f

#### Where does my line intersect (or not) my bounding box.

```Known: P and Q, two points on the infinite length line L.
Wanted: segment l=(p,q) of L that intersects a box specified by x,y,w,h.

All the algorithms I have seen so far require the (p,q) segment apriori, but
I only have P and Q.

L is to be a crease in a paper folding application (java).  I hope to use
segment l and the box to create java2d areas that are intersections or
subtractions of a subject area.  One of the areas would be flipped about the
crease to emulate a folding action.

--
Richard A. DeVenezia

``` 0 1/6/2004 5:08:00 AM comp.graphics.algorithms  6674 articles. 1 followers. 3 Replies 468 Views Similar Articles

[PageSpeed] 51

```"Richard A. DeVenezia" <radevenz@ix.netcom.com> wrote in message
news:btdfpk\$61m94\$1@ID-168040.news.uni-berlin.de...
> Known: P and Q, two points on the infinite length line L.
> Wanted: segment l=(p,q) of L that intersects a box specified by x,y,w,h.
>
> All the algorithms I have seen so far require the (p,q) segment apriori,
but
> I only have P and Q.

Is this stated correctly?  If you are looking for the intersection (p,q),
but are given (p,q) a priori, then there is nothing to do...

The points P and Q are the end points of a line segment.  You
need to compute the intersection of this segment with an axis-aligned
rectangle.  Source code for this is at
http://www.magic-software.com/Intersection.html
files WmlIntrLin2Box2.*, the first "FindIntersection" function.

--
Dave Eberly
eberly@magic-software.com
http://www.magic-software.com
http://www.wild-magic.com

``` 0 1/6/2004 5:35:18 AM
```"Dave Eberly" <eberly@magic-software.com> wrote in message
>
> "Richard A. DeVenezia" <radevenz@ix.netcom.com> wrote in message
> news:btdfpk\$61m94\$1@ID-168040.news.uni-berlin.de...
> > Known: P and Q, two points on the infinite length line L.
> > Wanted: segment l=(p,q) of L that intersects a box specified by x,y,w,h.
> >
> > All the algorithms I have seen so far require the (p,q) segment apriori,
> but
> > I only have P and Q.
>
> Is this stated correctly?  If you are looking for the intersection (p,q),
> but are given (p,q) a priori, then there is nothing to do...
>
> The points P and Q are the end points of a line segment.  You
> need to compute the intersection of this segment with an axis-aligned
> rectangle.  Source code for this is at
> http://www.magic-software.com/Intersection.html
> files WmlIntrLin2Box2.*, the first "FindIntersection" function.
>
> --
> Dave Eberly
> eberly@magic-software.com
> http://www.magic-software.com
> http://www.wild-magic.com

Dave:

Thanks for the pointer to our code.  Lots to eschew there.

I didn't state my problem clearly enough. Points big P and Q are different
than points little p and q, perhaps I should have said p^ and q^.

P and Q are given and must be used to compute the coefficients of L (the
crease)
p^ and q^ are the unknown intersections of L with the known box x,y,w,h.

My best case scenario is when the 'crease' L goes through two sides of the
box. One of these would occur...
p^x = x and q^y = y (L crosses left and bottom)
p^x = x and q^y = y+h (left and top)
p^x = x+w and q^y = y (right and bottom)
p^x = x+w and q^y = y+h (right and top)
p^x = x and q^x = x+w (left and right)
p^y = y and q^y = y+h (bottom and top)
I still have to contend with L not going through a box edge, L going through
one or two corners and matters of orientation.

My first thoughts were to solve L: y = mx + b and then solve for x or y when
replacing y or x with boxY and boxY+h or boxX and boxX+w.  Of these four
points, the two middlest (in x) should be the shortest segment intersecting
the box.

--
Richard A. DeVenezia

``` 0 1/6/2004 6:49:36 AM
```"Richard A. DeVenezia" <radevenz@ix.netcom.com> wrote in message
news:btdlo4\$5pgj9\$1@ID-168040.news.uni-berlin.de...
> My first thoughts were to solve L: y = mx + b and then solve for x or y
when
> replacing y or x with boxY and boxY+h or boxX and boxX+w.  Of these four
> points, the two middlest (in x) should be the shortest segment
intersecting
> the box.

The code at my site uses a parametric representation of the
line segment from P=(Px,Py) to Q=(Qx,Qy),
(x(t),y(t)) = (Px,Py)+t*(Qx-Px,Qy-Py)
for 0 <= t <= 1.  The intersection is computed by Liang-Barsky
clipping.  You clip the line segment against the x-faces, then
against the y-faces.  Along the way you find out that the segment
does not intersect the box, or that it does and the new parameter
interval is [t0,t1], a subset of [0,1].  Your "p" and "q" are
p = (x(t0),y(t0)) and q = (x(t1),y(t1)).

--
Dave Eberly
eberly@magic-software.com
http://www.magic-software.com
http://www.wild-magic.com

``` 0 1/6/2004 6:59:52 AM Similar Artilces:

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