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I have enjoyed reading the thread about recoil and shorter barrels. I wonder if it is possible to measure the "rocket effect". I am no expert, so there may be an obvious problem with my idea. Just take your favorite overbored rifle, pour your regular powder load into a case, plug the top of the case with whatever filler is used for big bore bullets, take the rifle to a safe place and pull the trigger. Whatever recoil you feel (or don't feel) would be the "rocket effect".
Am I oversimplifying the problem?
Hunt2
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I'll jump in and probably be wrong!
The configuration you suggest will not model the normal combustion process of typical gun powder due to the absence of the bullet. If I recall correctly, the powders we burn require a substantial amount of ambient pressure(pressure within the case-bullet system) to burn correctly.
So, if you don't have the bullet in front of the powder creating the necessary pressure, the powder doesn't burn the same and therefore the resulting volume of gas, maximum achieved pressure, and/or the pressure vs. time relationship is not real.
KwM
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KwM has it right.
In principle, what you want to do is cork the muzzle, pump the bore up to 10,000 PSI, and then pull the cork. That will give you a pure rocket effect.
The measurement I'm going to do will measure force vs. time, so we will be able to see recoil develops as the bullet moves, and finally exits. That's the least difficult to do approach that I've come up with. The last part of the curve that is generated will be just the rocket effect.
Be not weary in well doing.
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Campfire Tracker
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The gases don't leave the barrel until the bullet leaves, under 50,000 PSI or so of pressure.
The general assumption in calculating small arms recoil is that the propellant gases exit the muzzle at about 4,000 fps, and so that is used to figure the reactionary force of however many grains of powder were in the cartridge plus the weight of the projectile(s) at whatever speed they exited.
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Campfire Greenhorn
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There is one way to get a measurement of this effect. The precision might be a little limited, but it is simple. Suspend the rifle by two parallel strings from an overhead support so that the bore is horizontal. Fire the rifle and measure how high it swings in recoil. The momentum of the bullet is its mass times its velocity, and the rifle must have an equal and opposite momentum. Measure the mass of the rifle in the same units as the bullet, and get the rifle velocity that produces that same momentum. From v=sqrt(2gh) get the height of swing of the rifle produced by that initial recoil velocity. Presumably, the rifle will actually have swung higher than that. The difference is due to the additional effect of the powder gases.
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You could use a high speed camera and see if the recoil of the firearm accelerates after the projectile leaves the bore.
I don�t know if it will tell you a lot, but I'm just crazy for visuals aids. <img src="/ubbthreads/images/graemlins/crazy.gif" alt="" />
Last edited by Muddog; 02/08/06.
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Sounds like you are trying to find the part of recoil due to the exiting gas. Simplest emthod is to meaure the total momentum of the rifle and subtract the momentum of the bullet. The remainder is the momentum of the gas.
General rules of thumb. For a smokeless high power rifle say the gas has an average velocity of 4000 fps. For a handgun, say the gas has an average velocity of 1.5 times the velocity of the bullet.
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Volume, bore diameter and length will play a part in this. How much and how fast it can get out. The powder that is unburned when the bullet is gone will also play a part.
The way I picture it is. The cartridge is the motor that fills the barrel with pressurized gas (hot expanding air). Once the bullet leaves, the tube (barrel) acts as a governor. Or I could say the pressurized gas can move out faster in a .45 bore then a .224 bore. Thus a bigger differance in pressure from one end of the barrel to the other. Rocket Effect.
I'll have to come up with some sort of volume, bore, length, pressure squared. Thinggy.
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The posts suggesting using a ballistic pendulum, and subtracting the easily calculated and well understood parts from the total, to get rocket effect are, of course correct. You can do it exactly that way, and it will give you the total effect of the venting gas.
I think it is pretty well settled that the venting gas contributes about 30% of the total recoil momentum.
What we don't know is how quickly that occurs. Force, which is what your shoulder responds to, is d(mv)/dt, the first derivative of momentum. If the gas mostly vents in, say, 1/10 the time that the bullet is in the barrel, it may generate fairly large forces. That is what is yet to be discovered.
The measurement I am going to do puts a transducer behind the butt of the rifle, and measure force vs. time.... digital oscilloscope approach. When I'm done, we should be able to make some statement about how the peak force generated by the rocket effect compares with the portion that comes from accelerating the bullet. Right now, the data we have only gives a cumulative sum.
Be not weary in well doing.
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There is one way to get a measurement of this effect. The precision might be a little limited, but it is simple. Suspend the rifle by two parallel strings from an overhead support so that the bore is horizontal. Fire the rifle and measure how high it swings in recoil. The momentum of the bullet is its mass times its velocity, and the rifle must have an equal and opposite momentum. Measure the mass of the rifle in the same units as the bullet, and get the rifle velocity that produces that same momentum. From v=sqrt(2gh) get the height of swing of the rifle produced by that initial recoil velocity. Presumably, the rifle will actually have swung higher than that. The difference is due to the additional effect of the powder gases. +1. <img src="/ubbthreads/images/graemlins/laugh.gif" alt="" /> You would only be limited by your ability to get good measurement of the height and access to a chrono. Done right I think you'd have a answer in minutes. gmack
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Done right I think you'd have a answer in minutes. True, but it's the answer we already have. The many recoil calculators we have already tell us that part.
Be not weary in well doing.
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Homer Powley, the father of American ballistics, once calculated that the energy from firing a .308 was about 8000 foot pounds. Less than 3000 food pounds for the bullet and the rest "wasted" in accelerating the gases, recoil, etc. Some energy is used to spin the bullet but IIRC this was very small, about 1%.
Don't blame me. I voted for Trump.
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Campfire Kahuna
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Lee, Denton is more or less on track with the 10KPSI number. Pressure at the muzzle and chamber are vastly different, the range of the former running in the 5-17KPSI range, or maybe a bit higher for some of the cartridges such as the RUM family, or the STW's.
I am..........disturbed.
Concerning the difference between man and the jackass: some observers hold that there isn't any. But this wrongs the jackass. -Twain
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Less than 3000 food pounds for the bullet and the rest "wasted" in accelerating the gases, recoil, etc. Don't forget heat.
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Atmospheric pressure, Speed of sound, shock waves. Rocket Effect. I�m stuck in a loop. If I can�t turn it off it�s going to be a long time before I�ll get a goodnights sleep.
Have you ever needed a eureka moment? <img src="/ubbthreads/images/graemlins/help.gif" alt="" />
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I think Homer was probably about right.
Friction isn't THAT big a deal. Friction on a bullet is ROUGHLY 150 pounds. Using the work-energy theorem, friction robs about 300 foot pounds over a 2 foot slide.
The energy contained in the remaining gas is tricky. It matters a lot how you take the energy out. If you do it very slowly, so you don't have to account for the temperature change of a decompressing gas, you can account for another 3,400 foot pounds of waste energy as the gas vents (.308 bore).
Heat is important. 108 foot pounds of energy is about enough to raise one pound of iron .7 degrees C. So, raising 6 pounds of iron 2 degrees C costs 1,850 foot pounds.
Just quick and dirty, I get 300 + 3,400 + 1,850 = 5,550 foot pounds of lost energy that I can account for. That, plus maybe 2,500 carried by the bullet is very close to Homer's 8,000 foot pounds of total input energy. Recoil energy of the rifle is too small to worry about-- ~25 foot pounds.
Be not weary in well doing.
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Campfire Greenhorn
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denton, I am looking forward to your test results. Throughout this whole discussion about recoil, rocket effect, short barrels, etc., I have been wondering why someone with a strain gauge doesn't just MEASURE it. I would if I had the $. Thanks for doing this!
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Back to the subject of short barrels, the recoil from a short-barreled rifle with any drop to the stock will usually have more perceived recoil due to the muzzle rising faster and higher than a longer-barreled rifle. The primary reason is that the longer barrel has more weight farther away from the center of rotation at the buttplate contact with the shoulder. The force of recoil has several components, one of which is the angular momentum of this rifle rotation (muzzle rise).
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Interesting discussion. I own a 358 Norma, and it is my impression that it kicks less than a 338 Win with the same bullet weight and velocity. Who knows??
Anybody who seriously concerns themselves with the adequacy of a Big 7mm for anything we hunt here short of brown bear, is a dufus. They are mostly making shidt up. Crunch! Nite-nite!
Stolen from an erudite CF member.
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Denton; I think you are on the track to find out why some calibers have a "sharper" kick. A force/ time trace would be really interesting. Take Care!
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