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At 11:29 AM 6/8/2006, you wrote:
>Michael Dorl wrote:
>
>>Since one has a good idea of the course and speed between any two
>>observations,
>>it seems to me the problem is to determine points on the equal altitude
>>circles separated by that vector.

and Herbert Prinz replied

>Right. And this has to be done point for point along the circle, not by
>just moving its center by that vector. The "circle" isn't a circle on a
>mercator projection to start with and changes its shape when being moved
>north or south. Said vector translates to different delta longitude at
>different latitudes, so by how much do you shift the GP of the star? You
>know this only after you have the latitude of your fix.

Bear with me...

I'm not thinking of any kind of graphical solution but rather the spherical
trig behind it. Consider two observations A and B.  Call the zeniths ZA and
ZB and the great circle between them AB.  Now think of a great circle
passing through ZA offset from AB by some angle X. We can write equations
for coordinates of points on the equal altitude circle for A as a function
of X, ZA, and the observed altitude of A.  Those equations represent places
we could have been when we made observation A. Now since the equations give
us a starting position, can't we extend the starting point by the vector
representing the distance we traveled between observing A and B and write
equations for the ending point as a function of X, ZA, and the distance
sailed vector? I understand  this requires that we make some assumptions
about the shape of the Earth. Also we end up with whatever errors occur
when translating speed - time - bearing
data to great circle data but these errors will arise in any treatment of
this problem. Ok, so now we have equations for our ending position as a
function of X, ZA, the altitude of A, and the distance sailed vector.  Now
we can write equations for the observed altitude
of B from our ending position as functions of X, ZA, the altitude of A, the
distance sailed  vector and ZB.  Solve these equations for X knowing the
observed altitude of B.  There will in general be two solutions on opposite
sides of AB.  Knowing X, we can compute our
starting and ending positions.

   

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