Worlds Largest Combustion PVC Cannon?

[Engineer87]

We'll while this thread is hot I'd better post this two. I have 4 guns. The big one. An awesome mapp gas metered chamber, a hair spray chamber, and a mini. The gun I use in this vid is the common hair spray gun.

Def go high quality on this one, worth it.

[jackssmirkingrevenge]

6,000ft-lbs?!?!?!? i'll have what you're smoking bud!!! sched 40, at 8 inches? hell, you're lucky to get 100 PSI. i've seen some of the postings here, "children".

you guys are babes in the woods, 6000 ft-lbs would make a crater you couldn't dig out of! and certainly schedule 40 is not capable of 6000 ft/lb

With such a large calibre, even low pressures provide massive amounts of power - especially considering the instantenous and high flow pressure generated in a combustion. 40 psi would be pushing on the surface of an 8" projectile would give you over 2,000 lbs of force along the length of the barrel. 6,000 ft/lbs isn't even that unrealistic, say you had to load a 16lb bowling ball, at a modest 50 feet per second it's already doing over 600 ft/lbs, that's much more than a 45 ACP pistol bullet. Increase your velocity to 150 fps and you've almost hit 6,000. For this design, I would think a longer barrel would be needed (and a slightly larger diameter as apaprently bowling balls are 8.5") but it's not unachievable.

I can't imagine anymore power on that thing.

I would definitely include a chamber fan in the design, and also reconsider using towels as a sabot. This cannon has a lot of potential performance but is being let down by less than optimal firing conditions. If you just want it for fun, that's perfectly fine, however if you want to take it to the edge of what it can achieve, needs a bit more work. In any case, size is always impressive, good job

[starman]

you guys are babes in the woods, 6000 ft-lbs would make a crater you couldn't dig out of! and certainly schedule 40 is not capable of 6000 ft/lb .

Who are you callin' a "babe in the woods" big man??

Why do you question 6000 ft lb on that big cannon? It's not that difficult to approach 1000 ft lbs with a relatively small 1X advanced combustion/burst disk/golfball cannon.

I suggest you back off a bit MrBad. It's obvious you don't know what you're talking about. I thought your basic combustion gun last week was a decent specimen. Why don't you take a little more humble tone for a while until you've been around a while?

[Ragnarok]

6,000ft-lbs?!?!?!? i'll have what you're smoking bud!!! sched 40, at 8 inches? hell, you're lucky to get 100 PSI. i've seen some of the postings here, "children".

As a general rule, the average (that being the important thing, not peak) pressure in a well made combustion's barrel is about 30 psi, or 2 bar.

If you multiply the pressure by the volume of the barrel (being the area times the length) - with the right units of course - it converts it straight to an approximate muzzle energy. (Multiplying pressure by area gives force, and force times distance is energy)

The volume of this barrel is around 3000 cubic inches, multiply by 30 psi, you get 90,000 inch-pounds of energy. Convert to ft-lbs by dividing by 12, the number you get back is 7,500 ft lbs.

In other words: in³ * lbs-in^-2 / 12 in-ft^-1 = ft-lbs

6,000 is actually a pretty conservative estimate, which allows for possible blow-by of the projectile and other unseen factors. This cannon is easily capable of that.

you guys are babes in the woods

Sure, this site gathers some of the world's premier experts on these things with hundreds of years of combined experience between them - and me, who's judgement I'd like to think most of them consider worth listening to (or reading, whatever.)

I've seen more than enough cannons in my time that I can usually give a decent estimate of their muzzle energy potentials before they're even fired. I've seen this fired, so I stand by my numbers.

and certainly schedule 40 is not capable of 6000 ft/lb.

The pipe has nothing to do with the energy that can be generated. It affects the maximum pressure, which limits the energy generated per cubic inch of barrel, but like I showed above, even moderate pressures like 30 psi in such a large barrel create vast muzzle energies.

[mark.f]

@engineer87
I should probably direct your attention towards a special type of fan for this cannon.

They're available at almost every Wal-Mart in the US, they're Ozark Trails camping fans.

They're extremely cheap, and push a lot of air for their size. A few notable spudders such as SMOMW and DR have used them on large-bore launchers between 6-8" bores, with good results.

Just let me dig through the archives real quick...

DR's threads no longer have images, because his villagephotos account is dead, but the movies still work.

SMOMW's cannon - uses camping fan components (motor, blades).

DR's cannon - uses camping fan.


Cheap, easy way to get fuel mixing. No external wiring required.

@Mrbadexample
You're starting to sound like a whiny little [youknowwhat]. Is this cannon getting more attention than yours or something???

[TurboSuper]

Y'know, specialty electronics stores tend to sell industrial brushless cooling fans for pretty cheap. Looking through my catelogue, there's one here for $15(CAD) which moves about 120 CFM.

Looks like a sweet cannon, I bet it has a "shock" factor when people see it's sheer size.

[Engineer87]

If I do it I want to do it right. Especially on such am expensive gun. I would go with CPU fan, Plus 2 square batteries, and a switch. I've never see one of those cad fans before though.
Are there fans small enough to fin inside a 3" chamber?

[Ragnarok]

I've never see one of those cad fans before though.

The CAD referred to Canadian dollars I would imagine. You tend to get quite proficient at converting between international currencies on a forum like this.

[mrbadexample]

they have small DC fans, down to 40mm square for a realtively low price.

http://www.geeks.com/products.asp?cat=FAN



energy has nothing to do with volume, energy is determined by 2 factors, velocity and mass. my little 1.5" x 38" barrel (67ci) generates more energy than most handgun cartridges (1400 lb-ft) a 4 oz potato at 600 fps is 1397lb-ft.

[Ragnarok]

energy has nothing to do with volume, energy is determined by 2 factors, velocity and mass.

Energy can be calculated many ways, but they all boil down to force times distance - the velocity times mass equation is later derived from this.

Firstly, look at the units: "Foot pounds".
Most people use the unit but have no idea what it actually means. Units tell you a lot - here, it is pounds of force, times the number of feet over which it was applied. Now:
- Pressure times area gives force.
Pressure is pounds per square inch. In other words, force on each square inch. If you multiply by the number of square inches, you get total force.
- Force times distance gives energy.
As explained above. Pounds of force times feet of distance gives foot-pounds of energy.

We have multiplied pressure by area to give force, and then multiplied again by distance, this gives energy.
Distance times an area gives a volume, ergo, pressure times volume gives energy.

In equations:
Pressure * Area = Force
Force * Distance = Energy
Pressure * Area * Distance = Energy
Area * Distance = Volume
Pressure * Volume = Energy

Simple enough - the physics and maths support it entirely, and it applies to all cannons, as long as you remember to use the average pressure with respect to distance, which is about 30 psi in an advanced combustion.

~~~~~

Now, when you talk about generating 1400 ft-lbs from a 1.5" D x 38" L barrel, would this be from this combustion?

[Engineer87]

You guys are right but kinda wrong too.

Force * Distance = Work
And work is in units of energy.

Kinetic energy need to be included.

KE = (1/2)mv^2 = one half mass times velocity squared

The correct way to find the average force your gun exerts on any object down the length of the barrel niglecting friction in the barrel would be to shoot the gun parallel to the ground and measure the velocity of the projectile.
Then encorporate it into the equations.

At the point in time where the projectile just leaves the barrel the kinetic energy of the projectile equals the work done by the gun on the projectile.

KE = Work , and solve accordingly, but I'm sure some of you guys knew that already. Or solve the other way if your already know pressure.

@ mrbadexample
Thanks for the reference, and....
How can you dig yourself out of a crater?

[Ragnarok]

Then incorporate it into the equations.

That would be an ideal world situation, however there is little point in working out that force exerted - if you already have measured the muzzle energy, you don't then need to have the information to guess at it.

The practice as I talk about it, uses a typical internal pressure to generate a good estimate of the muzzle energy when the means to actually measure it are unavailable.

[Engineer87]

Where talking about the same thing. Because you can relate the internal pressure to Work and kinetic energy, and doing the way i described it is no less accurate then the way you described how to do it. It just all depends what you know and what your looking for.

[mrbadexample]

[quote="Ragnarok
Now, when you talk about generating 1400 ft-lbs from a 1.5" D x 38" L barrel, would this be from this combustion?[/quote]

actually it is derived from a simple ballistics calculation.
mass (in grains) x velocity(fps) x velocity(fps)/450,436=ft/lb there are 437.5 grains in a standard ounce. (i've been shooting real guns for about 30 years) i could probably tote out a shooting chrony, but i just use my computer and video editing software and shoot at a target at a known distance. the computer breaks down the video to thousandths of a second.

[Ragnarok]

actually it is derived from a simple ballistics calculation.

I'm afraid you've misunderstood me. Is the combustion I linked to the same cannon you are claiming produces 1400 ft-lbs?

If it is, I suggest getting out the shooting chrony and getting some figures from that rather than from your video software, as I have grave doubts about that cannon generating more than about a quarter of that energy.