Finding the normal to a 2d line
Author
Basically I've got two points: (x1, y1), (x2, y2) that form a line(duh) and I want to find a vector perpendicular to that line so i can turn it into a unit normal. I'm frightened because this seems like it should be real easy to figure out ,but I've searched up and down Gamedev and Google and can't find what I want. The closest I've gotten was something another thread suggested: finding the perp-dot product of the two vectors. But I don't know what to do with it once I've found it :(. Any suggestions?
There are two normals (if you stay in 2D), one into each half-space defined by the original line.
If the lines are ordered, and you want the counter-clockwise normal, then it's simple to see what to do.
First, generate the rotation matrix that will rotate a point by 90 degrees counterclockwise in 2D. I'm assuming row vertices on the left:
Then, generate the tangent for the line:
Last, rotate through the matrix and simplify:
(Watch the order!)
Normalize and you have the actual normal. The clockwise normal would be the 90 degree rotation the other way.
If the lines are ordered, and you want the counter-clockwise normal, then it's simple to see what to do.
First, generate the rotation matrix that will rotate a point by 90 degrees counterclockwise in 2D. I'm assuming row vertices on the left:
0 1 [x y] * = [Xn Yn] -1 0Then, generate the tangent for the line:
[Xt Yt] = [x2 y2]-[x1 y1] = [x2-x1 y2-y1]Last, rotate through the matrix and simplify:
0 1 [Xn Yn] = [x2-x1 y2-y1] * = [x1-x2 y2-y1] -1 0(Watch the order!)
Normalize and you have the actual normal. The clockwise normal would be the 90 degree rotation the other way.
Author
I should have been able to figure that out :P. Isn't what you showed above the clockwise rotation though? I mean if we have points (4, 0) and (0, 4) putting it into that equation gives
[x1-x2, y2-y1]
[4 - 0, 4 - 0]
[4, 4]
which is clockwise.
[x1-x2, y2-y1]
[4 - 0, 4 - 0]
[4, 4]
which is clockwise.
Quote:Original post by hplus06030 1 [Xn Yn] = [x2-x1 y2-y1] * = [x1-x2 y2-y1] -1 0
0 1 [Xn Yn] = [x2-x1 y2-y1] * = [y1-y2 x2-x1] -1 0 0 -1 [Xn Yn] = [x2-x1 y2-y1] * = [y2-y1 x1-x2] 1 0
If the line formula is defined as: y = m * x
where m = dy/dx = (y2-y1)/(x2-x1)
Then the perpendicular line is defined as:
y = m'*x
where m' = -1/m = -(dx/dy) = -dx/dy = dx/-dy
So you have two solutiond for m'
m' = -dx/dy
so m' = x1-x2/y2-y1 (1)
or m' = dx/-dy
so m' = x2-x1/y1-y2 (2)
Which are the solutions HPPlus presented
Luck!
Guimo
where m = dy/dx = (y2-y1)/(x2-x1)
Then the perpendicular line is defined as:
y = m'*x
where m' = -1/m = -(dx/dy) = -dx/dy = dx/-dy
So you have two solutiond for m'
m' = -dx/dy
so m' = x1-x2/y2-y1 (1)
or m' = dx/-dy
so m' = x2-x1/y1-y2 (2)
Which are the solutions HPPlus presented
Luck!
Guimo
i belive you could just:
v is the direction of the line
n1 is the first "normal"
n2 is the second
n1<-v.y, v.x>
n2<v.y, -v.x>
normalize them if necessary.
of course i'm assuming you're defineing the line as a segment with two endpoints or in terms of a vector and a point.
v is the direction of the line
n1 is the first "normal"
n2 is the second
n1<-v.y, v.x>
n2<v.y, -v.x>
normalize them if necessary.
of course i'm assuming you're defineing the line as a segment with two endpoints or in terms of a vector and a point.
Much easier way:
1) Find the direction of the line by subtracting one point from the other
2) Convert the direction into a 3D vector, leave z as 0.
3) do a cross product with (0,0,1)
4) normalize the result
Voila, you have the normal to the line!
1) Find the direction of the line by subtracting one point from the other
2) Convert the direction into a 3D vector, leave z as 0.
3) do a cross product with (0,0,1)
4) normalize the result
Voila, you have the normal to the line!
So you have two solutiond for m'
m' = -dx/dy
so m' = x1-x2/y2-y1 (1)
or m' = dx/-dy
so m' = x2-x1/y1-y2 (2)
Which are the solutions HPPlus presented
Careful with this approach, as you need to handle the case when y1-y2 is zero.
Much easier way:
1) Find the direction of the line by subtracting one point from the other
2) Convert the direction into a 3D vector, leave z as 0.
3) do a cross product with (0,0,1)
4) normalize the result
Voila, you have the normal to the line!
Not everyone working with 2D space has a 3D math library readily available. Hplus' approach makes the most sense (with an thorough explanation even!) and avoids 6 multiplications, 3 subtractions, and storage overhead for z component and the extra vector.
Not everyone working with 2D space has a 3D math library readily available. Hplus' approach makes the most sense (with an thorough explanation even!) and avoids 6 multiplications, 3 subtractions, and storage overhead for z component and the extra vector.
It's super cheap and easy to make your own cross product function that takes one 2d vector, assumes the z is 0 and crosses it with (0,0,1).
Here, let me give you the exact formula:
Given a 2d vector (a,b), the normal is (x,y):
x = b
y = -a
So basically, flip a and b, and negate the y. Two assignments and one negation. Can't be cheaper!
This topic is closed to new replies.
Advertisement
Popular Topics
Advertisement
