Lb/Ft vs Ft/Lb for Ballistics.

[ClayInTx]

Let’s please don’t let this turn into a ‘tis, ‘taint, ‘tis, ‘taint discussion. I’m asking a question and truly want to find out the answer.

To keep this gun related let’s consider the values given for various cartridges. Unless it’s a valid comparison for your example let’s leave the V-8 in your Corvette out of it.

Now I suppose it could have been decided either way by those who decide such things, but it was decided that pound feet would represent force, or torque, and foot pounds would represent work.

My understanding of the figures given in the ballistics tables is that these reference the available instantaneous force of the bullet when it strikes the target. This force can be likened to torque because time is not a factor of torque.

Now if the tables represent the mass displaced by the bullet during a period of time then the tables are correct in giving the value as foot pounds; and those speaking of foot pounds in reference to a bullet impacting the target are correct. This value is the actual work done by the bullet and time is a factor here.

I see a problem in measuring foot pounds in this case because the mass moved would have to be measured and also the time it took to move it. Now because the displaced mass moves at a constantly decreasing rate as the bullet’s energy is dissipated I suppose this displacement rate would have to be integrated over the time it takes the bullet and mass to come to a complete stop.

Do these tables actually give the value of the work, or foot pounds, accomplished by the bullet?

 

[Sport45]

A ft∙lb and a lb∙ft are the same thing. Be careful when you use terms like ft/lb or lb/ft as you did in the title as these literally mean "feet per pound" and "pound per foot" which are entirely different.

Work is force applied over a distance. If it takes 20 lbs(force) to push a box across the floor and you pushed it 20 feet you did 400 ft∙lbs of work. If you pull the end of an 18" wrench with 80 lbs(force) perpendicular to the handle you are applying 120 ft∙lbs of torque. By convention work is expressed as ft∙lb and torque is expressed in terms of lb∙ft.

Of course, if you look at the handle of the torque wrenches in my garage the scales are marked in ft∙lbs, go figure...

When we talk about the kinetic energy (of a bullet, for instance) the formula is 0.5*mass*velocity^2 (1/2 mass times velocity squared). The English unit of mass is a slug, but we tend to screw things up on our side of the pond by using pounds interchangeably when referring to mass and weight.

Weight is a force and force equals mass times acceleration. So the mass of a 150 grain bullet assuming a gravity acceleration of 32.2 ft/sec^2 is 0.000665 slugs or 0.000665 lb(force)∙sec^2/ft

When that bullet is traveling at 2700 ft/sec its kinetic energy is 1/2mv^2 or 0.5*0.000665*2700*2700 or 2425 ft∙lb after unit conversions.

I suppose we could say the energy is 2425 slug∙feet squared per second squared but folks would look at us funny.

Do these tables actually give the value of the work, or foot pounds, accomplished by the bullet?

Neither. The tables only express the kinetic energy of the bullet.

 

[ClayInTx]

Sport45,

Just got back inside after making some noise to let the neighbors know they can’t out do me. Made a quick check to see if you responded. LOL figured you couldn’t let that one go by.

Anyway, I’ll go over this later and, like Scarlett in GWTW, think about that tomorrow.

The reason I made the title that way was to duplicate how a lot of the tables show it. Yep, I know it isn’t correct, but...

Good going over this with you. Now where did I put that dang calculator? It’s 2+2=5 isn’t it?

Clay

 

[Sport45]

It's really kind of fun to get back into this. Would you believe I earned spending money in college teaching physics labs? That was in the early 80's so I'm more than a bit rusty.

 

[44 AMP]

We get ft/lbs because of the math

And the way we set up the equation. One sets up the equation to calculate the energy, and using what they used to teach as algebra, you cancel out units as you work the equation, and wind up with a number and ft/lbs.

You can set it up with the variables in different positions, but the math stays the same, and you wind up with the same number except it comes out lbs/ft.

 

[Mal H]

Now see, 44, you're falling into the same trap that Clay did with the units and the way they are written.

It's not feet divided by pounds or, more commonly, feet per pound. Nor is it pounds per foot. The final units for energy (or work) in the English system are foot-pounds or pound-feet.

Sport45 gave an excellent explanation.

 

[ClayInTx]

MalH,
I explained why I defined the units in the title of my post that way. It was to emulate how I've seen it in some ballistics tables. Yes, it's not correct.
Clay

 

[ClayInTx]

Those of you defining the math are correct in that. We know that A*B=C is the same as C=A*B.

My original contention: We all use some words incorrectly and should not make a big deal about it because we all also know what is meant. Another example is the word since when we should actually use because. No big deal there since (grin) we know what’s meant.

In the other thread I was a bit too subtle and should have expained my position more.

That led to this new thread which is becoming very informative in trying to determine the correct term, and the math examples are good, also. And my question is: Do the ballistics tables we normally see using foot pounds use the term correctly, or should it be pound feet? Regardless of the end result we will still know what’s meant by either term. (Grammarians might note I began a sentence with a conjunction, and that’s done often, also.)

I didn’t drag down Marks Handbook to reference because I would have had to retype it. Instead, I found this on the internet which pretty well explains the difference. I did correct some spelling typos and what I believe was a math typo; he calculated 550 where it seems it should have been 500—I changed it. (Corrected by edit [bolded].)
The article:

WHAT'S THE DIFFERENCE between foot-pounds and pound-feet?

Both consist of the same units-a force and a distance-multiplied together. The physical quantities being expressed, however, are quite different. Yet applying the terms interchangeably is one of the most common misuses of technical language.

The product of a force times a distance, as in "pound-feet" ("Newton-meters," for those metrically inclined), is torque. In electric machinery, we all know what torque means. It's the turning effort developed by a motor that causes a shaft to rotate. Based on the simple principle of the lever, torque is increased by either increasing the force itself (as by strengthening a magnetic field) or by increasing the distance--the lever arm--between the point of force application and the axis of rotation. (Anyone using a wrench is familial with that concept.)

Any physics text or engineering handbook will make the principle clear and will define force times distance, pounds times feet, or ounces times inches--as torque. To "say it right," remember the universal standard convention that the force comes first; the distance second.

So, is "foot-pounds" meaningless, then? Not at all. When the distance is named first, and the force second, the product represents something equally real but quite different: work.

Work measures the expenditure of energy. Thus, when a mass is lifted through a distance (against the force of gravity), or moved horizontally against a friction force, the amount of work done is measured by the product of the force involved and the distance moved. Increasing either one increases the amount of work.

How fast that work is done is a measure of power (horsepower or watts). For example, if a 100-pound weight is lifted 5.5 feet in one second, the work done is 550 foot-pounds, and the power involved is 550 foot-pounds per second-which equals one horsepower (746 watts).

Whether speaking of torque, then, as related to fastener tightening or motor capability, use the term putting force first (as in pound-feet). Putting the distance first (as in foot-pounds) is not equivalent; it represents something not at all related to torque.

And as we've said elsewhere and often, using technical terms incorrectly invites skepticism concerning your knowledge of technology.

By Richard L. Nailen, P.E., EA Engineering Editor
Copyright Barks Publications Sep 2004
Provided by ProQuest Information and Learning Company. All rights Reserved

 

[wmeSha]

I really appreciated this example. It just goes to show that paying attention to the units can get you there 90% of the time, but not 100% of the time. You still have to use your brain to see if you are making sense.

It had never occurred to me that torque has the same units as work despite being the two concepts being completely incomparable.

 

[Sport45]

As for the ft∙lb vs lb∙ft thing, I think we've got it right. The biggest violators of the convention are the automakers who claim their engines put out so many ft∙lb of torque. That and the tool makers who scale their torque wrenches in ft∙lbs.

The engine technicians on the other hand tend to get it right. At every motor and engine test I've attended the measured torque value was listed in lb∙ft if not in SI units.

 

[Blondie.357]

Ah nvm.

I have only smoked it once before, but can you pass that joint?

 

[ClayInTx]

Error correction

There is a typo error in the article by Richard Nailen which I posted, above.

In the sentence:
For example, if a 100-pound weight is lifted 5 feet in one second, the work done is 500 foot-pounds, and the power involved is 500 foot-pounds per second-which equals one horsepower (746 watts).

There was an odd symbol which was either a software error or a typo.

The sentence originally read:
For example, if a 100-pound weight is lifted 5ft feet in one second, the work done is 550 foot-pounds, and the power involved is 550 foot-pounds per second-which equals one horsepower (746 watts).

I thought that “5ft feet” was a typo and then corrected the “550” to “500” to match the apparent math. It probably should have been “5.5 feet”.

Just in case someone wants to copy and paste that article, or reads just it without noting this correction post, I’ll go back and make corrections now.

Keep those cards and letters coming.

 

[ClayInTx]

Are we ready for a final result?

I’ve mulled over this and here is my thinking:

When the gun is fired the gun powder changes form and becomes mostly energy. It imparts this energy to the bullet by increasing the pressure in the barrel to an extremely high pressure. The bullet leaves the barrel and is a mass traveling at some velocity. This means that the gun powder did some work. If it did work, then by the agreed upon definition, it has put foot pounds onto the bullet.

Some of this energy of the bullet is expended to overcome air resistance, but it reaches the target with energy left over.

It hits the target. It has velocity and mass. The mass remains but the velocity is absorbed either by the target or the backstop. Let’s just say the target is a 3/4” steel plate.

The steel plate surely must move a little, does it not? If this is so then it means work has been done to move the plate.

If it’s work then by definition it has expended foot pounds. And the term foot pounds of the ballistic charts is correct. Right?

Now before I decide this is what it is I want y’all to chime in with your thoughts because I don’t want to have to change my opinion again.

Now speaking of a tiny bullet hitting a big 3/4” steel plate reminds me of the conundrum of the fly hitting the freight train. To keep it gun related: If a bullet going south at 500 mph hits, head on, a freight train going north at 50 mph the bullet has to stop going south and begin going north. If the bullet stops then that means the freight train also has to stop. Right?

Oh hell, did I just provide grist for another Thread mill?

 

[wally626]

I think most of the confusion in the topic of ft·lb for energy comes from the use of pound to mean both mass and force (weight) in english units. A pound of force is defined as the force exerted by a 1 pound mass under the force of gravity (32.17405 ft/sec2) at the surface of the Earth and is defined as 1 pound-force = 32.17405 pounds-mass ·ft/sec2.

Pound of force (lbf) is used in the unit of energy for english units but pound of mass (lbm) is used in calculating the value. An example is shown below.

Example Case:
.40 S&W 165 grain bullet, 1150 ft/sec velocity

Convert grains to pounds mass, 7000 grains per pound

165 gr/7000 gr/lbm = 0.023571 lbm

Kinetic Energy = 1/2 x mass x veloctiy2

1/2 x 0.023571 lbm x 1150 ft/sec x 1150 ft/sec = 15,587 lbm·ft2/sec2

Convert from pound mass to pound force


32.17405 lbm·ft2/sec2 x 1 lbm/32.17405 lbf = 485 ft·lbf

 

[Tucker 1371]

Now I suppose it could have been decided either way by those who decide such things, but it was decided that pound feet would represent force, or torque, and foot pounds would represent work.

Just to clear things up the guy who decided this was Sir Isaac Newton, father of modern day physics and the inventor of Calculus. I think he had a good handle on what he was talking about . High school physics also explains it pretty well too.

Ft * Lbs are energy Lb/ft (as in pounds per foot of lever arm) is torque, which I as far as I know doesn't apply too much to in flight ballistics, maybe to recoil and muzzle flip though.

I think this is why we need to switch to the metric system, energy is just joules, makes things simple.

 

[ClayInTx]

We hear “go metric” a lot of times.

The only advantage of the metric system is the math.

However, consider these things about the English System:

The dozen: A low number divisible by more other numbers than any comparable value. Usually applied to things which are not handy to split. Apples, eggs, donuts, etc. You have a dozen donuts and some Boy Scouts: you have 12—each gets one; you have 6—each gets 2; you have 4—each gets 3, you have 3—each gets 4; you have 2—each gets six; you have one—one gets sick.

Liquids: Each whole unit in the English System is determined by halves. You have a gallon of milk and need a cup but you have no measuring cup: divide the gallon by half = 2 quarts; divide the 2 quarts by half = 1 quart; divide the quart by half = one pint; divide the pint by half = one cup.

Length is a also system of doubling or even multiples. The only bad one is dividing the foot by 12. Well, being given speed in furlongs per fortnight isn't all that easy to deal with.

BUT: the big thing is fasteners. Our screw thread system was developed over centuries of trial and error for what works better. The metric screw threads were determined by committee. They’re not all that bad but not as good as American. Note that I said American, in this case, not English.

Anecdote: some years back we were designing a plant to be built in Switzerland. We talked with the Italian contractor who was going to build it. We needed to know which pipe sizes were commonly used. This information isn’t as available for Europe as for the USA. He told us what the practice was, but added a zinger. He said, “But if you want to use American pipe, we are very familiar with it. Actually, if we have an application where we absolutely cannot have a leak we use American pipe.”

The only advantage of the metric system is the math.

 

[wally626]

GSUeagle1089
Ft * Lbs are energy Lb/ft (as in pounds per foot of lever arm) is torque, which I as far as I know doesn't apply too much to in flight ballistics, maybe to recoil and muzzle flip though.

Torque is measured in foot pounds, or other units but not Lb/ft. It is the application of 1 pound of force at a distance of 1 foot from the fulcrum. Pounds per foot would be a linear load. Pounds per square foot is a unit of stress or pressure.

Edit to add, pound · foot is often used in discussing torque to distinguish it from foot·pound energy units. In metric Newton Meter are units for torque which is also the units for energy but energy is normally referred to as a Joule ( 1 Joule = 1 N·m) while torque is not in order to keep the distinction.

 

[Tucker 1371]

pound · foot

I think that was what I was getting at, been about 2 years since I've had any physics, my bad .

 

[JohnKSa]

Without getting into units, here it all is in a nutshell.

Force is mass x acceleration. Force is required to accelerate an object.
Force is mass x deceleration. Force is applied when a moving object is decelerated.

Acceleration (or deceleration) is the same thing as velocity / time. (Useful later.)

Work is Force x distance. Work is done when a force moves an object over a distance.
Work can also be expressed as mass x acceleration x distance. Work is done when a mass is accelerated over a distance.
Work can also be expressed as mass x deceleration x distance. Work is done whan an object is decelerated over a distance.

Kinetic energy is also Force x distance. It has the same units as work because the two are sides of the same coin. An object gains kinetic energy when work is done on an object and a force accelerates it into motion
Kinetic energy can also be expressed as mass x acceleration x distance. An object gains kinetic energy when it is accelerated over a distance.
Kinetic energy can also be expressed as mass x deceleration x distance. Kinetic energy is lost/used/applied whan an object is decelerated over a distance.

Force is mass times acceleration
Kinetic energy is mass times acceleration x distance
Therefore Force can be expressed as kinetic energy / distance (or, said another way, as the change in kinetic energy/distance over which the change in kinetic energy occurred)

A moving object applies force when it hits an object. How much force is applied is determined by how much kinetic energy the moving object has and how much distance it takes to stop the moving object. The shorter the distance the more force is applied. The more kinetic energy the more force is applied.

Momentum is mass x velocity. Momentum relates to how hard it is to stop an object.

Force is mass x acceleration which is the same thing as mass x velocity / time
Momentum is mass x velocity
Therefore Force can be expressed as momentum / time (or, said another way, as the change in momentum/time over which the change in momentum occurred)

A moving object applies force when it hits an object. How much force is applied is determined by how much momentum the moving object has and how fast the moving object is stopped. The faster the moving object is stopped the more force is applied. The more momentum, the more force is applied.

 

[Sport45]

IMHO it's all about conservation of energy.

When a loaded cartridge is in the chamber it has potential energy (chemical). The amount of that energy is determined by the power of the primer and the weight and type of the propellant.

When the firing pin strikes the primer it detonates igniting the propellant. The potential energy (chemical) of the propellant is transferred to potential energy (pressure) and thermal energy (heat). The pressure acts on the bullet pushing it from the case and down the barrel. During this time the potential energy (pressure) is being transferred into kinetic energy (mass times velocity) and more thermal energy (friction heating of bore by bullet). As the bullet leaves the barrel the excess energy goes into the air as sound, heat, and displacement of the air's mass by the expanding gasses.

In flight, the bullet is slowed down by air resistance. It's initial kinetic energy is being transferred to the air it displaces and also into heat from friction.

When it strikes the target, say a 1" armor plate, the remaining energy is dissipated into the plate and bullet as potential energy (heat), and kinetic energy (movement of plate and bullet fragments, maybe sound?)

It takes energy to rip a bullet apart which is one reason retained weight is important in hunting bullets. It indicates the energy was used on the target rather than on the bullet.

I've never seen a ballistic pendulum or a drawing of one. But I think in the early days of ballistics testing they suspended a known weight from a pivot of a known height. The weight would have to be made of something that could catch a bullet without shedding weight of it's own, possibly wood. By measuring displacement of the pendulum after impact (swing height relates to potential energy) it's just a few steps to determine the initial kinetic energy delivered by the bullet and thus its velocity.