I needed to determine the shortest possible distance around the planet given my
course to make sure it meets the requirement of being 21,600nm. This lead
to a difficult calculation, the distance from Tristan De Cuhna and the South West
Cape of New Zealand. Its not directly traversable by great circle given the lower
limit used to calculate circumnavigation of antartica of 63 degrees south. It is
a composite path of two great circles and a segment along the longitude
of 63S shown below.
The question becomes, how to mathematically determine the shortest path.
First, we know there are three paths:
- Great Circle from Tristan to 63S
- Rhumb line segment along 63S
- Great
Circle from 63S to the South West Cape
Great Circle math is widely known, but for the reader, here is how to calculate
the distance of the great circle and the direction of travel.
Where R is the radius of the earth and λ and θ are the lat and longitude of the
initial (i) and final (f) positions respectively.
At any point X on the planet, there is a great circle path between you and a point
Y. The headig you need to follow at anypoint to get there is given by:
bearing = tan-1( sin(Δθ)cos(λf),
cos(λi)sin(λf) − sin(λi)cos(λf)cos(Δθ)
)
NOTE: this is the inverse tangent function normally found as arctan2(y,x) to
accomodate a continous function versus the piecewise function. Also note, MS Excel
expects the parameters in the reverse order acrtanw(x,y)
In order to determine the maximum latitude of great circle, we need to use Clairaut's
formula: This relates the latitude (λ) and true course (θ) along any great circle,
namely:
sin(θ)*cos(λ)=constant.
That is, for any two points on the GC:
sin(θ1)cos(λ1)=sin(θ2)cos(λ2)
Since at the highest latitude (λmax) reached the θ must be 90/270, we
also have:
λmax=cos-1(abs(sin(θ)*cos(λi)))
The final piece we need is the distance of the rhumb line segment at 63 south:
Putting this all together we can find the distance as function of two variables
(x,y) which are the longitude of the two points on the 63rd southern parallel which
form the end and start point of our two great circle segments:
Ok... this seems problematic to solve, requires differential equations and well,
I am rusty on that...
We know that the vertex of each great circle will be at -63degrees south, so if
we plug the bearing formula into Clairaut's formula, we can solve for x and
find the point, repeat this for the other great circle segment and be able to calculate
the distance. For this I used excel and an online version of Mathmatica available
at http://www.wolframalpha.com/
which gives a numerical analysis of 55.031. Similarly, solving for the other great circle, we find the position to be 111.079 resulting in a total distance of 6431.43NM