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_1</math> almost all of the time, as <math>d_{initial} = | + | Note that <math>d_{initial} + v_{retreat} t</math> is bigger than (his) <math>(v_{bullet} - v_{retreat})t_1</math> almost all of the time, |
− | (v_{bullet} - v_{retreat})t_1</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. | + | |
+ | as <math>d_{initial} = | ||
+ | (v_{bullet} - v_{retreat})t_1</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 16:08, 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_1</math> almost all of the time,
as <math>d_{initial} =
(v_{bullet} - v_{retreat})t_1</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.