Distance only triangulation

[Philip Underwood]

On Thu, 2003-01-23 at 06:46, Bill Frantz wrote:

> At 2:47 AM -0800 1/22/03, Sergey Sorokin wrote:
> >Hello!
> >
> >John Halleck wrote:
> >> > You must have at least three legs from each station to get all of them
> >> > fixed. If you have a station with 2 legs than it can be located in two
> >> > different places (unless it is directly on line between it neighbours),
> >> > so you will need some extra data to fix it at the single location.
> >>
> >>   Three legs from each station is ambigious, there are two solutions,
> >>   mirror imaged across the plane defined by the three reference points.
> >
> >Heh... I was thinking about 2d case... In 3d 3 legs is not enough for
> >sure...
> 
> Note that if all the legs are in the same plane, you don't get something
> that works for 3d, no matter how many legs you have.
But surely, if all the legs are in the same plane, you don't care about
the third dimension? Not sure I understand what you're getting at here.

For locating a point in 3d using distance alone, you need distances from
4 known points, or 3 known points and a protocol.

Distance from first point gives you a sphere upon whose surface the
unknown point must lie

Distance from second point gives you another sphere upon which the point
must lie. As the point must lie on both spheres, it must lie on the
intersection of those spheres (a circle)

Distance from third point again gives another sphere upon which the
point must lie; the intersection between those gives two points.

Here you either need a protocol for determining which of these points is
correct (further into cave, on the earth in the case of GPS), or a
fourth point to identify which is which. The fourth point cannot lie in
the plane formed by the first three points if it is to discriminate
between these last two choices. (Is that perhaps what you meant)
> 
> Good Caving - Bill
> 
>