Math wizardry

[Richard Miller]

Thanks to help from Jim Hurley, I can now compute the angles of 
separate, but interesecting, lines, each relative to the horizontal.

So I've got these two angles. But I'm not having any success figuring 
out (consistently and accurately) what the angle is between the two 
lines. I'm only interested in the less-than-180-degree angle. The 
problem is not knowing from which direction these lines were drawn by 
the user. There are four possible combinations:

- drawing from one endpoint to the intersecting point, then from the 
intersecting point to the endpoint of the other line;

- drawing from one endpoint to the intersecting point, then drawing the 
second line from its endpoint back to the intersecting point;

- drawing from the intersecting point to the end point of the first 
line, then from the intersecting point to the endpoint of the other 
line;

- drawing from the intersecting point to the end point of the first 
line, then from the end point of the second line back to the 
intersecting point.

All of these give different values for the resulting angles, depending 
on the direction from which the lines are drawn. How do I consistently 
determine the angle between the two lines?

Thanks.
Richard Miller
Imprinter Technologies


On Mar 2, 2005, at 12:25 PM, Jim Hurley wrote:

>>
>> Message: 1
>> Date: Wed, 2 Mar 2005 10:41:27 -0500
>> From: Richard Miller <wow at together.net>
>> Subject: Math wizardry
>> To: How to use Revolution <use-revolution at lists.runrev.com>
>> Message-ID: <2eadbdcb635a7d516daedb0bcc373e3c at together.net>
>> Content-Type: text/plain; charset=US-ASCII; format=flowed
>>
>> I've got two line graphics drawn on the screen. I need to find out if
>> they intersect and, if so, what the angle is that is formed by their
>> intersection. Any simple way to do this?
>>
>> Thanks.
>> Richard Miller
>> Imprinter Technologies
>>
>
> Richard,
>
> Here are a couple of functions which will should be helpful:
>
> function theLineAngle p1,p2
>   --Angle of line defined by the two points p1 and p2
>   put item 1 of p2 - item 1 of p1 into dx
>   put item 2 of p2 - item 2 of p1 into dy
>   put atan2(dy,dx) into tAngle
>   return tAngle
> end theLineAngle
>
> And
>
> function intersection line1,line2
>   --Intersection point of two lines defined by line1 and line2
>   --Where line1 is defined by its two end points x1,y1,x2,y2
>   put item 1 to 2 of line1 into p1
>   put item 3 to 4 of line1 into p2
>   put item 1 to 2 of line2 into pp1
>   put item 3 to 4 of line2 into pp2
>   put item 1 of p1 into x1
>   put item 2 of p1 into y1
>   put item 1 of p2 into x2
>   put item 2 of p2 into y2
>   put item 1 of pp1 into xp1
>   put item 2 of pp1 into yp1
>   put item 1 of pp2 into xp2
>   put item 2 of pp2 into yp2
>   if x1 = x2 and xp1= xp2 then add .0001 to xp2
>   if x1 = x2 or xp1 = xp2 then
>     if x1 = x2 then
>       return x1&comma&yp2 + (x1-xp2)*(yp2-yp1)/(xp2-xp1)
>     else
>       return xp2&comma& y2 + (xp1-x2)*(y2-y1)/(x2-x1)
>     end if
>   end if
>   if (y2-y1)/(x2-x1) = (yp2-yp1)/(xp2-xp1) then add .0001 to y2--if 
> lines are parallel
>   put yp2 - y2 + x2*(y2-y1)/(x2 - x1) - xp2*(yp2 - yp1)/(xp2 - xp1) 
> into numerator
>   put ((y2 - y1)/(x2 - x1) -( yp2 - yp1)/(xp2-xp1)) into denom
>   put numerator / denom into x
>   put y2 + (x-x2) *(y2-y1)/(x2-x1) into y
>   return x & comma & y
> end intersection
>
> You may wish to handle the special cases were the lines intersect at 
> infinity differently. In the above function I have chosen to slightly 
> alter the given points so that the lines intersect at a great distance 
> (close to infinity, so to speak.)
>
> The above intersection  function assumes that these are effectively 
> the infinite lines of Euclidean geometry. The end points are simply 
> two points which define the infinite lines. To find out whether the 
> intersection point lies between the end points of each line you will 
> need to test whether the distance between the intersection point and 
> ALL the end points is less than the length  of each line respectively. 
> (I can't imagine that sentence is clear.)
>
> Other functions which might be helpful are the perpDist function (the 
> perpendicular distance between a point and a given line, and 
> thePerpProj function (the perpendicular projection of a given point 
> onto a line.) You can see how these work in my recent Bouncing Ball 
> post which can be retrieved by putting this into the msg box:
>
> go url  http://home.infostations.net/jhurley/BouncingBallTools.rev
>
> Let me know if you have any  problems.
>
> Jim
>
> P.S. I've currently lost my  mind entirely and am working on a 
> simulation of pool. So far I've got the balls (just two balls) 
> colliding and rebounding so that they satisfy the physical laws of 
> collision dynamics. What fun.
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