04-01-2023, 08:57 PM (This post was last modified: 04-02-2023, 12:21 AM by EdCa.)
Hi gents. Here's the answer to this confusing situation. It's much simpler than you can imagine. Pub. No. 249 Vol. 2 page. xi does a pretty BAD job of showing how to get LHA and the Ap longitude used for plotting. But, at least they have examples in Eastern longitudes!
I'll use the example in the attached sheet at the bottom of this post. It's from Pub. No. 249
First, we'll get LHA the easy way.
In Eastern longitudes get LHA as follows;
Body- Moon
DR. Longitude- E 7° 28'
GHA- 323° 37'
Make it easy on yourself. IGNORE the minutes of GHA and minutes of DR. longitude.
Add only the whole degrees of GHA and DR. longitude
323° + 7° = 330°
Then add 1° to the sum above; 330° + 1° = 331°
So, in this case- LHA= 331°
Try the other examples on that sheet and your answers for LHA will agree with theirs.
Second, we'll get the Ap longitude which is needed when plotting. Please understand that Ap longitude means, "Assumed position longitude".
In Eastern longitudes get Ap longitude as follows (using the same GHA as above).
Use only the minutes of the GHA and the whole degree of the DR. longitude
Moon's GHA- 323° 37'
DR. Longitude- E 7° 28'
Always subtract the GHA minutes from 60
60' - 37' = 23'
Next....combine the DR. longitude whole degree with the 23 minutes found just above to get the Ap longitude.
Ap longitude= E 7° 23' (I left off the 2 leading zeros)
Yes, you can perform all of the steps to get LHA and Ap longitude as they show you in '249, if you can possibly understand the awful way they format it and explain it, but I presented it this way so you wouldn't get confused.
Frankly, I have NO IDEA why they make this process so wretchedly difficult for us mortals to understand?!
The following is supposed to be funny-
If you can't figure out how to get the LHA and Ap longitude in Eastern Longitudes the solution is easy- don't sail there!
An example of when the “short cut” method of finding your LHA in Eastern longitudes gets it wrong and why you should not rely on it.
This is taken from my copy of Pub.249 Vol II, commercial edition, page xi (see attached scan of the page which includes the pertinent part of the computation of the LHA.) I could not find a publication date in the volume but it offers correction tables from 1981 to 2016. It would no have been in my possession (purchased new from Celestaire) before about 2000.
Their Example:
1978 Jan 1 moon by bubble sextant at Greenwich Mean Time of 00h 53m 45s DR position: 23° 42' S 113° 25' E
GHA moon for 00h 50m 00s: 300° 08' inc for 03m 47s 0° 54' GHA moon for 00h 53m 47s 301° 02' AP lon (add because east ) 112° 58' (also the long. used for your plotting) LHA 413° 60' LHA (carry the 60') 414° 00' subtract 360° -360° 54° 00' (the whole degree LHA needed for tables)
Here is the short cut method:
whole degrees DR longitude 113° whole degrees LHA 301° add 1 degree more 1° LHA result 415° subtract 360° - 360° LHA by shortcut 55° <= this is the wrong answer! 54° is correct.
If you plug 55° LHA into the tables, or app, or calculator, you will get results for Hc and Z, just not the results the examiner got. If you were sitting for a test you just blew it!
Not only does the short cut method give you the wrong answer it also fails to provide you with a starting point of longitude for your plotting. So aside from being wrong it is also useless for navigation.
Now this particular example also demonstrates the requirement (for exams) to derive your AP longitude to within 30' of your DR longitude. Here is how that all goes when done rigorously:
DR long. was 113° 25' E
Starting from the GHA moon for 00h 53m 47s found earlier: GHA moon 00h 53m 47s 301° 02'
Subtract just the minutes of GHA from 60'
60' - 02' 58' (this is what you append to your whole degrees of DR long.)
AP long. is now: 113° 58' E
Now we check if we are within 30' of our DR longitude.
58' - 25' 33' <= too big for Mr. Examiner!
So now we adjust the whole degrees of AP long by one degree to 112° 58' E
Check again to see if we are not within 30' of our DR.
we need 02' to get back to 113° E from 112° 58' and then we add another 25' from our own DR
02' + 25' 27' difference in long. which is OK for Mr. Examiner.
So the correct AP longitude to use in this example is 112° 58' E
Scroll back up to the beginning of the example to see that they used 112° 58' E as the AP longitude. This resulted in a LHA of 54° NOT 55° as would be given by thte short cut.
My advice is to forget about the short cut. It isn't that hard to do it properly and you will not only get the right answer you also get the AP longitude needed for plotting.
04-03-2023, 08:18 PM (This post was last modified: 04-03-2023, 08:45 PM by EdCa.)
Uh oh.....I better go back to the drawing board. I thought I had it figured out!
Will study the problem.
I'm trying to find in Bowditch or '249 an explanation for the necessity of finding Ap longitude within 30' of your DR. longitude. Consider also that your DR longitude (for various reasons...current, poor taking of DR)might simply be really wrong thus messing up the final LHA.
The only reason you need to be less than 30' difference in longitude from your DR is to please the examiner and to get the same answer as he did for Z and Hc.
If you pick the "wrong" AP long. and use the "wrong" long. from that AP the plotted LOP will still be valid and will still pass through your actual position.
But your answers for Z and Hc won't match the examiner's nor pass the exam because they are from a different AP than he used.
I'm sorry that this subject is confusing.... I'm here to help.
I reduced and plotted the LOP using your 249 commercial attachment. Yes, you're correct that the LOP's with my method and yours do pass over top of each other. The problem with my method using a LHA of 55° is that the intercept was 79' (l° 19') in length which is way too long. My guess is a navigator looking at that LOP would say, "let's try a different LHA....how about 54° ?" (that will provide a "closer" intercept and not such a huge one).
Another idea I had to obtain the LHA and Ap longitude, and tell me if this isn't correct, using your example of DR longitude of E 113° 25' and Latitude of S 23° 42'
To get the Ap longitude using GHA 301° 02'
60 - 2= 58'
The value of 2' reduces your DR longitude from the whole degree of 113°. So the DR longitude (forgetting the minutes of arc) becomes 112° 58' which now become the Ap longitude.
04-07-2023, 09:58 PM (This post was last modified: 04-07-2023, 10:34 PM by EdCa.)
I've got an easier way to get LHA....just for us simple folks.
Get an approximate LHA using the "simple method"- one that's either plus or minus 1°. Compare the uncorrected Hc figure found in Pub. No. 249 for the correct Latitude and Name. The desirable LHA is the one closest to your Ho.
Using that method the 54° LHA result appears as the closest to Ho.
From the 2023 Air Almanac are examples of Eastern longitude problems. See the attached screen shot at the bottom herein.
Look at the examples for only Aries and Aldebaran.
In these cases you have to add an extra whole degree to the Ap longitude so that when comparing your DR longitude to your Ap longitude the LHA, when entering '249, results in a shorter intercept. Whereas in PeterB's example from Pub. No. 249 you are decreasing LHA by one degree which for the same reason of will give you a closer intercept.
For me the solution to understanding this was simply doing a reduction of PeterB's '249 attached page and plotting the LOP using a LHA of 112° (the correct method) and my "simple method" of using a LHA of 113°.
My intercept was way too long- 79' or so. I could've seen there was a "too long intercept" problem when comparing the uncorrected Hc to Ho in '249.
I am pleased that you actually did the sight reduction and plot from both AP's to see that the resulting LOP's are both valid. It is one thing to be told that is so and another to see it for yourself and to report your results to others. Thanks for that.
There are two other minor points that I'd like to add at this stage:
ONE:
The "...less than 30 minutes of longitude from your DR..." requirement also exists when working exam problems in Western longitudes. All for the same reasons -- which is to get an answer that matches the examiner's.
TWO:
In actual navigation you may have a DR that is way off for various reasons. In that case you could easily end up with an AP far from your actual position resulting in long intercepts. That's OK. If you find yourself in this situation -- and you are absolutely sure you didn't blunder in your sight reduction process -- simply use the new "fix" as an updated DR. Derive a new AP from that point and re-do the sight reduction from there. The updated fix obtained by this method probably won't be different from the first one by more than a handful of miles.
Is the process the same for getting a good Ap λ in Western longitudes? That is- subtract the GHA minutes from 60? (and subtract 1° if there's over a 30 minute difference?)
04-19-2023, 09:32 PM (This post was last modified: 04-19-2023, 09:35 PM by PeterB.)
Paul,
The short answer is "No, the process is not the same in western longitudes." The good news is that it is some what easier to do and less confusing.
The Greenwich Hour Angle (GHA) of the body is how far west the body's geographic position (GP) was from the Greenwich meridian at the time of the sight.
Your Dead Reckoning (DR) longitude is (approximately) how far eastward or westward you were from the Greenwich meridian at the time of the sight.
The Local Hour Angle (LHA) is how far westward the body's GP was from your longitude, or from your nearby chosen AP's longitude.
All "Hour Angles" of any sort are always only measured westward from their starting point all the way around to 360°
If you are east of Greenwich you have to add your eastern longitude to the GHA of the body to compute how far westward the body's GP was from yourself. Additionally, just for using tables or taking an exam, the LHA result has to come out to be a whole number of degrees. To work that out we first find the GHA of the body at the time of the sight. Then we subtract the minutes and tenths of the GHA that we found from 60.0' to find a value for the minutes and tenths to append to our DR whole degree of longitude to create an AP longitude that will make the addition and the resulting LHA come out to be a whole number of degrees. We know full well that we are not at the AP longitude that we then use. We only contrived it in order to make the tables work -- or to get the same answer as on the exam.
When you are in western longitudes you will be subtracting your longitude from the Greenwich Hour Angle to determine how far westward the GP of the body was from you at the time of the sight. Once again, even in western longitudes, for using the tables or taking and exam, you need the result to come out to be a whole number of degrees. But this time all you need to do is to take the minutes and tenths of the GHA as you found it and append that value unchanged to your whole degree of DR longitude to contrive your AP longitude. Now when you subtract the longitude of your AP position from the GHA of the body to find your LHA the minutes and tenths will disappear to 00.0' leaving you with a whole degree result.
The final check (only really important for taking exams) is to confirm in either case that your resulting AP longitude is within 30.0' of longitude from your DR position, and if not to adjust the whole degree value of your AP by one degree while leaving the minutes and tenths alone so that it is -- just so your results will match the exam question answer.
Hope this helps -- let me know if you have more questions.
The final check (only really important for taking exams) is to confirm in either case that your resulting AP longitude is within 30.0' of longitude from your DR position, and if not to adjust the whole degree value of your AP by one degree while leaving the minutes and tenths alone so that it is -- just so your results will match the exam question answer.
I understand about Ap and LHA in Western longitudes and because of your excellent explanation above, about Ap and LHA in Eastern longitudes and how to get it for exams.
But, it's the Western longitudes I'm now curious about and the need to be within 30 minutes of DR λ and Ap λ to subtract (or add?) 1°.
An example of finding the AP longitude in western longitudes including adjusting the AP lon to be within 30.0' of the DR lon.
A sight of the sun's lower limb was taken on July 15, 2022 at corrected watch time time of: 15h 01m 53s
DR position at 15 00 hrs was 37° 47.5 N 123 26.2W
The local time is Pacific Daylight Time so it is +7 hours to get UTC
UTC time was 22 01 53 on July 15, 2022
GHAsun for 22 00 00 148° 29.3' Inc for 00 01 53 1° 28.3' GHAsun 22 01 53 148° 57.4' <= the minutes that you append to the DR lon for the AP lon. AP lat 38° N AP long 123° 57.4' <= subtract to get LHA LHA 25° 00.0'
Now we check to see if the AP longitude is within 30.0' of longitude to the DR
AP lon 123° 57.4 DR lon 123° 26.2 <= subtract lon. diff. 0° 31.2' <= too big for the right answer on the exam !
So now we adjust our AP lon by 1° to bring the AP lon to within 30.0' of the DR and check again
DR lon 123° 26.2' Updated AP lon 122° 57.4 <= subtract lon. diff. 0° 28.8' <= less than 30.0' and what the examiner would use
Last adjust the LHA
GHAsun 22 01 53 148° 57.4' AP lat. 38° N AP long 122° 57.4' W <= subtract to get LHA LHA 26° 00.0' <= the LHA the examiner would use
Remarks:
If you did the entire sight reduction all the way through to plotting with 25° LHA instead of 26° you would still get a useful line of position, however your intercept and azimuth would not match the exam answers.
Also with slightly different numbers you might have to add a degree to the AP lon. to make it work out to less than 30.0' longitude from the DR position.
Ok, that makes it perfectly clear. I see it's a very simple process. It's interesting that I have never seen any direct reference to either the Eastern or Western Ap λ adjustments other than adding (or subtracting) DR λ to/from GHA.
It's alluded to in HO. 249 for the Eastern longitude but nowhere else. Strange....very strange.
Thank you for taking the time and effort in explaining this topic of confusion. Well done.
This is a very interesting topic and I've enjoyed watching the development of it, especially as PeterB explains the procedure.
For curiosities sake, I have several questions related to determining Ap λ in Eastern longitudes and am hoping you men can answer them.
1- Your DR λ can be way off in a bad case by several degrees because of weather, current, sea sickness, etc. So when you finally get the chance to make an observation you'll see your DR λ and resulting Ap λ (and LHA for the sight reduction to get Hc) are far away from the LOP. So, lets say your Ap λ shows that you're 3 degrees (180 nm) away from the LOP. Would you then adjust your Ap λ (and then LHA) complete the sight reduction process again to obtain an intercept that's much closer to the LOP?
2- Exam questions and the "right answer". Getting the right answer based on a difference of less than 30' is fine but in reality it's unimportant....unless the test requirement is to have an Ap λ that's 30' or less from the DR λ. If that's the case then I agree. In looking at the literature available, no one is observing this requirement or even cares about it. But, for a simpler and shorter intercept the 30' minute or less necessity is a very good idea. When using Ho. 249 we plot from whole degree of latitude and no minutes.
You most certainly can wind up with a DR and resulting AP that are a long way from where you actually were. However the resulting LOP is not entirely "bad" although it is not as "good" as it could be.
The azimuth line through the AP toward or away from the geographic position (GP) of the observed body is actually along a great circle track. The compass direction along great circle tracks changes constantly, so a very long intercept (a) implies that the azimuth line plotted on your approximately Mercator plotting sheet should show as a slight curve. Usually the amount of this curvature is very slight because we deliberately keep the intercept short but in an example such as you have suggested with 3 degrees of longitude difference it should curve to a small extent.
Also the Sumner line we plot as a LOP is really a circle of position, though it is usually not a great circle - so if your final fix is way out on one or more of your Sumner lines then this too will exhibit small errors because it really should be a curve. This situation is more acute for high observed altitudes since the true circle of position would have a smaller radius and exhibit more curvature.
When any of your plotted lines start to look extra long then it is good procedure to take the resulting fix as an updated DR. Redo the AP and sight reductions from that spot. Chances are the updated new fix will only differ from the first by a handful of miles, but it should be more accurate.
Since the math doesn't care where you say to start from there is no absolute requirement to have your AP long. within 30' of your DR. To say that it MUST BE SO would be incorrect. However on exams where you have to MATCH the examiner's answer you have to start from the SAME AP that the examiner used - and that is pretty much ALWAYS within 30' of longitude from the stated DR. If you use a different AP than the examiner used you will get a different azimuth and intercept than he did.
That makes perfect sense and I now understand that it's a good idea to have a shorter intercept for practical reasons. Of course, getting the correct answer on an exam is more consistent with good practice.
