[GRASSLIST:3252] re: help with theory of map projections

[chris2]

On Wed, Feb 27, 2002 at 04:04:45PM +0000, Glynn Clements wrote:
> 
> chris2 wrote:
> 
> > I have been trying to learn the theory behind map projections from the
> > book "Map Projections, A reference Manual" by Bugayevskiy and Snyder. 
> > I am stuck on page 2 in Chapter 1, if you can believe it, where they
> > take an ellipsoid of revolution, and a point on its surface at
> > latitude phi, and mention that the radius of curvature of the meridian
> > through the point is:
> > 
> > M= a*(1-e^2) / [(1-e^2*sin(phi)^2)^(3/2)]
> > 
> > Where e, a, and b are the eccentricity and semimajor and semiminor
> > axes of the ellipse of revolution.
> > 
> > I set out to verify this formula as a way of reviewing my vector
> > caclulus and getting started with the book.
> > 
> > I used the following parameterization of the ellipse:
> > 
> > x = a*cos(t)
> > y = b*sin(t)
> > 
> > My t, by the way, is equivalent to their phi, I am pretty sure. That
> > is, t is the angle between a normal line to the ellipse through the
> > given point, and the x-axis.
> 
> I don't think so. Differentiating gives:
> 
> dx/dt	= -a*sin(t)
> dy/dt	=  b*cos(t)
> 
> => dy/dx	= -(b/a).cot(t)		[cot(x) = 1/tan(x) = cos(x)/sin(x)]
> 
> => normal grad.	= (a/b).tan(t)		[normal grad. * tangent grad. = -1]
> 
> Clearly, the normal gradient is tan(phi), by the definition of phi.
> 
> If it's any consolation, I also got stuck right at the beginning of
> that book. My conclusion is that geodetic longitude sucks from a
> coordinate geometry perspective.

You are right.  I went over where I thought I had established that t=phi and I was mistaken.

The actual relationship is tan(phi) = (a/b)*tan(t).

I corrected this in my R script and the curves now match.

Thanks!

Chris Marshall