 01.0 Jupiter 51 28.1

02:38:02 at N27 50.9 W136 56.3 Schedar 51 11.7
02:38:12 at N27 50.8 W136 58.8 Schedar 51 11.7

02:43:30 at N27 51.6 W136 54.2 Altair 50 21.1
02:43:40 at N27 51.5 W136 56.7 Altair 50 21.1

We now have a perfect match between computed and observed angles.
However, some error remains in the time and position. I believe this is
mainly due to error in the dead reckoning between the afternoon lunar
and the evening sights. The navigator used 22 knots on true course 069
to come up with the estimated positions at dusk. I introduced an error
in both course and distance made good, equivalent to 3% of the distance
run. Thus, the same set of time and position corrections cannot be valid
for both the afternoon and evening observations.

The results were still pretty good. I could have injected errors in the
observed angles too, but there are limits to how hard I want to work!
Fortunately my HP 49G calculator makes it easy to solve simultaneous
linear equations. It automatically uses the method of least squares if
the solution is "overdetermined" (the number of equations exceeds the
number of unknowns). This is not a very modern calculator, either. It's

The problem in my first message, with three equations in three unknowns,
could actually be solved with a simple calculator, pencil, and paper.
Some readers may have solved similar sets of equations that way in high
school. Unfortunately, a least squares solution is not so simple. I'd
hate to work it with nothing but a simple calculator. It wouldn't be too
bad with something like an early 1980s HP-15C, which can invert and
multiply matrices. That's the hard part.

One refinement of the least squares solution is to put additional weight
on a lunar distance, because usually that angle is more accurate than an
altitude. In my example, that would be accomplished by multiplying the
coefficients of the first equation by say 1.5 or 2.0.

The same thing can be accomplished by expressing the residuals (observed
- computed value) and the partial derivatives in "sigmas", where one
sigma is the standard deviation of the measurement. This could simply be
the navigator's estimate.

--
I filter out messages with attachments or HTML.

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