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Interpreting Muzzle Energy?

A place to ask general spud cannon related questions.
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Interpreting Muzzle Energy?

Unread postAuthor: RBurke83 » Thu May 02, 2013 10:54 am

Hey guys, I searched for an answer to this query, but didn't find any definitive answer. So I use a muzzle energy calculator for my launcher, which for specificity is a standard piston pneumatic, muzzle velocity is approximately 580 fps with an approximately 160grain projectile. Seems easy, plug it in to energy calculator and come up with 120ft/lbs. The problem is that the projectile is a steel dart... math types will school me on this, but wouldn't the energy of a dart be focused at the tip? Are energy calculators estimating a chunk of lead as opposed to a tiny penetrator? I understand that xmass traveling at xvelocity produces xinertia, but is there a factor im missing? Pardon my physics ignorance please!
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Unread postAuthor: jackssmirkingrevenge » Thu May 02, 2013 11:12 am

Muzzle energy is what it is.

A bucket of water dropped on your head has about three times the energy as a bullet shot from a rimfire rifle. Which would you prefer to get hit in the head by?

As such, muzzle energy alone is a poor indicator of ballistic performance.

You should have a read through this thread: http://www.spudfiles.com/forums/penatra ... t7563.html
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Unread postAuthor: RBurke83 » Thu May 02, 2013 11:35 am

Thanks, I bookmarked that thread. From what I took from it was, Faster, Harder, Longer....hehe Debbie Does Dallas could teach shooters quite a bit about target damage. Seriously though, I suppose the best judge of terminal performance is depth and diameter of penetration into a uniform substance.?
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Unread postAuthor: jackssmirkingrevenge » Thu May 02, 2013 11:49 am

RBurke83 wrote:I suppose the best judge of terminal performance is depth and diameter of penetration into a uniform substance.?


Pretty much... but then again, it depends on what you're shooting at.

If you want to defeat an armoured target, a dense metal dart is an ideal projectile.

If you're shooting at a living target, a softer large diameter projectile is ideal.

This is why you have such a bewildering variety of bullets available. A hollow point is an excellent way to put a guy down, but if he's wearing body armour it's next to useless and you would prefer a steel cored round.
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Unread postAuthor: RBurke83 » Thu May 02, 2013 1:28 pm

I suppose a ballistic pendulum would simply mirror the muzzle energy calculator?
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Unread postAuthor: jackssmirkingrevenge » Thu May 02, 2013 2:08 pm

A ballistic pendulum is a poor man's chronograph.

What you probably want is this: http://www.myscienceproject.org/gelatin.html
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Unread postAuthor: Kilash » Thu May 09, 2013 10:59 am

It depends on what you're projectile was meant to do. A crossbow bolt has about 60 foot pounds of energy behind it when it hits its target. A 3.5 gallon bucket of water, dropped from two feet, also hits with 60 foot pounds of energy. Yes, in normal circumstance a bucket dropped on your head would hurt, but not lethal. But take that same bucket, put a spike on the bottom, and it would do about the same damage to your head as a bolt fired from a crossbow.

The tiny cross-section of a sharp crossbow bolt insures it transfer the maximum amount of force onto a small area to penetrate its target, and hopefully cutting a jugular or vital organ. A bullet does about the same thing, except it has a lot more energy and it also causes damage through hydrostatic shock to your vitals.


I use pressure setting and gun dimensions (volume, length...etc) to calculate my muzzle velocity and energy. If you can roughly account for any friction or flow efficiencies, the result is reasonably accurate (I have not confirm this myself seeing that I don't have a chronograph, but I have done the math for a few spudguns that have been chronographed by their builders, and using just their dimensions I come pretty close to their results).

First you need to find the area of the base your projectile, usually its just finding the area of a circle (unless you're using a square barrel :lol: ). Please note this only works for pneumatics.

Example:

You have a 2 inch barrel, so with a nicely snug flat bottom* projectile, the base area would be:

1 inch^2 x 3.14 = 3.14 sq.in

Now we multiply the pressure you'll be using by this base area, which assuming in our example was 100 psi, comes out to:

3.14 sq.in x 100 psi = 314 pounds (force)

This goes to show why we should use pressure rated material for our spud guns, seeing how the force exerted on the projectile alone is already the weight of a very large man/woman, at 100 psi we're already rivaling the draw weight of a Roman ballista.

At this stage you might be inclined to simply multiply your pounds of force by the length of your barrel to get your foot-pounds of energy, but we have to remember that as our air pressure expands into and down our barrel, the pressure goes down. So the force on our projectile is not constant throughout. 314 pounds is just our INITIAL applied force.

We need to account for this drop in pressure, but fortunately we won't need any calculus-level math for this. All we need is the volumes of our air chamber and barrel to calculate the pressure drop.

Going back to our example, let's say our 2 inch barrel is 4 feet long (48 inches), and behind our barrel and pilot valve, we have a 3 inch chamber that's 2 feet long (24 inches).

We calculate the volume of both:

Barrel Volume = 1 inch^2 x 3.14 x 48 inches = 150.72 cu.in

Chamber Volume = 1.5 inch^2 x 3.14 x 24 inches = 169.56 cu.in


We'll round these to 151 cu.in & 170 cu.in respectively. Now we have these volumes, we can use them to find the pressure exerted on our projectile just as it's about to exit the muzzle of our gun. We use Boyle's Law for this, which states that an increase in volume results in a proportional decrease in pressure. If we add the chamber and barrel volume together, we get the total volume that our air expands into at the last stages of the projectile.

Hence, we have these ratios:

Initial Pressure X Chamber Volume= Final Pressure X (Chamber + Barrel Volume)

Ex.

100 psi x 170 cu.in = P2 x (170 cu.in + 151 cu.in) <--- P2 being our final pressure.

17,000 = P2 x 321 cu.in

17,000 / 321 cu.in = P2 <--- we move over our total volume algebraically.

52.96 psi = P2

From this, we know that as our projectile reaches the end of the barrel, our pressure dropped to about 53 psi. Hence, the force exerted at that stage is:

53 psi x 3.14 sq.in* = 166.42 pounds

*Our projectile's base area from before.

So our applied force drops from 314 pounds to around 166 pounds. If we average that out:

( 314 pounds + 166 pounds ) / 2 = 240 pounds of average force.

240 pounds of force, that is quite nice - This is why when we have a larger pneumatic chamber relative to our barrel, we get better muzzle velocity. There's less of a pressure drop, so our projectile is accelerated faster.

Now we simply just take our averaged force and multiply it by the length of acceleration (or the length of the barrel) we get a theoretical value for our muzzle energy.

In our example that works out to:

240 pounds x 4 feet = 960 foot pounds of muzzle energy

960 foot pounds is close to the muzzle energy of a .221 Remington Fireball or a .44 black powder pistol. A reason why not to point a spud gun at any person you don't intend on maiming or killing. Of course, this is the theoretical maximum energy, we haven't accounted for friction in the barrel, flow restrictions, and a bunch of other things that cut down on the 100%, but usually I just multiply the number by a rough efficiency percentage (80%) and take that as my real muzzle energy in the field. That still works out to 768 ft.lbs of energy in our example, not too bad for a bunch of plumbing parts.

If you want to calculate your muzzle velocity, you can take your calculated muzzle energy and do something with it along with the weight of your projectile, but since I'm Canadian, I have no idea how to work around imperial system in that respect, but I can derive it for you at a later time if you do request it.

Time to go to the dentist for me, cheers!
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Unread postAuthor: RBurke83 » Thu May 09, 2013 11:57 am

Thanks! I kind of assumed there was something else going on that I wasn't accounting for. Math has never been my strong suit, but at your leisure could you tell me what energy performance you would anticipate from my launcher?

Chamber Vol. is 18.7 cubic inches
Barrel Vol. is 6.7 cubic inches
Initial Pressure is 250 psi
Projectile coned nail dart 8 grams
Muzzle velocity approximately 600 feet/sec (audio clocked)

The dart cone base is actually flat, and filled with foamboard.
Thanks so much!
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Unread postAuthor: Kilash » Thu May 09, 2013 12:24 pm

What's your barrel length?
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Unread postAuthor: RBurke83 » Thu May 09, 2013 12:32 pm

Barrel is 42 inches.
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Unread postAuthor: Kilash » Thu May 09, 2013 1:36 pm

One last thing, what's your barrel diameter.
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Unread postAuthor: RBurke83 » Thu May 09, 2013 1:42 pm

Its .55 inch ID, ty
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Unread postAuthor: Kilash » Thu May 09, 2013 1:56 pm

Nevermind, i derived your using half inch pipe.

Your theoretical muzzle energy at 250 psi is about 121 foot pound. Assuming 80% efficiency, its probably about 97 foot pounds, higher if you have no bottlenecks and your using a quick release valve. That energy is slightly higher than that of a high performance compound bow.
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Unread postAuthor: Kilash » Thu May 09, 2013 2:03 pm

My math tells me your muzzle velocity for a 8 gram projectile would be about 597 ft per sec. Thats with 97 ft.lb as your energy, so looks like it works out 8) .
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Unread postAuthor: RBurke83 » Thu May 09, 2013 2:12 pm

Thanks Kilash! I used a muzzle energy calculator on a different site entering just muzzle velocity and projectile weight and I think it came up with 115ft/lbs so your calculation is probably right on the money. It has a piston valve which is almost air tight while being very low friction, currently I'm trying to design a new pilot valve to fit inside of a ryobi cordless drill body, to use its 45 style grip and linear trigger.
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