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August 6[edit]

Wind speeds on gas and ice giants[edit]

I keep hearing about incredible wind speends on gas and ice giants, both inside and outside the solar system, and how destructive the winds could be, but something is bothering me about it. If the wind speed is relative to the average rotational speed of the bulk of the planet, then wouldn't you be just fine so long as you avoid the interaction zones of jet streams? What would turbulence be like within a Neptunian jet stream and would you even notice exciting one, if you had no visual point of reference? I've got the impression that you can easily traverse latitudinally without worrying about being ripped appart by sudden changes in wind direction, but you got little to no control over your latitudinal direction of traverse due to the magnitude of the wind speed. Perhaps, it's a case of 'one does not simply' travel from pole to pole. Plasmic Physics (talk) 10:07, 6 August 2018 (UTC)[reply]

You are referring strictly to wind speed. You can also consider wind force. There is a lot of documentation you can find on Martian wind storms, which you can use to consider the giants. Storms on Mars reach very high speeds, but there is very little atmosphere. So, there is very little force. To do damage, you will need high speeds and a thick atmosphere. Jupiter has that once you get down into the atmosphere. But, it is a gas giant. You should expect that for all gas giants, and looking at Jupiter and Saturn, there is no apparent "calm" spot. The ice giants are different. You need to check the atmosphere to see how thick it is. It could be thick methane. It could be thin hydrogen. 209.149.113.5 (talk) 13:56, 6 August 2018 (UTC)[reply]

You also need a place to stand. If you are traveling at the same speed as the wind you experience no wind. Some of these planets have no well-defined surface. --Guy Macon (talk) 20:00, 6 August 2018 (UTC)[reply]

This depends on wind shear, which may be high near edges of jet streams. Ruslik_Zero 20:04, 6 August 2018 (UTC)[reply]

From the description in the article, I think that wind shear is exactly what I'm refering to. What is the wind shear typically like in those kinds of planets, and is there a way to avoid shear zones when travelling with the air currents or is there too much turbulence? Plasmic Physics (talk) 03:07, 7 August 2018 (UTC)[reply]

Wind shear is a difference in speed and/or direction over a short space. Again, this is speed. You could safely stand on the Moon in the middle of an area with a wind shear of hundreds of miles per hour because there is not enough atmosphere for that wind shear to create substantial force. However, if you are in the thick atmosphere of Jupiter, a small wind shear would be enough to do significant damage. Nearly all of the time, wind speed, shear, and force are used with the assumption that we are discussing sea-level winds on Earth. When discussing other planets, the relationships change. An example many people know is the storm on Mars at the beginning of "The Martian." That wouldn't happen. You could get those winds and dust, but it wouldn't knock anyone over. It would just be an annoying little breeze. If it took place on Earth, it would be deadly. 209.149.113.5 (talk) 13:54, 8 August 2018 (UTC)[reply]

Wind speed doesn't really factor into it, which is the point I raised in the opening statement. Wind shear is measured in Hertz, so the acceleration you'd experience as you travel across an air current is equal to the product of your crossing velocity and the wind shear. Follows, F = ma. Plasmic Physics (talk) 20:52, 8 August 2018 (UTC)[reply]

Yes. F=ma, but it is common that as a increases, m shrinks. Therefore, when m is very small, wind shear doesn't matter. You simply won't notice it. When m is large, a small wind shear can be devastating. My point is that the concern needs to focus more on the content of the atmosphere and less on the difference in wind speeds. For example, trying to stand in a 5mph breeze on Jupiter with absolutely no wind shear would be very hard. It would be trying to stand up to a train as it slowly rolls over you. 209.149.113.5 (talk) 17:11, 9 August 2018 (UTC)[reply]

I get that, but how small an m is small enough for wind shear not to matter, and is this actually typical? Your Jupiter example should only be accurate if you're trying remain stationary, whereas I'm working with the condition of having a comparable velocity-component in the same direction as the air current. This means that you are travelling with the air current, while attempting to cross it, like swimming across a river. Plasmic Physics (talk) 06:25, 10 August 2018 (UTC)[reply]

Inverse flight[edit]

In a fixed wing aircraft the flying is made possible because of the wing configuration whereas the lower surface has a flat profile and the upper surface is convex. However there are numerous examples when military aircraft fly upside down. I saw an example on a AHC TV show Top Ten as a post WWII massive 4 engine jet bomber, B-47, flipped during a flight and flew half a minute inverse with the belly facing the sky. Japanese zero fighters during WWII did it routinely for the purpose of showing off as well as observing for a possibility of being attacked from below. Where is the aerodynamics in such cases? How do they do it? Thanks, AboutFace 22 (talk) 15:49, 6 August 2018 (UTC)[reply]

Where you say "made possible", it would be more accurate to say "made more efficient". Lift is generated when the wing forces air downward; it doesn't matter whether it does it because of the shape of the wing (the convex-on-top shape you refer to) or because of its angle of attack. The shape used for wings is one that will force the most air downward with the least friction (drag), so it doesn't need a high angle of attack to get enough lift in cruising flight. But if the airplane is pitched down (the nose is lowered), the angle of attack can reach the point where more air is forced upward for that reason than is forced downward by the shape of the wing, so the "lift" becomes negative. But if the airplane is inverted while doing this, the lift is upward and will keep the plane airborne. However, the drag in this position is much higher than in normal flight, so it requires a high-performance aircraft. --76.69.47.228 (talk) 16:42, 6 August 2018 (UTC)[reply]

Mandatory XKCD. The übernerds at explain-xkcd have a decent explanation of why that explanation is wrong. Tigraan^{Click here to contact me} 18:33, 6 August 2018 (UTC)[reply]

Lift (force)#False explanation based on equal transit-time. Also see: List of common misconceptions#Physics. --Guy Macon (talk) 20:05, 6 August 2018 (UTC)[reply]

There are so many details to consider during the study of inverted flight! Some airplanes have symmetric airfoils, and some also have symmetric dihedrals. There are symmetric gravity immersion oil systems for the engines; some airplanes use pumps so the liquid stuff like fuel and oil don't depend on gravity to go where they are needed. All of these engineering details change performance characteristics for the airplane in upright and inverted attitudes.

To specifically answer the original question, we should restate the question about the airfoil: how does a symmetric shape like the wing of an Extra 300 generate lift?

The best way I can understand these physics is to think of the momentum transfer model. My simplified model emphasizes that air is sticky when it's flowing. That means the airfoil's angle of attack - the angle that the chord makes against the incident wind - can be the primary physical mechanism to change the airflow's bulk motion. To first order, the actual shape of the airfoil isn't important for the generation of lift! A flat board, held at an angle to the incident wind, will deflect air, and that will generate a change in momentum. The actual shape of the airfoil becomes important when we look at the secondary details: how efficient is the deflection of air? (This is the L/D parameter, for any aeronautical physics nerds in our audience). And - when does the flow cease to be laminar? (This is the critical angle). Those important numbers can be estimated from fluid dynamics equations; but usually, they are derived experimentally. I can plug them into my mental model to set the boundaries of my operable range.

These ludicrous oversimplifications give me a good, workable, simplified linear-equation that is accurate enough for my purposes - which are, of course, to maintain aircraft control in all flight attitudes. Put another way - how hard do I need to work the controls to move the airplane where I want it to go? During inverted flight, I can't solve nonlinear aerodynamics equations while the blood is rushing to my feet - I need linear equations that only require me to consider one or two parameters!

It takes years of study to really grok airfoils; if you want a good book, Aerodynamics for Naval Aviators is available for zero cost and is recommended by the FAA for advanced flight students. This book pulls no punches. The physics and math are difficult and if you aren't a domain-expert, you'll easily get lost in the details. But it is one of the best, most thorough, practical physics books that doesn't simplify the real science of aerodynamic analysis.

Nimur (talk) 14:52, 7 August 2018 (UTC)[reply]

Most airplane wings have a convex upper surface and a less-convex, or even flat, lower surface but these two features are not essential for an airfoil; they are used to achieve the most efficient design possible, not to guarantee that lift can be generated. Many airfoils have a symmetric profile.

The essential feature of any airfoil is that it has a sharp trailing edge - see Kutta condition. Whether a wing is operating with its “usual side” up, or upside down, its trailing edge is sharp so it functions as an airfoil and generates lift. (If the airplane was falling backwards, the wing would not generate lift, for the obvious reason.) Dolphin (t) 22:05, 7 August 2018 (UTC)[reply]

Nonsense. I can take a 1" x 12" x 12" wood plank, hold it out of the window of a car going 60 MPH, and depending how I angle it, get plenty of lift.

Also see: Here's What It Takes To Fly a Model Plane Backwards]. --Guy Macon (talk) 21:19, 8 August 2018 (UTC)[reply]

Understandably, most people imagine an airfoil must have a rounded leading edge with a large radius. The fact that a flat plate will generate lift proves that a rounded leading edge is not required. The essence of an airfoil is not the rounded leading edge - that is provided on all subsonic airfoils to maximise the stalling angle and, therefore, the lift coefficient.

Designers of all fixed-wing aircraft would love to be able to use a wing with a generously rounded trailing edge. (The wing profile of a Rankine body!) It would provide so much extra volume for a deep rear spar, extra fuel storage space, and extra volume into which the undercarriage could be retracted. Unfortunately a wing with a generously rounded trailing edge will not generate much lift so designers must make use of airfoil sections with what is always called a sharp trailing edge.

One of the first successful attempts to analyse, mathematically, the properties of airfoil sections was the Joukowski airfoil which has a cusped trailing edge. A cusped trailing edge is not essential for a good airfoil - a so-called sharp trailing edge is sufficient. Dolphin (t) 06:20, 9 August 2018 (UTC)[reply]

"The essential feature of any airfoil is that it has a sharp trailing edge". Evidence, please. (I gave you a good counterexample of lift being generated with a squared-off trailing edge.) "If the airplane was falling backwards, the wing would not generate lift". Evidence, please. (I posted a video of a model airplane flying backwards with plenty of lift) "a wing with a generously rounded trailing edge will not generate much lift". Evidence, please. (The sharp trailing edge exists to minimize drag and only has a small effect on lift -- ultralight aircraft often have a tube at the leading and trailing edges with cloth stretched in between, and they generate lift just fine.) --Guy Macon (talk) 15:11, 9 August 2018 (UTC)[reply]

Aerodynamicists describe an individual airfoil by the radius of its leading edge; and the angle of its trailing edge - the angle between the upper and lower surfaces at the trailing edge. When we talk about an airfoil having a sharp trailing edge we don’t mean sharp like a scalpel, we mean having a measurable trailing edge angle like the trailing edge of the wing of a Boeing, Airbus etc. (The upper and lower surfaces of a flat plate are parallel so its trailing edge angle is zero.) Wikipedia has a suitable quote by Kuethe and Schetzer in the lead section of Kutta condition:

”A body with a sharp trailing edge which is moving through a fluid will create about itself a circulation of sufficient strength to hold the rear stagnation point at the trailing edge.”

I am happy to respond to Guy’s requests for evidence but this Reference Desk is not the appropriate place. I will do so elsewhere. Dolphin (t) 19:02, 9 August 2018 (UTC)[reply]

Actually, this Reference Desk is the appropriate place. You are posting false information on the reference desk. You are taking various aspects of airfoil design that are there to minimize drag and falsely claiming that without them there can be no lift or greatly reduced lift. --Guy Macon (talk) 09:42, 10 August 2018 (UTC)[reply]

It is the appropriate place for Users to respond to the question asked by User:AboutFace 22. It is also an appropriate place for Users to engage in dialogue with AboutFace. It is not the place for you and I to engage in dialogue - that would be hijacking AboutFace’s thread for our own purposes. If you want Ref Desk answers to questions of your own, start your own thread by asking your own questions. Other Users will then respond to your questions. If you want a dialogue with me, or you want me to respond to your requests, a User Talk page is the appropriate place.

AboutFace’s original question is about the ability of airplanes to fly inverted - he gave the Boeing B-47 and the Japanese Zero as his examples. The requests you have made of me are unrelated to the ability of these airplanes, or similar airplanes, to fly inverted.

As you know, I have offered to respond to your questions in another place. I will most likely respond on my Sandbox. When I have done so I will leave a suitable prompt at the thread I started on your Talk page. Dolphin (t) 16:28, 10 August 2018 (UTC)[reply]

Or, you could determine the appropriate dividing line within this section and then insert a sub-section header. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:57, 10 August 2018 (UTC)[reply]

Use of millimeter paper[edit]

Do people still use millimeter paper in a non-educational setting? Are there scenarios where it's a better choice than a laptop, tablet or the like?--Doroletho (talk) 15:59, 6 August 2018 (UTC)[reply]

Yes. Some technical professionals may use graphing software that is limited to particular subject matter or scales, and would use graph paper for graphs that aren't within the domain of their graphing software.

General purpose graphing software can be expensive, or hard to use, so a person who only needs to create an occasional graph may use graphing paper.

Also, sometimes it may be necessary to make a graph that is larger than the available printers. If this need only occurs once in a while, it may be more appropriate to use large paper. Jc3s5h (talk) 16:21, 6 August 2018 (UTC)[reply]

I've not come across the phrase "millimeter paper" before but the spelling is American. Knowing the aversion some Americans have to the metric system I would imagine they would use sheets printed with a grid of one-inch squares, each square divided into 100 smaller squares. 95.150.52.197 (talk) 09:49, 8 August 2018 (UTC)[reply]

1/100-inch graph paper is fairly unusable. Can one even visualize that small a space once the width of the line itself is accounted-for? 1/16 in or rarely 1/32 in is about the smallest ruler I've ever used in technical settings. But 1 mm is easily seen. "Millimeter paper" is a redirect where you can learn about it and the inch-fractions that are commonly found. DMacks (talk) 13:49, 8 August 2018 (UTC)[reply]

Or did "divided into 100" mean a 10x10 grid of 1/10-in squares? Those are available ("Engineering paper" entry on the graph-paper article). DMacks (talk) 13:51, 8 August 2018 (UTC)[reply]

Why are all notable nearby stars getting nearer?[edit]

Nearest Stars in a time range between 20,000 years in the past and 80,000 years in the future.

In this graph (original), every star shown has negative slope at time zero, implying that every one coming closer to Earth (before turning around sometime in the future), where I'd expect half of them to be approaching and half to be receding. Is there any reason this is so? Or are we just better at detecting or care more about stars coming towards us? Thanks, cmɢʟee⎆τaʟκ 17:39, 6 August 2018 (UTC)[reply]

Any object smaller than about a galaxy cluster would be under enough gravity to overcome the metric expansion of space, which means that their relative motions of those objects is determined primarily by non-Big Bang related motion (things like gravity and inertia). Many of the closest stars to our star are probably going to be ever-so-slightly attracted to each other due to gravity, while galaxies within our Local Group also moving together; the Andromeda Galaxy is rushing at us at a pretty good clip. There are going to be some stars which are moving away, but I would suspect that most of the closest stars are gravitationally bound in such a way as to be drifting generally closer. --Jayron 32 17:53, 6 August 2018 (UTC)[reply]

No need to invoke the big bang or expansion on the scale of a few light years... The figure is incomplete in the sense that it does not include all stars currently within 10 lyr, with Sirius at 8.6 lyr a notable omission. The figure on the right is more complete and shows that roughly half of the stars are indeed receding. The other figure was drawn after a figure in a German popular astronomy magazine, and seems to refer to this article, which deals with the successors to Proxima Centauri as the clostest star to the solar system. --Wrongfilter (talk) 18:13, 6 August 2018 (UTC)[reply]

Yes, to the original poster, sorry, but you've misinterpreted the graph. That graph depicts stars that will be nearest to the Sun in the astronomically-near future. List of nearest stars and brown dwarfs shows there are many more nearby stars, and not all of them are moving towards the Sun. --47.146.63.87 (talk) 00:11, 7 August 2018 (UTC)[reply]

Thanks, everyone. Now I understand that "Nearest Stars..." should be interpreted as "Stars that will hold the title of nearest star..." In this case, shouldn't Lalande 21185 be excluded? Thanks, cmɢʟee⎆τaʟκ 12:35, 8 August 2018 (UTC)[reply]

In 20,000 years Lalande 21185 will hold third place on the "nearest stars" chart, so I think it's worth a mention. — PhilHibbs | talk 13:57, 8 August 2018 (UTC)[reply]