MOC numerical, characteristics not on diagonal (25 cm)

Idea

When the characteristics do not lie on the diagonal of the grid, $a\Delta t<\Delta x$: the $C^+$ and $C^-$ lines through P (at $t_2$) meet the line $t_1$ at points R and S that lie between the grid points, not at A and B. $Q$ and $H$ at R and S are found by linear interpolation along the time line $t_1$.

 t2 :                P
                    /|\
               C+  / | \   C-
                  /  |  \
 t1 :   A--R---------C---------S--B
           |<-a dt ->|<-a dt ->|
        |<--- dx --->|<--- dx --->|

Here R lies on the $C^+$ line between A and C, and S on the $C^-$ line between C and B, each at a distance $a\Delta t$ from C.

Interpolation factor

$$\theta=\frac{a\,\Delta t}{\Delta x}=\frac{850\times1}{1000}=0.85$$

Point R (between A and C, $C^+$ foot):

$$Q_R=Q_C-\theta\,(Q_C-Q_A),\qquad H_R=H_C-\theta\,(H_C-H_A)$$ $$Q_R=0.64-0.85(0.64-0.6)=0.606\ \text{m}^3/\text{s},\qquad H_R=20.4-0.85(20.4-20)=20.06\ \text{m}$$

Point S (between C and B, $C^-$ foot):

$$Q_S=Q_C-\theta\,(Q_C-Q_B),\qquad H_S=H_C-\theta\,(H_C-H_B)$$ $$Q_S=0.64-0.85(0.64-0.65)=0.6485\ \text{m}^3/\text{s},\qquad H_S=20.4-0.85(20.4-20.6)=20.57\ \text{m}$$

Answer: at $t_1=1$ s, $Q_R=0.606$ m$^3$/s, $H_R=20.06$ m on the $C^+$ line, and $Q_S=0.6485$ m$^3$/s, $H_S=20.57$ m on the $C^-$ line. These replace $Q_A,H_A$ and $Q_B,H_B$ in the $C^+$ and $C^-$ equations

$$H_P=H_R-B_p(Q_P-Q_R)-R_pQ_R|Q_R|,\qquad H_P=H_S+B_p(Q_P-Q_S)+R_pQ_S|Q_S|$$

with $B_p=a/(gA_p)$ and $R_p=f\,a\Delta t/(2gDA_p^2)$ (the friction length is $a\Delta t=850$ m).