Gentlemen,

Quite recently  I've identified little (  maybe ;>) problem, I ignored for a long time. Let  me make it short and concise:

Working on LOP  for a star.  To find LHA to use in Pub.229, vol. 3  input,  I need to add GHA Aries to  SHA of the star and subtract DR longitude.  Well, as you know, to get  the whole number for LHA I need to  use the assumed longitude.  
And this is where I  found a little problem.:

Depending on the order of adding and subtracting,  the  magnitude of assumed longitude  is changing.    In minutes, of course.  But still is changing.

The example:

 GHA Aries-  212-05.7
SHA of the star 146-09.1
Dr Longitude- 41-50w.

Now,  to get LHA  I add  212--05.7 + 146-09.1= 358-14.8

LHA of the star= 358-14.8  - 41-14.8= 317.

Assumed longitude is 41-14.8.

But,  if  I do  it slightly different,

GHA Aries-  DR  Assumed longitude= 212-05.7  - 41-05.7= 171
171 + SHA of the star = 171+ 146-09.1= 317-09.1.=LHA.  I can round it to 317.

  But in this case  assumed longitude is 41-05.7.  Not 41-14.8  like in the above.

And my question is: What assumed longitude to take? Because the difference is NOT zero.

Please, enlighten me.

Thank you.

This might be "before your time" but I watched Camp Runamuck as a kid.  It was hilarious!

I came upon this happy scene;

Camp Runamuck


Craig

I used to use a small-ish TIMEX IronMan watch that had a metal band.  It was easy to read and served well but the band had a tendency to tear out of the lugs on the watch case, which was resin, after awhile.  I learned to live with it and to keep a spare on hand since they were not expensive.

A few years ago they discontinued that model so I was on a quest to find a suitable replacement before my last IronMan (of that specific model) gave up.

I must have looked at dozens of potential replacements. Casio is a very popular brand and makes a number of models that have the features I wanted:

Easily readable digital display to the second
Day and Date
stop watch
count down timer
back light
water proof sufficient for showers or swim, or occasional inadvertent dunking - not a dive watch
Attractive and unobtrusive

There must be at least a half dozen Casio models that look nearly identical to one another with various sorts of bands, face sizes, graphics, and lighting.  According to the reviews that I found most of them have inferior lighting. Also I did not want a plastic band as that is likely to break after a few months and I find them unattractive.  I finally decided to take a chance with:

Casio Model 3640 WD-1A  Men's silver watch with stainless steel band.

I have been wearing it for 20 months. It is holding up very well. The back light actually works as it should (unusual with many Casio similar models.)  I was concerned that the face (crystal) would scratch easily because it is quite exposed, but it has not, and this watch gets tough daily use.
It is attractive and not overly large so it would be equally at home on a lady's wrist as a man's.
It was not expensive so if you were really wanted to have reliable time off shore you can easily afford a spare to stow below. (The rate may be different when stowed than it is when you are wearing it on your wrist so you need to check each watch both ways)

Its rate on my wrist over the past 5 months has been 0.0186 seconds/day or approximately 54 days to loose a single second. Impressive!

If  you are seeking a timepiece for cel-nav this one is a good choice.  Be sure to copy that model number fully to get the correct version.
 
I have no affiliation with Casio or any seller of watches.

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Today is May 5th 2023 so this seems like a good time to post the following

A long while back on an obscure forum that I can't seem to find again I found this response to a poster seeking help on a rating exam question:

If the question mentions Arcturus the answer will be A
If the question mentions Denebola the answer will be D
And if the question mentions May 5th  the answer will be C  -- as in "Cinco de Mayo"


PeterB

USCG Exam Question Sidereal vs Solar (Tropical) Year

I was poking about and came across a USCG practice exam here:

https://www.dco.uscg.mil/Portals/9/NMC/pdfs/examinations/q154_nav_problems-oceans.pd

Question 8 on page 3 asks:

8. The Tropical year differs from which year by 20 minutes?
A) astronomical
B) sidereal
C) equinoctial
D) Natural

The correct answer is B) sidereal

To start off with I'm not even sure why they would ask this question. I don't know of any case where such knowledge would be useful to a navigator or required for his daily execution of his duties. I also do not recall it being taken up in any of the many texts on celestial navigation I have studied. If anyone else knows of a good example of when such knowledge would be useful or required please let me know.

What I do know is that the sidereal day is 4 min shorter than the mean solar day and this is information that is useful to celestial navigators. Particularly for planning meridian sights in advance or for general knowledge of what is happening in his sky day-by-day.

So it seemed to me that after 365 1/4 days of a mean solar year that the sidereal time reckoning should be 24h 21 min ahead of the solar time reckoning. So what happened to the extra day?

As it turns out the sidereal year is 20 min longer than the mean solar year. Some of that has to do with precession. I have not gotten into that part and I do not intend to, but I was still scratching my head on the whole missing day.

I think I have it worked out now, at least schematically. See the attached illustration of a planet that circles its sun in 4 solar days and how, during those 4 solar days, it actually experiences 5 sidereal days.

Still I wonder where is this curious bit of trivia useful to a navigator? Any ideas?

PeterB

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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!

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  PubNo249Vol2edit.pdf (Size: 37 KB / Downloads: 322)

Carlos Pindle had mentioned he did not understand why the Astron app allows you to use a "decide for yourself" latitude and longitude for an assumed position.

I will try to shed some light on that subject if I can.

When we derive a calculated altitude (Hc) for an  observed body what we are actually doing is finding a proxy value for the distance along a great circle from that assumed position (AP) to the geographic position of the body. The reason I say it is a proxy distance is because the actual distance from or AP to the GP of the body is found by
 (90° - Hc) = zenith distance.

The reason for using the altitude above the horizon is simply because it would be very difficult to base our observations from a selected spot on the celestial sphere that was directly over our heads (our zenith.) But since we know that the total distance from our zenith to the true horizontal is always 90° it is easy enough to figure out the zenith distance.

We take a measurement of how high a body appears to us above the horizon, which is a little below true horizontal because the Earth is sphere and the surface falls away from us in all directions.  We correct this for that dip of the horizon and for atmospheric refraction, and for some bodies parallax and semi diameter.  We now have Ho.   If we subtract Ho from 90° the result will be how far we are from the GP of the body expressed in degrees. Converting this to all arc-minutes converts that number into nautical miles.

Now suppose we wanted to know how far it is by great circle from New York to London?  It turns out we use the same math for that calculation. We input an AP near to New York that either IS the EXACT lat and lon of New  York for calculator methods, or we derive an AP nearby to New York as our starting position (usually called "the departure"
but I prefer "starting point") and we put in the location of London (usually called The Arrival, but I prefer "the end point") using its latitude in place of declination and its longitude as a Greenwich Hour Angle.  Then we do a perfectly run-of-the mill sight reduction.  The Zn is the initial great circle course (it changes throughout the track) and (90° - Hc) = the great circle distance.  Because it is a great circle every arc minute is one nautical mile, so converting Hc into all arc-minutes gives the great circle distance in nautical miles.

Now if  you are still with me consider the following:

Suppose you knew of a reef that  you wanted to give plenty of berth? Or maybe a restricted zone you did not want to enter?   If you can pick ANY AP you want you can pick a point on the reef's edge or on the demarcation line of the restricted zone and using that IN PLACE OF a "normal" AP you can get an HC from that spot to compare to the Ho you got of a body from your actual location-- and that result tells you how close  you are to the reef.

The one thing you have to keep in mind is that the lines of position (LOP's) plot perpendicular to the azimuth so you need a body that you see in the general direction of the reef; or its reciprocal.

By extension if  you want to figure out how far you progressed along your course you could pick an AP some where out on your desired track and use a body ahead or astern and by this means check your speed over the bottom.

It is perhaps for these purposes that the Astron app lets you pick your own AP.  It provides increased versatility all out of the same math.

Peter

Celestial navigation isn't particularly easy- especially when you're tired and cold and forgot even the simplest of things about it.  Sextant...whats a sextant?  Oh, I remember...he's a janitor in a church. (sexton)

Here are some words that'll help cheer you in your voyage-

The Great One himself- Nathaniel Bowditch- that great American navigator and mathematician who said the following words;

“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.”

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