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A Source of Confusion in Selecting an Assumed Position - PeterB - 03-26-2023
My apologies if this post is a little long, but if selecting an AP in the Eastern longitudes has you confused this might get to the heart of the reason why. I try to be complete in my discourse because browsers seem to pass through here looking for answers, and they may not be fully up to speed on the past conversations and rhetoric.
AhHa!
I now SEE a big source of confusion regarding Assumed Positions to which I was previously oblivious! It all started for me with the thread “LHA in Eastern Longitudes?”
https://thenauticalalmanac.com/Forum/showthread.php?tid=243
The problem I now see is that many students are studying TO THE RATING TEST, not to the underlying principles and the practical final plotted line of position (LOP.)
Let me explain:
I was reading older posts in this forum. One very popular thread in particular was "Celestial Problem I Can Not Solve." https://thenauticalalmanac.com/Forum/showthread.php?tid=190 It was here that I realized that in at least three of the celestial problems offered in the referenced video http://www.seasources.net/youtube%20videos/CompleteSightReduction.mp4 the student was NOT EXPECTED to plot the problem as a final LOP, instead only expected to find the true azimuth (Zn) and intercept (a) that are the correct answer to the problem.
Naturally most humans have the approach of "I don't want a lecture on theory here - just show me a quick way to get the right answer..." And of course this is more so if the student is not a fan of celestial navigation for any reasons, not the least of which is he or she may be of the opinion that celestial is totally obsolete.
In this special case for the EXAM you DO NOT NEED to know the precise longitude of the Assumed Position (AP) since you will not be plotting the result. You only need the AP latitude (easy); the declination of the body (you look that up and use it as found); and the Local Hour Angle (LHA) for which, rigorously speaking, you need the correct AP longitude.
...but...
By a "short cut" you can get the same LHA that was used for the solution, ...well, most of the time. If all you want is the right answer on the test and don't care about the LOP or to expend the time and effort necessary to fully understand what you are doing, the short cut certainly seems enticing.
An aside: I do not hold any "short cutters" in poor regard. I presume that if they are studying for one of these exams they have got a heck of a lot of other stuff to learn besides celestial, and all of that is in addition to seeing to all their other daily responsibilities. Since celestial may seem like an arcane backwater of that total body of knowledge a "quick fix" certainly is attractive. Also celestial can be confusing to learn and anyone who didn't know that this was a kind of short cut could easily be enticed by seeing a simpler way to do things. -- Myself included.
Let's examine how the short cut works by first looking at West longitude dead reckoning (DR) position:
The LHA is the difference in longitude between your AP and the Greenwich Hour Angle (GHA) of the observed body. All hour angles are ALWAYS measured ONLY westward so the GHA of the body is how far west its geographic position (GP) is from the Greenwich meridian.
Your own longitude is how far west YOU are from the Greenwich meridian.
To get the LHA, which is how far west the body's GP was from YOU, you subtract your DR longitude from the GHA of the body.
If you are using a calculator or an electronic app you can use your DR and the GHA of the body complete with their minutes and tenths as found. The resulting LHA will probably have some minutes and tenths in its result. You can use that LHA along with its minutes and tenths in the electronic app or calculator solution. The results will be an azimuth (Z); a true azimuth (Zn); and an intercept distance in nautical miles (a); to be plotted directly from your DR.
However the test uses tables to do the sight reduction. Usually Pub 229 for USCG. These tables and many others such as Pub. 249 require that you enter them with a whole degree of AP latitude and an AP longitude that results in a whole degree of LHA. You do not enter them with your DR position. Instead you “deliberately take up” a “position of convenience” nearby to where you are which we call and “Assumed Position.” Here “assumed” means “deliberately taken up.”
The reason for the "whole degrees" thing is just for compactness of the tables. It eliminates the huge number of possible solutions that would exist between every whole degree demarcation, making the printed books of tables physically small enough to carry aboard. It has NOTHING to do with the math. The math doesn't care about whole degrees at all.
In order to get the same answer as the exam you need the whole degree latitude and whole degree LHA that THE EXAM WRITER used to enter Pub. 229. If you use these, even with an app, a calculator, or different type of tables, you will come to their same answer - which is the immediate goal - even if you don't actually know what an AP is or how the heck they got one.
For western DR positions you subtract your AP longitude from the GHA of the body. To do this rigorously you first determine the GHA of the body including the minutes and tenths. Then you append those same minutes and tenths of the GHA to the whole degree of your longitude to get an AP longitude. In real life you will need this EXACT AP longitude later for plotting, so you would not short cut. But for the EXAM you don't plot so all you need is the LHA as a whole degree – and it has to be the same one the exam writer used or your answer will be different.
Here you can see if you simply ignore the minutes and tenths of BOTH the GHA and the DR when you subtract the DR longitude degrees from the GHA degrees you (usually) get the right answer. ... More on"usually" later.
When in eastern longitudes things change a bit. Here you must add how far east of the Greenwich meridian you are to how far west the GP of the body was from the Greenwich meridian to determine how far west the body's GP meridian is from YOUR meridian.
All the above discussion regarding using calculators or apps still applies. The math does NOT need a whole degree of anything, but the tables, and the exam question answer based on using the tables, DO need a whole degree AP latitude and a whole degree LHA. The whole degree latitude part is easy - just round the DR latitude to the nearest degree. The whole degree LHA is a little trickier.
Rigorously speaking since you will be ADDING your AP longitude to the GHA of the body the minutes and tenths of your contrived AP must sum with the minutes and tenths of the GHA to equal exactly 60.0' which is one whole degree.
EXAMPLE: GHA 34° 40.0' DR lon 15° 32.5' E and 34° 40.0' + 15° 32.5 E = 50° 12.5'
this doesn't give a whole degree LHA so instead you contrive an AP longitude of 15° 20.0' E.
Then 34° 40.0' GHA
+ 15° 20.0 E AP lon (contrived so LHA will comeout to a whole degree)
49° 60.0'
=> 50° 00.0' LHA for use with Tables after you carry over the 60.0'
The RIGOROUS result for LHA will ALWAYS have the format of a trailing " 60.0' "
and this will ALWAYS increase the degree value by ONE WHOLE DEGREE.
It is this trailing 60.0' that Chris Nolan mentions in his teaching video when he comments " ... how many times have I forgotten to add that 1 degree?" https://youtu.be/X6VokWcuonU?t=204 but if you watch that segment carefully you will clearly see that he isn't short-cutting. He actually derives the proper AP longitude rigorously. What he “forgets” is to to carry over the 60.0'
The trailing “ 60.0' ” is also the basis of the shortcut method that says: "add the whole degrees of the DR to the whole degrees of the LHA and then add 1 degree to that result."
Looking at the above example if we do the following we get the LHA for the exam question, though it leaves us without knowing the longitude of the AP necessary for plotting and therefore in real life would be useless - but it WILL get the correct answer for the exam ... most of the time.
It isn't the "correct" way to do it. It is NOT based on either the mathematical requirements nor on good practice, nor a requirement for the tables. Its just a trick, but it can work...for the exam ... but not for actual navigation because you don't have a real AP longitude from which to start your plot. You skipped that step and went directly to finding the LHA.
Short Cut Version:
34° LHA degrees
+ 15° E DR degrees
+ 1° short cut "add one degree"
50° LHA
Next: About that "...most of the time"
Examiners want you to pick an AP longitude within 30.0' of longitude to your DR. Sometimes the above shortcut won't do that, and if you have not applied the rigorous method you may not be aware of that flaw because you never actually derived the AP longitude.
Suppose that in the above example the DR longitude had been 15° 52.5' E instead of 15° 32.5' E ? Using the short cut still would have a result of 50° for LHA. However employing the rigorous approach the AP longitude of 15° 20.0'E is now 32.5' of longitude away from the DR longitude – too far for the examiner! Here the "correct" AP longitude is adjusted to be 16° 20.0' E bringing the AP longitude to 27.5 minutes of longitude from the DR. It also changes the LHA to 51° In this case the short cut gets you the wrong answer for the exam!
All of that rigorous adjusting to make the AP longitude within 30.0' of the DR longitude makes essentially NO DIFFERENCE to the resulting LOP in real life AFTER PLOTTING, but may make a small difference in the results for Z and Zn values, and it WILL make a big difference in the intercept length leading you to the wrong answer for the exam.
If you used the short cut for exams and your Zn looks pretty good but your intercept is way off, try adjusting the LHA by one degree. Or better yet just take the extra minute to derive your AP longitude the rigorous way - it's really not that difficult.
My conclusion is that learning how to use the short cut is no easier than learning the rigorous method;
That the short cut confuses students because the do not realize it is just a trick; That the short cut is useless in real life because you don't have an AP longitude from which to base your plots; And because of these reasons the short cut, like most short cuts in learning, is not a better choice in the end.
Peter
RE: A Source of Confusion in Selecting an Assumed Position - P.Rutherford - 03-27-2023
Peter,
The above is very useful but could you please give an example for Eastern longitudes that isn't for a the short cut method and gives an answer that provides the Ap longitude with minutes used during a LOP plot.
Thank you,
Paul
RE: A Source of Confusion in Selecting an Assumed Position - PeterB - 03-27-2023
Paul,
First thank you for your question. I know this is initially a point of confusion for many and I worried that no questions meant I did not explain it very well! Full disclosure: It confused the heck out of me when I was trying to learn it!
I go into the rigorous approach a bit more including an example in the thread titled "LHA in Eastern Longitudes" There is a link to that thread above but I'm working on a mobile device and its not getting the link here - sorry.
For finding an AP longitude from an eastern longitude DR by the rigorous method:
subtract the minutes and tenths of the GHA as found from 60.0'
append those resulting minutes to the whole degree of DR longitude to get AP longitude
add the AP longitude to the GHA getting a result with trailing 60.0' for LHA
carry over the trailing 60.0' as one degree to the LHA
for examinations: check that the resulting AP longitude is less than 30.0' from DR longitude
example: GHA 18 54.2' DR long. 20 12.1E DR latitude 15 54.2'N
start with 60.0' - 54.2' = 05.8' so the 05.8'' is what we append to our DR long degrees for our AP longitude. That result is an AP longitude of 20 05.8'E
Now we add 18 54.2' to 20 05.8'E = 38 60.0' as LHA
when we carry over the 60.0' as one full degree this updates to 39 00.0' as the LHA
This is an LHA you will find listed in the tables (because they only list whole degrees LHA and lat.)
Here you now have the whole degree LHA needed for most tables (39 00.0') AND an actual AP longitude of a starting point for your plot ( 20 05.8'E )
In this example 20 05.8'E is less than 30.0' of longitude away from the DR longitude of 20 12.0E
Since theDR latitude is 15 54.2'N you set the AP latitude to the nearest whole degree 16 00.0'N
The AP is put down on your chart or plotting sheet at:
16 00.0' N 20 05.8' E and it is from THIS point (not your DR) that you plot in the true azimuth and intercept distance to the line of position.
With a bit of luck the LOP will pass very close to your DR.
Hope this helps. I'm happy to answer any further questions because I'm pretty sure there are more.
Peter
A wrinkle:
H.O. 211 Ageton's tables and its various derivatives including a version in Bowditch as well as The S Tables by Mike Pepperday allow sight reduction and plotting directly from your DR. This is also true of most electronic apps and direct calculation methods.
However since all of them work from any position they ALSO work with traditional AP's per the above discussion. After all the DR is just a position that is somewhat close to where you think you are and in this regard the math and methods don't know the difference from a DR to an AP.
It is only the commonly used tables that require a whole degree LHA and Lat ... not the math.
If you DO use Ageton's or direct calculation then on exams you still must find the traditional AP that the examiner who wrote the question used. When you use that with your DR methods you will get the correct answers for the exam, the same as if you had used Pub. 229 tables.
Therefore you really ought to learn how to properly find the AP even if you do not use it day to day.
Peter
RE: A Source of Confusion in Selecting an Assumed Position - CarlosPindle - 03-28-2023
Peter,
That's a very good explanation and makes it much simpler to understand and easier to carry out.
I'm curious, do people still do Ageton reductions? I've given it a try several times and have enjoyed it....but a calculator makes the process easier in any case.
Thanks,
Carlos
RE: A Source of Confusion in Selecting an Assumed Position - PeterB - 03-28-2023
Carlos
I imagine some people still use Ageton's (H.O. 211) because it is readily available, often as a free download. In addition there are a number of variations of it available, also often for free. Aside from being available for free the other two "selling points" for Ageton's are that you do not have to learn how to derive an assumed position because you work from your DR position, and you don't have to interpolate.
I personally find all the additions, subtractions, and rules that must be applied as you work the sight using Ageton's are the source of many blunders even under ideal conditions (e.g. sitting at my desk after a good night's sleep and not having to worry about "sailing" my house.) I usually find it takes me several tries to get a correct sight reduction. This more than offsets the inconvenience of having to derive an assumed position -- which, once you "get it," is not hard to do. Others may not share my arithmetic "challenges" and may disagree with that assessment. To each his own...
Ageton's was (probably still is) available in Bowditch (American Practical Navigator) in what used to be Table 35. (Vol 2, circa 1981) The numbers of the Bowditch tables changed at one point when they went from the two volume format to all one volume -- they have since changed back to a two volume format but I have not researched if the numbers changed again or if they changed back to the old numbers.
In the original Ageton's it is said that a solution may be in error by 1 to 2 nmi when the meridian angle approaches 90° (this is the same as a LHA of 90° or 270°) The original also recommends that you toss out sights where the K value (an intermediate value you derive during the process) approaches 90°. This is often called "The Forbidden Zone."
In an article written in "The Astronomical Society of the Pacific (vol 56 No.3222 p. 149) titled "The Accuracy of Ageton's Method in Celestial Navigation" the author, S. Herrick, points out that errors of up to 6 nmi can be expected in "normal" ranges and up to 30 nmi may occur when you enter the "Forbidden Zone."
If you look in the "old" Bowditch Vol II in the front of the book they have an "Explanation of the Tables" section that addresses each table separately. Bowditch seems to agree with Herrick and states in the explanations section that although these are tables are generally used without interpolation that when the meridian angle, R, or K get to within 2° 30' of 90° then "interpolation of the B value of angle R..." (intermediate numbers used in the Ageton's process) "...alone reduces the errors to about 2 nmi. Therefore when K falls within 10° of 90° it is best to discard those sights or to interpolate the B value of R." I am paraphrasing here, the exact quote may differ slightly. Another commonly stated "positive attribute" of Ageton's was that you normally don't need to interpolate, but now that is negated.
By my own inspection of the tables the need to interpolate the B value of R occurs about when the A value used to look up R is less than 666 making it easy to recall a trigger of when you really should interpolate.
If you are a little lost on the alphabet soup here -- well that is the way Ageton's goes... Lots of A's and B's and K's and R's etc. most of which have "rules" associated with them, and since they are just intermediate values on the way to a final solution it is impossible to have any intuitive sense as to whether you are on the right track or not until you complete the process -- and if you are me, come up with a nonsensical answer 3 times out of 4. It's not Ageton's fault -- the method does work. The fault is entirely mine, but it still happens.
Frankly I find using the Ageton's tables to be tedious, and for me, prone to many blunders. Now that you can get Pub. 249 (or Pub. 229) as a free download right from from this site I see no reason to recommend Ageton's for sight reduction, and certainly not for new students.
And I also agree that direct calculation with an electronic calculator is a better approach these days, but it isn't a bad idea to learn how to use some sort of printed tables, especially if you expect to sit for an exam at some point.
But as before -- to each his own.
RE: A Source of Confusion in Selecting an Assumed Position - P.Rutherford - 03-28-2023
Peter,
Your explanation(s) are most helpful. By providing the complex and simple way to get LHA, as you did, there's more clarity to the overall problem.
Have you ever tried the Doniol method?
Thanks again....
Paul
RE: A Source of Confusion in Selecting an Assumed Position - PeterB - 03-28-2023
Paul,
I am glad that my post has been helpful.
I have never used the Doniol method. Perhaps I will give it a try one of these days.
Peter
RE: A Source of Confusion in Selecting an Assumed Position - CarlosPindle - 03-30-2023
Hi All,
I'm confused by the result I got using Astron here- https://friendsofthevigilance.org.uk/Astron/Astron.html
Note- Astron DOES NOT use DR Latitude or DR Longitude. It was Assumed Latitude and Assumed Longitude.
I used the following figures;
Date- March 29, 2023
UTC- 1300
Ap Latitude- N 40°
Ap Longitude- E 060° 30'
Body- Sun UL
Hs- 15° 00.0
He- 3 meters
Ho- 14° 37.5'
Hc- 14° 11.3'
GHA- 13° 47.7' (that's fine)
LHA- 74° 17.7' (if I were using paper it would be 74°)
Now, the problem is when you click the button at the bottom of Astron "Add sight" and then "Log/Plotter" the "Ass Long" reads
E 060° 30'
Isn't it the Ap Longitude supposed to be 60° 12.3' ????
I have no idea why Astron uses a "decide for yourself" Ap longitude instead of using a DR longitude whole degree figure.
Can anyone explain why there's a difference between Astron an the way Peter describes it above?
Infinite thanks,
Carlos
RE: A Source of Confusion in Selecting an Assumed Position - PeterB - 03-30-2023
Carlos
The confusion is the mislabeling of Assumed Position in the app.
That should have been labeled either " Dead Reckoning Position" or "Estimated Position."
Since it is an electronic calculation it does not require a whole degree of Local Hour Angle as do the printed tables.
You are correct that for a TABLES solution you would used AP lon of 60 12.3' because when you subtract the 47.7' of the Greenwich Hour Angle from 60.0' that is the result. Appending this to the whole degree value of the DR longitude gives you AP lon 60 12.3' E
And that added to the GHA results in a LHA of 74 00.0' which is listed in the tables.
But the app doesn't require a whole degree LHA so when they add the 60 30.0' E to the GHA of 13 47.7 they get a LHA of 74 17.7' That is not a whole degree and so would not be listed in the tables -- but the MATH doesn't require a whole degree LHA to work.
When using that app you can put any lat and lon you choose in under their field for "Assumed Position" so most of the time I suppose you would just put in your DR position.
A fun and free app to use for practice and checking your work is Sight Calc. Again for this one you don't NEED to use a whole degree of LHA but of course you CAN for checking your tables work.
Let me know if this doesn't clear things up.
Peter