Triangle with $3$ unknowns

Question

I have a situation where I am trying to calculate a leading shot for a character in a 2D top down game. The enemy character moves with a certain speed $s$, which is applied to its normalized direction vector each game tick--so the game ticks are discrete units of "time" as opposed to actual time.

I have my character, which shoots a bullet with speed $z$. During my leading shot calculation, I can aim my character towards the enemy character, and calculate $\phi$ between the line drawn directly between the enemy and my character.

I need to somehow calculate $\theta$, the angle I need to shoot the bullet in order to hit the enemy character...the unknowns are: $t$, $\theta$, and the third angle.

Can't remember how to do this.

Here is a picture of the situation:

Answer 1

By the Law of sines, we have $$ \frac{\sin \theta}{s\cdot t} = \frac{\sin \phi}{z\cdot t} $$ We can cancel the common factors of $t$ and isolate $$ \sin\theta = \frac{s}{z} \sin\phi $$ This should help you find $\theta$.

Answer 2

After seeing the answer above, I was able to mostly solve the problem. There was an issue, however, in that the sign of $\theta$ is always positive when you actually go through the whole solution--calculating the vectors, then $\phi$, then $\theta$ using the law of sines.

So the issue is: which side of the character face vector is the enemy direction vector (assuming the character is facing the enemy). There doesn't seem to be a trivial way to figure this out, given two arbitrary vectors, using the tools a computer has.

The simplest, most straight forward way that I found to accomplish this--without any kind of logic or pattern recognition--was to just use the sign of the determinant of the two vectors--so $\operatorname{sign}(x_1*y_2 - y_1*x_2)$. The asymmetry gives you a left and right with only 2 multiplies and a subtraction (important when you are in a game loop that is doing 120 fps).

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Anyhow, here is how it worked out:

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In order to calculate $\phi$ in the drawing above, I used the dot product:

$$ \mathbf{A}\bullet\mathbf{B} = ||A||\operatorname{cos}\phi $$

Where $\mathbf{A}$ is the unit vector pointing from the enemy to my character (in the drawing), and $\mathbf{B}$ is the unit vector pointing in the enemy's direction of travel.

Having solved for $\phi$ by taking the $\operatorname{arccos}$ of the dot product result, I calculated $\theta$ in the drawing above with the law of sines,

$$ z*t*\operatorname{sin}\theta = s*t*\operatorname{sin}\phi\\ \theta = \operatorname{arcsin}(\frac{s}{z}\operatorname{sin}\phi) $$

But there was a problem: $\theta$ was always positive. The issue was the range of the $\operatorname{arccos}$ function: $[0,\pi]$. As a result of the $\operatorname{arccos}$ range, $\phi$ was always positive, and the character would only lead the enemy when the angular velocity of the enemy relative to the character was positive.

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So, in order to recapture the "sign" of the angle, I think that what you need is the sign of the determinant of the two vectors (did not find an authority or proof...just thought about it and tested).

So,

$$ \theta' = \operatorname{sign}(\operatorname{det} \begin{bmatrix} A_x & A_y \\ B_x & B_y \end{bmatrix} ) * \theta $$

Will give you back the "sign" of the angle.