Re: arc or curve fitting program

Ballendo,

You lost me here....

What algebraic equation will find the center of an arc given three
points? That's what we're trying to find isn't it, the center of the
circle/arc that passes through three points?

Somewhere I have a VB routine I wrote that does exactly that. As I
recall it wasn't a complicated programming job. I could post it if
anyone's interested. IMO, the unwieldy part is, given the three
points and knowing which is the arc start, determining whether it's
G02/03.

thanks,

Doug

--- In CAD_CAM_EDM_DRO@y..., "ballendo" <ballendo@y...> wrote:
> Yes, but the "math" equivalent of that geometric "construction"
> technique is unwieldy, as one has to find distances, slopes, and
> inverse slopes, midpoints, etc.
>
> The algebraic equation is simpler by a long stretch...
>
> Ballendo
>
> P.S. The "perpendicular bisector" algorithm WILL come up in the
first
> few hits of a yahoo search on "three point circle"
>
> --- In CAD_CAM_EDM_DRO@y..., "Smoke" <smoke@t...> wrote:
> > > To find the center of a three point arc, all you need to do is
> construct
> > the
> > > perpendicular bisector of two lines connecting the points. The
> math is
> > > simply straightforward trig,
> >
> > PS the intersection of the bisectors is the center of the desired
> circle.
> >
> > Smoke.