Re: Pulsar Nav accuracy

Hello John L.,
  Thanks for the explanation below, but of course, I've more questions
to
ask then...  ;)

>Hal,
>      A roundabout and overly simplistic answer is;
>1 point establishes a sphere that you are on the 
>surface of, I presume the radius can be determined.
>2 points establish a line, and therefor you are on the
>
>'surface' of a circle defined by the intersection of 
>the two angles to the two ends of the line.
>3 points determine a plain, and only two locations
>can satisify the requirements for the required angles
>to the determined locations. (I.E. plus or minus
>angles)
>4 points are needed to determine the plain and
>determine
>if you are above or below the level of the plain.
>5+ points are better.

Here is my reasoning...

If you have Pulsar A and Pulsar B, you have two points establishing a
line.
 You know the distance from A to B as well, since before heading out,
you
established A's distance from Earth, and B's distance from Earth, and
thus,
A's distance from B.

>From point C, you are at the unknown location.  All you can get at C at
this point in time, is the bearing to A.  This gives you angular
co-ordinates in this 3 dimensional triangle problem.  This bearing would
be
described in both X and Y axis, but not in Z, because Z is unknown.

>From Point C, you are at the unknown location.  All you get at this
point
in time, is the bearing to B.  This gives you Angular co-ordinates for
your
X and Y, but not z because you don't know how far away you are from B.

Using the angles generated from your bearings to A and B, the only thing
you don't have is distance in your equation right?  Hmmm, what about the
distance from A to B which is a known quantity?  Now you have a triangle
with *all* angles known, plus one side of the triangle's distance known.
>From that, can you not determine your other two sides?  From that, can
you
not determine your location in a rough manner?

I must be missing something.  Either that, or I am right, but am
uncertain
enough to say why...