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
奇怪的是,我可以回答,但不能评论…… 我喜欢Multitaskpro的方法,它可以移动所有东西,使圆的中心落在原点上。不幸的是,他的代码中有两个问题。首先在平方根下的部分,你需要去掉双倍的幂。所以不是:
is underRadical = Math.pow((Math.pow(r,2)*(Math.pow(m,2)+1)),2)-Math.pow(b,2));
but:
under Radical = Math.pow(r,2)*(Math.pow(m,2)+1)) - Math.pow(b,2);
在最后的坐标中,他忘记把解移回来。所以不是:
var i1 = {x:t1,y:m*t1+b}
but:
Var i1 = {x:t1+c。x, y: m * t1 + b +陈守惠};
整个函数就变成:
function interceptOnCircle(p1, p2, c, r) {
//p1 is the first line point
//p2 is the second line point
//c is the circle's center
//r is the circle's radius
var p3 = {x:p1.x - c.x, y:p1.y - c.y}; //shifted line points
var p4 = {x:p2.x - c.x, y:p2.y - c.y};
var m = (p4.y - p3.y) / (p4.x - p3.x); //slope of the line
var b = p3.y - m * p3.x; //y-intercept of line
var underRadical = Math.pow(r,2)*Math.pow(m,2) + Math.pow(r,2) - Math.pow(b,2); //the value under the square root sign
if (underRadical < 0) {
//line completely missed
return false;
} else {
var t1 = (-m*b + Math.sqrt(underRadical))/(Math.pow(m,2) + 1); //one of the intercept x's
var t2 = (-m*b - Math.sqrt(underRadical))/(Math.pow(m,2) + 1); //other intercept's x
var i1 = {x:t1+c.x, y:m*t1+b+c.y}; //intercept point 1
var i2 = {x:t2+c.x, y:m*t2+b+c.y}; //intercept point 2
return [i1, i2];
}
}
其他回答
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;
}
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。
我只是需要它,所以我想出了这个解决方案。语言是maxscript,但是它应该很容易被翻译成任何其他语言。 sideA, sideB和CircleRadius为标量,其余变量为[x,y,z]。假设z=0在XY平面上求解
fn projectPoint p1 p2 p3 = --project p1 perpendicular to the line p2-p3
(
local v= normalize (p3-p2)
local p= (p1-p2)
p2+((dot v p)*v)
)
fn findIntersectionLineCircle CircleCenter CircleRadius LineP1 LineP2=
(
pp=projectPoint CircleCenter LineP1 LineP2
sideA=distance pp CircleCenter
--use pythagoras to solve the third side
sideB=sqrt(CircleRadius^2-sideA^2) -- this will return NaN if they don't intersect
IntersectV=normalize (pp-CircleCenter)
perpV=[IntersectV.y,-IntersectV.x,IntersectV.z]
--project the point to both sides to find the solutions
solution1=pp+(sideB*perpV)
solution2=pp-(sideB*perpV)
return #(solution1,solution2)
)
如果直线的坐标为A.x, A.y和B.x, B.y,圆心为C.x, C.y,则直线公式为:
x = A.x * t + B.x * (1 - t)
y = A.y * t + B.y * (1 - t)
0 < = t < = 1
这个圆是
(C.x - x)²+ (C.y - y)²= R²
如果你把直线的x和y公式代入圆公式,你会得到一个t的二阶方程,它的解是交点(如果有的话)。如果你得到的t小于0或大于1,那么它不是一个解,但它表明这条线“指向”圆的方向。
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!