Re: Pulsar Nav accuracy

--- hal@buffnet.net wrote:
> Hello John L.,
>   Thanks for the explanation below, but of course,
> I've more questions to
> ask then...  ;)
> 
....
> >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.

> 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,... and thus, A's distance from B.
-----
Agreed.   Now we need to do one thing to make it
more interesting, on the A-B line we place a compass
disk and define 'North' as the bearing from the 
centerpoint of the A-B line to earth, I.E. zero
degrees.

hals Logic portion removed...
> 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?
-----
Agreed, you will be able to define your location in 
terms of angle/dist fm A and angle/dist fm B.	What 
this information will not tell you is the difference 
between the A-B centerline 'North' and your current
angular location.
Or perhaps in a more simplistic case, knowing all the
angles and distances on a triangle will not tell you
if the triangle is standing up or lying down.

> I must be missing something.	Either that, or I am
> right, but am uncertain enough to say why...
------
  Perhaps, in this case the full extent of the problem
was not defined.

Please note that while I agree with all of the points
you made, I.E. your logic is correct, you would still
be completely lost in space because you did not have 
enough information to solve the problem.

Bye for now,
John L.

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