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




Re: Loxodromic midpoint
Posted:
Jan 6, 1999 4:32 PM


Robert Hill wrote:
> > 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.
Oh dear, oh dear! I am wrong and you are quite right. I cut'n'pasted an earlier equation and replaced the wrong bits of it with "M"s <sigh>.
> > 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)) >
Agreed!
A couple of asides...
1) There is, of course, a name for your "log(f(x))" = "log(sec(x)+tan(x))". It is the inverse Gudermannian function: gd^1 (x). gd(x) being that lovely function 2 tan^1 e^x  pi/2.
2) I was totally wrong in thinking I knew what Mercator's projection was! http://www.utexas.edu/depts/grg/ustudent/frontiers/fall95/meacham/meacham.html http://www.civil.buffalo.edu/cie/cie303/lect2.html
I had assumed that it was the same as the equalarea cylindrical projection: http://www.ahand.unicamp.br/~furuti/ST/Cart/ProjCyl/projc.html but it isn't, and loxodromes are far from being straight lines in that projection.
In fact, Mercator's projection does not seem to be a "geometrical projection" as such (if one attempts to imagine rays of light coming from some source, and shadows, etc).
Suppose our aim is to define a projection in which loxodromes become straight lines. If x and y are the coordinates on the map, then, we must have
k = dy/dx = sec v dv/du
So, we could choose (as Mercator seems to have done)...
x = const * integral du = const * u and y = const * integral sec v dv = const * gd^1 (v)
There are presumably uncountably many other projections where x is not linear in u...
 Clive Tooth http://www.pisquaredoversix.force9.co.uk/ End of document



