Question

我认为我的动力引擎存在数学问题，过去两天我一直试图解决这个问题。我对编程比较陌生，正在Unity中开发2D自上而下的台球游戏。

我的代码附在下方，但基本上玩家会点击球并向后拖动以产生力量。一旦它们达到所需的功率并放开，球将以所需的功率以所需的方向发射。为了这个例子，让我们说我将maxPower设置为100，并且玩家对象的线性拖动和平铺拖动保持不变。在下面的示例场景中，所有镜头都处于全功率状态，我从桌子的左下方拍摄：

我拍摄了一个纯粹的X轴镜头 - 球向右侧100次射击
我射击纯粹的Y轴射击 - 球以100次向上射击
我以45度角拍摄 - 我怎么能算出这里发生了什么？

如果你看下面的代码，我会释放鼠标后使用AddForce将球推到x和y轴上。我跟踪每个镜头在各个角度拍摄的距离，并尝试了以下电源管理：

在x轴和y轴之间分离功率 - 最终产生比纯x / y轴镜头（每轴50功率）弱得多的镜头
为每个轴提供全部功率 - 最终产生更强大的镜头（每轴100个功率）
感觉像金发姑娘一样，我试图让这个正确的计算机使用毕达哥拉斯定理进行计算，这就是我在努力的地方。

非常感谢任何帮助。

下面的代码 - 定义了所有变量，我刚删除了不相关的代码。

    public float maxForce = 100;
    public float forceMultiplier = 2


    void Awake() {

        startPosition = transform.position;

    }

    void Update () {     

        /// Tracks distance (start vs current)
        distance = Vector3.Distance(transform.position,startPosition);

        //On Click
        if (Input.GetMouseButtonDown (0)) {   

            startPosX = Input.mousePosition.x;
            startPosY = Input.mousePosition.y;
        }

        /// While being clicked
        if (Input.GetMouseButton (0)) {

            ///  Determines force to be applied based on mouse drag
            current_xForce = (startPosX - Input.mousePosition.x);
            current_yForce = (startPosY - Input.mousePosition.y);

            // Stores the original drag angle
            yCheck = current_yForce;
            xCheck = current_xForce;


            ///  if current x/y force is greater than maxForce, set to maxForce
            if (current_yForce > 0 && current_yForce > maxForce)
                current_yForce = maxForce;
            if (current_yForce < 0 && current_yForce < -maxForce)
                current_yForce = -maxForce;
            if (current_xForce > 0 && current_xForce > maxForce)
                current_xForce = maxForce;
            if (current_xForce < 0 && current_xForce < -maxForce)
                current_xForce = -maxForce;

            //  Determines the % of x/y while aiming
            current_xPercentage = Mathf.Abs (current_xForce) / (Mathf.Abs (current_xForce) + Mathf.Abs (current_yForce));
            current_yPercentage = Mathf.Abs (current_yForce) / (Mathf.Abs (current_xForce) + Mathf.Abs (current_yForce));

            //  Decides the Power bar% - power determined by highest powered axis (may need improvement)
            current_powerPercent = Mathf.Max (Mathf.Abs (current_xForce)/maxForce, Mathf.Abs (current_yForce)/maxForce);


            //  get angle from start position to relative mouse position... add relative mouse position   NOT WORKING... need degrees
            shotAngle = Vector2.Angle(startPosition,Input.mousePosition);
        }

        // Mouse button released
        if (Input.GetMouseButtonUp (0)) {               

            ///  Only shots with greater than % of total power count - allows you to cancel shots
            if (current_powerPercent >= 0.10) {

                ///  Increase force by public multiplier
                xForce = current_xForce * forceMultiplier ;    
                yForce = current_yForce * forceMultiplier ;



                ///  Use % of absolute xCheck to determine shot angle 
                float xForcePercent = Mathf.Abs (xCheck) / (Mathf.Abs (yCheck) + Mathf.Abs (xCheck));
                float yForcePercent = Mathf.Abs (yCheck) / (Mathf.Abs (yCheck) + Mathf.Abs (xCheck));

                ///  Adds force to x/y based off xForce and the shot angle
                float x_addForce = xForce * xForcePercent;
                float y_addForce = yForce * yForcePercent;

                ///  Attempt at a-squared + b-squared = c-squared.... this is supposed to return c
                powerVar = Mathf.Sqrt(Mathf.Pow (x_addForce*xForcePercent, 2) + Mathf.Pow (y_addForce*yForcePercent, 2));


                ///  Applies the force to the player object.. .for each axis I take powerVar * angle * sign (to adjust for negative values)
                GetComponent<Rigidbody2D>().AddForce(new Vector2(powerVar*xForcePercent*Mathf.Sign(x_addForce), powerVar*yForcePercent*Mathf.Sign(y_addForce)));


                yForce = 0;
                xForce = 0;
                startPosX = 0;
                startPosY = 0;   
                distance = 0;
                startPosition = transform.position;  /// resets Distance counter


            }
        }
    }

Answer 1

Pythagorean theorem表示在直角三角形中，以下情况属实：

c是斜边。现在，在您的情况下，100为c，a与b或45°之间的角度为c。这意味着你是平等的＆＃34;分裂＆＃34; a和b之间的a = b, therefore: c * c = a * a + a * a, which means c * c = 2 * a * a。好吧，让我们这样做：

c = sqrt(2) * a, and therefore
a = c / sqrt(2)

如果我们采取双方的平方根：

a = sqrt(2)/2 * c

您可能会看到以下形式写的相同关系：

你去吧。如果100为a，那么b和70,711将为d。

如果使用trigonometric functions（基于毕达哥拉斯定理），您可以对任何角度进行概括。对于c和a之间的任何角度a = c * cos(d) b = c * sin(d)，以下情况均属实：

 180 degrees = pi radians

如果您打算使用三角函数，请确保您的单位正确。角度通常以弧度给出，而不是度，两者之间的关系是：

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2d Billiard power physics

1 个答案: