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On 2016-06-14 9:38, Robert VanderPol II wrote:
> Here's an idea you might want to follow: what is average accuracy if values 
are tabulated at each 0.5° for the range 0°-80° and 0.2° or 0.1° for 80°-90°?

It's too much work to emulate a table with a non-uniform tabulation
interval. On the other hand, a .2 or .1 minute interval throughout is
trivially simple, since the code was written with that in mind. But
right now I'm reorganizing the program, since the proliferation of
features has made it untidy.

> Additionally does the data you have already allow you to plot the 95% or 90% 
error or whatever percentage error on a graph of Meridian Angle (t) vs. Dec? 
I envision a family of curves where each curve represents a 90% likelihood 
that the error is equal to or less than 0.5nm, or 1.0nm, 1.5, 2.0 . . .

I suppose it could be done, but that wouldn't address the large errors
in the outliers. For instance, if the tabulation interval is reduced to
0.2', 96% of altitudes are within 0.5' of the truth even with no
interpolation. Yet the max error is 24', scarcely better than standard
Ageton.

On the other hand, if we use the standard table, do not interpolate,
exclude all sights where t is within 8° of 90, and all declinations
greater than 75°, the max error decreases to 2.9'. Compared to the
preceding paragraph, root mean square altitude error is cut in half,
though the percentage of altitudes more than 0.5' in error more than
doubles.

That illustrates some of the complexity of choosing an effective
strategy for employing the Ageton table. I'm looking forward to testing
the Sadler algorithm.

   

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