Re: complex interscetions
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does anyone know a formula for line intersections through polygons. =
I can do a 2d line intersection and a 3d line intersection, but this is =
hurting my head. Any info would be helpful.
reddager at animalis.com=20
I believe that if you have an equation for each line of a polygon(you =
don't even need endpoints for this part) then you can figure out for =
each line which direction is inside the polygon and which side is out by =
using the > or < symbols. I remember in simple graphing in algebra I =
class, we would have three equations. All of these (2D) had a > or < =
instead of an =3D.
Example:
Graph the following:
y < 1/2x + 2
y < -3x + 4
y > -1x - 1
The area in the middle is all that can possibly exist using those three =
equations. I assume that this would be the same with 3 dimensional =
equations. I'm not sure exactly how to do this, but I think it's =
possible to use this to figure out if a 3d line intersects a 3d polygon =
made with this method. I've thought about this already a little, and I =
might more later. Hope this helps.
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<DIV> does anyone know a formula for line =
intersections=20
through polygons. I can do a 2d line intersection and a 3d line =
intersection,=20
but this is hurting my head. Any info would be helpful.<BR><A=20
href=3D"mailto:reddager at animalis.com">reddager at animalis.com</A> =
<BR></DIV>
<DIV> </DIV>
<DIV>I believe that if you have an equation for each line of a =
polygon(you don't=20
even need endpoints for this part) then you can figure out for each line =
which=20
direction is inside the polygon and which side is out by using the > =
or <=20
symbols. I remember in simple <FONT size=3D3>graphing in algebra I =
class, we=20
would have three equations. All of these (2D) had a > or < =
instead=20
of an =3D.</FONT></DIV>
<DIV><FONT size=3D3></FONT> </DIV>
<DIV><FONT color=3D#000000><FONT size=3D3>Example:</FONT></FONT></DIV>
<DIV><FONT color=3D#000000><FONT size=3D3></FONT></FONT><FONT =
size=3D3></FONT>Graph=20
the following:</DIV>
<DIV>y < 1/2x + 2</DIV>
<DIV>y < -3x + 4</DIV>
<DIV>y > -1x - 1</DIV>
<DIV> </DIV>
<DIV>The area in the middle is all that can possibly exist using those =
three=20
equations. I assume that this would be the same with 3 dimensional =
equations. I'm not sure exactly how to do this, but I think it's =
possible=20
to use this to figure out if a 3d line intersects a 3d polygon made with =
this=20
method. I've thought about this already a little, and I might more =
later. Hope this helps.</DIV>
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