Thread: shooting high
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Old March 8, 2000, 01:01 PM   #8
Join Date: January 26, 2000
Posts: 57
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Herodotus:
Won't speculate on all the rest, but recoil does indeed start from the time the primer ignites. Technically, I guess, it starts from the time the firing pin starts to move forward. You cannot drive the bullet down the barrel without a counter reaction.[/quote]

What you are referring to is considered by some as "primary recoil". This is a theory shared by very few.

As for the momentum of the hammer/firing pin moving forward, the energy required to complete this process is "stored" in the compressed state of the spring.

As for the force required to drive the bullet down the barrel, it is different depending on whether you are firing a revolver or a closed breech gun.

The gasses generated by a modern load have a tendency to travel at 17,000 fps. As we know, there is no modern cartridge capable of delivering this velocity. However, the pressure is still there, and it is expanded and expended in all directions. The bullet is forced out of the barrel simply because it is a movable object, and as such allows the gasses to flow in that direction (as long as all other factors are not compromised). On a revolver the cylinder gap releases part of that pressure in a manner which vents equally in all directions axially to the bore of the cylinder. (Look at the front of the cylinder, the throat of the barrel, and especially the bottom of the topstrap (where severe gas-cutting can be seen). On semi-auto and closed breech guns this escape is not possible. Therefore all products of combustion must exit the muzzle. This is where "secondary recoil" comes into play.

If you were to take a barrel which had a bore large enough to contain all the gasses, and the projectile, without venting to the outside world, and in which the bullet would come to rest 1" proximal from the muzzle. There would be no recoil.

Now, chop 2 feet off the end of that barrel. Under the same conditions fire another round. As the projectile and the gasses exit the bore you will notice movement of the barrel. This movement is "recoil".

As I stated previously, "((weight of the bullet in pounds x the velocity of the bullet in feet per second)+(4,700 x the weight of the powder in pounds))squared / (64.348 x the weight of the gun in pounds)".

Until the bullet actually leaves the barrel it "technically" has no weight of its own. Nor does the weight of the powder. The weight of the bullet and powder, as long as it is in the barrel is "technically" part of the weight of the gun. Therefore, with a bullet/powder weight of "0", the formula shows zero recoil. When the bullet and gasses exit the barrel they become separate entities and their values can be inserted into the formula.

For a practical demo of this, get yourself a long piece of pipe, some fuse, and a few tennis balls. Lay the pipe on the ground, place a very small charge of powder in the pipe, and lodge the tennis ball agains the charge. (Make sure the charge is not strong enough to expell the ball from the pipe.) Light the fuse and watch the pipe. If the tennis ball does not exit the pipe, the pipe will not move. Next, increase the charge enough to just barely force the ball to exit the pipe. You may find that the pipe will show some movement. Next increast the charge enough to move the tennis ball at a sufficient velocity from the pipe, and you will definately see movement of the pipe.

As you increase the amount of powder, or increase the number of tennis balls placed over the powder, you will see increased movement of the pipe. If you use too much powder, or place too many balls in the pipe, you will see catastrophic results. The pipe will burst.

This is not intended to be a "flame", but rather as evidence of proof of my earlier post, which some have taken wrongly.

(And I prefer two cups of coffee in the morning, not milk; and since I have a round bed, I'm not really sure "which side" I got out on...grin)
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