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 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/

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