Message: 7 Date: Fri, 4 Mar 2005 16:45:58 -0500 From: James Steiner <[EMAIL PROTECTED]> Subject: Re: Math wizardry To: How to use Revolution <[email protected]> Message-ID: <[EMAIL PROTECTED]> Content-Type: text/plain; charset=US-ASCII
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.)
That's not required, which is good, because the distance function is costly.
First, lets assume the purpose of the function is not only to return the intersection of the two *lines* but rather to return the intersection of the two *line segments*, and returns a special value if the *line segments* do not intersect.
Yes, but to write this new function you still need to be sure the point of intersection lies between the end points.
You might consider *two* useful functions (1) a point of intersection function and (2) a function to tell if the intersection lies between the end points.
For example, an intersection function which allows for any one of three different ways of describing the two lines might be:
function intersection p1,p2,p1', p2'
get the paramcount
if it is not 4 then
if it is 2 then
--Each geographic line is two Run Rev lines of 2 items each
if the number of lines in p1 is 2 then
put line 2 of p2 into p2'
put line 1 of p2 into p1'
put line 2 of p1 into p2
put line 1 of p1 into p1
else
--Each geographic line is 4 items, i.e. p1 and p2 combined
if the number of items in p1 is 4 then
put item 3 to 4 of p2 into p2'
put item 1 to 2 of p2 into p1'
put item 3 to 4 of p1 into p2
put item 1 to 2 of p1 into p1
end if
end if
end if
end if
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 p1' into x1'
put item 2 of p1' into y1'
put item 1 of p2' into x2'
put item 2 of p2' into y2'if x1=x2 and x1'=x2' then add .0001 to x1--Two vertical lines --The above will return something like: 200,121999738
if x2'<> x1' then
put (y2'-y1')/(x2'-x1') into m'
put y1' -m'*x1' into b'
end if if x1 <> x2 then
put (y2-y1)/(x2-x1) into m
put y1 -m*x1 into b
end ifif x1 = x2 then return x1,m'*x1 + b' if x1'= x2' then return x2',m*x2' + b if m = m' then add .0001 to m--Or whatever return (b'-b)/(m-m'),(m*b'-m'*b)/(m-m')
end intersection
(You may wish to treat the special case of parallel lines differently. In most game applications and simulations, you don't want to stop execution when the lines are parallel. Best to let the intersection point be at some great distance. )
And:
function betweenEndPointsOfLine pt, tLine
put line 1 of tLine into p1
put line 2 of tLine into p2
put dist(pt,p1) into s1
put dist(pt,p2) into s2
put dist(p1,p2) into s
if s1<s and s2< s then
return true
else return false
end betweenEndPointsOfLine
First, lets assume the purpose of the function is not only to return the intersection of the two *lines* but rather to return the intersection of the two *line segments*, and returns a special value if the *line segments* do not intersect.
Then, given the three points on a line (and we know that the calculated intersection point is on the lines, if Ax is between Bx and Cx and Ay is between By and Cy, then point A must be between points B and C.
Yes, this is quite true.
Also, as a side note, sometimes, like when comparing distances, it's OK to use a partial calculation, rather than the complete distance calculation. For example, if distance A is equal to distance B, then it is true that A^2 is equal to B^2. So, rather that performing the full distance calculation: sqrt ( ( x1 - x2) + (y1 - y2) ) , you can leave off the sqrt, and compare the partial result, which is really the sqaure of the distance, saving some calculation time (sqrt is a much more expensive function than multiplication). In applications (like games or physics-based simulations) where many of these calculations are occuring every second, the saved time can be used for other things.
Good point. [ I'm sure you meant sqrt((x1-x2)^2 + (y1-y2)^2)]
Jim
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