Tradiotional MEA proof

I came up with a proof for traiotional MEA which fits on a margin :)

First of all we notice that even the slowest bullet is faster than a bot at full speed. Thus any bullet eventually hits a bot. Now we assume that both the bot and the bullet are spherical points and move in vacuum :), also that there are no walls. We also assume that the bot can reach maximum speed instantaneously.

Let's put the firing bot in the origin and the target somewhere on the right of the x axis (no y displacement). Now we notice that in order to hit the bot, the bullet must have exactly the same y projection of the speed as the target bot.

Vbullet_y = sin(theta)*V_bullet = Vbot_y

Sin V_bullet is fixed in a given situation, the maximum theta (MEA) will be when Vbot_y is maximized, i.e. V_bot_y = V_bot. Substituting it above, we obtain traditional formula

MEA = asin( Vbot/Vbullet )

excellent proof! I traped myself in polar coordination and forgot the most important cartesian coordinates;)

I really like this proof!

There is another that you can do, with calculus, so find the escape angle that maximises theta. But that one is more complicated =)

Anyway, the less assumptions you made, the better ;) Do not assume constant speed & direction — use integral instead.

Assume the firing robot sits at the origin, and the target robot sits on the positive x-axis.

Since when hit, <math>y_{bullet} = y_{robot}</math>, we have

<math>\int_{0}^{t_1} v_{bullet} sin(\theta) dt = \int_{0}^{t_1} v_{robot} cos(\alpha) dt</math> where <math>\theta</math> is escape angle, <math>\alpha</math> is retreat angle (the heading of the robot relating to the y-axis).

Since <math>v_{bullet}</math> and <math>\theta</math> is constant, we have

<math>v_{bullet} sin(\theta) t_1 = \int_{0}^{t_1} v_{robot} cos(\alpha) dt</math>

assume <math>v_{robot}</math> is always the max — 8, therefore constant, we have

<math>v_{bullet} sin(\theta) t_1 = v_{robot} \int_{0}^{t_1} cos(\alpha) dt</math>

let <math>\alpha_{0}, \alpha_{2}</math> s.t. <math>cos(\alpha_{0}) \leqslant cos(\alpha) \leqslant cos(\alpha_{2})</math>

there exists <math>\alpha_{1}</math>, <math>cos(\alpha_{0}) \leqslant cos(\alpha_{1}) \leqslant cos(\alpha_{2})</math> s.t. <math>cos(\alpha_{1}) t_1 = \int_{0}^{t_1} cos(\alpha) dt</math>

then we have

<math>v_{bullet} sin(\theta) t_1 = v_{robot} cos(\alpha_1) t_1</math>

<math>v_{bullet} sin(\theta) = v_{robot} cos(\alpha_1)</math>

<math>sin(\theta) = \frac{v_{robot}}{v_{bullet} } cos(\alpha_1)</math>

then we have

<math>\theta_{max} = asin(\frac{v_{robot}}{v_{bullet} })</math>, which can be reached when <math>\alpha_1</math> = 0.

since <math>cos(\alpha) \leqslant cos(0) = 1</math>, <math>\alpha</math> must be always 0, which proved that retreat angle must be always 0.

And we proved that <math>\theta_{max} = asin(\frac{v_{robot}}{v_{bullet} })</math> is true.