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Re: Geometry problem (slightly OT)
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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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References: 
 >Geometry problem (slightly OT) (From: Graham Cox <email@hidden>)
 >Re: Geometry problem (slightly OT) (From: Nicko van Someren <email@hidden>)
 >Re: Geometry problem (slightly OT) (From: Graham Cox <email@hidden>)

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