Uniform Acceleration Motion:

If we can identify points moving with constant acceleration, Equations 8 and 9 would hold for both views for the same $\beta$'s. The time $t$ can be replaced with the frame number. From the image velocities of the projections of 4 points in 8 frames in view ${\bf A}$, $\beta$'s that characterize the point configuration can be computed. We want to identify the corresponding frame $k$ in view ${\bf A}$ for the frame $j$ in view ${\bf B}$. The image velocities of the projections of the four points in view ${\bf B}$ at time instant $j$ are ($v_{ixj}$,$v_{iyj}$), $1 \le i \le 4$. Therefore, the shift is $(k-j)$. We have

$$(\; \beta_0 + \beta_1 k + \beta_2 k^2 + \beta_3 k^3 \;) v_{1xj} +$$

$$(\; \beta_4 + \beta_5 k + \beta_6 k^2 + \beta_7 k^3 \;\;) v_{2xj} +$$

$$(\beta_8 + \beta_9 k + \beta_{10} k^2 + \beta_{11} k^3) v_{3xj} +$$

$$(\;\beta_{12} + \beta_{13} k + \beta_{14} k^2 + \beta_{15} t^3) v_{4xj} = 0$$ (20)

And a similar relation in $v_y$ values. These can be written as

$$\eta_1(k) v_{1xj} + \eta_2(k) v_{2xj} + \eta_3(k) v_{3xj} + \eta_4(k) v_{4xj} = 0$$

$$\eta_1(k) v_{1yj} + \eta_2(k) v_{2yj} + \eta_3(k) v_{3yj} + \eta_4(k) v_{4yj} = 0$$ (21)

where $\eta_i(k) = (\beta_{i*4} + \beta_{i*4+1} k + \beta_{i*4+2} k^2 + \beta_{i*4+3} k^3)$, $0 \le i \le 3$. We can solve for $k$ using a linear least squares solution technique by minimizing the sum of squares of the error functions

$$f_1(k) = \eta_1(k) v_{1xj} + \eta_2(k) v_{2xj} + \eta_3(k) v_{3xj} + \eta_4(k) v_{4xj}$$

$$f_2(k) = \eta_1(k) v_{1yj} + \eta_2(k) v_{2yj} + \eta_3(k) v_{3yj} + \eta_4(k) v_{4yj}.$$

Alternately, we can solve for the roots of a cubic polynomial of the form

$$\gamma_0 k^3 + \gamma_1 k^2 + \gamma_2 k + \gamma_3 = 0$$ (22)

where $\gamma_0 = (\beta_3 v_{1xj} + \beta_7 v_{2xj} + \beta_{11} v_{3xj} + \beta_{15... ...3 = (\beta_0 v_{1xj} + \beta_4 v_{2xj} + \beta_8 v_{3xj} + \beta_{12} v_{4xj})$.