Calculate intersect point / aim point

Question

I'm working on a simple game, and I need to calculate the position to aim.

Since my math days are a bit behind me, I put together a simple example to illustrate:

So given that I know the position of the two objects, and the velocity, how do I calculate the point at which they intersect (i.e. the point to aim for)?

EDIT: To clarify, the cannon is fired at time 0, there is no waiting.

Answer 1

I was able to solve this using trig instead of vector math. Here's how it looks as a triangle.

Notice that since this is meant for a computer coordinate system, the \$y+\$ axis is down. Also the angles are as such: \$x+ = 0°\$, \$y+ = 90°\$, \$y- = -90°\$, and \$x- = ±180°\$

Additionally, we know that line B's angle is \$∠B = -10°\$.

The speeds don't matter since they can be expressed as a ratio \$r = 120 / 25\$. So \$A\$ can be expressed as \$B * r\$, as shown.

We can calculate the length and angle of \$C\$ as:

$$ \begin{align} C &= \sqrt{(b.x - a.x)^2 + (b.y - a.y)^2} &&\approx 10.05 \\ ∠C &= atan2(b.y - a.y, b.x - a.x) &&\approx 84.29° \end{align} $$ therefore:

\$∠a = ∠C - ∠B ≈ 94.29°\$

Now we have all we need to use the Law of Cosine, which states:

\$A^2 = B^2 + C^2 - 2*B*C*cos(a)\$

Since we know that \$A\$ is a ratio of \$B\$, we can express it like so:

\$(B*r)^2 = B^2 + C^2 - 2*B*C*cos(a)\$

Solve for \$B\$ and we get:

\$B = ± \frac{ \sqrt{ C^2 * (cos(a)^2 + r^2 -1 ) } - cos(a)*C }{r^2 - 1} \approx 2.18\$

Since time is not a factor in this equation, we get two results, depending on if we're forward in time or backwards. We'd have to check that the \$B\$ we get is in the right direction.

Finally, we now know enough to calculate what we want:

$$ \begin{align} c.x &= a.x + B * cos(∠B) &&\approx 4.14 \\ c.y &= a.y + B * sin(∠B) &&\approx 1.62 \end{align} $$

For bonus points, if we wanted the shooting angle \$∠A\$, we can get it like so:

\$∠A = atan2(c.y - b.y, c.x - b.x) \approx -83.72°\$

Answer 2

I tried to solve it following @david van brink comment.

P: intersection point (unknown)
S1: car's starting point(2,2 here)
S2: canon's starting point(3,12 here)
|v1|: car's velocity length (25 here)
|v2|: canon ball's velocity length (120 here)
a: angle between the car's velocity and the X-axis (10 degrees here... or Pi/18)
b: angle between the canon ball's velocity and the X-axis (unknown)

(1): P.y = tan(a) * P.x + S1.x (This is true because P should be a point on the car's path, so it is on a straight line y = a*x + b)

(2): Distance(P,S1) / Distance(P,S2) = 25/120 (This is true because, for example, if the canon ball had double the speed of the car, then, at any given time, the distance covered from the car is half the distance covered from the ball)`

(3): b = arctan((P.y - S2.y)/(P.x - S2.x))

From (1) and (2) ==> {

A = (1 + tan(a)^2) * (|v1|^2 * |v2|^2)

B = |v2|^2 * (2 * tan(a) * (S1.x - S1.y) - 2 * S1.x) +
|v1|^2 * (2 * tan(a) * (S2.x - S2.y) - 2 * S2.x)

C = |v2|^2 * (2 * S1.x^2 + S1.y^2 - 2 * S1.y * S1.x) -
|v1|^2 * (2 * S2.x^2 + S2.y^2 - 2 * S2.y * S2.x)

(4): A * P.x^2 + B * P.x + C = 0
}

By solving (4) ==> you can find P.x,

From P.x and (1) ==> you can find P.y

From P.x and P.y and (3) ==> you can find b.

Answer 3

This is how I would pose the problem:

position equation: $$ \begin{align} P_{dest} &= P_{01} d_1 t_1 = P_{02} d_2 t_2 \\ t_1 &= P_{dest} / (P_{01} d_1) \\ P_{02} d_2 t_2 &= P_{01} d_1 P_{dest} / (P_{01} d_1) \\ t_2 &= (P_{01} d_1 P_{dest} / (P_{01} d_1) ) / P_{02} d_2 \end{align} $$

movement equation: $$ \begin{align} P_{01} + v_1 d_1 lapse_1 &= P_{dest} \\ P_{02} + v_2 d_2 lapse_2 &= P_{dest} \end{align} $$

Relationship: $$ \begin{align} v_1 d_1 lapse_1 = t_1 \\ v_2 d_2 lapse_2 = t_2 \end{align} $$

Then continue to substitute terms into each other until I find what I need.

Considering your question I think what you need is \$d_2\$. I guess the things you know are:

• \$P_{01}\$, \$d_1\$, \$v_1\$
• \$P_{02}\$, \$v_2\$

So try to make an equation that exhibits \$d_2=\$ only stuff you know from the set of equations in the drawing. That has to be possible.