Talk:Foot-pound (energy)

2006

You didn't set this page up, Greg. Don't be delusional. The talk page, sure, but the article it is tied to has been here for nearly three years.

I am not delusional. The original article I started got sucked up, as have others. Probably because the article I started was too much like this article. When it comes to physics, thou shall not tread on the purists.Greg Glover 22:09, 26 December 2006 (UTC)

SI didn't exist, and the newton and joule did not exist under those names, when the CGPM more than a century ago, defined the standard acceleration of gravity for what's in their province (the metric system, and as far as a coherent system for calculations goes, that was mostly various forms of cgs at the time) as "980.665 cm/s²". 3rd CGPM, 1901. That has been the proper value for g_n in any metric units application ever since. And it doesn't matter in the least whether you express it that way, or as "9.80665 m/s²".

You are absolutely correct Mr. Nygaard. I did, unlike most folks around here all 4 of the references you see below on the article page were posted by myself. One thing I have perceived since I have been here at Wikipedia (over a year) is that if you’re in insider (engineers of teacher) you don’t have to cite references. However us outsides (the rest of us) had better cite some reference or “poof”, gone is your work. I read your rather nasty and condescending comments in the history part of this page; shame on you.

Obert is your source for what, exactly? Gene Nygaard 14:59, 6 October 2006 (UTC

Huh?Greg Glover 22:09, 26 December 2006 (UTC)

Greg, your use of "UOM" as an abbreviation is a personal idiosyncracy that we can put up with here on the talk page, but please refrain from inserting the unfamiliar and unhelpful acronym into the article itself. Gene Nygaard 14:59, 6 October 2006 (UTC)

See Wikipedia:Manual of Style, to learn what is tollerated around here.Greg Glover 23:36, 28 December 2006 (UTC)

And, since you ask about perceived resistance to your editing, here's a tip. Invest in a spell-checker. There are many fairly intelligent people who have problems with spelling, or at least with spelling in English, but I suspect that unlike many Wikipedia editors, you don't have another language to fall back to in that regard. Gene Nygaard 15:06, 6 October 2006 (UTC)

Cited or sited my good man, spell check comes with Microsoft Word.Greg Glover 22:09, 26 December 2006 (UTC)

Ballistics calculation added

I'm sure this needs to be Wiki-fied, but this formula is much, much harder to find with a Google search than it ought to be, so lets put it here as this is the best home for it I can think of.--Solidpoint 02:41, 2 March 2007 (UTC)

Ballistics calculation discussed

It appears some folks do not want this calculation here. Someone put it here and I maintain it along with the rest of the article. The idea and concept of foot-pound force is mostly an archaic one now. It seems only to lend its self to small arms ballistics. But since foot-pound force falls strictly with the category of “Customary units in the United States”, I am opposed to any removal of information set forth within this article at this time.Greg Glover 03:48, 5 July 2007 (UTC)

Foot-pound vs Foot-pound force

According to Edward F. Obert, Thermodynamics, Mc Graw-Hill Book Co., 1948; a foot-pound force is equal to a 1 pound mass being accelerate at 32.1739 feet per second squared.

According to Webster’s Dictionary of the English Language Unabridged “Encyclopedic Edition”, J. G. Ferguson Publishing Company, 1977; a foot-pound is the energy required to lift a 1 pound mass the height of 1 foot.

I would like to have “foot-pound” redirected away from this article. The term foot-pound is an abbreviation of the correct term foot-pound force. Because engineers and mathematician abbreviate everything, so it does not make it correct within a Wikipedia article.

What say you?Greg Glover 04:40, 5 July 2007 (UTC)

I'm adding this section in for clarity way after the fact.

The above definition like all definitions for the foot-pound are misleading. Anyone who believes this definition is correct and pertains to foot-pound force just does not get it. Here is why: energy is that capacity to do work. All kinetic energy equations have one thing in common; velocity. Why because you must have motion to have kinetic energy. Can you read anywhere within the above definition a velocity? No you can not! Therefore you must use the number one rule of mathematics. Assume it to be there. So if you assume it to be there, the assumed velocity is 1 foot per second. If you plug in the numbers, 1 pound mass being lifted 1 foot in hight in 1 second you get; 1 x 1²/ 2 x 32.174 = .01554ft-lb_f

Again .01554 dose not equal 1 and the foot-pound dose not equal the foot-pound force.

Beam me up Scotty, no intelligent life here. Greg now banging his head against the keyboard. Just kidding.Greg Glover (talk) 02:27, 17 February 2008 (UTC)

I disagree. Foot-pound may be less accurate, but it is in more common usage (10 times as many google hits as foot-pound force). People are more likely to search for the more common term. Why make the information hard to find? Rracecarr 17:31, 5 July 2007 (UTC)

You are absolutely correct. I find it a lot easer to use abbreviations and slang. The most used adjective in my vocabulary is probably the “F” word, but dose that make it right here at Wikipedia?

Wikipedia is about disseminating the correct information. If you can get Mr. Webster to agree with yourself or Oxford for that matter, then maybe we should make a change.Greg Glover 19:56, 7 July 2007 (UTC)

What I think is the amiable thing to do as in other articles is to add this sentence, “Typically the vernacular term of ‘foot-pounds’ is used in place of linguistically unaffordable expression of foot-pound force.” I hope this will suffice as a compromise.Greg Glover 22:57, 7 July 2007 (UTC)

Expanded Thought for those who don't see the mathematics in the two definitions. Foot-pound force comes from the English engineering system which uses four Units of Measure. When the original kinetic energy equation is factored to the two remaining variables, force through distance is what’s left over. Force through distance is mathematical expressed as: ft • lb_f or from the written language; foot times the pound force. Conversely the foot-pound is mathematically expressed as: ft • lb_m or from the written language; foot times the pound mass.

As you can read the foot-pound and the foot-pound force are two separate Units of measure.

• 1 ft • lb_m = .015 406 ft • lb_f

Greg Glover 19:56, 7 July 2007 (UTC)

The above statement:

"According to Edward F. Obert, Thermodynamics, Mc Graw-Hill Book Co., 1948; a foot-pound force is equal to a 1 pound mass being accelerate at 32.1739 feet per second squared."

makes no sense.

However, the statement:

"According to Webster’s Dictionary of the English Language Unabridged “Encyclopedic Edition”, J. G. Ferguson Publishing Company, 1977; a foot-pound is the energy required to lift a 1 pound mass the height of 1 foot."

does make sense--and it uses the term foot-pound, not foot-pound force. More support for the former term. Rracecarr 20:15, 9 July 2007 (UTC)

Yes, feet per second squared is a unit of measure from these four systems: Absolute English, Absolute Metric, Technical English and Engineering English. Feet per second squared is known as acceleration as in: m • v² and m • c². I think Sir Isaac Newton and Professor Einstein might have something to say about this.

Unless you were apart of any of the sciences or engineering prior to about 1980 and specifically October 1960 you would not know of these systems. We do not use the English or metric systems anymore (I am aware of bench units). We use SI. However there are millions of people that use foot-pound force each and everyday. Because you cannot understand (and I understand why) the concept of foot-pound force it dose not mean others can’t.

Rracecarr, out of curiosity do you even use foot-pound force as a unit of measure? I see you are an Oceanographer. I though all the sciences had adopted SI? I would guess and this is only a guess, you are under that age of 40 (you can delete this last sentence).

Foot-pound force has been around for about 200 years. It is obsolete by now but it is a real unit of measure.

See this link under http://www.digitaldutch.com/unitconverter/ and look under ENERGY as a conversion and for foot pound-force (I did not put the hyphen in the wrong spot) in the drop box. The joule is the default unit of measure.

Please, let’s keep up the dialog until we can come to an understanding. There is too much in fighting around here.

Also, could you please use the colon to move your thought and responses to the right? It makes it easier for all wiki community members to follow the thread.

Thank you.Greg Glover 20:49, 9 July 2007 (UTC)

• a = d / t² or v / t Greg Glover 20:54, 9 July 2007 (UTC)

[Resetting indent to avoid all the white space] Yes, ft/sec^2 is a unit of acceleration. It is not a unit of velocity, as was stated in the ballistics section I removed (see below). Foot-pound and foot-pound force (wherever you put the hyphen) are not two different units, but two names for the same unit (an English unit of energy: one foot time one pound [force]), one more common, the other more descriptive. I am happy to discuss changes to the article with you, at least for a while, and if we seem to be making progress.

Here is the section I removed. At least the error I pointed out should be corrected before it is readded to the article.

Ballistics calculation

To calculate the foot-pound force (ft•lb_f ) of energy for a bullet in English units of measure; given bullet weight in grains and velocity in feet per second, use the following formula.

E_t = w • v² / ( 2 • g_c • 7000 )

Where as E_t is the transitional kinetic energy of the projectile in foot-pound force, (ft•lb_f); w is the weight of the projectile in grains, (gr]); v is the velocity of the projectile in feet per second squared,(ft/s²); 2 is the numeric coefficient that comes from the word "half" in the classic statement for kinetic energy; g_c is the dimensional constant of 32.1739 lb_m • ft / lb_f • s² and 7000 is the conversion factor to set the equation equal to the pound (lb).

Example: 2913 ft • lb_f = 180 • 2700² / ( 2 • 32.1739 • 7000 )

Rracecarr 21:18, 9 July 2007 (UTC)

Okay, I see where you are misunderstanding the equation, what’s given and the example. What’s given is the bullet weight and velocity. Can we agree on that? The equation shows, what appears to be a unit of measure for acceleration as in v² (the example as well). Can we agree on that? What I think you see as a discrepancy is that velocity is entered in to the equation but it is called feet per second squared as given; Correct? I think we can both agree a velocity entered as x feet per second is not equal to feet per second squared. That is a very astute observation.

1. Please stop telling me what I'm misunderstanding. This is simple physics and I understand it perfectly.
2. The units of v² are not the same as the units of acceleration. The former is length squared over time squared, the latter length over time squared.
3. It sounds like you're agreeing to remove the statement that the units of velocity are length over time squared. That is progress. Good. Rracecarr 14:14, 10 July 2007 (UTC)

Here is what’s going on. I will try to make this as short as possible but first some history. The original scientific experiment for falling bodies was done and correct me if I am wrong, by Sir Christopher Wren using Robert Hooke’s vacuum pump to create a vacuum within glass tube. The “vacuum tube” had a brass ball and a feather. Then the two objects were dropped within the vacuum tube. The idea was to observe air resistance. Sir Isaac Newton is first credited with this theory. The original equation for the energy (then called force) for both the brass ball and the feather from the experiment was: F = wz.

Now I would refer you back to the foot-pound force page for the entire equation and its factors but you have removed it. That section was called Origins of Foot-pound force. It was put there for a reason. The reason is that no one remembers where foot-pound force came from.

I did not remove a section called Origins of Foot-pound force. The only section I removed was called Ballistics. Rracecarr 14:14, 10 July 2007 (UTC)

I don’t think you intentionally removed the “Origins of foot-pound force” on purposes. You deleted the entire “Units of Work or Energy” section at 10:00 UTC July 10th 2007.Greg Glover 23:32, 14 July 2007 (UTC)

What you must understand is that when calculating energy or work a velocity must be entered into the equation. The original equation took two measurements for velocity. One measurement is at rest the other at impact or across a datum point. The two velocities are then averaged. What is left from the original equation within the “classic statement” for kinetic energy is the word half; as in “…half mass time the velocity squared”. You see the word half within the numeric equation as the numeric coefficient, .5 or below the divisor bar as 2.

Actually, the most basic definition of work does not include a velocity: work = force * distance. So, for example, the muzzle energy of a bullet is indeed half mass times velocity squared, but it is also average propelling force times length of barrel (force times distance). Rracecarr 14:14, 10 July 2007 (UTC)

What you don’t know is how the kinetic energy equation of E_k = .5 m v² is derived out of the original equation of wz and why a variable for velocity is used and not acceleration.

What must be given and entered into the any kinetic energy equation is a velocity; the variable v. You do not enter a value for acceleration; the variable a. The fact that v is squared dose not make it a value of acceleration even if the description (given) says so. Unfortunately numeric equations sometimes do not translate well in to the written language or spoken language.

If you translate them correctly, they do. Rracecarr 14:14, 10 July 2007 (UTC)

Believe it or not the equations of: F = wz; E_t = m v² / 2 g_c and E_k = .5 m v² are all equal. The first two equations were originally in English units and the third equation is now in SI units. All three equations can be set to the foot-pound force or joule.

Most of what you have removed from the foot-pound force page has been there for a year. I spent many days explaining all of this to several other folks. Just read the first part of this discussion.

Just because other people gave up on improving the page doesn't mean I can't give it a shot. Rracecarr 14:14, 10 July 2007 (UTC)

What you removed is all true and correct mathematics. I would respectfully request you put it back. Again because the numeric values do not translate well to the written language is not of my making. Can we discuss things and do the correct edits. It may be necessary to correct or compromise on the given or description of the given to be more accurate.

Finally the foot-pound dose not equal the foot-pound force. That is why I propose the two be separated. The foot-pound is not uses anymore and the foot-pound force seems to be used only in the United States for small arms ballistics and automotive sales. They are old leftovers. Greg Glover 00:20, 10 July 2007 (UTC)

What you call foot-pound force, most other people call foot-pound. So yes, they are the same thing. I refer you back to your own posting of the definition of foot-pound from Websters: "the energy required to lift a 1 pound mass the height of 1 foot". That is, one pound (force) times one foot. Rracecarr 14:14, 10 July 2007 (UTC)

One final opinion: I think the word "transitional" should be left out of this article. "Transitional kinetic energy" usually refers to molecular energies associated with transitions between different atomic states. Find me a ballistics reference not written by you that includes "Transitional kinetic energy" if you want to change my mind. Rracecarr 14:14, 10 July 2007 (UTC)

Rracecarr, first I wish to apologize for the word “misunderstand”. I am not trying to put you down or imply anything. I will refrain from any negativity.

The first definition for transitional kinetic energy not written by myself (I am only rewriting what I understand) comes from Wikipedia kinetic energy; 5.1 Kinetic energy of rigid bodies and 5.3 Kinetic energy of systems. The other is from Mc Graw-Hill encyclopedia of Science and Technology, volume ice-lev, 9th Edition, Mc Graw-Hill, 2002. That book can be found at most public libraries.

I do understand the changing states of the energy stored within liquid fuel as it coverts to mechanical energy in an internal combustion engine or the changing states of the energy stored within a powder charge as it is converted to kinetic energy for a small arm. Since a bullet is a ridge body and in motion in a straight line (as opposed to rotational kinetic energy [ E_r ]) it is then called transitional kinetic energy ( E_t ). This is the Wikipedia definition, not mine. If you can, go to the kinetic energy ( E_k )article here at Wikipedia and let me know if I have misunderstood the definition for “transitional kinetic energy”. But I don’t think so.

OK we are definitely getting somewhere. What you are talking about is translational kinetic energy, not transitional. I don't know where the word transitional first crept in, but it doesn't belong. Rracecarr 02:00, 11 July 2007 (UTC)

I am very dyslexic. I have been using the term “transitional” for 40 years. I reread the term in Wikipedia. The correct term is most definitely translational according to Wikipedia. Therefore “translational kinetic energy” it shall be.

Can we agree that a projectile or body in motion in a straight line is “translational kinetic energy”? Can we agree that the English units of measure are the foot ( ft) and the pound force (lb_f)? Can we agree that to calculate the foot-pound force of a projectile or a body in motion in a strait line, we use the kinetic energy equation of E_k = .5 • m v² (the appropriate conversions factors included to change the equation from SI to English units)? If so I think we are moving in the right direction.Greg Glover 02:21, 12 July 2007 (UTC)

I agree with most of what you've written. I would phrase it a bit differently: a body in motion has kinetic energy, which can be separated into two parts--the part due to the motion of the center of mass (translational), and the part due to rotation about the center of mass (rotational). Also, the expression E_k = .5 • m v² is correct for any self-consistent set of units, not just SI. If you measure mass in pennies, distance in cubits, and time in months, then the expression is correct, so long as you measure energy in penny cubits squared per month squared. In particular, the expression is correct for English units if you measure mass in slugs, distance in feet, time in seconds, and energy in foot-pounds. That is because a slug ft^2/s^2 is the same thing as a foot-pound.Rracecarr 22:05, 12 July 2007 (UTC)

When I read the Webster’s definition for a foot-pound, I believe the definition conveys that by lifting a one pound mass, one foot off the ground that is equal to one foot-pound. Do you concur? If so can you please explain why or if not, can you explain why not?

I concur, almost. I would say it this way: lifting a one pound mass one foot (in Earth's gravitational field) requires an energy of one foot-pound. That is because a force of one pound-force is required to lift a mass of one pound, so in lifting the mass one foot, you have a force of one pound-force acting through a distance of one foot.Rracecarr 02:00, 11 July 2007 (UTC)

You hedged a bit. But that’s okay. I would say, “You don’t concur” and that you understand kinetic energy as I do.

As I will respond below, everything we discuss or write about, we must be careful; to use your words. Sometimes we are communicating in English and sometimes in physics by the written word.Greg Glover 02:21, 12 July 2007 (UTC)

My answer to the above questions is this: Webster’s definition for energy and force are synonyms. Would you agree?

No, I don't agree. I'm sure Webster's has different definitions for energy and force. But I can't be absolutely certain, because you haven't posted a definition of force.Rracecarr 02:00, 11 July 2007 (UTC)

Here is the on line definitions for both “force” and “energy”: http://www.m-w.com/dictionary/force and http://www.m-w.com/dictionary/energy.

There are 37 entries for “force” and 13 entries for “energy”. The problem comes when a definition for an entry dose not specify its application. An example of an application from my dictionary for “force” as a noun would be: 1. strength; 2. intensity; 3. power (human physical strength); 4. power (human intellectual strength); 5. power (control); 6. power (military); 7; organization; 8 organization (human) 9; law; 10; physics, with synonyms of strength. power, might and energy.

You are absolutely right that there are many definitions for "force," "energy," "power," etc. The ones we're concerned with here are the definitions within physics. You are also right that the typical wiki user can't be expected to know exactly what these words mean within physics. That is why this article links to other articles (like energy), which explain the concepts more fully than is appropriate in a article about the foot-pound.Rracecarr 22:05, 12 July 2007 (UTC)

In my dictionary and the on line dictionary the definitions for “foot-pound” http://www.m-w.com/dictionary/foot%20pound and “foot-poundal” http://www.m-w.com/dictionary/poundal do not specify the application. Both are specified as nouns from the on line link. Now, read down to my next response concerning precise language.

By the why you are extremely articulate. I wish I had 1/5 your ability to communicate using the written word as you.Greg Glover 02:21, 12 July 2007 (UTC)

Okay, that was a rhetorical question. The answer is that in the written language of English, both force and energy mean the same thing. But in physics force and energy are different. Webster’s implies, if you read all the definitions for: force; energy; kinetic energy; foot-poundal and foot-pound that a “foot-pound” is equal to a one pound mass being lifted off the ground one foot. I infer this as do many. Conversely, Webster’s definition for the foot-poundal implies that a foot-poundal is equal to kinetic energy by the acceleration of a one pound mass, one foot. I infer this as well. The calculation of a one pound mass being accelerated one foot is equal to .0155 foot-pound force.

Have to be a little careful here. A poundal is the amount of force required to accelerate a one-pound mass at 1 foot per second squared. (Note that the language I've just used is quite precise--it isn't clear what you mean when you say "the acceleration of a one pound mass, one foot"). A poundal is a lot less than a pound-force, because a pound-force accelerates a one-pound mass at about 32 feet per second squared. So a foot-poundal is much less energy than a foot-pound. Rracecarr 02:00, 11 July 2007 (UTC)

Okay, here is where I put all my eggs. If you don’t buy my argument, than foot-pound stands as the short version for foot-pound force. Again as you pointed out above this is where precise language must be used to convey the correct meanings.

You stated, “A poundal is a lot less than a pound-force, because a pound-force accelerates a one-pound mass at about 32 feet per second squared.” I fully concur and agree with that statement as a correct and true statement of physics (see http://en.wikipedia.org/wiki/Poundal ). I ask you, where do you read in the on line definition for the “foot-pound”, “…accelerates a one-pound mass at about 32 feet per second squared?”

A pound-force is the amount of FORCE that accelerates a one pound mass at 32 ft/sec/sec. A foot pound is a unit of ENERGY. It is equal to a pound-force multiplied by a foot.Rracecarr 22:05, 12 July 2007 (UTC)

I would say that you will not see your above quote in the on line dictionary. It certainly dose not exist in my dictionary. Also, neither my dictionary nor the on line dictionary relates the foot-pound to the foot-poundal. Meaning a foot-pound is about 32 times more force than a poundal. I would wager, you will not see your above quote in any dictionary and maybe only in specialized encyclopedia like Wikipedia, if you know what you are looking for.

A pound (force) is about 32 times more force than a poundal. A foot-pound is not a unit of force. It is a unit of energy. So it doesn't make sense to say that a foot-pound is equal to so many poundals. Rracecarr 22:05, 12 July 2007 (UTC)

The problem is the common person or folks with no basic back ground in physics, will not use the correct definitions for words like force, energy, foot, pound, pound force ect… Therefore many people will not understand what foot-pound force is.

What I am trying to convey is that many people read the definitions and all the words within the definitions as plan English. So a definition for foot-pound is read as the force it takes to move a one pound mass through the distance of one foot. The average person thinks that means that the word force is equals to the pound mass time the foot. You know and I know in physics that is not true.

The “foot-pound” or foot-pound force is the force required to accelerate a one-pound mass at 32.1739 feet per second squared. Because a foot-poundal is the force required to accelerate a one-pound mass at 1 foot per second squared. Therefore a foot-pound force is 32.1739 times more force than a foot-poundal. Would you agree?

A pound-force and a poundal are both units of force. A foot-pound and a foot-poundal are units of energy. They are not units of force. Rracecarr 22:05, 12 July 2007 (UTC)

If you can agree or some what agree with my argument that the average person dose not distingue between the application of word meanings (i.e. plan English vs. physics), than maybe we can start cleaning up this article, making it clear, concise and concrete. I would suggest a lot more information.Greg Glover 02:21, 12 July 2007 (UTC)

I think we can agree that 1 dose not equal .0155; therefore a foot-pound ( ft • lb_m ) dose not equal a foot-pound force ( ft. lb_f ). Again I will refer you back to the conversions website: http://www.digitaldutch.com/unitconverter/ . Also there are 12 types of “force” and 28 types of “energy”.

I know of no use for the unit which you have written as (ft • lb_m). It is certainly not a unit of energy. Length times force (or, equivalently, length squared times mass divided by time squared) gives you a unit energy. Length times mass does not. The unit "foot-pound" is not equal to one foot times one pound (mass), it is equal to one foot time one pound-force. It is the same thing as "foot-pound force", just a shorter (and much more common) name.

The number of forces/kinds of energy depends on the context. Fundamental physics has only 4 forces: strong, weak, electromagnetic, and gravitational. Electroweak theory combines the electromagnetic and weak forces, leaving only 3. Then the strong force can be folded in as well, leaving two. And physicists are working on a grand unified theory that will combine gravitation with the others, so that there may really only be one type of force.Rracecarr 02:00, 11 July 2007 (UTC)

The now defunct systems of measure: Absolute English and Engineering English use the pound mass (lb_m) and foot (ft) as units of measure. The foot-poundal is derived from the Absolute English System using three units of measure: pound mass (lb_m); foot (ft) and second (s). The foot-pound force is derived from the Engineering English System using four units of measure: pound force (lb_f); pound mass (lb_m); foot (ft) and second (s).

For the most part the old metric system was replaces at the General Convention of Weights and Measures in 1954 and effectively in October 1960 conference by international agreement SI was adopted making all other systems obsolete. We here in the United States have just not followed the law. I vaguely remember Congress agreed several time to phase in SI and that was back in the 1980’s.

What I am getting at is within the context of “foot-pound force” and “customary units of the United States” it is perfectly appropriate to use obsolete and defunct units of measure. The SI units of measure can be related to the English units to then bring the article into conformity with Wiki etiquette. Other wise we are mixing apples and oranges; as in “foot pounds” and foot-pound force.Greg Glover 02:21, 12 July 2007 (UTC)

Perhaps this is what is causing all the confusion: a pound can be a couple of different things: it can be a unit of force, or a unit of mass. The word pound-force is used to make it explicit that what is being used is the force unit, not the mass unit. The unit of energy is the unit of force times one foot. So whether you call it a foot POUND or a foot POUND-FORCE it is the same thing--you're just using different words for the unit of force. Rracecarr 22:05, 12 July 2007 (UTC)

Yes the problem is specifically with the word POUND. The pound can not be a unit of force. It never has been a unit of force and never will be a unit of force. The pound is a unit of mass. Mass [(m) pound mass] and force [(F) pound-force] are neither equal nor equivalent.

I am quite aware of many word or word combination used in our every day lives. Most of the time those “words” if used legitimately are defined in a dictionary or encyclopedia. The dictionary or encyclopedia can also be specific to a particular discipline.

I went by the library to see if I could find the word POUND as a synonym or as defined by the word FORCE. I was not successful. I would ask you as you have asked me, can you show a reference for the definition of the word POUND as FORCE?Greg Glover 23:32, 14 July 2007 (UTC)

http://www.unc.edu/~rowlett/units/dictP.html Look at the entry for pound. It says, "A traditional unit of mass or weight". Pound-force simply makes the 'force' definition explicit.72.10.124.205 (talk) 14:34, 7 June 2011 (UTC)

Thank you for this most stimulating and civil discourse. Again please accept my apologies.Greg Glover 01:20, 11 July 2007 (UTC)

At this point in time I have neither the time nor the energy to play the redo and undo game. I will assume that there are many others that found the Foot-pound force article in need of work. I see most of the suspect words are linked. That’s a good thing. It’s also a good thing if the folks at Wiki Physics Project do not mess with this article. They don’t ask questions, they delete, lock and block. If one of those folks starts working the article it will become instantly incoherent.Greg Glover 23:32, 14 July 2007 (UTC)

Torque

Can someone help me with the definition of “torque”, as in the torque wrench? If I put a 32 pound mass atop a 1 foot long lever that is parallel to the ground, am I exerting the force equivalent to 1 foot-pound force or 32 foot-pound force?

32 foot pounds of torque. Force (32 pounds) times distance (1 foot).

I ask this because I have used torque wrench for years when building engines. A stock big block Chevy head bolt is torque to about 65 foot-pounds. It sure doesn’t feel like I am pushing 2080 pounds.

My understanding is that a foot-pound force and the Slug (mass) are equivalent in value. A slug is equal to that acceleration of a 32.1739 pound mass, 1 foot per second squared. What am I missing here?Greg Glover 04:07, 12 July 2007 (UTC)

Your understanding is wrong. Slug is a unit of mass and foot pound is a unit of torque or moment (force x distance, akin to Newton meter in the metric world). Where the heck are you getting your information? The confusion in this article/talk page is no different to the kooks who think pound is a mass and then drop the ball when they start trying to apply laws of physics. 203.129.23.146 (talk) 13:23, 5 July 2013 (UTC)

A slug is a unit of mass (equal to about 32 pounds). A foot pound-force is a unit of energy, so cannot be equivalent to a slug. Rracecarr 22:09, 12 July 2007 (UTC)

The answer: the “torque” as in torque wench, measures force by means of weight which can be expressed by the unit of measure called the slug. “Torque” as in the torque of and engine is a measurement of energy or mechanical work which can be expressed by the unit of measure called the foot-pound force.Greg Glover 23:25, 14 July 2007 (UTC)

That is wrong. Torque of an engine is exactly the same thing as torque of a torque wrench. It is a measure of how hard something is twisting. Slug is a unit of mass. It has nothing to do with torque. The units of torque are foot-pounds (or newton meters, or any other unit of force multiplied by a unit of distance). Though the units are the same as those of energy, torque and energy are not equivalent. Energy is a scalar, and torque is a vector. Rracecarr 12:53, 16 July 2007 (UTC)

I’ve spent quit a bit of time pondering how to express the difference between the torque of an engine and the torque of a wrench. I think the best way is use the two different units of measure for torque; one is the foot pound force and the other is the foot pound. I’ll use as much plain English as possible but for the purpose of semantics I’m going to use some mathematically terms to support and codify the written language.

Mathematically the foot pound force can be expressed as the term wz. When wz is factored it yields the two units of measure Fd or as expressed in English: force through distance. Mathematically the foot pound can be expressed as wd. When wd is factored it also yields the two units of measure Fd or as expressed in English: force through distance. But make no mistake about it; the Fd from the measurement foot pound force is not the same as the Fd from the measurement foot pound.

The original term of wz express kinetic energy. This is a measurement of an object in motion. The object in motion could be spinning crank shaft, a Ferris wheel or train. Any one of these three things can have their kinetic energy measured. In the case of the spinning crank shaft or the Ferris wheel we can call the measurement foot pound force, torque or rotational kinetic energy. In the case of the train we can call the measurement foot pound force or translational kinetic energy. The measurement of spinning crankshaft or moving train is derived form the ability of the earth’s gravity to attract any object at the rate of 32.174049 ft/s² or mathematically d equals 32.174049 feet.

The term of wd express weight through a distance; torque. This is a measurement of an object at rest. The object at rest could be a head bolt. When we measure the torque of a bolt we are measuring the static resistant of a bolt as its threads wedge against the treads of the head. The bolt is motionless and the measure of its resistant to be moved is derived form distance through which the force is being applied or mathematically d equals 1 foot.

In short 32.174049 feet dose not equal 1 foot. The foot pound force dose not equal the foot pound and the torque of an engine dose not equal the torque of a wrench.Greg Glover (talk) 20:15, 15 February 2008 (UTC)

I already answered this. See above. It doesn't matter whether you're talking about a wrench or an engine--torque is the same thing and it's measured in the same way. Here are a few corrected misconceptions:

• Foot-pound and foot-pound force are not different things. They are two names for the same unit.
• An object does not have to be at rest to experience a torque.
• Torque is not the same thing as rotational kinetic energy. The former is a vector and the latter is a scalar. See above.
• The kinetic energy of an object (translational or rotational) does not depend on the force of gravity.

Hope that helps.Rracecarr (talk) 21:41, 15 February 2008 (UTC)

Good evening Rracecarr,

Well I see we agree on three of your four bullet points. For your first bullet point, I think we can add a third name for the same unit; pound-feet or is the pound-foot.

Your second bullet point, you are absolutely correct an object dose not have to be at rest to have torque. My wife’s 2006 GT Mustang proves that on the week ends.

Your number three bullet point, I think we will have to agree to disagree on that one. For a bullet to fly straight, it needs to spin. That spinning motion, perpendicular to the axis is called both rotational kinetic energy and torque. It just depends on what definition you are using. However you still use the same equation: 1/2Iw². This rotational kinetic energy is set to SI. The English form is a long and nasty equation.

As for your forth bullet point, again you are absolutely correct. Gravity has never stopped me from squeezing a trigger. The bullet still comes out the end of the muzzle. When I chronograph a load I can compute the bullet's translational kinetic energy right up to the time it stops.

Having said that the original equation of wz was derived for an experiments for air resistance by Sir Christopher Wren at Boyle’s laboratory. All good experiments need a hypothesis, some include equations. That equation was wz. The w factors to m×g/g_c. g is is local acellaration of gravity. Now, tell me and everyone else who reads this post that gravity has nothing to do with foot pound force.

For the edification of others, the factor, g_c within the w term is the dimensional constant. The dimensional constant is derived from F=ma and has a numerical value of 32.174049 here on earth. All kinetic energy equations have the dimensional constant with in it. The most commonly used numeric value is 32.163. Example: A 180 grain bullet with a muzzle velocity of 2700 feet per second; 180×2700²/ 2×32.163×7000 = 2914.17378 =2914ft-lb_f. Even the SI equation of 1/2mv² has a dimensional constant attached to it. You can't see it because it's buried within the conversions form the original units of measure of English to the units of measure in SI . There are 2.204 pounds per kilogram, 3.281 feet per meter and 1.356 Joules per foot pound force: 2.204×3.281²×1.356 = 32.17.

As always you forget there are many definitions for a word (i.e. pound and pound). The world dose not perceive everything form physics. I don't mind verbally sparing with you here on the discussion page but your edit warring is beginning to wear (ware and where) on me. Rewriting an article for punctuation, grammar and syntax is one thing. I have also tolerated your deleting and undoing of work. But to completely ignore 275 years of established science is intolerable.Greg Glover (talk) 03:53, 16 February 2008 (UTC)

ft·lbf ?

Is it correct to write the unit for foot-pound force as ft·lbf , i.e. with the second f not subscript? AJHW 15:06, 4 October 2007 (UTC)

Shigley uses lb and lbf interchangeably. 203.129.23.146 (talk) 13:32, 5 July 2013 (UTC)

In short the answer to your question is yes. Traditionally the abbreviation for "force" as in pound force (lb_f) is subscript and italicized. This is also true of the abbreviation for "mass" as in the pound mass (lb_m).

However, SI is the Wikipedia standard. Therefore, as some people may imply, SI standards should apply to non-SI units of measure or be the standard within an article that explains an old English unit of measure. Therefore the pound force is (as I have been told) abbreviated within Wikipedia as: lbf.Greg Glover 02:34, 16 October 2007 (UTC) Correction made to this post for clarification as cited below.Greg Glover (talk) 18:38, 9 December 2007 (UTC)

The SI has nothing to say about how to abbreviate foot-pound force: it's not an SI unit. Jɪmp 23:23, 6 December 2007 (UTC)

Actually SI dose have something to say about non-SI units of measure; see http://physics.nist.gov/cuu/Units/rules.html.

However, I think I understand what you are conveying. I was not clear that it is Wikipedia that is enforcing SI upon non-SI units of measure. I shall correct my post from 16 OCT 07.Greg Glover (talk) 18:30, 9 December 2007 (UTC)

The NIST is American. It doesn't control the SI. JIMp _talk·cont 05:14, 13 July 2010 (UTC)

Why use "ft-lbf" when that abbreviation is just as ambiguous as "ft-lb"? I find the designation of a ballistic (kinetic) energy unit as a "ft-lbf" or "ft-lb_f" (as for example here) unhelpful. While distinguishing between lbf and lbm is important when the unit is simply "pound", when the unit is "foot-pound" the distinction to be made is between a unit of energy and a unit of torque (force), and "ft-lbf" does not resolve that ambiguity. I have seen some sources go to the abbreviation "fpe" for "foot-pound (energy) and to "fpt" for "foot-pound (torque)". Icammd (talk) 17:00, 21 November 2016 (UTC)

lb-ft ?

In the U.S. automotive field, torque is always specified as "pound-feet" with the abbreviation "lb-ft" -- yet U.S.-specific automotive pages on Wikipedia are constantly being systemmatically edited to change any reference to "lb-ft" to the psuedo-SI designation "ft·lbf". What is the justification for this? If you open any U.S. automotive magazine, book, brochure, spec sheet, or shop manual, you will never see an engine's torque output specified as "ft·lbf"; you always see "lb-ft". Why should psuedo-SI designations take precedence over industry convention, especially in a nation which doesn't use SI anyway? I don't see any such forced standardization of units when it comes to horsepower: American-specific pages use HP, British-specific pages use BHP (brake horsepower), German-specific pages use PS, etc.... 72.79.86.182 (talk) 22:31, 27 July 2008 (UTC)

I was suspicious of the assertions made about “pound-feet”, but could only verify it today. I looked in a few of my shop manuals that have been in storage since I moved. As I supposed (the hand full I thumbed through), none of them use “lb-ft”.

I suspect the use of pound-feet only goes back 10 years or so when the manufacturers had already changed their nomenclature from cubic inch displacement to liters. The use of pound-feet is purely an advertisement ploy and the words mean nothing.Greg Glover (talk) 02:56, 15 October 2008 (UTC)

reverting good faith edits by greg glover

I have reverted all recent changes. They were all for the worse.

This one is flat wrong--we want force, not mass; this one is ungrammatical; these add incorrect/cryptic statements about the origins of energy and torque, and put horsepower under the heading Torque, which is wrong.

Greg, you seem to think that I revert you for fun, or competition, or something. That's not true at all. I just don't want false information in the article. I have spent quite a while trying to explain torque and energy to you, but you seem impervious to reason. As long as you continue to add stuff that's wrong, I will continue to take it out. It's not personal.Rracecarr (talk) 02:22, 25 February 2008 (UTC)

Mile-stone

Umm. Mile-stones, right? Are we having a unitspoofer here? —Preceding unsigned comment added by Giese (talk • contribs) 08:58, 18 September 2008 (UTC)

Conversion deleted as is an incorrect use for mile stone.Greg Glover (talk) 03:09, 15 October 2008 (UTC)

A vote to let "pound" denote a unit of force

I suggest that the page be edited such that the word "pound" denotes a unit of force - and the term "pound force" and "foot-pound force" be relegated to history. I am a BSEE who started as a physics student - and am therefore conversant with SI and its conventions.

I understand that the English system existed before SI was developed and that the English system is in use in the USA, however seeing the article titled "foot-pound force" caused me to think that the author(s) conflate(s) forces with torques - and makes me doubt everything that follows.

It might be useful to survey a bunch of engineering and physics professors (who would tend to be more rigorous than we engineers) and get their reactions.

Jsusky (talk) 20:05, 1 December 2008 (UTC)

Pound is a force, but you will never win that battle here. There are way too many armchair experts with nothing better to do that engage in edit wars and flout their precious policies. Best to just leave it and treat Wikipedia as the source of interesting but questionable information that it is. 203.129.23.146 (talk) 13:39, 5 July 2013 (UTC)

Hi Jsusky,I disagree for two reasons. The first is that Foot-pound force is history. The second is you are correct in that “author(s) conflate(s) forces with torques”.

Specifically, to address my difference with your thought let me say this. I spent many hours writing and giving the history and origins of Foot-pound force. Foot-pound force is an obsolete unit of measure here at Wikipedia. That is because Wikipedia serves the world. There are maybe two countries in the world that still use 'foot pounds”. If you are not an engineer or scientist you will have no idea that foot-pound force, foot-pounds, pound (force), pound (avoirdupois), torque (vector) and torque (energy of an internal combustion engine converted from mechanical unit of measure) are differing concepts. I believe this article and its terms should stay as "Foot-pound force" and stay as a matter of historical value. As for the second reason, your experts, “engineering and physics professors” have posted here and it would appear they disagree with each other. To me force and torque (as it appears here) are not the same thing. I tried to argue (when it was put in a year or so ago) that torque should be removed from this article. However I have lost that argument (just read a few posts above).

I would ask you if there's some way we can make the article better? Not more abbreviated. I am sure you are highly educated and have forgotten more about translational kinetic energy than I will ever know. But don't you think the average, non-engineering, educated person deserves to know whole story of foot-pound force, what it is and where it comes form? Should we not always use the complete and correct terms when writing here at Wikipedia?Greg Glover (talk) 03:54, 6 December 2008 (UTC)

P.S. Dose BSEE stand for Bachelors of Science, Electrical Engineering? I just Googled it. You might want to read the first post, seven paragraphs down signed Gene Nygaard 14:59, 6 October 2006 and see how some folks feel about abbreviations.Greg Glover (talk) 04:07, 6 December 2008 (UTC)

“exactly”

Two values that were called “exact” forms in this version of the article are not. The value was given as 1.3558179483314004 (and a decimal multiple of it). Carried out to a precision of 100 digits, the value is 1.355817948331400399999999900459473744029708329461580973429590812884271144866943359375. It will appear to be equivalent to a certain number of digits to 15 to 17 digits on most computers that simply rely upon double-precision math, which is built into most microprocessors’ math libraries. I’ve corrected this. Greg L (talk) 21:35, 24 September 2009 (UTC)

Greg, I'm not sure where you get your value but I do get 1.3558179483314004 N·m exactly. If a foot is 0.3048 m, a pound is 0.45359237 kg and standard gravity is 9.80665 m/s², this is what we get ... exactly ... or should we not be using the CGPM value for g? JIMp _talk·cont 05:27, 13 July 2010 (UTC)

Confusing Sentence

"The pound-foot, as in the ability to lift a one pound rock, one foot or tighten a bolt, one pound using a one foot long wrench is a measurement of weight or torque and should not be confused with the foot-pound which is a measurement of energy; foot-pound force." This statement is confusing both as a clarification and grammatically. However, I don't feel confident rewording it myself.

This sentence is grammatical incorrect if you don’t understand that the foot-pound, foot-pound force and pound-foot are three completely different things.

Here are the definitions:

Foot-pound is the ability to raise a one pound rock in weight, one foot in height. It is also the ability to apply one pound of weight to a one foot long wrench. Both are measurements of weight. The latter is called torque. Mathematically this is mass times distance: dm (foot-pound). Specifically the “foot” and “pound” in this definition means one foot in distance and one pound mass in weight. The term foot-pound; ftlb_m is not recognized as a unit of measure within any system.

Foot-pound force is a measurement of energy (also called "torque" but more corectlly called translational torque ) based in the standard acceleration of gravity. Mathematically this is force times distance: dF (foot- pound force). Specifically the “foot” and “pound force” in this definition means one foot in distance and one pound force in force.

Pound-foot is a made up word by an unnamed mechanical engineer in 1978 to differentiate the misused word of “foot pounds” for both dm and dF to denote torque. The pound-foot has no recognized definition outside of wikipedia. Mathematically this is mass times distance: md (pound-foot). Since m times d is the same as d times m, there is no difference between the "pound-foot" and "foot-pound".

Pound is a measurement of weight also known as the pound mass.

Pound mass is a measurement of weight also known as the pound, is defined mathimatically by a mass times standard acceleration of gravity divided by the dementional constant.; F = mg/g_c→ 1.

Pound force is a measurement of force, defined mathimatically by a 1 pound mass times standard acceleration of gravity.; F = mg→ 32.174 049.

Foot is a measurement of distance.

Torque is momentum of force. This quote is from the torque article #4 Unit , "Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings." This is probably why there are three definitions for the words “foot-pound” and “torque”.

Torque is rotational energy. Probably a miss use of the word.

Torque is translational energy; ft-lb_f. Typically this is the chemical energy of fule, converted to linear energy (translational) and transfered through a single axis, like an internal combustion engine. Defiantly a miss use of the word. The energy of the engine is created from the fuel not the rotating crankshaft.

*Please note that the grammar does not exactly translate to the math.

This is just sad, but still oh so funny :) 203.129.23.146 (talk) 13:43, 5 July 2013 (UTC)

So once again, the foot-pound is not the same as foot-pound force. And the made up word of "pound-foot" is a now uesd by advertisers to sell cars becuase poeple don't understand the corret word usage of torque.

P.S. I redid the above sentance.Greg Glover (talk) 15:20, 23 July 2010 (UTC)

I though I would add "torque" list of definitions just to make things more interestingGreg Glover (talk) 20:08, 24 July 2010 (UTC)

Concepts

Let me expand my thoughts from above.

Part of the disagreement here, is in the understanding of force, energy, torque and weight. All of these concepts are derived from F = ma. The operating word here is “concept”. Factually, as a mathematical equation; force, energy, torque and weight are all equal to “F”. Conversely; force, energy, torque and weight are all equal “ma”. Why? Because “F” is equal to “ma”.

The problem within this discussion, comes when one person is implying “F” and another person is inferring “ma”. Within the context of foot-pound force as defined by the English Engineering System force (F) is a term and energy (E_t) is the product; dF = E_t → ma = F. If you are defining force (F), force is the product; F = mg → F = ma. It is clear that “dF” (energy) can not equal “F” (force) and yet mathematically both are equal. This is because all four of these concepts are derived from F = ma. Again, mathematically force, energy, torque and weight are all equal to “F” (the same), but as a concept; force, energy, torque and weight are derived from F = ma (similar).

When each of these words are used together it is important to understand what the concept is. If the concept is energy, then torque is equal to energy; foot-pound force. If the concept is torque then torque (a vector quantity) is not equal to energy (a scalar quantity ); torque is t and energy is E_t. If the concept is weight, then weight is equal to force; (W = mg) = (F = mg). If the concept is force then force (pound force) is not equal to weight (pound mass); lb_f and lb_m.

Are you begin to see the pattern. I can do these comparisons to a total of 16 times, maybe more if I include the different systems of measure. If you get stuck on the math then everything becomes a circular argument. Each concept (subject) must be explained within its own context.

Come to think of it if I add the words; pound, slug, mass and foot you can add a another 32 comparisons. Greg Glover (talk) 20:10, 29 July 2010 (UTC)

I’d like to add this.

In light of the above math which can not be argued. Would it not be extremely rude if not out right vandalism to undo a major portion of the communities work or redirect it without discussion? Could I not redirect the Wiki pages for weight, pound, pound force and kilogram to the force page? And if not why not? It is a fact that weight, pound, pound force and kilogram are all mathematically equal to: F = mg? Why are there so many Wiki pages for the same thing?

The answer of course is, I can not. So why is such behavior happening here? Greg Glover (talk) 20:35, 29 July 2010 (UTC)

Based on your writings above ("force, energy, torque and weight are all equal to F"?), one has little confidence that you understand the concepts clearly enough to write a coherent encyclopedic article about them. There is little reason why Foot-pound (energy) should be structured significantly differently from Joule; a lot of the content you have added is completely irrelevant to the topic of the article, and borders on original research. Hqb (talk) 15:16, 30 July 2010 (UTC)

Good afternoon hqb,

I’m sorry you have little confidence in my ability to write an encyclopedic article. Read here to see why I’m absolutely correct that [force], energy , torque and [weight] are mathematical equal but as a concept... different!

I have been perplex here at Wikipedia over the last 5 years or so at the level of resistance to the proper explanation of foot-pound force. This in light of the fact that foot-pound force is from two subsystems of the long defunct Foot-pound-second systems of units (I see some one has redirected that artical to Pound force). Also this article falls under the WikiProject Measurements. Most of unsupportive comments that I have been able to identify are from folks with a heavy math or engineering background. I have no idea what your background is and have not identified anyone who might be a member of WikiProject Measurements.

If you are a member of WikiProject Measurements I look forward to your thoughts and ideas on how to expand this (article) stub in an informative and interesting manner. If some of the text is perceived as irrelevant please let me know what it is. I may agree. However, some of what is written is preemptive to keep people from removing cogent information because it is not part of a wiki standard. The most glaring example is that It is a wiki standard to ues only SI units of measure. If you don't, then poof, there goes your work. You can not use SI units of measure to explain English Engineering Units of measure. This is like using the pound to explain the pound.

And for thoughs reasons I do believe that foot-pound force can not be discuss within the “structural” context like the Joule. Maybe all the forgoing is why WikiProject Physics has not exerted its presents here.

I must say, I am at a loss as to people who don’t even use foot-pound force in their every day lives (I‘m not saying you don‘t) as I do, want to take a stub with several paragraphs and some math and reduce it to, two sentences. My greatest resentment to the two sentence explanation is this phrase, …“a force of 1 lb through a displacement of 1 ft.”

Again there is that naughty word; "pound". And yes, I know that we both know the "pound" in this context means the pound force. I just wonder, if you poled 1000 people on the street and asked them what 1 lb means, how many would say, "One pound force." I think a reasonable guess would be about 1 percent.

As for original works, I supose you could say this article is a plagiarism of Obert’s first twenty four pages. But as a stub, I think there are enough wiki links to reference most if not all of the text. As you read above in the first paragraph, I used a wiki page (Joule) as proof to my Complete understanding of this subject.

Am I a good writer? No, far from it. Do I understand the concepts of force, energy, torque and wight (spacificly foot-pound force)? Yes. I would hope we can work together rather than just erase what has existed in one form or another for many years. I would say the "Joule" page has more than two sentances. Can we also have as much information here on the "Foot-pound (energy)" page as the "Joule" page? Greg Glover (talk) 23:34, 30 July 2010 (UTC)

P.S. I would be more than willing to xerox off the first twenty four pages of Obert's University text book (570pg. total)and send it to you (or any one else for that matter). I think you would be very suprised, at how much differant the terms from 1948 are from Wikipedia 2010. At the time of the writing of the text, Edward F. Obert was an Associate Professor of Mechanical Engineering, Northwestern Technological Institute, Northwestern University. This chap knew his stuff.

P.P.S. There is no reference existing within Wikipedia that I can find, that tells where or from what reference the fps system comes from. Also, that is the same criticism to the subsystems of Gravitational, Engineering and Absolute as found in Wikipedia.

I can tell you this, on page 8 of Obert’s text, the names (including their quantities) of the subsystems are: English Absolut (mdt); Metric Absolut (mdt); English Technical (Fdt); English Engineering (Fmdt). So, has Obert correctly named the systems; considering these systems were the law in 1948? Or, have the unknown references within Wikipedia correctly named the systems here in 2010? Greg Glover (talk) 02:32, 31 July 2010 (UTC)

First, I still cannot make any real sense of your assertion that "force, energy, and torque are all mathematically equal". (I agree, of course, that "weight" is a particular instance of "force"). Energy, or work, corresponds to the application of a force through a distance, or the application of a torque through an angle. That is why, in SI, torque can be alternatively expressed with the unit "J/rad". In other words, applying one pound-foot (torque) for one radian (~57.3°) of rotation does one foot-pound (energy) of work. This is no different whether one uses imperial or metric units; it's just that the English variant (1) often doesn't make a proper formal distinction between pound (mass) (metric: kilogram) and pound (force) (metric: kilopond), and (2) lacks a single named unit for (this amount of) energy. This means that ft·lb has to do the work of both the joule and the newton-metre in SI, and the crutch of calling the torque unit "pound-feet" doesn't really carry through to the notation. It would be quite appropriate to have a section in Foot-pound (energy) entitled "Confusion with pound-foot (torque)", mirroring the one in Joule; but the general relationship between energy, torque, and power is perfectly adequately and clearly explained in Torque#Relationship between torque, power and energy (including its subsections). There is absolutely no reason to bring in general discussions of, and equations for, kinetic energy in an article about a specific unit, other than perhaps briefly in a single example.

Second, you shouldn't put any particular weight on whether someone is a "member" of WikiProject Measurement, or any other project for that matter. Membership is entirely self-declared, and conveys no special privileges or authority. It is not surprising that you encounter the most resistance to your edits from the engineering contingent; most physicists today, even American ones, use SI exclusively, and probably don't care much more about technical accuracy in Foot-pound (energy) than in Cubit. Still, it does a much better service to the general Wikipedia readership to explain the various English/imperial systems in terms of the modern SI concepts, rather than in terms of their own internal logic – just like Aristotle's term logic has been almost universally replaced by predicate logic (but the former can still be explained in terms of the latter), or Newton/Leibniz's infinitesimal calculus has been superseded by the general concept of limits. It's not that those systems were wrong per se (and they were certainly towering achievements for their time); but that focusing on the right primitive concepts gives a much better organized system. Hence the switch from "weight" to "mass" as the fundamental notion in physics (long ago), and in most engineering disciplines (somewhat later). Wikipedia articles should reflect the modern understanding of English Engineering Units as a mostly-historical system, rather than rely on a pre-SI textbook (however popular it might have been at the time) as a structuring framework. Hqb (talk) 08:04, 31 July 2010 (UTC)

What’s not to understand: ${\rm {1\ J=1\ N\cdot m}}$; the Joule is equal to the Newton meter; energy is equal to torque. I didn’t make this up. Someone else on the Joule page wrote it. If you have any disagreements please take it up with them at the Joule discussion page.

And please no response that I am conflating energy with torque. I was the first (a figure of speech) to say, “Although energy and torque are equal, they are different concepts.” Or in the words of another, “While it is dimensionally correct to express joules as newton metres or N·m, such use is discouraged by the SI authority to avoid confusion with torque.”

Also, many of the folks here at Wikipedia believe in a unit of measure called the... And in your words, “…the crutch of calling the torque unit "pound-feet"…”. Maybe you would like to critique the pound-foot (torque) page. My opinion is that the page should be removed as the notion of the “pound foot” is nonsense. But hay, that’s just my opinion. Greg Glover (talk) 15:36, 31 July 2010 (UTC)

Yes, 1 J = 1 N m; no, torque is not "equal to" energy. They are two different quantities that both happen to be measured using the same units, as the Joule page clearly says. The foot-pound (energy) page should be emphasizing the distinction (as the Joule page does), not perpetuating the confusion, as in Foot-pound (energy)#Related units of power, where you have written, ft-lb_f is the energy (also called "torque"). As for "pound-foot", using a crutch is certainly better than falling over, and the term has quite a respectable origin, going back to at least 1920 – not "made up by an unnamed mechanical engineer in 1978" as you claim above. Hqb (talk) 17:05, 31 July 2010 (UTC)

Good, now we are getting somewhere.

Please don’t feel that I’m being I’m antagonistic towards you. I agree!!! As a concept, or as wiki-linked above, there should be a distinction between energy and torque.

And yes, the term pound foot may go back to the 1920’s. The unnamed engineer was “Herb” but I can spell his last name: phonetically pronounced; Bair-i-fee. He was my fathers hunting partner and inventor of the modern day washing machine. However he will never get that credit, as he work for Westinghouse during the 20’s and 30’s. Therefore, all patents are in the name of Westinghouse. Herb was well into his 80’s on our last hunting trip that I remember in 1974. I was 15.

By the by my father was a senior engineer at Hughes Air Craft Company. He was the publisher for the Surveyor program. The pictuer of the foot pad I linked (Figure 14) is the narrow angal (croped) of Figure 12 from a report names "Surveyor 1, a Preliminary Report" (NASA SP-16). Both are found on sheet 10 of 12 form the press release. (side bar: the picture of the Surveyor in shadow at the wiki artical is upside down and transposed; sheet 4 of 12)

Also, in may hand as I write this is the first picture (200 line scan picture) taken by the Surveyor 1. Surveyor 1 landed on the moon 11:17:37 Pacific Daylight time, June 1, 1966. The first photo was transmitted 37 minuets later. The press release was written by my father at approximately 4:00am at JPL.

Just so that we are clear, in the context of this discussion, a photograph is an image exposed on photographic paper. I have the first one. A picture is an ink reproduction of a photograph on writing paper.

I don’t make claims and I will be more than glad to fax or mail the press release, photos and report to you or anyone else that is interested.

Image from Surveyor 1 of its footpad in order to study soil mechanics in preparation for the Apollo manned landings.

Surveyor 1's shadow against the lunar surface.

But that aside do you want to write a distinction between energy and torque for the Foot pound (energy) page?

Another question. You seem to be criticize my uses or writings of a known and accepted unit of measure but support a made up word; pound foot. Can you explain why?

I really have no idea why you are bringing "Herb" or your father into this; the person I'm referring to as the originator of the pound-foot (torque) is the British physics professor Arthur Mason Worthington (who also coined the slug) in a 1920 textbook; I just added the relevant citation (with a full-text link) to the pound-foot article. As for choosing between pound-foot vs. foot-pound as the canonical name for the unit of torque, I note that, in Google Books, a substantial fraction of the publications using "foot-pounds" seem to be car-repair manuals and the like, whereas those using "pound-feet" are more general engineering textbooks and journal volumes; so I think it's quite appropriate for Wikipedia to use the latter term (even though the former seems a bit more common overall), especially in the interest of emphasizing the distinction to the unit of energy. Hqb (talk) 19:35, 31 July 2010 (UTC)

Engineers don't use the word "feet" in any unit of measurement, because "feet" is plural of "foot" and a unit is always referenced singularly. Engineers generally use either "pound foot" or "foot pound". I think "pound foot" is probably more technically correct because generally mass/inertia units come first when you're writing symbols (lb.ft). I think it comes from the whole F=ma thing. Same reason why you'll probably never see Newton meter written as meter Newton. 203.129.23.146 (talk) 14:10, 5 July 2013 (UTC)

Kudos Hdb you have referenced the first known use for the word "pound foot" here at wiki. However, pound foot is not and never has been a recognized unit of measure. Therefore, it should be and is demanded by Wiki standards to be removed form all articles if cited as a unit of measure. Sorry that’s just a fact.

If pound foot has never been a recognized unit of measure, someone had better tell the FAA because they use it for limit hinge moment in Federal Airworthiness Regulation 23.415(a)(2) here: http://www.ecfr.gov/cgi-bin/text-idx?c=ecfr&rgn=div5&view=text&node=14:1.0.1.3.10&idno=14#14:1.0.1.3.10.3.71.32 203.129.23.146 (talk) 14:10, 5 July 2013 (UTC)

I am also suspicious of Worthington's use for the word slug “this we shall call…” . I think it has been around since the 1700's. However, kudos again. You are the first to reference the known use for the slug here at wiki.

Please see Worthington’s use for the word “pound” and “lb”.

Not to say, I told-you-so, but may I remind you of your own words, "Wikipedia articles should reflect the modern understanding of English Engineering Units as a mostly-historical system, rather than rely on a pre-SI textbook (however popular it might have been at the time) as a structuring framework."

And not to be too smug, but I think Worthington’s book, “Dynamics of Rotation” is about Rotational energy as defined by the first word Rigid Body on page 1 of his text. If you visit the Rotational energy page here at wiki, it specifically cites the kinetic energy equation. If you visit the kinetic energy page it cites the rigid body. Can you see the relationship in Energy and torque? Can you see the relationship of foot-pound force in energy and foot-pound force of torque? Greg Glover (talk) 20:13, 31 July 2010 (UTC)

I reference may dad and Herb as this is a discussion page for the puposes of credibility. Discussion and discord is and tends to be informal. This is not the article page. I will not engage in the type of discussion promoted by Gene Nygaard.]].

You seem to be a very reasonable and knowledgeable person. Please, I pled with you to apply your ability to write, to this stub (article). I wish not to engage in arguing with you for the sake of argument. This is counterproductive. All of the energy that has been wasted here in the last few days could have been put into this stub.

May I remind you, it was not until today at 12:15 had anyone referenced the “slug” let alone “pound foot”. Conversely I have been arguing (the technical use for the word not an emotional reaction) that the pound foot is nonsense for months. Greg Glover (talk) 20:46, 31 July 2010 (UTC)

Foot Notes and Referances

To whom it may concern, I will start foot noting as much as I can. I do not want any one to think this is my own original work. Also, thanks to Hqb I can also use Worthington as a foot note. If you look over on the Pound force page, I have put in a few foot notes already. Greg Glover (talk) 02:50, 2 August 2010 (UTC)

Edit warring

Greg, let's talk about this. I think it's important to keep this article focussed on the topic. Foot pound is a unit of energy. This needs to be stated up front (it isn't included in your first paragraph). Besides this basic definition, it is important to prominently include the conversion factors to other units of energy. Take a look at newton metre as an example of a good, short article on a similar unit. It is entirely appropriate, if sources can be found, to include a "History" and/or "Usage" section as well, where the prevalence of the unit in ballistics in the United States (among other things) can be discussed. I'm not sure examples are useful here, but they certainly need to be accurate. The discussion of the automobile which I removed is wrong. In this example, you are confusing torque and energy. Please discuss rather than replacing. Rracecarr (talk) 15:16, 2 August 2010 (UTC)

Hi Racecar. Ya, let’s talk.

Let me first point this out, “While it is dimensionally correct to express joules as newton metres or N·m, such use is discouraged by the SI authority to avoid confusion with torque.” This qoute is from the Joule page. Also as I remember, was it not you that add the text, “energy (a scalar), and torque (a vector)” several years ago (back today). Also, this kind of writing appears in the Newton meter page, “…torque but not work, or work but not torque, or both, or neither.” So if you feel like busting someone’s chops, go over to Jouel page and tell them they are confusing energy with torque.

Okay, now that that’s off my chest, How can we make this a better stub without confusing everybody?

You believe that within the first paragraph the word energy should appear. Well okay. Could you have just done that before deleting every thing else?

It took a lot of time to foot note all of the important work? Was the page not clear that the a know quantity of velocity must be present in order for an object to poses translation kinetic energy (Foot-pound force or Joule)? Is it not clear that the a know quantity of Force must be present in order to apply torque (Foot-pound force or Newton meter)?

This page is know as Foot-pound (energy). Cool, I can live with that. It makes sense. There is another page known as Pound-foot (torque). Not cool. It makes no sense.

So what do you purpose? I purpose we make the article clear with out the reader having to know university level Calculus or Physics. The obscene definition of “the force of one pound through the distance of one foot” requires no less than a solid knowledge of Algebra 1 and Physics 101. This to me as an averagely educated person is unacceptable. Now, can that type of writing be present? Sure, as long as the conversation is brought back down to the none engineer or mathematician.

The Number one problem is that people see things in different ways. Some view foot pound force from the prospective of energy, some from torque, some form the Gravitational system and others form the Engineering or Absolute System. By the way there is no such thing as the foot-pound force ether within the SI system or Absolute system; only equivalents. So when some one writes “F = mg”. What are they talking about? F = mg is absolutely correct for SI but not for the Gravitational (F = ma ) or Absolute (F = ma ) Systems. Do you see my point. The subtle difference is the base units like pound force, kilogram force, pounded and slug. Thoughs are the little details that should be present to help the read understand foot-pound force. Greg Glover (talk) 18:48, 2 August 2010 (UTC)

I don't understand everything you have said, but I'll try to respond to each thing I do understand.

I don't think anybody's chops need to be busted at Joule. There does not appear to be any confusion between energy and torque there. Possibly the sentence you quoted from could be rewritten more clearly, but it is not wrong.

Velocity need not be nonzero for energy to exist. Foot-pounds can be used to measure any form of energy, not just translational kinetic energy.

There are some valid arguments for retaining the name Pound-foot (torque) for the other article, although I agree that foot-pound is more commonly used. The unit is often abbreviated lb-ft for one thing. I don't feel strongly one way or the other.

I don't think there is anything too advanced in the current definition, but if you want to add that it is approximately the amount of energy required to lift a one-pound mass a distance of 1 foot (in Earth's gravity) that seems reasonable. Rracecarr (talk) 19:46, 2 August 2010 (UTC)

Big Question!!!

The first stanza of this article states that Foot-Pound force is, “energy transferred on applying a force of 1 pound-force (lb_f) through a displacement of 1 foot (ft). ”

So why is torque, dot product and cross produce a part of this article????????????? Greg Glover (talk) 18:54, 27 September 2011 (UTC)

Force multiplied by displacement is work, not energy. Force multiplied by an orthogonal distance is torque or moment. I dunno what the big deal is but clearly some people who are involved in these articles shouldn't be. 203.129.23.146 (talk) 14:31, 5 July 2013 (UTC)

Article Name Change AGAIN

So we have chaneged the name of this artical for the third maybe forth time!!!

Foot-Pound force is still Foot-Pound force and has been for the last 275 years.

What are you people trying to accomplish? Greg Glover (talk) 18:44, 27 September 2011 (UTC)

One person made the change without discussing with other editors. The title of "Foot-pound (unit of force)" is confusing to readers and wrong. The article is about a "unit of energy" or work, and is a dot product of a displacement ("foot") and a force ("pound force"). The unit is called "foot-pound force" and is sometimes abbreviated to "foot-pound" or "ft · lbf".

Shall we ask an admin to move it back to "Foot-pound (energy)"? That is the title it was moved to, from Foot-pound force. Since then, it stood by consensus for nearly two years.

--Hroðulf (or Hrothulf) (Talk) 15:35, 28 September 2011 (UTC)

The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the move request was: moved by NuclearWarfare (talk · contribs) (non-admin housekeeping closure). Jenks24 (talk) 22:01, 1 October 2011 (UTC)

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Foot-pound (unit of force) → Foot-pound (energy) – As the unit is a unit of energy not a unit of force. --Hroðulf (or Hrothulf) (Talk) 16:47, 28 September 2011 (UTC)

• Support: The foot-pound force is a unit of energy not force; E_k = dF = ft·lb_f. That is F does not equal dF. Greg Glover (talk) 18:20, 28 September 2011 (UTC)

The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.