XNA矢量數學運動

我在c＃xna上做了一個自頂向下的2D rpg遊戲，用於我在大學的遊戲設計課。我正在嘗試創建一個簡單的AI，將敵人移向玩家。目前我的代碼如下XNA矢量數學運動

/// <summary> 
    /// method to move the enemy 
    /// </summary> 
    /// <param name="target">the position of the target</param> 
    /// <returns>the new position to be moved to</returns> 
    public virtual Vector2 move(Vector2 target) 
    { 
     Vector2 temp = (target - Position); // gets the difference between the target and position 
     temp.Normalize();     // sets the vector to unit vector 
     temp *= moveSpeed;     // sets the vector to be the length of moveSpeed 
     float x = temp.X; 
     float y = temp.Y; 
     float xP, yP; 
     double angle = Math.Acos(((x * direction.X) + (y * direction.Y))/(temp.Length() * direction.Length())); //dot product finds the angle between temp and direction 
     angle *= agility;                      //gets the angle to move based on agility 
     xP = (float)(Math.Cos(angle) * (x - direction.X) - Math.Sin(angle) * (y - direction.Y) + x); 
     yP = (float)(Math.Sin(angle) * (x - direction.X) - Math.Cos(angle) * (y - direction.Y) + y);   // these lines rotate the point x,y around the direction vector by angle "angle" 
     return new Vector2(xP, yP); 
    }

在正確的目標是通過在更新方法：

/// <summary> 
    /// updates the enemy 
    /// </summary> 
    public void update() 
    { 
     this.Position = move(Game1.player.Position); 
    }

但敵人並沒有移動。我將代碼添加到構造函數中，確保敏捷性和移動速度不爲0.更改這些值什麼也不做。

感謝您的任何幫助。

來源

2012-10-03 Mike Eaton

爲什麼不使用'Dot'方法？ – phoog

您也可以使用'Math.Atan2（（PlayerPosition.Y - EnemyPosition.Y），（PlayerPosition.X - EnemyPosition.Y））;'來計算角度。我發現它更易於閱讀和理解。 –

在你的代碼返回這個（代碼爲方向的角度）：

xP = (float)(Math.Cos(angle) * (x - direction.X) - Math.Sin(angle) * (y - direction.Y) + x); 
yP = (float)(Math.Sin(angle) * (x - direction.X) - Math.Cos(angle) * (y - direction.Y) + y);   // these lines rotate the point x,y around the direction vector by angle "angle" 

return new Vector2(xP, yP);

但你需要返回本作的舉動：

temp *= moveSpeed;     // sets the vector to be the length of moveSpeed  
float x = temp.X;  
float y = temp.Y; 
... 

return new Vector2(x, y);

來源

2012-10-03 19:11:36 marcel

XNA矢量數學運動

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