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Gentlemen,

here are the results of a study re: Bygrave error, scale length, etc.

I was wondering why Bygrave chose his particular sight reduction
formula and his design. Since nobody on the list answered

I decided to look into it myself.

I started out by calculating the lengths of a
log10(sin(x)) - scale and the length of a log10(tan(x)) - scale

both with a "1 mm per 1 arcmin"  resolution.

This resolution was chosen because I assumed it being close to what I need for
a comfortable reading. However, I will not argue against "2mm per arcmin"

or thereabouts.
 
The lengths came out as:
log10(sin(x))  64 meters   log10(tan(x)) 28 meters.

The reason for this large difference is the simple fact that the

step size of the first varies much more than the one of the second.

G. LaPook's Flat Bygrave's (FB)  tan() - scale stretch over only ~ 380" or ~ 10 meters.,
with the cos() - scale about half as far or less. Why, then, can the FB yield - occasionally at least - a result with an error of  2 arc min or less as G. LaPook reports?

BTW: If the FB were twice as long this error would be, I assume, half of that.

That chimes roughly with my assumption of "1mm per 1 arcmin" of a readable scale.
In fact, on the copy of the FB in my possession, at ~45 deg on  the tan() scale you can at discern 1 arcmin with ease - outside much less.

Getting closer to the real question: In the FB, the cos() scale has been shrunk far below that "1mm per 1 arcmin" threshold. Look at the 5 deg area! Why is the result still as good as Gary et al. claim? In other words, why does that very rough cos() scale not prevent the overall results to be so close to the correct one? That is odd!

Two components to the answer:

-  H is read on the scale with the highest resolution, the tan() - scale. ( Actually, that is the case for the azimuth as well.)

-  In Bygrave's formula, the cos() and its errors have a limited control, so to speak.

The 2 attached graphs demonstrate this. I will not bother you with the details of the calculations. The first shows the error dH of H given a correct Y and a grossly wrong azimuth A. Its offset dA  is assumed to be as much as 10 arc min. Same for the second graph with dA only 100 arc min. ( Both are interesting cases in their own right.)

We conclude for

dA = 10 arcmin: Unless the pair (Y, A)  falls in to red zone dH will never exceed 1 arcmin.

The good area (white) is a very wide area. Most sight reduction cases will fall into it.

dA = 100 arcmin: Despite this large error of the azimuth the Bygrave will produce good results for all values of Y if the azimuth itself is below 20 deg. If we however limit the range of admissible Y's  to ~ 20 deg the azimuth may be as large as 50 or 60 deg and the Bygrave will still deliver good results. And for the application of this case: there are often other, perhaps simpler, means to determine the azimuth with an error of  up to 100 arcmin. In those cases, one can take the value found with those and proceed to finding H on the Bygrave.

I hope this has found your interest.  I admire the insights of Bygrave that he put to use in his great little slide rule!

Regards

Hanno

On Wed, Jun 11, 2014 at 12:16 PM, Francis Upchurch <NoReply_Upchurch@fer3.com> wrote: