Math wizardry

[Jim Hurley]

>
>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.