Difference between revisions of "Thread:User talk:Dsekercioglu/MEA/And to make it even faster/reply (11)"

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Note that <math>d_{initial} + v_{retreat} t</math> is bigger than (his) <math>(v_{bullet} - v_{retreat})t</math> almost all of the time, as <math>d_{initial} + v_{retreat} t =
 
Note that <math>d_{initial} + v_{retreat} t</math> is bigger than (his) <math>(v_{bullet} - v_{retreat})t</math> almost all of the time, as <math>d_{initial} + v_{retreat} t =
  (v_{bullet} - v_{retreat})t</math>. Therefore his integral (<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{(v_{bullet} - v_{retreat})t_1}dt</math>) is bigger.
+
  (v_{bullet} - v_{retreat})t</math>. Therefore his integral (<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{(v_{bullet} - v_{retreat})t_1}dt = \frac{v_{lateral}}{v_{bullet} - v_{retreat}}</math>) is bigger.

Revision as of 15:06, 24 September 2017

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>


Note that <math>d_{initial} + v_{retreat} t</math> is bigger than (his) <math>(v_{bullet} - v_{retreat})t</math> almost all of the time, as <math>d_{initial} + v_{retreat} t =

(v_{bullet} - v_{retreat})t</math>. Therefore his integral (<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{(v_{bullet} - v_{retreat})t_1}dt = \frac{v_{lateral}}{v_{bullet} - v_{retreat}}</math>) is bigger.