Hello

[Mike1843]

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

[c_davidson]

Mike,

We are all trying to master this skill and make it as easy as possible and it does get confusing.

Firstly, Chris Nolan is very good at explaining celestial navigation.  But sometime there's complexity we miss.

The following formulas are found in the Almanacs at TheNauticalAlmanac.com and also here;
https://thenauticalalmanac.com/Formulas.html

Arriving at Hc is very simple and I can get it in about 10 seconds using a calculator, of course (and I'm no math whiz-bang).

To get Hc

Here's the proper formula;

Hc = sin-1[sin(declination) x sin(Latitude) + (cos(Latitude) x cos(declination) x cos(LHA)]

Here's the formula when using the easy Natural Math calculators;

Hc= Sin-1(Sin(Latitude) x Sin(declination) + Cos(Latitude) x Cos(declination) x Cos(LHA)

To get Z and then Zn;

Z proper formula;

Z  = cos-1[(sin(declination) – sin(Latitude) x sin(Hc)) / (cos(Latitude) x cos(Hc))]

Z formula when using the easy Natural Math calculators;

Cos-1((Sin(declination) – Sin(AP Latitude) x Sin(Hc)) / (Cos(AP Latitude) x Cos(Hc)

Once you obtain Z you got to get Zn using the following Rules;

To obtain Zn apply the following rules for Northern and Southern latitudes;.

In Northern Latitudes

LHA greater than 180°....Zn=Z
LHA less than 180°...........Zn=360° – Z

In Southern Latitudes

LHA greater than 180°....Zn= 180° – Z
LHA less than 180°...........Zn= 180° + Z

Please let us know if that solves your problem.

Welcome aboard.....

Clen

[Mike1843]

Thanks Clen and thank you for your time posting this lengthy reply.

I have much to learn!.

There are certainly similarities between Chris Nolans formulas and the ones you posted above. I will have a study and see what I can make of it. I have a dozen or so worked examples which use AP3270 for Z/Zn and Hc so I will judge my efforts using the formulas against these answers.

I will certainly get back to you with a progress report in due course but it could be a while.

Mike

[c_davidson]

Mike,

In the formulas above press the Arc Sin key when you see Sin-1 or press the Arc Cosine key when you see Cos-1

Sorry about that.

Clen

[PeterB]

I use the following:  here " t " is the meridian angle either east or west

Sin(Hc) = Sin(L)Sin(d) + Cos(L)Cos(d)Cos(t)

Tan(Z) = Sin(t) / [Sin(L)Cos(t) – Cos(L)Tan(d)]

When I do sight redcution by calculator I find it useful to begin by making a small table.  I put " L, D and t" across the top row and "sin cos and tan" down the first column. I then populate the table with values to five decimal places. Tan(L) and Tan(t) are not used.  It may seem like extra busy work but it sure makes it a lot easier to ferret out blunders later on!

[CarlosPindle]

Peter,

Will you please explain the meridian t angle stuff. I've read about it but it has always been too difficult to understand. Why not just use LHA?

Thanks,

Carlos

[PeterB]

Peter,

Will you please explain the meridian t angle stuff.  I've read about it but it has always been too difficult to understand.  Why not just use LHA?

Thanks,

Carlos

Carlos

Both the local hour angle (LHA) and the meridian angle represent the angular difference in longitude between yourself (or your chosen assumed position) and the longitude of the geographic position (GP) of the body.  This is the angle at the pole vertex of the navigation triangle.

The only difference is that LHA is ALWAYS measured ONLY westward from 0 to 360 degrees
A meridian angle ( t ) can be measured either eastward or westward.  Usually they are from 0 to 180 degrees. In some well respected texts this ONLY WESTWARD requirement of all Hour Angles is dealt with a bit loosely which can add to confusion.

A westward LHA of 270 from the AP would ALSO be a meridian angle of 90 eastward.

If the meridian angle just so happens to be measured westward it is the same as the LHA.
A meridian angle of 45 degrees westward would be the same as a LHA of 45 degrees.

Some forms of sight reduction use meridian angles instead of LHA simply for compactness. H.O. 211 Ageton's being a good example. They can do this because the trig values used to solve the nav triangle repeat as you progress from 0 to 360 degrees -- but the plus and minus signs change -- so you need to apply certain "rules" to keep track of things properly when using those methods.

When working by calculator you must apply some simple deductive logic instead -- usually to figure out how to get a proper true azimuth (Zn) from your calculator result.

e.g. If the AP lat is north of the body's dec your Zn will be in a generally southward direction because the GP is south of your AP; AND if the LHA is less than 180 the Zn will be in a generally westward direction. Note that here I revert back to the LHA for reference, not the meridian angle since the meridian angle can go either east or west.

When I convert a LHA into a meridian angle I always make a note with the meridian angle if that is now being measured eastward or westward to save future confusion.

I hope this helps. I'm happy to answer more questions if you have them.

Peter

[CarlosPindle]

Peter,

Ah, now I understand. So then using meridian t is only for certain sight reduction methods. That makes sense.

Now I FINALLY understand it (it has been years of confusion).

Thank you,

Carlos

[Mike1843]

Hi,

Just to let you know I have not disappeared.

Pleased to see I have sorted Carlos's "years of confusion" out due to creating this thread. I will get round to the "t" stuff in due course.

Back to Clen's formula for Hc which is probably the same one as Chris Nolan's but slightly rearranged and correctly presented?.

First question that comes to mind is what if DR Lat is 35 degrees North and Dec is 16 degrees South? (different names)

Perhaps it is obvious but I have not spotted it yet.

Mike

[c_davidson]

Mike,

If the heavenly body's declination is Contrary name to your DR Latitude enter a negative sign before it.

Pretty easy.

Clen

[Mike1843]

Mike,

If the heavenly body's declination is Contrary name to your DR Latitude enter a negative sign before it.

Pretty easy.

Clen

Hello again,

Apologies for the tardy response, I have just got around to having a proper look at the "proper formula" and as I think I have got it sorted I thought I would post my calculations.

As I indicated above I do have a dozen practice sights already worked out using sight reduction tables AP3270.

First one as follows

DR  49 17.4N     Adjusted to 49 degrees N
    017 14.3W

Dec 15 38.2N     Converted to 15.6366 degrees

LHA 302 09.5     Adjusted to 302 degrees

As I am using a calculator I am aware I do not need to adjust DR Lat and LHA but to get a proper comparison with AP3270 I will use the same figures.

Hc using AP3270 is 32 degrees 33 minutes

Using proper formula------

Hc = sin -1       sin dec x sin lat          +                cos lat  x cos dec x cos LHA

Hc = sin -1       0.2695 x 0.7547          +               0.6560 x 0.9629 x 0.5299

Hc = sin -1           0.2033                    +                         0.3347

Hc = sin -1                                    0.538

Hc =                                       32.5475 degrees      (inverse sin) ? On my calculator (2nd Function) (sin) 

                                                  0.5475 x 0.6 = 32.8

Hc = 32 degrees 32.8 minutes (Just 2 tenths different to AP3270)

Having adjusted DR Lat to 49 degrees this will become the Lat of my AP. but what is the longitude? As I have adjusted LHA by 09.5 minutes surely I should also adjust DR longitude by 09.5 minutes. In the moderate amount of study I have undertaken into celestial navigation I cannot remember this being made clear, it just seems logical. Therefore my DR longitude of 017 14.3 needs 09.5 adding = 17 degrees 23.8 minutes. Is this correct???. 

Final AP ready to plot is 

 49 degrees N

017 degrees 23.8 minutes W

Delighted to have any of my misunderstandings corrected.

Thanks

Mike

[PeterB]

Mike,

The purpose of an assumed position (AP) has nothing to do with the underlying math.The purpose is to make the tables compact enough to carry aboard. The  AP is simply a contrived "trial position" somewhere close to where we expect to find ourselves which ALSO is chosen to match the limited number of solutions for Hc and Z listed in the tables.  The number of tabulated solutions is diminished by only listing those solutions for starting points (AP's) that have a whole degree of latitude and a whole degree for the Local Hour Angle (LHA). This eliminates thousands of solutions that one would find with the partial degrees of these values between those listed.

  For direct calculation by computer app or scientific calculator, and some tables that do not require an AP (H.O. 208 Ageton's and its derivatives) we use our dead reckoning  (DR)  position as a starting point but it amounts to the same thing: it is just a trial starting point that we think is somewhere close to where we will eventally find our fix.

  Thus when using a calculator to do your sight reduction or an app such as Sight Calc you don't need to derive an AP, although you certainly can.

When using most tables you need to "Assume" as in "Deliberately take up" a contrived position to suit the tables which is also close to where you think you are.

The latitude part is easy: You simply use the whole degree of latitude nearest to your DR. If your DR is exactly half way between two whole degrees of latitude it doesn't matter if you pick the whole degree larger or smaller -- either will work equally well.

The AP longitude is not dependent upon the AP latitude or the change from your DR latitude to the AP latitude whatsoever. Instead it is only dependent on the body's LHA which is how far west that the body's geographic  position is from your DR measured in degrees of longitude.  It is highly unlikely that this LHA will compute to be a whole degree, so to use most tables we need to contrive an AP longitude to use where that will be be the case.

If you are in Western longitudes, to find the LHA from the desired AP you will be subtracting the AP longitude from the Greenwich Hour Angle (GHA) of the body calculated for the time of the sight using almanac data.  Since you need this to come out to a whole degree (no remaining minutes and tenth's) you make up your new AP longitude by using the whole degrees of your DR longitude but then append to that the minutes and tenths of the GHA.  Now when you subtract the AP longitude from the GHA to get the AP's LHA it will be a whole degree with no minutes and tenth's -- which is necessary for the tables.

Things are different in Eastern longitude because you will now ADD your AP longitude to the GHA of the body. Therefore you start with a whole degree of longitude near to your DR and append to that the necessary minutes and tenth's so that the addition comes out to be a whole degree for LHA from the AP. These minutes and tenth's are found by subtracting the minutes and tenth's of the GHA of the body from 60.0'  the result being what you need to tack onto the whole degree of Eastern longitude from your DR to get an AP that has  a whole degree LHA.

There has been a pretty extensive discussion in the General Forum lately on this including examples as well as covering some further nuances. If you start with "LHA iin Eastern longitudes ?" you should get a lot of your questions answered.

If not, then ask away.  We are here to help one another.

PeterB