I have a few questions on how to figure the recoil a gun produces. Well, actually, I have question about the formula.
It has been stated here and other places that the cartridge with the least amout of power produces less recoil even if the rest of the factors are the same.(gun/velovity/caliber ect).
From my schooling there are two laws of physics that I apply to this idea and each seem to disprove the idea listed previously. So, my question is, Why does the amount of powder that is used to "push" a projectile to the same velocity matter? To me it would seem that no matter how much powder used, if the projectile is pushed to the same velocity, then there can be no more positive "force" used to produce that velocity, therefore there can be no more negative "force" produced which can be felt be the shooter.
I am really looking for a good explination or example of this so I can present it back to my friends at school, because four of us have not been able to figure it out and all of us are curious. Even the formula would help, because I might be able to de-solve it and break it into a form that I can describe to the others.

GTR

Hmmm...Ok, I just thought of one variable which is the mass of the powder used, which would tend to lean towards the most amount of powder doing less recoil because its mass it heavier than the lighter load which would lessen the recoil a bit (I know we are splitting hairs talking 10 grains ect)

I believe that your instincts are correct. The "equal and opposite reaction" thing is in effect here. (If I were well schooled in physics, I could tell you who proclaimed that law)! But I'm not...

What you're basically looking at is pressure versus recoil. If you load a 30-06 hot enough to drive a 150 grain bullet to 2900 fps, you're going to get a specific amount of recoil.

If you load a .308 win hot enough to drive that same bullet to that same velocity, you're going to get less felt recoil.

The larger case of the 30-06 requires more powder to produce the pressure needed to make that bullet go that fast. (The '06 case is larger, hence the extra powder needed for equal pressure).

However, within the same cartridge, I don't believe that a fast powder will produce less recoil than a slow one, or vice versa. This may be the perception merely because the fast powder develops the pressure quicker, and may expend the full force of its recoil over a shorter time constant than the slower powder does, but the sum total of the energy dump into your shoulder won't be any different.

My thoughts, my opinions...

Dan Newberry
green 788

Because the powder weight has to be added to the projectile weight in order to obtain the total weight of the projected mass.

try using conservation of momentum

some of the powder/gas is propelled along with the bullet the entire length of the barrel. Some of it is not. I guess the equations have made a determination/assumption to account for this.

The total mass of the ejecta would be less with the lighter powder charge; therefore, it would have lower recoil.

George

also, the velocity of the gas is not the same as the velocity of the bullet, if you believe this reference

http://www.loadammo.com/Topics/August01.htm

The designer of the M107, M110 and M158 guns [and chief engineer of the vehicle design] told me he uses 1.5 times the velocity of the projectile for calculating the velocity of the gas to calculate recoil. The M110 has a 480,000 pound reaction.
I think he worked from the text "Hayes Elements of Ordinance".

Thanks for the responces, I see my missing variable. I had not taken into account the velocity of the powder charge and I had a memory lapse and was thinking that the powder charge was ignited just behing the projectile, which I see my error now. (I should of done better analysis on paper, I just could not see it in my head). Again, I appreciate the help and I have bookmarked the link, I believe I can clear it up for everyone around me now.

Thank you
GTR

wouldnt burning rates come into play ( although in a very small way it seems) like a fast burning powder like bullseye can pus a bullet to say 800 fps, is short order....but you would have to use more 2400 to push the bullet the same speed, and I thinks it takes a longer process to burn.....Bob

The same velocity but different in powder speed would be easier to feel in a revolver than a well tuned semi auto.

I am convinced that the hand can feel the individual pulses of recoil in a handgun. The semi auto averages the force over time as the recoil spring is compressed and the slide accelerated forward. This is less abrupt than the bullet acceleration felt in a revolver. But, if the recoil spring in the semi auto is inadequate, the slide can bounce off the frame faster than a bullet can accelerate, and a spike of acceleration goes into the hand.

I once worked up to 460 Rowland loads in a 20 ounce Patriot [now Cobra]. The recoil was so horrible, I developed a flinch. I would push my right shoulder forward with every shot. I made a 42 pound triple recoil spring assembly and the flinch went away.

GTR,
This is a restatement of several of the comments from my learned brethren. You must consider both the mass and velocity of the powder. Even thought the powder is converted from a solid to a gas, it has the same weight, and it is headed out the barrel. It's velocity is actually faster than that of the bullet. Therefore, in the overall equation, the mass times the velocity of the powder charge has its own contribution to the recoil. Ku-dude