# And to make it even faster

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:

$\theta = \int_0^{t_1} \frac{v_{lateral}}{d_{initial} + v_{retreat} t}dt$

where $\theta$ is the escape angle, $t_1$ is the total time (from bullet fired, to bullet hit). (this formula is true even if $v_{lateral}$ or $v_{retreat}$ is not constant)

Note that $d_{initial} + v_{retreat} t$ is bigger than (his) $(v_{bullet} - \overline{v_{retreat}})t_1$ almost all of the time,

as $d_{initial} = (v_{bullet} - \overline{v_{retreat}})t_1$.


Therefore his integral ($\theta = \int_0^{t_1} \frac{v_{lateral}}{(v_{bullet} - \overline{v_{retreat}})t_1}dt = \frac{v_{lateral}}{v_{bullet} - \overline{v_{retreat}}}$, assume $v_{lateral}$ is constant) is bigger.

Xor (talk)15:02, 24 September 2017