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
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。
其他回答
也许有另一种方法来解决这个问题，使用坐标系的旋转。
通常，如果一个线段是水平的或垂直的，这意味着平行于x轴或y轴，交点的求解很容易，因为我们已经知道交点的一个坐标，如果有的话。剩下的显然是用圆的方程找到另一个坐标。
受此启发，我们可以利用坐标系旋转，使一个轴的方向与线段的方向重合。
让我们以圆x^2+y^2=1和线段P1-P2为例，P1(-1.5,0.5)和P2(-0.5，-0.5)在x-y系统中。下面的方程提醒你旋转的原理，其中是逆时针方向的角度，x'-y'是旋转后的方程组:
x'=x*cos () + y*sin () y' = - x*sin () + y*cos ()
和反向
X = X ' * cos - y' * sin Y = x' * sin + Y ' * cos
考虑P1-P2方向(用-x表示为45°)，我们可以取=45°。将第二个旋转方程转化为x-y系统中的圆方程:x^2+y^2=1，经过简单的运算，我们得到x'-y'系统中的“相同”方程:x'^2+y'^2=1。
利用第一个旋转方程=> P1(-根号(2)/2，根号(2))，P2(-根号(2)/ 2,0)，线段端点变成x'-y'系统。
假设交点为p，在x'-y'中，Px = -根号2 /2。使用新的圆方程，我们得到Py = +根号(2)/2。将P转换成原始的x-y系统，最终得到P(-1,0)
为了实现这个数值，我们可以先看看线段的方向:水平，垂直或不垂直。如果它属于前两种情况，很简单。如果是最后一种情况，应用上述算法。
为了判断是否有交集，我们可以将解与端点坐标进行比较，看看它们之间是否有一个根。
我相信只要我们有了它的方程，这个方法也可以应用于其他曲线。唯一的缺点是，我们应该在x'-y'坐标系下解方程，这可能很难。
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")
如果你找到了圆心(因为它是3D的，我想你是指球体而不是圆)和直线之间的距离，然后检查这个距离是否小于可以做到这一点的半径。
碰撞点显然是直线和球面之间最近的点(当你计算球面和直线之间的距离时，会计算出这个点)
点与线之间的距离: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html
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。
这里是一个用golang写的解决方案。这个方法和这里发布的其他一些答案类似，但不完全相同。它易于实现，并已经过测试。以下是步骤:
Translate coordinates so that the circle is at the origin. Express the line segment as parametrized functions of t for both the x and y coordinates. If t is 0, the function's values are one end point of the segment, and if t is 1, the function's values are the other end point. Solve, if possible, the quadratic equation resulting from constraining values of t that produce x, y coordinates with distances from the origin equal to the circle's radius. Throw out solutions where t is < 0 or > 1 ( <= 0 or >= 1 for an open segment). Those points are not contained in the segment. Translate back to original coordinates.
这里导出了二次曲线的A、B和C的值，其中(n-et)和(m-dt)分别是直线x坐标和y坐标的方程。R是圆的半径。
(n-et)(n-et) + (m-dt)(m-dt) = rr
nn - 2etn + etet + mm - 2mdt + dtdt = rr
(ee+dd)tt - 2(en + dm)t + nn + mm - rr = 0
因此A = ee+dd, B = - 2(en + dm)， C = nn + mm - rr。
下面是函数的golang代码:
package geom
import (
"math"
)
// SegmentCircleIntersection return points of intersection between a circle and
// a line segment. The Boolean intersects returns true if one or
// more solutions exist. If only one solution exists,
// x1 == x2 and y1 == y2.
// s1x and s1y are coordinates for one end point of the segment, and
// s2x and s2y are coordinates for the other end of the segment.
// cx and cy are the coordinates of the center of the circle and
// r is the radius of the circle.
func SegmentCircleIntersection(s1x, s1y, s2x, s2y, cx, cy, r float64) (x1, y1, x2, y2 float64, intersects bool) {
// (n-et) and (m-dt) are expressions for the x and y coordinates
// of a parameterized line in coordinates whose origin is the
// center of the circle.
// When t = 0, (n-et) == s1x - cx and (m-dt) == s1y - cy
// When t = 1, (n-et) == s2x - cx and (m-dt) == s2y - cy.
n := s2x - cx
m := s2y - cy
e := s2x - s1x
d := s2y - s1y
// lineFunc checks if the t parameter is in the segment and if so
// calculates the line point in the unshifted coordinates (adds back
// cx and cy.
lineFunc := func(t float64) (x, y float64, inBounds bool) {
inBounds = t >= 0 && t <= 1 // Check bounds on closed segment
// To check bounds for an open segment use t > 0 && t < 1
if inBounds { // Calc coords for point in segment
x = n - e*t + cx
y = m - d*t + cy
}
return
}
// Since we want the points on the line distance r from the origin,
// (n-et)(n-et) + (m-dt)(m-dt) = rr.
// Expanding and collecting terms yeilds the following quadratic equation:
A, B, C := e*e+d*d, -2*(e*n+m*d), n*n+m*m-r*r
D := B*B - 4*A*C // discriminant of quadratic
if D < 0 {
return // No solution
}
D = math.Sqrt(D)
var p1In, p2In bool
x1, y1, p1In = lineFunc((-B + D) / (2 * A)) // First root
if D == 0.0 {
intersects = p1In
x2, y2 = x1, y1
return // Only possible solution, quadratic has one root.
}
x2, y2, p2In = lineFunc((-B - D) / (2 * A)) // Second root
intersects = p1In || p2In
if p1In == false { // Only x2, y2 may be valid solutions
x1, y1 = x2, y2
} else if p2In == false { // Only x1, y1 are valid solutions
x2, y2 = x1, y1
}
return
}
我用这个函数进行了测试，确认解点在线段内和圆上。它创建了一个测试段，并围绕给定的圆进行扫描:
package geom_test
import (
"testing"
. "**put your package path here**"
)
func CheckEpsilon(t *testing.T, v, epsilon float64, message string) {
if v > epsilon || v < -epsilon {
t.Error(message, v, epsilon)
t.FailNow()
}
}
func TestSegmentCircleIntersection(t *testing.T) {
epsilon := 1e-10 // Something smallish
x1, y1 := 5.0, 2.0 // segment end point 1
x2, y2 := 50.0, 30.0 // segment end point 2
cx, cy := 100.0, 90.0 // center of circle
r := 80.0
segx, segy := x2-x1, y2-y1
testCntr, solutionCntr := 0, 0
for i := -100; i < 100; i++ {
for j := -100; j < 100; j++ {
testCntr++
s1x, s2x := x1+float64(i), x2+float64(i)
s1y, s2y := y1+float64(j), y2+float64(j)
sc1x, sc1y := s1x-cx, s1y-cy
seg1Inside := sc1x*sc1x+sc1y*sc1y < r*r
sc2x, sc2y := s2x-cx, s2y-cy
seg2Inside := sc2x*sc2x+sc2y*sc2y < r*r
p1x, p1y, p2x, p2y, intersects := SegmentCircleIntersection(s1x, s1y, s2x, s2y, cx, cy, r)
if intersects {
solutionCntr++
//Check if points are on circle
c1x, c1y := p1x-cx, p1y-cy
deltaLen1 := (c1x*c1x + c1y*c1y) - r*r
CheckEpsilon(t, deltaLen1, epsilon, "p1 not on circle")
c2x, c2y := p2x-cx, p2y-cy
deltaLen2 := (c2x*c2x + c2y*c2y) - r*r
CheckEpsilon(t, deltaLen2, epsilon, "p2 not on circle")
// Check if points are on the line through the line segment
// "cross product" of vector from a segment point to the point
// and the vector for the segment should be near zero
vp1x, vp1y := p1x-s1x, p1y-s1y
crossProd1 := vp1x*segy - vp1y*segx
CheckEpsilon(t, crossProd1, epsilon, "p1 not on line ")
vp2x, vp2y := p2x-s1x, p2y-s1y
crossProd2 := vp2x*segy - vp2y*segx
CheckEpsilon(t, crossProd2, epsilon, "p2 not on line ")
// Check if point is between points s1 and s2 on line
// This means the sign of the dot prod of the segment vector
// and point to segment end point vectors are opposite for
// either end.
wp1x, wp1y := p1x-s2x, p1y-s2y
dp1v := vp1x*segx + vp1y*segy
dp1w := wp1x*segx + wp1y*segy
if (dp1v < 0 && dp1w < 0) || (dp1v > 0 && dp1w > 0) {
t.Error("point not contained in segment ", dp1v, dp1w)
t.FailNow()
}
wp2x, wp2y := p2x-s2x, p2y-s2y
dp2v := vp2x*segx + vp2y*segy
dp2w := wp2x*segx + wp2y*segy
if (dp2v < 0 && dp2w < 0) || (dp2v > 0 && dp2w > 0) {
t.Error("point not contained in segment ", dp2v, dp2w)
t.FailNow()
}
if s1x == s2x && s2y == s1y { //Only one solution
// Test that one end of the segment is withing the radius of the circle
// and one is not
if seg1Inside && seg2Inside {
t.Error("Only one solution but both line segment ends inside")
t.FailNow()
}
if !seg1Inside && !seg2Inside {
t.Error("Only one solution but both line segment ends outside")
t.FailNow()
}
}
} else { // No intersection, check if both points outside or inside
if (seg1Inside && !seg2Inside) || (!seg1Inside && seg2Inside) {
t.Error("No solution but only one point in radius of circle")
t.FailNow()
}
}
}
}
t.Log("Tested ", testCntr, " examples and found ", solutionCntr, " solutions.")
}
下面是测试的输出:
=== RUN TestSegmentCircleIntersection
--- PASS: TestSegmentCircleIntersection (0.00s)
geom_test.go:105: Tested 40000 examples and found 7343 solutions.
最后，该方法很容易扩展到射线从一点开始，经过另一点并延伸到无穷远的情况，只需测试t > 0或t < 1，而不是两者都测试。
