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#1 tyronejk

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Posted 11 December 2012 - 06:19 PM

So what if you had a frictionless shake-n-shoot with 200 0.689-inch frictionless paintballs. What would be the minimum amount of time required to empty the loader without shaking it?

#2 madsnipes

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Posted 13 December 2012 - 05:40 PM

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#3 XxJellyFilledxX

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Posted 13 December 2012 - 05:51 PM

yay pron.

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#4 ExcellentStudent

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Posted 13 December 2012 - 05:52 PM

Mod shens

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#5 PieCars

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Posted 13 December 2012 - 08:16 PM

Depends on how fast you can pull the trigger

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#6 Red Infinity

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Posted 13 December 2012 - 09:32 PM

now i'm dissapointed. :(
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#7 tallsmallboy44

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Posted 13 December 2012 - 10:22 PM

now i'm dissapointed. :(


fuck yolo
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#8 sonicx059

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Posted 14 December 2012 - 12:51 AM

200 ball capacity
-----------------
8 balls per second(gravity fed)

equates to a minimum of 25 seconds providing that the balls say screw Murphy's Law.

first time i played with an electro I couldn't figure out why the gun would shoot all the time. Took it to the refs and it work. They watched me play a game and told me I was out shooting the hopper... At that moment I fell in love with the vibe and felt as though I transcended into godly hood. Then they let me use a real loader.

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#9 tyronejk

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Posted 14 December 2012 - 08:56 AM

200 ball capacity
-----------------
8 balls per second(gravity fed)

equates to a minimum of 25 seconds providing that the balls say screw Murphy's Law.
...


But that's a real-world scenario. What about the frictionless one I mentioned?

And nice title, mods. I see what you did there ;)

#10 ExcellentStudent

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Posted 15 December 2012 - 11:57 AM

What was the title before this?

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#11 TechPB-Mike

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Posted 16 December 2012 - 12:48 AM

wait... what?

ibtl

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#12 erg993

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Posted 16 December 2012 - 03:08 AM

yes

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#13 cockerpunk

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Posted 16 December 2012 - 09:23 PM

wait... what?

I farted.

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#14 Eskimo

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Posted 17 December 2012 - 03:03 PM


200 ball capacity
-----------------
8 balls per second(gravity fed)

equates to a minimum of 25 seconds providing that the balls say screw Murphy's Law.
...


But that's a real-world scenario. What about the frictionless one I mentioned?

And nice title, mods. I see what you did there ;)



that is frictionless. its assuming you have 200 paintballs falling free in a straight down line with nothing to stop or slow them.

if friction was involved shit gets real.

IE we need to take the integration of all the points of a paintball making contact with the sides of the feedneck for the height of the feedneck to find the total work done by friction then remove that energy from the potential that gravity gives it, some math later and you find that the acceleration of the ball down a feedneck is slightly less then 9.81 m/s2

from there then you take 200 many balls travaling at said acceleration pass through the breach, then account for the time it takes for the bolt to move foward and backwards (IE shooting the ball below it) and find out if the balls are actually stacking ontop each other waiting for the next one to be shot some more math and you have the most paintball you can shoot per second, X 200 paintballs and you have X seconds of non stop shooting with Y delay between each shot.

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#15 tyronejk

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Posted 17 December 2012 - 04:37 PM



200 ball capacity
-----------------
8 balls per second(gravity fed)

equates to a minimum of 25 seconds providing that the balls say screw Murphy's Law.
...


But that's a real-world scenario. What about the frictionless one I mentioned?

And nice title, mods. I see what you did there ;)



that is frictionless. its assuming you have 200 paintballs falling free in a straight down line with nothing to stop or slow them.

if friction was involved shit gets real.

IE we need to take the integration of all the points of a paintball making contact with the sides of the feedneck for the height of the feedneck to find the total work done by friction then remove that energy from the potential that gravity gives it, some math later and you find that the acceleration of the ball down a feedneck is slightly less then 9.81 m/s2

from there then you take 200 many balls travaling at said acceleration pass through the breach, then account for the time it takes for the bolt to move foward and backwards (IE shooting the ball below it) and find out if the balls are actually stacking ontop each other waiting for the next one to be shot some more math and you have the most paintball you can shoot per second, X 200 paintballs and you have X seconds of non stop shooting with Y delay between each shot.


So a vertical tube of 200 paintballs would fall at a rate of 8 bps?
200 paintballs * 0.689 inch/ball = 137.8 inches = 3.500 meters
Assuming that every single ball will start to fall as soon as you release the bottom one:
1/2*9.8t^2 = 3.5
t = 0.845 s
So a vertical, frictionless tube of 200 incompressible paintballs would empty in 0.845 seconds, which is 236.7 bps. Not 8.

#16 dertydood

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Posted 17 December 2012 - 05:17 PM

wait... what?

I farted.

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#17 Eskimo

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Posted 17 December 2012 - 07:50 PM

So a vertical tube of 200 paintballs would fall at a rate of 8 bps?
200 paintballs * 0.689 inch/ball = 137.8 inches = 3.500 meters
Assuming that every single ball will start to fall as soon as you release the bottom one:
1/2*9.8t^2 = 3.5
t = 0.845 s
So a vertical, frictionless tube of 200 incompressible paintballs would empty in 0.845 seconds, which is 236.7 bps. Not 8.


its easier then that.
200 balls located in the same place, (think a hopper. in you calc the 200th ball is 3.5m from your 0 point, which means its going REALLY fast when it his the breach)
every ball falls at 9.81m/s
so if it takes 0.0175006m to fall into the breach (0.0175006 is the hieght of a 0.689inch ball)
then you need to find how long it takes 1 ball to fall [0.0175006m + the height of the feedneck (say... 3cm = 0.03m)]

then its just the time it takes for 1 ball to fall 0.0475006m, thats assuming the ball falls from the top of the feedneck at rest, into the breach. X 200 paintballs
if you ignore the feedneck its just the time required to fall the distence of itself by 1 time (think balls stacked ontop each other)

0.0475006 = 1/2(9.81)(t2)

(0.0475006)(2)/(9.81) = t2

(0.009684118)1/2 = t

0.0984(seconds/ball) for a ball to fall that very short distance under just gravity.
0.0894(s/ball) x 200 (ball) = 19.68 seconds to empty a hopper if every ball falls from the feedneck and is instantly shot out.

Round up becuase of errors in continous calc and about 20 seconds is a fair quickie ignoring many many real world situations.

to get your Balls per second, simply inverse 0.0984(seconds/ball)

1/0.0984 = 10.16BPS <- seems odd, but remember each ball is falling a short distence and when you increase the distence it falls, its velocity is higher, and then the BPS becomes much much larger. thats why you got an absurd BPS at 3.5m up.

Edited by Eskimo, 17 December 2012 - 07:57 PM.

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#18 tyronejk

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Posted 18 December 2012 - 01:04 PM

Oh, I see your logic. So you're assuming each ball comes to a complete stop before being fired (with a dwell of zero). That makes more sense than the "drop test"-type scenario I was think of. Thanks.




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