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Re: Bygrave range of angles?
From: Hanno Ix
Date: 2015 Apr 21, 16:01 -0700
As for the angle range on a Byrave: Even though you might have done more
From: Hanno Ix
Date: 2015 Apr 21, 16:01 -0700
Francis,
you are THE expert in making cylindrical SR so, of course, stick with that.
As for me, there are some practical issues I wouldn't know how to deal with.you are THE expert in making cylindrical SR so, of course, stick with that.
than plenty of reductions you don't know my luck: Zip!
So I need a rationale, not just luck, for choosing that range. Gary's line of thought
seems to point in the right direction.
H
On Tue, Apr 21, 2015 at 3:50 PM, Hanno Ix <hannoix@gmail.com> wrote:
Gary,that is the answer I needed. Thx!I will study your website.HOn Tue, Apr 21, 2015 at 3:01 PM, Gary LaPook <NoReply_LaPook@fer3.com> wrote:The scale on my Flat Bygrave covers 0° 55' to 89° 16' and is all that is needed for all latitudes up to 90° .There is nothing to be gained by trying to extend it further in either direction, either closer to 90° or to 0°.From my website:====================================================================================SPECIAL RULES
There are some unusual cases that require slightly different procedures and all of these special cases are described on the form. If "H" is less than 1º or greater than 89º simply assume a longitude to bring "H" within the range of the scales. The intercept will be longer but perfectly usable for practical navigation.
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.
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.
Another rare possibility is that "Y" will be less than one degree or that it will exceed 89º after adding "W" to co-declination so it won't fit on the scale. The simple way to handle this situation is to assume a different latitude so the "Y" does fit on the scale even though the resulting intercept is longer but still usable.
An extremely unlikely case (I only mention it to be complete) is that "W" exceeds the range of the cotangent scale, 89º15', so cannot be computed in the first step of the process. This can only happen when shooting one star, Kochab, which has a declination of 74º 07' north and then only if "H" exceeds 87º 20', an extremely unlikely event.
The attached form steps you through the computation and contains the rules for the special cases. The special cases are likely to come up only very rarely in practice.
The first rule for H < 1º or H > 89º only involves LHAs covering 5 degrees out of 360º (LHA in the ranges of 0 -1, 89-91, 269-271, and 359-360) so only occurs by chance very rarely and these can be avoided if sights are preplanned as is the normal procedure for flight navigation. In the worst case, you have to change the time of the observation by four minutes.
Rule 3 covers the case when Y is less than 1 or when Y exceeds 89º which covers a range of three degrees out of a possible 180º so is also very rare. Co-lat is in the range of 0º-90º and W is also in the same range so X comes in the range of 0º -180º. If X is less than 89º then Y is also less than 89º. If X is greater than 91º then Y is also less than 89º. Only in the case of X between 89º and 91º will Y exceed 89º. This situation can't be avoided in advance because you can't predict what the value of W will be but if it occurs then just assuming a latitude that differs at most by one degree solves the problem which will result in a longer intercept but one that is still usable.
The fourth rule deals with cases of bodies bearing almost directly east or west and this situation can be avoided by choosing a different body to shoot or, if only the sun is available, by waiting a few minutes to allow the azimuth to change out of this range.
The remaining situation covered by rule two (declinations less than one degree) concerns only bodies in the solar system since none of the navigational stars have declinations less than one degree. Obviously the most important body is the sun and its declination is between 1º north and 1º south for five days in March and again in September so this situation can't be avoided and this is the most important special case. The special rule handles it nicely and the Hc is completely accurate. The computed azimuth is an approximation but is never more than one degree different than the actual azimuth and is usually much closer. Since you can use your D.R. for the A.P. for this situation the intercepts can be kept short and the slight inaccuracy in the azimuth will not make a noticeable difference in the LOP.gl
From: Hanno Ix <NoReply_HannoIx@fer3.com>
To: garylapook---.net
Sent: Tuesday, April 21, 2015 11:16 AM
Subject: [NavList] Re: Bygrave range of angles?
I am asking myself about the angular range of the log-cot() scale.Gary,thanks for the response..What is the upper limit of angle it should go to?Robin Stuart's magnificent scale I have in my hands goes up to89deg 40min. But note: Given a certain physical length of the scale,the closer you go to 90deg - unreachable itself - the morecompressed the scale becomes around 45deg where the more frequentlyused angles appear. Robin's scale takes about 25% of its lengthfor angles >= 89 deg - a big percentage for angles rarely used.Or so I think.Now, then again, what should this upper limit be trading off the variousneeds of an active blue water sailor with a 40' boat operating up to70deg latitude?HOn Tue, Apr 21, 2015 at 5:11 AM, Gary LaPook <NoReply_LaPook@fer3.com> wrote:There is no limit on the latitudes that can be handled with the Flat Bygrave. There is a special case when the declination is less than one degree on the latitude is zero then you must use one degree for the assumed latitude but it still computes an accurate Hc and Z. From my website:===================================================================================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.===========================================================================================.See the complete explanation on my websitegl
From: Francis Upchurch <NoReply_Upchurch@fer3.com>
To: garylapook---.net
Sent: Monday, April 20, 2015 10:51 PM
Subject: [NavList] Re: Bygrave range of angles?
Hanno,I don't know about the flat Bygrave, but with my cylindrical version (about the same size nearly as the original), I have done hundreds of reductions and rarely out more than 1-2' compared to calculator. the lat range seems to cover +/-nearly 90° as far as I can tell and for some reason i do not understand, why when you end up on the "compressed" side of the scales and are interpolating between fairly small scale marks, the accuracy stays the same. Maybe Gary knows why? (Interestingly, I have not found this with the Fuller, which is definitely more accurate in the "expanded" range of the scales. I get 4-7 places with the big Fuller depending on where you are in the scales)Anyway, good luck. I'm sticking with cylindrical rules. I find them easier to make and use accurately using pre-manufactured, precision tubes, felt friction brakes, no pivot holes to drill etc. But each to their own I guess.Best wof luckFrancis