Difference between revisions of "Thread:User talk:Dsekercioglu/MEA/And to make it even faster/reply (11)"
Jump to navigation
Jump to search
m |
m |
||
Line 5: | Line 5: | ||
The correct one is: | The correct one is: | ||
− | <math>\theta = \int_0^ | + | <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 = | 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 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</math>) is bigger. |
Revision as of 15:05, 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</math>) is bigger.