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Theoretical Maximum Efficiency of any Marker


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#1 tyronejk

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Posted 29 August 2012 - 05:58 PM

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.

#2 brycelarson

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Posted 31 August 2012 - 08:28 AM

that seems to be in line with previous calculations.

#3 cockerpunk

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Posted 31 August 2012 - 08:32 AM

good work, this is about where other calculations have put it as well.
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#4 tyronejk

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Posted 31 August 2012 - 12:48 PM

that seems to be in line with previous calculations.



good work, this is about where other calculations have put it as well.


Thanks!

#5 Egomaniacal

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Posted 31 August 2012 - 01:36 PM

Since 4500 psi is gauge pressure, there's no need to subtract atmospheric to get total energy in the tank. It won't make much difference because atmo is so small in comparison to 4500 psi, but since it's gauge pressure the energy of atmo is already accounted for/neglected.
eiπ = − 1

#6 Cookybiscuit

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Posted 31 August 2012 - 04:32 PM

No idea how Bob Long gets away with claiming 3k shots off the G6R.

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#7 cockerpunk

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Posted 31 August 2012 - 04:41 PM

No idea how Bob Long gets away with claiming 3k shots off the G6R.


only way to do that is a slower chrono speed. at 260fps i believe it.
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#8 tyronejk

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Posted 31 August 2012 - 04:48 PM

Since 4500 psi is gauge pressure, there's no need to subtract atmospheric to get total energy in the tank. It won't make much difference because atmo is so small in comparison to 4500 psi, but since it's gauge pressure the energy of atmo is already accounted for/neglected.


Yes! You are correct. I'll have to make some corrections.

#9 Cookybiscuit

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Posted 31 August 2012 - 05:01 PM

only way to do that is a slower chrono speed. at 260fps i believe it.

Well he claims 3k at 280FPS (according to that tour video put up a few days ago) from a 70cu. It must be to do with the tolerances on BL's I hear so much about. :rolleyes:

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#10 andrewthewookie

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Posted 31 August 2012 - 05:58 PM

Don't forget that all the angles are at angles for perfect air flow.

Edited by andrewthewookie, 31 August 2012 - 05:58 PM.

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#11 IhasAcellular

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Posted 31 August 2012 - 06:59 PM

and they cross! Its like porting and polishing engine heads!

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#12 510waffles

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Posted 31 August 2012 - 07:40 PM

Im going to give the benefit of the doubt that Bob long is not lying and he does have the gun get 3k shots, its just that there are some underyling things that hes not talking about. hes probably shooting at a very low BPS, with a 70CI,cold fill,probably chrono around 260-270,dwell at 4.Hell, probably not even 3k, might be like 2950 or something.


Using 260 as average chrono speed, you can get 3.2k shots according to my calculations, so 3k shots at 280 isnt all that preposterous

after using all the forumlas and back solving,His average chrono must be under 274 and thats with a 68 calculation. add those 2 cubic inches and youre at a reasonably close to 280

Edited by 510waffles, 31 August 2012 - 07:48 PM.


#13 Cookybiscuit

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Posted 31 August 2012 - 08:48 PM

Im going to give the benefit of the doubt that Bob long is not lying and he does have the gun get 3k shots, its just that there are some underyling things that hes not talking about. hes probably shooting at a very low BPS, with a 70CI,cold fill,probably chrono around 260-270,dwell at 4.Hell, probably not even 3k, might be like 2950 or something.

That or hes using the brand new 2013 Bob Long pod, featuring the appearance of a 140, with half the capacity.

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#14 Troy

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Posted 31 August 2012 - 11:08 PM

Im going to give the benefit of the doubt that Bob long is not lying and he does have the gun get 3k shots, its just that there are some underyling things that hes not talking about. hes probably shooting at a very low BPS, with a 70CI,cold fill,probably chrono around 260-270,dwell at 4.Hell, probably not even 3k, might be like 2950 or something.


Using 260 as average chrono speed, you can get 3.2k shots according to my calculations, so 3k shots at 280 isnt all that preposterous

after using all the forumlas and back solving,His average chrono must be under 274 and thats with a 68 calculation. add those 2 cubic inches and youre at a reasonably close to 280


I wouldn't say that he's lying... but I've seen many people, even scientists... hell, even me, WANT a result so badly, that they fool themselves into getting it. If you round up, you don't pack your pods all the way, the chrono is somewhere around 280, etc. you can probably fudge 3k rounds. It doesn't mean he is lying, it just means that he's not objective.
\m/

#15 tyronejk

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Posted 01 September 2012 - 06:57 AM

At 260 fps and with a 70 cut inch tank, the theoretical maximum number of shots is 3544.919. At 280 fps and with a 70 cut inch tank, it's 3056.588 shots.

So Bob Long could've gotten 3k shots off of a 70 cu in. I think it probably was a cold tank.

#16 Egomaniacal

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Posted 01 September 2012 - 01:50 PM

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

Not to mention you're completely neglecting the air required to cycle the marker, and when you're talking about efficiency numbers as high as these, that volume becomes non-negligible.

Edited by Egomaniacal, 01 September 2012 - 01:54 PM.

eiπ = − 1

#17 mr.satire

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Posted 01 September 2012 - 01:52 PM

Yeah i dont buy it. If you include losses from friction, and other factors its safe to assume a 20 to 30 percent loss, so if you have the theoretical value of 3544 shots at 260 then a real world guess at 80% efficiency (very unlikely) is 2,835 shots and at 70% it is 2480 shots. So bob longs claim of 3K really is impossible but 2.5K at 260 could happen.

#18 firebirdjimbo

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Posted 03 September 2012 - 11:16 PM

He states in his video that he was shooting above 260fps so probably 265 +/-5ish. In his other tests he shoots at about 8bps so that it can all recharge correctly, I believe that he gets 3000 shots off of a cold and full 68/45 shooting 8bps at 265fps.
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#19 misterkyle

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Posted 04 September 2012 - 01:50 AM

He states in his video that he was shooting above 260fps so probably 265 +/-5ish. In his other tests he shoots at about 8bps so that it can all recharge correctly, I believe that he gets 3000 shots off of a cold and full 68/45 shooting 8bps at 265fps.


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Edited by misterkyle, 04 September 2012 - 01:51 AM.


#20 Troy

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Posted 04 September 2012 - 09:01 AM

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.
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#21 dertydood

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Posted 05 September 2012 - 10:48 PM

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.


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#22 Egomaniacal

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Posted 09 September 2012 - 09:46 AM


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.

Edited by Egomaniacal, 09 September 2012 - 09:46 AM.

eiπ = − 1

#23 thycalmesuperman

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Posted 10 September 2012 - 11:51 PM

I got 5000 shoots off my tippman 98.

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

#24 Pump Player

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Posted 15 September 2012 - 02:06 AM

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

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#25 timekeeper

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Posted 16 September 2012 - 09:02 PM

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


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#26 tjmartin

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Posted 26 October 2012 - 04:00 PM

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

#27 andrewthewookie

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Posted 26 October 2012 - 05:33 PM

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.

Edited by andrewthewookie, 26 October 2012 - 06:22 PM.

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#28 tjmartin

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Posted 26 October 2012 - 06:19 PM

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

#29 Blade of grass

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Posted 27 October 2012 - 10:44 AM

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.

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.

Edited by Blade of grass, 27 October 2012 - 10:45 AM.

all my legos are stored at my parents hose... so that wont be happening....

48fhih.png

#30 andrewthewookie

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Posted 27 October 2012 - 07:36 PM

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.

Edited by andrewthewookie, 27 October 2012 - 07:39 PM.

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#31 Blade of grass

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Posted 27 October 2012 - 08:32 PM

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.

all my legos are stored at my parents hose... so that wont be happening....

48fhih.png

#32 andrewthewookie

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Posted 27 October 2012 - 08:52 PM

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.

1/2mvi2 + mghi  + Wexternal = 1/2mvf2 + mghf  + Wloss

That equation is how we can find the energy required to move the paintball. Since the paintball is not moving initially, 1/2mvi2 = 0 and can be removed from the equation. Because there is no change in potential energy due to height change, we can remove mghi. On the other side, we are keeping 1/2mvf2 because the paintball does have a velocity now, and mghf is removed for the same reason it was not included on the left side. Because this is a theoretical exercise, we're not including Wloss, 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 Wexternal (which is the air) = 1/2mvf2. 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.

65761e9c7ec650ec33b3f3af5f7124fd.png


Edited by andrewthewookie, 22 April 2014 - 05:59 PM.

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#33 Cookybiscuit

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Posted 27 October 2012 - 10:53 PM

Maths.

*yawn*

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#34 REDCOBRA

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Posted 28 October 2012 - 02:54 AM

Posted Image

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#35 tyronejk

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Posted 28 October 2012 - 05:27 PM

Andrew for the win

#36 Blade of grass

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Posted 31 October 2012 - 04:09 PM


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.

1/2mvi2 + mghi + Wexternal = 1/2mvf2 + mghf + Wloss

That equation is how we can find the energy required to move the paintball. Since the paintball is not moving initially, 1/2mvi2 = 0 and can be removed from the equation. Because there is no change in potential energy due to height change, we can remove mghi. On the other side, we are keeping 1/2mvf2 because the paintball does have a velocity now, and mghf is removed for the same reason it was not included on the left side. Because this is a theoretical exercise, we're not including Wloss, 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 Wexternal (which is the air) = 1/2mvf2. 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.

65761e9c7ec650ec33b3f3af5f7124fd.png

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.

Edited by andrewthewookie, 22 April 2014 - 06:00 PM.

all my legos are stored at my parents hose... so that wont be happening....

48fhih.png




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