Question

我希望以30˚的角度和30 m / s的速度发射我的箭头 GameObject。在脚本中，我为此箭头添加了一个刚体。但是，我也试图在 3D场景中向玩家（远离敌人）方向发射箭头。我无法弄清楚如何插入这些变量来获取“arrowRigidbody.velocity”的Vector3

//THE VARIABLES REFERENCED ABOVE APPEAR LIKE SO:
Rigidbody arrowRigidbody;
Transform playerTransform;
Transform enemyTransform;
float angle = 30f;
float velocity = 30f;

//How do I use these variables in order to shoot the projectile at a speed
//of 30 m/s and an angle of 30˚ in the direction of the player in 3D scene
arrowRigidbody.velocity = /*????*/;

感谢您的时间和耐心：）

Answer 1

使用某些几何，知道向量的幅度（m）为1，y分量为m / 2，x分量为m *（3 ^ .5）/ 2。这将使你的最终价值：

arrowRigidbody.velocity = new Vector2(Mathf.Pow(3, .5f)/2, 1/2) * velocity;

对于变化的角度，你知道x分量将是m * cos（角度），y分量将是m * sin（角度），留下你：

float velx = velocity * Mathf.Cos(angle * Mathf.Deg2Rad);
float vely = velocity * Mathf.Sin(angle * Mathf.Deg2Rad);
arrowRigidbody.velocity = new Vector2(velx, vely);

Answer 2

假设你只拍'前锋'，你可以使用简化版：

    var targetDirn = transform.forward;
    var elevationAxis = transform.right;

    var releaseAngle = 30f;
    var releaseSpeed = 30f;

    var releaseVector = Quaternion.AngleAxis(releaseAngle, elevationAxis) * targetDirn;
    arrowRigidbody.velocity = releaseVector * releaseSpeed;

如果您需要拍摄'离轴'，可以替换前两行：

    var targetDirn = (target.transform.position - transform.position).normalized;
    var elevationAxis = Vector3.Cross(targetDirn, Vector3.up);

Unity在角度发射刚体

2 个答案: