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Dear George,

> An observer sees a celestial body (which is at dec1, GHA1) at an
> altitude of alt1, which is not near his zenith. Presumably all will
> accept  that he is then somewhere on a position circle, radius (90 -
> alt1), centred on the geographical position below dec1, GHA1. But we
> know nothing about where he may be on that circle.
>
> Now he travels through a known course and distance. Wherever he may be
> starting from on that circle, the course and distance are the same.
> The course is defined as an angle relative to true North, the normal
> definition of a course that mariners use. No complication about that.
> The distance is in miles, measured over the Earth's surface. The Earth
> is assumed to be, for our purposes, spherical.
>
> The question is; what is his locus then? Is it a circle? I claim not.

You are right. I tried to explain this in my previous message.

> Even then, his
> final locus is not a circle.

Right, in general. (For SOME original circles, the new position line
will be a circle, but these initial circles are exceptional).

> Is the problem now stated well enough to suit Alex? Not "the same
> direction", but "the same course and distance"

This is exactly how I interpreted it in my perevious message.
And tried to explain that there is no other reasonable interpretation.

> Each observer travels due North through a distance of 60
> miles.

Sure. If you travel 60 miles, the new position line will differ
from a circle by a few miles.

> Now A is at
> 61N, 0W, B is at 1N, 60W, C is at 46N, 45W. Do
> those 3 new positions lie on a circle?

George, you are right in principle,
but your example is not good. Of course, these three points lie on a
circle:-)
Because EVERY triple of points belongs to SOME circle.

This is a theorem of elementary geometry, and here is its simple proof:
Think of our sphere as a sphere in 3-space. Through every triple of points
in the 3-space a plane passes. The intersection of this
plane with our sphere is a circle.
Which evidently contains our three points.

If you really want to illistrate your statement
by a numerical example, you have to pick at least 4 points.

If you really want to include such illustration to your paper,
let me give you a hint how to prove that 4 points are not on the
same circle: you compute their coordinates in the ambient 3-space.
And use the simple fact of elementary geometry that I mentioned above:
points on the sphere belong to a circle on the sphere
if and only if these points belong to a plane in the ambient space.

Because every circle on the sphere is an intersection of this sphere
with some plane.


> centre and
> what is its radius? I guarantee that nobody will provide such a centre
> or radius, because those positions are no longer on a circle.

Sorry I don't have time now to compute the center and the radius,
but I will be glad to do this after I return from my trip
(unless you figure it yourself, following the suggestions above).

> discrepancy from a circle is not by an infinitesimal amount, either.
> but by something over 14 miles, after a shift of only 60 miles.

I can compute the discrepancy if you wish me to,
for your particular circle, or in general,
but in general, it is much smaller than the distance traveled,
from the original position line. (Unless the original circle was close to
a pole).

Alex.

P.S. In my example in the previous message, I intentionally
choose a small circle, because this is the ONLY case when the use
of position CIRCLE (as opposite to position LINE) is justified.
For large position circles, you can safely replace them
with position lines, and a shift of a position line is a line,
very nearly. (Though, mathematically speaking, this shift is neither
a line nor a circle).

   

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