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Topic:
Loxodromic midpoint
Replies:
8
Last Post:
Jun 17, 2003 6:12 AM




Re: Loxodromic midpoint
Posted:
Jan 6, 1999 9:15 AM


I wrote:
> dv/du = k cos u
Clive Tooth wrote:
> Did you mean dv/du = k cos v ??
Er, yes, sorry. And even if my ODE had been right, my subsequent statement about arc length would have been wrong, as I accidentally wrote down the arc length in a Cartesian (u,v) plane rather than that on the sphere!
> Clive Tooth went on to write in another article: > > > dv/du = k cos v > > Anyway, this is the way I see it... > u is the longitude and v is the latitude. > Let r be the radius of the Earth. > Let a be the angle made by the loxodrome with each parallel of latitude. > Let tan a = k. > Let the initial and final latitudes be Lat0 and Lat1. > Let the initial and final longitudes be Lon0 and Lon1. > Let the latitude and longitude of M be LatM and LonM. > > Now, > ds/dv = r cosec a > > Integrating from P0 to P1 gives: > > s = (Lat1Lat0) r cosec a > > In other words, the distance is _linear_ in the latitude. > So > LatM = (Lat0+Lat1)/2 (1)
Yes. I realised this on the way home last night, but have no computer at home so I couldn't post a correction.
> Again, > k = sec v dv/du (2) > > Integrating from P0 to P1 gives: > > k(Lon1Lon0) = log((sec Lat1 + tan Lat1)/(sec Lat0 + tan Lat0)) > > So > k = log((sec Lat1 + tan Lat1)/(sec Lat0 + tan Lat0))/(Lon1Lon0) > (3)
Yes. > Integrating (2) from P0 to M gives: > > k(LonMLon0) = log((sec LatM + tan Lat1)/(sec LatM + tan Lat0))
Where do these expressions
sec (one angle) + tan (a different angle)
come from? I get
k(LonMLon0) = log((sec LatM + tan LatM)/(sec Lat0 + tan Lat0)).
However I haven't checked the ways you can transform this using (1), so it may be equivalent to yours.
> giving > > LonM = Lon0+log((sec LatM + tan Lat1)/(sec LatM + tan Lat0))/k > (4) > > Where LatM and LonM are already know from (1) and (3) above.
I don't see why (4)'s right (see comment above).
I'd like to get a symmetric expression for LonM. To reduce the clutter, let's write f(x) = sec x + tan x. (As far as I can see, we gain nothing from the fact that f(x) can also be expressed as tan (pi/4 + x/2) or whatever it is.) Then the equation of the curve is
log(f(v)) = ku + const
so g1  g0 = k(Lon1  Lon0) (5) gM  g0 = k(LonM  Lon0) (6)
where g0, g1, gM stand for log(f(Lat0)), log(f(Lat1)), log(f(LatM)).
(5) is just a restatement of your (3). Dividing (5) by (6) and rearranging, we get LonM = (gM(Lon1Lon0) + g1 Lon0  g0 Lon1)/ (g1g0),
or, expanding the g's and assuming monospace font,
(Lon1Lon0)log(f(LatM)) + Lon0 log(f(Lat1))  Lon1 log(f(Lat0)) LonM = . log(f(Lat1)/f(Lat0))
 Robert Hill
University Computing Service, Leeds University, England
"Though all my wares be trash, the heart is true."  John Dowland, Fine Knacks for Ladies (1600)



