Hi Gary
If we are to keep the resolution of the MHR1 scales, then we can use the actual length of the MHR1 scales to calculate new lengths based upon the input range in degrees.
We will not expand the scale length to be able to precisely see each arc minute graduation on Tube I (outer), in which we see 0 to 5 degrees incredibly compressed. We will not compress the scales such that a new range of inputs uses the existing length, leading to loss of resolution. We could do that, keeping the same tubes, but the resolution would suffer.
I will use the cosine function for Tube I (outer) and the tangent function for Tube II (inner). You can easily switch these identities to secant // cotangent respectively, but for simplicity I will stick to cosine // tangent.
From Ronald van Riet's paper, we have been given the tangent scale length equal to 0.758 meters and the cosine scale length of 4.143 meters.
The length of the tangent scale is directly related to
Span1=log(abs(tan(89 deg 40 min)))  log(abs(tan(0 deg 20 min)))
The length of the cosine scale is directly related to
Span2=log(abs(cos(0 deg 0min)))  log(abs(cos(89 deg 40 min)))
So for the MHR1, we have Span1(MHR1) = 4.470478 and Span2(MHR1) = 2.235246
If we substitute the values of the Flat Bygrave, we can see the required length of the new scales! Span1(FlatBygrave)=3.674547 and Span2(FlatBygrave)=1.837319
So Span1(FlatBygrave)/Span1(MHR1)*0.758meters length = 0.623 meters for the Flat Bygrave
Span2(FlatBygrave)/Span2(MHR1)*4.143meters length= 3.405 meters for the Flat Bygrave.
By starting and stopping 50 minutes from the singularity, instead of 20, you have saved considerable scale length! Well done Gary.
++++++
Now lets consider if we are to go to within 1E15 of an arcminute of the singularities. That is instead of 0 deg 0 min, use 0 deg 1E15 min. This will yield results well within the noted 1 arc minute accuracy of the MHR1. So instead of 90 deg 0 min, use 89 deg 59.999999999999999 min. (that's 15 nines). The practical navigator will never notice anything in manipulation. Only we the creators of this tweaked scale will know.
Span1(TotalRange)=18.050115
Span2(TotalRange)=16.212841
Span1(TotalRange)/Span1(MHR1)*0.758 meter length= 3.062 meter length. The additional 2.25 meters is above 89 deg 40 min. It took 2.25 meters to cover the remaing 20 minutes whilst retaining the vaunted 1 arc minute accuracy of the Bygrave.
Span2(TotalRange)/Span2(MHR1)*4.143meter length= 30.050 meter length. The additional 26 meters in length is to be split, with 13 meters for below 20 arc minutes and 13 meters for above 89 deg 40 min. So a total of 26 meters to gain 40 arc minutes of input range.
+++++++++
So as a practical matter, Bygrave saved a lot of scale in exchange for small loss of angular input range. I suspect that this is why the Flat Bygrave extended that small loss to 50 arc minutes from 20.
But this also precisely why the "official instructions" are not to use the MHR1 in the restricted ranges. There simply isn't enough physical tube and scale length to "estimate" the proper Tube I to Tube II relationship. The restricted ranges required significantly more length to be shown at the current resolution.
I would be very curious to see what the scale lengths and operating ranges of the MHR2 table top calculator. Ronald  Do you know these values?
++++++
I'd also like to hazard a guess as to why the original models did not use the clever extension that is advertised by Gary. Its pure speculation and conjecture on my part. Here goes:
Gary's clever solution is advertised to have an error of 0.9 degrees in azimuth. The Bygrave is advertised to be accurate to 1 arc minute. Thats a 54 arc minute error as compared to a one minute arc error.
So why advertise degraded accuracy? Even if they did think of it ...
Again, pure speculation on my part. Its binding on no one!
+++
Wolfgang  if you want any other dimension off of the MHR1, feel free to ask. Its not a bother at all. After seeing your first model, I know the second will be fantastic. Its worth my time.
Regards
Brad Morris
On Jan 28, 2013 1:44 AM, "Gary LaPook" <
garylapooknet> wrote:
I find it surprising that neither Bygrave or Dennert & Pape came up with a method for dealing with the situation of the body's declination being within 20 minutes of zero since this situation exists for the sun (the most important body) for two days in the spring and two days in the fall, four days out of 365 days a year. My method was not hard to discover and provides accurate results for this situation so it s a mystery to me why similar instructions were not provided with these instruments. Without providing such a procedure navigators would need to learn and carry a complete other celestial computation method to be able to operate each day of the year.
gl
 On Sun, 1/27/13, Hewitt Schlereth <hhew36com> wrote:
From: Hewitt Schlereth <hhew36com>
Subject: [NavList 22143] Re: Re: Dennert & Pape MHR1 Dennert & Pape MHR1 HR2 To: garylapooknet Date: Sunday, January 27, 2013, 6:48 PM
Brad 
I will defer to Wolfgang, but my translation of
Bei stundenwinkel und deklination zwischen 0 und 0 20 sowie bei
stundenwinkel 90 ist der Hohenrechenschieber nicht zu verwenden
is:
For hourangle and declination between 0 and 0 20' as well as for hourangle of 90, the altitude slide is not to be used.
The two dots above the o are called an umlaut (Höhenrechenscheiber). The ö sound is made by shaping your lips as though you were going to say 'oh', then trying to say 'ee'.
Hewitt
Sent from my iPad
On Jan 27, 2013, at 4:11 PM, "Brad Morris" wrote:
> Hi Gary
>
> On the inner tube of my MHR1 (noted as tube II in the German instructions &
> also the scale upon which we set the declination), the minimum angular
> dimension is 20 min (179 deg 40 min). At the other end of this same scale
> I find 89 deg 40 min (90 deg 20 min).
>
> Tube I (outer tube, upon which in the first step we set zero, or in the
> third step, upon which we set azimuth) starts at zero. The other end of
> tube I is marked 89 deg 40 min (90 deg 20 min). The double markings begin
> at 35 deg (145 deg). Below 35 deg, there is only a single mark, obviously
> just due to limited room.
>
> I thought to mention this as your email indicates that the smallest angular
> dimension on this scale is 30 min. Not true for mine. I simply don't know
> about others.
>
> There ARE special instructions (in German) for when "stundenwinkel" (t)
> and "deklination" are between 0 and 0 deg 20 min. I cannot read German. I
> include the text following. All typo's are mine.
>
> Bei stundenwinkel und deklination zwischen 0 und 0 20 sowie bei
> stundenwinkel 90 ist der Hohenrechenschieber nicht zu verwenden.
>
> The first 'o' in Hohenrechenschieber has the two dots above it.
>
> Regards
> Brad
> On Jan 27, 2013 6:39 PM, "Gary LaPook" wrote:
>
>> 
>> Do you have any actual instructions for using your MHR1 for dealing with
>> the special case when the declination is less than the lowest declination
>> marked on the tangent (or cotangent) scale, 20' for the Bygrave and 30'
>> for the MHR1 and 50' on my flat Bygrave? I have never been able to find
>> instructions for this situation. I am attaching a description for the use
>> of this instrument (in French) which says it is "inutilisable" for this
>> situation and the original Bygrave instruction manual is silent as to this
>> situation, also attached.
>>
>> I came up with a method of dealing with this situation and it is posted on
>> my website at:
>>
>> https://sites.google.com/site/fredienoonan/otherflightnavigationinformation/modernbygravesliderule
>>
>> (In my notation "W" is the same as "y" in Bygrave's notation and "x" on
>> the MHR1.)
>>
>> "When declination is less than one degree you can't begin the computation
>> the normal way to find "W" because you have to start the process with
>> declination on the cotangent scale and this scale doesn't extend below 1º.
>> So in this case you just skip the computation of "W" and simply set "W"
>> equal to declination. Using this method you arrive at an azimuth that is
>> not exact but is a close approximation and in the worst case I have found
>> the azimuth is still within 0.9º of the true azimuth but most are much
>> closer. If the declination is less than one degree and the latitude is also
>> less than one degree, follow this procedure and also assume a latitude
>> equal to one degree. After you have computed the Az you then follow the
>> same procedure discussed above for azimuths exceeding 85º by interchanging
>> the latitude and declination and then computing Hc which will produce an
>> exact value of Hc."
>>
>> Also see my prior post at:
>>
>>
>>
>> http://fer3.com/arc/m2.aspx/ComputingazimuthwithBygravespecialcasesLaPookfeb2010g11826
>>
>> gl
>>
>>
>>  On *Sun, 1/20/13, Marc Giordan * wrote:
>>
>>
>> From: Marc Giordan
>> Subject: [NavList 22013] Re: Dennert & Pape MHR1 Dennert & Pape MHR1
>> HR2 Cylindrical Slide LineRuler
>> To: garylapooknet
>> Date: Sunday, January 20, 2013, 11:08 AM
>>
>> 
>>
>> Sir,
>> I am French and also quite familiar with the use of the MHR 1 at sea (I
>> used it for 2 atlantic crossings by sail)
>> Would any one wish to be trained in using it, I am ready to help !
>> The MHR 1 was bought by my father some 30 years ago at the flee market in
>> Paris
>> Yours
>> Marc Giordan
>> 26 rue de Paris
>> 78560 le Port Marly
>> (0033)675210703
>> 
>> NavList message boards and member settings: www.fer3.com/NavList
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>> 
>>
>> View and reply to this message: http://fer3.com/arc/m2.aspx?i=122013
>>
>> *Attached File: 122138.calcul de navigation, v2 (courante).pdf
>> *
>> *Attached File: 122138.bygrave manual.pdf
>> *
>>
>> View and reply to this message: http://fer3.com/arc/m2.aspx?i=122138
>>
>
>
>
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