Wikipedia:Reference desk/Archives/Science/2016 October 23

= October 23 =

Triplet paradox
Let there is a triplet of A, B, and C on an asteroid initially. A stays on an asteroid while B and C set out for a long space journey with high speed (say 0.5c and 0.9c) at the same time in the same direction relative to A. Assume each 10 years old at the time of departure. B and C are gone for 60 years relative to A. Afterward, B and C return home at the same time and reunited with A on an asteroid.

What would be the age of A relative to B and C? What would be the age of B relative to C and A? What would be the age of C relative to A and B?

Since each person can have only one physical appearance and one age. Thus who would be right?2001:56A:739C:6D00:D418:247C:CE19:4D81 (talk) 03:07, 23 October 2016 (UTC)eek
 * What do you mean by "right"? ←Baseball Bugs What's up, Doc? carrots→ 04:04, 23 October 2016 (UTC)
 * I assume in this case it's "in agreement with observed reality", not "in agreement with the official opinion of the great leader". --Stephan Schulz (talk) 07:55, 23 October 2016 (UTC)
 * So each one would be "right". ←Baseball Bugs What's up, Doc? carrots→ 13:27, 23 October 2016 (UTC)
 * Since B and C move out and back again, the two that move are not in an inertial frame of reference - they need to accelerate up to speed, decelerate, turn around, and reaccelerate to go back (and decelerate to come to a stop again). Thus, the three are not in equivalent situations. If you want to take the acceleration into account in detail, you need the general theory of relativity, but you can approximate the result with the special theory of relativity, which will tell you that A will be older than B, which will be older than C. This is just a generalisation of the Twin paradox, which has a fairly good article. --Stephan Schulz (talk) 07:54, 23 October 2016 (UTC)
 * Yes, including formulas that should enable the OP to calculate a quantitative answer to his question. ←Baseball Bugs What's up, Doc? carrots→ 13:25, 23 October 2016 (UTC)
 * Agreed, but I'm not sure we've understood the question properly. As Bugs said, it's important to figure out what the OP means by "right". There is no "right" age that would be superior to the other two - each of the three would experience idiosyncratic amounts of time away from their siblings. We may prefer to think of things in terms of comparisons to "A" because they occupy the same inertial state as we do, but all three are equally valid. Matt Deres (talk) 13:31, 23 October 2016 (UTC)
 * Each person actually has an age, not just various ages as measured by other travellers. If they carried a reliable clock with them, it has a specific time on it at the end.  So there is no paradox in relative differences in ages; they should all work out right.  The specific ages vary due to acceleration as described above. Wnt (talk) 23:19, 23 October 2016 (UTC)

I didn’t include many things in order to keep the question short as i presumed that it was understandable from pundits pov. Since everything is ideal / theoritical therefore B and C are also traveling in the ideal rocket ship. Their rocket ships can accelerate up to their desired speed within second or minute in the time frame of A. Further they can stop just for a movement and come back through the same route - no turn around.So I think acceleration or declaration ( two or three seconds / minutes) should not be the problem while calculating the ages. Regarding “Right”

1. B and C have disagreement on the age as well as physical appearance of A

2. A and B have disagreement on the age as well as physical appearance of C

3. C and A have disagreement on the age as well as physical appearance of B

Although we can add infinite number of clowns of the aforementioned leader to the scenario but for simplicity, 96 clowns are sufficient to understand if it is difficult to guess who is right on the age as well as the physical appearance of A, B and C. Clown# 96 stays on an asteroid while the rest take off at the same time with the following speeds relative to Clown#96, in the same direction for their long synchronized space journey. Assume each 10 years old at the time of departure. All 95 clowns gone for 90 years relative to clown 96. Afterward, 1 to 95 return home at the same time and reunited with clown 96 on an asteroid.

Speed of clown 1 is 0.01c, 2 is 0.02c, 3 is 0.03c, 4 is 0.04 c, ..., 10 is 0.1c, ......, 20 is 0.2c, ......, 90 is 0.9c, ....,95 is 0.95c

Again: One clown can have only one age and one physical appearance therefore who would be right on the age as well as the physical appearance of 96 clowns.

Make the above triplet paradox quadruple

Our solar system revolves around another celestial mass of our galaxy. Assume D the fourth brother / sister of aforementioned A, B and C is on this celestial mass. Since D experienced no time dilation so D finds B and C aged at the same rate as A via his special binocular. D is in disagreement with A, B, and C.2001:56A:739C:6D00:D77:CB3B:5A2B:EBDD (talk) 03:51, 24 October 2016 (UTC)eek


 * If each observer takes into account general relativity as well as what you call the "time dilation" of special relativity, then there is no paradox. When they finally meet up, your clowns clones are all observed to be different ages (as measured by their biological deterioration or by the accurate clock that each has carried), but each correctly calculates the age of each of the others when they take into account everything that has happened.    D b f i r s   09:51, 24 October 2016 (UTC)
 * Perhaps the OP means to talk about clones because "clowns" sounds funny. AllBestFaith (talk) 15:36, 24 October 2016 (UTC)
 * Apologies to the OP for not realising this. At the end of their journeys, the clones would no longer be identical, nor would the accurate clocks they carried show the same time, but each clone would be able to predict the exact age of each other clone from knowledge of their journey through space-time, and all clones would get the same answer for a particular brother.     D b f i r s   16:36, 24 October 2016 (UTC)

Thank you AllBestFaith for correcting me. My mind is very scattered nowadays due to some inexplicable reasons, therefore, I apologize for the mistake. Further, doesn’t space dilate for moving frames instead of time? 2001:56A:739C:6D00:E867:777:70B1:58FF (talk) 02:12, 25 October 2016 (UTC)Eclectic Eccentric kamikaze


 * You should be aware that neither space nor time dilate for a moving object. Only the observation from a different inertial frame appears to show this.   D b f i r s   07:29, 25 October 2016 (UTC)

The word time dilation in a sentence itself explains that observations are made from different inertial frames. Since time dilation is the main crux of the discussion so it’s not necessary to mention inertial frames all the time impov.


 * The so-called "time dilation" is symmetrical. A moving observer sees the stationary clock to be running slow.    D b f i r s   19:57, 25 October 2016 (UTC)


 * With physics, time is what clocks measure, see time in physics. For this reason, time dilation is more than just an appearance or simply just a change in reference frame. So let's be clear here when we give answers and discuss this, for we do measure time dilation as a difference of elapsed time between two events due to either relative velocity or a difference in gravitational potential. Of course relative velocity is symmetric, and in accordance with relativity, time dilation is symmetric too. Certainly both variables are real and their magnitudes do matter. For example, with the Hafele–Keating experiment which tested time dilation, instead of the airport as a fixed reference point from which the planes took off, they must calculate their velocity relative to the Earth's center instead, which yields their relative motions not only with respect to each other, but also their rotational velocity with respect to the universe at large because the Earth's surface is not an inertial frame. Whether the clocks slowed down or sped up with the experiment depended entirely on which direction the planes took off. -Modocc (talk) 01:15, 26 October 2016 (UTC)

Defining the Kg in terms of (Atmospheric) Pressure
There seem to be many attempts at redefining the value of the kilogram, especially since it remained the only SI unit still dependable upon a physical artifact. Are there any practical reasons for trying to (re)define the kilogram in terms of atmospheric pressure, or height of column of mercury under specific temperature and gravity conditions (such as, for instance, [re]defining the pressure exercised by a column of mercury $3/4$ m tall at 0°C and standard gravity as 105 N exactly) ? I ask this because I don't know how practically feasible it would be to extract mass from pressure, rather than the other way around. — 79.113.203.205 (talk) 04:23, 23 October 2016 (UTC)


 * It is not likely to be feasible, because atmospheric pressure changes all the time, so how do you define a standard atmosphere? You suggest the height of a mercury column, but not only does gravity vary from place to place on the earth, it also varies with time, depending on the tide, and how much water is on the soil.  It would be simpler for you if you tried to set a kilogram by the mass of mercury in a cubic meter.  But even this precise value will depend on temperature, atmospheric pressure, isotopic composition (dependent on mine site and processing), amount of noise and motion in the liquid, electromagnetic fields present, number of cosmic rays impacting, mass of the neutrino etc etc.  Graeme Bartlett (talk) 05:17, 23 October 2016 (UTC)
 * So what you're saying is that, although we do have a precise theoretical value for standard gravity, actually achieving that exact value with great accuracy, even in a laboratory setting, is —from a purely practical perspective— unfeasible ? So much so, that we're back to square one, only this time with mercury instead of Vienna Standard Mean Ocean Water ? — 79.113.203.205 (talk) 06:49, 23 October 2016 (UTC)


 * I'm not sure what exactly you're trying to get at with your question. As far as I know, the desired end goal is to define the kilogram, and all the SI base units, in terms of invariant physical constants. That way no one has to worry about their calibration standard varying with the environment. See Proposed redefinition of SI base units. --47.138.165.200 (talk) 09:26, 23 October 2016 (UTC)
 * Their desired goal is to rationalize all/many fundamental physical constants... which would make some sense if they would be nice and round, such as in the case of c ≈ 3·108 $m⁄s$; unfortunately, this is not the case, so... For purely aesthetic reasons, I hope they fail. :-) — 79.113.203.205 (talk) 10:28, 23 October 2016 (UTC)


 * Seems completely the wrong way, IMHO. Firstly it's technically difficult - you simply can't do this by using "the atmosphere" as a standard, you'd have to construct and calibrate a simulated standard atmosphere.
 * Secondly the atmosphere's standardisation is based on pressure, which is based on force, thus on mass. So the standard conditions themselves are recursively self referential. If you redesigned the whole of metrology so that pressure became a fundamental unit, then you would no longer be trying to make a mass standard, you'd want to make a pressure standard instead.
 * Thirdly, the movement in metrology is to simplify the definitions of standards in terms of the fundamental unit they're based upon. So the intention is to supersede the platinum mass standard, defined through its weight, and replace it with a silicon sphere, based on geometry and crystalline properties. Andy Dingley (talk) 09:52, 23 October 2016 (UTC)
 * No, they aren't. We have a fixed height of $3/4$ m; a specific substance, mercury; a specific time unit, the second, used to measure local gravitational acceleration and compare it to the exact value 9.80665 $m/s^{2}$. Mass times gravitational acceleration yields force, which, when distributed over a specific unit of surface (the thickness of the mercury tube), gives pressure. We then `baptize` this pressure with the name 0.1 mega Pascal, and then use Pa = kg·m-1·s-2 to define the kilogram. Somewhat forced, perhaps, but definitely not convoluted. (Unless I'm missing something). — 79.113.203.205 (talk) 10:28, 23 October 2016 (UTC)


 * So, if I understand correctly, you aren't really trying to define an "atmosphere" so much as you are trying to define the kg in terms of the weight of a column of mercury of known height? Once you dispense with the distractions, this would seem equivalent to simply weighting a known amount of mercury and adjusting for local gravitational acceleration.  To be an improvement over current methods one would need at least ~1 in 100 million precision in the purity of the mercury, your ability to measure its volume, and your ability to correct for deviations in local gravity.  All of that seems impractical while offering no obvious benefits compared to the current approach of using an arbitrary physical artifact.  Dragons flight (talk) 14:57, 23 October 2016 (UTC)

Defining the kilogramme in terms of pressure is actually a fairly good idea, as I see it; the trick is not to use atmospheric pressure (which, as has been noted above numerous times, highly variable), but rather some pressure the value of which is completely invariant, such as the pressure at the triple point of a pure substance, or the critical pressure of a pure substance. For instance, the vapour–He-I–He-II triple point of helium-4 (which is actually one of four triple points for 4He—see the table in triple point) has a pressure of 5.048 kilopascals (0.04982 atmospheres); if one set the pressure for this triple point at exactly 5.048 kPa, one could then define the newton in terms of the pascal and the metre, and, then, in turn, define the kilogramme in terms of the newton, the metre, and the second, totally eliminating the need for the International Prototype Kilogramme, which could then be consigned to a museum as a historical artifact, just like the International Prototype Metre has been. Or, if one desired, one could take the critical pressure of oxygen (5.043 megapascals), define it as exactly 5.043 MPa, and use that to define the newton, and, by extension, the kilogramme. Whoop whoop pull up Bitching Betty &#124; Averted crashes 16:02, 23 October 2016 (UTC)
 * Do I remember rightly that there's an approximate relationship between length, mass, and volume? I was thinking that the original design was for a cubic decimetre to be a litre, and a litre of water to have a mass of a kilogramme.  Of course the numbers have been redefined over the centuries, but I don't see a reason why they can't define a kilogramme to be the mass of a litre (or a slightly different fraction thereof, to keep the current mass unchanged) of water at a certain temperature and pressure, with the litre's definition to be the volume of a cubic decimetre (or a slightly different fraction thereof, to keep the current volume unchanged).  We already have the metre being defined in terms of the speed of light during a specific period of time, and time being defined in terms of how long it takes for a specific atom to produce a specific number of oscillations.  Presumably they already would have done this kind of definition if it were as simple as I'm suggesting, so where have I gone wrong?  Nyttend (talk) 23:34, 23 October 2016 (UTC)
 * The original definition would have been like that, but it has the same problems as the mercury standard. Can you get the water 100.0000000% pure? Can you avoid water evaporating in your weight comparison? The density of water changes with air pressure, temperature, and other disturbances too, so it is not such a practical standard. Graeme Bartlett (talk) 01:51, 24 October 2016 (UTC)
 * One of the fundamental reasons why the SI was even introduced in the first place is reproducibility, since -without it- we are back to medieval times. If you cannot tell another person from the other side of the planet how to obtain the unit for themselves, without physically transporting some random artifact, then it has all been in vain. — 79.113.203.205 (talk) 05:20, 24 October 2016 (UTC)
 * Reproducibility is a separate question from practicality. Defining the metre as one fourty-millionth of the Earth's diameter is even less practical than moving a sample around. Of course, neither has to do with the key selling points of the metric system at the time, which were (1) decimal conversions and (2) aggressive marketing. Tigraan Click here to contact me 16:11, 24 October 2016 (UTC)
 * You are precisely wrong. The mere fact that the meter or the kilogram even have a definition in the first place shows a change in perspective and mentality. Until then, all units -other than those of time- were basically "random", varying from region to region and form town to town. (The only ones that had an actual definition were the units of time). So the SI did for weights and measures what until then was possible only for time, by endowing them with logic, and universality, and replicability. Not that premetric units of length were completely devoid of any logic, i.e., the fathom is the distance spanned by a man's outstretched arms; but the question arises: which man ? Because they all vary. So some sort of universal standard had to be sought. For instance, defining a foot as the distance traveled by light in a nanosecond is a different thing altogether than creating some random artifact based on a random person's walking step, and then proclaiming it as norm. — 79.113.255.194 (talk) 04:17, 25 October 2016 (UTC)
 * As far as Napoleon himself is concerned, I highly doubt that Russians (and other Slavs), along with Germans, Nordics, Italians, the Spanish, and the Portuguese, were particularly huge fans of his, or his empire, to the extent to which they might completely forsake their own rich historic tradition of weights and measures, and replace it with that of a foreign power. — 79.113.255.194 (talk) 07:38, 25 October 2016 (UTC)
 * We have an article History_of_the_metric_system. During the late 18th century, all of Europe was in the process of unifying measurements at the time; it is just that the Revolution sped up the things in France. But having a replicable definition of the units was a secondary concern - otherwise, "the length of a pendulum beating the second" (i.e. fixing $$g=\pi^2$$) would have won over "one fourth of meridian". The important thing was to have the same system everywhere and with reasonable divisibility properties. It was only in the second half of the 19th that deeper questions of metrology became important (see Metre Convention for instance), even if a posteriori one can see that the seeds were already there in the post-Revolution reports.
 * As for the history of adoption, Napoleonic wars brought new management with civil law and metric system (see History_of_the_metric_system). You could argue that the metric system would have been adopted anyways (the few attempts to reject it post-1815 were fairly quickly reversed) but that is speculation. Tigraan Click here to contact me 08:41, 25 October 2016 (UTC)
 * Regardless of whether we take g = &pi;2, or 1 `decimal minute` of a gradian = 1 km, or c = 3·108 $m⁄s$, the idea is the same: finding a universal reference point. (The point being that neither gravity, nor the Earth, nor light itself are going away anytime soon, whereas randomly created artifacts can be easily lost or destroyed). — 79.113.255.194 (talk) 11:03, 25 October 2016 (UTC)
 * Not true. Bazza (talk) 11:33, 25 October 2016 (UTC)
 * The variations you're referring to are insignificant. For any significant change to occur, millions of years have to pass. — 79.113.203.12 (talk) 14:57, 25 October 2016 (UTC)
 * First of all, $$g$$ was already known to vary with latitude at the time, and "g, but at the equator" is less universal than one would like it. (Of course, we now know it varies also with longitude, but much less and for reasons other than the Earth's rotation and non-spherical shape.) Second, even if the definition of the metre did not rely on a physical artifact (or on one big enough to not lose), the meridian length measurement was done once and all subsequent calibrations ultimately relied on physical artifacts (the mètre des Archives and its copies). From a practical point of view, the metre and kilogram definitions were equally contingent on a physical artifact.
 * As I already wrote with links to our articles on the subject, at the start of metrication, universality of the definition was at best a minor concern. When it became important, there was another issue, namely to pick a definition that allows the most precise measurement possible; as has already been pointed out, that is what makes definitions as pressure at triple/critical point or density of a pure substance hard to use. Tigraan Click here to contact me 15:51, 25 October 2016 (UTC)
 * I am well aware of g's variation with latitude and altitude, primarily. As I am equally aware that that was one of the main reasons why its initial `definition` was not adopted officially. As also were you, apparently. Which only begs the question why you even brought it up in the first place. Secondly, choosing g45 = &pi;2 at sea level would have been a pretty acceptable choice, regardless. Thirdly, the whole approach was `philosophic` and `universal` from its very inception: just read the article you yourself (repeatedly) linked to. Regardless of whether we are talking about the endeavors of John Wilkins, Christiaan Huygens, and Christopher Wren (on one hand), or of that of Gabriel Mouton (on the other), they all have one thing in common. The same can be said about its latest (re)definition, relating it to the value of c. Fourthly, the kilogram was defined with regard to the meter, and to water, at a meaningful temperature (as opposed to, say, grabbing some random rock, and proclaiming it to be `da norm`). Fifthly, even if from a practical perspective no one is crazy to (repeatedly or periodically) sponsor expeditions for measuring the length of the earth, it can be done if need be (i.e., if someone shoots all scientists and steals all copies of the meter). Furthermore, yes, no one denies that the main reason it was ultimately adopted was to serve a practical purpose; i.e., if humanity would have possessed an already-universally-agreed-upon common measuring unit for length and weight, the introduction of the meter would never have taken place in the first place, and the fact that the length of a second pendulum is incredibly close to both the ten-millionth part of a quadrant of a terrestrial meridian, and to that of a third of the hundred-thousandth part of the length traveled by light in the span of a second would have been one of these little curiosities confined to the realm of `popular science`. — 79.113.203.12 (talk) 18:05, 25 October 2016 (UTC)

Feynman Lectures. Lecture 33. Formula 33.2
$$\frac{A \cos\,(i - r)}{a}=\frac{-1}{-1}$$ I don't understand why does Feynman set aside reflected wave, deriving the formula. He says "since the two amplitudes on the left side of Eq. (33.2) each produce the wave of amplitude −1.". Even geometrically it is not true PNG dwg. Formula must be next: $$\tfrac{A \cos^2\,(i - r)+B\cdot cos^2(\pi-2i)}{a+b}=\tfrac{-1}{-1}$$

Username160611000000 (talk) 12:22, 23 October 2016 (UTC) Later Feynman says : Quote It is possible to go on with arguments of this nature and deduce that b is real. To prove this, one must consider a case where light is coming from both sides of the glass surface at the same time, a situation not easy to arrange experimentally, but fun to analyze theoretically. If we analyze this general case, we can prove that b must be real, and therefore, in fact, that b=±sin(i−r)/sin(i+r). It is even possible to determine the sign by considering the case of a very, very thin layer in which there is reflection from the front and from the back surfaces, and calculating how much light is reflected. We know how much light should be reflected by a thin layer, because we know how much current is generated, and we have even worked out the fields produced by such currents. Unquote I do not recall that Feynman ever showed the formula for current. It seems he uses 31.17 to show that field which is generated by glass has inverse sign of source field. But I can't understand why should we use a case when light is reflected from both sides of the glass plate? And second, why should we prove that $$b$$ is real value? $$\tfrac{\sin^2\,(i-r)}{\sin^2\,(i+r)}$$ is real.

Username160611000000 (talk) 17:51, 23 October 2016 (UTC)

Earth ocean rifts total gas output
Earth ocean rifts total gas output

Is there a source that provides Earth's ocean rifts total gas outputs to the oceans and to Earth's atmosphere? Terry D Welander (talk) 18:35, 23 October 2016 (UTC) [Redacted] — Preceding unsigned comment added by 173.21.166.135 (talk) 18:22, 23 October 2016 (UTC)


 * I don't know about total gas output, but this paper from 1998 gives an estimate for CO2. This 2013 paper updates the result (on overall emissions it's mostly narrowing the margin of error). The newer estimate is around 7-8x1011 mol/year, or (unless my math is off) around 35 megatonnes. For comparison, human emissions are around 29 gigatonnes/year. --Stephan Schulz (talk) 20:42, 23 October 2016 (UTC)

Storm surge
If the Americas were pushed out to sea about how fast would the land have to move to repeat (more or less) Hurricane Sandy's surge in New York Harbor? With several minutes of acceleration time. How long would it take the new sea level to reach equilibrium? Would a few minutes acceleration time be enough for it to be more like a surge than a tsunami? Yes, yes the power to cut a continent off the Earth and move it and keep it from sinking or causing earthquakes is beyond ludicrous. I just wonder about the wrong frame of reference where the land moves instead of the water. Sagittarian Milky Way (talk) 19:15, 23 October 2016 (UTC)
 * You mean like if the Americas were on gigantic pontoons? ←Baseball Bugs What's up, Doc? carrots→ 20:30, 23 October 2016 (UTC)
 * Wouldn't pontoons affect the sloshing? Cut off all crust above the elevation of the continental shelf edge, turn off friction at the cut, hold it up so it doesn't sink or tilt and start moving it. Sagittarian Milky Way (talk) 21:12, 23 October 2016 (UTC)
 * That sounds like it's in the same league with how Will Rogers proposed to get rid of the German U-Boats in World War I: "Boil the ocean!" ←Baseball Bugs What's up, Doc? carrots→ 22:37, 23 October 2016 (UTC)
 * I just want to know how fast we have to go to make a significant bow wave from Greenland to Cape Horn. Helm: let's start with 3 knots southeast. Sagittarian Milky Way (talk) 22:58, 23 October 2016 (UTC)
 * We do have an article, Floating island, but the Americas are not one. Alansplodge (talk) 19:29, 24 October 2016 (UTC)

Orion spaceship
Does anyone happen to know what is the peak G-force that the Orion Spaceship experiences during launch and/or atmospheric reentry? 2601:646:8E01:7E0B:40E5:EEF0:65B8:D229 (talk) 23:14, 23 October 2016 (UTC)


 * Which "Orion spaceship"?


 * Several different designs of the 1958 nuclear pulse-propelled spacecraft of that name existed, at least one made to be carried through the atmosphere on a Saturn V chemical first stage. Our article Saturn V indicates the maximum acceleration during ascent powered by that first stage was 4 Gravities, kept down to that stage by cutting off the central of five rocket engines, limiting total thrust, thus acceleration. The nuclear pulse propulsion engine was capable of a range of accelerations from 2 to 4 g (in a crew vehicle coupled to the drive unit by mechanical dampers - shock absorbers, essentially) to 100 g for unmanned vehicles.  That Orion concept is still being considered as one of several ways of preventing massive asteroids from impacting the Earth owing to that awesome acceleration capacity.


 * Our article Orion (spacecraft) which refers to NASA's Orion Multi-Purpose Crew Vehicle (Orion MPCV) currently in development and testing has no information on the maximum acceleration experienced during either ascent or re-entry of the craft.
 * The NASA paper "Orion Capsule Handling Qualities for Atmospheric Entry" reports "Some extreme off-nominal evaluation runs exceeded 8g‘s as predicted by the ANTARES simulation".
 * The Orion spacecraft as planned has the unprecedented capacity of "skip entry," in which the spacecraft can control its re-entry into the atmosphere, "skipping" across the atmosphere to dissipate its momemntum in a controlled fashion before re-entering.
 * The possible Orion re-entry flight profiles are shown on Figures 12-14 on the graph, showing a "sensed acceleration" experienced by crew ranging from free-fall to peaks above and below 4 G (depending on the re-entry profile studied), to Earth-normal acceleration. loupgarous (talk) 03:17, 27 October 2016 (UTC)


 * Thanks! So for a normal reentry, the peak G-force is about 4 G (about 30 seconds before the "roll to lift-up"), right?  (And yes, I did mean that Orion, not the nuclear one!)  Related question: Would this level of acceleration be acceptable for carrying civilian passengers (I mean without requiring extensive training and medical evaluations) into space? 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 05:28, 27 October 2016 (UTC)

It's a simple question of weight ratios!
So it's been established that a five-ounce bird cannot carry a one-pound coconut. But are there any aircraft (non-experimental) that can carry at this ratio, with the maximum safe weight of cargo being more than three times the manufacturer's empty weight? I note that two prominent US military cargo aircraft, the Lockheed C-5 Galaxy and the Lockheed C-130 Hercules, both have manufacturer's empty weights that are greater than their cargo capacities, so I'm guessing that any such aircraft would be much smaller, since the USAF might well not use their current aircraft if smaller aircraft existed that could carry similar amounts of freight or if similarly sized aircraft could carry much greater amounts of freight. Nyttend (talk) 23:41, 23 October 2016 (UTC)
 * ...Probably not in the categories you're thinking about. Athough my personal reservoir of knowledge is not exhaustive, I don't know of any mainstream aircraft in the fixed-wing or rotorcraft category that would satisfy what you're looking for.  Depending how you choose to define "useful load" - in aviation, we include fuel as part of that figure - you might find the weight ratio you want in the payload fractions in the rockets used for launching cargo in to space.  And again, if you consider fuel as part of the useful load, then the GlobalFlyer (a one-of-a-kind experimental aircraft) or the Rutan Voyager might meet your requirement.  (You might read about Burt Rutan's other aircraft for his experimental airplanes that cater to a "slightly" more mainstream audience).  And of course, if you consider lighter-than-air aircraft, including airships, the cargo mass can be considerably greater than the aircraft mass.
 * Nimur (talk) 00:10, 24 October 2016 (UTC)
 * Sorry I wasn't clear. I was definitely meaning   divided by   is at least 3, so I was imagining an out-of-fuel aircraft for the first one.  My idea was either a fixed-wing aircraft or a helicopter (but at least in my imagination, they don't carry as much weight for their size; their advantage is maneouvreability, not mere carrying capacity) flying in the atmosphere; I wondered about a space rocket, but in the end I was only thinking about atmosphere-restricted aircraft.  I've never heard of rocket airplanes (except experimental ones), but if any rocket airplanes have ever gotten out of the experimental stage and become "normal" aircraft, they'd definitely be eligible for what I was looking for.  Nyttend (talk) 01:16, 24 October 2016 (UTC)
 * Indeed, you are using the terminology correctly - the basic empty weight is defined as the aircraft's weight with empty fuel-tanks. There are a handful of technicalities - some mass of fuel can never be removed from some types of airplanes, because of the way real-world, non-"spherical-cow" tanks and fuel-lines work... that's the opposite of the usable fuel.  ...So the definition of the term "basic empty weight" is "the weight of the airframe, engines, all permanently installed equipment, and unusable fuel" plus the weight of installed optional equipment.  The definitions are carefully laid out, e.g., in the Aviation Maintenance Technician's Handbook and the Weight and Balance Handbook; and of course, each individual aircraft in the United States carries its own paperwork documenting its official weight and balance - that stuff has to be on board each aircraft; it's easy to remember, as it's the final item on the ARROW Checklist of required paperwork.  Always make sure you measure your fuel, your weights, and all other important parameters correctly for your specific aircraft.  Nimur (talk) 21:21, 24 October 2016 (UTC)
 * This aircraft Weighs 220 Lbs and lifts 2,423 Lbs... Also see: --Guy Macon (talk) 01:35, 24 October 2016 (UTC)


 * If you are looking at max takeoff weight / empty weight, the B-52 gets somewhat close at 2.6. The Lockheed U-2 is even better at 2.8, though in that case it's almost entirely fuel and almost no cargo.  Dragons flight (talk) 02:10, 24 October 2016 (UTC)


 * About rocket planes, there WAS in fact one "normal" rocket plane, but its weight ratio was only 2.26. 2601:646:8E01:7E0B:A0FD:B2DF:3F45:5E5B (talk) 20:36, 24 October 2016 (UTC)
 * But still, that was rather experimental; it wasn't used for a significant period of time, for one thing. I don't care what's in the storage tanks; I was just interested in the ratio of parts to non-parts.  The B-52 and the U-2 are more what I was looking for: they demonstrate that it's possible to create an aircraft that has such a weight ratio, and it's demonstrably possible to keep such an aircraft flying for a significant period of time (i.e. such a design isn't necessarily unstable enough that it won't last long in service).  Thanks!  Nyttend (talk) 20:42, 24 October 2016 (UTC)
 * The Embraer KC-390 military transport is stated by its manufacturer Embraer to have a maximum payload of 23.6 tons, a wing fuel load of 23 tons, a maximum normal take-off weight of 74.4 and a "logistical" take-off weight of 81 tons. These figures agree roughly with airforce-technology.com's estimate of the KC-390's empty weight of 23.6 tons. So, depending on how you look at it, a KC-390 carries roughly its empty weight in payload, or twice its weight in payload and fuel. loupgarous (talk) 03:45, 27 October 2016 (UTC)