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Barrel Length

Post questions and info about combustion (flammable vapor) powered cannons here. This includes discussion about fuels, ratios, ignition systems, safety, and anything else relevant.
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Barrel Length

Unread postAuthor: fastfox89 » Thu Mar 29, 2007 11:40 am

If I use a 4" piece of PVC that is 2 feet long what is a good length for the barrel.

thanks kris
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Unread postAuthor: keep_it_real » Thu Mar 29, 2007 12:02 pm

I would use a cb ratio of 2:1. so calculate the volume of the chamber.


h*pi*r^2 so in your case that would be 24"*pi*2^2. (I don't feel like finding my calculator). Now it depends on what size barrel you use.

x = barrel length
r = barrel radius
.5(chamber volume) = x*pi*r^2

Now just do some calculations.
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Unread postAuthor: fastfox89 » Thu Mar 29, 2007 12:05 pm

sorry i am using a 2" barrel
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Unread postAuthor: fastfox89 » Thu Mar 29, 2007 12:15 pm

that would make it like 8 ft correct :shock: or did i do the math wrong
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Unread postAuthor: keep_it_real » Thu Mar 29, 2007 12:23 pm

If thats what the calculations said. It sounds good.
technically, if you filled the chamber to 100 psi you could have a crazy cb ration of 1:7 and the projectile would still be accelerating (does not account for friction in the barrel so make it 1:6) but that would just be retarded.

Edit: that would be a 55 foot barrel hahahahaha.
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Unread postAuthor: Freefall » Thu Mar 29, 2007 12:26 pm

2:1? That's going to be a loud cannon.
I've seen recommendations from 0.8:1 for an efficient, quiet cannon to 1.5:1 for a loud cannon.

If you want a plug-in-the-numbers answer, here goes...

Chamber volume Vc:
Vc = pi*Rc^2*Lc = pi * 2^2*24 = 96 pi in^3

Barrel volume Vb:
Vb = pi*Rb^2*Lb ; Rb & Lb unknown

C:B ratio:
C:B = Vc/Vb
Just drop the ":1" portion.
a 1.5:1 ratio = 1.5
a 2:1 ratio = 2

So: choose a C:B ratio and barrel size, then solve for barrel length.
C:B = Vc/Vb
C:B = (96*pi)/(pi*Rb^2*Lb)
(pi terms cancel)
C:B = (96)/(Rb^2*Lb)

You want to solve for Lb:
- multiply both sides by Lb
C:B*Lb = (96*Lb)/(Rb^2*Lb)
Lb cancels on the right side
C:B*Lb = (96)/(Rb^2)

Divide both sides by C:B
(C:B*Lb)/C:B = (96)/(Rb^2*C:B)
C:B cancels on the left side, and there's your answer.

Lb = 96/(Rb^2*C:B)

So just pick a C:B ratio and barrel diameter, then plug in the numbers.
Example: let's say you have a barrel diameter of 1.5" and you want a 1.5:1 C:B ratio:

Rb = barrel diameter / 2 = .75
Lb = 96/(.75^2*1.5)
Lb = 113"

For the actual inner diameter of your chosen barrel size, look here:
http://www.harvel.com/pipepvc-sch40-80-dim.asp
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Unread postAuthor: fastfox89 » Thu Mar 29, 2007 12:44 pm

thanks for the info guys, i think i am just going to go with 8 ft, that way i can still move it around without having to take it apart
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Unread postAuthor: ArticWolf » Fri Mar 30, 2007 4:06 am

If you are going to use 4" pvc for your chamber at 24 inches long, than your barrel would be 2" pvc at 61 inches long, for a 1.5:1 C:B ratio which is good. This might work if you need to find out for youself check this out http://www.advancedspuds.com/SpudToolonline.htm This will also help you with propane metering too. I hope I don't get yelled at for this though LOL
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Unread postAuthor: Cannonus Potatus » Fri Mar 30, 2007 11:37 am

I agree with ArcticWolf, with a lot of trial and error I find that I like a
1.5:1 ratio. I think that a combustion cannon should be loud as all HELL! :twisted:
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Unread postAuthor: fastfox89 » Mon Apr 02, 2007 10:21 am

Cool thanks guys, I also feel they should be as loud as you can get them.
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Unread postAuthor: sandman » Mon Apr 02, 2007 11:16 am

is there such as a dangerous ratio?
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Unread postAuthor: Scope » Mon Apr 02, 2007 11:41 am

not if you use pressure rated parts and nothin crazy like use pure oxygen
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Unread postAuthor: SpudBlaster15 » Mon Apr 02, 2007 11:50 am

An extremely long barrel can be hazardous to any components housed in the chamber, such as the ignition electrodes and chamber fan. The vacuum created when the gases behind the projectile begin to cool can suck the projectile back into the chamber at high speed, with disastrous results.
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Unread postAuthor: paaiyan » Mon Apr 02, 2007 12:13 pm

fastfox89 wrote:Cool thanks guys, I also feel they should be as loud as you can get them.


The volume of a cannon is directly proportional to the value of the so-called "cool factor", but the efficiency of said cannon is inversely roportional. Ideally, once the projectile reaches the tip of the barrel, the pressure should be at its maximum point and should then immediately begin to contract due to cooling. If that was the case, there should be very little sound emanating from the cannon other than the whooshing of the object escaping and a small thud due to the inherent sound of an explosion.

The loud bang is created when the gasses continue to expand rapidly after the object leaves the barrel, creating a large pressure-related shockwave.

So we can represent this proerty of a spud cannon as follows, assuming that µ represents some constant that has yet to be calculated, v will represent volume in decibels, and e represents efficiency,

e = µ/v or v = µ/e.

The more volume you have, the lower the efficiency or vice versa.

Though really I just have nothing better to do at the moment than spew a bunch of nonsense, albeit true nonsense.

BANG! is directly proportional to mighty.

EDIT: By the way spudblaster, I love that quote of mine.
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Unread postAuthor: jimmy101 » Mon Apr 02, 2007 1:30 pm

paaiyan said: Ideally, once the projectile reaches the tip of the barrel, the pressure should be at its maximum point and should then immediately begin to contract due to cooling


Not really. The projectile will continue to accelerate after the pressure in the gun has peaked. The spud doesn't start to slow down until the pressure in the chamber drops below the net force retarding the movement of the spud; atmospheric pressure + dynamic friction.

Ballpark figures would be that the spud continues to accelerate until the chamber pressure drops below about 20 PSIA (5 PSIG).
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