LISP、.NET计算两个圆的交点(附带EXCEL)2024.6.11更新
本帖最后由 fangmin723 于 2024-6-11 10:40 编辑2022.5.9更新：优化当X轴相同时报错！
;;说明:求两圆交点
;;参数:cen1:圆心1
;;参数:r1:半径1
;;参数:cen2:圆心2
;;参数:r2:半径2
;;返回:有交点侧返回交点列表，没有则返回nil
(defun 2ci(cen1 r1 cen2 r2 / a1 a2 a3 b2 b3 c3 cx1 cx2 cy1 cy2 delta x1 x12 x2 y1 y12 y2)
(setq cx1 (car cen1) cy1 (cadr cen1) cx2 (car cen2) cy2 (cadr cen2))
(cond
((and (= cx1 cx2) (/= cy1 cy2))
(setq y12 (/ (+ (- (* r1 r1) (* r2 r2)) (- (* cy2 cy2) (* cy1 cy1))) 2.0 (- cy2 cy1)))
(setq a3 1 b3 (* -2 cx1) c3 (+ (* (- y12 cy1) (- y12 cy1)) (* cx1 cx1) (* -1.0 r1 r1)))
(setq delta (- (* b3 b3) (* 4.0 a3 c3)))
(cond
((> delta 0)
(setq x1 (/ (+ (* -1.0 b3) (sqrt (- (* b3 b3) (* 4.0 a3 c3)))) (* 2.0 a3)))
(setq x2 (/ (- (* -1.0 b3) (sqrt (- (* b3 b3) (* 4.0 a3 c3)))) (* 2.0 a3)))
(list (list x1 y12) (list x2 y12))
)
((= delta 0) (list (list (/ (* -1.0 b3) (* 2.0 a3)) y12)))
(t (princ "\n没有交点！") nil)
)
)
((and (/= cx1 cx2) (= cy1 cy2))
(setq x12 (/ (+ (- (* r1 r1) (* r2 r2)) (- (* cx2 cx2) (* cx1 cx1))) 2.0 (- cx2 cx1)))
(setq a3 1 b3 (* -2 cy1) c3 (+ (* (- x12 cx1) (- x12 cx1)) (* cy1 cy1) (* -1.0 r1 r1)))
(setq delta (- (* b3 b3) (* 4.0 a3 c3)))
(cond
((> delta 0)
(setq y1 (/ (+ (* -1.0 b3) (sqrt (- (* b3 b3) (* 4.0 a3 c3)))) (* 2.0 a3)))
(setq y2 (/ (- (* -1.0 b3) (sqrt (- (* b3 b3) (* 4.0 a3 c3)))) (* 2.0 a3)))
(list (list x12 y1) (list x12 y2))
)
((= delta 0) (list (list x12 (/ (* -1.0 b3) (* 2.0 a3)))))
(t (princ "\n没有交点！") nil)
)
)
((and (= cx1 cx2) (= cy1 cy2))
(cond
((= r1 r2) (alert "\n同一个圆求交点，怕不是个傻子吧你！"))
(t (alert "\n同心圆求交点，你没毛病吧！"))
)
nil
)
(t
(setq a1 (+ (- (* r1 r1) (* r2 r2)) (- (* cx2 cx2) (* cx1 cx1)) (- (* cy2 cy2) (* cy1 cy1))))
(setq a2 (/ a1 2.0 (- cy2 cy1)))
(setq b2 (/ (* 1.0 (- cx2 cx1)) (- cy2 cy1)))
(setq a3 (+ 1.0 (* b2 b2)))
(setq b3 (* -1 (+ (* 2.0 cx1) (* 2.0 (- a2 cy1) b2))))
(setq c3 (- (+ (* cx1 cx1) (* (- a2 cy1) (- a2 cy1))) (* r1 r1)))
(setq delta (- (* b3 b3) (* 4.0 a3 c3)))
(cond
((> delta 0)
(setq x1 (/ (+ (* -1.0 b3) (sqrt (- (* b3 b3) (* 4.0 a3 c3)))) (* 2.0 a3)))
(setq x2 (/ (- (* -1.0 b3) (sqrt (- (* b3 b3) (* 4.0 a3 c3)))) (* 2.0 a3)))
(setq y1 (- a2 (* b2 x1)))
(setq y2 (- a2 (* b2 x2)))
(list (list x1 y1) (list x2 y2))
)
((= delta 0) (list (list (setq x1 (/ (* -1.0 b3) (* 2.0 a3))) (- a2 (* b2 x1)))))
(t (princ "\n没有交点！") nil)
)
)
)
)
(2ci (list 100.0 100.0) 50 (list 120.0 30.0) 70.0)
((149.569 93.4483) (61.3745 68.2499))
2024.6.11新增：.NET方式求交：
/// <summary>
/// 数学方式求两圆交点
/// </summary>
/// <param name="center1">圆心1</param>
/// <param name="radius1">半径1</param>
/// <param name="center2">圆心2</param>
/// <param name="radius2">半径2</param>
/// <returns>两圆焦点集合</returns>
public static List<Point3d> CircleInsert(this Point3d center1, double radius1, Point3d center2, double radius2)
{
//两圆同心
if ((center1.X - center2.X).Abs() <= 1e-8 && (center1.Y - center2.Y).Abs() <= 1e-8 && (center1.Z - center2.Z).Abs() <= 1e-8)
return [];
//两圆相离或大圆包含小圆
if (center1.DistanceTo(center2) > radius1 + radius2 || (center1.DistanceTo(center2) + radius1.Min(radius2)) < radius1.Max(radius2))
return [];
var centerX1 = center1.X;
var centerY1 = center1.Y;
var centerX2 = center2.X;
var centerY2 = center2.Y;
if (centerX1 == centerX2 && centerY1 != centerY2)
{
var resultY = (radius1.Pow() - radius2.Pow() + (centerY2.Pow() - centerY1.Pow())) / 2.0 / (centerY2 - centerY1);
var argA = 1.0;
var argB = -2 * centerX1;
var argC = centerX1.Pow() + (resultY - centerY1).Pow() - radius1.Pow();
var delta = argB.Pow() - (4.0 * argA * argC);
return delta switch
{
> 0 =>
[
new(((-1 * argB ) + delta.Sqrt()) / 2.0 / argA, resultY, 0),
new(((-1 * argB ) - delta.Sqrt()) / 2.0 / argA, resultY, 0)
],
0 => ,
_ => [],
};
}
else if (centerX1 != centerX2 && centerY1 == centerY2)
{
var resultX = (radius1.Pow() - radius2.Pow() + (centerX2.Pow() - centerX1.Pow())) / 2.0 / (centerX2 - centerX1);
var argA = 1.0;
var argB = -2 * centerY1;
var argC = centerY1.Pow() + (resultX - centerX1).Pow() - radius1.Pow();
var delta = argB.Pow() - (4.0 * argA * argC);
return delta switch
{
> 0 =>
[
new( resultX,((-1 * argB ) + delta.Sqrt()) / 2.0 / argA, 0),
new( resultX,((-1 * argB ) - delta.Sqrt()) / 2.0 / argA, 0)
],
0 => ,
_ => [],
};
}
else
{
var argA1 = radius1.Pow() - radius2.Pow() + (centerX2.Pow() - centerX1.Pow()) + (centerY2.Pow() - centerY1.Pow());
var argA2 = argA1 / 2.0 / (centerY2 - centerY1);
var argB2 = (centerX2 - centerX1) / (centerY2 - centerY1);
var argA3 = 1 + argB2.Pow();
var argB3 = -2 * (centerX1 + (argA2 - centerY1) * argB2);
var argC3 = centerX1.Pow() + (argA2 - centerY1).Pow() - radius1.Pow();
var delta = argB3.Pow() - (4.0 * argA3 * argC3);
if (delta > 0)
{
var x1 = ((-1 * argB3) + delta.Sqrt()) / 2.0 / argA3;
var x2 = ((-1 * argB3) - delta.Sqrt()) / 2.0 / argA3;
var y1 = argA2 - (argB2 * x1);
var y2 = argA2 - (argB2 * x2);
return ;
}
else if (delta == 0)
{
var x1 = (-1 * argB3) / 2.0 / argA3;
var y1 = argA2 - (argB2 * x1);
return ;
}
else return [];
}
}
.NET方式求交所用扩展方法：
/// <summary>
/// 获取最小值
/// </summary>
/// <param name="value">数值参数</param>
/// <param name="ints">数值参数，可变</param>
/// <returns>最小值</returns>
public static int Min(this int value, params int[] ints)
{
return Math.Min(value, ints.Min());
}
/// <summary>
/// 获取最大值
/// </summary>
/// <param name="value">数值参数</param>
/// <param name="ints">数值参数，可变</param>
/// <returns>最大值</returns>
public static int Max(this int value, params int[] ints)
{
return Math.Max(value, ints.Max());
}
/// <summary>
/// 获取最小值
/// </summary>
/// <param name="value">数值参数</param>
/// <param name="doubles">数值参数，可变</param>
/// <returns>最小值</returns>
public static double Min(this double value, params double[] doubles)
{
return Math.Min(value, doubles.Min());
}
/// <summary>
/// 获取最大值
/// </summary>
/// <param name="value">数值参数</param>
/// <param name="doubles">数值参数，可变</param>
/// <returns>最大值</returns>
public static double Max(this double value, params double[] doubles)
{
return Math.Max(value, doubles.Max());
}
/// <summary>
/// 返回双精度浮点数字的绝对值。
/// </summary>
/// <param name="value">大于或等于 Double.MinValue 但小于或等于 Double.MaxValue 的数字</param>
/// <returns>双精度浮点数 x，使 0 ≤ x ≤ Double.MaxValue</returns>
public static double Abs(this double value) => Math.Abs(value);
/// <summary>
/// 返回指定数字的指定次幂
/// </summary>
/// <param name="x">要乘幂的双精度浮点数</param>
/// <param name="y">指定幂的双精度浮点数,默认：2</param>
/// <returns>数字 x 的 y 次幂</returns>
public static double Pow(this double x, double y = 2) => Math.Pow(x, y);
/// <summary>
/// 返回指定数字的平方根
/// </summary>
/// <param name="d">将查找其平方根的数字</param>
/// <returns>
/// <code>
/// d为零或正数 => d 的正平方根
/// d为负数 => NaN
/// d等于NaN => NaN
/// d等于PositiveInfinity => PositiveInfinity
/// </code>
/// </returns>
public static double Sqrt(this double d) => Math.Sqrt(d);
5月9日优化更新 不知有何用处，谢谢大神分享， 不错的构想，谢谢大神的分享！下载来学习一下 宁静港湾 发表于 2022-5-10 07:53
不知有何用处，谢谢大神分享，
何用，用处大着呢，当你想求两个圆的交点时，你一般会怎么做，先通过函数绘制出两个圆，然后再用交点函数处理，然后再转表，如果不需要圆的时候还得删除，在不需要圆的情况下，这种直接纯数学的方式更为简便！ 感谢大佬共享！ 感谢大佬分享，学习了
2024.6.11：新增.NET方式求交：
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