Re: Geometry problem (slightly OT)
Re: Geometry problem (slightly OT)
- Subject: Re: Geometry problem (slightly OT)
- From: Frederick Cheung <email@hidden>
- Date: Tue, 21 Sep 2004 17:18:03 +0200
On 21 Sep 2004, at 16:59, 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. Normally trigonometry is not something I have a problem
with, but in this case I haven't been able to come up with a geometric
construction that yields readily to being worked on with trig
functions. Intuitively I can see that I have all the info I need to
solve the problem, but deriving the solution with only that info is
proving a bit tough. Maybe I'm just too tired...
well you are missing the height of the rectangle unless that is some
known constant
if a side makes an angle theta with the horizontal then an equation is
(sin theta)x+(cos theta)y=k where k is some constant. You can find k by
plugging in the coordinates of a single point on the line, which you
can find by taking the centre of the rectangle and adding on a vector
of magnitude 1/2 the width (or height depending on which side you are
working on) and angle theta or 90 +theta with the x axis (again
depending on which side of the rectangle you are working on
Fred
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