APPENDIX A

Let

$$\tau_0 = \tau(x_r,0)={z_t \over v_t}\left({1 \over \cos \alpha} + {1 \over \cos \beta}\right). \eqno(A.1)$$

Substituting x and z from equation (4) into equation (6), we have

$$\sqrt{(x_r-x_s-\lambda \gamma \sin \beta)^2 + \lambda^2 (1-... ... \sin^2 \beta + \lambda^2(1-\gamma^2 \sin^2 \beta)}. \eqno(A.2)$$

Solving this equation for $\lambda$ yields:

$$\lambda = {v^2_m \tau^2_0 - (x_r-x_s)^2 \over 2[v_m \tau_0-(x_r-x_s) \gamma \sin \beta]}. \eqno(A.3)$$

According to the geometry shown in Figure 2,

$$x_r=x_s+z_t(\tan \alpha + \tan \beta). \eqno(A.4)$$

Substituting equation (A.4) into equations (A.3) and (4), we obtain equations (7) and (8):

$$\lambda ={z_t \over \gamma \cos \beta}[1+(\gamma^2-1) {\cos^2 ({1 \over 2}(\beta -\alpha)) \over \cos \alpha \cos \beta }]$$

and

$$\begin{array} {lll} x & = & x_t + z_t \tan \beta - \lambda \... ...a \\ z & = & \lambda \sqrt{1-\gamma^2 \sin^2 \beta}.\end{array}$$