我有一条从a到B的直线和一个半径为R的圆。
用什么算法来检查直线是否与圆相交?它在圆边的哪个坐标上?
当前回答
我知道自从这个帖子被打开以来已经有一段时间了。根据chmike给出的答案，经Aqib Mumtaz改进。他们给出了一个很好的答案，但只适用于无限线，就像Aqib说的那样。所以我添加了一些比较来知道线段是否与圆接触，我用Python写的。
def LineIntersectCircle(c, r, p1, p2):
#p1 is the first line point
#p2 is the second line point
#c is the circle's center
#r is the circle's radius
p3 = [p1[0]-c[0], p1[1]-c[1]]
p4 = [p2[0]-c[0], p2[1]-c[1]]
m = (p4[1] - p3[1]) / (p4[0] - p3[0])
b = p3[1] - m * p3[0]
underRadical = math.pow(r,2)*math.pow(m,2) + math.pow(r,2) - math.pow(b,2)
if (underRadical < 0):
print("NOT")
else:
t1 = (-2*m*b+2*math.sqrt(underRadical)) / (2 * math.pow(m,2) + 2)
t2 = (-2*m*b-2*math.sqrt(underRadical)) / (2 * math.pow(m,2) + 2)
i1 = [t1+c[0], m * t1 + b + c[1]]
i2 = [t2+c[0], m * t2 + b + c[1]]
if p1[0] > p2[0]: #Si el punto 1 es mayor al 2 en X
if (i1[0] < p1[0]) and (i1[0] > p2[0]): #Si el punto iX esta entre 2 y 1 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i1[1] < p1[1]) and (i1[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i1[1] > p1[1]) and (i1[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
if p1[0] < p2[0]: #Si el punto 2 es mayor al 1 en X
if (i1[0] > p1[0]) and (i1[0] < p2[0]): #Si el punto iX esta entre 1 y 2 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i1[1] < p1[1]) and (i1[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i1[1] > p1[1]) and (i1[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
if p1[0] > p2[0]: #Si el punto 1 es mayor al 2 en X
if (i2[0] < p1[0]) and (i2[0] > p2[0]): #Si el punto iX esta entre 2 y 1 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i2[1] < p1[1]) and (i2[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i2[1] > p1[1]) and (i2[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
if p1[0] < p2[0]: #Si el punto 2 es mayor al 1 en X
if (i2[0] > p1[0]) and (i2[0] < p2[0]): #Si el punto iX esta entre 1 y 2 en X
if p1[1] > p2[1]: #Si el punto 1 es mayor al 2 en Y
if (i2[1] < p1[1]) and (i2[1] > p2[1]): #Si el punto iy esta entre 2 y 1
print("Intersection")
if p1[1] < p2[1]: #Si el punto 2 es mayo al 2 en Y
if (i2[1] > p1[1]) and (i2[1] < p2[1]): #Si el punto iy esta entre 1 y 2
print("Intersection")
其他回答
' VB.NET - Code
Function CheckLineSegmentCircleIntersection(x1 As Double, y1 As Double, x2 As Double, y2 As Double, xc As Double, yc As Double, r As Double) As Boolean
Static xd As Double = 0.0F
Static yd As Double = 0.0F
Static t As Double = 0.0F
Static d As Double = 0.0F
Static dx_2_1 As Double = 0.0F
Static dy_2_1 As Double = 0.0F
dx_2_1 = x2 - x1
dy_2_1 = y2 - y1
t = ((yc - y1) * dy_2_1 + (xc - x1) * dx_2_1) / (dy_2_1 * dy_2_1 + dx_2_1 * dx_2_1)
If 0 <= t And t <= 1 Then
xd = x1 + t * dx_2_1
yd = y1 + t * dy_2_1
d = Math.Sqrt((xd - xc) * (xd - xc) + (yd - yc) * (yd - yc))
Return d <= r
Else
d = Math.Sqrt((xc - x1) * (xc - x1) + (yc - y1) * (yc - y1))
If d <= r Then
Return True
Else
d = Math.Sqrt((xc - x2) * (xc - x2) + (yc - y2) * (yc - y2))
If d <= r Then
Return True
Else
Return False
End If
End If
End If
End Function
You can find a point on a infinite line that is nearest to circle center by projecting vector AC onto vector AB. Calculate the distance between that point and circle center. If it is greater that R, there is no intersection. If the distance is equal to R, line is a tangent of the circle and the point nearest to circle center is actually the intersection point. If distance less that R, then there are 2 intersection points. They lie at the same distance from the point nearest to circle center. That distance can easily be calculated using Pythagorean theorem. Here's algorithm in pseudocode:
{
dX = bX - aX;
dY = bY - aY;
if ((dX == 0) && (dY == 0))
{
// A and B are the same points, no way to calculate intersection
return;
}
dl = (dX * dX + dY * dY);
t = ((cX - aX) * dX + (cY - aY) * dY) / dl;
// point on a line nearest to circle center
nearestX = aX + t * dX;
nearestY = aY + t * dY;
dist = point_dist(nearestX, nearestY, cX, cY);
if (dist == R)
{
// line segment touches circle; one intersection point
iX = nearestX;
iY = nearestY;
if (t < 0 || t > 1)
{
// intersection point is not actually within line segment
}
}
else if (dist < R)
{
// two possible intersection points
dt = sqrt(R * R - dist * dist) / sqrt(dl);
// intersection point nearest to A
t1 = t - dt;
i1X = aX + t1 * dX;
i1Y = aY + t1 * dY;
if (t1 < 0 || t1 > 1)
{
// intersection point is not actually within line segment
}
// intersection point farthest from A
t2 = t + dt;
i2X = aX + t2 * dX;
i2Y = aY + t2 * dY;
if (t2 < 0 || t2 > 1)
{
// intersection point is not actually within line segment
}
}
else
{
// no intersection
}
}
编辑:增加了代码来检查所找到的交点是否实际上在线段内。
这里你需要一些数学知识:
假设A = (Xa, Ya)， B = (Xb, Yb)， C = (Xc, Yc)。从A到B的直线上的任意一点都有坐标(*Xa + (1-)Xb， * ya + (1-)*Yb) = P
如果点P的距离是R到C，它一定在圆上。你想要的是解决
distance(P, C) = R
这是
(alpha*Xa + (1-alpha)*Xb)^2 + (alpha*Ya + (1-alpha)*Yb)^2 = R^2
alpha^2*Xa^2 + alpha^2*Xb^2 - 2*alpha*Xb^2 + Xb^2 + alpha^2*Ya^2 + alpha^2*Yb^2 - 2*alpha*Yb^2 + Yb^2=R^2
(Xa^2 + Xb^2 + Ya^2 + Yb^2)*alpha^2 - 2*(Xb^2 + Yb^2)*alpha + (Xb^2 + Yb^2 - R^2) = 0
如果你将abc公式应用到这个方程来求解，并使用alpha的解来计算P的坐标，你会得到交点，如果存在的话。
好吧，我不会给你代码，但既然你已经标记了这个算法，我认为这对你来说无关紧要。 首先，你要得到一个垂直于这条直线的向量。
y = ax + c是一个未知变量c是未知变量 为了解决这个问题，计算直线经过圆心时的值。
也就是说, 将圆心的位置代入直线方程，解出c。 然后计算原直线与其法线的交点。
这样就能得到直线上离圆最近的点。 计算该点到圆中心之间的距离(使用矢量的大小)。 如果这个小于圆的半径，看，我们有一个交点!
下面是JavaScript的一个很好的解决方案(包括所有必需的数学和实时插图) https://bl.ocks.org/milkbread/11000965
尽管该解决方案中的is_on函数需要修改:
函数is_on(a, b, c) { return Math.abs(距离(a,c) +距离(c,b) -距离(a,b))<0.000001; ｝
