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

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<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{d_{initial} + v_{retreat} t}dt</math>
 
<math>\theta = \int_0^{t_1} \frac{v_{lateral}}{d_{initial} + v_{retreat} t}dt</math>
  
where <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)
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where <math>\theta</math> is the escape angle, and <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)
  
  

Revision as of 15:31, 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>

where <math>\theta</math> is the escape angle, and <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.