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?
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
采取
E是射线的起点, L是射线的端点, C是你测试的圆心 R是球面的半径
计算: d = L - E(射线方向矢量,从头到尾) f = E - C(从中心球到射线起点的向量)
然后通过…找到交点。 堵塞: P = E + t * d 这是一个参数方程 Px = Ex + tdx Py = Ey + tdy 成 (x - h)2 + (y - k)2 = r2 (h,k) =圆心。
注意:我们在这里将问题简化为2D,我们得到的解决方案也适用于3D
得到:
Expand x2 - 2xh + h2 + y2 - 2yk + k2 - r2 = 0 Plug x = ex + tdx y = ey + tdy ( ex + tdx )2 - 2( ex + tdx )h + h2 + ( ey + tdy )2 - 2( ey + tdy )k + k2 - r2 = 0 Explode ex2 + 2extdx + t2dx2 - 2exh - 2tdxh + h2 + ey2 + 2eytdy + t2dy2 - 2eyk - 2tdyk + k2 - r2 = 0 Group t2( dx2 + dy2 ) + 2t( exdx + eydy - dxh - dyk ) + ex2 + ey2 - 2exh - 2eyk + h2 + k2 - r2 = 0 Finally, t2( d · d ) + 2t( e · d - d · c ) + e · e - 2( e · c ) + c · c - r2 = 0 Where d is the vector d and · is the dot product. And then, t2( d · d ) + 2t( d · ( e - c ) ) + ( e - c ) · ( e - c ) - r2 = 0 Letting f = e - c t2( d · d ) + 2t( d · f ) + f · f - r2 = 0
所以我们得到: T2 *(d·d) + 2t*(f·d) + (f·f - r2) = 0
求解二次方程:
float a = d.Dot( d ) ;
float b = 2*f.Dot( d ) ;
float c = f.Dot( f ) - r*r ;
float discriminant = b*b-4*a*c;
if( discriminant < 0 )
{
// no intersection
}
else
{
// ray didn't totally miss sphere,
// so there is a solution to
// the equation.
discriminant = sqrt( discriminant );
// either solution may be on or off the ray so need to test both
// t1 is always the smaller value, because BOTH discriminant and
// a are nonnegative.
float t1 = (-b - discriminant)/(2*a);
float t2 = (-b + discriminant)/(2*a);
// 3x HIT cases:
// -o-> --|--> | | --|->
// Impale(t1 hit,t2 hit), Poke(t1 hit,t2>1), ExitWound(t1<0, t2 hit),
// 3x MISS cases:
// -> o o -> | -> |
// FallShort (t1>1,t2>1), Past (t1<0,t2<0), CompletelyInside(t1<0, t2>1)
if( t1 >= 0 && t1 <= 1 )
{
// t1 is the intersection, and it's closer than t2
// (since t1 uses -b - discriminant)
// Impale, Poke
return true ;
}
// here t1 didn't intersect so we are either started
// inside the sphere or completely past it
if( t2 >= 0 && t2 <= 1 )
{
// ExitWound
return true ;
}
// no intn: FallShort, Past, CompletelyInside
return false ;
}
其他回答
在此post circle中,通过检查圆心与线段上的点(Ipoint)之间的距离来检查线碰撞,该点表示从圆心到线段的法线N(图2)之间的交点。
(https://i.stack.imgur.com/3o6do.png)
在图像1中显示一个圆和一条直线,向量A指向线的起点,向量B指向线的终点,向量C指向圆的中心。现在我们必须找到向量E(从线起点到圆中心)和向量D(从线起点到线终点)这个计算如图1所示。
(https://i.stack.imgur.com/7098a.png)
在图2中,我们可以看到向量E通过向量E与单位向量D的“点积”投影到向量D上,点积的结果是标量Xp,表示向量N与向量D的直线起点与交点(Ipoint)之间的距离。 下一个向量X是由单位向量D和标量Xp相乘得到的。
现在我们需要找到向量Z(向量到Ipoint),它很容易它简单的向量加法向量A(在直线上的起点)和向量x。接下来我们需要处理特殊情况,我们必须检查是Ipoint在线段上,如果不是我们必须找出它是它的左边还是右边,我们将使用向量最接近来确定哪个点最接近圆。
(https://i.stack.imgur.com/p9WIr.png)
当投影Xp为负时,Ipoint在线段的左边,距离最近的向量等于线起点的向量,当投影Xp大于向量D的模时,距离最近的向量在线段的右边,距离最近的向量等于线终点的向量在其他情况下,距离最近的向量等于向量Z。
现在,当我们有最近的向量,我们需要找到从圆中心到Ipoint的向量(dist向量),很简单,我们只需要从中心向量减去最近的向量。接下来,检查向量距离的大小是否小于圆半径,如果是,那么它们就会碰撞,如果不是,就没有碰撞。
(https://i.stack.imgur.com/QJ63q.png)
最后,我们可以返回一些值来解决碰撞,最简单的方法是返回碰撞的重叠(从矢量dist magnitude中减去半径)和碰撞的轴,它的向量d。如果需要,交点是向量Z。
只是这个线程的一个补充… 下面是pahlevan发布的代码版本,但针对c# /XNA,并做了一些整理:
/// <summary>
/// Intersects a line and a circle.
/// </summary>
/// <param name="location">the location of the circle</param>
/// <param name="radius">the radius of the circle</param>
/// <param name="lineFrom">the starting point of the line</param>
/// <param name="lineTo">the ending point of the line</param>
/// <returns>true if the line and circle intersect each other</returns>
public static bool IntersectLineCircle(Vector2 location, float radius, Vector2 lineFrom, Vector2 lineTo)
{
float ab2, acab, h2;
Vector2 ac = location - lineFrom;
Vector2 ab = lineTo - lineFrom;
Vector2.Dot(ref ab, ref ab, out ab2);
Vector2.Dot(ref ac, ref ab, out acab);
float t = acab / ab2;
if (t < 0)
t = 0;
else if (t > 1)
t = 1;
Vector2 h = ((ab * t) + lineFrom) - location;
Vector2.Dot(ref h, ref h, out h2);
return (h2 <= (radius * radius));
}
以下是我在TypeScript中的解决方案,遵循@Mizipzor建议的想法(使用投影):
/**
* Determines whether a line segment defined by a start and end point intersects with a sphere defined by a center point and a radius
* @param a the start point of the line segment
* @param b the end point of the line segment
* @param c the center point of the sphere
* @param r the radius of the sphere
*/
export function lineSphereIntersects(
a: IPoint,
b: IPoint,
c: IPoint,
r: number
): boolean {
// find the three sides of the triangle formed by the three points
const ab: number = distance(a, b);
const ac: number = distance(a, c);
const bc: number = distance(b, c);
// check to see if either ends of the line segment are inside of the sphere
if (ac < r || bc < r) {
return true;
}
// find the angle between the line segment and the center of the sphere
const numerator: number = Math.pow(ac, 2) + Math.pow(ab, 2) - Math.pow(bc, 2);
const denominator: number = 2 * ac * ab;
const cab: number = Math.acos(numerator / denominator);
// find the distance from the center of the sphere and the line segment
const cd: number = Math.sin(cab) * ac;
// if the radius is at least as long as the distance between the center and the line
if (r >= cd) {
// find the distance between the line start and the point on the line closest to
// the center of the sphere
const ad: number = Math.cos(cab) * ac;
// intersection occurs when the point on the line closest to the sphere center is
// no further away than the end of the line
return ad <= ab;
}
return false;
}
export function distance(a: IPoint, b: IPoint): number {
return Math.sqrt(
Math.pow(b.z - a.z, 2) + Math.pow(b.y - a.y, 2) + Math.pow(b.x - a.x, 2)
);
}
export interface IPoint {
x: number;
y: number;
z: number;
}
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; }
我写了一个小脚本,通过将圆的中心点投影到直线上来测试相交。
vector distVector = centerPoint - projectedPoint;
if(distVector.length() < circle.radius)
{
double distance = circle.radius - distVector.length();
vector moveVector = distVector.normalize() * distance;
circle.move(moveVector);
}
http://jsfiddle.net/ercang/ornh3594/1/
如果需要检查与线段的碰撞,还需要考虑圆心到起点和终点的距离。
vector distVector = centerPoint - startPoint;
if(distVector.length() < circle.radius)
{
double distance = circle.radius - distVector.length();
vector moveVector = distVector.normalize() * distance;
circle.move(moveVector);
}
https://jsfiddle.net/ercang/menp0991/