Talk:G-force

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Contents

[edit] contradictions

  • Any exposure to around 100 g or more, even if momentary, is likely to be lethal.
  • Formula One race car driver David Purley survived an estimated 179.8 g in 1977 when he decelerated from 172 km·h−1 (107 mph) to 0 in a distance of 66 cm (26 inches) after his throttle got stuck wide open and he hit a wall.

--

Not necessarily: The first means "Don't try this at home", the second means "but you might get lucky" (roughly). I'd imagine that repeating the experiment is unlikely to be healthy. It is conceivable that restraints reduced the g force somewhat or otherwise made it more likely than usual to be safe. Ealex292 03:56, 1 December 2006 (UTC)

74.129.197.37 (talk)

There are so many factors as to make it nearly impossible to determine the g-force a particular person can survive. Think of the 100g's as an 'LD50' - that is, at least 50% of people would die after being exposed to greater than 100g's. —Preceding unsigned comment added by 74.129.197.37 (talk) 05:59, 15 December 2007 (UTC)

[edit] GeForce

People looking for the graphics card GeForce may end up on this page, as gforce redirects here. Just a note JayKeaton 15:01, 29 November 2006 (UTC)

[edit] Incorrect Physics

The article claims that the variations in the acceleration due to the gravity of the earth are due to so-called centrifugal forces. This is a poor explanation. I believe a better one would be:

One reason involves the difference in the distance from the centre of the Earth between the two positions due to the equatorial bulge – this leads to a variation in the gravitational field strength. The equator is further away from the centre of the Earth than the poles leading to a difference of about 0.05 m s–2. The second reason is due to the rotation of the Earth. The person on the equator experiences a centripetal acceleration. Given that the scales read the normal (or reaction) force N, in this case N = m(g–ac). Therefore there is a slight reduction (of the order of the first effect) —Preceding unsigned comment added by 132.181.7.1 (talk • contribs)

[edit] g-force a force?

The article states in the introduction

g-force or g-load is a force-equivalent, equal to 9.80665 N/kg

but N/kg is a unit of acceleration, not of force, so I have to conclude that 1 g = 1 g-force. Is that intended? AxelBoldt 18:09, 19 December 2006 (UTC)

[edit] g divided by time

Shouldn't this article include the difference of g force divided by time?

There are great difference of experiencing 10 g's for a millisecond and for ten seconds. A person have experienced and survived over 46 g's in a certain amount of time, but it is lethal to experience for example 25 g's over a minute. This difference may not be understandable in this article.

[edit] Equivalent versions of Helmert's equation

It might be helpful to note that the version of Helmert's equation in the article is equivalent to the following (discounting rounding errors and expression of lengths in cm instead of meters):

http://lists.nau.edu/cgi-bin/wa?A2=ind0101&L=phys-l&P=39244

g = 980.616 - 2.5928*cos(2*phi) + 0.0069*cos^2(2*phi) - 3.086x10^(-6)*H

where: phi = latitude H = elevation (in cm)

This version is on several web pages and in a CRC Chemistry and Physics Handbook, where it can be difficult to find as noted here:

http://www.lhup.edu/~dsimanek/scenario/labman1/accel!.htm In a CRC handbook like mine, there's no index entry for Helmert's equation. Look up "acceleration due to gravity at any latitude and elevation, equation." It is in the "definitions and formulas" section.

The version above may be calculated more quickly because only one trig value is needed: cos(2*phi). Once that trig value is known, it can be squared for the third term. The version in the article requires calculating two trig values separately: sin(phi) and sin(2*phi).

It may be of interest that there is another, equivalent, version of this formula (with no double-angle trig arguments) which may be obtained by applying trig identities in yet another way. It has the form: g = g_0 * (1 + a * (sin(phi))^2 + b * (sin(phi))^4)

This latter version is not found as frequently on the web, but it is here: http://galitzin.mines.edu/INTROGP/notes_template.jsp?url=GRAV%2FNOTES%2Flcorrect.html&page=Gravity%3A%20Notes%3A%20Latitude%20Variation%20Corrections Ac44ck 07:33, 28 January 2007 (UTC)

[edit] Helmert's equation superceded?

There seems to be a more recent (1984) formula for the variation in g with latitude: http://earth-info.nga.mil/GandG/publications/tr8350.2/tr8350.2-a/Chapter%204.pdf

g = 9.7803267714 * (1 + 0.00193185138639 * (sin(L))^2) / sqrt(1 - 0.00669437999013 * (sin(L))^2)

This formula is on page 3 of the pdf file.

Ac44ck 07:38, 28 January 2007 (UTC)

[edit] strongest g-forces survived by humans

Probably there's someone with higher g-force survived http://en.wikipedia.org/wiki/Kenny_Brack#Return_to_IRL --84.234.42.68 21:33, 20 June 2007 (UTC)

[edit] Roller coasters

I wondered about the paragraph on amusement rides, where it is said that they usually don't pull over 3 g with some listed exceptions. However, according to "rcdb.com" and other coaster-related sources, almost every looping coaster on the world pulls about 4-5 g on entering the loop (e.g. the Vekoma Boomerang which is found in many parks around the world is said to pull 5.2 g on its first inversion).—Preceding unsigned comment added by 141.203.254.65 (talk • contribs)

[edit] Proposed changes

To me, the first couple of sections are a bit confusing. I was going to tweak a few things, but in fact ended up effectively rewriting the intro and "Explanation" sections, and incorporating the "Convenient definition" section. Because it's a relatively significant change I feel I should give others the opportunity to comment before applying it to the article. This is what I've got. Matt 02:53, 25 July 2007 (UTC).

g-force (also gee-force, gee-loading) is a measurement of an object's acceleration expressed in gees. The "gee" (pronounced [dʒiː]; symbol g) is a non-SI unit equal to the nominal acceleration due to gravity on Earth at sea level, defined as 9.80665 m/s2, or approximately 32.174 ft/s2. More precisely, g-force measures the difference between the acceleration that an object actually experiences and the acceleration that gravity is trying to impart to it, as explained further below. The symbol g is properly written in lowercase and italic, to distinguish it from the symbol G, the gravitational constant, which is always written in uppercase; and from g, the symbol for gram, which is not italicized.

Although strictly a measurement of acceleration, the term g-force, as its name implies, is commonly understood to refer to the force that an accelerating object "feels", expressed as a multiple of the force that it "feels" when resting stationary on the Earth's surface. The relationship between force and acceleration stems from Newton's second law, F = ma, where F is force, m is mass and a is acceleration.

These so-called "g-forces" are experienced, for example, by fighter jet pilots or riders on a roller coaster, and are caused by changes in speed and direction. For example, on a roller coaster high positive g-forces are experienced on the car's path up the hills, where riders feel as if they weigh more than usual. This is reversed on the car's descent where lower g-forces occur, causing the riders to feel lighter or even weightless.

Because of the potential for confusion about whether g-force measures acceleration or force, the term is considered by some to be a misnomer. Scientific usage prefers explicit reference to either acceleration or force, and use of the appropriate units (in the SI system, metres per second per second for acceleration, and newtons for force).

Calculating g-forces [level 2 heading]

Unlike simple acceleration, g-force is a measure of an object's acceleration relative to the local gravitational acceleration vector, rather than being compared to an inertial reference frame. In other words, it is the (vector) difference between an object's actual acceleration and the acceleration that it would experience if it were falling freely. It is this difference, rather than the actual acceleration of the object, that gives rise to the feeling of force ("apparent weight"), and hence to the feeling of heaviness and lightness in high and low g-force environments. For further details, including examples of conversion between acceleration and apparent weight force, see apparent weight.

In a simplified scenario, where accelerations are assumed to act only downwards (positive) or upwards (negative), calculating this difference simply amounts to subtracting the object's actual acceleration from the gravitational acceleration. For an object on or near the Earth's surface, gravitational acceleration is for practical purposes equal to 1 g. (For more precise measurements, the variation of Earth's gravity with location and altitude must be taken into account.) So, for example:

  • A non-accelerating object experiences a g-force of 1g − 0g, or just 1g ("normal weight").
  • An object in free fall (accelerating downards at 1g) experiences a g-force of 1g − 1g = 0g ("weightless")
  • An object accelerating upwards at 1g experiences a g-force of 1g − (−1g) = 2g ("twice normal weight")
  • An object accelerating downwards at 2g experiences a g-force of 1g − 2g = −1g ("negative g").

More generally, the object's acceleraion may act in any direction (not just vertically), so in a fuller treatement it must be considered as a vector quantity. The "difference" in acceleration that g-force measures is found by vector addition of the opposite of the actual acceleration and the local gravitational acceleration vector (about 1 g downward on or near the Earth's surface).

In cases when the magnitude of the acceleration is relatively large compared to 1 g, and/or is more-or-less horizontal, the effect of the Earth's gravity is sometimes ignored in everyday treatments. For example, if a person in an car accident decelerates from 30 m/s to rest in 0.2 seconds, then its deceleration is 150 m/s2, so one might say that it experiences a g-force of about 150/9.8 g, or about 15.3 g. Strictly speaking, due to the vector addition of the gravitational acceleration, the true g-force has a slightly larger magnitude and is pointing slightly downwards (intuitively this is because the car is already experiencing 1 g just by sitting on the road).

[edit] Comments on proposed changes

  • I don't think example calculations should be included. Explained some but no great detail. It looks like a tutorial thing to me. What about the human tolerance & experience stuff? Do you want to get rid of that? I wouldn't mind that, at least cut it back. -Fnlayson 14:01, 25 July 2007 (UTC)
    • Sorry, perhaps it wasn't clear. This is a replacement just for the current intro, "Explanation" and "Convenient Definition" sections - the first part of the article. I wasn't proposing at this stage making any changes to the rest of the article. I think that the examples are important, but I take your point that it seems a little too detailed. I have amended the above to cut down on the detail. Matt 13:08, 26 July 2007 (UTC).
  • I've now replaced with the above, overwriting a recent change claiming:
g-force (also gee-force, gee-loading) is a non-SI vector measure of force, where 1 g (pronounced [dʒiː]) is defined to be the force an object would experience by gravity at the surface of the earth (the units are newtons or pound-force).
This is wrong. The "gee" (symbol g) is a unit of acceleration, not force. In any case a unit can't be defined as the force "an object" would experience -- it depends on the object. Matt 12:39, 28 July 2007 (UTC).
Agreed. 1 g is the acceleration an object experiences on the Earth's surface. -Fnlayson 16:12, 28 July 2007 (UTC)
If we are talking about g-force, then it is a force. It is described by the acceleration which causes that force. Lets try to get this right. Rlsheehan August 12, 2007
  • Resulting force has nothing to do with it. That's a reaction force, internal force. G-force is an acceleration divided by acceleration of gravity so it is in g's. Having force in the name does not mean anything, since its units are g's not force units. -Fnlayson 03:17, 13 August 2007 (UTC)
  • I agree that "g-force" is a misnomer. Using this name confuses many people about force vs acceleration. If we call it a force, then it should be a force. It can still be described by the accleration which causes it. Rlsheehan Aug 14, 07
    • G-force is simply described as an acceleration in g's now. What do you suggest? -Fnlayson 15:17, 13 August 2007 (UTC)
*It is certainly acceptable to report accelerations in the non-SI unit of g-s: see also Shock (mechanics). The confusion is that we are trying to call this a force. Perhaps a new first paragraph like this might help:
"g-force is a comomonly used term which can have two different but related meanings: 1) the acceleration on an object or person expressed in gs and 2) the reaction force resulting from an applied acceleration, expressed in gs." Rlsheehan Aug 13, 07
I've removed statement 2 since a force cannot be "expressed in g's" — any more than velocity can be expressed in kilograms, or time expressed in litres. The g is a unit of acceleration and that's that. Force per unit mass could be expressed in g's though. I think I mentioned that in the article at one point, but it got taken out for some reason. Matt 00:09, 30 August 2007 (UTC). —Preceding unsigned comment added by 86.134.55.114 (talk)
The item we are discussing is g-force, thus we must include the force interprretation. Yes, the term is probably a misnomer, but ,it is here and we can discuss it. It can be a force described by the acceleration which causes it: the causing acceleration can be described in terms of gs. Rlsheehan 01:56, 5 September 2007 (UTC)
  • Yea and that is explained in the "Connection with force" section. No need for that to be in the Lead too. -Fnlayson 02:43, 5 September 2007 (UTC)
Yes, it is important to start with a correct lead paragraph. The subject is g-force so force must be included in the lead paragraph. Otherwise, change the title. Rlsheehan 14:12, 5 September 2007 (UTC)

[edit] weight or force

in the 'human g-forces' section it says "a weight of 1g" a few times.

g is in N/kg and a weight is in kg, so this seems incorrect to me.

Any suggestions?

Mushlack 20:00, 4 August 2007 (UTC)

  • Weight is a force by scientific and tehnical definations. The SI unit for force is Newton (N) and the SI unit for mass is kilogram (kg). But the acceleration of gravity is not a weight either. So the wording should be adjusted/corrected. -Fnlayson 20:39, 4 August 2007 (UTC)

[edit] Italicization

Is the g in this article really properly italicized? Robert K S 07:42, 21 September 2007 (UTC)

  • I had the same thought, assuming you're talking about the "g" in "g-force" (rather than the standalone unit symbol). I have not found any dictionary that italicises the g, and unless anyone can provide some authority that says it should be italicised I propose the italics should be removed throughout. Matt 00:43, 23 October 2007 (UTC). —Preceding unsigned comment added by 86.136.195.74 (talk)
    • I've also looked at a selection of pages thrown up by Google, and basically I can't find anyone who italicises the "g" in g-force, so I've removed it. If you disagree and think it should be italicised then we need a solid authority I think... Matt 02:05, 31 October 2007 (UTC). —Preceding unsigned comment added by 86.150.100.198 (talk)

[edit] Eyeballs

Just a quick point: in the part about human tolerances, various references are made to difference tolerances with "eyeballs-in" or "eyeballs-out". Perhaps there should be an explanation of the term? I'd put it in if I knew what it meant myself, but I really don't... anyone else have a clue? Is it about eyes being open or closed? Eyes being slightly popped out of their sockets? Thanks. Hagger 19:09, 17 October 2007 (UTC)

Yeah, an eyeballs-out g-force 'tries' to pop the eyes out, an eyeballs-in g-force 'tries' push them in. It must be emphasised that it doesn't succeed in either case in any survivable situation ;-)- (User) WolfKeeper (Talk) 01:04, 17 December 2007 (UTC)

[edit] Formula One data

I'm skeptical about the Formula One data must be wrong. On a level road, with no airfoil, the greatest acceleration a car can have is mu*g, where mu is the coefficient of static friction, and g is the acceleration of gravity. Since mu for rubber on asphalt is no more than about 0.5, the greatest acceleration or deceleration you can have is about 0.5 gees. Unless these figures are referring to banked turns, or the cars have airfoils that can generate a downward force equal to 10 times the cars' weight, I don't think the article can be correct. Is there a source for these numbers?--76.81.160.198 (talk) 03:44, 10 December 2007 (UTC)

Your figures are way off. Standard cars can pull about .92 g on a dry road. Motorbikes with relatively sticky tires can pull over 1.2 g. Formula one cars have sticky tyres and downforce at maximum speed of about 5 or 6 times their own weight or more due to groundforce (predominately), but also the wings help some as well.- (User) WolfKeeper (Talk) 04:16, 10 December 2007 (UTC)
Oh yeah, and formula 1 cars go fast enough that just lifting off at high speeds gives about 1g deceleration (with the 'help' of the rear wing); putting the brakes on as well adds another 4g or so, even though they have relatively small front tyres.- (User) WolfKeeper (Talk) 04:20, 10 December 2007 (UTC)
Okay, if the airfoils really do produce a force many times greater than gravity, then the figures in the article are plausible.--76.81.160.198 (talk) 22:03, 16 December 2007 (UTC)
Oh yeah, massive downforce. Somebody calculated that a formula one car could theoretically drive continuously upside down from about 100 mph, but fortunately (or unfortunately), there's no upside down tracks anywhere to try this ;-)- (User) WolfKeeper (Talk) 01:01, 17 December 2007 (UTC)
I noticed that there is citation for the g-forces that Formula One drivers usually experience. I think that the figures are okay but they should be verified by a reliable source. --Cryonic07 talk ° contribs 21:36, 9 June 2008 (UTC)