• 2D空间中求线段与圆的交点


    出处: https://answers.unity.com/questions/366802/get-intersection-of-a-line-and-a-circle.html

    测试脚本(返回值为交点数量):

    using System.Collections;
    using System.Collections.Generic;
    using UnityEngine;
    
    public class LineCircleIntersect : MonoBehaviour
    {
        public Transform a;
        public Transform b;
    
        public Transform circleCenter;
        public float radius;
    
    
        void OnDrawGizmos()
        {
            if (a == null || b == null || circleCenter == null) return;
    
            var intersect1 = default(Vector2);
            var intersect2 = default(Vector2);
            var intersectCount = BetweenLineAndCircle(circleCenter.position, radius, a.position, b.position, out intersect1, out intersect2);
    
            if (intersectCount > 0)
                Gizmos.DrawWireSphere(intersect1, 0.1f);
    
            if (intersectCount > 1)
                Gizmos.DrawWireSphere(intersect2, 0.1f);
    
            Gizmos.DrawLine(a.position, b.position);
            Gizmos.DrawWireSphere(circleCenter.position, radius);
        }
    
        int BetweenLineAndCircle(
         Vector2 circleCenter, float circleRadius,
         Vector2 point1, Vector2 point2,
         out Vector2 intersection1, out Vector2 intersection2)
        {
            float t;
    
            var dx = point2.x - point1.x;
            var dy = point2.y - point1.y;
    
            var a = dx * dx + dy * dy;
            var b = 2 * (dx * (point1.x - circleCenter.x) + dy * (point1.y - circleCenter.y));
            var c = (point1.x - circleCenter.x) * (point1.x - circleCenter.x) + (point1.y - circleCenter.y) * (point1.y - circleCenter.y) - circleRadius * circleRadius;
    
            var determinate = b * b - 4 * a * c;
            if ((a <= 0.0000001) || (determinate < -0.0000001))
            {
                // No real solutions.
                intersection1 = Vector2.zero;
                intersection2 = Vector2.zero;
                return 0;
            }
            if (determinate < 0.0000001 && determinate > -0.0000001)
            {
                // One solution.
                t = -b / (2 * a);
                intersection1 = new Vector2(point1.x + t * dx, point1.y + t * dy);
                intersection2 = Vector2.zero;
                return 1;
            }
    
            // Two solutions.
            t = (float)((-b + Mathf.Sqrt(determinate)) / (2 * a));
            intersection1 = new Vector2(point1.x + t * dx, point1.y + t * dy);
            t = (float)((-b - Mathf.Sqrt(determinate)) / (2 * a));
            intersection2 = new Vector2(point1.x + t * dx, point1.y + t * dy);
    
            return 2;
        }
    }
  • 相关阅读:
    响应式布局
    bootstrap--前端开发框架
    ADO.NET Entity Framework
    dns
    自动完成脚本
    一个Banner广告收缩效果
    对联广告2
    蓝色经典的对联广告代码
    Js弹性漂浮广告代码
    jquery悬停tab2
  • 原文地址:https://www.cnblogs.com/hont/p/8991751.html
Copyright © 2020-2023  润新知