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Author
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Topic: Just one more:use of co-angles
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Capt Ahab
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posted January 31, 2013 10:42 PM
Hate to be a pest, just like to understand why things work...if you can steer me somewhere to an explanation.......The navigation triangle is comprised of co-angles. E.G. co-declination, co-altitude, co-latitude. We enter the tables with latitude & declination, .......now I can understand the table guy knew this, did the subtraction for us, and the answers reflect the fact the table maker did our work for us. Thus if I wanted to solve an ordinary spherical triangle for whatever reason using my tables, I would have to remember to enter the complimentary angles...not the actual ones..if I have this right so far..or not.?
But for the calculator: When I use the law of cosines in its two mods for Hc and A ("A" rather than table "Z")...it all gets magically taken care of without my doing extra additon and subtraction for the "90 -"
How is the law of cosines formula compensating for the complimentary angle issue to get Hc and the A directly. We always claim we are solving for two sides and included angle for HC... but the included sides are 90 - lat and 90 - dec not lat and declination... I'm probobly missing the obvious, as I am almost as good at doing that as dumb arithmatic mistakes doing sight reductions with tables...
From: Noank, Ct
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David Burch
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posted January 31, 2013 11:22 PM
Here are notes on your questions:
The navigation triangle is comprised of co-angles. E.G. co-declination, co-altitude, co-latitude.
---yes, that is the way the triangle is defined in terms of terms we use for navigation
We enter the tables with latitude & declination, .......now I can understand the table guy knew this, did the subtraction for us, and the answers reflect the fact the table maker did our work for us.
---- Exactly.
Thus if I wanted to solve an ordinary spherical triangle for whatever reason using my tables,
---- key here is what do you mean by "my tables." if your tables are the same as mine, ie the sight reduction tables, then you enter those with dec, lat, lha
I would have to remember to enter the complimentary angles...not the actual ones..if I have this right so far..or not.?
---- i would say, NO. you use the normal values for the sight reduction tables. i do not know any tables that are set up for entering with co-angles
But for the calculator: When I use the law of cosines in its two mods for Hc and A ("A" rather than table "Z")...it all gets magically taken care of without my doing extra additon and subtraction for the "90 -"
----- you never have do the co-angle computation. I would skip directly to the solutions we give in the glossary for the navigation triangle, then you use regular values.
--- if you are using law of cosines, then you have to enter the proper co-angles to the formula, but this is not the best approach. use the formulas for Hc and Z that we give in the glossary.
How is the law of cosines formula compensating for the complimentary angle issue to get Hc and the A directly. We always claim we are solving for two sides and included angle for HC... but the included sides are 90 - lat and 90 - dec not lat and declination... I'm probably missing the obvious, as I am almost as good at doing that as dumb arithmetic mistakes doing sight reductions with tables...
---- when using such formulas (ie those in the glossary under navigational triangle) the only thing you need to recall is that south lat and south dec are negative. the rest will all work out for you.
I do not know where you are in the progress of the course, but i think it is always best to proceed through the traditional tables method first, then come back to direct computations at the end of the course.
sometimes focusing on details early in the game distracts from the over all process. once the overall process is under stood, then many of these details will be clear.
--david
From: Starpath, Seattle, WA
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Capt Ahab
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posted February 05, 2013 10:45 AM
I don’t always word my thoughts clearly, especially when working through a confusion. I have worked problems exclusively with 229 and 249, and the calculator solutions are the new stuff to me.
I futzed around over the weekend and made a few discoveries I think help answer my own poorly worded question..
My first wonder was how we get away with entering the complements from outside the nav triangle, and magically get the right answers for the navigation triangle itself. I hate to enter numbers into a canned formula without knowing why….
Then I realized!:
Our first formula is “Sin HC=Sin lat Sin Dec + Cos lat Cos Dec Cos LHA”
When I look up the law of Cosines I find: “Cos c =Cos a Cos b + Sin a Sin b Cos C”
I notice that Sin and Cos are switched from the formal identity for all but the LHA in the format.
That’s how the formula works… The fact that Cos a = Sin 90 – a…. so we enter with the compliment, and get the result for the side of the triangle because we switched from cos to sin between the identity and our formula.
By swapping the sin for cos in the trigonometric identity, we adjust for the issue of entering complimentary angles to get the actual sides of the triangle.
We notice the local hour angle does not flip-flop sin/cos as we are using the actual included angle of the triangle or (360 – included angle) in the relationships…which in the beauty of math…these have the same numerical Cos value, except it switches the +/- from positive or negative, which accounts for the subtraction requirments.
Solving for A works exactly the same way from another rewrite of the law of cosines.
cos A= (Sin Dec - Sin Lat Sin Hc)/(Cos Lat Cos Dec)
And the spherical trig identity again cos A= (cos a – cos b cos c)/(sin b sin c)
All of the sin/cos functions are reversed so the complementary value entry value adjustments hold firm….and we still get "A" for a result.
When we generalize the only real spherical trig used to solve sun sights is the law of cosines, it is very true, but they forgot to tell us how they futzed with the written identities to make it work just for funky data entry on the globe. THat's how I got lost, as it wasn't obvious.
This means in the tables that for any navigation triangle with sides (90 –a), (90 – b), and (90 – c) then entering the tables with the a (= lat), b (= dec), and the included angle C (LHA) we will get me the actual missing side value. As the tables have the same beauty as the above formulas noted, and adjust for complimentary angle entry.
Solving a “non-navigation triangle” with the true sides known measuring a, b, and one included angle “C” can be done with the tables, but I must enter with 90 –a, 90 – b, and C. so that the complimentary value accomodation is intact. If i know the true sides, and the table is set for entry of the complimentary angle of the true side to give the value of the third side, then I must do the subtraction myself to enter the tables to preserve th "a" and "90 -a" relationship.
I got started on this apsect playing around with great circle routes, and it is I suppose a lot of unnecessary insanity on my part, but it does help in understanding how the trig all hangs together and works.
From: Noank, Ct
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David Burch
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posted February 05, 2013 03:11 PM
Have you had a look at the NavList group. There is an active group of experts on cel nav that discuss these matters daily, which i am sure you would enjoy.
You can find the list at www.fer3.com. then you can look through what they have discussed and decide if you want to sign up and post yourself. They are a very friendly and helpful group, and again, experts in the field who will have all answers.
From: Starpath, Seattle, WA
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