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Slide rule sight reduction accuracy
From: Paul Hirose
Date: 2009 Jun 15, 21:07 -0700

In a recent message I described a method to reduce celestial navigation
observations with an ordinary slide rule using rectangular coordinates
instead of spherical trigonometry. At the time I omitted some of the
mathematics. Here is the full procedure.

Convert LHA to an angle "theta" which is -90° (or 270) at the meridian,
increasing east. This step isn't really necessary, but omitting it puts
the rectangular coordinate frame into an unconventional orientation.
It's no problem mathematically, but I find it awkward to visualize.

theta = -90 - LHA

Convert the body's local hour angle and declination to rectangular
coordinates in a frame whose +z axis is directed to the north pole and
+y axis directed to intersect Earth's axis.

x = cos(dec) * cos(theta)
y = cos(dec) * sin(theta)
z = sin(dec)

Rotate the coordinate frame about the X axis by the complement of
latitude. This orients the +z axis to the zenith and +y north. In the
second equation, y is the old y, not the new y computed in the first
equation.

y = y *  cos(90-lat) + z * sin(90-lat)
z = y * -sin(90-lat) + z * cos(90-lat)

Find azimuth. Note that x and y are swapped from their usual positions
so azimuth will be zero at north, increasing east. This formula yields a
value in the range -90 to +90. If y < 0, add 180 degrees.

az = arctan(x / y)

Compute the body's distance from the z axis.

r = sqrt(x*x + y*y)

Compute elevation.

el = arctan(z / r)

I implemented this in a computer program which simulates slide rule
accuracy. At each place a slide rule would be used, the result is
multiplied by a number of the form (1 + x), where x is a random value,
centered on zero, with Gaussian distribution and .001 standard
deviation. In other words, the simulated slide rule has .1% accuracy.
That's the figure commonly quoted for 10 inch slide rules, and in a test
with one of my own rules I confirmed it.

Sight reduction problems are automatically generated, starting with
a random azimuth and elevation. In order to evenly distribute the
targets about the sky, elevation is the arc sine of a random number
between 0 and 1. (If you simply distribute elevations evenly between 0
and 90 degrees, the band of sky from 0 to 10 degrees will have as many
targets as the band from 80 to 90, though the latter is much smaller.)

A random latitude is obtained with the same arc sine method. The program
can restrict elevations and latitudes to specified limits; I restricted
elevations to 5 - 80 degrees and latitudes to 0 - 70.

Declination and LHA are then computed from azimuth, elevation, and
latitude. All these values are, for practical purposes, perfectly
accurate.

Declination, LHA, and latitude are submitted to the sight reduction
routine, and the returned azimuth and elevation compared to the correct
values. This occurs in a loop which runs any desired number of problems
and tabulates the statistics.

With this Monte Carlo simulation program I've found the slide rule sight
reduction method outlined above is accurate in elevation to 3.1 minutes
(square root of the mean squared error). About 95% of the results are
within 6.2 minutes. The worst case results are about 15 minutes off.
These appear to be due to unfavorable combinations of the random errors;
I can't see any pattern in the azimuths and elevations where they occur.

Azimuth RMS error is about 3.3 minutes. Worst cases are nearly one
degree, and always occur when the problem is near the upper elevation
limit (80 degrees in this test).

My program is designed in a modular fashion so different sight reduction
algorithms can be plugged in easily. I plan to implement others. If
anyone has a burning desire to see a certain method put to the test,
speak up. I'll move it to the head of the list.

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