Interesting: Recoil due to powder weight.

[Macchina]

Don't want to beat this to death, but I think it's important, and interesting to a lot of folks. I posted above (#18), but was a little vague. Here is the formula for calculating free recoil energy. The k here is a constant applied to the velocity of the powder gasses. Different constants are used because of the different dynamics of firearms. Shotguns with a long barrel use a low of 1.25 due to the relatively large bore, small powder charge, and low velocity. HP rifles use a high of 1.75, again relatively small bore, large charge and high vel. The reason it's not an 'exact' is that the powder gas column is expanding and the front of that expanding column will be different than the rear.

R.E. = [0.5 / M(gun)] * [M(bullet)*V(bullet) + M(powder)*k*V(bullet)]^2

Of course, you need to keep all the units in check; lbs, ft., sec.

Also don't forget to take the acceleration due to gravity out of the weight (pounds) of an object to get its mass (slugs). Acceleration due to gravity is obviously 32.2 ft/s^2.

I.E. A 180 grain bullet weights 0.0257 pounds (weight). The mass of the object is 0.0257/32.2 = 0.000798 slugs (mass) for use in the equations above. By squaring the velocities in the equations above, you are converting the slugs back into pounds.

Note: Pounds Mass (Lbm) are not to be confused with Pounds Force (Lbf)... The beauty of the English System...

 

[jstein650]

Quote:
Originally Posted by Haxby
The online recoil calculators work well to calculate recoil.
And if you check you will see that powder weight is in the equation used to calculate recoil.
__________________
In God we trust, all others pay cash!



Of course, and little else! No mention of chamber pressure, time in barrel, length of barrel, powder burn rate... or anything else.

The reason I wanted to avoid any automatic on-line calculators, was to get to the heart of the matter as to what we're really looking at in the world of real world physics.

 

[jstein650]

"Also don't forget to take the acceleration due to gravity out of the weight (pounds) of an object to get its mass (slugs). Acceleration due to gravity is obviously 32.2 ft/s^2.

I.E. A 180 grain bullet weights 0.0257 pounds (weight). The mass of the object is 0.0257/32.2 = 0.000798 slugs (mass) for use in the equations above. By squaring the velocities in the equations above, you are converting the slugs back into pounds."



michaelmcgo:

That kind of pertains to relative ballistics in 1g, doesn't it? More of a trajectory thing... Now there I will use a computer. I'll pass on an enormous multi point by hand calculus, thank you.

 

[1911Tuner]

re:

The momentum imparted to the firearm; M=vm , which is why my (Lyman's) formula neatly takes into account the momentum of the mass of the firearm, as well the ejecta, based upon the VELOCITY of the ejecta

Yes. Momentum is a factor of mass and velocity, but momentum isn't recoil by the Newtonian definition. What we perceive as recoil is mostly momentum. The actual recoil...by Newton's definition...is over so quickly that our brains don't have time to process it until after it's over.

Again:

Recoil is the reaction side of an action/reaction system and it's only in play while the system is being accelerated. Once the accelerating force has been removed from the system...action AND reaction end, and any motion of the components are due to conserved momentum.

Analogy:

The boxer's sharp left jab snaps his opponent's head back, but the actual punch only exists while his glove is in contact with his opponent's face. When the glove loses contact, the punch ends.

In an alternate universe, where the known laws of physics don't exist and momentum isn't conserved...motion would stop the instant that an accelerating force is removed.

 

[brickeyee]

You use the momentum of the ejects to compute the momentum imparted to the gun.

Momentum MUST be conserved.

You then use the momentum of the gun to determine a velocity of recoil.
(dived by the mass of the gun, watch out for units).

The velocity of recoil is used with E=(mv^2)/2 to compute a free recoil energy of the gun.

There are more complicated ways of doing all this based on impulse that take into account how fast the acceleration occurs.

They are rarely needed since FELT recoil is dominated by other things anyway (stock shape, etc.).

Get a rifle stock cut at the wrong angle and get smacked in the face with every shot and the felt recoil is worse.

 

[Njal Thorgeirsson]

The recoil impulse would naturally be less due to ifferent force requirements to achieve the rate of acceleration.

I disagree with this statement (I'm 99% sure this is false).

Here is why:

Things we can establish:
-The force felt by the bullet is at any given time the same as the force felt by the back of the case and subsequenty, the bolt face and gun.
-The Force vs. Time (Pressure vs. Time) profiles (of forces felt by the bolt gun and bullet) WILL differ between the two loads. HOWEVER, the integral of these graphs (which is momentum) will be identical, which should make sense, as both bullets will have the same momentum upon exiting the gun.
-Now, as there will be a different F vs. T profile for each gun, the recoil impulse will differ for any given increment of time during which and only during which the bullet is in the barrel. Once the bullet leaves the barrel, the total change in momentum (net impulse) will be the same for both guns. Both guns will have identical velocities into your hand as a result.

Yes, the impulse will differ during times in which the bullet is actually in the barrel, but the net change in gun velocity will be the same.

Unless you can feel slight variations in the gun's acceleration during the tiny fraction of a second that the bullet is accelerating down the barrel (which you can't), the recoil will feel identical, because a) both guns will accelerate to the same velocity in the same amount of time, b) suring that amount of time, average acceleration of both guns will be the same, and c) the increment of time during which the acceleration profiles of both guns differ is imperceptibly small.

But of course the one with the heavier powder load will be slightly heavier in recoil due to more mass being pushed out of the gun, but that has nothing to do with the above explanation.