How to solve Great Circle Sailing Calculations with intermediate points?? Distance, courses, vertex!

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How to solve Great Circle Sailing Calculations with intermediate points?? Distance, courses, vertex!

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Using an example, this video shows how to calculate the distance, initial course, final course, and the position of the vertex along a great circle course.
This video also shows how to calculate the position of the intermediate points located along a great circle track.
If you liked this video, you have the opportunity to become an exclusive member of our channel "Steering Mariners". For AUD 9.99, the membership will provide you with multiple benefits such as extra videos, live streams, and live interaction with me:) Link to exclusive membership: / @steeringmariners
Contents of this video will benefit mariners preparing for exams (written and oral examinations).
Mariners will also benefit by watching the following videos:
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Пікірлер: 35

@waelsailor5414 2 жыл бұрын

Dear Sir We can never thank you enough for these videos You are a true benefactor

@oceankiwi100 4 жыл бұрын

Thank You. suddenly its all making more sense. Can i ask what program you are using on your ipad to present these videos. It looke great.

@SteeringMariners 4 жыл бұрын

Thanks Geoff. I am glad the videos are helping. I am just using the screen recorder function on my ipad and laptop. Nothing fancy:)

@SatishKumar-mk2ky 5 жыл бұрын

Thanks sir for discussing this great circle

@Cadetgautam 3 жыл бұрын

Thank you sir 🙏🙏

@mohitsaxena9641 5 жыл бұрын

Sir can u please solve this question Find the initial course, final course, distance and the position of vertex along the gc A: 49°50'N 005° 15'W B: 32°29'N 064°00'W and find the longitude in which the great circle track crosses the parallel of 40°N.

@SteeringMariners 5 жыл бұрын

mohit saxena will do buddy 😊

@SteeringMariners 5 жыл бұрын

Done:) kzfaq.info/get/bejne/fqtomqV4vLHSqmw.html

@nalinkashyap6008 11 ай бұрын

Thank You Sir ❤

@narensharma6813 5 жыл бұрын

Thanks alot for this:)

@SteeringMariners 5 жыл бұрын

Naren Sharma you are most welcome 😊

@pranavraj6399 7 ай бұрын

Sir, I tried a question but couldn't solve a problem asked in written exam. Que. A VESSEL SAILED ON A GC TRACK FROM A POSITION 45° North 160° E,where would the vessel cross the equator,if its initial course was 270° T. Also find the course when the vessel cross the equator.

@nikchoweller 4 жыл бұрын

Sir you said that the same quadrant vertex will lie outside and towards the higher latitude. So if it is North and South latitudes which one will we choose?

@SteeringMariners 4 жыл бұрын

Look at the initial and final courses. If both are in the same quadrant (e.g. NE and NE or SE and SE) then vertex will lie between A and B. If the courses are different quadrants (NE and SE or NW and SE) then vertex will lie somewhere between A and B but towards the higher latitude.

@jomaricornel5682 4 жыл бұрын

Sir can u illustrate the great circle sailing in the globe?

@paramsingh9169 2 жыл бұрын

Sir after locating the vertex... Can we use cosine formula.. If a spherical triangle is having 90' angle

@muzammil_navi 5 жыл бұрын

Sir am solving this problem by abc formula which is Subramanian book i am able to solve upto initial and final course . Problem is i am unable to find suffix of initial and final course. Can you help me??/plz

@arcenaivanr.855 3 жыл бұрын

what is the solution for solving ground course by rhumbliine and distance by rhumbline sir? Thanks

@shaneburas1124 4 жыл бұрын

Can u make a video r direct me to one that solves this problem with intermediate miles.

@SteeringMariners 4 жыл бұрын

kzfaq.info/get/bejne/ap-pjdCUz-C-fqs.html

@elmerares1540 4 жыл бұрын

My solution for track cut at 30d W & 50d N which in my opinion is easier & convenient than Napiers Track cut at 30d W: Tan Lat = Cos P x Tan Lv = Cos (37d -14.49974117d) xTan 51.48557916 = Cos 22.50025883 x Tan 51.48557916 = Tan-1 (1.160873728) Lat = 49.25772806 N = 49d 15m 27.82s N track cut at 30d W Checking: Cos P = Tan Lat / Tan LatV = Tan 49d 15m 27.82s / Tan 51.48557916 = Cos-1 (0.923877794 P = 22.50026021 W +/1Long V = 14.49974117 W Long = 037d W Track cut at 50d N: Cos P = Tan Lat / Tan LatV = Tan 50 / Tan 51.48557916 = Cos-1 (0.948453449) P = 18.47654866d W +/- LongV = 14.49974117 W Long = 32.97628983 W Long = 032d 58m 34.64s W Checking: Tan Lat = Cos P x Tan Lv = Cos 18.47654866 xTan 51.48557916 = Cos 22.50025883 x Tan 51.48557916 = Tan-1 (1.191753593) Lat = 50d N

@anandgupta2588 4 жыл бұрын

sir would you pls help me to know how to find lat at which gc track croses 180 deg meridian

@ChandraKumar-ee4gi 4 жыл бұрын

Sir I have one doubt that can you tell why we use tooo many methods like greatcircle ,mercator sailling, plane sailling like this one method is not enough

@SteeringMariners 4 жыл бұрын

Plane sailing is for less than 600 miles; Mercator and Great Circle is for more than 600; Great circle is the shortest distance but dues to less gnomonic charts, we draw great circle courses and transfer them as rhumb line course on mercator charts

@ChandraKumar-ee4gi 4 жыл бұрын

Awesome reply sir you're good teacher and also you have to do one chart plotting video

@mohitsaxena9641 5 жыл бұрын

Sir can you please slove this question On 23 sept 2008 in DR 23°40' N 161°56'E compute the sextant meridian altitude of the sun's LL if the IE was 2.3' on the arc and HE was 10.5m. And more questions like this

@SteeringMariners 5 жыл бұрын

mohit saxena will do buddy. No worries 😊

@mohitsaxena9641 5 жыл бұрын

@@SteeringMariners ok thank you sir

@danzhokhov56 5 жыл бұрын

Could you explain something: the angle opposite PV is A and the angle next to PV is P. So if I am no mistaken instead 90-P must be 90-A the angle next to PV. Or have I misunderstood something

@SteeringMariners 5 жыл бұрын

Hi Dan, the angles opposite to the side must be written in the Napier's diagram. the angle opposite PV is A. so I have written 90-A opposite to side PV. similarly, the angle opposite to AV is P, so I have written 90-P opposite side AV. Then the remaining side (PA) comes at the bottom because that is the side opposite the right angle V. does this make it clear?

@danzhokhov56 5 жыл бұрын

@@SteeringMariners I do not clearly understand what you mean when you wrote 90-A. Is this 90-A=P(As the sum of angles in triangle is 180. V=90 and P=90-A)???

@SteeringMariners 5 жыл бұрын

@@danzhokhov56 The sum of angles in a spherical triangle is not 180. in plane triangles, the sum of the internal angles is 180. Napier's rules are for spherical triangles. so don't get confused with 90-A. its a formula Napier devised for solving the spherical triangles. his formulaes are based on cosine rules. let me know if its still not clear.

@SteeringMariners 5 жыл бұрын

the sum of internal angles in a spherical triangle is between 180-540 degrees. the sum of the three sides is always less than 360 degrees

@danzhokhov56 5 жыл бұрын

@@SteeringMariners Thanks a lot finally I've got the gist f it