Great Circle Sailings - Thursday Night Live Navigation Problems

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Great Circle Sailings - Thursday Night Live Navigation Problems

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Practical Navigator

Күн бұрын

Пікірлер: 27

@mokudesigns914 4 жыл бұрын

Thank you. Best celestial navigation videos and instruction out there, hands down.

@NavigationTraining 4 жыл бұрын

Glad it was helpful!

@siddharthchaudhary7753 Жыл бұрын

Really good for a mariners for a quick revision.!

@Elfazka 3 жыл бұрын

Please keep spreading the knowledge, i really like the way you explain these formulas and how to operate them. Really easy to understand

@MarinaMatic 3 жыл бұрын

Hi there. Marina, lead instructor in Celestial navigation for over 15 years; from Croatia. Due to quarantine, I started uploading my live lectures online and now YT is recommending me your videos :D. It is great to see in your eyes that emotion "This is so easy, but how do I explain this to them" :D I have that a lot. Regarding 09:41, I was excellent in math and when the captain tried to explain great circle it was confusing so I was looking and asked him: "Is that a triangle?" And he was like yes, but... "No, professor, is that a triangle? With your fancy names lat, long, mer, but bottom line, triangle?" "Yes, Marina". Perfect. I know all about the triangle, all I have to do is use your names for sides and angles. :D For me minimum is 5 numbers after the decimal point. Formula at 23:17, what I do is flip it so cos is last, in calculator I can just swap 10 with 20, 30 and get results faster. I have so much more to say but lets stay at this one. Thank you.

@postholedigger8726 2 жыл бұрын

The professor was correct when he said "yes, but". Navigational problems are solved with a branch of mathematics called spherical trigonometry. It is different from plane trigonometry. in several ways. One major difference is the sides of spherical triangles are measured in degrees instead of linear measurements. Another difference is the sum of the angles of a spherical triangle can exceed 180 degrees. A spherical triangle can have three right angles. The triangle side lengths are defined as angular arcs whose vertex is located at the center of a sphere (the earth). Solving spherical triangles is different than solving triangles on a flat plane. Spherical trigonometric formulas are derived from the intersection of a trihedron, whose vertex is at the center of the sphere, intersecting the sphere's surface. . david

@iputudharmayoga9922 3 жыл бұрын

very good explanation and detail, keep going sir on your videos. ieasy to understand

@Chris-ux1ij 4 жыл бұрын

This question is a bit off topic. There was Captain Cook day last week and so I was wondering how did he do the navigation (probably one of the first to do so with the use of a Nautical Almanac?). I found out that on his first voyage of exploration (1768-1771), Captain Cook was able to take with him only the "The Nautical Almanac and Astronomical Ephemeris" for 1768 and 1769 (the very first published book was for 1767). At the time of the first voyage there appears to be the auxiliary books on "A table of proportional logarithms to be used with the Astronomical and nautical Ephemeris" (1766) and "Tables requisite to be used with the Astronomical and Nautical Ephemeris" (1766) that look like they're needed also. These early Nautical Almanacs from google books seem to have some of the familiar basics but also some other unfamiliar stuff on Jupiter for land based work. The novel part at the time was the new moon table (from Tobias Mayer) for the lunar distance method but I don't know how you would be able to extract the longitude from said tables in practice? and how accurate could one expect the results to be compared with the modern form?

@NavigationTraining 4 жыл бұрын

The two main methods of longitude in that era were using the moons of Jupiter eclipsing (which you can see through a pair of binoculars), and the lunar distance, which measures the rate of change of the moon in the sky compared to the background stars. That is still a valid method today and hobbyists do it for fun. But it is hard work. The idea is that the moon moves faster against the backdrop of the stars and versus the sun. So if you can measure the distance between the moon and a star or the sun, you can compare that to tables to give you your time. Once you know your time, you can extract longitude. Highly recommend Dana Sable's book Longitude or the equivalent TV series. Thanks for the question!

@Chris-ux1ij 4 жыл бұрын

@@NavigationTraining Thanks for the reply. I'll have another look at the book you recommend.

@BillAbel3 2 жыл бұрын

@Practical Navigator Two questions. 1. When you reference a formula as "Bowditch 31" where is that found? I can find the formula in vol 2 on page 236, but it's not labeled as 31, instead it's part of section 405. 2. Where do you get the small format Bowditch you have. I have Vol 1 2019 in a larger format, but I don't see a small format shown on the NGA website.

@causewaykayak Жыл бұрын

I have a link to a useful online copy you can print pages - the whole is some 900 sides- be warned 😏

@delruner 2 жыл бұрын

Thought this was a fun exercise, concerning the calculation of waypoints c.What is the latitude of a point 10 degrees west of the Vertex? how would the formula change if you needed easterly points, is their rules ?

@josephlai9759 2 жыл бұрын

At 32:00 mark you worked out formula 36 with a '+' but earlier you had it as '-' when introducing it. I have no access to Bowditch so would like to clarify if you had written a typo error. If I am not mistaken, the operative '+' should be in the formula, not '-'. Thank you.

@pql42 Жыл бұрын

I think so too, may be it was just mistake. Bowditch states "plus"

@jefrymatusala.d9118 3 жыл бұрын

Hello sir, my name jefry from Indonesia. Can you explain , the navigation use of SIN OF LAW and COSINE OF LAW. Thankyou Sir , God bless you.

@NavigationTraining 3 жыл бұрын

Certainly, here you go! en.wikipedia.org/wiki/Spherical_law_of_cosines For problem solving this law of cosines for spherical trigonometry is the theory, but for passing exams, you can simply memorize or look up the formula to solve the problems. Hope that helps! - Chris

@jefrymatusala.d9118 3 жыл бұрын

@@NavigationTraining Thankyou Sir.

@causewaykayak Жыл бұрын

At 8:30 sinL1 x sinL2 you have negative 0.35 (-0.35) for a result - - why negative? These two angles are surely in the quadrant where the ratios all have + values ...

@causewaykayak Жыл бұрын

I've rechecked and the attribution of sign Negative materially affects the outcome here. I guess Chris is out there on the ocean somewhere so if anyone else beach-combing U Tube. comes across this and can explain why its suddenly negative it would be great to hear from you.

@causewaykayak Жыл бұрын

To myself (in case of an audience less of than One) . The matter of the attribution of negative sign. Bowditch 1981 vol 2 Page 612. (and the cheap Amazon reprint of the 2017 edition page 348) says clearly "The sign convention and course angle by equations 36 & 37 respectively, is that the latitude of the destination is treated AS A NEGATIVE QUANTITY when the latitudes of departure and destination are OF CONTRARY NAME" this fits, everything works. For those like me trailing along in the dust behind the academics, I hope this is helpful. Clearly this vid is no longer monitored. (John in Ireland)

@delruner 2 жыл бұрын

Picked up a copy of Bowditch 2002 bicentennial edition and the trig functions for computing Great Circle sailing was not included. Has this been replaced by table lookups?

@NavigationTraining 2 жыл бұрын

Yes indeed, the two-volume version is the one available in CG testing centers, you would be looking for Bowditch Volume 2. Thanks!

@delruner 2 жыл бұрын

@@NavigationTraining found it in the 2002 version pg351 chapter calculations and conversions. Paragraph Calculations of the Sailings. thanks... fun topic