And to make it even faster
I also agree that it's wrong. But why is it bigger than the correct one? Can you clarify that?
I understood that the problem with this is that it assumes that it's optimal to move in a fixed retreat angle (like moving in a perfect orbit), when it's actually true that the best retreat angle isn't a function only of the initial position, but it changes as you move around the wave. Is that the real issue?
I knew it's wrong because I had made the same mistake in my previous research.
It's bigger, simply because it uses some smaller distance to calculate the radians.
The correct one is:
<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{d_{initial} + v_{retreat} t}dt</math>
where <math>\theta</math> is the escape angle, <math>t_1</math> is the total time (from bullet fired, to bullet hit). (this formula is true even if <math>v_{lateral}</math> or <math>v_{retreat}</math> is not constant)
Note that <math>d_{initial} + v_{retreat} t</math> is bigger than (his) <math>(v_{bullet} - \overline{v_{retreat}})t_1</math> almost all of the time,
as <math>d_{initial} =
(v_{bullet} - \overline{v_{retreat}})t_1</math>.
Therefore his integral (<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{(v_{bullet} - \overline{v_{retreat}})t_1}dt = \frac{v_{lateral}}{v_{bullet} - \overline{v_{retreat}}}</math>, assume <math>v_{lateral}</math> is constant) is bigger.
- Actually, I'm not sure that Traditional MEA is correct. It assumes that the bot doesn't change it's move angle until the wave hits. Because of that you can't get a MEA higher than Pi / 2 with Traditional MEA formula. When you move orbitally, lateralVelocity / (bulletSpeed + advancingVelocity) is the formula that will give you the EA so you can get a EA higher than Pi / 2.
You can't get escape angle higher than Pi/2 by the traditional formula simply because it is impossible.
If your formula can, it must be wrong.
- Found it!
- It should be sin(a) / (v / 8 - cos(a) / 2).
- I found a formula higher than Traditional MEA and this one should be correct.
Math.asin(Math.sin(angle) / (bulletSpeed / 8 - Math.cos(angle) / 2))
I don't think that this one is wrong. I only added advancing velocity to the Traditional MEA which shouldn't break anything with the calculations.
I don't need integral. I can get the average distance.
distance - (advancingVelocity * timeToHit / 2) = bulletFloatTime - advancingVelocity / 2
No you can't use average distance, as distance is used like x / distance, not x * distance.
- It is equal at infinity.
(8 / 5 + 1 + 8 / 11) / 3 = 1.109090909... (8 / 5 + 8 / 6.5 + 1 + 8 / 9.5 + 8 / 11) / 5 = 1.080029444...
- This goes closer to 1 every time I decrease the step size.