Login    Register
User Information
Username:
Password:
We are a free and open
community, all are welcome.
Click here to Register
Sponsored
Who is online

In total there are 37 users online :: 4 registered, 0 hidden and 33 guests


Most users ever online was 218 on Wed Dec 07, 2016 6:58 pm

Registered users: Bing [Bot], Google [Bot], MSNbot Media, Yahoo [Bot] based on users active over the past 5 minutes

The Team
Administrators
Global Moderators
global_moderators.png CS

How to Determine the Number of Shots you can get from a Tank

Post questions and info about pneumatic (compressed gas) powered cannons here. This includes discussion about valves, pipe types, compressors, alternate gas setups, and anything else relevant.
Sponsored 
  • Author
    Message

How to Determine the Number of Shots you can get from a Tank

Unread postAuthor: webgeek » Sat Jun 13, 2009 4:19 pm

Since this community has been so helpful, I wanted to give back what little I could. This might be obvious to all of you but it wasn't to me so I figure I'd post it for people to see. I'd love to claim credit for the math but I had one of my friends help me with it. I'm just posting it as he explained it to me.

Let's say you are using a reservoir air tank to fill up your chamber for each shot and you want to know how many shots you can get. The math to solve this at a basic level is called the "Ideal Gas Law" and can be found here. It looks like this:

pV = nRT

p = pressure
V = volume
n = "moles" usually
R = gas constant
T = absolute temperature

This is a nasty bit of work normally but we can ignore most of it for our purposes. Since we are going air to air, the n is ignored. The gas constant can be ignored as well. For the sake of keeping this simple, we will assume the temperature while filling doesn't change - this is the biggest assumption and might or might not have a big impact on the results depending on the differences in pressure.

So as a concrete example, let's say you want to know how many shots you can get out of a 4500 PSI, 68 cubic inch HPA tank the paintball guys use. Your launcher is fired at 100 PSI and has a 3' chamber of 2" schedule 40 PVC. If you go here you can see what the ID of that happens to be (2.047").

To determine the volume of your chamber, use the area of a cylinder equation: V = pi * radius * radius * height. So:

radius = 2.047 / 2 = 1.0235
V = 3.14 * 1.0235 * 1.0235 * 36 = 118.415 cubic inches (rounded)

This doesn't take into account fittings, elbows, the valve and the like so it's on the low side. Depending on your configuration, you will need to pad this or calculate more.

-- the air tank
pV = 4500 * 68 = 306000

-- the launcher
pV = 100 * 118.415 = 11841.5

Number of shots = 306000 / 11841.5 = 25.84

Not too shabby, though each shot is a bit expensive when you count refilling. Let's try a different example but use the same launcher. Say this 11 gallon, 125 PSI tank from Harbor Freight.

First we need to get all of our units the same. I let Google do the work for me. Just search for "11 gallons in cubic inches" and it will tell you the answer is 2541. You can see this here. So if we run the math again, it looks like this:

-- the air tank
pV = 125 * 2541 = 317625

-- the launcher
pV = 100 * 118.415 = 11841.5

Number of shots = 317625 / 11841.5 = 26.82

A tiny bit better, but significantly cheaper as you could use your own air compressor at home to do the filling. This does NOT take into account the tank cooling as you fill for each shot. In the HPA scenario, this could add up fast. In the larger tank scenario, this is a negligible difference. I hope this helps people out a bit in calculating their shots.

Thanks!

--Mike

P.S. Based on this math, you can see that the 10 gallon, 200 PSI tank here is a MUCH better deal at $34.99 with 39 shots available.
  • 0


webgeek
Private
Private
 
Posts: 17
Joined: Sun May 31, 2009 9:47 am
Reputation: 0

Unread postAuthor: MountainousDew » Sat Jun 13, 2009 4:46 pm

Thanks for the info, but I think this is a more appropriate topic for the "How-to" section...
But thanks again for putting all the info together in one post...
  • 0

<a href="http://s709.photobucket.com/albums/ww95/JKazkid/?action=view&current=MDSig.jpg" target="_blank"><img src="http://i709.photobucket.com/albums/ww95/JKazkid/MDSig.jpg" border="0" alt="Photobucket"></a>
User avatar
MountainousDew
2nd Lieutenant
2nd Lieutenant
 
Posts: 230
Joined: Tue Apr 14, 2009 6:06 pm
Reputation: 0

Unread postAuthor: webgeek » Sat Jun 13, 2009 4:53 pm

I put it here because it's pneumatic only but if how-to makes more sense, I'm sure a mod will go ahead and move it :)
  • 0


webgeek
Private
Private
 
Posts: 17
Joined: Sun May 31, 2009 9:47 am
Reputation: 0

Unread postAuthor: mobile chernobyl » Sun Jun 14, 2009 5:53 pm

The only issue with using that formula is it doesn't take into account the pressure change in the main tank when refilling the receiving vessel. (If you consider that a problem, which I'd assume most of us on here would)

Basically you won't get 25 or 26 shots at the initial filling pressure of the main tank... and the last shot will be rather.... weak? lol

Good post tho, and your writing style is nice and clear.
  • 0

User avatar
mobile chernobyl
Colonel
Colonel
 
Posts: 732
Joined: Sun Dec 03, 2006 11:53 am
Reputation: 9

Unread postAuthor: THUNDERLORD » Sun Jun 14, 2009 10:07 pm

I'd like to look this over/ plug it in, when I have the time.
Thanks! :)
The issue mobile chernobyl brought up is the reason I prefer gasses that are stored in liquid form.

I dream of dry-ice though. :P
I wish someone (yourself???) could come up with info and math for dry-ice.
It costs a $ per pound near me.
Guess I'll search the net on it.
Ragnorok had a little info before.

BTW, (just got an idea)...Maybe it's possible to 24 hr fed-ex the stuff???
Personaly i wouldn't want to be involved in that idea due to liability issues that could arise but... 8)
  • 0

-----SPEED,STRENGTH, AND ACCURACY.-----
"Procrastination" is five syllables for "Sloth".
Theopia 8)
Born To Be Alive!

THUNDERLORD
Major General
Major General
 
Posts: 1264
Joined: Fri Mar 28, 2008 1:42 pm
Reputation: 0

Unread postAuthor: jimmy101 » Mon Jun 15, 2009 2:35 pm

Building on WEbgeek's post...

Starting from PV=nRT you can also get to another handy form of the Ideal Gas law;

P<sub>1</sub>V<sub>1</sub>/(n<sub>1</sub>RT<sub>1</sub>) = P<sub>2</sub>V<sub>2</sub>/(n<sub>2</sub>RT<sub>2</sub>)

Since R is a constant, that simplifies to;

P<sub>1</sub>V<sub>1</sub>/(n<sub>1</sub>T<sub>1</sub>) = P<sub>2</sub>V<sub>2</sub>/(n<sub>2</sub>T<sub>2</sub>)

This relation is true even if there is nothing in common between the two gas samples. (Basicaly, it just says that PV/(nRT)=1). From this form of the Ideal Gas Equation you can derive the three common sub-equations.

If the Pressure is constant (P<sub>1</sub> = P<sub>2</sub>)
and the number of moles is constant (i.e., the same sample of gas under different temperature and volume conditions)
then you get Charles's Law:

V<sub>1</sub>/T<sub>1</sub> = V<sub>2</sub>/T<sub>2</sub>


If the Temperature is constant (T<sub>1</sub> = T<sub>2</sub>)
and the number of moles is constant (i.e., the same sample of gas under different pressure and volume conditions)
then you get Boyle's Law:

P<sub>1</sub>V<sub>1</sub> = P<sub>2</sub>V<sub>2</sub>


If the Volume is constant (V<sub>1</sub> = V<sub>2</sub>)
and the number of moles is constant (i.e., the same sample of gas under different temperature and pressure conditions)
then you get Gay-Lussac's Law:

P<sub>1</sub>/T<sub>1</sub> = P<sub>2</sub>/T<sub>2</sub>

You don't have to remember these three laws since it is easy enough to derive them from the original P<sub>1</sub>V<sub>1</sub>/(n<sub>1</sub>T<sub>1</sub>)=P<sub>2</sub>V<sub>2</sub>/(n<sub>2</sub>T<sub>2</sub>) equation.

These four law's typically work well enough for many spud gunning calculations. There are a couple places where they will have problems. Air isn't an "ideal gas", propane, water, CO2 etc are even less ideal. That causes problems mostly when you are dealing with changes in energy or in energy transfer from the gas to the ammo. There are corrections for the non-ideality but it is a bit more involved than most spudder's care for.

These laws assume an adiabatic system, no heat is lost or gained in going from state-1 to state-2. This is almost true in a fast process (like a pneumatic gun firing). It is always false in slow processes, like pressurizing a chamber with a compressor (heat is lost) or pressurizing a chamber from a resevoir (heat is gained).
  • 0

Image

jimmy101
Lieutenant General
Lieutenant General
 
Posts: 3130
Joined: Wed Mar 28, 2007 9:48 am
Location: Greenwood, Indiana
Country: United States (us)
Reputation: 7

Unread postAuthor: Solar » Thu Jun 18, 2009 2:08 am

ok, so I tried this and it seemed to work out good with my 68 and 48 cubic inch tanks, but when i tried to convert using an online converter, I got 138240 cubic inches from a SCUBA aluminum 80 cubic foot tank. I wound up with something like 9216 shots. after calculating the launcher pressure at 300psi with a 15 cubic inch ballast. 3000psi in the aluminum 80. What did I do wrong? The numbers I got the other day worked good for the smaller tanks and seemed correct.
  • 0

User avatar
Solar
Colonel
Colonel
 
Posts: 512
Joined: Tue May 17, 2005 11:53 pm
Country: United States (us)
Reputation: 11

Sponsored

Sponsor
 


Return to Pneumatic Cannon Discussion

Who is online

Registered users: Bing [Bot], Google [Bot], MSNbot Media, Yahoo [Bot]

cron
Reputation System ©'