A way to fill it with a single stroke of a hand pump?

[boyntonstu]

Is there any way to fill an airgun it with a single stroke of a hand pump?

Lets calculate:

Atmospheric pressure at sea level is about 14.7 psi.

2 Atmospheres means twice 14.7 psi or 29.4 psi.

If you take any size piston in a 12" long cylinder and you push it in 6" you halve the air volume and therefore double the pressure.

We have a rifle that needs 300 PSI in a 1/2" x 6" chamber.

1/2" x 6" is about 1.25 cu in.

300 PSI/14.7 = 20 Atmospheres.

IOW We need a 20:1 compression ratio with a final volume of 1.25 cu in.

Lets say that we can supply 150 pounds of force to the pump whose piston is 1 sq in.

(A car engine has about 150 PSI or so)

If we push our 1" piston a closed cylinder we can achieve 150 psi.

OK We are at 10 Atmospheres, half way to 20 Atmospheres.

We need to compress 12.5 cu in into 1.25 cu in. to get 300 psi.

If we take the 150 PSI and push a free floating 1/2 sq in piston (300 psi) in a ~12.5 cu in cylinder, we would reach our goal.

I calculate that we need about a 60" cylinder to achieve 300 psi.

Are my calculations correct?

BoyntonStu

[Hotwired]

Single Stroke Pneumatic systems have been around for a fair while in air rifles. Pretty consistent things if built well. However they do only compress sufficient to propel a pellet and not at high power either.

I really wouldn't do it if the volume you're after requires a 60" (~120"+ with rod extended) cylinder attached to the cannon. A separate pump would be far easier.

Also, hand pump and 150lb of force (at end of travel) doesn't seem right. If it was a floor pump that would be manageable by using body weight. But again... back to whether its worth the effort to have such a large pump on the cannon.

[spot]

I'll try to calculate it too I'm not used to using inches though... So I could be wrong

Volume of the chamber of the gun:
1/4" x 1/4" x PI = 0.19634375 square inches
0.19634375 x 6" = 1.1780625 cubic inches

Piston diameter?
Compression ratio of 20:1 (like you calculated)
=> volume cilinder must be 1.1780625 x 20 = 23.56125 cubic inches
Let's say we can supply 150 pounds of force: 300psi x surface of piston = 150
=> surface of piston = 0.5 square inches
=> radius: 0.5 = r x r x pi => r x r = 0.5 / pi => r x r = 0.159 => r = 0.396 " => diameter of piston = 0.793 "

Length of cilinder?
we needed a volume of 23.56125 cubic inches => 0.396 x 0.396 x PI x length = 23.56125 => length = 23.56125 / 0.159xPI = 47.1698 inches

According to my calculations you would need a 47.1698 inch cilinder with a diameter of 0.793"

[boyntonstu]

Single Stroke Pneumatic systems have been around for a fair while in air rifles. Pretty consistent things if built well. However they do only compress sufficient to propel a pellet and not at high power either.

I really wouldn't do it if the volume you're after requires a 60" (~120"+ with rod extended) cylinder attached to the cannon. A separate pump would be far easier.

Also, hand pump and 150lb of force (at end of travel) doesn't seem right. If it was a floor pump that would be manageable by using body weight. But again... back to whether its worth the effort to have such a large pump on the cannon.

The pump would not be part of the rifle.

The rod would be on only the first pump; use free pistons on the rest.

A 3 or 4 stage pump would make it more compact.

BoyntonStu