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#1 cockerpunk

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Posted 14 January 2009 - 12:20 AM

hey guys, this might be a thread for some of our more educated members.

so, all this compression testing i did today had me thinking. while compression testing is certainly interesting, impact loading would be better for describing most paintball breaks.

this got me thinking, exactly how much force is acting on a paintball during a typical break. not an exact answer, just something in the ball park, factor of 10 accurate kinda thing.

simplifying it down to constant acceleration makes it easy, and some other value picking like -

.1 inch break distance, meaning the paintball from start to failure moves a bit under an eigth of an inch, between deformation of the ball, and the movement of the object it strikes.

assume an impact velocity of 250 feet per second.

assume .1 ounces for the weight of the ball.

now, using something like this equation - x = V(t) + (.5)at^2 to find the acceleration, then using F=mA to find the force that would produce such an acceleration. that would the be the force acting on the paintball.

the problem i am having is finding the time to use in that equation. we need time to find acceleration, and i tried using some kind of average velocity assumption to find the time, but thats just not cutting it.

any high speed we can use? or possibly someone has a good idea for a formula to figure out an average velocity?


if i knew a formula, it would not be hard to write an excell macro to calculate if you are going to get a break or not. or the probability you are going to get a break.

Edited by cockerpunk, 14 January 2009 - 12:35 AM.

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#2 Leftystrikesback

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Posted 14 January 2009 - 12:58 AM

solving x = v(t) + .5 a * t^2 for a in terms of V_f (exit velocity) and x (acceleration distance) i get a = v_f^2/(x*2)

I can post a derivation if you want.

I'll put that into matlab and see what I get.

edit: don't need it, simple equation to solve for what we have, here's my derivation:

x = .5 at^2 + v_i*t
v_i = 0
assume constant acceleration, a:
v(t) = V_final / t_final * t (that's the slope of v(t) under a = const and v_i = 0 multiplied by time)
v(t) = dx/dt
=> x = int(v(t)) from 0 to t_f = v_f * t_f / 2 * t^2 evaluated from 0 to t_f
x = v_final * t_final / 2
t_f = x*2/v_f

x = .5 * a * t_f^2
plug in for t_f and solve for a
a = x*2/t_f ^2 = x*2/(x*2/v_f)^2 = v_f^2 / dist*2

is this what you are looking for?

Edited by Leftystrikesback, 14 January 2009 - 01:23 AM.

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#3 Lord Odin

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Posted 14 January 2009 - 01:02 AM

In the videos posted by Jack, doesn't he say what frame rate he was using to record the video? The cycle time will probably vary from marker to marker but perhaps you could use that as a reference. Or you could make it easier and just PM him or the manufacturer of your gun. Just some ideas.

#4 cockerpunk

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Posted 14 January 2009 - 01:05 AM

In the videos posted by Jack, doesn't he say what frame rate he was using to record the video? The cycle time will probably vary from marker to marker but perhaps you could use that as a reference. Or you could make it easier and just PM him or the manufacturer of your gun. Just some ideas.


the issue is that we would need to know the velocity of the balls, and i don think jack has posted anything like that.

depending on a few things i get between 25 and about 400 lbf which is a bit larger range then i wanted to ...

Edited by cockerpunk, 14 January 2009 - 01:06 AM.

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#5 Leftystrikesback

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Posted 14 January 2009 - 01:13 AM

so If I use v_f as 300 ft/s
x = .5 ft

a = 300^2 = 90,000 ft/s^2

f = m*a = 17.48 pounds

Edited by Leftystrikesback, 14 January 2009 - 01:34 AM.

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#6 cockerpunk

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Posted 14 January 2009 - 02:14 AM

the trouble is that the average velocity forumla solved for time is -

t = X (in) / 12 / vo / 2

but in the constant acceleration formula we have x/12 = vo (t) + .5at^2

quickly substituting we see the problem -

x/12 - vo (X (in) / 12 / vo / 2) (2) = at^2

so everything cancels at that point. we might as well have not even used the average velocity equation, becuase it all cancels.

what im fundamentally looking at is that the average acceleration formula uses essentially the midpoint velocity (and thus time), so we have really 2 equations, but only 1 independent equation.

i need either a value, or another equation to make the system consistent. otherwise we have to many values and not enough equations.
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#7 Leftystrikesback

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Posted 14 January 2009 - 02:26 AM

are you using vo = initial velocity at t = 0 ? in other words, vo = v(0)?
because in paintball accelleration I think we can safely assume v(0) = 0

so x(t) = .5*a*t^2

and I'm not quite sure what you're talking about as "average velocity equation". To me that is v(t) = v_final / t_final * t under constant accelleration. v_final = v(t_final) = velocity as ball stops accellerating

Edited by Leftystrikesback, 14 January 2009 - 02:31 AM.

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#8 cockerpunk

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Posted 14 January 2009 - 02:37 AM

no, im saying that the average velocity is the mid point between vo and 0 fps (0 is of course and assumption, and may bot be correct, as the ball tends to fail before it has stopped moving). unfrotuently, the acceleration formula seems to make that same assumption too, which means we have only 1 independent equation, with two unknowns, t and a.

now, if we could get time from a highspeed video, then we could get everything and know what kind of force the paintball is subjected to.
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#9 Leftystrikesback

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Posted 14 January 2009 - 02:53 AM

I'm not sure I understand what you're saying. The acceleration equation is x(t) = .5* a * t ^2 + v_i * t
where v_i is the initial velocity (assume it's zero). It doesn't assume anything about an average velocity.

x varies with time, however we know that it is the bore length at t = t_f (the time it takes a ball to travel the length of the barrel)
so we know x(t_f) = bore length
now, the equation is x(t_f) = 0.5 * a* t_f^2
we need to know a and t_f
we can derive t_f by taking the integral of the velocity equation v(t) from t = 0 to t_f where v(t) = (v_f/t_f ) * t (average velocity)

I get t_f = x(t_f) *2 / v_f (I think this is the part you have a problem with?)

we know v_f (300 ft/s)
we know x(t_f)
so we know t_f

substitute this back in:
x(t_f) = 0.5 * a * ( x(t_f)*2/v_f )^2

a = v_f^2 / ( x(t_f)*2 )
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#10 Jack Wood

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Posted 14 January 2009 - 05:30 AM

Gordon, what exactly are you trying to find?

The force used to accelerate a paintball in a barrel?

Don't we know that from the pressure in the barrel?

Can you just clarify for us what exactly you are trying to calculate.
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#11 Snipez4664

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Posted 14 January 2009 - 09:53 AM

I believe what he wants to do is find the force it takes to create a fracture in the ball - the impact force. Doing this from momentum is going to be tough, and I wouldn't expect kinematics in the absence of high speed footage or other experiments to be able to do it - it's too heavily dependent on shell compliance.
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#12 cockerpunk

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Posted 14 January 2009 - 01:07 PM

I believe what he wants to do is find the force it takes to create a fracture in the ball - the impact force. Doing this from momentum is going to be tough, and I wouldn't expect kinematics in the absence of high speed footage or other experiments to be able to do it - it's too heavily dependent on shell compliance.

yup, thats the issue right there. from high speed we could estimate v0 and vf and find the time, from that we could then see the force.
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#13 Leftystrikesback

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Posted 14 January 2009 - 01:17 PM

this got me thinking, exactly how much force is acting on a paintball during a typical break. not an exact answer, just something in the ball park, factor of 10 accurate kinda thing.

simplifying it down to constant acceleration makes it easy, and some other value picking like -

.1 inch break distance, meaning the paintball from start to failure moves a bit under an eigth of an inch, between deformation of the ball, and the movement of the object it strikes.


Ok re-reading this part it looks like I answered the wrong question, you're looking at force during a typical paintball impact with a person? or at least an equation to calculate it's deceleration?

I tend to agree with snipez that it won't tell you much, the impact toughness of paintballs varies widely between brands, with temperature and all that. and estimating the distance traveled until a ball breaks is really what it all depends on and it's not such an easy quantity to measure without high speed video.

but, the deceleration calculations should be similar to acceleration out of a barrel except with an initial velocity term.
a = -x - .5* (v_i^2 / x^2)

x= .1 inches, v_i = 250 ft/s

a = -21701 ft/s^2
m=1 ounce
F = 42.2 pounds force

Edited by Leftystrikesback, 14 January 2009 - 01:23 PM.

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#14 Leftystrikesback

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Posted 14 January 2009 - 01:29 PM

This just occured to me:

impact loading would be better for describing most paintball breaks.


If this is true, it's because of strain hardening or other nonlinear material behavior, and in that case you're not going to describe whats happening by simple acceleration equations and F=ma. You're actually looking for the force plus the rate that the force is applied at. And then you need to characterize the (nonlinear?) behavior of the paintball material.

Edited by Leftystrikesback, 14 January 2009 - 01:32 PM.

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#15 HeroForADay

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Posted 14 January 2009 - 02:14 PM

I attempted to work this out in my head using a slightly different method, but as i don't have anything to work with ill just try and describe how i was going about it. I was attempting to use the theories of how much energy is put into the motion of accelerating the paintball t how much is lost at say a point at the tip of a 14 inch barrel. I know that we lost some to heat(friction), sound, and air resistance. Then using the changes to find from point A to point B

#16 Jack Wood

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Posted 14 January 2009 - 04:02 PM

So, if I gave you some footage like the first one I posted, with balls dropping on a surface, gave you the frame rate and , what else? Would that do it?

I can do that easily. I then have probably 10-12 clips showing fractures of balls dropping on a surface (actually thrown at a surface) so you can see if there is any repeatability.

Is that what you need?

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#17 cockerpunk

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Posted 14 January 2009 - 04:10 PM

So, if I gave you some footage like the first one I posted, with balls dropping on a surface, gave you the frame rate and , what else? Would that do it?

I can do that easily. I then have probably 10-12 clips showing fractures of balls dropping on a surface (actually thrown at a surface) so you can see if there is any repeatability.

Is that what you need?

Jack


yeah, if that is a high enough velocity (say above 200fps) to model and in game hit.
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