Rhumb line track calculation

[thibautg78]

Hi all,

I would like to know how do we calculate the rhumb line track between 2 points on the earth?

Say between A (45°00N ; 010°00'W) and B (48°30'N ; 015°00W).

Thank you very much for your help.

TG

[IrishJetdriver]

I think this is what you want but it's hard to remember (or remember what I was talking about in my notes anyway)

Convergency is for Great Circle and is d.long x sin of MEAN LAT
Conversion Angle is half of convergency and is for Rhumb Line 1/2 x d.long x sin MEAN LAT

If you want distances as well then 2 examples are shown below.

1/ 47N 3543W to 32N 3543W
d.lat is 15
No other difference
15 x 60 is 900nm

2/ 49S 1225E to 49S 2545E
d.lat is 0
13 x 60 is 780
780 + 20 is 800
COS 49deg is 0.656
800 x 0.656 is 524.8nm

If this is utter rubbish I apologise, I've spent too long pressing "Direct To" and then marvelling at how destination magically appears in front.

[BackSeatDriver]

The key points to note in this question are 'Rhumb Line Track' and 'approximate'. You are given no track direction information, so you cannot apply Conversion Angle to convert between Great Circle and Rhumb Line Track; you are not expected to apply 3-dimensional spherical trigonometry; so, there MUST be a way to approximate.

First, draw the problem out as if it were on a Direct Mercator (straight and parallel Meridians), because then a straight line represents a Rhumb Line. If you work out the N/S distance, it comes to 210nm; if you work out the East/West Departure distance at 45N it comes to 212nm-ish. You therefore have an approximate right angled isosceles triangle where the right angle is at 45N 15W. The Rhumb Line track from 10W to 15W along 45N would be 270; add the angle from the isosceles triangle (45degs); result approximately 315T.

[dawsonj1]

True enough, but this is not an approximation. The answer is a precise trig calculation.

tan(angle) = opposite/adjacent

opp= 210nm and adj = 212nm

Now invert the tan function, you get...

(angle) = arctan(210/212) = 45 degrees

Add this to the 270 degees and you get 315 degrees precisely!

Sorry for being pedantic, I rarely get to use my Maths degree these days!

[L'aviateur]

For small distances, you can use:

tan course = (Dlong x cos Mean Lat) / Dlat
distance = d'lat / cos course


For larger distances, you would require a table of Meridional Parts, (parts of the meridian in Mercator's projection, corresponding to each minute of latitude from the equator up to 70 or 80 degrees; tabulated numbers representing these parts used in projecting charts, and in solving cases in Mercator's sailing). Then the calculation would be tan course = dlong / difference in meridional parts.

Then for great circle, you simply move into the spherical trig.