Help Me Understand FPE (And Why My Shoulder Doesn't Blow Off)

[Mouse hunter]

Admittedly this is a total newbie question involving bad math. I know I am wrong about something here, I just want to be shown where I am wrong.

My understanding of FPE is that it is the measurement of the energy needed to move a one lb object a distance of one foot. My 30-06 gets to right around 3,000 FPE depending on ammo.

So, by my way of thinking, that bullet coming out the muzzle ought to be able to hit a one pound object and (assuming it expends all its energy into the object) it ought to be able to propel that object 3,000 feet. I could be wrong, but I don't think it could possibly throw a 1 lb weight even 100 feet, and certainly not 3,000 feet. The alternative is that it is enough energy to move a 3,000 lb object one foot. Again, I don't think my 30.06 is capable of that, not even close.

Likewise, since there is an equal and opposite reaction, my rifle ought to recoil with 3,000 FPE. My rifle and I combined weigh 200 lbs exactly, so 3,000 FPE divided by 200 lbs should mean that me and the rifle fly 15 feet backwards, or, alternatively, my shoulder just rips right off.

Basically, my understanding of FPE does not jive with reality at all. In reality I feel a solid recoil and then work the bolt and fire again, no bruise, no flying backwards, no dislocated shoulder.

So can someone please explain how FPE works? Otherwise I am left in a Hollywood world in which .45 pistols pick people off their feet and propel them through walls. I don't want to live in that world.

Thanks!

 

[rob_benj]

FPE is not the measurement as you described, but rather the amount of force exerted on a one foot square area (144 sq in). Your .30-06 having 3000 FPE only has a cross section of 0.0745 sq in (area of .308" circle) and is moving a projectile of 160gn weight. Your rifle impacts your shoulder in an area of 15 sq in and has a weight of 49000 gn (estimated 7lb rifle).

So, a 3000 lb force hitting a 144 sq in target is the equivalent of a 312 lb force hitting a 15 sq in target (your shoulder). The actual effect of this however is altered by the rate that the force is traveling (accelerating). The rifle in the example above is 306 times heavier than the projectile so it should be safe to assume that the rifle will be accelerating toward you at 1/306th the rate. The perceived recoil will be much less than the 312 lb and will probably be closer to 30 lb.

I know I am missing something in this as well and perhaps someone else can explain it better than I.

 

[denton]

One foot-pound of energy is what is required to lift a one pound object one foot, subject to gravity at the Earth's surface. (The object then has an additional one foot-pound of potential energy.)

Do not confuse energy with force. Force, momentum, and energy are all different things which are intimately connected mathematically.

When a bullet strikes an object, most of the energy goes into shredding the object. It's what's called an inelastic collision. Momentum is conserved, but kinetic energy definitely is not. If the object traps the bullet (no exit) then the total momentum of the bullet + object = the momentum of the bullet before impact. If you work through the math, this says that a monster elk does not move much under the influence of the bullet.

Kinetic energy does not split evenly between the bullet and the rifle... not even close.

During discharge, a 30-06 rifle will experience up to 500 g of acceleration for about a millisecond. The force you feel on your shoulder is the rate of change of the rifle's momentum as your shoulder and body brings it to rest. Bring it to rest slowly by inserting a recoil pad, and the force is much less. Bring it to rest quickly with an steel butt plate and limited shoulder motion, and the force is much greater.

 

[Mouse hunter]

Great answers, both of you. This is very helpful and informative. Now I know why I still have my shoulder, and why my deer always fall down instead of flying way up in the air and landing on other side of the woods.

I am one of those guys who really likes physics, but knows only enough about it to be dangerous.

 

[230RN]

Hatcher's Notebook has an excellent and authoritative discussion of recoil and "kick" in his theory of recoil chapters.

Denton is 100% right in that the terms used in informal discussion are sometimes mixed up in peoples' minds. As examples, weight and mass are commonly confounded, as is momentum and energy, resulting in confusion and disagreements.

http://www.info.com/search?qcat=web&qkw=hatcher's+notebook

While it is out of date (late 1940s) for many things like the latest and greatest cartridges and firearms, having a copy of this book (it's also available on-line, I am told) would provide answers to a large majority of questions asked on firearms forums.

If you decide to buy a copy, make sure you spring for the hardbound one because you will refer to it very frequently. I've gone through two copies and I'm almost ready to get my third.

Terry, 230RN

 

[Arizona_Mike]

Force and amount of substance need to be distinguished. There have been various pedagogical methodologies over the years. The current is to use "weight" and "mass", a previous scheme was to use pound-force and pound. At one time we used slugs.

Because of it's pedagogical use for a couple generations, the weight/mass distinction have escaped into the common vernacular as a form of ahistoric pedantic invention. People who don't use it are assumed to be wrong even if they are making the distinction between amount of substance and force.

On to the original question, momentum is what is conserved when objects collide. Two billiard balls of equal weight will exchange energy but only because the weight is the same. If the cue ball were heavier it would continue forward regardless of the english.

Ignoring the momentum of the powder (which is significant--can add another 20-30% or so) let's look at the momentum of a 180gr .30-06 projectile at 2700 fps. You have a momentum of 486,00 gr-ft/sec. Ignoring the human holding the gun, a 9lb (63,000 gr) gun resting on a frictionless slide will recoil 7.7 fps.

Mike

 

[Sunray]

"...why my deer always fall down instead of..." If Bambi did fly off, you would too. That Newton fellow passed a law about it. There are multiple parts of a shot that absorb or help absorb recoil. Your shoulder is one of 'em.
Buy a copy of Hatcher's Notebook anyway. It should be required reading for anybody who gets into the shooting sports. Good read too. Runs about $30 at your local gun shop or Amazon.

 

[Panzerschwein]

And the reason why you can fire a 5 pound projectile at 1000 FPS+ off the shoulder from a recoiless rifle is because the rounds has a charge in the back that blows out in order to cancel out the rearward recoil of the projectile.

 

[45_auto]

It's all basic physics. English units can be confusing. You're thinking of the ft-lb energy number from your rifle as WORK, which is Force x Distance. But it's not Work, it's Kinetic Energy. Energy is NOT the same thing as Work. First thing you have to do is understand the difference between Work, Energy, and Power. They're not the same thing.

http://hyperphysics.phy-astr.gsu.edu/hbase/work.html

Energy is the capacity for doing work.

Work refers to an activity involving a force and movement in the directon of the force.

Power is the rate of doing work or the rate of using energy, which are numerically the same.

To put the 3,000 FPE of your .30-06 into perspective, the AA batteries in your TV remote have about 7,000 FPE each. The D cell batteries in your flashlight have about 55,000 FPE each.

A bullet impact is called an "inelastic" collision because all of the energy is NOT transferred into the target. Some goes into heat, some goes into deforming the bullet, etc. Kinetic energy is NOT conserved during the collision.

You can see how much energy is dissipated into other forms by using a ballistic pendulum, like this web site example:

http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

Although kinetic energy is NOT conserved, momentum (which is what you're feeling as recoil) IS conserved, which is why a ballistic pendulum works. A perfectly elastic collision (where energy IS conserved) involving 3,000 FPE would raise a 3000 pound weight 1 foot. That would be a direct conversion of Kinetic Energy into Potential Energy. But that ain't gonna happen with your bullet.

Plugging the values for a 180 grain bullet (11.7 grams) at 3000 FPS (2045 MPH) into a ballistic pendulum calculator we can see that a 180 grain bullet at 3000 FPS would raise the 3000 pound weight (1,360,780 grams) by a height of .0000103 feet, or .00012 inches. That's just over 1/10 of 1/1000 of an inch. A human hair is about .003", or 3/1000 of an inch, so you're raising the weight about 1/30 the thickness of a hair.

The same impact would raise a 200 pound weight .028 inches. About 1/3 of 1/16 of an inch. Should be obvious why people don't fly through walls as shown in Hollywood!

I'll try to go over recoil later if no one else does.

 

[denton]

Launching a bullet out of a rifle is essentially an inelastic collision in reverse. Momentum before the event = momentum after.

We use this as our recoil calculation because it is convenient.

Unfortunately, it does not give a very useful answer. First, a rifle fired from the shoulder soundly violates the assumptions of the calculation, which require a freely recoiling rifle. Second, the calculations produce a summation of a property you can't perceive. What matters--what you can sense--is force (or more properly, pressure, which is force/area), and how it varies over time.

The standard recoil calculation is based on momentum. What matters at your shoulder is not momentum itself, but how fast the rifle is shedding the momentum it gains as the bullet discharges. That's force.

 

[45_auto]

OK, let's try recoil.

Likewise, since there is an equal and opposite reaction, my rifle ought to recoil with 3,000 FPE.

You are correct about the equal and opposite reactions part, although that applies to forces, NOT Kinetic energy as you stated. The FPE that you are referring to is the KINETIC ENERGY of the bullet. You're not going to understand it at a basic level until you understand the difference between FORCE and ENERGY.

To use equal and opposite reactions you'll have to calculate the FORCE which is pushing the bullet down the barrel, which is equal to the FORCE pushing the rifle backwards. Once you know the force on the rifle, you can calculate it's acceleration and velocity. And if you have the velocity and mass of the rifle, you can calculate it's recoil FPE, which is NOT going to be anywhere near the bullet's FPE.

As I said earlier, a AA battery contains 7,000 FPE, just as your recoiling rifle on your shoulder contains 3,000 FPE. Do you think that if you put an AA battery up to your shoulder that you should recoil with 7,000 FPE just because it contains that much energy?

I'll go over the recoil math, but you're going to have to be pretty bored to follow it! Most people won't like it!

First, check out this table of recoil energy. Note that a .30-06 firing a 180 grain bullet at 2700 FPS only has 20 FPE of recoil energy, although if you do the calculations for the bullet it has 2900 FPE. Why is that?

http://www.chuckhawks.com/recoil_table.htm

Hopefully you know that Muzzle Energy (Kinetic Energy) of a bullet is Mass times Velocity Squared divided by 2. Sometimes written like this:

KE(bullet) = M(bullet) * V^2 / 2

The little "^" sign means raised to that power, squared in this case.

Likewise, KE(rifle) = M(rifle) * V^2 / 2

Ok, it's easy to find the mass of the rifle, but we also need the recoil velocity of the rifle.

Using Conservation of Momentum would be much simpler, but we'll go all the way down to the basic F = MA (Force = Mass * Acceleration) level.

Weight of bullet = 180 grains.

Since Force = Mass * Acceleration, we need the mass of the bullet. We'll use English units, and the English unit of mass is the "slug". Weight(lbs) = Force = Mass * Acceleration. On Earth, acceleration due to gravity = 32.2 ft/s^2

Knowing that good stuff we can figure out that Mass(slugs) = Weight(lbs) / Acceleration, so Mass(bullet) = (180 grains / 7000 grains/lb ) / 32.2 ft/s^2 = .0008 slugs.

Now we know the Mass(bullet), we need the Acceleration of the bullet so we can use F = M * A to calculate the Force on the bullet.

We're going to say that we have an 18" barrel on our rifle. We know that the bullet starts at a Velocity of 0, and accelerates to a velocity of 2700 ft/s in 18". So we can calculate the acceleration required to make it go from 0 to 2700 ft/s in 18".

The basic position equation is:

X = X(initial) + ((Velocity(initial)) * time)) + ((acceleration * time^2) / 2)

Usually written: X = X(i) + V(i) * T + (1/2) * AT^2 (This will be Equation 1)

The first derivative of position with respect to time is velocity, and second derivative of position (therefore the first derivative of velocity) is acceleration. I'm not going to try to teach you calculus, either take my word for it or stop reading now!

What this means is:

V = AT (This will be Equation 2)

In other words, Velocity = Acceleration * Time.

We currently have two unknowns for our bullet, Acceleration and Time. But the good thing is, we have two independent equations we can apply to our bullet. If you remember your basic algebra, you can solve for any number of unknowns if you have the same number of independent equations.

So, Equation 1:

X = X(i) + V(i) * T + AT^2 / 2

We're going to say that the zero point (that's X(initial), or X(i)) for our bullet is it sitting in the chamber. The end point "X" will be the end of the barrel. So X(i) = 0, and X = 18". We need to use the same units, so since we're using ft/sec we need to use feet for the barrel. 18" = 1.5 feet.

We also know that the initial velocity of our bullet is zero, that means that V(i) = 0.

So we can write:

X = X(i) + V(i) * T + AT^2 / 2 (which is Equation 1)

or

1.5 = 0 + 0 * T + AT^2 / 2 (which is still Equation 1)

which, if you remember algebra, is:

1.5 = AT^2/2 (which is still Equation 1)

Now we've got one equation and two unknowns. We need another equation, we can use Equation 2:

V = AT (Equation 2)

We know the velocity of our bullet at the end of the 1.5' barrel, it's 2700 ft/sec. That means Equation 2 is:

2700 = AT (Equation 2)

or

A = 2700 / T (still Equation 2, both sides divided by T)

Now we have two equations and two unknowns. We solve for A and T.

We'll plug our value for A from Equation 2 into Equation 1

1.5 = AT^2/2 (Equation 1)

1.5 = (2700 / T) * T^2 / 2 which reduces to:

3 = 2700 * T

T = 3/2700 = .00111 seconds

So our bullet goes from 0 to 2700 ft/sec in 18 inches over a time of .00111 seconds.

We can now plug this value for T back into Equation 2 to find the Acceleration:

V = AT (Equation 2)

2700 = A * .00111

A = 2700 / .00111 = 2,432,432 ft/sec^2

As a side note, if we want to know how many G's it's pulling, that would be

=2,432,432 / 32.2 = 75,541 G's

If you're designing some fancy electronics (self guiding, maybe?) for a .30-06 bullet, you better make them capable of withstanding a force of over 75,000 times their weight!

Back to recoil:

Now we know the bullet's acceleration, so we can finally calculate the average force over the length of the barrel on it:

F = MA

F = .0008 slugs * 2,432,432 ft/s^2 = 1,946 pounds

Finally! The Force on the bullet is 1,946 pounds.

As a side note, this is the same way you calculate the rocket burn if you're trying to rendezvous with the space station. Slightly different numbers but it works the same way. Figure out the difference in velocity you're trying to compensate for, the thrust(force) of your motor, and weight of your rocketship. A motor exerting 1,946 pounds of force on the base of a 180 grain rocketship for .00111 seconds will accelerate it to 2700 FPS in 18 inches.

Anyway, since we know about equal and opposite reactions, we know that this means that the Force on the rifle from the bullet also equals 1,946 pounds.

But to balance the Forces, there's also an additional Force shoving the rifle back. There's also the mass of the powder times the acceleration of the burning powder that acts on the gun.

Velocity of the gas is usually taken to be 1.5 times the velocity of the bullet. If we say that we used 50 grains of powder in our .30-06, we can figure out the Force due to the acceleration of the burning powder just like we did for the bullet. You can do it yourself, or just accept my calculations that 50 grains accelerated to 4050 ft/s in 18 inches requires a Force of 800 pounds.

This means that the total Force on the rifle is 1946 + 800 = 2746 pounds

2,746 pounds shoving on my .30-06 you say! I bet you figure I'm pretty stupid to claim that. But if you've made it this far, keep going another few minutes and you'll see!

Now we can calculate the Acceleration of the rifle.

We know that Force = Mass * Acceleration, and we have Force acting on the rifle. It's easy to get Mass, same way we got the bullet mass. We'll say that we have an 8 pound rifle. We need rifle mass in slugs which is:

M = F(weight) / A(gravity) = 8 / 32.2 = .25 slugs

Now it's easy to find the rifle's Acceleration:

F = M * A

2746 = .25 * A

A = 2746 / .25 = 10,984 ft/sec^2

This means that our rifle accelerates into our shoulder at 10,984 ft/sec^2 for the .00111 seconds that the bullet is in the barrel.

Remember Equation 2, the Velocity Equation? Now we can figure out the recoil velocity of our rifle:

V = A * T

V = 10,984 * .00111 = 12 ft/sec

So our rifle is recoiling at 12 FPS when the bullet leaves the barrel at 2700 FPS.

Look at the recoil velocity of the 8 pound .30-06 in the Recoil Table. Notice how they show 12.8 FPS. The difference between the 12.0 FPS we calculated and their 12.8 FPS is due to round-off and/or assuming a different powder charge.

http://www.chuckhawks.com/recoil_table.htm

Now that we know the recoil velocity of the gun, it's easy to compute the recoil energy:

E(gun) = (Mass(gun) * Velocity(gun)^2) / 2

E(gun) = .25 * 12^2 / 2 = 18 ft-lb of energy.

The recoil table shows 20.3 FPE, due to the fact that our V(gun) was 12 FPS and their V(gun) was 12.8 FPS.

Bottom line is your 180 grain bullet at 2700 FPS has 2900 FPE, and your 8 pound gun at 12 FPS recoil velocity has 18 FPE.

Told you that you wouldn't like it!

All this stuff is over 400 years old. Gun materials, strengths, and thicknesses as well as rocket trajectories, structural loads, power and fuel requirements, burn times, etc, are all based on Newton's Laws from 1687.

 

[45_auto]

Note: If you REALLY want to be precise about this stuff you would need to integrate the powder's pressure curve over time to get the force and acceleration of the bullet. I used the average acceleration thus an average force of 1,946 pounds on the bullet over the entire length of the 18" barrel.

Figuring 60,000 PSI SAAMI max pressure for a .30-06 and a .308 diameter bullet gives a max force on the base of the bullet of 60,000 * (.308^2*PI)/4 = 4,470 pounds.

Obviously it takes a given amount of time for the powder to build to this force, then it falls off as the bullet travels down the barrel. There's an example of an M16 pressure curve here:

http://www.frfrogspad.com/intballi.htm

Note that the bullet exits the barrel in 1.1 milliseconds, which is .0011 seconds, which happens to be the same thing we calculated in the example above. Always good to get some confirmation that your calculations are at least in the ballpark!

 

[Jim Watson]

One correction.
Energy is fully conserved, just not as kinetic energy. Most of it ends up as heat energy by various routes of friction, bullet deformation (Ever pick up a bullet that had JUST flattened against a hard target?), and "shredded" tissue relaxing into final form.

 

[Nom de Forum]

To put the 3,000 FPE of your .30-06 into perspective, the AA batteries in your TV remote have about 7,000 FPE each. [/B]

Now you've done it! It will only be a matter of time before someone reloads .50" diameter AA batteries into their .50 caliber BP rifle attempting to get an instant increase of 7,000 FPE.

 

[45_auto]

Energy is fully conserved, just not as kinetic energy.

True, changed it to say "Kinetic energy" instead of just energy.

 

[230RN]

^ Thank you ! Bless your heart for being less lazy than I.

I wonder if your explanations could be combined into one post and used as a sticky somewhere.

TNX,

Terry, 230RN

 

[Arizona_Mike]

Thank you 45 Auto for taking the time you did.

Mike

 

[Mouse hunter]

Thanks so much everyone. These were some very thorough answers, and yes, 45 Auto, I did read the whole thing. Its a bit over my head, but I followed it and it makes sense.

Thanks to everyone

 

[Chemistry Guy]

This has already been covered, but I'll add a bit more. Because kinetic energy includes a velocity-squared term and momentum only contains a velocity to the first power term, the bullet will contain more kinetic energy than the recoiling body when momentum is conserved. The greater the ratio of the difference in mass between the bullet and the recoiling body, the greater the asymmetry in how much KE is imparted into each.

Practically, this means that the tighter that your rifle is kept to your shoulder, or the more that you and your rifle recoil as a single mass, the less kinetic energy will be imparted. If you were to loosely hold the stock of a powerful rifle that fires a heavy bullet a few inches from your body and pull the trigger, your shoulder would take quite a bruising because the rifle would recoil like a much lighter object than when it is firmly grasped.

 

[jmr40]

Short answer, your bullets may impact with roughly 3000 ft. lbs of energy at the muzzle, but as the bullet slows that energy is much less. At 200 yards it will be down to roughly 2000 ft lbs.

Your 30-06 will recoil with somewhere between 15-20 ft lbs of energy depending on the exact load and the weight of the rifle.

The bullet is moving at 2900-3000 fps, your much heavier rifle will only be coming back at around 14-15 fps.

All the math above is probably right, but gives me a headache. I don't attempt to understand exactly why, I just plug the numbers into a program and let them do the math for me.

 

[Jim K]

One interesting effect of energy is that when a bullet moving down a barrel hits an obstruction (like a stuck bullet), there is a massive and instant heat dump as the kinetic energy of the bullet is converted to heat. It is that heat that softens the barrel steel and allows pressure to bulge or split it. The heat is generated and dissipated so fast that usually the bluing will not be affected!

The same heat dump causes the melting and splashing effect when a bullet strikes a steel plate. It is also why armor-piercing bullets work; the energy converted to heat when the bullet strikes melts the steel and the AP core goes through.

Jim