Tradiotional MEA proof

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 target robot relating to the y-axis, the direction from y-axis to x-axis is positive).

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.

I'm still sure that getting slightly away is better both with orbital and traditional one.

Better for what? It does not increase your MEA. Though it decreases the target angular size, so it reduces the probability to be hit. Calculating probability to be hit is yet another beast, which we probably should attempt to solve analytically.

There is another issue, the traditional MEA is good only for a single isolated wave, the real question is how to increase the MEA in a stream of constantly fired bullets.

For wave surfers, I think my work is useful — User:Xor/Better_MEA_formula, as the movement is constant with time, it is naturally suitable for multiple waves ;)

The problem with your formula is that it does not match traditional MEA in the border cases.

The second problem that I not fully understand it. It looks like you are integrating angular velocity to an undefined time t.

it matches orbital MEA in border cases though, as always moving perpendicular is impossible, so you have to fallback to orbital ;)

I'm integrating angular velocity from fire time to hit time, which is obviously the escape angle ;) sorry for not saying that clearly in advance, the article is still not finished ;)

It increases also. If the formula below is correct, it is true.

asin(sin(a) / (v / 8 - cos(a) / 2))

You can try brute forcing.