Intersection of two moving objects

Solution 1:

We can describe the trajectory of the first object as $$ r_1(t) = A (1 - \lambda) + B \lambda \quad U_1 T_1 = \lvert AB \rvert \quad \lambda = t / T_1 = \frac{U_1}{\lvert AB \rvert} t $$ Similar we get $$ r_2(t) = C (1 - \mu) + D \mu \quad U_2 T_2 = \lvert CD \rvert \quad \mu = t / T_2 = \frac{U_2}{\lvert CD \rvert} t $$ for the second object moving from $C$ to some point $D$, needing time $T_2$.

Interesting it gets if $D$ is the collision point: $$ D = r_1(T_2) = r_2(T_2) $$ for $T_2 \le T_1$. This gives $$ D = A (1-\lambda) + B \lambda = A + \lambda (B-A) \quad \lambda = \frac{U_1}{\lvert AB \rvert} T_2 $$ so with the unit vectors (the first known, the second sought): $$ e_A = \frac{B-A}{\lvert AB \rvert} \quad e_D = \frac{D-C}{\lvert CD \rvert} = (\cos(\phi), \sin(\phi)) \\ D = C + e_D \lvert CD \rvert = C + e_D U_2 T_2 = A + e_A U_1 T_2 $$

Per component we get: \begin{align} C_x - A_x &= (e_{A,x} U_1 - \cos(\phi) U_2) T_2 \quad (*) \\ C_y - A_y &= (e_{A,y} U_1 - \sin(\phi) U_2) T_2 \end{align} where we equate for $T_2$: $$ \frac{C_x - A_x}{e_{A,x} U_1 - \cos(\phi) U_2} = \frac{C_y - A_y}{e_{A,y} U_1 - \sin(\phi) U_2} \quad (**) $$ Assuming $A \ne C$ this is one equation in one unkown $\phi$, so it could be solvable, if the given $A$, $C$, $U_1$, $U_2$ permit it. Alternatively one could use $$ \cos(\phi) = e_{D,x} \quad \sin(\phi) = e_{D,y} \quad e_{D,x}^2 + e_{D,y}^2 = 1 $$ in equation $(**)$ and try to solve for $e_{D,x}$ or $e_{D,y}$.

Changing the coordinate system can simplify the above. E.g. one might translate and rotate such that $A_x = C_x$. This would simplify $(*)$ to $$ \cos(\phi) = \frac{U_1}{U_2} e_{A,x} $$

Solution 2:

This diagram shows a geometric solution to finding the desired direction of $\vec U_2$. The idea here is that we have changed the entire frame of reference by adding vector $-\vec U_1$ to all velocities. In this frame, point $A$ is stationary and point $C$ has been given a velocity of $-\vec U_1$. We want to add an additional velocity vector $\vec U_2$ to point $C$ to aim it at stationary point $A$.

As the diagram shows, there may be two different directions of $\vec U_2$ that would suffice. This will happen if the angle $\theta$ between vectors $-\vec U_1$ and $\overrightarrow{CA}$ is acute and $|\vec U_1|\sin\theta<|\vec U_2|<|\vec U_1|$. It should be clear what conditions must hold for there to be no solutions. (This discussion is very similar to explaining the cases in solving the triangle in SSA for trigonometry class.)