[dertydood]

tyronejk, on 29 August 2012 - 05:58 PM, said:

This is a brief calculation of the theoretical maximum efficiency of any 0.68 cal paintball maker off a 68/4500 tank.

Feel free to correct any errors!

mass of an average paintball: 3.2 g = 0.0032 kg
300 fps = 91.44 m/s
68 cu inches = 0.0011143 cu meters
4500 psi = 31026407.8 pascals
atmospheric pressure at sea level: 101325 pA

kinetic energy of a single paintball at full velocity:
E = 0.5mv^2
= 0.5 * 0.0032 kg * 91.44^2 = 13.3780 J

stored energy in a 68/4500 tank:
E = pV
= (31026407.8 pA - 101325 pA) * 0.0011143 m^3 = 34459.820 J

number of paintballs fired from the energy in one tank:
34459.820 J / 13.3780 J = 2575.857 paintballs

=================================================

In conclusion: a marker with a bolt of zero grams, zero friction, zero internal volume (including regulators, hoses, and chambers) firing at 300 fps, at sea level, with an outdoor temperature of 0 degrees celsius or 32 degrees fahrenheit, and firing each ball with enough of a delay to allow the tank pressure to reach equilibrium in between every shot (approx. 5-10 seconds), you can get up to 2,575.857 shots!

note: The 2,576th shot will be a break.

http://cdn.memegener...50/24516400.jpg

[Egomaniacal]

Troy, on 04 September 2012 - 09:01 AM, said:

Egomaniacal, on 01 September 2012 - 01:50 PM, said:

I seriously doubt Bob's valves are good enough to be getting the efficiency numbers of a perfectly ideal valve.

I think that "seriously doubt" is an obscene understatement... I believe you know that as well, but I just wanted to underline that point.

In my business "seriously doubt" is about on par with "DAMN LIAR!!!"


At any rate, you should note the effect of paintball mass on your calculation. Bryce and co. have stated they routinely see paint with a mass of 2.7g, and paint just seems to be getting smaller and smaller these days. Another thing to note is that nobody actually chronos at 300 fps. Here's the calculation with 2.7gram paintballs and a velo of 290 fps - I get 3278 shots per fill with those numbers. Feel free to play around with them.

http://www.wolframal...29+to+joules%29

All that said, we know from experience Bob's valves aren't 3k efficient in normal playing conditions.

This post has been edited by Egomaniacal: 09 September 2012 - 09:46 AM

[thycalmesuperman]

I got 5000 shoots off my tippman 98.

IDK what a bob long is... other than brand of marker lol.

[Pump Player]

I'm in 7th grade and I don't know any of that but it appears your right if Brycelarson and cockerpunk say so

[timekeeper]

Pump Player, on 15 September 2012 - 02:06 AM, said:

I'm in 7th grade and I don't know any of that but it appears your right if Brycelarson and cockerpunk say so

[tjmartin]

is the kinetic energy of the paintball measured at the tip of the barrel or down range? that would effect the calculation wouldn't it? because air resistance will slow the paintball the further it goes

[andrewthewookie]

Well, we chrono the paint right as it comes out of the gun, so it would be pretty silly to chrono downrange. Besides, we're calculating how much energy it takes to move the paintball from 0fps to 300fps (and dividing that into the max energy capacity of the tank), which takes place in the length of the barrel, not the barrel plus flight distance.

This post has been edited by andrewthewookie: 26 October 2012 - 06:22 PM

[tjmartin]

ahh. see i was picturing the force being measured from something via force plate down range and then finding the energy from there but the way you are doing it makes much more sense haha

[Blade of grass]

tjmartin, on 26 October 2012 - 04:00 PM, said:

is the kinetic energy of the paintball measured at the tip of the barrel or down range? that would effect the calculation wouldn't it? because air resistance will slow the paintball the further it goes

What he calculated was the TRANSFER of kinetic energy.

andrewthewookie, on 26 October 2012 - 05:33 PM, said:

Well, we chrono the paint right as it comes out of the gun, so it would be pretty silly to chrono downrange. Besides, we're calculating how much energy it takes to move the paintball from 0fps to 300fps (and dividing that into the max energy capacity of the tank), which takes place in the length of the barrel, not the barrel plus flight distance.

But 1/2(Weight)(speed^2) is the calculation for transfer of energy. Not how much energy it take to go from 0 FPS to 300 FPS.

This post has been edited by Blade of grass: 27 October 2012 - 10:45 AM

[andrewthewookie]

To calculate the energy needed to move a paintball from rest to 300fps, all we have to do is convert to metric and see what 3.2g at 300fps is. Since we don't have to worry about potential energy or spring energy, it's initial energy (0) plus external work equals the final energy in the paintball. Since the initial energy of the paintball is 0, we can equate the final energy of the paintball to the work required to move it. We are also ignoring any friction or other forces acting to remove energy from the system (it is theoretical maximum after all).

Simply put, ¹/₂(3.2g)(300fps)² = Energy needed per paintball.

This post has been edited by andrewthewookie: 27 October 2012 - 07:39 PM

[Blade of grass]

andrewthewookie, on 27 October 2012 - 07:36 PM, said:

To calculate the energy needed to move a paintball from rest to 300fps, all we have to do is convert to metric and see what 3.2g at 300fps is. Since we don't have to worry about potential energy or spring energy, it's initial energy (0) plus external work equals the final energy in the paintball. Since the initial energy of the paintball is 0, we can equate the final energy of the paintball to the work required to move it. We are also ignoring any friction or other forces acting to remove energy from the system (it is theoretical maximum after all).

Simply put, ¹/₂(3.2g)(300fps)² = Energy needed per paintball.

I don't think that's right, I don't know why.
Should we have to calculate the transfer of energy of the bolt? Not of the paintball? The paintball isn't what moves to be fired, the bolt moves to fire the paintball.

[andrewthewookie]

The bolt only moves to seal the breech and open the valve to let the air move the paintball. The only work to move the paintball from 0-300 is being done by the expanding air. We are not counting the energy needed to move the bolt, or the friction on the system. We are merely finding the energy required to move a 3.2g object from 0fps to 300fps.

¹/₂mv_i² + mgh_i + ¹/₂kx_i² + W_external = ¹/₂mv_f² + mgh_f + ¹/₂kx_f² + W_loss

That equation is how we can find the energy required to move the paintball. Since the paintball is not moving initially, ¹/₂mv_i² = 0 and can be removed from the equation. Because there is no change in potential energy due to height change, we can remove mgh_i. There is also no spring in the system of paintball plus air, so we can remove ¹/₂kx_i². On the other side, we are keeping ¹/₂mv_f² because the paintball does have a velocity now, and mgh_f is removed for the same reason it was not included on the left side. The spring energy ¹/₂kx_f² is also removed, because again, no spring. Because this is a theoretical exercise, we're not including W_loss, or any of the friction or work needed to move the bolt and open the valve. Once all those are removed (either because their value is 0 or they are not involved in the first place), we're left with W_external (which is the air) = ¹/₂mv_f². Mass is in Kg, and velocity is in meters per second, which is squared, so we have Kg * m²/s² which is the units for a Joule.

This post has been edited by andrewthewookie: 27 October 2012 - 09:07 PM

[Cookybiscuit]

Maths.

*yawn*

[REDCOBRA]

[tyronejk]

Andrew for the win

[Blade of grass]

andrewthewookie, on 27 October 2012 - 08:52 PM, said:

The bolt only moves to seal the breech and open the valve to let the air move the paintball. The only work to move the paintball from 0-300 is being done by the expanding air. We are not counting the energy needed to move the bolt, or the friction on the system. We are merely finding the energy required to move a 3.2g object from 0fps to 300fps.

¹/₂mv_i² + mgh_i + ¹/₂kx_i² + W_external = ¹/₂mv_f² + mgh_f + ¹/₂kx_f² + W_loss

That equation is how we can find the energy required to move the paintball. Since the paintball is not moving initially, ¹/₂mv_i² = 0 and can be removed from the equation. Because there is no change in potential energy due to height change, we can remove mgh_i. There is also no spring in the system of paintball plus air, so we can remove ¹/₂kx_i². On the other side, we are keeping ¹/₂mv_f² because the paintball does have a velocity now, and mgh_f is removed for the same reason it was not included on the left side. The spring energy ¹/₂kx_f² is also removed, because again, no spring. Because this is a theoretical exercise, we're not including W_loss, or any of the friction or work needed to move the bolt and open the valve. Once all those are removed (either because their value is 0 or they are not involved in the first place), we're left with W_external (which is the air) = ¹/₂mv_f². Mass is in Kg, and velocity is in meters per second, which is squared, so we have Kg * m²/s² which is the units for a Joule.

Ah, I get it, I was just thinking about the transfer of energy from the paintball to ____, not the fact that it requires the SAME amount of energy to move set paintball, thanks for the clarification.