Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Re: Loxodromic midpoint
Posted:
Jan 6, 1999 4:32 PM
|
|
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 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(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 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 equal-area 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
|
|
|
|
