Guaranteed Hit?

hit rate of max escape angle based random targeting is roughly bot width in radians / max escape angle

max escape angle never changes for the same bullet power, and bot wdith in radians is roughly 18 / distance. Then, as long as the distance is close enough, random hit rate can be close enough to 1, making a hit guaranteed.

note that the actual hit rate formula is harder to calc when edge cases are considered, but the main factors affecting hit rate are still covered.

I use asin(botWidth / (distance - 18)) which takes bot width and the square shape of the bot into account(Isn't it better than botWidth / distance) but I am pretty sure that everybody else uses the formula which calculates the exact width.

asin(botWdith / (distance - 18)) is incorrect because when distance = 18 it gives NaN instead of PI.

if you assume a bot is a circle then asin(botWidth / distance) is correct; if you assume a bot is a square then atan(botWidth / (distance - 18)) is correct as long as absolute bearing from source to target is 0, 90, 180 or 270 degrees.

however, a moving bot is neither a circle nor a square, that’s why precise intersection is used.

Anyway, botWidth / distance is fine as long as distance is far, and the result is almost the same as the asin/atan version.

It gives a NaN but a distance less than 36 is impossible in robocode. botWidth / distance is always incorrect, asin(botWidth / distance) is always incorrect too but it less incorrect however, asin(botWidth / (distance - 18) is correct in four cases and gives higher results than all other imprecise formulas which always results in closer results to the precise one.

a distance less than 36 is impossible in robocode, but a correct formula should give correct result in this case. So I prefer the atan approach.

asin approach is incorrect by geometry as well.

however trigs are always expensive, so using 18 / distance is acceptable since it gives similar result.

I haven't checked it in Java but asin(positive infinity) results in positive infinity which is true for the bot width formula if we assume that the bullet starts from the robot location. The same case also happens with Traditional MEA when bullet speed is 0.

Wrong. Infinity means nothing in radians. Valid values are 0~2pi or -pi~pi or so.

An angle should never be infinty radians.

If it means nothing can't we assume that the traditional formula is completely wrong since with a bullet velocity of 0, it will return positive infinity.

Bullet velocity = 0 is another geometrical situation. But in this case, distance < 36 should still fit in to your image. Or, it must be wrong.l

Ah, thanks. I like to think of "bot width" as a "radar shadow" or "cross section" in this sort of scenario since "width" already has an established (and constant) meaning.