```Date: Jan 6, 1999 4:32 PM
Author: Clive Tooth
Subject: Re: Loxodromic midpoint

Robert Hill wrote:> >     k(Lon1-Lon0) = log((sec Lat1 + tan Lat1)/(sec Lat0 + tan Lat0))> >> > So> >     k = log((sec Lat1 + tan Lat1)/(sec Lat0 + tan Lat0))/(Lon1-Lon0)> > (3)> > Yes.> > > Integrating (2) from P0 to M gives:> >> >     k(LonM-Lon0) = 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(LonM-Lon0) = 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 anearlier 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(Lon1-Lon0) + g1 Lon0 - g0 Lon1)/ (g1-g0),> > or, expanding the g's and assuming monospace font,> >        (Lon1-Lon0)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 projectionwas!http://www.utexas.edu/depts/grg/ustudent/frontiers/fall95/meacham/meacham.htmlhttp://www.civil.buffalo.edu/cie/cie303/lect2.htmlI had assumed that it was the same as the equal-area cylindricalprojection:http://www.ahand.unicamp.br/~furuti/ST/Cart/ProjCyl/projc.htmlbut it isn't, and loxodromes are far from being straight lines in thatprojection.In fact, Mercator's projection does not seem to be a "geometricalprojection" assuch (if one attempts to imagine rays of light coming from some source,andshadows, etc).Suppose our aim is to define a projection in which loxodromes becomestraight lines.If x and y are the coordinates on the map, then, we must havek = dy/dx = sec v dv/duSo, we could choose (as Mercator seems to have done)...x = const * integral du = const * u  andy = const * integral sec v dv = const * gd^-1 (v)There are presumably uncountably many other projections where x is notlinear in u...--  Clive Toothhttp://www.pisquaredoversix.force9.co.uk/End of document
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