Talk:G-force/Archive 4
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Expert feedback
I’ve contacted someone at Honeywell. I’ve asked for a high-fidelity illustration of the actual internal workings of an accelerometer that they can release into the public domain. If I can’t find a great place on g-force for it, I’ll add it to accelerometer. More importantly, I asked Honeywell if they could make available an expert engineer in the design of accelerometers to assess the accuracy of the article. They have my contact info.
So far, the individual I spoke stated that, whereas he is not an engineer, he does, however, have a degree in physics. We carefully went through the first thee paragraphs of the intro. He said, yes, it is entirely correct. I explained the nature of the dispute (how a couple of editors are saying accelerometers to not respond to gravity and can only respond to inertial accelerations (I'm sitting at 1g but my acceleration is zero. Acceleration is m/s^2; m/s = zero.) After a pregnant pause, the Honeywell contact responded with “Oh.” But, yeah, notwithstanding all that, he finds the first three paragraphs to be accurate. He did ask about the nature of Wikipedia by asking “can’t just anyone come there and change what you wrote?” I informed him that certainly the case but these sort of issues normally aren’t a big problem.
The next step is for this physics guy at Honeywell to go hunt down an expert engineer there who is willing to review the entire article—at least the parts I wrote (I haven’t gotten to the physiological effects yet). When all is said and done, I expect to have an e-mail sent from the engineer, to anyone interested in receiving it. I will also post here, those portions of it he is willing to put into the record. I’ll ask that he respond with an e-mail that can be fully quoted into the public domain here on a Wikipedia talk venue.
Until we can get this more comprehensive review and response, I strongly encourage SBHarris and Wolfkeeper to exhibit restraint and refrain from editwarring. It would be most unfortunate if we hear from two volunteer Wikipedia editors that they have the physics all figured out and a Honeywell engineer isn’t a reliable source. Greg L (talk) 16:19, 23 January 2009 (UTC)
- I'm sorry, this simply isn't credible. And even if it was credible, it's not verifiable. We have no way to ensure that you didn't ask a dozen people and cherry picked the one you liked the sound of, so even if it was verifiable there's absolutely no way of checking for undue weight. And it's not a reliable source. It's none of these things. Why are you even bothering to waste our time? This crap doesn't fly here.- (User) Wolfkeeper (Talk) 16:56, 23 January 2009 (UTC)
- I’m sorry, your protestations are absurd and your disruption of the Wikipedia process is getting tedious. We have source after source—all highly reliable ones—that say inertial acceleration and gravitational acceleration are indistinguishable to an accelerometer, yet you refuse to all this simple fact into your belief system.
The only source you like is you. And that just doesn’t fly here because you and everything else comprising the Wikipedia community is not a WP:reliable source; only outside reliable, notable sources with direct expertise on the subject are.
The world knows that gravity is not an electromagnetic force. I am not trying to convince you and SBHarris of that. I am no longer trying to convince you of anything, nor is it necessary for me to do so. No editor can edit against consensus. This citation and proof-reading is for everyone else here so we can finally establish a clear consensus and move on—with or without you. If you persist at editing against consensus, then it is up to others to deal with you.
As I stated above, the reference is verifiable because anyone who wants to directly receive the engineer’s e-mail can do so by signing up here. Anyone who choses to read the results here, need to do nothing more. Greg L (talk) 18:37, 23 January 2009 (UTC)
- I’m sorry, your protestations are absurd and your disruption of the Wikipedia process is getting tedious. We have source after source—all highly reliable ones—that say inertial acceleration and gravitational acceleration are indistinguishable to an accelerometer, yet you refuse to all this simple fact into your belief system.
Stability for outside review
As I stated above, there is a dispute with a couple of intransigent editors over whether or not acceleration, as perceived by devices that measure acceleration, includes gravity in addition to inertial acceleration.
Two recent edits [1] were (thankfully) not contrary to this fact and were a nice attempt to clarify. However they fell short of the goal by stating that 1 g of vertical acceleration due to gravity would be added or subtracted from by riding in a car. Clearly, unless we are talking about very steep hills, a ride in a car would change the vertical component of acceleration—not unless the car has rockets and is accelerating upwards. An automobile is a poor choice to illustrate the concept of adding to gravitational acceleration.
Further, there is no need to try to make the lead of an article into a complex battle ground that address subtleties; like all good Wikipedia leads, it just needs to state the simple facts. Expanding on nuances belongs in the main body text.
Can we have stability for a day or two as we have the factual accuracy of the most fundamental point of all settled? The first paragraph:
The measurement of g-force (or g-load) is the measure of an object's acceleration—gravitational and inertial.[1] The unit of measure is informally but commonly known as the “gee” (symbol: g), /ˈdʒiː/.
…speaks straight to the heart of this issue. So too does the first paragraph of the Gravitational and inertial acceleration section, which reads as follows:
An accelerometer measures acceleration in one or more axis. It responds to both gravity and inertial acceleration.[1] If you orient a stationary, single-axis accelerometer so its measuring axis is horizontal, its output will show zero gee. Yet, if you rotate the accelerometer 90° so its axis points upwards, it will read +1 g upwards even though still stationary. If you mount the accelerometer in an automobile with its axis aligned forward with the vehicle’s direction of travel, and drive down the road at a constant speed, it will read 0 g. Yet, if you hit the brakes, it will read about −0.9 g. Accelerometers respond equally to gravity and inertial acceleration.
- ^ a b MEMSIC: ACCELEROMETER PRIMER
In order to resolve these issues, we need stability for a day or two for the review by an outside expert at a company that makes accelerometers. It is unfortunate that it has come to this, but editwarring has never been a desirable or acceptable solution to anything. It has also become clear that debate and reasoning here on this talk page is utterly futile with a couple of editors here who have got some WP:refuse to get the point-issues in the face of some clear-as-glass citations. Further, their protestations seem to have caused confusion with some of the other editors here, who aren’t quite sure what to think now. We need to establish some facts via this outside review and require some stable, crystal clear statements in the intro.
Once a consensus on the facts has been established (and a “consensus” does not mean 100% of editors in complete agreement and it never has), wen we can all move on. Greg L (talk) 18:54, 23 January 2009 (UTC)
- The idea that you need a stable version for outside review is entirely spurious. You can (and should) point anyone unfamiliar with wiki at a particular version from the history, rather than relying on the current version to be stable. That said, please be aware of the obvious: external opinions are mere opinions, and non-binding. If you can find someone with valuable comments to make, that would of course be valuable William M. Connolley (talk) 19:54, 23 January 2009 (UTC)
- The question whether or not acceleration, as perceived by devices that measure acceleration, includes gravity in addition to inertial acceleration is meaningless, like asking "Did you stop beating your wife?" to someone who has no wife or has never beaten her. You can't ask whether it includes gravity in addition to inertial acceleration because it doesn't include inertial acceleration.
- Inertial forces are forces which are artifacts of a choice of a frame of reference, as opposed to interactions with other objects. Of course, an accelerometer can't "measure" accelerations which are due to a choice of a frame of reference, as it neither knows nor cares which frame of reference you are using to describe its motion. It can only measure accelerations due to interactions with other bodies, that is, electromagnetic interactions (assuming it is too big to be affected by weak interactions and strong interactions, that is, bigger than an atomic nucleus). I won't repeat what accelerometers measure because you know it (1 g upwards when sitting on a table, 0 when free-falling, etc.), but I just want to point out that the phrases "gravitational acceleration" and "inertial acceleration" don't mean what you probably think. (SBHarris has been, as far as I can tell, completely correct about physics throughout these discussions; you just misunderstood what he said.) -- Army1987 – Deeds, not words. 20:49, 23 January 2009 (UTC)
- The primer you linked, I guess, by "inertial acceleration" means "acceleration relative to an inertial frame of reference". That is not the same meaning in the sentence "According to GR, inertial acceleration and gravitational acceleration are equivalent." It is also, I guess, using the word "gravity" meaning apparent weight (I've seen some distinction between gravity to mean that vs gravitation to mean the downwards force which according to GR is just another inertial force, though I consider that usage to be confusing.) -- Army1987 – Deeds, not words. 21:10, 23 January 2009 (UTC)
- Thanks for pointing that out. Yes, the primer is being non-standard and confusing, here, in using the phrase "inerital forces." If it means by that term "mechanical forces" it's wrong to lump gravity in with them. If it means "inertial forces" in the conventional sense of "fictitious forces that arise in accelererated frames," then acclerometers don't measure those either-- just the conventional reactions to them when they cause contact with an an object, and a resulting rective mechanical force. I was amused last night to find a physics board where somebody explained that even gravimeters don't actually measure gravity, but rather the mechanical force pushing upward on the device, because gravity-alone is not being allowed to act on it (if it were, the device would be falling and registering zero). Just so. And as you said, all these peope who think they'd score a point in pointing out that gravity can't be distinguished from an inertial (fictitious force) ala Einstein, have actually shot themselves in the foot, because inertial forces are merely products of chosen frame. You can choose any (acclerated) frame you like that produces "inertial forces," and it won't affect an accelerometer in the frame, if it is connected to nothing. It continues floating around, doing as it likes, reading zero. You'll just see it accelerate, but it won't read that acceleration. The only way the accelerometer knows what acclerated frame you've chosen, and that you want to read a fictitious force in, is if you bolt it to the frame! (bolt it to something at rest in the frame). And then, the device doesn't feel inertial force, it feels the mechanical force from the mounting bolt. That's so conventional and boring, but true. SBHarris 21:43, 23 January 2009 (UTC)
Greg L, please read the section above, where I explain how the term "gravitational acceleration" means two different and contradictory things. Given that, can you try to avoid that particular phrase in the introduction?
Better yet, would everyone be happy with this intro, with the car replaced by an upward-accelerating rocket? It seems weird to throw out my whole proposed introduction because of one easily-fixed sentence. --Steve (talk) 21:20, 23 January 2009 (UTC)
- The problem is you're all fixated on the idea that "push" is from gravity and acceleration. Any push on an accelerometer from gravity is not felt or registered. Any "push" on it from being in an acclerated vehicle like an elevator or rocket or car, is not felt, either, until it runs smack into a surface. So then, the push is from the surface, and certainly not from any inertial or fictitious force produced by the accelerated frame (or vehicle).
You'll be so much happier if you lose the notion that acclerometer are magic and can "see" gravity or fictitious forces. They can't.
If you hang an accelerometer in space, it will read zero. If you poke it with a finger, that's mechanical, and it will read that. If you bring a planet up on it (perhaps it's sitting in the planet's orbit, and the planet arrives), then it accelerates like mad toward the planet, but it doesn't read that accleration: it reads zero.
If you remove the planet, and put the accelerometer out in space in a rocket, and fire the rocket, it still reads zero as it's accelerating toward the carpeted rocket crew cabin. When it contacts that floor, suddently it reads an acceleration! Congrats. It has managed to feel the push from the floor, and read it out as a "fictitious or inertial force, suitable for artificial gravity". Whoa, check it out! But mundane. That happens on Earth, too, of course. No push from floor or ground, no reading. What's hard to understand about this? All readings are due to mechanical pushes, in all cases, everywhere. No reading from anything else. That is all. SBHarris 22:01, 23 January 2009 (UTC)
- The problem is you're all fixated on the idea that "push" is from gravity and acceleration. Any push on an accelerometer from gravity is not felt or registered. Any "push" on it from being in an acclerated vehicle like an elevator or rocket or car, is not felt, either, until it runs smack into a surface. So then, the push is from the surface, and certainly not from any inertial or fictitious force produced by the accelerated frame (or vehicle).
- No, the accelerometer reading will change due to it being affected by the gravity due to the planet being there, as my previous posts on this subject continue to tell you that still means you are wrong. You are incorrectly fixated on the accelerometer being connected to something before it is able to produce a reading. WorkingBeaver (talk) 16:20, 24 January 2009 (UTC)
- No, look if you're referring to the planet suddenly appearing, that can't happen, but consider a case where you go past a planet at high speed, one moment, no planet, planet, no planet. Would the accelerometer register it? No. You're in free-fall the whole time! The acceleration due to the planet is the same at all times for the test mass inside the accelerometer as for the casing of the accelerometer, as for the spaceship, at all times. That means that there's no forces on the mass it can register. It's only when you accelerate the casing and not directly accelerate the test mass that you get a force on the mass due to the spring having to push the test mass to make it keep up with the case that the accelerometer reads non zero.- (User) Wolfkeeper (Talk) 16:37, 24 January 2009 (UTC)
- It's like when you read old Jules Verne about travelling to the moon, he said that you get positive g near the Earth, zero g in between the Earth and Moon and negative g when you get near the moon. In fact, with Apollo it was zero-g the whole time the engines weren't on!- (User) Wolfkeeper (Talk) 16:37, 24 January 2009 (UTC)
- Firstly it doesn't matter if the planet cannot suddenly appear or not because the point still stands, that is the accelerometer is affected by changing gravity either when the direction of the gravity vector changes or when the object producing gravity appears and disappears. You both seem to contend that accelerometers are not affected by gravity, you are both wrong. In the examples I've given the acceleration due to the planet's gravity is not constant for the mass in the accelerometer, inverse square law applies, and the object is not what you mistakenly think free-fall is either. Since the object in the examples I've given is initially at rest and then moves due to gravitational forces changing upon it then obviously it is not always in your mistaken idea of what free-fall really is. I notice you keep on writing "free-fall" as if you think it magically supports your point of view, but it doesn't since the examples I give don't apply to your mistaekn idea of how it really applies. Since the assumptions you make are wrong then your conclusion is also wrong. So I vhallenge you, answer the question in the situations I have previously described instead of trying to claim they are not possible. Once you do that then you will have to concede the acceleromoter is affected by the changing gravity and then you will have to concede your point of view is wrong. WorkingBeaver (talk) 18:35, 24 January 2009 (UTC)
- If you toss an accelerometer (supposing it is not sensible enough to measure the viscous friction of air, or that you toss it in vacuum), until it hits the ground it will read 0 g. What would you expect it to read, if instead of tossing it near Earth's surface, you threw it near Jupiter's surface? I would expect the reading to be that on Earth, multiplied by the ratio between some property of Jupiter and the corresponding property of Earth. And 0 g times any finite number is 0 g. And, if you put a spaceship around the accelerometer, of course, if the accelerometer is freely floating within it (not pushed to its walls), it will behave exactly as if the spaceship weren't there. That is, it would continue to measure 0 g. How could it do anything else? -- Army1987 – Deeds, not words. 19:50, 24 January 2009 (UTC)
- Working beaver does have a point though- he's really talking about tidal forces. If the accelerometer isn't located at the center of mass of a body, then it can read g-force even in free fall. There are even differential accelerometers that can literally feel the tidal force due to your fist. But these tidal g-forces are microscopic in most circumstances and are due to the curvature of space, rather than the gradient of space. Being able to read the curvature of space in an absolute way is in no way banned in GR.- (User) Wolfkeeper (Talk) 17:43, 25 January 2009 (UTC)
- If so, he's right
(although I'd guess that very few accelerometers smaller than a house would be sensible enough to feel tidal forces except deep down inside the event horizon of some relatively small black hole). -- Army1987 – Deeds, not words. 19:33, 25 January 2009 (UTC)
- If so, he's right
- No, no, it's routinely done in satellite work in fact, and tidal forces due to the Earth are quite noticeable by astronauts aboard the ISS (free floating things move around noticeably over an orbit or so). And it's even been used on planes.- (User) Wolfkeeper (Talk) 19:51, 25 January 2009 (UTC)
- The current article is still right though; it's just a special case of deviation from freefall- you can only really understand freefall in terms of the motion of the center of mass of the whole structure, everything to one side or other of the plane of the CofM is not perfectly in freefall. I didn't want to bring it up earlier, because if you don't have the basics down, it really does just confuse matters.- (User) Wolfkeeper (Talk) 20:10, 25 January 2009 (UTC)
- I'm not writing about tidal forces. If you cannot figure out a way that an acceleromoter can be affected by gravity (and mass) then you just have to say so and I'll explain it, but you have had enough hints based on what I've previously written to figure it out. WorkingBeaver (talk) 22:28, 25 January 2009 (UTC)
- Well if you are referring to: The claim "The ONE AND only thing that makes an accelerometer register, is to poke it or push it, and that with a conventional mechanical force (like a push from the hand or ground" is completely false. Think about a spring mass accelerometer floating in free space and a very large mass unconnected and a very long distance away from the device which then approaches at a large speed very close but doesn't touch the device and proceeds to pass by the device. The mass in the device will be attracted by the gravitational attraction to the very large mass obviously with varying strength due to inverse square law. This is turn will cause the accelerometer to register varying acceleration caused by the very large mass. The device of course is free floating so nobody is pushing it or holding it. The accelerometer therefore measures that which is directly linked to the effect of the changing gravity at the device caused by the large mass. Obviously this refutes the "one and only" claim made above. WorkingBeaver (talk) 04:05, 23 January 2009 (UTC) then it's tidal. Tidal is when you have significant variations in the strength of the gravitational acceleration over the diameter of your structure due the inverse square law. That's exactly what tidal forces are. If you're referring to something else then maybe not. Tidal forces have the effect of giving you accelerations either side of the center of gravity of the vehicle.- (User) Wolfkeeper (Talk) 01:39, 26 January 2009 (UTC)
- Indeed, if WorkingBeaver isn't talking about tides, I have no idea what he's talking about. Nothing else makes an accelerometer measure anything when it whizzes around a large mass, then back into space again. When the busted-up Apollo 13 whizzed around the moon to get them approximately back on the Earth-return course, they felt no g-forces. When delicate planetary probes like the Voyagers or New Horizons perform the gravity-well maneuver by whizzing around Jupiter to give them an extra kick and velocity away from the Sun (very much like our thought experiment) it doesn't warp their delicate antennas or do anything else to them. This is entirely a free-fall maneuver, all the way. SBHarris 04:49, 26 January 2009 (UTC)
- Well if you are referring to: The claim "The ONE AND only thing that makes an accelerometer register, is to poke it or push it, and that with a conventional mechanical force (like a push from the hand or ground" is completely false. Think about a spring mass accelerometer floating in free space and a very large mass unconnected and a very long distance away from the device which then approaches at a large speed very close but doesn't touch the device and proceeds to pass by the device. The mass in the device will be attracted by the gravitational attraction to the very large mass obviously with varying strength due to inverse square law. This is turn will cause the accelerometer to register varying acceleration caused by the very large mass. The device of course is free floating so nobody is pushing it or holding it. The accelerometer therefore measures that which is directly linked to the effect of the changing gravity at the device caused by the large mass. Obviously this refutes the "one and only" claim made above. WorkingBeaver (talk) 04:05, 23 January 2009 (UTC) then it's tidal. Tidal is when you have significant variations in the strength of the gravitational acceleration over the diameter of your structure due the inverse square law. That's exactly what tidal forces are. If you're referring to something else then maybe not. Tidal forces have the effect of giving you accelerations either side of the center of gravity of the vehicle.- (User) Wolfkeeper (Talk) 01:39, 26 January 2009 (UTC)
- If you've really got no idea then perhaps you both shouldn't be commenting on the subject? WorkingBeaver (talk) 09:31, 3 February 2009 (UTC)
E vs N
I removed the Newtonian section [2] on the grounds that this article is trying to do too much. This article is about g-force, which is nothing other than acceleration, except people tend to use it in particular circumstances (planes, cars, rockets). Duplicating stuff from the forces page, or whereever it comes from, is a Bad Idea.
G restored it [3] on the grounds This article is titled g-“force”. It needs a proper discussion of the role of forces. “Editing” does not mean 100% deletion unless it is 100% wrong. I disagree: entirely correct material may be removed, and often is, simply for being aduplication of stuff elsewhere. Fine. But as an admin, your opinion is but one of many. “Community consensus” didn’t go out the window when you arrived. I find many of your edits quite valuable. I find a wholesale amputation with a chainsaw without bothering to see how other editors feel is somewhat presumptuous. I suggest that if you are going to be helpful here, that you exercise a lighter touch. Greg L (talk) 21:06, 23 January 2009 (UTC)
This [4] is also duplication. In my opinion, we should have the plane or the car, but not both William M. Connolley (talk) 20:51, 23 January 2009 (UTC)
- Clearly the better approach to wholesale deletion is to see whether the other editor who worked on that, John, agrees with you. That would be a courtesy. Army1987 too seems to have thought it valuable enough to have made some edits to it. Better yet, we might see how the many others here feel here about such a radical amputation, even though they might not have had a hand in crafting it. Greg L (talk) 21:02, 23 January 2009 (UTC)
Just fix the first sentence
The g-load is not the product of all acclerations on an object, just the mechanical ones. If you have an object in space and bring a planet near it, it will acclerate rapidly toward the planet. But it will feel no g-load whatsoever. Whether it feels a "g-force" depends on your definition of g-force. It's not really a well defined concept, as you see here. I would say, it doesn't. People in a rocket in such a situation feel zero-g. SBHarris 22:22, 23 January 2009 (UTC)
- Citation needed, including the part about “mechanical” accelerations. Greg L (talk) 22:29, 23 January 2009 (UTC)
- You want a citation that there is no g-load when you're being pulled on by a mass, and experiencing your acceleration that way? Your citation is Physics !01. Alternately, you can re-do Einstein's thought experiment with him. But his reviewers didn't ask him for a citation, either, since the fact that you find you have a g-load on you in an elevator before the elevator cable snaps (called your "weight,") but not after (even though you're acclerating), is self-evident to anybody who knows anything about physics. SBHarris 01:24, 24 January 2009 (UTC)
- I don't think the term "mechanical forces" is an adequate synonym for "every force in the world but gravity". For example, a reader might not be sure whether the force applied by an electromagnet is "mechanical" or not. Can you think of a better term? For example, I think "acceleration relative to free-fall" is a better option along these lines. --Steve (talk) 02:30, 24 January 2009 (UTC)
- Some good thinking, there, Sbyrnes; you have a point. I can for example, live with "A g-force is a measurement of the total effective "push" that one feels related to gravity and acceleration." Actually it doesn't matter what the push is related to-- what matters is whether it is uniform. It's the non-uniform push that causes the strain, and also these "g-forces" that do damage if they are too high. For our diamagnetic frog floating in the divergent field of a very powerful electromagnet, he feels no stresses or strains, presumably (or only a little bit of internal strain, and then only because he's not purely made of one substance). A magnet which was made of pure material floating in a magnetic field, or pulled by a field at some acceleration, would presumably feel no g-forces, even though it would have an acceleration in the last case. Similarly a homgenously charged drop (if such a thing existed) accelerated by an electric field. Even mechanical forces can be distributed homogenously in odd circumstances; for example bouyancy forces. An ideally homogenous and ideally neutrally bouyant jellyfish could be suspended in water and accerated at 100 g's (external acceleration of the water container) and the creature would feel NO "g-forces" and be fine. Put it on a waterbed and repeat, and you get jellyfish jelly.
Thus, I'm afraid that in terms of the stresses ("g-forces" are stresses by definition, yes?) that this article attempts to quantitate, with the term it uses, I think it's also nsufficient to define the relevent acceleration which produces g-forces MERELY as "that which deviates from free fall." Example is the jellyfish in the tank of the 100-gee rocket deviates quite substantially from freefall, but feels no g-forces. And would feel to be in freefall. That's the sense I meant by "mechanical". I also mean "mechanical strain" caused by internal force and load redistribution. That's typical of g-forces caused by external mechanical acceleration, but as in the cause of the buoyancy-distributed force on the neutrally dense creature, even that's not a careful enough definition. So this really is a hard one to define.
Material added later on re-edit-- actually the purely mechanical force must causes some kind of strain, even for our jellyfish. He/she/it would survive 1000 g's in a rocket, because jellyfish survive more than 1000 atm of pressure in the deep ocean. BUT you couldn't arbitrarily increase this to 1 million g's, because the bottom of the jellyfish would be subjected to 100,000 bar or atm of pressure (assuming we have him in a tank 1 m deep, and he's at the bottom; it's 1 bar per 10 meters of water per g), and that would produce chemical changes incompatible with life-as-we know it. By contrast, people and jelly fish falling into a supermassive blackhole (we neglect tides) in the middle of a galaxy would pass the point at which their acceleration as seen by flat space observers as having passed 1 million g's (I seen to remember "g" at the surface of a neutron star is larger than that, even), but the rocket and passengers would not feel any "g-force" stress, free falling even in such an intense field. Until they impacted the surface of the neutron star, of course. (Tide problems long before that, of course). So there's something about mechanical acceleration that produces g-force stress, or some analogy of it (pressure = energy density/volume) that can't be gotten away from. SBHarris 03:22, 24 January 2009 (UTC)
- (after e/c, dbl-indent and ignoring differential/tidal forces, since they're not necessarily applicable to the real-world concept of g-force) Well, to start with (stupid questions here) - why is "gravity" and "acceleration" conflated? Acceleration to me is actually sitting in the car while it's speeding up. That's not that same as sitting in a chair, I don't think of that as "acceleration". Whereas if you pull the chair away, I'd call what happens next acceleration. If you leave me alone on the chair, I'd call that a "force". So yes, what is this article trying to do? Franamax (talk) 03:36, 24 January 2009 (UTC)
- g-force is the vector sum of the inertial and gravitational forces
acceleratingresisted by a mass, expressed as a ratio to the same action of gravity on a mass at rest at the surface of the Earth. This can also be thought of as the "push" on an object, compared to the same push the object feels from the normal force of gravity. - I won't necessarily suggest that for the lede, but is that an accurate statement of reality? Franamax (talk) 03:57, 24 January 2009 (UTC)- Not a supid question, but a key one. The reason acceleration got all dis-entangled with dv/dt (which is what it used to be-- very straightforward) was that after relativity, we could never tell if "v" was changing without looking out the window, and even then, we weren't too sure. So now, do we have "a" or not? You can sit on a chair and say you're not accelerating, but how about if your whole room is, and the force you feel from the chair is related to that? Except for the vibration, you have to look outside the porthole to tell if the rocket containing your chair is sitting on the ground or blasting though space at 1 g.
The idea of using accelerated reference frames (like your room, where you notice everything falls to the floor) caused people to introduce the idea of funny "forces" which made them all do that, or else there would be nothing to "explain" the odd behavior. These were called "inertial forces" and "ficticious forces", and when divided by mass: F/m, they got to be regarded as inertial force fields producing inertial accelerations. Even if nothing was moving, much like the gravitational acceleration which is said to be acting on you, even though the chair keeps you from acclerating downwrd. Then it occured to Einstein one day (long before the days of rockets or the Vomit Comet, so this was a non-obvious throught) that inertial forces could not be told from gravitational forces. They both act the same way. And that's why you have to look out of your window to tell if your "gravity" is "artificial" or not.
There's no point in trying to do a vector sum of inertial forces and gravitatinal forces, because why separate them, as you're trying to do? They both need to be opposed by mechanical forces like your chair provides, if you want to be at rest in your cozy accelerated frame (wherever it is-- might be in a plane, rocket, train, rollercoaster, or at home during a large earthquake). Opposing the vector sum of these forces requires a vector mechanical force from your chair. I think that's what the article is trying to say is the g-force, but I can't tell. The problem is, I don't think the editors can even agree on what it is that we're talking about. I vote for this horrid force to be the total stressful force acting on your body, trying to rip the aorta off your heart and make your breasts sag. What say you all? SBHarris 04:26, 24 January 2009 (UTC)
- OK, I can follow all that and agree, but isn't g-force (this article) by its very definition then defined by the local reference frame? g-force is what I experience, not you the neutral observer in a different frame passing by in a relativistic spacecraft. It's my blood draining down into my feet - regardless of how you decide the equivalence of inertial and gravitational forces partition from your side, from my point of view, I started off with 1g of downward force then coughed wrong and jerked the joystick to one side whilst accidentally piloting a jet fighter. From the POV of my own heart, I now have a vector product, 1g in -Z, 4g in +X - so I experience a g-force of 4.24, do I not? And yes, as you say, it's the opposing force that causes the problem, if you would only pull my pilot seat (and the jet) away at the same time, my blood will stay in the same place it was and I won't experience a higher g-force - actually I suppose I will then experience zero g-force for a short while.
- Anyway, my vector sum terminology is an attempt to recognize that in the context of "g-force", as opposed to gravitation in general, the "force" is defined in the local reference frame, in which it's valid to consider gravitational and inertial forces as distinct. In particular, 1g is defined as the force of local gravity for an object at rest at sea level, and g-force is a ratio derived from that constant. Franamax (talk) 05:01, 24 January 2009 (UTC)
- If you're asking if g-force is frame dependent or frame independent, it's frame independent. If we hand you a weight on a spring with a ruler to measure the displacement, anyone can read that off correctly in any frame of reference. In GR it's the proper acceleration divided by g_n and again, it's frame independent.- (User) Wolfkeeper (Talk) 10:28, 24 January 2009 (UTC)
- It can't really be any other way, g-force has an LD-50, you can't really have half the people dying in one frame of reference and not another! g-force has real, physical consequences.- (User) Wolfkeeper (Talk) 10:32, 24 January 2009 (UTC)
- Not a supid question, but a key one. The reason acceleration got all dis-entangled with dv/dt (which is what it used to be-- very straightforward) was that after relativity, we could never tell if "v" was changing without looking out the window, and even then, we weren't too sure. So now, do we have "a" or not? You can sit on a chair and say you're not accelerating, but how about if your whole room is, and the force you feel from the chair is related to that? Except for the vibration, you have to look outside the porthole to tell if the rocket containing your chair is sitting on the ground or blasting though space at 1 g.
- Some good thinking, there, Sbyrnes; you have a point. I can for example, live with "A g-force is a measurement of the total effective "push" that one feels related to gravity and acceleration." Actually it doesn't matter what the push is related to-- what matters is whether it is uniform. It's the non-uniform push that causes the strain, and also these "g-forces" that do damage if they are too high. For our diamagnetic frog floating in the divergent field of a very powerful electromagnet, he feels no stresses or strains, presumably (or only a little bit of internal strain, and then only because he's not purely made of one substance). A magnet which was made of pure material floating in a magnetic field, or pulled by a field at some acceleration, would presumably feel no g-forces, even though it would have an acceleration in the last case. Similarly a homgenously charged drop (if such a thing existed) accelerated by an electric field. Even mechanical forces can be distributed homogenously in odd circumstances; for example bouyancy forces. An ideally homogenous and ideally neutrally bouyant jellyfish could be suspended in water and accerated at 100 g's (external acceleration of the water container) and the creature would feel NO "g-forces" and be fine. Put it on a waterbed and repeat, and you get jellyfish jelly.
- (e/c+o/d)Yes! Apparent weight describes it well I think. With ref to your comments just above, while I agree that the "g-force" (i.e. the gravitational force, aka sitting inside a box undergoing uniform acceleration in empty space) is frame-independent in general - in the context of making a Wikipedia article, we have to decide what exactly the article is about. Sbharris asked the same question, but may have changed on his latest addition.
- To me, the colloquial concept of g-force is necessarily in the local (rotating-Earth system) frame of reference. g-force is most commonly used to describe the forces on automobiles, jet-fighter pilots and astronauts in near-Earth boost and local zero-g. Each one of these references the local gravitational field as an implicit part of the explanation (or not - depends on how they do the vectors).
- Yes, of course you could replicate the exact same forces in another gravitational system or in hypothetical empty space, but then 1g becomes meaningless. g-force is defined by the gravitational force you feel standing at the surface of the earth - which is precisely your weight!
- So can the article proceed from that practical description, and slowly descend into equivalencies? The whole point of g-force numbers are a comparison to how much you weigh, right? Franamax (talk) 11:27, 24 January 2009 (UTC)
- Unfortunately the term 'apparent weight' is usually used to denote cases where the weight of something varies due to buoyancy; so we probably shouldn't use it. But the relation of g-force to your effective weight is real- at 2g your effective weight appears to double.- (User) Wolfkeeper (Talk) 11:40, 24 January 2009 (UTC)
- Henry Spencer claims (but I haven't seen a reference to it yet, but he's very, very rarely wrong) that G-force denotes multiples of g_n, whereas g-force denotes multiples of local gravity. that would at least make some sense of the usage that I've seen, but the chances of everyone getting it right must be vanishingly small ;-)- (User) Wolfkeeper (Talk) 11:47, 24 January 2009 (UTC)
<-outdent] With defining g-force as "measure of any acceleration which results in departure from free fall," we've made progress. I agree. That, by the way, is the quanity which accelerometers measure. It's also more or less the quantity which organisms feel, since there's no way to depart from free fall for an organism, except to mechanically nudge it. Gravitational fields and inertial-fields (whatever those are) don't do anything to move you from free-fall, and thus they don't produced g-stress, g-forces, nor affect an accelerometer. Well, what does presence of a g-field (from a mass) in the area do, if it doesn't affect accelerometers nor g-forces? All it does is provide a correction you need to tell where you're going with respect to your initial position, if you're measuring g-forces. If there's no mass and no g-field, the g-forces tell you all by themselves what your accleration from your initial position is. But with a gravitational field in the picture, you have to subtract out the g-force you spend resisting it, to tell where you're going. None of this has to do with the g-force; it's only relevent to where you're moving. In space, a 1 g-force of 1 gee tells you that you're going somewhere. Hovering over the Earth (or sitting on the ground), a g-force of 1 gee tells you that you're going nowhere. But in neither case does the grav-field affect the g-force, which is why I'm glad we got it out of the defintion. The grav-field only tells you there's a "tax" to the g-force you want to put up with, if there's a path you want to keep on, which isn't in the direction of the gravity (i.e., the direction of the mass making the grav-field). Like a linear acclerated path in a racecar: without the grav-field, your only g-force would be spent in giving you linear acclereration (racecar would be a rocket). With it added in, only some of the g-force you feel does that (accerate you linearly), and the rest is spent in resisting gravity. But gravity doesn't contribute to g-force unless you LET it, by resisting it because you don't want to GO that way. But it's your (mechanical) resistance that does this, not the poor grav-field. So blame the mechanical resistance (rocket, wheels, feet, whatever).
Now, bouyancy is interesting. Ultimately, it's a mechanical support, caused by the differential pressure on an object due to imersion in a fluid of density, in a g-field. It's basically a waterbed, but one that acts on every surface, so you never get any surface stresses, and even something as delicate as a waterballoon can take 1000-g's if it's totally immersed in a tank of water. BUT, that's not to say there's no g-force on a waterballoon under acceleration. What happens instead is that pressures start from zero at the top of the balloon (we assume the top is exactly at the water surface in the tank), and increases as you go down toward the bottom of the balloon. Any fish inside the balloon would feel the pressure, and a fish at the bottom of a 10 cm waterballon at 1000 g's would feel the same pressure as being under 100 meters of water-- trivial for most fish but not for a human diver!. Scale that up 100 times (100,000 g's) and the balloon is still intact, but organisms at the bottom of it would be subjected to the pressure at the bottom of the Challenger deep. So this is no joyride, and will eventually kill. Needless to say, none of this happens if the 100,000 g's are due to freefall acceleration. But they would happen in a rocket, because bouyancy, despite its gentleness from being evenly distributed, it still a mechanical force, and is thus felt as a mechanical pressure. So our definition of g-force as what you can feel is still correct, with the caveat that what you feel may be transformed in fluids from a 1-d "force" or a 2-D "surface pressure" into a 3-D "internal fluid pressure." SBHarris 01:48, 25 January 2009 (UTC)
"g-force" carries the connotation of a mechanical strain in a compound object, does it not?
Although we don't really talk about it, the reason that an accelerometer measures "g-force" (or, to put it another way, why we define g-force as what an accelerometer measures), is because g-forces all involve internal strain within an instrument. The reason being that the force accelerating the instrument is not evenly distributed, but the pushes and pulls are directed at the outside of the device, or selected points inside of it, like mounts, and then the force which accelerates the object as a whole is transmitted by mechanical stresses within the object, to all parts of it. An acclerometer is built to literally sense that transmitted mechanical strain, by looking at the differential strain, as one little test-body inside the accelerometer is mechanically pushed into the same acceleration as the rest of the device, and that push is measured by a strain gauge or something of the type. In fancy acclerometers like the one in the GOCE satellite, the test bodies are held in place by little electric fields, but the instrument would not work if test body experienced the same gravitational field as its surroundings in the instrument, because they it would all fall together: GOCE senses microgravity differences. Similarly, if GOCE were being accelerated between two giant capacitor plates in a non-divergent field, the test bodies in its accelerometer would feel no different field than their surroundings, so the probe would not be able to measure accelerations due to that.
Now for the same reasons, macro g-forces don't happen when an accelerometer is dropped down a uniform gravity well (neglect tides and microgravity) and that's why there is no g-force in that case (say, there's a monster black hole so large the tides can't be measured, but the accleration going into it, is still large). We can pretend that this failure to measure the acceleration of gravity with an inertially free accelerometer, is because gravity is not really a force, in Einstein's view. But that doesn't get us out of the problem, because we can accelerate objects in other ways as I've mentioned-- in fact any way which isn't mechanical will do it. Also, objects with no internal structure would seem impervious to the idea we're trying to get across here.
Here's a thought experiment which gets to the core of the problem: suppose we have an electron accelerated at a million g's in an electron gun, such as the one in an old TV image tube. Does the electron experience any "g-force" in that situation, any more than it would if it accelerated at a million g's while falling into a black hole? I would tend to say "no" in BOTH cases. I think when we made up the term "g-force" we intended to talk about the internal stresses and strains of acclerated objects, and NOT really their objective accelerations as measured by their paths. Thus, "g-force" as we coined the term, is a sort of a macro-concept about inhomogeneities that we really don't fully declare the significance of, in this article.
To sum up, and I think this should be in the LEAD, a "g-force" is not the force that acts to accelerate an object-- rather it is merely the commponent of the accelerating force which can BE SENSED by the object. And that might be quite an insignificant part of the force total force acting on the object. SBHarris 02:37, 17 June 2009 (UTC)
Negative g
The example given of negative g is misleading; we need another one. The current statement is: A classic example of negative Gs being pulled is in a dive bomber. In this case, the plane can actually accelerate toward the ground faster than gravity when transitioning into the dive. The objects (and any occupants) feel a net pull in the opposite direction of gravity. The dive bomber only experiences -g during the nose-over into the dive (the transition as stated) and not necessarily even there, during the dive its vertical speed may or may not increase. If it were to accelerate vertically at more than 10m/s2 it would soon be supersonic. In fact most divebombers, the Stuka in particular deployed airbrakes to stop it accelerating. Speed was never the point of dive bombing, it had to do with the trajectory of the bomb, the poor quality of bomb sites and the target presented to AA gunners. A more appropriate 'classic' -ve g might be either a). the top of a loop performed by an aerobatic aircraft or the Vomit comet though the latter only goes for zero not neg or b). the humpbacked bridge effect that leaves your stomach behind when a car is driven over it. c). a fully inverted roller coaster ride Ex nihil (talk) 07:30, 28 July 2009 (UTC)
This article should be removed
The whole concept of referring to an acceleration as a force is wrong. The term, g-force, is a slang term used by people who are not specialized in principles of dynamics. This article creates more confusion about understanding fundamental principles of mechanics. —Preceding unsigned comment added by RHB100 (talk • contribs) 18:18, 28 July 2009 (UTC)
- Much preceding discussion on this and of course from a physicist's point of view you are absolutely right, every high school physics student has been told that there is no such thing as g-force. From a physicist's point of view black is not a colour either, try telling that to an artist. G-force is very common, if not universal, parlance when referring to the apparent, sunjectively felt force that we experience under this kind of acceleration and the least confusing way to handle this is to place it under the common term but explain its real nature in the text. Which is sort of what has happened, although perhaps as it stands it's a bit garbled, feel free to simplify it although there have probably been too many cooks already. Ex nihil (talk) 04:22, 29 July 2009 (UTC)
- More thoughts on this. Although it is an old topic I don't think we have handled it very well in the article. How about a new definition in line 1? My suggestion is:
The g is a unit of acceleration equal to the force exerted by gravity and used to indicate the apparent force experienced by a body subjected to acceleration. Ex nihil (talk) 04:37, 29 July 2009 (UTC)
I am in complete disagreement with your statement, "The g is a unit of acceleration equal to the force exerted by gravity". You are equating an acceleration to a force. This is a horrible sin. RHB100 (talk) 20:23, 29 July 2009 (UTC)
- There is nothing "apparent" about g-forces at all. They are simply mechanical forces, like being hit by a truck. Nothing apparent about it. It produces a stress by a certain means (mechanical stress transmitted from one part of the body to the other, but NOT like the influence of gravity). It will produce an acceleration, unless opposed. It is conventient to measure g-forces in terms of the mechanical stress your body feels when you use your feet against the floor to RESIST gravity (or for that matter, when you lie on the floor, doing the same). Here gravity is resisted by mechanical force, so there is no acceleration (change in location, dx^2/dt^2). The comparison of any mechanical force to the natural mechanical force that is needed to resist gravity and stay in same location, is natural, since the two effects of these two forces are similar, and produce similar stresses and strains in the body. Perhaps a better term would be "mechanical force in units of standard g-resistive-force" but history has given us what it's given us.
Perhaps the real problem is that g-force IS a force (or rather a force per mass), but is measured in (equivalent) units of specific acceleration, a = f/m. However, it does not necessarily produce an acceleration, since it's often countered by some other force or acceleration. SBHarris 07:00, 29 July 2009 (UTC)
- There is nothing "apparent" about g-forces at all. They are simply mechanical forces, like being hit by a truck. Nothing apparent about it. It produces a stress by a certain means (mechanical stress transmitted from one part of the body to the other, but NOT like the influence of gravity). It will produce an acceleration, unless opposed. It is conventient to measure g-forces in terms of the mechanical stress your body feels when you use your feet against the floor to RESIST gravity (or for that matter, when you lie on the floor, doing the same). Here gravity is resisted by mechanical force, so there is no acceleration (change in location, dx^2/dt^2). The comparison of any mechanical force to the natural mechanical force that is needed to resist gravity and stay in same location, is natural, since the two effects of these two forces are similar, and produce similar stresses and strains in the body. Perhaps a better term would be "mechanical force in units of standard g-resistive-force" but history has given us what it's given us.
The question of whether gravity is best described as an acceleration or force, and how to describe centripetal / centrifugal forces, is an interesting and open one between physics and engineering. Both fields use both definitions in areas, and the best one can hope for is that any particular subfield is consistent and any given source is consistent.
Wikipedia is an encyclopedia, not a strict physics textbook. This is how people look for it and understand it, and a part of how people use it professionally in engineering and the sciences. As long as that's true - and it is, clearly - we have an article here under that name. That's the Wikipedia standard.
If you want to help make the article as consistent and clear as possible that's great. But we need to have an article and to contain the fundamental information for both approaches to understanding its physical significance. Georgewilliamherbert (talk) 21:01, 29 July 2009 (UTC)
- Yes. That is why I suggested emphasizing mechanical stress in the LEAD, so people will understand that it is the stress that gives this phenomenon its name, hence the use of "force" when technically it's measured in units of force/mass = acceleration. However, one does not feel (most of) the acceleration due to gravity itself. Which is why g-forces (save for tidal ones) are (confusingly) not produced by the action of gravity-alone (free-fall in a g field). The term was invented, and meant to be applied, to a force you can FEEL-- something that also can crush or kill you, if too strong. It probably remains in use, because you can always feel a "g-force" (even though it may or may not cause an acceleration, depending on whether it's opposed or not); but you can NOT always feel an acceleration (your spaceship being drawn toward a planet-- your only clue is tides). For an interesting demo: consider the frog suspended in an intense magnetic field: see magnetic levitation in the diamagnetism wiki. Does the frog feel a 1-g force, here? No. Floating in water, yes. Terminal velocity in skydiving, yes (in both cases the fluids support you with a mechanical force). But here, no, because the support doesn't come from the surface of the frog. And yet does the frog feel exactly zero-gee? No, also. The frog is not made of pure water, so the more watery tissues are here supporting the less watery bones, etc. There's a little g-force here, but not much. By comparison, if you took 50-gees while floating in water in a rocket, your aorta would be in just as much danger of ripping off your heart as if you were lying on a crash couch. All that would change is that your external support would be better distributed in water, like one of those air-beds or a Tempur-pedic mattress. SBHarris 21:22, 29 July 2009 (UTC)
A Way Forward
The g-force is deliberately not intended to be a very technical term it’s for laypeople; if we make it into a precise and complex concept we destroy the nature of it. The beauty of the g-force concept and the reason it is so popular is that it provided a simple label that could be intuitively understood by non technical people expressed in terms of how they might perceive gravitational forces acting on them should they wish to fly a loop or stand on the Moon or ride the Vomit Comet. One good reason it isn’t an SI unit is that although it’s handy to be able to tell the pilot: “don’t exceed +6g and -3g or we’ll have to ground the plane”, it isn’t much use to the aeronautical engineer in designing the airframe other than as a performance specification that has implications for how it needs to be designed, the engineer will use other means and other units in doing this. I used to fly an aircraft with +6g and -3g limitations, it had a tamper-proof recorder which was basically a spring balance and a tell-tale indicator I couldn’t zero. The g-force I, as an ignorant pilot, could understand; how that translated to the stresses in my wing spar I was happy to leave up to an engineer with maths. In this article we have made this simple, crude imprecise but extremely useful concept into a monster. I think it would improve if we could:
- settle on a logical progression of headings that tell the basic story;
- drastically simplify it;
- reduce the length by at least 30%;
- eliminate repetition and
- outlink any reference to interesting but arcane and detailed theories of Einstein, microgravity, tidal forces and mechanical stress to relevant articles that handle those concepts.
Let’s not make a sow’s ear into a silk purse. Ex nihil (talk) 08:50, 2 August 2009 (UTC)
Tidal g forces
Tidal forces are perfectly good g forces, and they are not negligable. They are why all astronauts speak of "microgravity" not "zero-gravity." Their influence can been seen in minutes in any free floating object in the space shuttle. On larger scales, of course they produce ocean tides-- not a trivial effect here on Earth. The frictional force is 3.75 terrawatts, and these forces cause the Earth's rotation to measurably slow, the moon to measurably recede in its orbit, and to be locked with one side facing us (which wasn't always the case). They prevent accretion of ice and dust that are too near a primary body into moons, and are why Saturn has rings, why Io has volcanos, and Europa liquid water under its surface. They are strong enough that near bodies such as neutron stars and small black holes, that they would cause spaghettification of anything that fell in by "free fall". The only way this would not happen is if black hole and neutron stars don't exist-- otherwise the effect is guaranteed.
Now, perhaps not all of this belongs in the LEAD, but enough of it does to make the lead correct, which right now it simply is not: it states objects in free fall experience zero-gee; wrong-- they don't if they are in "free fall" in a gravitational field.
Later on, the article goes into a great deal of detail about how gravity is a "fictious force" according to Einstein. But this is only true at a single point in an object in free fall in a gravitational field. All other points in the object (thus in any extended object) will NOT be in free fall in such a case, and acceleromters at those other points will easily detect tidal forces as mechanical strains. In fact, these g-forces on test masses in accelerometers, arrising from local g-gradients, are the only things that an instrument like GOCE, which is in a trajectory as near free-fall as humans can give it, measures. If they didn't exist, GOCE would have no data at all. SBHarris 15:31, 29 July 2009 (UTC)
- Tidal forces are pseudo-forces, as is the so-called g-force. As to Einstein's categorization of gravity as a "fictitious force", that's just Einstein's theory, and quantum theory disagrees with him on that point. If you can resolve the fundamental disagreement between the two theories, there's a Nobel prize waiting for you in Stockholm.RockyMtnGuy (talk) 15:47, 1 August 2009 (UTC)
- There's an argument about tidal forces, but none whatever about g-forces. They are certainly NOT always pseudoforoces or fictitious forces. The force of the chair against the seat of your pants is a g-force. It's mechanical-- molecule vs. molecule and electromagnetic in nature. It counts as a real force if anything does. It's the only kind of force acclerometers can "see". SBHarris 21:18, 5 August 2009 (UTC)
- Steven, please by careful- it's an acceleration, not a force. G-force is never due to a pseudo-force, it is acceleration you have after you have accounted for the accelerations due to the pseudoforces. Accelerometers are completely insensitive to pseudo-forces.- (User) Wolfkeeper (Talk) 23:02, 5 August 2009 (UTC)
- Sigh. It's not technically an acceleration, either, though it certainly is expressed in units of acceleration. Acceleration is dx^2/dt^2 = dv/dt (x and v are position and velocity vectors). When you're sitting in your chair looking at your computer as you read this, your acceleration is zero. Really, it is! Your position and velocity are not changing so don't argue with me on this (note that acceleration is frame-dependent). You do, however, feel a force on your bottom which is NOT frame-dependent. Which force you may interpret using acceleration units of F/m, since it would be doubled if your mass were doubled, but which you don't really measure as an acceleration, any more than an accelerometer does (accelerometers measure forces, and calculate accelerations, using a known test mass). We all know the physics, but it's the percieved mechanical force only (what your pants seat, or your accelerometers "sees" due to being under real mechanical strain) which is "felt". So the idea of a perceived force is really key to the idea we'd like to put over here in this article. Which has been said by several editors in several ways. G-force is a mechanical force per unit mass. Yes, it has units of acceleration, but that doesn't mean it somehow IS an acceleration. It may be, it may not be. It's a force/mass that sometimes produces a physical accleration, and other times, no. Whether a = dv/dt = 0 or not, however, "g-force" does always signify F(mechanical)/m. SBHarris 03:06, 6 August 2009 (UTC)
- No, it is actually an acceleration, which is why it has units of acceleration. It's the proper acceleration in units of 'g'. You're thinking of coordinate acceleration. I agree that it's not a coordinate acceleration.- (User) Wolfkeeper (Talk) 09:32, 6 August 2009 (UTC)
- I was sticking with Newtonian physics, where "coordinate acceleration" is the only kind there is. Yes, we are indeed talking about proper acceleration, and because of the exact congruence, this might merit a mention in the LEAD as the formal connection of this concept to relativity theory. But no more than that, because we've been getting many complaints above that the article is becoming technically jargonish, at the cost of clarity for the average reader. Let me add a link to my proposed new LEAD below. SBHarris 18:22, 6 August 2009 (UTC)
- No, it is actually an acceleration, which is why it has units of acceleration. It's the proper acceleration in units of 'g'. You're thinking of coordinate acceleration. I agree that it's not a coordinate acceleration.- (User) Wolfkeeper (Talk) 09:32, 6 August 2009 (UTC)
- Sigh. It's not technically an acceleration, either, though it certainly is expressed in units of acceleration. Acceleration is dx^2/dt^2 = dv/dt (x and v are position and velocity vectors). When you're sitting in your chair looking at your computer as you read this, your acceleration is zero. Really, it is! Your position and velocity are not changing so don't argue with me on this (note that acceleration is frame-dependent). You do, however, feel a force on your bottom which is NOT frame-dependent. Which force you may interpret using acceleration units of F/m, since it would be doubled if your mass were doubled, but which you don't really measure as an acceleration, any more than an accelerometer does (accelerometers measure forces, and calculate accelerations, using a known test mass). We all know the physics, but it's the percieved mechanical force only (what your pants seat, or your accelerometers "sees" due to being under real mechanical strain) which is "felt". So the idea of a perceived force is really key to the idea we'd like to put over here in this article. Which has been said by several editors in several ways. G-force is a mechanical force per unit mass. Yes, it has units of acceleration, but that doesn't mean it somehow IS an acceleration. It may be, it may not be. It's a force/mass that sometimes produces a physical accleration, and other times, no. Whether a = dv/dt = 0 or not, however, "g-force" does always signify F(mechanical)/m. SBHarris 03:06, 6 August 2009 (UTC)
- Steven, please by careful- it's an acceleration, not a force. G-force is never due to a pseudo-force, it is acceleration you have after you have accounted for the accelerations due to the pseudoforces. Accelerometers are completely insensitive to pseudo-forces.- (User) Wolfkeeper (Talk) 23:02, 5 August 2009 (UTC)
Layperson's suggestion
g-force is confusing because g-force is a force (if an object experiences a "g-force" then it is because of a reactive force, and it's this that we're interested in), and yet the unit "g" is a unit of acceleration. Thus, the units of "g-force" are not "g" (or not the same "g" that measures acceleration, at any rate). Thus, statements such as "The g-force acting on a stationary object resting on the Earth's surface is 1 g" are dubious without further clarification of the sense in which "g" is being used. I wonder if the article could somehow make this clear then some of the problems might go away. 01:54, 5 August 2009 (UTC). —Preceding unsigned comment added by 86.152.244.43 (talk)
- You are right, it is very unsatisfactory, I know it's technically wrong, but nobody has been able to articulate it better. Why don't you try. The following attempt links closely with articles on earth's gravity, apparent weight. Below is an attempt at an alternative Introduction, for comment. You might like to pick it apart.
New Introduction
The current introduction is probably not correct and is attracting doubt. The following is offered. For comment:
g-force /ˈdʒiː/ is the apparent weight of an object subject to both Earth's gravity and the acceleration of the object in any direction. The concept of apparent weight inherent in g-force is extremely useful to describe the subjective experience of an object undergoing acceleration, for example: the experience of a pilot flying aerobatics.
" g" here denotes the local acceleration due to gravity and refers to the acceleration that the Earth imparts to objects on or near its surface; "-force" refers to the reactive normal force created by the resistance of the ground to the object's acceleration, equal to 1 g-force, more commonly written 1 g. Any acceleration in addition to gravity is expressed in multiples of g the force being the reaction created by resistance to that acceleration. The resultant g-force acting on an object under acceleration is the combined vector result of the reactive inertial force arising from an object's acceleration plus the gravitational component.
There is no International System of Units symbol for g-force, but in practice the shorthand g is used to represent g-force meaning "times the the normal force of gravity" as in: "2g is twice the normal force of gravity".
Both the object's acceleration and gravity are measured in m/s² (metres per second per second, equivalently written as m·s−2 or N/kg) and are mathematically identical. Earth's gravity on the surface at the equator has an approximate value of 9.8 m/s², which means that, ignoring air resistance, the acceleration of an object falling freely near the Earth's surface increases by about 9.8 metres per second every second. This quantity is informally known as "little g" (contrasted with G, the gravitational constant, known as "big G"). There is a direct relationship between gravitational acceleration and the downwards weight force experienced by objects on Earth (see Conversion between weight and mass); the same relationship can be applied to the apparent weight arising from inertial acceleration in other directions.
- The g-force acting on a stationary object resting on the Earth's surface is 1 g (upwards)
- The g-force acting on an object in any weightless environment such as free-fall in a vacuum is 0 g.
- The g-force acting on an object under acceleration can be much greater than 1 g, for example, the dragster pictured right can exert a horizontal g-force of 5.3 when accelerating.
- The g-force acting on an object under acceleration downwards can be negative, for example when fully inverted in a rollercoaster loop.
Measurement of g-force is typically achieved using an accelerometer (see discussion in Measuring g-force using an accelerometer) below or in certain cases even suitably calibrated scales. Ex nihil (talk) 06:23, 5 August 2009 (UTC).
- I think the old (i.e. current) Introduction is bad, but there are also problems with this proposed New Introduction in my opinion. Rather than inventing new terminology, we should go back and review fundamental principles of engineering dynamics, primarily Newton's Laws. An understanding of these laws provides a far better understanding than is provided by the invention of new terminology. RHB100 (talk) 19:33, 5 August 2009 (UTC)
- G-force is a term that back to the 40s. It's probably not best understood in Newtonian terms, 'proper acceleration' is a general relativistic concept in SI units, and there's a deeper understanding why it's relative to the gravitational acceleration. OTOH we can't really expect the readers of this article to have that background ;-).- (User) Wolfkeeper (Talk) 23:20, 5 August 2009 (UTC)
Dubious and Lacks Source
The following statement in the first paragraph is vague, ambiguous, and dubious,"Technically, it is not a force but an acceleration, ... but is commonly treated as if it were a force." RHB100 (talk) 19:05, 5 August 2009 (UTC)
There is no source document provided for the opening statement, "The g-force (pronounced /ˈdʒiː/) experienced by an object is its acceleration relative to free-fall." The pronunciation information is more harmful than helpful. There appears to be no reputable engineering handbook which even mentions g-force. I have several books on rigid and elastic body dynamics and none of them mention g-force. RHB100 (talk) 19:05, 5 August 2009 (UTC)
- It's predominately a human factors term in aerospace/vehicle engineering. The nearest concept to it in formal science is proper acceleration which is measured in m/s^2 rather than g's, but is otherwise the same thing. It's not simply coordinate acceleration; it's the type of acceleration measured by accelerometers. The fact that you seem so adamant that it doesn't exist if anything shows that this article is needed.- (User) Wolfkeeper (Talk) 23:02, 5 August 2009 (UTC)
- Yes, I like Wolfkeeper's expression "predominately a human factors term", that's exactly it, can we get that into the intro? I also agree with RHB100, it is indeed hard to find in engineering books because it isn't an engineering term and that is why I feel uncomfortable making this article too technical. It is just a really useful, simple handle to put on what is otherwise an awkward effect to describe. Without it, instead of telling the pilot "Don't pull more than 6g" we would have to say something like "Don't induce a force normal to the airframe of more than 60m/s^2". I know it's not kosher to edit other people's discussion comments but feel free to edit my suggested Intro above if anyone thinks there is any merit in it. Ex nihil (talk) 02:05, 6 August 2009 (UTC)
Another Intro suggestion, using lots of suggestions from above, and also the present one
g-force /ˈdʒiː/ is a human factors term used in aerospace and vehicle engineering. It refers to the weight per mass of an object, and is often applied to objects undergoing dynamic motion, and not simply resting on the surface of the Earth. However, the term may also be applied to objects at rest on Earth, in which case the weight per mass is the standard “gee” or " g". This quantity denotes the mechanical force per mass imparted due to the mechanical resistance exerted by the Earth's surface to the object’s natural inertial-path motion, which would otherwise be dictated by gravity (causing the object to fall toward the center of the Earth, during which it would feel no g-force).
The weight per mass of objects is a mechanical force per mass, and thus is expressed in units of acceleration. Weight (a mechanical force) which is a function of an object’s mass always induce strains on the object, and this is why the term “force” was historically used to describe the types of accelerations which produce weight, rather than the general term “acceleration.” The term “g-force” requires that the accelerations are such that they produce forces and stresses that are tangible and measurable as weight.
Measurement of g-force is typically achieved using an accelerometer (see discussion in Measuring g-force using an accelerometer below) or in certain cases, even by suitably calibrated scales which measure weight simply. Since mechanical-type forces are generally needed to cause masses to deviate from inertial (free-fall) paths, accelerometers automatically measure the mechanical force per mass which is required to cause deviation from an inertial path (that is, acceleration relative to free-fall). In the theory of relativity this type of acceleration is called proper acceleration.
The physical-displacement acceleration of objects (i.e., their change in velocity) is a quantity which depends on the observer and is sometimes referred to as coordinate acceleration. This type of acceleration may or may not follow the “g-force” proper acceleration which is measured by an accelerometer. The accelerometer always tracks the weight of the object-- the mechanical force needed to keep the object on its path through space-time, even when observed to be at "rest." For example, the airplane sitting at rest on a runway has no acceleration relative to the ground, yet an accelerometer and weight-scale in the craft will both indicate 1-gee, since this is the acceleration relative to free fall (the mechanical force-per-mass in this example is supplied by the runway pushing up on the airplane). Similarly, objects that are in free fall, such as objects in orbit or following any inertial path, experience no g-forces (to first approximation) and their accelerometers read zero. Thus, accelerations (velocity changes with time) on objects produced by gravitational fields may be large, but these coordinate acclerations do not (by themselves) cause objects to deviate from inertial paths, and thus they do not cause proper acceleration, nor contribute to g-forces.
Very high g-forces may be produced on objects (vehicles and riders) by various situations in which mechanical means are used to force them vigorously from their natural inertial paths. These means include centrifuges, where the push is from a mechanical arm, to vehicles in which the push may come from friction or from the pressure of fluids. These g-forces may be produced in any direction, and because of their inherently stressful nature in objects, g-forces always produce stress to some extent on the human body. Large g-forces (for example produced in a vehicle crash) may be many times the g-force associated with resisting "one standard gravity," and may be fatal.
SBHarris 04:31, 6 August 2009 (UTC)
- OK, maybe something in between, some comments for what they are worth: Needs to be shorter for an intro and needs to be pitched at a lower level to address an appropriate audience, not dumb but simple and capable of immediate understanding by a Yr 10 kid; they can read on if interested. Weight per mass is, I think, a very tricky concept unless you are in the middle of doing your high school physics, as is proper acceleration and coordinate acceleration, apparent weight is at least dealt with in Wikipedia. Wiki isn't meant to be a college physics text, an appropriate level to pitch it is along the lines of the following, although I think we can do better than this: from Wisegeek: G-force refers to either the force of gravity on a particular celestial body or the force of acceleration anywhere. G-force is measured in g's, where 1 g equals the force of gravity at the Earth's surface (9.8 meters per second per second). As Einstein realized, the force of gravity and the forces of acceleration are mutually indistinguishable on the subject; a person in an opaque box experiencing a g-force would be unable to tell whether its origin lies in acceleration through space or a gravitational field unless they had some way of peeking outside the box. Analysis of g-forces are important in a variety of scientific and engineering fields, especially planetary science, astrophysics, rocket science, and the engineering of various machines such as fighter jets, race cars, and large engines. And from How Stuff Works: One g is the force of Earth's gravity -- it is this force that determines how much we weigh. At 5gs, a driver experiences a force equal to five times his weight. For instance, during a 5g turn, there are 60 to 70 pounds of force pulling his head to the side. Ex nihil (talk) 04:18, 7 August 2009 (UTC).
- For one thing, the whole article on apparent weight should be scrapped, since it involves a distinction (which involves original research and original nomenclature, so far as I can see) the essence of which is silly and unneeded. Somebody has gotten the foolish idea that there is something more "real" about the weight of an object at 1-g and at rest, than its weight in any other circumstances (in a rocket or on the moon, etc), and wants to call the 1-g weight the "weight" or "real weight" (LOL) or "actual weight" or "really-truly-true weight (foot stomp)." This is the reason we have "rest mass and invariant mass" in physics. If you want an object's weight you can measure it on a scale, or you can calculate it as the product of its rest mass and the proper acceleration in its rest frame (as measured by an accelerometer). This will provide a number which is Lorentz invariant (as both rest-mass and proper acceleration are invariant, thus so is their product). Which is not suprising because the weight of an object is the same for all observers. But there is nothing more "real" about the weight of an object on the Moon than on Earth. Why confuse the reader? SBHarris 19:31, 7 August 2009 (UTC)
- Urrrrr.. isn't apparent weight a net force whereas g-force is an acceleration? And apparent weight subsumes the concept of buoyancy, since it's a net weight; if you're floating you have a net weight of 0 Newtons. I can see where you're coming from here Steven, but it seems to me that they are different and well-known concepts.- (User) Wolfkeeper (Talk) 20:30, 7 August 2009 (UTC)
- G-force is a weight/mass, which indeed is units of acceleration. However, it isn't always a "coordinate acceleration", as has been pointed out. It is always something that results in a weight hence the idea of a force or stress. Objects supported by bouyancy (forces) are no different from hovering aircraft supported by reaction forces (like a helicopter or Harrier jump jet). They have weight, but you can't measure it anymore with a small scale underneath, because their weight is now distributed over a very wide area. That doesn't mean it's zero. It's not a net weight of zero newtons unless you cheat and don't put all your scales down all the places where it's pressing down. It's still mg, exactly as expected and the same as if it were concentrated. But the mg is spread out, since all these forces are transmitted by Newton's third law to the ground anyway, eventually.
Example: if you put a water bed on a scale and then climb on, the bed will support you, and the scale will read more, by the amount of your weight. But the same will happen if you put a tank of water on a scale, and then climb into it and float! Moreover, you could put scales covering the bottom on the tank, where they would (if sealed) together all measure the total weight of the water, via their area and the pressure on them due to depth (total water weight = pressure x area). Climb in the tank and every scale would measure a slightly higher pressure, and if you integrated the change over the tank floor area (total scale area), the increase would be the force of your weight. Same with a swimming pool, where the bottom does see your weight after you get in and float, though the change for each scale gets smaller and smaller the larger the pool. All this serves to confuse the issue, since it seems as though (say) balloons have no weight (any more than helicopters or airplanes in flight). Of course they do, but it's distributed over the Earth's surface as a tiny pressure increase, everywhere. The total weight of the atmosphere is Earth's surface area times pressure integrated over all of it. Add a balloon or airplane (or any mass) as something coming in from outer space (for example the Huygens probe parachuting into Jupiter), and that weight would add to the atmosphere, and if you had scales all over the planet, you'd see it.
Okay, so how does all this tie in to g-force? Well, bouyancy doesn't make g-forces go away; they're just easier to take due to being distributed. Whether you're supported by the pressure of water on your body while freely floating, or just on a waterbed (same idea, but not quite as uniform) you still weigh the same, and an accelerometer on your chest still reads one g. Floating in water may vaguely simulate zero-gee because the forces on your body are better spread out, but they are not zero-- they are still exactly equal to your weight. Under high accelerations in a water tank, the pressure differential along your body in the direction of acceleration would result in the same kinds of pressure you'd get from a bed (though not quite as large, because better diffused). Still, eventually with enough acceleration, you'd build up enough rho*g*h pressure to stop blood flow in your skin at "depth= h (distance from the higest part of your body to the lowest)" and give you bedsores (people in floatation beds, like sand-floatation beds, still get pressure sores that they would not if they were in orbit in a space station). Zero-gee, it's not. It's just very good support. SBHarris 00:24, 8 August 2009 (UTC)
- (Another layman's comment.) "G-force is a weight/mass, which indeed is units of acceleration." Hmm. Weight (force), mass and acceleration are fundamentally different types of quantity with fundamentally different units. As far as I understand it, g-force is understood to be a force. Numerically, it is the reactive force per unit mass experienced by a object that is accelerating and/or prevented from falling in a gravitational field, expressed as a multiple of the reactive force that the same unit mass would experience when resting on the earth's surface. (By the way, I agree with the earlier comment that the Apparent weight article is unsatisfactory. However, because it's a commonly used term I suggest reworking and trimming (probably drastically) rather than deleting.) 86.133.242.249 (talk) 19:23, 27 August 2009 (UTC).
- I can only reiterate, it's not a force, it's an acceleration; and it's measured by accelerometers - there's a big clue in that! Accelerometers measure the same type of acceleration (a proper acceleration)- the type of acceleration you actually feel, and not the coordinate acceleration (that's why you're standing on Earth, you get 1-g, even though your position is fixed).- (User) Wolfkeeper (Talk) 19:54, 27 August 2009 (UTC)
- While "g" is a unit of acceleration, in every application of the term "g-force" that I have ever come across, the quantity of importance is the reactive force "felt" by an object, due to gravity and/or acceleration. It's called "g-force" -- there's an even bigger clue in that! 21:48, 27 August 2009 (UTC). —Preceding unsigned comment added by 86.133.242.249 (talk)
- A reactive force is measured in Newtons or lbf, not multiples of 'g_0'.- (User) Wolfkeeper (Talk) 22:34, 27 August 2009 (UTC)
- Yes, I'm aware of that. Strictly speaking, the unit "g" in "a g-force of 2g", for example, would have to be interpreted differently compared to "an acceleration of 2g". Really, "a g-force of 2g" means something like "a g-force of 2 g-forces" (i.e. twice the normal force of gravity). Despite this unfortunate notational muddle, I think that when most people use the term "g-force" they are most interested in the forces that an object experiences, not acceleration per se. 86.133.242.249 (talk) 00:19, 28 August 2009 (UTC).
- It may well be that that is the case, that everyday people are thinking in terms of forces. However, in terms of units, in terms of relativity, in terms of the method of measurement, in terms of the usage in the technical community with regards centrifuges and regards acceleration of vehicles such as rockets and aircraft engaged in aerobatics and in terms of crash testing- in all these cases this is a measure of acceleration. Now you can argue that acceleration is a measure of specific force, but the deeper insight is that both gravity and g-force are an acceleration not a force. In fact I strongly think that encyclopedia articles have a responsibility to explain things as they really are, rather than just regurgitate what may be commonly supposed.- (User) Wolfkeeper (Talk) 00:29, 28 August 2009 (UTC)
- Yes, I'm aware of that. Strictly speaking, the unit "g" in "a g-force of 2g", for example, would have to be interpreted differently compared to "an acceleration of 2g". Really, "a g-force of 2g" means something like "a g-force of 2 g-forces" (i.e. twice the normal force of gravity). Despite this unfortunate notational muddle, I think that when most people use the term "g-force" they are most interested in the forces that an object experiences, not acceleration per se. 86.133.242.249 (talk) 00:19, 28 August 2009 (UTC).
- A reactive force is measured in Newtons or lbf, not multiples of 'g_0'.- (User) Wolfkeeper (Talk) 22:34, 27 August 2009 (UTC)
- While "g" is a unit of acceleration, in every application of the term "g-force" that I have ever come across, the quantity of importance is the reactive force "felt" by an object, due to gravity and/or acceleration. It's called "g-force" -- there's an even bigger clue in that! 21:48, 27 August 2009 (UTC). —Preceding unsigned comment added by 86.133.242.249 (talk)
- I can only reiterate, it's not a force, it's an acceleration; and it's measured by accelerometers - there's a big clue in that! Accelerometers measure the same type of acceleration (a proper acceleration)- the type of acceleration you actually feel, and not the coordinate acceleration (that's why you're standing on Earth, you get 1-g, even though your position is fixed).- (User) Wolfkeeper (Talk) 19:54, 27 August 2009 (UTC)
- (Another layman's comment.) "G-force is a weight/mass, which indeed is units of acceleration." Hmm. Weight (force), mass and acceleration are fundamentally different types of quantity with fundamentally different units. As far as I understand it, g-force is understood to be a force. Numerically, it is the reactive force per unit mass experienced by a object that is accelerating and/or prevented from falling in a gravitational field, expressed as a multiple of the reactive force that the same unit mass would experience when resting on the earth's surface. (By the way, I agree with the earlier comment that the Apparent weight article is unsatisfactory. However, because it's a commonly used term I suggest reworking and trimming (probably drastically) rather than deleting.) 86.133.242.249 (talk) 19:23, 27 August 2009 (UTC).
- G-force is a weight/mass, which indeed is units of acceleration. However, it isn't always a "coordinate acceleration", as has been pointed out. It is always something that results in a weight hence the idea of a force or stress. Objects supported by bouyancy (forces) are no different from hovering aircraft supported by reaction forces (like a helicopter or Harrier jump jet). They have weight, but you can't measure it anymore with a small scale underneath, because their weight is now distributed over a very wide area. That doesn't mean it's zero. It's not a net weight of zero newtons unless you cheat and don't put all your scales down all the places where it's pressing down. It's still mg, exactly as expected and the same as if it were concentrated. But the mg is spread out, since all these forces are transmitted by Newton's third law to the ground anyway, eventually.
- Urrrrr.. isn't apparent weight a net force whereas g-force is an acceleration? And apparent weight subsumes the concept of buoyancy, since it's a net weight; if you're floating you have a net weight of 0 Newtons. I can see where you're coming from here Steven, but it seems to me that they are different and well-known concepts.- (User) Wolfkeeper (Talk) 20:30, 7 August 2009 (UTC)
- For one thing, the whole article on apparent weight should be scrapped, since it involves a distinction (which involves original research and original nomenclature, so far as I can see) the essence of which is silly and unneeded. Somebody has gotten the foolish idea that there is something more "real" about the weight of an object at 1-g and at rest, than its weight in any other circumstances (in a rocket or on the moon, etc), and wants to call the 1-g weight the "weight" or "real weight" (LOL) or "actual weight" or "really-truly-true weight (foot stomp)." This is the reason we have "rest mass and invariant mass" in physics. If you want an object's weight you can measure it on a scale, or you can calculate it as the product of its rest mass and the proper acceleration in its rest frame (as measured by an accelerometer). This will provide a number which is Lorentz invariant (as both rest-mass and proper acceleration are invariant, thus so is their product). Which is not suprising because the weight of an object is the same for all observers. But there is nothing more "real" about the weight of an object on the Moon than on Earth. Why confuse the reader? SBHarris 19:31, 7 August 2009 (UTC)
<outdent>Having units of acceleration doesn’t guarantee that something “is” an acceleration in some deep philosophical sense. No more than the fact that the “action” of an object has the same dimensional units as angular momentum, makes action a kind of angular momentum.
It seems to me that you’re unfairly defining the types of acceleration you accept as “proper acceleration” to be measured in units of “g-force”, and you’re ignoring the other kinds of acceleration which have the same units, but which aren’t (necessarily) measured by accelerometers. But the more I think about it, the more I think that this isn’t necessarily so. For one thing, accelerometers measure different things, depending on their design. One type measures net force on a test mass (see relative gravimeter); another type may actually measure coordinate departure of a test mass from its inertial path (see absolute gravimeter), and thus actually directly measure proper acceleration. What’s the difference? Well the difference is that if you constructed your accelerometer so that the test mass is held in place by a Lorentz force (like our floating frog, or better yet, a floating sphere of pure water) the device will, strictly speaking, measure no acceleration (although if the mass is allowed to move, the device can measure acceleration gradients).
But a drop suspended in a magnetic field is spherical, and thus tension free, and thus (to as good approximation as you like) feels no g-force even though it may have a proper acceleration (as when it sits on the surface of the Earth). Proper acceleration is the same number as the g-force only with the correct sorts of mechanical forces acting to induce the proper acceleration.
It seems to me that (despite what I wrote about g-force being always due very simply to proper acceleration), this is not exactly so. The whole point we’re trying to attain with the idea of “g-force” is that it’s always felt and measured as a mechanical force (per mass), and therefore this mechanical force always induces a mechanical strain. The strain-making type of acceleration is what we’re making reference to, in “g-force.” We’re not interested in kinds of forces, or in accelerations that may be due to them. And our accelerometers sometimes measure one kind of reactive force on a test mass (this is the force/mass due to g-force “acceleration”, if they use springs), and other times another type of force (some type of a restoring EM force) which produces no g-force in the test mass. And in yet other cases, an absolute gravimeter measures test mass motion directly, which is proper acceleration directly, and from which a sum of non-gravitational forces acting on the object to move it away from inertial behavior may be inferred, but is certainly not directly measured. So our problem is that coordinate acceleration, proper acceleration, and g-force “acceleration” are three different things, and while they may overlap or be the same in some situations, in others they are not, so no two are identical.
To summarize: I don’t think there can be said to be a g-force accleration on an electron suspended in an electric field and gravity field, which cancel. I don’t think the concept was intended to cover that situation. A proper acceleration, yes. A g-force accelation, no. SBHarris 01:09, 28 August 2009 (UTC)
- [Response to Wolfkeeper's last post]. Echoing some of the comments that others have made above, I get the impression that "g-force" is a slightly informal term that wouldn't be much used by physicists interested in thorough conceptual precision. I think the "everyday people" meaning might actually be the meaning (predominantly, at least). 86.133.242.249 (talk) 01:17, 28 August 2009 (UTC).
- Yes to the above. G-force doesn't belong to engineers; engineers work with mass, acceleration, stresses and strains but none of that tells us what it feels like. Language evolves to do a job; the term g-force arose to describe the human experience not to do the science behind it. Felt experience is indistinguishable from force, everybody intuitively understands that, they just need a label to describe what they actually live through. It was never needed to do the engineering and by describing it up front primarily in engineering terms we destroy its simple power to communicate the original idea. Every 12 year old boy knows exactly what g-force is until they read our intro on it, then they're mightily confused. Let's get this out of the physics lab and into the world we experience. Ex nihil 01:26, 28 August 2009 (UTC)
- Nah. Unfortunately this isn't physics versus common understanding, this is physics versus wrong. G-force is not, and has never been, a measure of force.- (User) Wolfkeeper (Talk) 02:09, 28 August 2009 (UTC)
- The main reason that physicists avoid the term is because it's a misnomer; and they understand that this is so.- (User) Wolfkeeper (Talk) 02:09, 28 August 2009 (UTC)
- It's a misnomer if you want it to measure acceleration. For people who are happy that "g-force" measures force, it isn't. (The "misnomer" is, in fact, measuring g-forces in units of "g" -- or, at least, expecting this "g" to be the same unit as is used to measure acceleration.) 86.133.242.249 (talk) 02:44, 28 August 2009 (UTC).
- Thanks for elegantly showing why physicists avoid the term.- (User) Wolfkeeper (Talk) 14:29, 28 August 2009 (UTC)
- With all due respect Steven, yes, an electron being suspended against gravity, say with a magnetic field, is accelerating in the sense of g-force, and will emit cyclotron radiation due to this acceleration. It really does feel the acceleration. (This would be a violation of the equivalence principle otherwise)- (User) Wolfkeeper (Talk) 02:09, 28 August 2009 (UTC)
- No. Look at the equations for radiation reaction (as you'd see in a linac) and (non-relatistic) cyclotron radiation, and you see they both depend on the jerk (da/dt). Since acceleration is constant at the Earth's surface, there is no jerk and no radiation, and the equivalence principle is not operative, since constantly accelerated electrons (constant positionally-accelerated electrons) don't radiate, either. Where would the energy for it come from, anyway, if it happened from just sitting still in a balanced g and E field? Do you think the Earth would radiate and lose mass? And no, before you suggest it, there is no Hawking/Unruh effect radiation from low g environments either, since no Rindler horizon is visible to radiate to. SBHarris 22:32, 28 August 2009 (UTC)
- In the case of the magnetic field the energy would come from the motion of the electron in the magnetic field.- (User) Wolfkeeper (Talk) 22:48, 28 August 2009 (UTC)
- I'm not particularly aware that cyclotron radiation is due to the jerk; so far as I am know you get cyclotron radiation even in circular motions which have no jerk.- (User) Wolfkeeper (Talk) 22:48, 28 August 2009 (UTC)
- See Cyclotron radiation and Abraham-Lorentz force (the radiation reaction force). SBHarris 03:18, 29 August 2009 (UTC)
- The cyclotron equation given there is not dependent on jerk; it's proportional to V^2 which in turn is proportional to the acceleration.- (User) Wolfkeeper (Talk) 12:27, 29 August 2009 (UTC)
- Sorry for the delay. Yes, the cyclotron radiation there is given in terms of V^2, but that's only true if you're not in the accelerating frame, and it doesn't take into account the radiation reaction force (which in a real cyclotron would cause V to be not-constant, as radiation would induce a retarding force in the direction of the electron motion, and thus cause an Euler force which would make it lose momentum and kinetic energy and spiral outward). The correct equation is in the cyclotron section of special relativity: http://wiki.riteme.site/w/index.php?title=Special_relativity&action=edit§ion=16. So the other is an equation that doesn't take this effect into account, and that's your missing jerk, which accounts entirely for the work of radiation, and without which there is NO radiation. In the constant angular-velocity accelerated frame where you must feel a "fictious" centrifugal force if you go at constant V in a circle, you'll see the cyclotron electron at a much smaller V, for sure, but not a ZERO V. Or a zero a. For keeping up with a radiating electron also requires you to Euler accelerate tangentially to its circular motion, which gives you a jerk-like dw/dt component, and so we preserve the notion that any time you see an electron radiate (to you), you see work being done on it, and an acceleration of it, RELATIVE TO YOU. If you put yourself in the doubly-accelerated frame where the electron is always at rest, all you see is its static field-- no radiation. Radiation is Lorentz-invariant across inertial frames, not invariant across any accelerated frames. Anybody always keeping up with a charge never sees radiation from it. The only equivalence principle involved is that I wouldn't see an electron radiate in an frame where it's at rest for me-- whether we're both in the same inertial frame, trickily accelerated frame, or whatever. And I don't see it when we're both in the same accelerated frame, in a g-field.
Now, obviously an electron hanging in space will look like it radiates to me if I shake myself back and forth in the static field, with an acceleration to-and-fro. But the work in that case comes from work done on ME to make me do that silly motion, not the electron. The work to make radiation (which may appear in one frame but not another, if the frames are accelerated relative to each other)-- that work has to come from somewhere, and that work involves a force (mass*acceleration) acting through a distance. An electron sitting still in a g field is not traversing a distance. SBHarris 03:52, 19 September 2009 (UTC)
- Whether there's radiation or not, there's definitely is a Lorentz force acting on it from the point of view of all frames of reference, and this causes an acceleration.- (User) Wolfkeeper (Talk) 02:08, 23 September 2009 (UTC)
- Surely! But this Lorentz force is not a constant in an electron moving freely in a B field because the v in v X B isn't constant, and neither is the acceleration from it. To the approximation that it is constant, if you make that your only acceleration, you see no radiation (example, an electron in a centrifuge held down, and NOT allowed to move inertially in a magnetic field, would not radiate to somebody sitting in the centrifuge next to it and subject to the same "g-force" (actually here a centrifugal acceleration). But an electron spiralling in a magnetic field would radiate to anybody not moving precisely as it does. It's the retarding radiation reaction force and retarding force (which is at right angles to the centripetal and centrifugal forces) which is associated with the radiation. Any time you replace the total force vector with gravity alone and sit next the electron feeling the gravity, no radiation do you see.SBHarris 02:31, 23 September 2009 (UTC)
- Nevertheless there's still an acceleration, just as there's an acceleration due to an electrostatic charge acting on an electron also. In fact an electron held by a charged set of plates is one form of accelerometer if you control/measure the electron position. I don't see that you can argue that it's not accelerating. The same setup would work on the surface of the Earth or in a 1g accelerating rocket in exactly the same way.- (User) Wolfkeeper (Talk) 03:11, 23 September 2009 (UTC)
- Surely! But this Lorentz force is not a constant in an electron moving freely in a B field because the v in v X B isn't constant, and neither is the acceleration from it. To the approximation that it is constant, if you make that your only acceleration, you see no radiation (example, an electron in a centrifuge held down, and NOT allowed to move inertially in a magnetic field, would not radiate to somebody sitting in the centrifuge next to it and subject to the same "g-force" (actually here a centrifugal acceleration). But an electron spiralling in a magnetic field would radiate to anybody not moving precisely as it does. It's the retarding radiation reaction force and retarding force (which is at right angles to the centripetal and centrifugal forces) which is associated with the radiation. Any time you replace the total force vector with gravity alone and sit next the electron feeling the gravity, no radiation do you see.SBHarris 02:31, 23 September 2009 (UTC)
- Whether there's radiation or not, there's definitely is a Lorentz force acting on it from the point of view of all frames of reference, and this causes an acceleration.- (User) Wolfkeeper (Talk) 02:08, 23 September 2009 (UTC)
- Sorry for the delay. Yes, the cyclotron radiation there is given in terms of V^2, but that's only true if you're not in the accelerating frame, and it doesn't take into account the radiation reaction force (which in a real cyclotron would cause V to be not-constant, as radiation would induce a retarding force in the direction of the electron motion, and thus cause an Euler force which would make it lose momentum and kinetic energy and spiral outward). The correct equation is in the cyclotron section of special relativity: http://wiki.riteme.site/w/index.php?title=Special_relativity&action=edit§ion=16. So the other is an equation that doesn't take this effect into account, and that's your missing jerk, which accounts entirely for the work of radiation, and without which there is NO radiation. In the constant angular-velocity accelerated frame where you must feel a "fictious" centrifugal force if you go at constant V in a circle, you'll see the cyclotron electron at a much smaller V, for sure, but not a ZERO V. Or a zero a. For keeping up with a radiating electron also requires you to Euler accelerate tangentially to its circular motion, which gives you a jerk-like dw/dt component, and so we preserve the notion that any time you see an electron radiate (to you), you see work being done on it, and an acceleration of it, RELATIVE TO YOU. If you put yourself in the doubly-accelerated frame where the electron is always at rest, all you see is its static field-- no radiation. Radiation is Lorentz-invariant across inertial frames, not invariant across any accelerated frames. Anybody always keeping up with a charge never sees radiation from it. The only equivalence principle involved is that I wouldn't see an electron radiate in an frame where it's at rest for me-- whether we're both in the same inertial frame, trickily accelerated frame, or whatever. And I don't see it when we're both in the same accelerated frame, in a g-field.
- The cyclotron equation given there is not dependent on jerk; it's proportional to V^2 which in turn is proportional to the acceleration.- (User) Wolfkeeper (Talk) 12:27, 29 August 2009 (UTC)
- See Cyclotron radiation and Abraham-Lorentz force (the radiation reaction force). SBHarris 03:18, 29 August 2009 (UTC)
- No. Look at the equations for radiation reaction (as you'd see in a linac) and (non-relatistic) cyclotron radiation, and you see they both depend on the jerk (da/dt). Since acceleration is constant at the Earth's surface, there is no jerk and no radiation, and the equivalence principle is not operative, since constantly accelerated electrons (constant positionally-accelerated electrons) don't radiate, either. Where would the energy for it come from, anyway, if it happened from just sitting still in a balanced g and E field? Do you think the Earth would radiate and lose mass? And no, before you suggest it, there is no Hawking/Unruh effect radiation from low g environments either, since no Rindler horizon is visible to radiate to. SBHarris 22:32, 28 August 2009 (UTC)
Oh, another comment on the proposed wording at the start of the section. I think the idea of starting off by essentially defining g-force as force per unit mass is a good one. Then we can subsequently explain that force per unit mass resolves to units of acceleration, so technically the units are those of acceleration. I find it easier to understand that way. 81.152.169.67 (talk) 23:36, 28 August 2009 (UTC).
- It's specific force, but specific force is acceleration.
Lead section
I have restored the following text erroneously removed with the comment "flat wrong":
- "However, in practice the concept of "g-force" is usually concerned with the forces that arise during acceleration, and g-force may be interpreted as a measurement of force per unit mass – a quantity that resolves to the same units as acceleration."
1. It is self-evident that, when referring to "g-force", most people are interested in forces, not acceleration per se. That is why the term is called "g-force".
2. It is correct that force per unit mass resolves to acceleration, and this is the sensible way to reconcile the "misnomer". 86.150.102.78 (talk) 14:21, 14 November 2009 (UTC).
- It's not a force, it has units of acceleration. It's measured by a device called an accelerometer, that measures acceleration (specifically it measures the proper acceleration). Force per unit mass also has units of acceleration; that's because that too is an acceleration, not a force.- Wolfkeeper 14:47, 14 November 2009 (UTC)
- Your edit also removed references that point out that it's acceleration. The Wikipedia is based on reliable sources, not upon editors opinions.- Wolfkeeper 14:47, 14 November 2009 (UTC)
- Yes, but all accelerometers measure force, and then divide by mass to get acceleration. (Actually they measure analogs to force like deflection, resistivity, etc. Point is, a mass is necessary.)
- Also, editors opinions are bad of course, but editors knowledge is GOOD. Don't muddle the two. Editors' knowledge is THE SOURCE for judgment about reliability of references. Without knowledge, bad references become ensconced, and worse, promulgated! Your dismissive arrogant "flat wrong" commentary is destructive to cooperation. Collegial style is favored.
- You seem to be confusing different levels. An accelerometer is a system that measures proper acceleration. Whether the system to do that involves dividing something by something else or calculating quaternions... is simply irrelevant. The article is about the system, and the article subject defines what the system does.- Wolfkeeper 22:38, 19 December 2009 (UTC)
- Some accelerometers work by dropping a test mass and timing it, and don't measure forces at all.- Wolfkeeper 22:38, 19 December 2009 (UTC)
- True (about the dropping things). I thought of that and tried to cover the possibility with the phrase "..they measure analogs..." thinking of distance/time as an analog, but that wasn't clear I suppose. Distance and time are analogs. Distance isn't linear, and time is inverse, neither is a continuous measurement, but they fall under the "analog" umbrella I think. Anyway, that is a sidetrack.
- I'm more concerned about the idea that if something "has units of X" then that something is X. I recalled Specific Impulse which is traditionally given in units of seconds (by using units of force as if they were units of mass - the old "pound-mass"). Of course, Isp is a physical idea nothing like time. So, "Having units of X" doesn't mean that something is X. Similarly, "G-force" is traditionally given in units of "Specific Force" (force normalized to mass), which as you know, resolves to units of acceleration. Again, that doesn't mean that "G-force" is an acceleration. It is and it isn't. It's just more complex than that to be dismissed as "misnomer". It's just more complex than that to justify with such strong assertions that it so most assuredly is only acceleration and nevermore anything other than acceleration. 108.7.11.97 (talk) 05:25, 4 January 2010 (UTC) (same guy, different IP)
- By the way, on quaternions. They're nifty and fun and often more suitable than vectors or matrices. Best of all, they're fun because they're mind stretching! Beware though, mentioning the Q-bomb usually makes a person look like that want to look smart more than it actually makes them look smart. 108.7.11.97 (talk) 05:36, 4 January 2010 (UTC)