标枪各弹道点切线斜率的标枪投掷仿真计算

Question

我试图模拟标枪在安卓上扔。计算标枪弹道各点切线的斜率。用弹丸运动方程计算弹道坐标

x = (int) (x0 + v0 * t * Math.cos(radians));    //for coordinate x 

和

y = (int) (y0 - v0 * t * Math.sin(radians) + 0.5 * g * t * t);

为了计算切线到标枪轨迹的斜率，我导出了关于x的这个方程：

y = Math.tan(radians) * x - g / (2 * Math.pow(v0, 2) * Math.pow(Math.cos(radians), 2)) * x^2
dy = Math.tan(radians) - (g * x) / (Math.pow(v0, 2) * Math.pow(Math.cos(radians), 2))

问题是，当仰角< 60°时，它能正确地工作，当仰角较大时，不计算正确的斜率。

以下是代码：

public class ThrowJavelin extends ImageView {
    private Context mContext;
    int x0 = -1;
    int y0 = -1;
    int x = x0;
    int y = y0;
    private Handler h;
    private final int FRAME_RATE = 5;
    private double t = 0;
    private float g = 9.81f;
    //initial velocity
    private int v0;
    //elevation angle in radians
    private double radians;
    //javelin current angle in degrees
    private double javelin_angle;



    public ThrowJavelin(Context context, AttributeSet attr)  { super(context, attr); }
    public ThrowJavelin(Context context, AttributeSet attrs, int defStyleAttr){ super(context, attrs, defStyleAttr); }

    public ThrowJavelin(Context context, Bundle args)  {
        super(context);
        mContext = context;
        h = new Handler();
        //input values
        v0 = args.getInt("velocity");
        radians = args.getDouble("radians");
   }

    private Runnable r = new Runnable() {
        @Override
        public void run() {
            invalidate();
        }
    };

    protected void onDraw(Canvas c) {

        Bitmap javelin = BitmapFactory.decodeResource(getResources(), R.drawable.jav);
        DerivativeStructure alpha = null;
        if (x < 0 && y < 0) {
            x0 = 0;
            y0 = c.getHeight() - 200;
            x = x0;
            y = y0;
            javelin = rotateBitmap(javelin, (float) Math.toDegrees(radians));
        } else if (y > y0) { //reset to beginning
            x = x0;
            y = y0;
            t = 0;
            javelin = rotateBitmap(javelin, (float) Math.toDegrees(radians));
        } else {
            //calculate current coordinates (depends on t)
            x = (int) (x0 + v0 * t * Math.cos(radians));
            y = (int) (y0 - v0 * t * Math.sin(radians) + 0.5 * g * t * t);

            if (x == 0) {
                javelin_angle = Math.toDegrees(radians);
            } else {
                // dy of 3rd equation 
                javelin_angle = Math.toDegrees(Math.tan(radians) - (g * x) / (Math.pow(v0, 2) * Math.pow(Math.cos(radians), 2)));
            }
            javelin = rotateBitmap(javelin, javelin_angle);
            t += 0.3;
        }
        c.drawBitmap(javelin, x, y, null);

        h.postDelayed(r, FRAME_RATE);

    }



    public Bitmap rotateBitmap(Bitmap image, double angle){
        float alpha = (float) angle;

        Matrix mat = new Matrix();
        System.out.println(-alpha);
        mat.postRotate(-alpha);
       return Bitmap.createBitmap(image, 0, 0, image.getWidth(), image.getHeight(), mat, true);
    }
}

我真的不明白，为什么不能在更大的角度上正确地工作。有什么想法吗？

Answer 1

首先，您的y(x)解决方案似乎减少了一些变量(例如，x0)。以下是完整的解决方案：

y(x) = y0 + (0.5 * g * (x - x0)^2)/(v0^2 * cos(radians)^2) - (x - x0) * tan(radians)

关于x的导数是：

dy/dx = (g * (x - x0)) / (v0^2 * cos^2(radians)) - tan(radians)

你的解看起来非常相似，只是它的y轴是倒转的，它忽略了初始位置。

与这个导数对应的角度是它的弧切线：

double c = Math.cos(radians);
javelin_angle = Math.toDegrees(Math.atan((g * (x - x0) / (v0 * v0 * c * c) - Math.tan(radians)));

我想，你换y轴是有原因的。所以你可以在这个公式中再做一次。

您的公式适用于小角度的原因是，弧切线接近于小角度的恒等式(红色的恒等式，蓝色的弧形)：

​

​