User talk:Greg Glover/Archive 28 FEB 08

WikiProject Firearms

Welcome to the WikiProject Firearms. I hope you enjoy being a member.--LWF 00:57, 27 March 2007 (UTC)[reply]

Infobox Drive for the Firearms Wikiproject

Hello Greg Glover. The Firearms Wikiproject is having an infobox drive. The purpose of this is to ensure that most (if not all) of the articles within our scope have the relevant infoboxes. The start date will be May 28^th. If you choose to participate, go to our project page and pick an article under the To-do list's Infobox section or look for firearm articles that need an infobox. Before you start editing an article, please cross it out on the list so that we don't have editor's work clashing. The drive will last for five days. If you are interested, please RSVP to LWF. Thank you, the Firearms Wikiproject. --Seed 2.0 09:17, 21 May 2007 (UTC)[reply]

Foot-pound force

I am happy to discuss changes to the foot-pound force article. Though ballistics is only one subject of many that could be included, I think a ballistics section is fine. However, it is important that the information there is correct. The section I have removed twice has errors. For example, as I pointed out earlier, ft/sec^2 is not a unit of velocity (in any system of units). Rracecarr 20:08, 9 July 2007 (UTC)[reply]

No, ft/sec² (ft s^-2) is acceleration and that's not what the kinetic energy equation is relating, even if it looks like acceleration to you. Come on now, this is beginning to get silly. You understand math. Most of math is about abbreviations and the assumed coefficient, as in: a(2x-2y). The a is assumed to be a positive 1a, right? Well okay then. Within the kinetic energy equation ft/sec² [ft s^-2] is assumed to be: (ft/sec)² [(ft s^-1)²]. Or in plan English, “the velocity of the object squared”. Never has the velocity in the kinetic energy equation ever been thought of as acceleration even if you see it or read it that way.Greg Glover 17:55, 4 August 2007 (UTC)[reply]

SI added above to help with explanation.Greg Glover 02:44, 16 October 2007 (UTC)[reply]

M-1 Carbine Revert War

The M1 carbine article is currently on lock down. An administrator has requested some discussion from memeber of the Firearms Wikiproject. Can you take a look? Sf46 (talk) 19:12, 6 January 2008 (UTC)[reply]

reverts

Hi Greg,

I seem to be reverting your edits a lot lately, especially at foot pound. If this bothers you, maybe we can discuss changes on the talk pages? Rracecarr (talk) 00:52, 16 February 2008 (UTC)[reply]

My edit summaries weren't nasty, just straightforward. There are only two uses for the unit foot-pound force. One is torque, and the other is energy. There are not many different definitions. Rracecarr (talk) 15:33, 16 February 2008 (UTC)[reply]

Torque

Let me start off by explaining the meaning of torque. I understand this is rhetorical to you but as I said, I will put the burden of proof upon myself.

Torque has three definitions from my Webster's Unabridged Dictionary. We can dispense with the first and third and go straight to the second meaning. Physics: torque is moment of force; a measurement of force and the tendency to produce tension and rotation about an axis; a vector force. Okay what is torque to Greg Glover… a force. Force mathematically can be expressed as a factor using the variable F. A torque wrench measures the force of a bolt at a single moment in foot-pound force and Newton-meter.

No, torque is not force, torque is force times distance. It is the cross product of the force vector with the displacement vector from the pivot point to the point where the force is acting. Note that in this context, moment does not mean "a short amount of time", but something totally different.

Yes, torque is force and no, “torque is force cross distance”. Energy is force times distance. As for the rest of your response, I meant just that. Please hold on for a moment; "a short amount of time". I was not making a funny of a lever, perpendicular or otherwise.

No, torque is NOT force. "Times" is not a well-defined word when you're dealing with vectors. There are different ways of multiplying vectors: the cross product and dot product are both forms of multiplication. Energy is the dot product of the force vector with the distance vector, and torque is the cross product of force with distance. So both energy and torque are Force "times" Distance. With "moment" I was pointing out, that when you copied from the dictionary "Physics: torque is moment of force," the meaning is entirely different than in your statement "A torque wrench measures the force of a bolt at a single moment..."

YES torque is force and since you wrote, Times is not a well-defined word when you're dealing with vectors, you have just proven my case. However torque is also a measurement of energy. It's all in how we use the words.Greg Glover (talk) 14:50, 24 February 2008 (UTC)[reply]

Good Lord. Torque is not force. Torque is not energy. All three are completely different from each other. If anyone ever uses "torque" to mean force or energy, they are completely misusing the word. You do not have to take my word for it. Look it up. Here on Wiki, or in any physics book.

Incoherent and confusing.Greg Glover (talk) 00:44, 26 February 2008 (UTC)[reply]

So if you followed that last paragraph to its logical conclusion, I think you can see a glaring error. If force is F then why are we measuring torque by Fd? Let's hold on to that thought for a moment (no pun intended).

Again, force is F, torque is Fd.

Restated: force is F; torque is Fd; Newton-metre is Fd; kinetic energy is F times d and the Joule is F times d.

Fd is the same thing as F times d. In algebra, when you stick to variables next to each other, multiplication is implied.

Torque can be used as a descriptive word for force when addressing the physical concept of a force being applied to right angle of an axis. Torque can also be used as a descriptive word when addressing the energy output of a rotating body.

Torque does not just describe the force tending to rotate an object, it also describes the distance of that force from the pivot point. You can get more torque EITHER by adding more force OR by moving the same force further from the pivot point.

Torque is never the same thing as energy output (power). Torque times angular velocity is power.

You have misunderstood my term energy output. The energy output of a rotating body is the measured foot-pound force of that rotating body. I was not implying horsepower.

The energy (not energy output) of a rotating body (or a nonrotating body) can be measured in foot pounds. There may also be a torque on the body, which can also be measured in foot pounds, but the torque and energy are not the same thing, and neither is the energy output, by which, whether you realize it or not, you mean power. To get the power of an engine (or other rotating body) you multiply the torque (in foot pounds, if you like) by the angular velocity (rpm).

Again incoherent and confusing.Greg Glover (talk) 00:44, 26 February 2008 (UTC)[reply]

Interesting. Before, you said "By the why you are extremely articulate. I wish I had 1/5 your ability to communicate using the written word as you."

Again, I think I can say because you wrote, Times is not a well-defined word when you're dealing with vectors, you have proven my case. I ask you Rracecarr, if times is not a so well-defined word when you're dealing with vectors, then why are you so insistent that torque and foot-pound force are the exact same thing? I am baffled by your rigidity and sense of competition.Greg Glover (talk) 14:50, 24 February 2008 (UTC)[reply]

Torque and foot-pound force are not the same thing: torque is measured in foot-pounds force (a unit) in the same way that distance is measured in feet (another unit). Distance and feet are not the same thing: distance can be measured in feet; torque can be measured in foot-pounds. It is not me that's rigid, it's classical physics. There's no room for interpretation. Also, I certainly am not competing with you. I would be happy for you to edit the articles however you wanted if you didn't add misinformation. I am sorry if you think I'm a jerk. I would be nicer if you would try to learn from me instead of argue with me. I am pretty good at teaching physics, and I can help you understand, if you let me. Rracecarr (talk) 18:38, 24 February 2008 (UTC)[reply]

I’m not learning anything here. I went though this course 30 years ago. So I know, I’m not misinformed. I would think the above thoughts were better applied when using a shovel.Greg Glover (talk) 00:44, 26 February 2008 (UTC)[reply]

That's too bad. You are completely lost. You don't understand the first thing about mechanics. I have spent hours over the course of the last year trying to help you understand, but, possibly because you feel threatened, you refuse to even consider the possibility that you could learn anything from me. Well, I'm finished wasting my time. I should have given up long ago, because I have told you the same things umpteen times, without ever making a dent. Rracecarr (talk) 01:22, 26 February 2008 (UTC)[reply]

Now let me explain energy. Again this for me I know you know what energy is. Again we will dispense with definitions first through fourth and go straight to the fifth. Physics: energy (kinetic) is the capacity to do work; the energy possess by a body due to its motion. Okay what is kinetic energy to Greg Glover… well it’s both translational and rotational. Kinetic energy can be mathematically expressed as factors, using the variables F and d. An internal combustion engine’s, release of chemical energy to mechanical energy is measured in foot-pound force and Joule.

Right, but usually you are interested in the energy produced by an engine per unit time. In the US, we usually use the unit horsepower for the energy put out by an engine.

Most folks find it easier to comprehend power rather than energy; like a 100 watt bulb or 130hp Honda Civic. Cognitively, I have already been able ingrate the differences in force, energy and power. After having been thoroughly indoctrinated in the propulsive systems of lunar landing modules it becomes second nature. However, using horsepower for the energy output of an engine is wrong and I always explain that to folks who are interested in more knowledge.

The only thing wrong with using horsepower for energy output is that it should be called power, not energy. Power is energy divided by time. No one cares that a Civic can put out as much energy as a Viper if you give it three times as long. They care that the Viper has 3 times as much power. (390 hp instead of 130, say).

So what do I gain from re-explaining to myself the definitions and meanings of torque and kinetic energy; clarification. It is clear that the miscommunication for the concept of a measurement of torque lies in both the linguistically meaning of the word and the application of F as defied by Fd. Even though torque has one meaning as it pertains to physics, someone a long time ago (and it was probably and engineer) changed the meaning to a slang word for foot-pound force. Is it not true that the “foot-pounds” and the torque of an engine are the same? However F and Fd are not.

Torque is the same for everyone. Foot-pound force is a unit of torque for everyone, engineers, physicists, mechanics, everyone.

I guess we will have to agree to disagree. I guess I’m not everyone. Hopefully I’m a someone.

Since you clearly care about these things, you should not agree to disagree, you should learn (in my opinion).

If we have come this far I think we can also agree that the Newton-meter and Joule is not the same thing. To borrow from Wikipedia: the Newton-meter and Joule are dimensionally equivalent. Wikipedia goes on to say, “in a Newton-meter, the force and the distance are normal to each other (badly worded) while in a Joule, the force and distance are co-linear.” Therefore foot-pound force (torque) is not he same as foot-pound force (energy).

When a Newton-meter is used as a unit of energy, it is exactly the same as a Joule. They are just two names for the same thing. In the statement above (which is not badly worded) normal means perpendicular.

You see, perpendicular is a better way of wording the sentence.

Conceded.

And here is how. Your use of torque is the correct use and application for force; a vector. The word torque as it applies to the energy output of an internal combustion engine is slang for the words, foot-pound force. Also because someone decided to apply the words foot-pound force to both a vector quantity and a scalar quantity (because both are correctly defined as Fd) dose not make them the same. I believe SI is correct in using two separate words to describe the two different quantities. I also believe that is why there is this new name for torque call (again borrowed for Wikipedia) Pounds-feet. Mathematically this is what I understand (going back to where said to hold on for a moment). The true and correct equation for the above definition for torque is: t = r × F (I have no greek symbol for torque so I will put it and others in italics). For the purposes of describing the output of what t (torque) is equal to, we use these two units: distance (r) and force (F). When the two are put together mathematically to convey an output they are written as, N • m (SI) and ft-lb_f (English). However, the output units in dimension (N • m) and name (Newton- meter) are not conveying force through distance. The units dimension and name are conveying a quantity of force through distance; N × m = t.

Example: we don't say, “What is the Newton-meter specs for that bolt?” Or, “What is the foot-pound specs for that bolt?” We say, “What is the torque specs for that bolt?” Then we follow along the matrix and find the torque spec; X Newton-meters or X ft-lb_f. Again we are not after a measurement of force though distance we are after a measurement for the quantity force through distance: torque.

Despite what the math is telling you, forget it. Use the number one rule of mathematics; its assumed to be there. What is assumed is that we are talking about force. Torque is the force to twist or rotate something not the energy: t = r × F. Torque has no capacity to do work; 1/2Iw². Torque is force, it is quantity;1(r × F ) not a term; Fd. Torque is not “co-linear”. Energy is “co-linear”; force through distance. That is why torque is a vector quantity and rotational energy is scalar quantity, despite the fact the both torque and rotational energy convey something rotating about an axis.

I feel torque within physics has no place within the foot-pound force page. The slang words of torque, foot pound, pounds feet or any other vernacular/techno jargon you can come up with is another matter. Those slang uses can be address as such in a subsection.

My intent for the Foot-pound force page is to give a clear, concise and concrete description of what foot-pound force is and what its origins are and explained it in a way the anyone can understand.Greg Glover (talk) 23:10, 16 February 2008 (UTC)[reply]

Opinion

Foot-pound force is from the original and now defunct English systems of measure. As far as I am aware, foot-pound force only applies the small arms ballistics here in the United States. I see there is a resurgences by the auto makers (again in the United States) in using “foot pounds” as a marketing tool for the energy output of their vehicles. The foot-pound force page has no relevant value to anyone that uses SI. I am perplexed at the resistance I perceive from folks with a major math back grounds like teaches, scientist or engineer. Why are such people concerned about system of measure not used anywhere else in the world and not used by any current engineering or scientific discipline here in the U.S.? My youngest is in college and here engineering curriculum is SI based. Go figure. I would think that the Kinetic energy page would be the place where such folks would prefer to hang out.Greg Glover (talk) 23:10, 16 February 2008 (UTC)[reply]

People don't like to see misinformation go on any page, even esoteric ones like foot-pound force. And it's not as esoteric as you make out. Lots of people use foot-pounds (same thing as foot-pounds force)--as a torque unit it is used to specify the torque output of automobile engines, the tightness of bolts, etc. It has current uses as an energy unit as well as you point out, such as the energy of a cartridge.

One thing that may be confusing you is the difference between torque and horsepower in engines. Torque is measured in foot pounds (or pound feet, or foot pounds force---all the same thing). Horsepower is power--it is the torque of the engine multiplied by the rate at which the engine is spinning (rpm). Because the energy output of an engine is of interest, it may seem that people talking about foot-pounds in the context of engine output are talking about the energy of the engine, but they're not--they're talking about the torque. Rracecarr (talk) 13:48, 18 February 2008 (UTC)[reply]

Nope, I’m not confused about the difference in torque and horsepower. I’ve spent way too many hours behind a dynamometer. I don’t know the social or professional world in which you move, but back in the days when I was running an engine shop no body talked about torque (unless it were the heads or mains). Energy output is and always will be the number one objective of an engine builder. How we get that energy output and where we want to apply that energy within the torque band is dependant on the type of motor sport. Folks that spend time spouting off about horsepower are called “fans”.

People do care about the torque of an engine, though. My last motorcycle had a 600cc inline 4, which made 100 hp, which sounds like a lot for a 400 lb bike, but it only made 100 hp at 12,000 rpm. You couldn't suddenly accelerate hard by opening the throttle all the way at 4000 rpm -- the engine didn't have much low-end (when the engine is running slowly) torque. Now I have a 1000cc V twin. Its peak horsepower is no greater (maybe even a bit less) but the power is much more accessible, because it has a much flatter torque curve--more low-end torque means you get the power without having to race the engine all the time.Rracecarr (talk) 21:43, 18 February 2008 (UTC)[reply]