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Nice diagrams. I noticed the insensitivity at the low end of the cosine scale 
and it gives you a feeling about what is going on in the computation. The 
sensitivity when the azimuth angle is near 90 degrees is well known and that 
is why there was the special procedure for cases within 5 degrees of 90. I 
wrote:

" If the computed azimuth angle is greater than 85º the computed altitude will 
lose accuracy even though the azimuth is accurate because, with azimuth 
angles in this range, even rounding the azimuth angle up or down one half 
minute can change the Hc by ten minutes. So you use the azimuth you have 
calculated but you compute altitude by interchanging declination and latitude 
and then doing the normal computation a second time. You discard the azimuth 
derived during this second computation of altitude and use the original 
azimuth."

see:

https://sites.google.com/site/fredienoonan/other-flight-navigation-information/modern-bygrave-slide-rule

The accuracy of the Bygrave has been confirmed and is due to the choice of formulas. 

See:

http://fer3.com/arc/m2.aspx/Bygrave-formula-accuracy-on-10-inch-slide-rule-Hirose-jul-2009-g8985


gl


--------------------------------------------
On Wed, 6/11/14, Hanno Ix  wrote:

 Subject: [NavList] Re: Longhand Sight Reduction
 To: garylapook@pacbell.net
 Date: Wednesday, June 11, 2014, 9:11 PM
 
 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 
 wrote:
 
 
 Thanks
 Hanno.Good challenge. Hospital next
 week. Out of action maybe 4 weeks, then ok for go.
 !
 My slide rules cost about $5 a
 piece so far, but plenty of work!
 
 Can be
 done.Keep up the interesting work. My
 money is on Bygrave, but I’m biased!
 
 Francis 
 From: NavList@fer3.com
 [mailto:NavList@fer3.com]
 On Behalf Of Hanno Ix
 
 Sent: 11 June 2014 17:01
 To:
 francisupchurch---.com
 Subject:
 [NavList] Re: Longhand Sight
 Reduction 
 Francis,
 
 Given the choice Danioli vs Bygrave:
 What would Chichester have chosen?
 Assuming, of course, he had access
 to and enough experience with
 both.
  I celebrated Greg's RIC /
 Danioli yesterday by playing around with it. 
 
 It indeed works and is fast. Comparison with the standard
 formula is attached below.
 
 Greg seems to think one can do that single
 multiplication with a 10" slide rule. 
 I am skeptic. In praxis, ten inchers do not
 yield correct 4 digits consistently, 
 
 and that's what I need for accuracy over the useful
 ranges of L,D,t.Now, there might be a challenge
 especially for you: a Fuller < = 10" that
 can
   - do 4 digit multiplication,
 i.e. yields vwxy = ABCD * abcd;  v, w, x, y being
 correct digits. 
   - can be built with
 standard and garage tools plus a PC and
 printer.  - in not more than, say, a
 week.
   - for about $50 or less,  $100
 max. 
 It need not look like an exhibit
 in a museum, although it should be sturdy enough 
 to survive a 1-week sailing trip in the Virgin
 Islands.  ( Where and when can I sign up?
 )
 Re: formulas. One example, with
 good or flawed results, is not really sufficient
 tojudge a formula or method. You
 need to show it yields accurate 4 digit results consistently
 
 
 for the full useful ranges of L,D,t. I am unsure, though,
 what "useful" means for our CelNav
 friends.Any ideas out
 there?
 I am studying the Bygrave in this
 respect right now. Stand by please.
 
 
 Hanno
 _____________________________________________________________________________________
 For the record, Danioli
 claims:
   
    
 sin(h) = n - ( n + m ) * a ;   n:
 cos(L-D);    m: cos(L+D);   a:  [1 -  cos(t) ] / 2 or
 hav(t);   
 
 Let's see. By
 inserting:     sin(h) =  cos(L-D)  
 -  [ cos(L-D) + cos(L+D) ] *  [ 1 - cos(t) ] /
 2;
 which is in more
 detail:    sin(h)
 = 
 cos(L-D)  - cos(L-D) * [ 1 - cos(t) ] / 2  -  cos(L+D) *
 [1 - cos(t) ] / 2;
 and more detail
 yet:    sin(h) =  cos(L-D) 
 -  cos(L-D) / 2  +   cos(L-D)*cos(t) / 2   -  cos(L+D)
 / 2 + cos(L+D)*cos(t) / 2;
 Collecting:                    
            sin(h) =  cos(L-D)/2 -
 cos(L+D) / 2    +     [ cos(L-D) / 2 + cos(L+D) / 2 ]
 * cos(t);
              
 =         sin(L) * sin(D)             
 +              cos(L) * cos(D) *
 cos(t);
 which is correct.
 
  
 On Tue, Jun 10, 2014 at 11:11 PM,
 Francis Upchurch 
 wrote:
 Oh dear. Is it time to put my
 beloved Bygrave away? Cant wait to here more details of the
 Bygrave maths.Chichester said he preferred the Bygrave when
 flying single handed, because he made mistakes with log
 tables. (Perhaps he did not have Haversines?)  But, could
 someone explain the main difference/advantages/disadvantages
 of the versine method (Vers ZD=Vers LHAxCos
 Latx Cos Dec+Vers(Lat+/-Dec) and the
 Haversine method? My versine method (Reeds
 Astro Nav Tables) uses tables of natural and log versines
 and log cos (total 11 pages).Does not need
 sines.
 Versine
 methodlog vers LHA  
  9.9019log cos Lat    
  9.9177
 log cos Dec    
 9.9642add            
  29.7838Nat Vers of 9.7838=      
  0.6081
 Lat-Dec=11013' Nat
 vers=0.0191. Add= 0.6272=6806'. =ZD.
 900-6806'=
 21054'Not a lot in it I would say?
 quicker for me than reduction tables and I understand what
 we are doing.
 Please correct me and
 explain the advantages of the Haversine over the versine. (I
 do not have haversines but do have versines! Where do I get
 haversines?)Bygrave.
 H=3600-LHA=78021',
 co-lat=55050', y(w)=64031',
 X=colat+y(w)=120021',
 Y=1800-X=59039', > Az
 =76024'> Hc
 21054'
 No contest! Took a
 fraction of the time and no mistakes from looking up 4
 figure logs etc. And I've got Az (OK
 done hundreds of Bygrave LOPs and only a couple of
 Versines!)
 I'll stick to my
 Bygrave!  
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 Attached File: 
 
 (img/127997.bygrave da 10.gif: Open
 and 
 save)
 
 
 Attached File: 
 
 (img/127997.bygrave da 100.gif: Open and 
 save)
 
 
 
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