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/sho...hp?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/sho...hp?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%20vide...uction.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

Hello,

Delighted to be accepted as a member of this exclusive forum.

I have been boating for more than 45 years and have decided it is high time I learned the art of Celestial Navigation. Having studied 2 books Tom Cunliffe and Mary Blewitt I feel I have a reasonable understanding of the subject, however both books use tables (AP3270) to determine the Intercept and to continue my studies I would like to be able to calculate Zn and Hc directly using a scientific calculator.

I turned to youtube and found Cram Daily PH

Formula for missing side

cos xz = cos px  cos pz  +  sin px sin pz cos P

After much ado I discovered how to use this formula and arrived at the same answer as the example given.

Formula for angle

cos Z = - cos P cos X + sin P sin X cos Px

Not attempted this formula yet 

I then discovered Chris Nolan

Formula for side

sin Hc = sin L sin D + cos L cos D cos LHA

Due to my success with the previous formula I had little trouble with this arriving at a very similar answer to the example given although I tried it to a different number of decimal places on the calculator which reveals quite different results

Chris Nolans formula for Z

cos z = sin D - sin L x sin Hc / cos L x cos Hc

Not tried this yet

I am at a very early stage regarding studying this direct method and hope to learn more from this forum.

If anyone is interested I can post my exact workings for the above

I do have a specific question but I guess this is enough of my ramblings for now.

Mike

I'd like to set up a UPS based on a distance of 60 seconds and am having a difficult time of it.   It's probably because I've been working at it too long and it's late!

Here are my assumptions;

- The scale for minutes in the middle of the UPS can also be considered to be just 60 seconds.
- In setting the correct meridian distance for say, N 35°, is done exactly as one would set the meridian distance for 35°.

Any ideas about this?

Thanks,

Craig

Dear member of the forum,
Does anybody know where in the USA I can send my sextant mirrors to be re-silvered?
My mirrors are starting to look a little sad.
Regards,
Steamburn

I've just added a new page where you can get everything you need for 2024 Celestial Navigation.


Click here- Everything you need for 2024


Almanacs
2024 Nautical Almanac- regular format
2024 Nautical Almanac- compact format

2024 Sun only- regular format
2024 Sun only- compact format


Sight reduction
PUB. NO. 249 (download individual Latitudes or Volumes) (Vol 1 courtesy of Celestaire) Epoch 2025
PUB. NO. 249 (download individual Latitudes or Volumes) (Vol 1 courtesy of Celestaire) Epoch 2020
PUB. NO. 229 (download individual volumes)

Stars
2024 Stars- SHA & Declination


Corrections
Increments & Corrections Table (yellow pages)
Increments & Corrections Sun only on 2 pages

ALTITUDE CORRECTION TABLES 10°--90°—SUN,STARS,PLANETS
Altitude Correction Tables for the Moon
Polaris- Correction for (Q) 2024


Conversion of Arc to Time
Meridian Passage and Declination of the Sun at 12 h UT
(from The Air Almanac 2024)
Equation of Time curve for the Sun


Astronomical Phenomena


Day of week, Week Number and Day of Year for 2024
ECLIPSES- Solar and Lunar- 2024
Visibility of Planets, Moon phases and Select Stars
(from The Air Almanac)
Moon Phases for 2023 in graphic form (GMT/UTC time based)
WORLD MAP OF TIME ZONES


The USNO hasn't yet released the Astronomical Phenomena for the year 2024 in pdf

I'm at a loss to understand how to calculate LHA while sailing in Eastern longitudes.

What I know;

LHA in Eastern longitudes=  GHA + ApL (Assumed position longitude) (minus 360° if necessary)


Example;

Date-  November 25, 2022
GMT- 10:10:59

ApL=  E 025° 45'

GHA= 336° 00.7'

LHA=  336° 00.7' + 25° 45'= 361°45.7' - 360° = 1° 45.7'

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)

Eastern longitudes, LHA= GHA (whole degree's) + ApL (whole degrees) + 1

The problem above could be easily solved like this-  LHA= 336° + 25° + 1° (minus 360°) = 2°

Any ideas about this?

Gentlemen, 

maybe something of an interest for you. I  kinda wrote a formula  which  maybe  useful.  MAybe;>  It can  be used to get a running  fix from bow  and beam relative bearings. It is well-known  procedure.  But  to speed up  calculations,   there are so-called method  of double angles and 0.7, and also distance travels equalls distance off.  Plus a few so-called special  angles , like 36 deg  bearing1 and 69 deg.  bearing 2.  In the last,  particular case distance travelled is equal  distance oiff the beam to  the object you are  taking bearing on.

Anyway, instead of using  just a few angles  why not  to use any angle at the first  bearing  and calculate what  angle should be  at the second bearing  if  you want  to  have distance travelled between bearings to  be equal  to  distance offf beam of that specific  object you took  bearings at.

a very simple drrawing is following.( see attachment).

Again,  nothin  revolutionary,  just maybe helpful.  Thank  you

Attached Files

Thumbnail(s)
   

Hi
 
I'm interested in the history of star finder models as an aid to celestial navigation and have a small selection of plan models (list below). My favourite is the Italian Sferoscopio del Pino (image enclosed). Earliest found to date by Rude 1921 leading to the 2102-x models, the 2102-D and British Admiralty NP323 still current.
 
I've recently extended to researching the celestial starfinder globes which predate the plan models  by some margin into the 19thC. Russian and Chinese models made up until the 70s at least come up on eBay. Freiberger also still offer one new. Image of my Russian 1978 example shown.
 
Although there are plenty of resources describing current use of the plan models, notably the 2102-D with David Burch's 'The Star Finder Book' recently into a 3rd edition 2019, have found no reports of the globes being used for navigation, past or present. The globes also have many more stars than the almanac tabulates (the Russian has 160, almanacs typically 57) and unclear how useful these extra stars would be without supporting tables.
 
Can anyone comment on how widely used the globes were or are being used.
 
Thanks.
 
David

Star finder models owned.
2102-B
Sferoscopio del Pino 1971 facsimiles of the 1937+ model (inserts in pocket of book 'La Navigazioni Astronomica' by Mario Sacchetti (1971) to support exercises)
USAF CP-300/U
2102-D vintage and current Weems and Plath
British Admiralty current NP323
Russian celestial star finder globe model 3Г 6.6 inches diameter.

April 30, 2022

Here is a link to a popular and reputable supplier of celestial navigation equipment having a sale on the simple but remarkable Davis Mk3 Lifeboat sextant.  I have no affiliation with the seller or Davis whatsoever other than to have purchased a few items from each of them over the years.

These simple MK3 instruments are perfectly adequate for real navigation. They make a great first sextant due to their simplicity and low cost -- yet you can do some real navigating with them and expect perfectly acceptable results. IMHO they are a better choice than the more expensive plastic sextants that are made to look like "real metal sextants." Those fancier plastic sextants are notorious for never holding their adjustment from one shot to the next in a round of sights which only frustrates beginners. The humble Mk3 doesn't suffer from that shortcoming.
 
If later on you decide to treat yourself to a "fancy metal sextant" you will still keep your MK3 forever because The MK3 is also great for backup; emergency; teaching; and for use on days when the spray is flying and you would prefer not to expose your fancy metal sextant to a salt water dousing.

In fact even if you already own a "fancy metal sextnat" you should consider buying a Mk3 as a backup for those reasons stated above. I did and I am very happy I did so.

On the other hand if you decide that celestial navigation is not for  you, you will be able to sell the MK3 used for about what you paid for it. Or you can just give it to a friend or stash it in the ditch bag. It is a rare "fancy metal sextant" that sells as used for about its original purchase price because people are wary of any unknown history of the instrument and possible hidden defects.

As for myself I started out with an (expensive) fancy metal sextant and turned my nose up at the MK3, although it was suggested to me to start with that model.  Many years later (and a few more "fancy metal sextants" in my collection) I bought a MK3 used.  I can't say enough good things about it. It is one of my favorite sextants. Sure, I'm not going to do lunar distances with it, but I can get position accuracy to within 2 nmi fairly regularly -- and as I said, no worries about eager students dropping it or salt spray ruining it.  In fact just recently I was able to bring down Venus in daylight for practice with the MK3 when my fancier Tamaya failed due to a much smaller field of view.

I love this thing even if it is humble and simple. I will keep mine forever for the reasons stated above, and if it gets somehow destroyed I'd buy a replacement in a minute. I think you would love it too if you would give one a fair try.  In fact the biggest down side to the MK3 is the ribbing you are likely to get about your "inadequate" or "toy" sextant from wanna-be navigators and arm-chair pirates who simply do not know any better.  But after you go out and do a round of sights with it you will be convinced  that they are only displaying their lack of experience.

I have seen it stated in reviews that occasionally the non-adjustable horizon mirror may not be perpendicular to the frame. To check for perpendicularity of the horizon mirror cut about 2mm off the corner of a paper business card at a 45° angle and gently present the card to the mirror as you would a carpenter's square. You need to cut the corner off the business card because there is a tiny plastic ridge at the base of that mirror where its mount meets the frame and  you must create clearance for this ridge. If your horizon mirror is not square you must request a replacement instrument -- it can not be adjusted.  Meanwhile expect to see some of the frame in the right hand side of the horizon mirror even if it is properly perpendicular to the frame. That is normal.

https://www.celestaire.com/product/davis...3-sextant/

A digital sextant. Very interesting.
No sight reduction methods or almanac required.

They claim it can get you a line of position from a single shot of the sun. Or it can give you a position fix with two bodies, or a running fix from two sun sights take some time apart.

What I don't' see is an input for a dead reckoning or assumed position, so how can it work?

A built in ephemeris would give it the geographic position of the observed body and the digital observed altitude would give it a zenith distance, so it would have a circle of position -- but it still needs an azimuth or you could be anywhere on that circle of position.

It must have a flux compass to get an azimuth to the observed body but it doesn't mention it.

Link is below. Skip ahead to about 1:52 to see the digital part described.

I would be curious to hear what others think.

https://www.youtube.com/watch?v=2C9-LMNNzH4