i have a 3d world where i have several 2d circles laying on the ground facing to the sky.
how can i check if a line will intersect one of those circles frop top-to-down?
i tried to search but all i get is this kind of intersection test: http://mathworld.wolfram.com/Circle-LineIntersection.html
but its not what i need, here is image what i mean: http://imageshack.us/m/192/8343/linecircleintersect.png
If you are in a coordinate system, where the ground is given by z = c for c some constant, then you could simply calculate the x, y coordinates of the line for z = c. Now for a circle of origin x0, y0 and radius R, you would simply check if
(x - x0)^2 + (y - y0)^2 <= R^2.
If this is true, the line intersects the circle.
In a 3D sense you are first concerned with not with a circle but with the plane where the circle lies on. Then you can find the point of intersection between the ray (line) and the plane (disk).
I like to use homogeneous coordinates for point, planes and lines and I hope you are familiar with vector dot · and cross products ×. Here is the method:
Plane (disk) is defined by a point vector r=[rx,ry,rz] and a normal direction vector n=[nx,ny,nz]. Together they form a plane W=[W1,W2]=[n,-r·n].
Line (ray) is defined by two point vectors r_A=[rAx,rAy,rAz] and r_B=[rBx,rBy,rBz]. Together they form the line L=[L1,L2]=[r_B-r_A, r_A×r_B]
交叉点由<代码>P=[P1,P2]=[L1 ×W1-W2*L2,-L2-W1]界定,或扩大为[P1,P2]。
P=[ (r_B-r_A)×n-(r·n)*(r_A×r_B), -(r_A×r_B)·n ]
The coordinates for the point are found by r_P = P1/P2 where P1 has three elements and P2 is scalar.
如果有坐标,请通过<代码>d=sqrt(r_p-r)和核对d<=R,其中
如果你知道圈子位于地面(r=[0,0])和面上(n=[0,0,1]),那么你可以对上述一般情况作许多简化。
Update:
在将地面(与+Zup)作为平面(圈子)时,使用<代码>r=[rx,ry,0]和n=[0,1],上述交叉点简化
r_p = [ rBy-rAy, rAx-rBx, 0] / (rAy*rBx-rAx*rBy)
其中你可以检查与循环中心之间的距离。
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