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HeliTorque :: View topic - Rhumb line tracks
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HeliTorque Forum Index » Ground School

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Rhumb line tracks
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paddywak
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PostPosted: Thu Nov 17, 2011 4:49 pm    Post subject: Rhumb line tracks Reply with quote

I have reached a stage where I am stumped with navigation and I wondered if any one has an explanations or easy to remember tips for the problem.

Here goes: The rhumb line track between 45 00 N , 010 00 W and 48 30 N ,
015 00 W is approximately ?

Is the answer derived from half the chart convergence?


Any have any idea's ?

I know that convergence is (change in longitude x sine mean latitude )

and that the conversion angle is half of the convergence. I am barking up the wrong tree or just barking?
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flip2
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PostPosted: Thu Nov 17, 2011 7:03 pm    Post subject: Reply with quote

A picture would paint a thousand words for this one, but hopefully you can wade through the explanation.

I would use trigonometry to get the answer:

Firstly you need to calculate the length of two of the sides of the right-angle triangle made by moving from point A to point B.

Point B (48 30N 015W)
|\
|..\
|....\
|......\
|........\ Point A (45N 010W)

The change in longitude and the change in latitude will give you 2 sides of the triangle:

Side 1: The Change in Longitude from 010W to 015W
You are moving "across" 5. To make 5 at 45N into a distance use the following formula: departure = change in longitude in minutes x cosine latitude... (5 x 60) x cos 45 = 212nm. This is the length of the bottom, horizontal line.

Side 2: The change in Latitude from 45N to 48 30 W
You are moving "up" 3 30'. Each degree of latitude is worth 60nm, so 3.5 x 60 nm/ = 210nm. This is the height of the left, vertical line.

We are not interested in the actual distance between the two points (the slanted line). But we are interested in the angle of the "Point A" corner.

There is basically no difference in the length of the 2 sides, so without using trigonometry you can make a good guess that the angle is going to be 45... but lets do the maths anyway:

If you remember your trigonometry formulae, you need arctan(210/212)... which is 45 to the nearest degree.

The track is therefore west, 270, plus the angle of that corner, 45.

So the answer is 270 + 45 = 315


Last edited by flip2 on Thu Nov 17, 2011 10:10 pm; edited 2 times in total
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LoachBoy
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PostPosted: Thu Nov 17, 2011 9:33 pm    Post subject: Reply with quote

Welcome to CPL(H)!!! Wink
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paddywak
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PostPosted: Fri Nov 18, 2011 11:05 am    Post subject: Reply with quote

Thanks Flip, a super explanation and 10 out of 10 for the ingenious diagram

I did wake up at 3 am this morning with a eureka moment thinking of a right angled triangle with its base on a parallel of latitude and switched the light on and took out my calculator ( much to the disappointment of my wife )

Thanks again for the help , thats cleared it up nicely.

Judging by the content in the CPL , I suspect I may be asking more questions in the future..... and I certainly take my hat off to all you CPLs out there after seeing what you have all gone through.

It all changes next year, the last date for the current CPLH exams are March 2012. Anyone starting the CPLH on the new syllabus has to take 13 exams instead of 9 so I have a limited window just to add to the stress.

Thanks again for the help,
Leigh.
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