Re: Geometry problem (slightly OT)
Re: Geometry problem (slightly OT)
- Subject: Re: Geometry problem (slightly OT)
- From: Nicko van Someren <email@hidden>
- Date: Tue, 21 Sep 2004 15:26:25 +0100
On 21 Sep 2004, at 14:46, Graham Cox wrote:
This has got to be straightforward enough, but it's late, I've had an
18 hour day, and I just can't get my head around it...
Given two rectangles lying at arbitrary angles but crossing over one
another, how do I calculate the four points of the rhomboid
intersection formed?
Obviously as the angle becomes closer and closer to 0, the
intersection area extends rapidly to infinity... also the 90° case is
trivial. It can be assumed that the intersection never includes any
corners or ends of the rectangles - in fact they are thick lines that
extend far beyond the area of intersection.
The knowns are:
* A single point representing the centre of the intersection - both
rects are centred widthwise at this point.
* The angle between the rects and their angles on the plane
* The widths of both rects, which can differ
This is easy. Given the way you have described the rectangles they can
each be considered to consist of two parallel sides which are important
and two ends which are unimportant. If the first rectangle has
important sides A and B and the second has important sides C and D then
the vertices of your rhomboid are the intersection point AC, AD, BC and
BD.
Incidentally is there anything for finding arbitrary intersections,
unions and diffs of paths in Cocoa?
What the union or intersection of two paths means really depends on
which winding rule you are using (especially when you're looking at
figure-of-eight paths). If you're using non-zero winding rules then
the union is easy because you simply compose a path made out of the two
original paths and it just works. If you want the intersection under
non-zero rules then essentially you're asking for the region where the
winding count is less than -1 or more than 1 for the composition of the
two paths.
I have some code which does this as part of solving another problem but
it's fairly ugly. Finding the intersection of two Bezier curves is a
pain since it involves finding up to nine roots of a nasty equation.
The bezierPathByFlatteningPath message come to your rescue but means
that you're not really solving the real problem. If you're interested
I can send you some of the code, though it's nasty to look at and
suffers in a few pathological cases.
Nicko
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