Can anyone tell me the formula's that I need to calculate the output air pressure from a cylinder/piston/spring setup.

IE, If I have a cylinder that has a diameter of .5 inch, with a length of 1.5 inches. How strong will the spring need to be to output X pressure from an nozzle with a diameter of 3mm.

((The numbers can and will need to vary, but I'd like to know the formula's so that I can calculate the desired air pressure.))

Also, I'd like to know how to calculate the pressure needed to move an object at a specific weight to a specific velocity within a certain distance.

I hope someone can help out. I vagulely remember doing similar calculations long ago in school.

Thanks.

Here is what you might need http://www.team116.org/2003/lessons/Lesson7-Pneumatics.pdf
Al.

Very close to what I need.

I actually need to calculate the air pressure generated by a spring powered piston. Not the force generated by an air driven piston.

It could be that the formula's on the page will do what I need. However, it's making my head hurt thinking of how convert them for use.

It has been a while for me too, but here goes:

.500" piston bore
area of a circle: pi(r*r)

A=3.1415(.250*.250)
A=3.1415(.0625)
A=.1963 square inches

100 pounds of applied force to this piston will give you 19.63 pounds of force. The amount of force will be the same whether it is mechanical or pneumatic. As for the amount of pressure at the nozzle, it will be the same no matter what size it is. All a nozzle does is regulate/restrict the amount of flow thru it.

In moving an object, a few other factors are involved. One is the amount of friction between the object and what it sits on. Otherwise Force = Mass times Acceleration.

Sorry, that is what comes to mind as of right now, hope this helps you out some.

To figure the amount of force of a spring driven piston, you need to know the spring's constant. It is figure that tells you how much force to compress the spring a specified distance. Take that figure and multiply it by the distance of the spring's compression, that will give you the amount of force. Now multiply that by the square inches of the piston face to get the amount of air pressure.

I don't think that the nozzle has zero impact on the the output force of the air.

Example, if you have a syringe with the needle on, and you squeeze the piston, you will spray water across the room. Again the same syringe without a cap on the end would cause the water to splash on the floor.

I'm sure it has something to do with the compression factor and the ratio of the piston diameter to the area the the energy is actually applied to.

Ah well, maybe some MIT student that breaths physics will jump in on this thread and set is all right.

There are similar products that use both spring and compressed air for their power supply. Both have similar results. The compressed air one use a supply of 80-150psi or so, while the spring gun isn't using anything close to an 80 pound spring.

Deviant,

What you need is to use Bernoulli’s Equation. Do a google search on it – there will be lots of hits and info on how to use it. Here is one link:

http://www.grc.nasa.gov/WWW/K-12/airplane/bern.html

With out going through numbers in your example, if your nozzle is releasing to the atmosphere, then you will only have dynamic pressure at this point (rV*V/2, r = density of fluid, V = Velocity). This will be equal to the static pressure in your system, which was explained how to get in the above posts.

Try working through it, if you are still having trouble, give a shout and I’ll show you some more.

Hope this helps. . .

plm

Definately something close to what I need.

According to what I've read so far. Bernoulli's equation applies to a steady flow of incompressible fluid.

I'm not sure how that relates to a short burst of air, which can be and probably is compressed, at least some what.

I'll keep reading.

If I may so bold as to ask, what are you trying to build.

Deviant,

They are referring to the compressibility property of the fluid. You can assume the air is an incompressible fluid and also disregard pressure losses due to friction etc. – you will be close enough. Also, the terms in Bernoulli’s Equation are in units of pressure heads – so you will convert your psi to a head of pressure.

Keep in mind; I’m recalling some of this stuff from fluid dynamics and a thermo course from over 30+ years ago – I’m a little rusty.

plm

Quick reference is


F
-----
P | A

Where :
F = force
P = pressure
A = Area
You will need to determine spring rates for the spring itself.

Question, Are you looking to find the force generated by the rod, or the amount of air pressure coming out the port on the cylinder when the spring is pushing on the piston?
just trying to clarify.

I don't think that the nozzle has zero impact on the the output force of the air.

Example, if you have a syringe with the needle on, and you squeeze the piston, you will spray water across the room. Again the same syringe without a cap on the end would cause the water to splash on the floor.

I'm sure it has something to do with the compression factor and the ratio of the piston diameter to the area the the energy is actually applied to.

Ah well, maybe some MIT student that breaths physics will jump in on this thread and set is all right.

There are similar products that use both spring and compressed air for their power supply. Both have similar results. The compressed air one use a supply of 80-150psi or so, while the spring gun isn't using anything close to an 80 pound spring.

When you take the needle off the syringe, you cannot apply the same force because you can't push fast enough. The volume flow of a gas through an orifice is determined by the properties of the gas, the size and share of the orifice, an by the upstream pressure (assuming that it is several times the downstream pressure).

Ken

This is going to be used in building a bolt action airsoft gun.

It fires a 6mm sphere, that weighs about .2 grams.

I need to generate enough pressure to move it at around 500-550 fps.

You will need to know the velocity the air is moving. Which is based on the spring/piston velocity. Theoretically if the spring moves the 1.5 inches instantaeously then the pressure will be infinite. The spring/piston does not move instantaeously thus the pressure will be less. Use Bernoulli's equation disregarding compression and friction. Use the continuity equation to get the velocity of the air moving through the 3mm orifice. Assume the starting pressure is zero.

Can anyone tell me the formula's that I need to calculate the output air pressure from a cylinder/piston/spring setup.

IE, If I have a cylinder that has a diameter of .5 inch, with a length of 1.5 inches. How strong will the spring need to be to output X pressure from an nozzle with a diameter of 3mm.

((The numbers can and will need to vary, but I'd like to know the formula's so that I can calculate the desired air pressure.))

Also, I'd like to know how to calculate the pressure needed to move an object at a specific weight to a specific velocity within a certain distance.

I hope someone can help out. I vagulely remember doing similar calculations long ago in school.

Thanks.

3mm is a huge diameter. This will not work unless the pellet you are shooting is a close fit in the bore.

Ken

You will need to know the velocity the air is moving. Which is based on the spring/piston velocity. Theoretically if the spring moves the 1.5 inches instantaeously then the pressure will be infinite. The spring/piston does not move instantaeously thus the pressure will be less. Use Bernoulli's equation disregarding compression and friction. Use the continuity equation to get the velocity of the air moving through the 3mm orifice. Assume the starting pressure is zero.

If the pressure were infinite, the spring wouldn't move at all. In fact it would move backwards.

The solution to the problem requires solving a differential equation. Someone suggested that the air could be treated as incompressible. I doubt that is true. In order to get velocities in the range of half the speed of sound, significant pressure will be required. Even a pressure as little as 15psi would compress the air to half its volume.

I'm not familiar with the details of the physics involved, but my guess is that the problem requires solving a non-linear differential equation that has no analytic solution. That means there is no simple formula that can be applied.

Ken

This is going to be used in building a bolt action airsoft gun.

It fires a 6mm sphere, that weighs about .2 grams.

I need to generate enough pressure to move it at around 500-550 fps.

Just to make sure I'm understanding this correctly: We have a spring pushing a piston in a .5 inch bore of 1.5 inch length compressing air which will exit a 3mm dia. orifice at the end of the bore to drive a 6mm x .2gm spherical projectile out of a 6mm+ dia. muzzle of unknown length.
We've got the projectile size and weight and the muzzle velocity and the dimensions of the air chamber. Do you have an intended muzzle length? With the muzzle length we can calculate the force necessary to push the ball at 500 fps. Then we can calculate the spring requirement and finally determine what might be going on in the compression chamber.

Chris

As Chris said, I think we need some more details such as is this thing is completely manual? i.e. no miniature compressed air cyl? I am assuming that's the case and that you are going to have some kind of built in mechanism to recoil the spring and lock it in position ready for release by the trigger mechanism.
I would attack this in another manner. Firstly, assuming the 3mm orifice isn't actually a valve (to stop/start air pressure to ball) that you didn't tell us about, then I believe the only reason you need the step (orifice) at all is so that the ball can't roll backwards and drop down into the air cyl. Therefore make the step 5mm diameter as this will minimise the pressure drop and resulting energy loss that would otherwise be caused by a 3mm orifice.
What comes to mind for me is the following:
I assume the ball doesn't get loaded by just dropping it down the barrel but instead gets loaded by some kind of loading mechanism (i.e. from a hole in the side of the barrel and that hole gets covered up (sealed) by this loading mechanism) If that's the case, I would have a screw on barrel and just ahead of the ball I would machine a groove in the barrel and fit an O-ring. This will stop the ball from dropping out of barrel and more importantly provide initial resistance when fired.
So the spring and piston in the air cyl you described are recoiled (i.e. spring fully compressed), load ball, pull trigger, piston is released and during that microsecond, air pressure rises to overcome the resistance offered by the O-ring acting against the ball, and finally ball is fired from barrel.

Personally I wouldn’t waste my time doing any theoretical pressure calculations. Volume calcs yes, but pressure no. Work out the volume of the compression chamber from the face of the recoiled spring and piston assembly to the back of the ball. I would make the barrel volume (from the front side of loaded ball to end of barrel) which translates to length, about 2/3 of compression chamber volume. Why? Because theoretically they should be of equal displacement but in reality (A) a bit of extra volume (in the comp. chamber) is used up during the compression phase and (B) faster projectile speed will be obtained if we don’t stop air flow at the exact moment the ball leaves the barrel. A little bit of extra air behind the ball at this time helps.
Now, the reason I said not to worry about the pressure calculations is that put simply and assuming the volume calculations in relation to barrel length have been done correctly, if you go to your local industrial supplies place and buy a range of comp. springs it will cost you about $10 and take about 30 minutes to find out which spring works best. Test it firing in hard foam and see which fires deepest from a given distance. Starting from weak springs going to strong, you will find that the speed will increase quite rapidly with spring tension increase but then there’ll be a point where it won’t get faster. This point is determined by factors such as: the restriction size at the smallest point (= 5mm) cannot flow air any faster, the weight of the spring and piston assembly trying to accelerate, etc. The factors affecting the equation are so many that it’s not worth attempting.

Maybe I’ve assumed wrongly and the 3mm restriction is required because it’s actually an air valve. If this is the case then I would still use the same volume calculations to determine barrel length except that comp. chamber is measured to back of valve not ball. And determine your spring requirement by testing as per above.

The barrel is 6.03mm inside diameter.

The bb's are roughly 6mm.

The barrel length will be about 650mm, no more than that.

As for the reasons on the calculation, I'd like to know that is possible to do what I want.

If the calculations show that I need a 30lb spring to reach 500 fps. It wouldn't be worth the effort.

However, if the calculations show that I can do it with a 8lb spring. I'd be more willing to give it a go.

As for the 3mm openning. The reason for this, as stated above, is for loading of the bb.

On the end of the cylinder will be a stem roughly15mmx6mm with a 3mm through hole. The stem will also be turned on the outside for a small oring to seal the barrel to keep the pressure from escaping back along the stem.

The stem is used for pushing the bb into the barrel from a magazine. As for stopping the bb from rolling out the end of the barrel. I planned on using a small detent, but an oring could serve the same purpose.

Thanks for all the help.


*edit*

Added picture of existing cylinder. I'm looking to make one smaller than this and recreate the feeding mechanism. This is identical to what I plan on making, but should give a better idea.

Just a comment, your interpretation may vary. . .

There has been a lot of good information given on the static aspects of this problem – the pneumatics portion. However, there seems to be some speculation on solving the dynamic aspects of the problem. Not that speculation is bad but, in this case it is not necessary. This problem is rather easy to solve and is a “classic” problem that is covered within the first few weeks of a Fluid Dynamics Course.

With regards to the second part of Deviant’s question involving the dynamic aspects, it has been stated more than once that the key to solving this problem is the use of Bernoulli’s Equation with the use of the idea that the flow rate at any part of the system will be the same (Area1 * Velocity1 = Area2 * Velocity2). It has also been stated more than once to assume an incompressible fluid and disregard the friction. Using this method will give you a good ballpark answer quickly. To be any more anal is at your discretion.

BillPSu said it best:

“Use Bernoulli's equation disregarding compression and friction. Use the continuity equation to get the velocity of the air moving through the 3mm orifice.”

When I originally responded to Deviant’s question, my intentions were not to initially solve the problem for him but, to point him in the “correct” direction in order to show him how to do it – it’s more beneficial to show someone how to catch the fish than to catch the fish for them.

Other than a follow up question by Deviant to clarify some aspects in the use Bernoulli’s Equation, I never heard back from him that he was having problems and assumed that he was on his way with the solution.

Good luck Deviant and stay on course.

plm

I corrected this posting and it is now message #36.

Chris

"As for the 3mm openning. The reason for this, as stated above, is for loading of the bb".
Actually it wasn't stated anywhere in this thread.

Also I made an error by stating "Work out the volume of the compression chamber from the face of the recoiled spring and piston assembly to the back of the ball" It should have read "work out the displacement volume of the air cyl" not the total volume of the compression chamber to back of ball. Not that that's relevent now but I just needed to correct it.

Thanks for adding some details of the project and good luck with it.

In follow up.

I haven't had a chacne to devote alot of time to this project from my original post. I did however open cad this weekend to try generate better numbers for the calculation.

I also searched the web for tips on applying Bernoulli's formula or online calculators.

I've been looking at this more today, and I've become a little irriated at trying to apply the formula.

So before I really start attempting to apply the formula, I need to know how the pressure from the spring is generated? Correct?

It would seem that if you have 8lbs of force applied to 1 inch of surface, you'd have 8psi. Although that doesn't seem "right". If you applied it to .5 inch how does that translate because now you have 8lbs of force on a small area, but it's still the same force.

How does the movement of the piston effect the end pressure. Ei, if you have 8lbs of force from a spring that moves 2.5 inchs. Do you end up with a 20psi from the compression?

The Bernoulli’s formula assumes that you know certain aspects, but I'm not sure how you calculate those.

ie, in this example. How do you calculate the velocity of a spring?

(Area1 * Velocity1 = Area2 * Velocity2)

All in and all, I guess I need a better lesson in physics.

The lesson continues.

We assume that pressure = force/area.

So we need to solve for area first.

We have a circle .5 inch in diameter.
Which has a radius of .25 inch.

Area equals

radius * radius * 3.14

Which is 0.196 square inches for area.

Now need the force.

I'll assume that I have a 8lb spring.

Pressure = 8/0.196

Which would be 40.8 psi?

((going to try and break this up via steps.))

"Which would be 40.8 psi" yes, that pressure will be reached if the outlet of the cylinder is blocked off but not if the air has an easier escape route such as the barrel with a light weight ball sitting in it. If the detent or O ring can offer a decent resistance you might get close to that pressure. Just one of the many factors that are just about impossible to calculate and haven't up to now been taking into account.

Next, assuming that, due to having the correct detent force in combination with the resistance offered by the piston during acceleration, the chamber happens to reach the theoretical figure of 40.8psi, if you now work out how far the piston has travelled down the cylinder bore due to compression of air and next look at the amount of cylinder displacement still left available, take the figure and convert it into cu/ins or whatever you're working in and compare that with the barrel volume and you will see that your barrel is considerably too long. I've already done the maths but I'll leave you with it.
cheers

Well the product exists now, I'm just trying to make one that is smaller.

The barrel length is truely up to me, but the longer the barrel the more stable the bb path.

If I need to lower it to 550mm vs 650, that will be exceptable. 650 is the maximum that the precision barrel length comes in.

I think the project is starting to lean toward one of the build it and see if it works type things. I was just hoping to get a "rough" idea of what dimensions and spring force I needed to reach the goal. Ideally, I wanted to know the formula's and how to apply them also.

Deviant, I don’t know if I said this earlier but I think this is a really cool project and I certainly understand when you say "I was just hoping to get a "rough" idea of what dimensions".

"The barrel length is truely up to me, but the longer the barrel the more stable the bb path. If I need to lower it to 550mm vs 650, that will be exceptable."

I’ve just realised I screwed up my calculations on the displacement volumes. (sorry to talk in mm3, you'll have to convert back to cu/ins)
3.143 x 6.35 x 6.35 x 38.1mm(0.5” x 1.5”) = 4828mm3 displacement for the cylinder and 3.143 x 3 x 3 x 550mm = 15,560mm3 displacement volume for the barrel. That’s a disaster as the ball will be under vacuum as it passes through the second half of the barrel! Assuming the barrel should ideally have a maximum of about 80% of the displacement volume of the cyl then using a 550mm barrel we will need a cyl with a displacement of 19,450mm3 divided by 127mm2 (surface area of ½” cyl) = 153mm = a six inch cylinder required. Whoops, can someone else check my maths for me (in inches of course) because this is serious!

"I think the project is starting to lean toward one of the build it and see if it works type things." I agree once we get the volume equations correct. Once again, sorry about the metric as that's what I started in.

Deviant, I just checked all my calculations and also added a few more.

(B1) Barrel at 650mm x 6mm = 1.122cu/ins = 18386mm3
(B2) Barrel at 550mm x 6mm = 0.9495295cu/ins = 15560mm3
(C1) Cyl: 0.5"D x 1.5"L = 0.2946562cu/ins = 4828mm3
(C2) Cyl: 0.5"D x 6.0"L = 1.178625 cu/ins = 19354mm3
(C3) Cyl: 0.75"D x 2.0"L = 0.884 cu/ins = 14486mm3
(C4) Cyl: 0.75"D x 2.5"L = 1.105 cu/ins = 18107mm3
(C5) Cyl: 0.75"D x 2.75"L = 1.2155 cu/ins = 19918mm3
(C6) Cyl: 0.75"D x 3.0"L = 1.326 cu/ins = 21729mm3

Remembering that you'll need a bit more cyl displacement than barrel displacement, you'll see that B2 will go happily with C2 OR in the 0.75inch cyls, you'll see that B1 will go with C5 but preferably with C6. For me the B1 to C2 match is a bit too marginal. Hopefully you understood what I was saying about if the barrel displacement is less than cyl displacement the ball probably won't even exit the barrel or if it does it will be awfully weak.

Lastly, assuming you are thinking about building these things as a sideline or business, are you able to get your hands on one of the competitor's guns and pull it apart and measure it, then try and transpose those measurements to the size you'd like?

Assuming I go with a different size cylinder/barrel combination to make the displacement in my favor.

What power spring would I be looking at? As we agreed on the previous post, the 8lb spring on a .5" piston generated 40psi.

However, if I use the same 8lb spring on a .75" piston.

I'm only getting 18 psi.

Am I missing something?

Does the pressure multiply by some value when it hits the .125" opening at the end of the .75" cylinder?

This is a solid bb correct? I just want to know what you are making this for and where you are planning on traveling? What you could make with this is a little scary.

This is a solid bb correct? I just want to know what you are making this for and where you are planning on traveling? What you could make with this is a little scary.

The bb is a 6mm solid plastic round that weighs .2grams. At 500fps, it provides 2.32 joules of energy on impact. Of course the bb doesn't carry the velocity very long.

And for your reference, a paintball weighs around 3.2 grams and travels around 300fps with an output energy of 13.3 joules.

So no, the idea or the implementation isn't scary. Although, you might get a small red welp if your shot up close.

Thats what people use them for is to poison people in a crowd. It is a hollow ball with poison in it. They walk up to someone shoot it and walk away, know onw knows a thing.

Thats what people use them for is to poison people in a crowd. It is a hollow ball with poison in it. They walk up to someone shoot it and walk away, know onw knows a thing.

It's nothing that doesn't exist already or could't be made better/cheaper without all the physics questions. Assuming the purpose is milacious.

Truth be told, it would be alot easiler to pick up a pellet gun/paintball gun from walmart than to spend the time trying to machine something quality.

Anyway, back on topic.

"However, if I use the same 8lb spring on a .75" piston I'm only getting 18 psi" Yes, unfortunately that's correct. You didn't answer my question about whether "you are thinking about building these things as a sideline or business" as opposed to just making a one-off for yourself so I'll assume the former. That being the case, I'll also assume that you want to perfect this thing as best you can on paper and then make a prototype or two for testing (just using of course, the standard balls for the prototype testing not the #706 poison version) so that the minimum amount of money is wasted on prototypes.

Ok, the way I see it is that you have a few possibilities so I'll toss some ideas into the ring and you can take it from there as to which direction to head. I'll assume that we'll be somewhere within the ballpark if we can apply 40psi to the back side of the 6mm ball. I don't know if that will be enough or not but I suggest we use that as a starting point and change springs at testing time. I'll also assume that 8 - 10lbs is about the maximum amount of spring tension that you want to pull against when having to recoil the spring. (both the 8lb spring and the 40psi were your figures and I'm just using them as base figures to work from)

(a) 0.5" x 6.0"L cyl with 550 x 6mm barrel with 8lb spring giving 40psi. This one seems the simplest one to do. I don't know if the length of the barrel combined with the length of the cyl will be too long or not but if it is you can always have the cyl mounted beside or on top of the barrel as it doesn't necessarily have to be behind it.

(b) 0.5" x 3.0"L cyl with a short barrel (and the 8lb spring giving 40psi), say 275mm long (haven't done the calcs) however you said the accuracy wouldn't be good being so short.

(c) 650mm barrel with 0.75"D x 3.0"L cyl with 17.68lb spring giving 40psi. If the 18lb spring is too much to comfortably recoil can you use a rachet or other mechanism for gaining a mechanical advantage? (such as pin and lever)

Now here's an idea that is so crazy it just might work.
As you are talking about a 650mm barrel, I have a mental picture of something like a rifle so can you make it with the rifle barrel at top, and a hand pump running parellel underneath the barrel (like a pump action shotgun) with a one-way valve to stop back passage of the air. (maybe even one of these new high efficiency bike pumps = cheap, with a normal bicycle air valve acting as the one-way valve?) The air is pumped into the compression chamber which is nothing more than an air reservoir (= much cheaper than a cyl/piston/spring set up). Ahead of the air reservoir is a 0.25inch "T" ball valve to which the trigger is attached. The balls are inserted in a tube with screw on lid and they drop down the tube and lowest one drops into the ball of the ball valve. (I said a 0.25" ball valve so you are guaranteed of the ball dropping into the ball of the ball valve) At this point in time the ball valve is holding back the 100psi or whatever of air. Trigger is pulled and ball valve rotates, ball pellet has nowhere to go but forward. This system doesn't even need a detent to stop ball rolling out of barrel. You would probably need a 5mm diameter insert to be fitted to the back side of the ball valve so that a ball pellet can not roll back into air reservoir when there is no air in system. Sure sounds cheap to make and try. You'd probably also need some kind of overcentre spring actuation of the ball valve by the trigger in order to turn the ball valve very quickly when trigger is fired.

Anyway, just an idea to get you thinking.

Pressure intensification.
Your pressure as a force is acting on the area of the bore restrained by the pellet.
So 40.8lb-force (your number is way too big, force equal to pressure x area of piston) on 6mm bore area (0.0438 in2) is like 931.5 psi (way too high) equivalent, pressure drops off real quick but has transferred its energy into the acceleration of the pellet. Work this problem backward from the energy conservation angle ie how much energy is in a 6mm pellet at 500 fps. My 6mm airsoft at 200-300fps use tiny springs with short throws. A 20 (500mm) inch barrel would require 0.876 in3 of air volume (1.99 in stroke x 0.75 in piston).
There are a whole range of airsoft rifles out there, just take one apart and beef up as necessary.
mikejkd there is a whole world of airsoft rifle pistols and machine guns out there based on the 6mm plastic pellet. There used to be photo realistic of all sorts of replica weapons but now they have an orange muzzle.

Originally Posted by Deviant
The barrel is 6.03mm inside diameter.

The bb's are roughly 6mm.

The barrel length will be about 650mm, no more than that.

As for the reasons on the calculation, I'd like to know that is possible to do what I want.

If the calculations show that I need a 30lb spring to reach 500 fps. It wouldn't be worth the effort.

However, if the calculations show that I can do it with a 8lb spring.


The formulas for calculating the force required to shoot the bb at a certain muzzle velocity over a particular barrel length are as follows:

V^2=Vo^2+2Ao(S-S1) where Ao is assumed to be a constant acceleration.

This reduces to V^2=2(Ao)(S) when the initial velocity Vo and the intial distance S1 are both zero. You want to solve this for Ao.

V=Vo+Ao(t-to) Use Ao from above and solve for t. This will give the time required for the bb to travel the length of the barrel. Again Vo and to are zero.

(t2-t1)F=mV2-mV1 Here the force is equal to the change in momentum divided by the time of travel. t1 and V1 are zero. Use t and m from above.


This gives the average force exerted on the bb over the length of the barrel.
The work done is F x d. The pressure in the barrel necessary to impart that force is P=F/A where A is the area of the bore.

Using a 1/4" bore x 24" length barrel and a .2 gm bb I get the following results:

Ao=62500 ft/sec^2
t=.008 sec
F=.856 lb
P=17.4 psi
W=1.7 ft-lbs or 2.3 joules

The force is the average force and the pressure is consequently the average pressure. I know that the acceleration looks like it's off the charts. If I'm missing something here, I hope someone will point it out to me.

It seems like the next part of the problem has to do with the acceleration rate of the spring/piston and the flow rate of the compressed air from the cylinder into the barrel. If the flow rate is low then it seems like some kind of constriction in the barrel ahead of the bb would be necessary to allow the pressure to build.

Chris

Skippy,

Thanks for the advice, but I'm looking to make a firing mechanism that is alot easier to reload in the field. Keep in mind that this will be used in a wargamer simulation.


carlnpa,

In response to taking apart an existing gun. I have owned most of the popular and not so popular models. I'm looking to improve on the short comings I see in them.

Most current airsoft guns are electric which takes away from the gun in my eyes. Most of the classics are gas and work on the o-ring pressure build up, however parts are next to impossible to get.

If I have to make parts, I might as make something new and better.

The bolt action guns, aps, vsr etc... all have bolts that are misplaced in regards to where the magazine would be on the real steel. Also the pull length/effort always seemed a little excessive which always bugged me. I had a KS model APS that had a really light pull, with extremely reliable velocity. This is one of the main reasons that I believe a gun can be made within the specs I'm looking for. And I'd guess that the weight pull on the KS Gun was around 5lbs. Of course I'm never hooked a scale up to see.

Which is to make small version of a bolt action similar to the sv-99. Since I'm decreasing the bolt size, I need to know whether it's feasible or not. If I cut the bolt volume in half, do I need to double the spring power. Etc.

((I don't have one of the guns currently to measure, hence all the speculation))

I've already begun working on a gas gun that should be very interesting, however alot of people shy away from gas do to it's unreliable nature in cold weather. *shrugs*


OCNC

Your equations give me hope, however I'm not following how to apply them 100%

Hopefully you can correct me.

Let's say I start with a 24" barrel which I want to shoot a bb at 500fps..

Now, if I follow your formula.

V^2=2(Ao)(S)

V = velocity which is 500fps

s = barrel length? 24" converted to 2 feet.

so it should be 500^2=2(Ao)2

Which if I do my algebra right.

Should be 250000/2 = 2(Ao) that then reduces to 62500 = Ao

But should it be ft/sec^2? That doesn't seem right.. I don't even think my rifles fire at that speed? Should something else be applied here? Maybe taking the square root or something? ((tosses idea in the air))

I believe my .270 is only around 3000+/- fps, at least it's muzzle velocity. Which isn't really the acceleration.

I guess I'm just having a little bit of trouble wrapping my brain around those numbers. I know that from a compressed gas standpoint. Most airsoft guns that fire 500fps, do so at around 80 psi. And the time for the bb to leave the barrel is definately something small like that.

As for the flow rate, I don't think that will be a problem. 3mm hole in the air nozzle is a pretty good for the low volume of the piston chamber.

I'm going to need to double up on the advil before this problem is solved. (lol)

I believe my .270 is only around 3000+/- fps, at least it's muzzle velocity. Which isn't really the acceleration.
(lol)

I think that's the key point. Velocity is not the same as acceleration. Acceleration is the rate of change of velocity and velocity is the rate of change of distance. Velocity is measured as ft per second per second which is written ft/sec^2. If you use the example of a baseball pitcher and plug in the corresponding numbers you might be more trusting of the formula. For an 80 mph pitch (117 ft/sec muzzle velocity) and a barrel length of 6 ft (the apporximate distance the ball moves in the pitchers hand from the time he cocks his arm until he releases the ball as he steps off the rubber) I get an acceleration of 1140 ft/sec^2. The time it takes the pitcher to throw the ball (time in the barrel) is about 1/10 sec. The average force on the ball while "in the barrel" is about 11.5 lbs. based on a 5 oz. baseball weight ( which has a mass of .0098 slugs). By the formula given the acceleration is proportional to the velocity squared and inversely proportional to the barrel length. The higher the velocity and the shorter the barrel the greater the acceleration will have to be. The velocity term, being squared, dominates. In terms of the acceleration it really doesn't matter what it is that you're trying to motivate. If you wanted to get an elephant up to 500 fps in a distance of 2 ft you'd have to use the same acceleration of 62500 fps. The force involved however would be much different.

Chris

Hi Devian,
The formulas you need to solve your problem are here: http://www.atnet.it/lista/easne/mce263-3.html
I would set up spreed sheet and make a graph of velocity vs. length of barrel.
Additional formulas you will need are F=ma and v^2=2ax.
Colin

Colin,

That's an awesome website, and almost exactly what I'm looking at doing. Cool find.

F=ma, Force = Mass x acceleration?

v^2=2ax, Velocity^2 = 2 x acceleration x X?

What is X?

Thanks

x is a variable and is the distance the pellet travels down the barrel.
From F=ma solve for a: a=F/m
plug into v^2=2ax: v^2=(2Fx/m)

Using the formula for force from the link you will have a relationship of velocity and distance the pellet travels down the barrel.
When you make your graph you will also be able to optimize the length of the barrel.

Right away it is apparent that all of the energy initially present in the system is in that compression spring behind the pistion. The energy in the cocked spring must be at least as large as the kinetic energy of the ball as it leaves the barrel, but in reality it must be considerably greater due to all the losses in the system.

If you want to only require 5 pounds of force to cock the spring, and a travel of 4 inches (a guess on the travel here), the spring can hold no more than 2.25 joules of energy. So, even with a perfect system you won't quite reach your goal of 500 ft/s. If we can use an 8 lbf spring, then the energy in the spring can be as great as 4.5 joules. If you can reduce the energy lost as kinetic energy of the piston, and optimize the barrel length so that a minimium of energy stored in the compressed air behind the ball is lost, then you might be able to get the muzzle velocity you are after. It would be tough though.

Just a follow up to everyone involved.

I bit the bullet and started building the system last night.

I figured it would be cheaper to by the aluminum than advil.

I'll post more as I get a working or not so working product.

Started on the piston tonight, still a little bit to go on it.

Definately looks small.

http://www.legendarylands.com/piston_s1.JPG

Solid model for the part

Since the photo is kinda sucky.

http://www.legendarylands.com/piston_s1b.JPG

Deviant, three cheers for taking "Nike's" advice (Just do it!) then see how it works and change/modify as required.

As I've never seen inside one of these guns I don't recognise the pieces. In your original photo (pg2) it seems like the cylinder is the long item at top (centre & right) and the piston is at top left. The piston appears to be about 1/4" long with a small spigot to locate the coil spring. Your picture above is nothing like the page 2 photo unless I'm not understanding it correctly? If this is the piston above then it's going to weigh substantially more than the other type which will negatively effect the performance of the gun but as I say, maybe I'm not understanding it.

EDIT Now that I've looked some more it seems that the bit to the right encases the coil spring and the slots are to hold the spring retainer and the piece at right is what you actually grab hold of the recoil the spring. Am I on track?

Anyway good luck with it.

pg2, the smaller black part is the piston I've started making. It slides inside the larger chrome cylinder.

My piston is about half the size of the one on pg2. Also, a slightly different design.