How much air?

[Thorpydo]

How much air and at what pressure does it take to propel a ball?

Anyone know how big the chamber is on a matrix or mag? + plus the volume of empty space between the chamber and ball?

Doesn't have to be super accurate, just a good approxiamtion.

Thanks

[Cristobal]

From this thread: http://www.automags.org/forums/showt...threadid=18939

With a barrel length of 10 inches, it takes about 28psi of constant pressure behind the ball to shoot 280fps

Of course that means that there's a lot of wasted pressure left in the air in the barrel when the ball leaves... so most guns dump a lot of air at the begining of acceleration and then let it expand as the ball moves down the barrel, such that the air is about at atmospheric pressure by the time the ball leaves. Looking at the graphs in the data thread, this corralates to an initial peak of around 90-100 psi behind the ball.

[the electrician]

Well I've been reading, and thinking about this.
I've been looking at a designing a paintgun just for the fun.

I know Pbjosh was looking at it from a 68cc X 3000 psi = 204000 energy units perspective. but I don't get this. the units are different, and you can't use all 3000 psi anyway. but energy is the thing here.

well I started looking at air consumption and tried to work up some numbers.

I've seen it said that an automag dump chamber is .55 cu in. and let's say it's got 400 psi in it. figuring the density of air at 400 psi is 2.18 ft/lbs^3 that's 15.75 standard cu. in. of air. seems like alot. if that's all actually propelling the ball. I don't know that it is. in an automag, is there any pressure left in the dump chamber after the bolt spring pushes the bolt back? I don't know.

just going further for the heck of it,if you use this 15.76 cu. in. of air to calculate shot per tank, theoretically of course, it seems that this might be close.

a 68cc/3000 tank has 8 cu ft of air in it as per the luxfer specifications. 8 cu ft of air converts to 13,824 cu in. of standard air. 68cc converts to 4.15 cu in. (not much huh?)

now you can't count on using 400 psi of the air because that the operating psi we assumed earlier. so using the 400 psi density agian and the volume of the cylinder in cu in. we can't use 119 cu in. of the standard air.
so 13,824-119=13,705. now divide this by the 15.76 cu in. used per shot, and you get 869 shots.

this of course is theoretical, but seems pretty close huh?

I know these numbers only apply at that pressure and for that gun. I wish I could find some correlation between this and the energy per shot used.

[bjjb99]

Originally posted by the electrician
I know Pbjosh was looking at it from a 68cc X 3000 psi = 204000 energy units perspective. but I don't get this. the units are different, and you can't use all 3000 psi anyway. but energy is the thing here.

The product of pressure and volume is a unit of energy.

Pressure is a force acting on a unit area.
Volume is, well... volume, an area times a distance.

The product of the two cancels out the area unit:
Pressure * Volume = (force/area) * (area*distance) = force * distance.

Energy is defined as a force acting through a distance, which is exactly what the product of pressure and volume gives us.

Originally posted by the electrician
[BI've seen it said that an automag dump chamber is .55 cu in. and let's say it's got 400 psi in it. figuring the density of air at 400 psi is 2.18 ft/lbs^3 that's 15.75 standard cu. in. of air.
[/B]

I used a slightly different method of calculating the number of standard cubic inches of air in the dump chamber and came up with 15.51 standard cubic inches. I took the 0.55 in^3 value you used, multiplied it by the absolute pressure in the tank (414.7 psia), and divided it by the absolute pressure of one atmosphere (14.7 psia). I think we're close enough that the method doesn't make too much of a difference for a ballpark estimate.

Originally posted by the electrician
seems like alot. if that's all actually propelling the ball. I don't know that it is. in an automag, is there any pressure left in the dump chamber after the bolt spring pushes the bolt back? I don't know.

I vaguely remember one of the older threads here stating that there was approximately 60 psig left in the dump chamber after firing... I suppose that means we're only using 85 percent of the dump chamber air when we pull the trigger. I would say we're using 340 out of the 400 psig in the chamber, and would use this 340 number instead for the remainder of the calculations.

Originally posted by the electrician
just going further for the heck of it,if you use this 15.76 cu. in. of air to calculate shot per tank, theoretically of course, it seems that this might be close.

a 68cc/3000 tank has 8 cu ft of air in it as per the luxfer specifications. 8 cu ft of air converts to 13,824 cu in. of standard air. 68cc converts to 4.15 cu in. (not much huh?)

Um... the tank is a 68 ci (cubic inch) tank, not a 68 cc (cubic centimeter) tank. The internal volume of the tank is 68 cubic inches.
Using the same method as above for calculating the number of standard cubic inches of air contained in a volume, I came up with just under 14,000 standard in^3. Again, our numbers are probably close enough for a ballpark figure.

Originally posted by the electrician
now you can't count on using 400 psi of the air because that the operating psi we assumed earlier. so using the 400 psi density agian and the volume of the cylinder in cu in. we can't use 119 cu in. of the standard air.
so 13,824-119=13,705. now divide this by the 15.76 cu in. used per shot, and you get 869 shots.

The 119 number is based off your 4.15 in^3 value which assumed a 68 cubic centimeter tank instead of a 68 cubic inch tank. You're actually looking at more like 1900 standard in^3 being unusable at 400 psig.

Keep in mind that the mag needs around 600-ish psig (I think) input for its internal regulator to be happy, so let's set the bottom end pressure in the 68 in^3 cylinder to 600 psig. This means we have around 2840 standard in^3 of air left in the 68 ci cylinder which we can't really use effectively.

14000-2840 = 11160 standard in^3 of usable air. Dividing that by 15.51 standard in^3 per shot gives me right around 720 shots per 68/3K tank. The rule of thumb for shots per tank is around 10 shots per cubic inch at 3000 psi, so 720 seems to be within reason.

Originally posted by the electrician
I know these numbers only apply at that pressure and for that gun. I wish I could find some correlation between this and the energy per shot used.

Theoretically there is a correlation. However, for any particular gun there will be an efficiency factor somewhere in the equation which makes determining this correlation a pain.

BJJB

[the electrician]

Okay first off, sorry. I don't know where I came up with the cc's.(I'm a dumba**)
I went back to recheck my total air in the tank and wondered why it didn't jive. now I know.
I went through and re-did my numbers.it does make more sense when you use the right ones. :)

thanks for the explanation of the cancellation of units, I should know this too.

so if we use the 340 psi as 85% of the 400, how much air is that? I'm gonna say 15.76 - 2.8 = 12.96. 2.8 being the volume of air for the 60 psi we didn't use.

so if the ball is done accelerating in what 5 or 6 inches,(you tell me) at what point does this pressure not have enough force to accelerate the ball with any real significance?

and how much of this air is actually used? some of it must be wasted.

[bjjb99]

[edit: Yes, it would help to actually type something instead of just quoting the previous post... doh!]

Originally posted by the electrician
Okay first off, sorry. I don't know where I came up with the cc's.(I'm a dumba**)

Don't worry about it. Everybody's brain farts from time to time. :)

Originally posted by the electrician
so if we use the 340 psi as 85% of the 400, how much air is that? I'm gonna say 15.76 - 2.8 = 12.96. 2.8 being the volume of air for the 60 psi we didn't use.

so if the ball is done accelerating in what 5 or 6 inches,(you tell me) at what point does this pressure not have enough force to accelerate the ball with any real significance?

I suppose you can treat the barrel with a paintball in it as a piston, and see how long the piston's stroke is before the pressure inside and outside the cylinder equalize.

Let's start with the 0.55 cubic inch volume. We use about 85% of this volume during firing, so we're looking at 0.4675 cubic inches of useable volume at 400 psig (414.7 psia).

The outside air pressure is 0 psig (14.7 psia).

The "piston" is 0.68 inches in diameter.

The volume of this piston is dependent on the position of the paintball, so we're looking at (pi/4 * 0.68^2 * X) where X is the ball's distance in inches from its starting point.

If we assume the propelling gas is an ideal gas (it's not, but we're doing ballpark calculations here), then we can use a simple equality of pressure and volume to determine the point where the air pressure behind the ball equals the air pressure outside the "piston".

P1 * V1 = P2 * V2

P1 = 414.7 psia
V1 = 0.4675 in^3

P2 = 14.7 psia
V2 = pi/4 * 0.68^2 * X

Plugging in all our numbers and solving for X gives us:

X = 36.3 inches, assuming I did my math correctly.

So in a meter long barrel the ball is no longer accelerated when it just passes the 3 foot mark, assuming zero friction, of course.

Originally posted by the electrician
and how much of this air is actually used? some of it must be wasted.

From the calculations above, it should be pretty clear that there is a significant portion of that air going to waste. The problem one faces is how to get the ball up to speed in the 12 or so inches of barrel on has to work with. Doing this in an unported barrel without having residual air pressure behind the ball is a challenge. I have some thoughts on the matter, but to test them I need to design, build, and rigorously test a valve. At the rate I'm going, expect first results in, oh, 2007 or thereabouts.

BJJB

[the electrician]

I've been looking at the design of some medium sized 2-way and 3-way solenoid valves. some use a small 4-way solenoid valve as a pilot to the main valve. the main valve stem is basically the center of a double acting ram type of set up. the pilot valve control the flow of air to the two side of the valve stem to open and close it.
I've been working on applying this to a paintball valve. just a simple crude design for now. this means no valve springs, no sear, no hammer and hopefully less recoil. but I worry about air consumption and when in the cycle it is being consumed. I don't think it will be as efficient as an automag, but hopefully it will work well, and be fun to make and shoot.

[pbjosh]

electrician,

We went round on this on the tinker's guild a while back. When it came to gun effecientcy I remember that the AKA guns, at 150 in*lbs of enegry per shot average @ 285 was about 85% effecient. I don't have the time to go back over this in all the detail, but I think a 100% effeceint gun uses 114-120 in*lbs or so per shot. (that is why when somebody said they got 1800 shots from a 45/45 I cried BS, their claimes were for a 110% effeceint gun)

Oh, and on HPA and N2 the Mags need a bit more pressure if I remember. 425 or so I thought. I used HPA for my calculations, and the numbers ended up close.

Josh

[the electrician]

very cool, thanks for comin' in on this one.
150 in*lbs to me seems amazing. I am impressed with AKA's efficiency.

hey thanks for the info.