[aprssig] Position Ambituity in APRS!

[Robert Bruninga]

> If I have my Kenwood at home 
> (35 57.458N  083 33.430W) set to 
> POS AMB 4, it would transmit 35  .  N\083  .  W.
> 
> Using your logic, my actual location could 
> (probably would) be outside of your
> 60 mile circle, right?

The kenwood implementation for transmitting position ambiguity
is not the recommended way to transmit position ambiguity.  It
truncates.  So it will transmit a 60 mile ambiguous position at
35N and 83W.  But you are actually 57 nautical miles north of
that position, and so you will just barely be inside the 60 nm
ambiguity circle.

If Kenwood had rounded instead of truncating, then you would be
very near to the center of the circle which would have been at
36N and 83W.

But even talking about being inside or outside the circle means
you still do not understand the concept.  The circle is not a
discrete boundary.  It is a special graphical process intended
to convey to you that the position is not well known and so it
could be anywhere in that generatl area within 60 miles or so...

 Talking about positions "within that circle" is meaningless...

Bob

> William McKeehan
> KI4HDU
> http://mckeehan.homeip.net
> 
> On Wed, January 9, 2008 12:36 pm, Robert Bruninga wrote:
> >>> No, it is preferred to be a circle of the
> >>> right ambiguity size centered on approximately
> >>> 35N and 83W.  The word "approximately"
> >>> is because it should be randomized there.
> >>> Probably within say half or less of the
> >>> radius of ambiguity...
> >>
> >> If you eliminate the "randomized" portion,
> >> where would you plot a station that
> >> gave you 35N and 83W as it's position?
> >> 3400.00N and 08300.00W ?
> >
> > If the guy is saying his best estimate of his position to
the
> > nearest whole degree is at 35N I don't see any reason why
you
> > would suggest plotting him at 34N.  If he says he is at 35N
83W,
> > then plot him at 35N and 83W with a 60 mile ambiguity
circle.
> >
> > Your software then has to use whatever internal math it
needs to
> > be able to activate the right pixels on the screen.  This
> > includes offsets, scale, mercator projection distrotion
> > elimination, etc.  But in my code, the entry for for "35"
> > degrees is just 35.
> >
> > Good luck.
> > Bob
> >
> >> >> So if my APRS packet says 35  .  N\083  .  W,
> >> >> I could actually be at 3467.85N\08272.45W ?
> >> >
> >> > Unfortunately, neither of those is a valid posit.
Minutes
> > can
> >> > only be 00 to 59.  But assuming you meant
> > 3457.85N\08253.45W,
> >> > then yes, the coordinates of  35N and 83W are closest to
> > that
> >> > position.
> >>
> >> Yes, I was thinking in base 100, not 60 when I made up
those
> > numbers.
> >>
> >> [big snip]
> >>
> >> --
> >> William McKeehan
> >> KI4HDU
> >> http://mckeehan.homeip.net
> >>
> >>
> >
> >
> >
> 
>