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

 Messages: [ Previous | Next ]
 Robert Hill Posts: 529 Registered: 12/8/04
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 = (Lat1-Lat0) 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(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.

> 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))

--
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)

Date Subject Author
1/1/99 Axel Harvey
1/5/99 Bob Street
1/5/99 Brian Skinner
1/5/99 Robert Hill
1/5/99 Clive Tooth
1/5/99 Clive Tooth
1/6/99 Robert Hill
1/6/99 Clive Tooth
6/17/03 anonymous