Most excellent!! Thank you!
On Sun, Jul 27, 2008 at 11:57 PM, Marten Svanfeldt (dev) <
[EMAIL PROTECTED]> wrote:
> The normal way of computing it is this (and requires no sqrts ;)
>
> Notation: A and B are the end points, C is the arbitrary point, D is the
> closest point on the segment
>
>
> u = (C-A) <dot> (B-A) / ||B-A||^2
> if u < 0
> D = A
> else if u > 1
> D = B
> else
> D = A + u * (B-A)
>
> -M
>
> Amir Taaki wrote:
> > aha,
> >
> > AB is the line segment
> > ^ is unit vector (we use this in england above vector for unit vector)
> >
> > C is the point
> >
> > On Sun, Jul 27, 2008 at 11:05 PM, Marten Svanfeldt (dev)
> > <[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>> wrote:
> >
> > Hi,
> >
> > What is A, B, C and AB? I assume you use ^ to indicate unit vector (a
> > bit strange imo ;)
> >
> > -M
> >
> > Amir Taaki wrote:
> > > Hi!
> > >
> > > So right now to find the closest point on a line segment I
> > project the
> > > vectors. Then to see if the point is outside the segment I do,
> > >
> > > c = C - A
> > > if (c^ = AB^)
> > > {
> > > if (|c| < |AB|) => lies inside segment
> > > else
> > > return B
> > > }
> > > else
> > > return A
> > >
> > > Is there a way to do it without a sqrt? the one needed to
> > calculate the
> > > unit vectors?
> > >
> > >
> > >
> >
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