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 16:30:07 +0100
On 21 Sep 2004, at 15:59, Graham Cox wrote:
On 22 Sep 2004, at 12:26 am, Nicko van Someren wrote:
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.
Yes, this is easy ;) However, they aren't really rectangles, they're
lines that go off for a long way and in fact are part of a curve, but
in the local intersection area they can be considered to be
rectangles. Given this, I don't have the points A B C and D, I only
have the specific information listed above. Generating the points A B
C D from this information is as difficult as the solution I'm looking
for it seems.
OK, it seems odd that you only have angles rather than vectors for the
line, but if that's the case it's still fairly easy.
Reset the origin to the intersection point; we'll add it back in later.
Your lines have a thickness of W1 and W2 respectively and angles a1 and
a2 relative to the X axis.
V1 and V2 represent the vectors from the middle of each line to the
edge:
V1 = { -0.5 * W1 * sin(a1), 0.5 * W1 * cos(a1) }
V2 = { -0.5 * W2 * sin(a2), 0.5 * W2 * cos(a2) }
In co-ordinates centred on the intersection the vertices of the
rhomboid are:
V1+V2, V1-V2, -V1+V2, -V1-V2
Adjust these by adding back the intersection point to get your final
points.
Nicko
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