Foot lbs for dummies...

B

Biff Tannen

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Socrates said that the man who admits he knows nothing is wise because he opens himself up to learning what he does not know...
With this said...
Can someone please explain to me, in the simplest terms, what foot lbs are and how they affect ballistics?
How Is the power distributed on the numerical scale? For example, is 400 foot lbs necessarily twice as powerful as 200 foot lbs, or is there some other ratio that comes into play?
Does a 9mm bullet with 1000 foot lbs and a .45 bullet with 1000 foot lbs do equal damage, even though the .45 is larger?
Thank you!!!
 
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Foot pounds are a measure of the kinetic energy of an object, in this case a bullet. "Kinetic" refers to motion, and "energy" in physics means the capacity to do work, which often means moving something, so kinetic energy is the ability of a bullet to move something else due to its (the bullet's) motion. The more ft-lbs a bullet has, the more energy it has. One ft-lb translates to the amount of kinetic energy it takes to move a 1-lb object one foot. As you know, rifle bullets typically have kinetic energy of a few thousand ft-lbs while handgun bullets are typically in the several hundred ft-lb range (with lots of variability in both cases for different calibers).

One thing the kinetic energy of the bullet moves is the firearm and shooter, producing what we call recoil. Downrange, it expends its kinetic energy, or much of it, in the target and/or backstop. In theory, and in a perfect world, a bullet that has 2,000 ft-lbs of energy could move a 2,000-lb moose (which I guess would be a very large moose indeed) one foot when it strikes it. However, the energy is used up in several other ways (generating heat, fragmentation, penetration, etc.) so actually very little of it remains to move the target, which is why "knockdown power" in the sense of physically knocking the target down is a myth and why animals are not actually knocked over by the impact of a bullet.

The general formula for calculating kinetic energy is mass (weight) of the bullet times the square of its velocity - it's the squaring of the velocity term that gives small but very fast bullets their high kinetic energy.

Kinetic energy in itself doesn't affect the bullet's ballistics in the sense of the path it takes from the muzzle to target. That's a function of velocity and ballistic coefficient, which is a measure of the relative ability of the bullet to slip through the air, along with a few other factors.

ETA: Sorry, forgot to respond to your two questions. In a theoretical sense, kinetic energy as measured in ft-lbs is linear, which means that a bullet (any object) with 400 ft-lbs of kinetic energy has the ability to do twice as much work as one with 200 ft-lbs. However, in the real world, and particularly in determining bullet performance, kinetic energy is only one of many factors and not necessarily the most important. That's pretty much the answer to your other question as well, although there's such a difference between 1000 and 100 ft-lbs of energy that I think I'd bet on the 9mm bullet in that case, but again many other factors (bullet size, shape, and construction, for example) come into play.
 
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Here's a much more simplified explanation 'cause I'm a simple guy. :D

Drop a 1 oz steel ball bearing on your foot from a height of 4 feet. Did that hurt? Probably not too much, eh?

Drop a bowling ball on your foot from the same height. Did you have to go to the hospital?

Not taking into account air resistance, both objects reached the same speed and reach its target at the same time, except one crushed your foot, while the other just made an ouchie. :D

Same thing with a 125gr 9mm -vs- 230gr 45ACP. If the 45ACP were 125gr, it would have little difference in ft lbs. Now if you're talking about displacement/tissue damage that's a different thing. A hollow out .45ACP will cause more damage than a 9mm of the same weight given the same ft-lb.
 
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Onward Allusion's example reminded me that I should mention that ft-lbs are also a measure of momentum, which is a quantity related to kinetic energy but not the same thing. In the case of momentum, the velocity term is not squared so differences in mass are much more important. Both kinetic energy and momentum are important in evaluating bullet performance.
 
Just a few errors to point out:

First of all there is an important difference between a force and a mass. In English units, this is very confusing, as the term lb (pound) is often used for both. To be pedantic about it, you should really use the full abbreviations: lbf for a pound-force, and lbm for pound mass. Now on earth, gravity pulls with a force of one lbf for every lbm, which is how we have come to see the units as equivalent. However, on, for example, the moon, this relationship is not true. The easiest way to calculate energy from work performed is that it is a distance multiplied by a force. In English units then, this is lbf*ft. In metric, this is N*m (Newtons, the SI unit of force, and meters here). The calculation of the energy (capacity to do work) carried by a moving object, i.e. kinetic energy is (1/2) mass * (velocity)^2. The units used in calculating this must be of mass e.g. lbm.

So this. . .
One ft-lb translates to the amount of kinetic energy it takes to move a 1-lb object one foot.
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is only true for moving an object against one pound of force, regardless of it's mass. For a 1 lbm object, you could move it 1 ft. vertically, because then you're acting against the 1 lbf of gravity on that object. Moving it horizontally 1 ft. would not necessarily be 1 ft*lbf.
 
In answer to your last question 9mm vs 45. if they are both fmj the 45 makes a bigger hole. The rest depends on bullet construction and a bunch of other variables.

In practical sense the more foot pounds of energy exerted the more it hurts.
 
Does a 9mm bullet with 1000 foot lbs and a .45 bullet with 1000 foot lbs do equal damage, even though the .45 is larger?
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I forgot this part...

The answer is 'sorta'...

Think of a counter sink punch or a sledge hammer...

Both hit you in the chest with 1000 ft pounds of energy...The pencil is light, small and going very fast to acgieve that 'energy'...The sledge is slow and heavy to achieve the same 'energy...

The punch pokes a hole in and out, and the sledge makes your ribs meet your spine and squishes everything in between...

There is a method of quesstimating the terminal 'balistics' of a projectile that is called the 'Taylor KnockOut Formula' (TKO) that tries to give a better real world estimate of 'killing energy' at the target than simple 'muzzle energy' alone...

http://www.dave-cushman.net/shot/tkochart.html

This takes into account the 'momentum' mentioned elsewhere in the thread...
 
One thing the kinetic energy of the bullet moves is the firearm and shooter
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No, that is momentum not kinetic energy.

The momentum of the ejecta from the barrel produce the recoil, not their energy.
 
I want to refer people to Robert Rinker's excellent book "Understanding Firearms Ballistics" for a more detailed discussion of the subject from a number of angles.

Basically it is this...

A foot pound is a measurement of kinetic energy. All objects posses energy for our purposes this is either potential energy (a brick sitting on a ledge) or kinetic energy (the brick in motion).

One foot pound is literally the amount of energy it takes to lift a one pound weight one foot off the ground at a certain altitude (it varies a small amount depending on altitude). 300 ft. pds. is literally a measure of how much energy it takes to lift a 300 pound weight one foot off the ground or a one pound weight 300 feet off the ground. It is a measurement of the energy that an object in motion possesses.

All bullets require energy to work. The ft. pd figure is a measure of a bullets ability to do work. They expend energy (work) to travel down the muzzle, to buck the wind, energy is spent in the muzzle flash and report, as heat, energy is spent in penetrating an object and expanding-or not- etc. All bullets loose energy as they travel down range. At greater distances they gradually lose their capacity to perform as intended as they lose energy. Energy figures are often more important in rifle rounds than in handgun rounds for this reason.

Yep a bullet with 400 ft. pounds of energy at the muzzle has twice as much to work with as one with 200 ft. pds. of energy at the muzzle. Other things to look at would be the weight of the bullet, caliber, bullet construction and what you want the bullet to do and at what ranges.

Energy figures are one factor to take into consideration in selecting a round. Energy figures should not be confused with how hard a bullet hits or how effective it will be for a given purpose. Momentum is something different as well. The relation between energy and velocity is an interesting one as well.

tipoc
 
1 Ft/Lb is the amount of energy needed to raise a 1 Lb object 1 ft in 1 second.

Here is an even simpler explanation, if a hair inaccurate on minor details.
Make a pyramid-shaped weight out of something fairly dense, say lead. Make the weight of the finished pyramid exactly 1 Lb.

Now raise it exactly 1 foot above your toes.

The possibility of released energy is 1 pound in weight moving through 1 foot in distance. This is 1 foot/Lb of potential energy.

Now drop it flat base down.:eek:

As it falls gravity accelerates it allowing the potential energy to become kinetic energy. It is at 1 Ft/Lb of kinetic energy when it arrives. The impact is caused by the release of the kinetic energy when it decelerates to a standstill as it hits the foot. This causes bruising & pain as a crushing injury.

Now we're going to do the same thing again, but with 2 differences. We're using the other foot & the pyramid is now pointy top bit down.:eek::eek:

We drop it again, exactly the same 1Lb & the same 1ft. It delivers exactly the same 1 Ft/Lb of force, but that force is concentrated into the pointy tip, instead of the flat base on contact.

This time it will penetrate more causing a penetrating injury & causing a cut & bleeding. Why? Because the 1 Ft/Lb has been focused. The tip has less area than the base so the Lbs/Sq inch are much higher (cuz there are less sq inches.) but the crushing energy of the flat base has converted into penetrating energy at the tip it is still 1 Ft/LB of energy, but the pounds per square inch of pressure applied have changed. A bit extreme, but that's part of the difference between the bullet diameters, but exaggerated for clarity.:)

So how does this relate to ballistics? Easy.
As you increase the energy available you increase the power that can be delivered to the target. Imagine either changing to a 10 Lb pyramid (bigger, heavier bullet), or raising it to 10 feet, (increasing the velocity.) you get the same effect by for example dropping a 5 Lb weight 2 feet, or a 2 Lb weight 5 feet. Its linear, no fancy math needed. 10 Ft/Lbs is 10 times as much as 1 Ft/Lb.

The details that are "off" to make it easier?
a weight won't fall 1 foot in 1 second, I completely ignored external ballistics, concentrating on terminal ballistics instead & I ignored supersonic Vs subsonic impact damage to keep it simple.
 
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Foot pound: A unit of work equal to the work done by a force of one pound acting through a distance of one foot in the direction of the force
 
This is simple guys.

A 1 lb weight is suspended 32 ft above your head and dropped.
It hits your head w/ 32 ft-lbs of energy ...
no matter you calculate it, either by distance dropped or terminal velocity when it hits you.

Potential energy + Kinetic energy are constant in a closed system
 
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Hmmm.. some esoteric explanations, most of them (e.g., flyfish) correct.
Kinetic energy is linear - 400 ft-lb has twice the energy of 200 ft-lbs.
The equation KE = 1/2m(mass) X V squared indicates that increasing velocity will have a greater effect on total energy than will increasing the mass. Thus, the relationship between the 45 ACP (lower velocity, heavier projectile) and the 9 mm (higher velocity, lighter projectile.
The essential problem however is that the effects on a biological target (say, a Whitetail deer or a boar) are not solely a function of kinetic energy. The effects depend upon penetration (light bullet breakup may = inadequate penetration, and complete through & through penetration means some of the energy wasn't expended in the target but in the tree behind it). Also, the anatomy of the biologic target (muscle mass, bone density and bone thickness, etc.) will affect the wounding/killing potential.
However, in general, fast, heavy bullets will have more damaging effects than light & slow bullets.
 
I was a bit confused about this as well, showing 45acp having less energy as 9mm in some cases.
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Yep that is true. It is also true that some loads for the 22 Magnum at 40 gr. of weight have more energy than some 230 gr. loads of the 45acp. But it means little when it comes to effectiveness of the two rounds when they strike an object.

The amount of energy a certain load may have is largely a paper figure. If a certain load produces 400 ft. pds of energy at the muzzle it does not mean that the bullet will strike with 400 ft. pds. of force and knock a 200 pound person off their feet or strike with the impact of 400 ft.pds. of force in a tiny spot. The 400 pt. pds. is a measure of the ability to do work. A man or a deer will receive only a small amount of that 400 ft. pounds of kinetic energy in the form of a push.

As mentioned the work that a bullet has to do is bucking the wind, heating up the barrel as it fights the resistance caused by friction, penetrating, expanding etc. is important but only one factor to look at. Bullet construction is another. A bullet needs both enough energy to do it's job and do it at the distances required of it but also needs to be shot in a caliber, bullet weight and bullet construction to get the job done. If the bullet expands too soon when it strikes the target more energy will be used to penetrate to a useful distance. If not enough energy is left to the bullet for this, the round may wound rather than kill the game.

tipoc
 
Can someone please explain to me, in the simplest terms, what foot lbs are and how they affect ballistics?
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Listing ft pounds is the easiest way a single common unit of measurement can be applied to two dissimilar things and display some sort of figure common to both.

A .45 acp and a 9mm may both produce 350 ft pounds of energy, but, they do it differently.
The .45acp uses a heavy slug at moderate velocity.
The 9mm uses a lighter slug at higher velocity.

Having a common denominator - foot pounds - allows a rough comparison to be made.

Back in the days of black powder, everything had pretty much the same velocity no matter what the caliber was. The only was to get more energy (foot pounds) was to increase the weight of the projectile. Since all the projectiles were round, the only way to increase the weight was to increase the diameter.
Double the weight and you double the energy (measured in foot pounds).

Smokeless powder changed all that.
W/smokeless, it became possible to increase the velocity.

Double the velocity and you quadruple the energy (measured in foot pounds).

Here's a link to a good calculator:
http://billstclair.com/energy.html

You can plug in a bunch of different figures and play to your heart's content.

BTW - don't get all hung up on energy figures as a lot of people do. In and of itself, it's a fairly meaningless thing where handguns are concerned.
Sectional density is a whole lot more important in the grand scheme of things. ;)
 
1 Ft/Lb

It is NOT "Ft/Lb"

It is ft-lb.

Feet time time force (a force over a distance), not divided by force.

And there is no time limit.

it is energy, not power.

Power is how fast you can supply energy (perform work).
 
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