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 13: | Line 13: | ||
as <math>d_{initial} = | as <math>d_{initial} = | ||
− | (v_{bullet} - v_{retreat})t_1</math>. | + | (v_{bullet} - v_{retreat})t_1</math>. (assume constant <math>v_{lateral}</math> and <math>v_{retreat}</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} | + | 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 16:21, 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>t_1</math> is the total time (from bullet fired, to bullet hit).
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>. (assume constant <math>v_{lateral}</math> and <math>v_{retreat}</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.