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Old 11-16-2012, 06:08 PM   #49 (permalink)
FooFighter
Just call me "Pistol Pete
 
Join Date: Oct 2012

With any physics problem, always simplify it to a reasonable extent to explain it at first. Then make it more complicated by adding real-world effects. Explaining a physics problem with major simplifications is still a valid way of analysing it. The differences between ideal and real-world is often very small, and it would be in this case as the time interval we are analysing is very small.

Note: This example only illustrates the possibility of clearing a line of fire. It does not address the probability of successfully clearing the line of fire.


So......

Put some numbers for 0.68 cal, 3 g paintballs into Chairgun Pro. I put in a ballistic coefficient of 0.005, a very big assumption, as I don't know the BC of a paintball, but it is assumed to be quite crappy, so I picked the lowest limit available in this program.

Assume paintballs are travelling parallel to the ground in the short time period of this example. Also disregard deceleration during this time period. In reality paintballs are arcing and constantly decelerating once in the air.

Situation: You are trying to cross a lane 30 feet from your opponent shooting 15 bps.
Initial velocity = 280 ft/sec
Paintball velocity @ 30 feet = 218 ft/sec (from ChairGun Pro)
15 bps = 1/15 = 0.0667 sec time interval between paintballs

The first question to ask is will a person fit between two paintballs in this setup.

Distance travelled in 0.0667 sec by a 218 ft/sec paintball before next paintball is shot:
d = [v(initial) + v(final)] / 2 x t assume no deceleration in this time period
d = v x t
= 218 x 0.0667 = 14.5 feet

Therefore, there is a gap of about 14.5 feet between paintballs. All humans will therefore fit between the gap (I measure about 2 feet left to right).

Next question: how fast does a person have to run in order to clear the gap?

I measure about 1 foot wide front to back. So how fast do I have to run, to clear 1 foot in 0.0667 sec?

d = v x t
v = d / t
= 1 / 0.0667 = 15 ft/sec
= 10.2 mph

I have to be already running at that speed when I cross the line of fire, and of course time it so the first ball just misses the front of my big fat tummy, and the second ball misses the back of my big fat butt. Also, we are not taking into account flailing arms. It's as though I ran with my arms by my sides.

Now if you're fatter and measure 1.5 feet wide, you'd have to run 15.3 mph, so it definitely pays to lay off those Twinkies!!!!!

To put this into perspective, average walking speed is 3-4 mph. Usain Bolt is touted to top at around 27 or 28 mph.

At 20 bps, t = 1/20 = 0.05 sec. The gap is only 10.9 feet. In the above situation, I would have to run 13.6 mph and have better timing. The fat person has to run 20.4 mph!!!

Note that the further the stream of balls travel, the shorter the gap becomes between paintballs. Why? Because if we put back in deceleration effects, the ball behind is travelling ever so slightly faster than the ball directly in front, at any point in the flight path. So the back ball shortens the gap over time. However, the frequency of balls shot doesn't change, so the time I have to run is the same. Also, the velocity of the paintballs are a lot slower. It would actually be quite difficult to try to cross a hail of paintballs shot from 200 feet away, even though they might be only travelling 50 ft/sec at that distance.
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