thanks a lot - I will test your solutions later on.

-olli


2006/12/21, Andy Herrman <[EMAIL PROTECTED]>:
For an arbitrary rectangle you'll have to do some math.  Of the top of
my head, here's a way to do it:

First you'll need to be familiar with vector math, including dot products:
http://mathworld.wolfram.com/DotProduct.html

So, pick an arbitrary point on the rectangle and get the vectors for
the 2 sides.  So, for a rectangle at:

p0----------p1
  |            |
  |   x        |
  |            |
p2----------p3

calculate:
v1 = p1 - p0 // This is the vector from p0 to p1
v2 = p2 - p0 // This is the vector from p0 to p2
vX = x - p0  // This is the vector from p0 to your point

Now, do:
d1 = dot(vX, v1)
d2 = dot(vX, v2)

Now, if the point is in the rectangle then:

0 <= d1 <= 1
0 <= d2 <= 1

That is, assuming I got my math right (doing this all in my head w/o any paper).

Try that.  And if anyone sees any problems with my math, let me know.

   -Andy



On 12/21/06, slangeberg <[EMAIL PROTECTED]> wrote:
> >
> > The rectangle is not parallel to the axes, instead its rotating all the
> > time.
> >
>
> I don't believe that using the bounds of a clip will account for any
> rotations.-Scott
>
> -Scott
>
> On 12/21/06, Janis Radins <[EMAIL PROTECTED]> wrote:
> >
> > Check whteher your point is iniside boundries by something like this:
> > if(point.x > rect.minX && point.x < rect.maxX && point.y > rect.minY &&
> > point.Y < rect.maxY)
> >
> > guess something like that should work.
> >
> > 2006/12/21, Oliver Müller <[EMAIL PROTECTED]>:
> > >
> > > Hi all,
> > >
> > > I'm looking for a mathematic way to check if a point resides within a
> > > rectangle.
> > > The rectangle is not parallel to the axes, instead its rotating all the
> > > time.
> > > It's not a movieclip so hitTest is out of the question.
> > > Any suggestions how I could solve this problem ?
> > >
> > > cheers,
> > > olli
> > > _______________________________________________
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> --
>
> : : ) Scott
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