I was checking out some practice questions for great circle calculations at this site:

https://www.starpath.com/cgi-bin/ubb/ult...6;t=000452

I do not know if the questions are U.S.C.G. or Starpath generated, but in the thread David Burch offers a link to a YouTube video where he shows their solutions by a plotting program called QTVLM  and there he points out that his answers vary slightly from those of the practice questions because QTVLM uses an ellipsoid model of the Earth while the Coast Guard uses a spherical model for these questions -- so I suppose the questions are U.S.C.G.

In the forth question, 5-615 , they list a starting point, an end point, and a distance between the two along a great circle track. They then give you a zone time of the start of the journey and zone time descriptors for both the start and end locations.  You are asked to figure out the estimated local zone time date and time of arrival at the destination based on a steady speed of 13 knots.

In the YouTube video Burch uses the automated program to generate times in UTC and then converts the UTC time of arrival into a local zone time. It seemed to be a fairly involved "work around" with the software to get the software to do what he wanted, but he did get to an acceptable answer. One that was close enough to the "correct" answer that you would definitely pick that one.

When I tried the question (before viewing the YouTube video) I simply took the provided great circle distance and divided it by 13 knots to get the time en-route in hours. Then I converted that into days/hours/ minutes.  This I added to the UTC time of the start of the journey, derived a UTC time of arrival at the end, and finally converted that to a local zone time -- and that didn't work out to be close to any of the provided answers.

I scratched my head, checked my math and it still didn't work out.

Knowing the "correct" answer I was able to back calculate the time en-route for that answer and see how much distance that would cover. The answer was far less than the distance of 4245 nmi offered as a given information in the question.

I then went ahead and solved the great circle problem for initial course and distance using H.O. 208 tables and found the distance by that method to be 4163 nmi.

Lastly I used MarineWaypoints.com great circle calculator to cross check and got a distance of 4164 nmi

So the upshot is the information listed in the question was incorrect. Using that incorrect information there is no way to get the correct answer except by tossing out their numbers and starting from the very beginning.

I attempted to reply to the thread at Starpath to point this out since it seems even David Burch isn't aware of this error, but even though the forum is a "public discussion" I was unable to log into the discussion to post.

PeterB

Hello All 

My first post, time to dip my foot in the water. 

I'm returning to Celestial Navigation after some decades.  The Mary Blewitt book Celestial Navigation for Yachtsmen gave me nightmares for years and years!

Working my way through David Burch's Celestial Navigation: A Complete Home Study Guide and am struggling to find the Polaris (Pole Star) Tables he mentions there are three corrections a0, a1 and a2 (T-22 in the book).  I've spotted a Polaris (Pole Star) Table 2024 A157 that uses a Q correction, then states that you need to add a figure for refraction. 

Have things moved on since David's book or am I looking in the wrong places?

Thanks 

Sandy

Hi
In the current nautical almanac daily page for June 20 to 22,
can anyone explain the two times given for moonset for latitude 62 degrees north?
(See the image)

regards
Russell

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David Burch of Starpath School of Navigation has just released a new book "ENC Essentials" dealing with the transition from paper charts to electronic navigation charts.
It is available from Amazon as a Kindle book or paperback.
I have no association with Amazon, Starpath, or David Burch other than having been an occasional customer of each.

How do you like the ENC's vs. paper charts?

My experience with ENC's is minimal but so far I much prefer the paper version. Much easier to see the "big picture" and then focus in on the details you need.  With ENC's some of the elements appear and disappear with the zoom level. So zoomed out you might see the entrance to a destination harbor in relation to your departure point but not some of the rocks or buoys that are there until you zoom in on them.

Also the zoom level doesn't always change the size of the rendered elements meaning that zooming in doesn't change the size of printed information.  It seems as if  you need a huge electronic screen to render some of this stuff in sizes that are readable at a glance..

I sometimes use raster e-charts which are scans of paper charts on an iPad mini or even my iPhone and with those you can increase print size by zooming in, or see the big picture by zooming out.

I guess it is age related ;)

Peter

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: 359)