But, here's the confusion, LHA in Eastern longitudes is supposed to be rounded up or something like that and I'm not really sure if it is supposed to be rounded up. Information on the web is vague and not explained well enough making too many assumptions. Another possibility is that very few who speak English are sailing, doing CN, in the Eastern hemisphere.
So, my estimate to solve the problem is simply this (tell me if I'm wrong)
But, here's the confusion, LHA in Eastern longitudes is supposed to be rounded up or something like that and I'm not really sure if it is supposed to be rounded up. Information on the web is vague and not explained well enough making too many assumptions. Another possibility is that very few who speak English are sailing, doing CN, in the Eastern hemisphere.
So, my estimate to solve the problem is simply this (tell me if I'm wrong)
The problem above could be easily solved like this- LHA= 336° + 25° + 1° (minus 360°) = 2°
Any ideas about this?
Greetings, CelNav57
I Think you are right. I would just make LHA a round number by making assumed lloongituda 26 deg. Therefore 336 +26 =362. 362-360=2 deg. I saw this method ina few books. When you take 25 deg instead of 26 , you make LHA smaller, because to round 25.45 to 25 is much more "rounding" than from 25.45 to 26.
Anyway, your assumption is correct the way I see it. Again, I saw the examples of calculating LHA in Lat.E and it is exactly how it was done. And, by the way, you may use LHA of 1 deg 45.7 for your calculations but then you need to go through correction tables which anyway will round up the final result.
This is the way I see it.
Thank you!
This little attachment shows much more clearer what I meant, that my previous post ;>
But, here's the confusion, LHA in Eastern longitudes is supposed to be rounded up or something like that and I'm not really sure if it is supposed to be rounded up. Information on the web is vague and not explained well enough making too many assumptions. Another possibility is that very few who speak English are sailing, doing CN, in the Eastern hemisphere.
So, my estimate to solve the problem is simply this (tell me if I'm wrong)
The problem above could be easily solved like this- LHA= 336° + 25° + 1° (minus 360°) = 2°
Any ideas about this?
Greetings, CelNav57
I think you are right. I would just make LHA a round number by making assumed lloongituda 26 deg. Therefore 336 +26 =362. 362-360=2 deg. I saw this method in a few books. When you take 25 deg instead of 26 , you make LHA smaller, because to round 25.45 to 25 is much more "rounding" than from 25.45 to 26.
Anyway, your assumption is correct the way I see it. Again, I saw the examples of calculating LHA in Lat.E and it is exactly how it was done. And, by the way, you may use LHA of 1 deg 45.7 for your calculations but then you need to go through correction tables which anyway will round up the final result.
This is the way I see it.
Thank you!
This little attachment shows much more clearer what I meant, that my previous post ;>
That was my conclusion as well. I was watching a video on youtube by the eminent sailor and USCG officer, Chris Nolan, in which he explains about LHA in Eastern longitudes. LHA eastern longitudes
In the video he says that LHA must add up to a whole degree and he explains the procedure. To me it seems simple;
Add the GHA whole degree (no minutes) and add the Ap longitude whole degree (no minutes) then just add 1° to the result. Simple. Even still the St. Hillaire method will get you the same basic LOP.
Oh, below is a quote I found on Chris Nolan's site, PracticalNavigator.org and look who said it!
“Whenever I meet with the words 'Thus it plainly appears,' I am sure that hours and perhaps days of hard study will alone enable me to discover how it plainly appears.”
11-30-2022, 01:20 AM (This post was last modified: 11-30-2022, 01:34 AM by P.Rutherford.)
I'd like to know why there's so much difficulty placed in determining LHA in Eastern longitudes when all you do is add the whole degrees and then add 1 degree to it? Chris Nolan's many excellent videos, as Craig mentions, goes through a procedure which seems superfluous and unnecessary in getting LHA. Since the minutes of apL are ignored in getting the result why even bother- just add.
Bowditch in Chapter 19- Sight Reductions page 313 reads;
LHA (Local Hour Angle): The LHA is the hour angle of the observed body at a λ . The LHA is GHA - a λ , for west
longitudes and GHA + a λ for east longitudes. Note that this should be a whole degree, else you have chosen the a λ incorrectly.
12-31-2022, 07:39 PM (This post was last modified: 01-01-2023, 05:37 PM by PeterB.)
East Longitude Hour Angles Reply
To find a Calculated Altitude (Hc) and True Azimuth (Zn) to compare to our observed altitude (Ho) we need to solve the navigation triangle.
We also need the exact spot from which we solved the navigation triangle to use as a starting place for plotting. The above short cuts might give you a useable LHA but they do not address the question of exactly where you begin your plotting from.
To solve the navigation triangle and get a useable line of position (LOP) we need to know three things.
ONE: The declination of the body at the moment we took it. This is the latitude of the body's geographic position (GP.) We look this up in an almanac.
TWO: Our approximate latitude. It doesn't have to be exact – that error will be resolved in the plotting process.
THREE: The angle between our longitude and the longitude of the GP of the body at the time of our sight measured in a westward direction from our meridian. This is the Local Hour Angle (LHA.) The LHA is dependent upon the body's Greenwich Hour Angle (GHA) at the exact moment of the sight, which is how far west it was from the 0° meridian, and our approximate longitude. Once again our approximate longitude doesn't have to be exact because the errors will be resolved in the plotting process.
The declination we use is always the exact degrees, minutes, and tenths as found in the almanac.
If we use direct computation with a scientific calculator, or one of the many electronic apps available, then the latitude and LHA can also be whatever numbers in degrees, minutes, and tenths that we derive using dead reckoning (DR.)
However, to make look-up tables that were compact enough to carry aboard even a small vessel the publishers omitted everything except the solutions computed for whole degrees of both latitude and LHA. This eliminated the nearly infinite number of solutions that exist between those whole degree demarcations and greatly reduced the size of the tables. As previously stated the fact that the position used to enter the tables wasn't the exact position of the vessel, only a position nearby, automatically gets accounted for when we get to the plotting.
In order to use the tables the navigator “takes up” a position that he knows is not his DR position nor is it his actual position. Instead it is one that (hopefully) is fairly close to where he thinks he is and which is contrived to work with the tables. This is called an “Assumed Position” or AP. In this case “assume” means to “take up” as in the sentence “She willingly assumed the responsibilities of Captain when she took command.”
Many generations of navigators would have been saved a great deal of confusion if they had named this position something else, such as “Starting Point for Calculations” or “Trial Position” but they did not, and so it is often confused as meaning the same thing as a DR position – which it certainly does not mean at all.
The characteristics of the AP is that it must have a whole degree of latitude and a longitude that results in a whole degree of LHA when combined with the GHA of the body. A lesser requirement is that it should be close to the DR position, but if your DR is off and you eventually find yourself somewhere far away from the DR it won't matter.
We start with the latitude. We simply pick the nearest whole degree of latitude to our DR position and use that without any minutes or tenths.
To find the LHA when in in western longitudes we subtract our longitude from the GHA of the body. The result is how far westward the GP lies from our meridian. We want a result with no minutes or tenths, so whatever the minutes and tenths are of the GHA of the body we append that to the whole degree of longitude of our DR. The result will be LHA with no minutes and tenths after we do our subtraction.
example: GHA 24° 31.2' DR lon 10° 22.0' W we use an AP lon 10° 31.2' W
Now 24° 31.2' – 10° 31.2' W = 14° 00.0' LHA which we will find listed in the tables.
We are fully aware that we are not at long 10° 31.2 W It is just a contrivance.
To find the LHA when in eastern longitudes we add our longitude to the GHA. The result once again is how far westward the body's GP lies from our meridian. In this case we need the addition of the minutes and tenths to come out to be exactly 60.0' which is one degree. We find out what the minutes and tenths of our AP will be by subtracting the minutes and tenths of the GHA from 60.0' This is the number we will append to our DR longitude's whole degree value to get our AP.
example: GHA 18° 54.2' DR lon 20° 12.1' E
we start with 60.0' – 54.2' = 05.8' so the 05.8 is what we need to append to our whole degree of longitude to get an AP long of 20° 05.8' E
Now when we add the GHA and the AP lon we get a whole degree:
18° 54.2' + 20° 05.8' = 38° 60.0' => 39° 00.0' LHA which is listed in the tables.
One wrinkle: In some cases your AP longitude may work out to be more than 30.0' of longitude away from your DR position. In actual practice this will not throw off your navigation any significant amount. However, some examiners may mark you off for not selecting the AP longitude such that the AP is less than 30.0' longitude from the DR position. Don't fret – there is an easy fix. Find the AP just as stated above. If the resulting AP is more than 30.0' from the DR simply adjust the whole degree value of the longitude of the AP by 1° while leaving the minutes and tenths alone. Then check that you are now within the 30.0' of longitude to the DR – all this just to please the examiner.
example: GHA 22° 58.2 DR lon 10° 46.0 E
60.0' – 58.2' = 01.8' so the AP lon is 10° 01.8' E and the LHA is 33°
But this is 44.2' of lon from the DR. Although it will work just fine the examiner might not like it. So add 1° to the AP lon to make it 11° 01.8' E. Now the AP lon is 15.8' from the DR lon and update the LHA to 34° From a practical standpoint it is unnecessary but it makes the examiner happy and it shows that you have some insight into the process.
One last comment: If you have picked a DR and subsequent AP that results in very long intercepts to a fix far from your DR double check all your work. If you are sure that you have not made a mistake in the sight reduction process re-work the sights using the new fix as an updated DR assigning new values to the AP lat and lon. Chances are the updated fix won't change from the first one by more than a handful of miles. Moitessier, the famous solo circumnavigator, wrote in his book “The Long Way” how he had experimented a few times with AP's that were deliberately hundreds of miles away from his DR and that in no cases did he have to iterate more than twice to get a good fix.
Also as a confidence building exercise you should try deliberately plotting a set of observations from a “good” AP and from an AP with the “wrong” latitude or longitude by a whole degree on the same practice sheet. You will find that either way the resulting fix is pretty much the same. That is good to know if you are running out of space on your plotting sheet and still want to fit the observations onto it.
