Unity 如何实现游戏中技能的扇形攻击范围

CoderZ

发布于 2022-08-29 15:41:04

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文章被收录于专栏：用户10004205的专栏用户10004205的专栏

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代码可运行

假设人物A向正前方释放一个技能，攻击范围为一个扇形，如何判断人物B是否在该范围内受到攻击。

1. 向量A的正前方forward 与 A指向B的向量direction 的角度offsetAngle小于扇形角度的1/2

2. 向量direction的模长magnitude，即A到B的距离小于半径

满足上面这两个条件即可判断B在扇形区域内，涉及的运算：点乘积、反余弦

/// <summary>
/// 判断target是否在扇形区域内
/// </summary>
/// <param name="sectorAngle">扇形角度</param>
/// <param name="sectorRadius">扇形半径</param>
/// <param name="attacker">攻击者的transform信息</param>
/// <param name="target">目标</param>
/// <returns>目标target在扇形区域内返回true 否则返回false</returns>
public bool IsInRange(float sectorAngle, float sectorRadius, Transform attacker, Transform target)
{
    //攻击者位置指向目标位置的向量
    Vector3 direction = target.position - attacker.position;
    //点乘积结果
    float dot = Vector3.Dot(direction.normalized, transform.forward);
    //反余弦计算角度
    float offsetAngle = Mathf.Acos(dot) * Mathf.Rad2Deg; //弧度转度
    return offsetAngle < sectorAngle * .5f && direction.magnitude < sectorRadius;
}

扇形范围由扇形的角度和半径构成，定义变量：

//扇形角度
[SerializeField] private float angle = 80f;
//扇形半径
[SerializeField] private float radius = 3.5f;
//物体B
[SerializeField] private Transform b;

定义bool类型变量flag，用来记录判断结果：

private bool flag;

private void Update()
{
    flag = IsInRange(angle, radius, transform, b);
}

完整测试脚本：

using UnityEngine;
using UnityEditor;

public class Foo : MonoBehaviour
{
    //扇形角度
    [SerializeField] private float angle = 80f;
    //扇形半径
    [SerializeField] private float radius = 3.5f;
    //物体B
    [SerializeField] private Transform b;

    private bool flag;

    private void Update()
    {
        flag = IsInRange(angle, radius, transform, b);
    }

    /// <summary>
    /// 判断target是否在扇形区域内
    /// </summary>
    /// <param name="sectorAngle">扇形角度</param>
    /// <param name="sectorRadius">扇形半径</param>
    /// <param name="attacker">攻击者的transform信息</param>
    /// <param name="target">目标</param>
    /// <returns>目标target在扇形区域内返回true 否则返回false</returns>
    public bool IsInRange(float sectorAngle, float sectorRadius, Transform attacker, Transform target)
    {
        //攻击者位置指向目标位置的向量
        Vector3 direction = target.position - attacker.position;
        //点乘积结果
        float dot = Vector3.Dot(direction.normalized, transform.forward);
        //反余弦计算角度
        float offsetAngle = Mathf.Acos(dot) * Mathf.Rad2Deg;
        return offsetAngle < sectorAngle * .5f && direction.magnitude < sectorRadius;
    }

    private void OnDrawGizmos()
    {
        Handles.color = flag ? Color.cyan : Color.red;

        float x = radius * Mathf.Sin(angle / 2f * Mathf.Deg2Rad);
        float y = Mathf.Sqrt(Mathf.Pow(radius, 2f) - Mathf.Pow(x, 2f));
        Vector3 a = new Vector3(transform.position.x - x, 0f, transform.position.z + y);
        Vector3 b = new Vector3(transform.position.x + x, 0f, transform.position.z + y);

        Handles.DrawLine(transform.position, a);
        Handles.DrawLine(transform.position, b);

        float half = angle / 2;
        for (int i = 0; i < half; i++)
        {
            x = radius * Mathf.Sin((half - i) * Mathf.Deg2Rad);
            y = Mathf.Sqrt(Mathf.Pow(radius, 2f) - Mathf.Pow(x, 2f));
            a = new Vector3(transform.position.x - x, 0f, transform.position.z + y);
            x = radius * Mathf.Sin((half - i - 1) * Mathf.Deg2Rad);
            y = Mathf.Sqrt(Mathf.Pow(radius, 2f) - Mathf.Pow(x, 2f));
            b = new Vector3(transform.position.x - x, 0f, transform.position.z + y);

            Handles.DrawLine(a, b);
        }
        for (int i = 0; i < half; i++)
        {
            x = radius * Mathf.Sin((half - i) * Mathf.Deg2Rad);
            y = Mathf.Sqrt(Mathf.Pow(radius, 2f) - Mathf.Pow(x, 2f));
            a = new Vector3(transform.position.x + x, 0f, transform.position.z + y);
            x = radius * Mathf.Sin((half - i - 1) * Mathf.Deg2Rad);
            y = Mathf.Sqrt(Mathf.Pow(radius, 2f) - Mathf.Pow(x, 2f));
            b = new Vector3(transform.position.x + x, 0f, transform.position.z + y);

            Handles.DrawLine(a, b);
        }
    }
}

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原始发表：2021-10-12，如有侵权请联系 cloudcommunity@tencent.com 删除

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