High Pressure QEV Air Gun

[MCRKilljoy]

I improved my air gun! I made the chamber twice as big, and used galvanized pipe instead of black iron. The ammo is AA batteries, metal bar sections, paper darts, mini marshmallows, steel balls, marbles, potatoes, or anything that fits in the barrel. At 120 PSI, I can get a muzzle energy of 100 joules with a steel bar. AA batteries get about 80 joules. I shot a computer case with an AA battery, and it punched through. The battery was heavily deformed. Steel bars don't deform, and they can go through 1" MDF. The launcher's tank is made of a 1" by 12" galvanized pipe nipple. The main valve is a 1/2" QEV rated to 150 PSI. The barrel is 1/2" copper or 1/2" SCH 80 PVC. The pilot valve is a 1/2" brass ball valve i had lying around, and the fill port is a 1/4' male air compressor quick disconnect plug. The back of the air tank acts as a stock. Since there is a 90 degree elbow for the QEV, I cut a strap out of a 2 by 4, and used a hose clamp to secure it.

[mrfoo]

foot/pounds are a measure of torque, not velocity. Are you sure you don't mean feet per second?

A quick calculation assuming around 20g for an AA battery gives around 3⅓ Joules.

Or maybe you meant foot pound *force*, which is a measure of energy and not velocity, in which case that's around 81 joules. Which is rather more respectable.

[MCRKilljoy]

foot/pounds are a measure of torque, not velocity. Are you sure you don't mean feet per second?

A quick calculation assuming around 20g for an AA battery gives around 3⅓ Joules.

Or maybe you meant foot pound *force*, which is a measure of energy and not velocity, in which case that's around 81 joules. Which is rather more respectable.

Thanks. I converted all the units to joules.

[Anatine Duo]

Quite appealing in its simplicity. The spacer between the reservoir and upper tube is a nice touch.

If you fire this from the shoulder does the pilot give you a blast of air?

[Ragnarok]

foot/pounds are a measure of torque, not velocity.

Possibly slightly late and off-topic to answer this, but what the heck.

While this part is sort of correct, both torque and kinetic energy use units of foot-pound force, being a force multiplied by a distance. (So foot-pounds, not foot/pounds. They're multiplied, not divided).

That said, Imperial or the related US customary units (please don't call them "English units" - a lot of the US unit values are different from any that were ever used in England. Somehow they came up with an even more nonsensical system where fluid ounces lost any sensible relationship to ounces of weight) just are bonkers, with pounds being both weight and force.

In metric, torque and energy are still both units of Newton-metres, it's just that the derived unit of Joules is often used in the context of energy. (Of course, Newtons are also derived: kg m s^-2)

The relationship is that a torque of 1 Nm applied through an angle of one radian requires 1 Nm of work. As radians are dimensionless, being a ratio between an arc of circumference (a length) and the radius (another length), both torque and energy have the same units.