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October 16

2 questions: Human powered flight and looking for the name of a theory

While watching the video in this article from NPR about the Sikorsky Prize for human powered flight, it made me wonder. Why would they build a system where the pilot has to use both his arms and legs? Wouldn't he be able to put more power into his legs if he could use his hands to brace himself? It seems to me that it would also be more fluid of a motion.

Also, it brings to mind a theory/principle/something that I read about here years ago. Basically it said that if you have a group of people pulling on a rope, the combined pull will have less force than if you were to add up their individual pulling forces when pulling one person at a time. Can anyone name what I'm thinking of? Our article used tug of war as an example but I can't find the term that I'm looking for when I look through what links to that article. Dismas|^(talk) 00:15, 16 October 2012 (UTC)[reply]

Doing a bit more reading, University of Maryland Gamera II Human Powered Helicopter says that "Up to 20 percent additional power for the 60 second runtime is achieved using this more complex method rather than pedal power alone." But it doesn't say why or cite a reference. Dismas|^(talk) 00:35, 16 October 2012 (UTC)[reply]

The reason is simple: By using both arms and legs, more muscles are brought into use, so greater power. For sustained effort, finite lung capacity comes into effect - evolution has provided us with lungs that are insufficient for maximum output from all muscles simultaneously, but for 60 seconds or so each muscle can run on oxygen held in the blood, and CO2 levels don't rise too much. Wickwack124.182.43.72 (talk) 01:13, 16 October 2012 (UTC)[reply]

Erm... doesn't swimming use both arms and legs at the same time? Alansplodge (talk) 12:53, 16 October 2012 (UTC)[reply]

I must admit I'm not a good swimmer, however I can swim. It seems to me that one's arms get to be used at something like full power, but while the legs contribute, they don't operate at maximum effort. Something that does use both, plus other muscles as well, at high effort is rowing in a racing skiff. I've never done that, but such races seem to be very short in duration. Wickwack 124.182.187.189(talk) 14:56, 16 October 2012 (UTC)[reply]

The contribution of different muscle groups depends on style and speed. For freestyle, about 80% of the force comes from the arms. For sprints, legs are important, but for long-distance swimming, you mostly only use the legs to keep the body in a good horizontal position. Those big leg muscles eat up oxygen like there is no tomorrow... --Stephan Schulz (talk) 21:14, 16 October 2012 (UTC)[reply]

Law of diminishing return / utility?GeeBIGS (talk) 00:57, 16 October 2012 (UTC)[reply]

I found the answer to my second question!! I trolled through my contributions a bit and it is... The Ringelmann effect! Dismas|^(talk) 07:05, 17 October 2012 (UTC)[reply]

Awesome. I added that to the article I wrote on diseconomies of scale. StuRat (talk) 07:22, 17 October 2012 (UTC)[reply]

Combined energy source vehicle ?

Extending the above question, wouldn't a plane that was neutrally buoyant (mostly filled with helium, or, if you dare, hydrogen), and covered with solar cells, in addition to being human powered, do better than one which is human powered alone ? (I realize this is likely against the rules of the competition.) StuRat (talk) 06:04, 16 October 2012 (UTC)[reply]

A lot better, though the solar cells are not going to help much. You're going to need some way of getting down though. See Larry Walters.--Shantavira|^{feed me} 07:22, 16 October 2012 (UTC)[reply]

StuRat's idea has been proposed before, but it's not a good idea. It depends on what you define as better. Sure you can use helium or hydrogen to make a craft neutrally boyant in air, but the required volume means it will have a big surface area and will want to go where the wind takes it, so you'll need a bigger power input to make it go where you want. This is why you see conventional winged airplanes carrying freight, but not airships. The power to payload ratio of winged aircraft overtook airships many decades ago. Wickwack 124.182.187.189(talk) 10:31, 16 October 2012 (UTC)[reply]

Good idea or not; it is in serious development as we speak; Not only do we have Martin’s HALE-D airship takes to the air but also Solar Ship: The hybrid airship with a low-carbon twist. Alansplodge (talk) 12:51, 16 October 2012 (UTC)[reply]

When they go into commercial production that might mean something. Meanwhile I classify them with nuclear aircraft engines (http://en.wikipedia.org/wiki/Nuclear_aircraft), Reagan's StarWars technology, and Rover gas turbine engines for cars - all of which got very serious funding, and all of which were totally dumb things technically to do, and consequently were scrapped. Wickwack120.145.19.16 (talk) 01:35, 17 October 2012 (UTC)[reply]

I realize this idea isn't practical as far as a commercial way to transport people or cargo. However, it might have recreational possibilities. That is, it might be a reasonable alternative to things like hang gliders, hot air balloons, and ultralights. And, just like those, you would be restricted to days with low wind, and, due to the use of solar panels, you'd also need bright sunlight. StuRat (talk) 02:17, 17 October 2012 (UTC)[reply]

Fair enough. But what we should do here is ask does combining gas buoyancy and power give us a synergistic advantage (or the best of two worlds) or does it just combine disadvantages? There's no doubt that hot air ballooning has considerable followers - I've tried it myself and its great fun, especially when you turn that great noisy burner off. Forget hydrogen - one bit of static discharge and your gone. Would helium catch on? Well, its pretty darn expensive - accepable for niche commercial purposes but too expensive for recreation. Now, does adding propulsion power to helium buoyancy give an advantage? Well yes it does, but once you have implemented propulsion power, regardless of the power source (gasoline engine, solar cells or whatever) you are better off with wings, as explained. That is, combine human power with solar power not in a buoyant lift craft but in a winged craft. But that's for chaps with olympic-standard fitness. What might have some synergistic recreational merit is a winged craft with solar cells covering the wings for cruising power, plus a small gasoline engine for take-off power. It should be possible as technology advances to get the weight of the solar cells and electric motor lower than the fuel weight for cruising range, and stay aloft for a long time that way. Wickwack124.182.144.228 (talk) 02:46, 17 October 2012 (UTC)[reply]

A gasoline engine would add weight, fumes, and noise. They could jump off of hills like hang-gliders do, to get started. And yes, my idea was hydrogen or helium filled inflatable wings with flexible solar panels on top (hydrogen is only flammable when mixed with air). I believe an unmanned vehicle was made like this, and stayed aloft for many days. But, my version wouldn't have batteries for night storage of energy and would add a pilot who could hopefully control the flight direction with a leg/solar-powered, turnable fan. The price of helium is a concern, especially if it was all just vented to deflate the aircraft for storage. Is there a way it could be sucked out and recompressed back into the tanks ? StuRat(talk) 06:39, 17 October 2012 (UTC)[reply]

I'm sure there is, though it might cost some significant energy. The problem is that if you have neutral buoyancy at some altitude, to descend you must either vent some helium (the usual way back in the 1930's), or compress it on board, which has been proposed before. The power to weight ratio of constant-speed gasoline engines is excellent, however maybe you should calculate how much cranking a human would need to do to control altitude by manual onboard gas compression. Don't be too concerned about the noise of an IC engine - you should hear the noise of a hot air ballon gas burner in full cry just above your head. My usual statement for this type of discussion on Ref Desk applies: Should a hand compressed helium lift wing work out, with or without photo-voltaic boosting, don't bother patenting it, folks, as by posting all these response StuRat & I have established prior art. Wickwack 124.178.143.72 (talk) 08:38, 17 October 2012 (UTC)[reply]

Perhaps make the vehicle slightly heavier than air, so that modest pedaling or sunlight on the solar panels provides sufficient thrust to provide the extra lift needed, via the airfoil. StuRat (talk) 08:52, 17 October 2012 (UTC)[reply]

I wondered if you'd come back with that. You'd need a lot more than modest pedalling if you got caught in a quite modest thermal. Wickwack124.178.143.72 (talk) 10:19, 17 October 2012 (UTC)[reply]

Just like a rip tide, you don't fight against a rising thermal, you get out of it first, then land. StuRat (talk) 01:34, 18 October 2012 (UTC)[reply]

Re hydrogen only flammable when mised with air (or oxygen) - yes, that's obviously true. Trouble is, it is the leakiest stuff imaginable, that tiny molecule can permeate such a range of things. Its pretty hard to guarantee it doesn't get mixed with air somewhere, it is combustible down to very low concentrations, and ignitable by the tiniest static discharge spark. Best avoided. Wickwack 124.178.143.72(talk) 08:38, 17 October 2012 (UTC)[reply]

Well, there is some risk, but I do feel the risk has been overstated as the result of the Hindenburg disaster (where the source of ignition was likely a flammable surface treatment). After all, most aircraft contain flammable fuels, it's just a matter of properly containing them to prevent explosions. Isn't there any flexible material which is impermeable to hydrogen ? StuRat (talk) 08:52, 17 October 2012 (UTC)[reply]

As the WP article covers quite well, many and various are the theories why the Hindenburg burned, and we cannot really know, but the official and most likely cause is a static electricity discharge. While aircraft fuel (gasoline or kerosene) is regarded as highly inflamable, there is a very considerable difference between these fuels and hydrogen. For a start, aircraft fuels being hydrocarbon liquids, they must first be vaporised and pyrolised (ie the molecules split up) before combustion can occur. Once vaporised they can all be ignited by a spark, but the required spark energy is very high in comparison. To ignite gasoline you need a clearly visible spark at least about 1 mm from any (cold) surface; for hydrogen you need only a spark too way too weak to be seen, and it can be right up against a surface. It's true that the extensive media coverage at the time including the live voice coverage turned the public off airships. But that doesn't change the fact that hydrogen filled airships, certainly of that design, weredangerous. Wickwack 124.178.143.72 (talk) 10:57, 17 October 2012 (UTC)[reply]

And yet, none of the 119 other Zeppelins suffered mysterious mid-flight explosions -- even shooting them with incendiary or explosive anti-aircraft rounds wasn't a reliable way of setting them on fire. --Carnildo (talk) 01:22, 18 October 2012 (UTC)[reply]

Neither did the Hindenburg suffer any mid-flight problem. The hazards come when coming down to land, and when tethered. Wickwack121.221.89.122 (talk) 02:42, 18 October 2012 (UTC)[reply]

Will canned food become hard to taste before the rust level becomes harmful?

Though if you eat it without the liquid the majority of the rust taste disappears. Hey I'm poor, I'm gonna eat it if up to 60% of the flavor is rust. Otherwise I throw away a dollar. Of course it also matters how often you eat from rusty cans and how many a lifetime. Why do most of them not taste like rust but a few taste bad? And I can't imagine how crappy the unlucky ones taste like by the time they're the "best by" age, which are like 12 times older than the ones in the supermarket). 96.246.59.161 (talk) 05:03, 16 October 2012 (UTC)[reply]

Whatever you're smoking, can I have some of it? Looie496 (talk) 05:40, 16 October 2012 (UTC)[reply]

I assume that English is a second language for you, and you meant to ask "Will I be able to taste that canned food is off before it becomes harmful, due to age ?". Most of the time, yes, but occasionally it might develop something nasty with no bad taste. I also don't think it's typically rust that you taste, it's more the coatings on the inside of the can and the oxidation of the contents. StuRat (talk) 05:53, 16 October 2012 (UTC)[reply]

I suspect this is getting a bit close to medical advice. FWIW from experience as a student it always seems that acid tinned stuff (grapefruit segments, tomatos) does get rusty. But although people have eaten rusty stuff no one is going to advise you its ok. The main ingredient of rust isIron(III) oxide which is insoluble and not going to do much and if the manufacturers of the tin had you in mind there might not be seriously bad stuff. --BozMo talk 05:55, 16 October 2012 (UTC)[reply]

It's worth remembering that as StuRat hinted at, the rust may not be your biggest concern with the tins. It's whatever has happened to the food due to whatever is growing on it thanks to the rusting/compromised tins. Nil Einne (talk) 11:49, 16 October 2012 (UTC)[reply]

Cans are always lined, and they don't rust on the inside. If they are damaged enough to expose the interior to rust, that's a very dangerous situation -- but it hardly ever happens. Most cans nowadays won't rust at all, unless exposed to something corrosive. The OP here is just making stuff up. Looie496 (talk) 14:57, 16 October 2012 (UTC)[reply]

Sure but if you can actually taste the rust as alleged by the OP or are going to be eating rust because of a rusting can as discussed by BozMo, then the dangers of the rust should not be high on your list of concerns. Nil Einne (talk) 15:02, 16 October 2012 (UTC)[reply]

What's in the coating that you taste it? Tin? It doesn't stop the 1.5 mm wide seam from rusting. Clearly if the can is damaged or has enough rust to eat it should be thrown out but I never saw one like that. 96.246.59.161 (talk) 23:22, 16 October 2012 (UTC)[reply]

Not sure what the metallic taste is. I've taken a bite from inside a can before, only to spit it right out due to a metallic taste, then check the expiration date and wonder if the can came with bell bottom pants and a rainbow colored disco wig. :-) StuRat (talk) 02:10, 17 October 2012 (UTC)[reply]

Beer in aluminium cans develops a metallic taste eventually, so rust isn't the cause in that case. Alansplodge (talk) 19:08, 17 October 2012 (UTC)[reply]

Why don't protons repel each other ?

As we know, protons are positively charged particles and it is also known that like charges repel each other. But protons occur in nucleus and they don't seem to repel each other. Why it is so ? What make protons and neutrons stick to each other ? Sunny Singh (DAV)(talk) 10:41, 16 October 2012 (UTC)[reply]

See nuclear force and strong interaction. Sean.hoyland - talk 10:51, 16 October 2012 (UTC)[reply]

(ec) Protons do repel each other by electromagnetic forces, but at the same time, the Nuclear forces hold the protons and neutrons together. These nuclear forces work only at very small distances. Even then, neutrons are necessary to stabilize the nucleus because these contribute to the binding by means of nuclear forces, but they are electrically neutral do they do not repel each other or protons. - Lindert (talk) 10:57, 16 October 2012 (UTC)[reply]

Actually neutrons stabilize the nucleus by reducing its Fermi energy - that's why stable nuclei have about as many protons as they have neutrons. Only for large nuclei does the electric energy becomes somewhat important - that's why heavier nuclei have an excess of neutrons (compared to their number of protons). Dauto (talk) 14:02, 16 October 2012 (UTC)[reply]

You can imagine the nucleus (especially heavy nuclei) as a balance of contradictory forces. When it gets out of balance, little bits can shoot out (radioactive decay) or, in extreme cases, the whole thing can rupture into two pieces (nuclear fission). Understanding that protons do repel each other, except at very small distances, also helps make sense of why nuclear fusion is so hard — you have to get the protons close enough to each other for the nuclear force to win out over the electromagnetic force (the Coulomb barrier — illustratedhere). --Mr.98 (talk) 14:54, 16 October 2012 (UTC)[reply]

If someone give further explanation of my first question, I'll thankful to him.
We know that protons and neutrons are made up of up and down quarks. I want to know which particles constitute electrons ? Sunny Singh (DAV) (talk) 02:00, 17 October 2012 (UTC)[reply]

I believe that the Standard Model posits that electrons (and other leptons) are elementary particles, not composed of other particles. --Trovatore (talk) 02:04, 17 October 2012 (UTC)[reply]

What particle contains three down quarks? Plasmic Physics (talk) 02:12, 17 October 2012 (UTC)[reply]

The Δ^- baryon. Dragons flight (talk) 02:21, 17 October 2012 (UTC)[reply]

See also the list of baryons, for the composition, names, and symbols for all of the baryons. TenOfAllTrades(talk) 02:26, 17 October 2012 (UTC)[reply]

What is unclear to you about the examples already given to you to the first question? The basic answer (the nuclear force) is discussed above, along with some more complicated refinements (e.g. Fermi energy). Protons do repel one another because of the electromagnetic force, but at short distances the nuclear force is dominant, which causes nucleons to attract to one another. The linked-to articles above go into nearly endless technical details. Just let us know what is unclear and someone will happily elaborate or translate into laymen's terms. --Mr.98 (talk) 02:45, 17 October 2012 (UTC)[reply]

Yes, you are right Mr.98, linked articles provided enormous information of my first question. But, what about my second question. Sunny Singh (DAV) (talk) 11:16, 17 October 2012 (UTC)[reply]

Re-read Sean Hoyland's answer. --Jayron32 11:57, 17 October 2012 (UTC)[reply]

They form a nucleus by falling in love, which is a stronger force than electromagnetism yet operates over a larger distance.Minorview (talk) 02:52, 23 October 2012 (UTC)[reply]

LOL! Double sharp (talk) 13:36, 23 October 2012 (UTC)[reply]

Thermodynamics of melting salted ice

Hello,

Considering a bath of ice at -10°C to which NaCl salt is added on surface, hence forming a layer of eutectic ice NaCl•H₂O (melting at -22°C instead of 0°C), the Gibbs free energy of the melting eutectic ice in a 25°C thermostated-environment will roughly have the following form:

$\Delta G_{\text{fusion}}=\int \limits _{\text{-10°C}}^{\text{-22°C}}c_{p}({\text{salted ice}}),\mathrm {d} T+\Delta H_{\text{fusion}}+\int \limits _{\text{-22°C}}^{\text{25°C}}c_{p}({\text{salted water}}),\mathrm {d} T+Entropy>0$

Experimentally, the use of a thermometer will show that the salted ice goes to -22°C when melting, hence showing that the first term of the equation (integral from -10°C to -22°C) describes the first physically occuring response of the system when heated to 25°C. This integral being the only negative term of the equation (entropy aside), my question is: which energy is transferred to the environment during the -10°C → -22°C transition? Would it be a heat transfer? Thanks, 188.194.48.183 (talk) 11:10, 16 October 2012 (UTC)[reply]

The temperature drop that occurs when the salt is added to the water comes from the net endothermic solvation process. The breaking of the Na-Cl ionic bonds is more endothermic than the formation of the solvent-ion bonds is, so it's just an endothermic heat of solution. That's why the temperature of the solution drops: the energy is being used to cause the NaCl to dissolve. --Jayron32 21:44, 16 October 2012 (UTC)[reply]

That makes a lot more sense put that way. Thank you! 188.194.48.183 (talk) 20:25, 17 October 2012 (UTC)[reply]

Likely heat situation of the ~26-mile BASE jumper yesterday

Regarding the man who went up in a helium balloon and jumped from what I'm hearing is about 26 miles altitude and reaching about 834 mph, I saw in a brief shot on the news that he was wearing a sort of space suit that seemed to be made of flexible fabric material. I wonder, given the composition of the atmosphere at the point at which he was going 834 mph, what was the likely temperature (ballpark figure) due to friction?20.137.2.50 (talk) 13:37, 16 October 2012 (UTC)[reply]

That's a good question. The suit Felix Baumgartner was wearing was a heated pressure suit, intended to keep him at a comfortable air pressure, and nice and warm, in the near vacuum of such high altitudes. That 834 mph was possible because, at that time, he was falling through so little air that there was little friction acting on him. The terminal velocity at low altitude is about 120 mph. So Baumgartner will have slowed from 800 to maybe 100 mph before he deployed his parachute. That speed will all have been lost due to friction (most in the last minute or so). Handwavingly assuming Baumgartner, suit, chute, et al weighed in at 200kg, I figure his kinetic energy was 12.7 MJ at the fastest point, down to about 0.2 MJ when he was at ~100 mph - meaning he'd have lost 12.5 MJ of kinetic energy. Not knowing much about the aerodynamics of falling spacesuits, I don't know how much of that went to eating up the air, and how much to heating up the suit. -- Finlay McWalterჷTalk 14:08, 16 October 2012 (UTC)[reply]

Hopefully most of that energy heated up the air! I estimated a little bit differently, assuming that all the gravitational potential energyneeds to convert to heat (well, most of it, anyway); so I got about 35 MJ using the standard "PE = mgh" formula. To order of magnitude, it's the same as Finlay's estimate. And, assuming a human is mostly water, with a specific heat capacity of 4.18 J/g·K, that's enough energy to raise a 90 kg human by almost 100 degrees centigrade! That's probably fatal! Of course, the rate that heat is conducted and radiated to the surrounding air is very rapid; and a human in a space suit is not well-approximated by an equal mass of water; but to first order, we have to assume that the majorityof that thermal energy is lost to air and/or ablation of the suit - hopefully in an engineered fashion. Nimur (talk) 15:56, 16 October 2012 (UTC)[reply]

[1] says that a failure of the heater in his face plate was not an issue in the capsule but would be on jumping. This confirms your impression that the rapid cooling effect from the passing -60 degree air is overwhelming the heat from reentry. Reentry from space involves much more lost potential energy and much less cooling by contact... in any case, I think your estimate and answer puts us in the lead! :) Wnt (talk) 17:06, 16 October 2012 (UTC)[reply]

I am sure there was some heating effect but I doubt it was a large one. 800 mph is fast, but is still very much less than the speeds of spacecraft re-entering the atmosphere from low Earth orbit - about 16,000 mph - or the speeds at which meteorites enter the upper atmosphere - at least 25,000 mph. Gandalf61 (talk) 14:55, 16 October 2012 (UTC)[reply]

One of our competitors has taken up the question, but isn't doing any better than we are so far. For the glory of the Wikipedia Refdesk, this is now officially a contest. :) Wnt (talk) 16:58, 16 October 2012 (UTC)[reply]

When he was going 800 mph, that was the "local" terminal velocity, which gradually became less as the atmosphere became more dense. So, when he was going 800 mph, the air resistance was equal to his weight, therefore the dissipated power at that time and later was his weight times his velocity and that was maximal right at that time when he reached maximal velocity. So, if he weighs 80 kg, the maximal dissipated power he experienced would be about 290 KW. This is way more than he could have sustained on the long run, but then this only lasted for a matter of seconds.

Another way to see that 800 mph is a problem, is to compare the velocity at which the air molecules hit you at that velocity with the how fast they hit you at rest and compute how high the temperature needs to be for this to be the same. You can roughly estimate this as follows. At rest at -40°C, <v^2>is about (455 m/s)^2 )(using the formula 3 k T/m), the component in the direction of motion is a third of this, so sqrt[<vz^2>] = 263 m/s. If we add to this the 800 mph, we get vz = 623 m/s, so relative to Baumgartner, sqrt[<v^2>] = (sqrt[623 m/s)^2 + 2 * (263 m/s )^2] = 725 m/s. This is the same as being at rest at a temperature of 317 °C. In reality it isn't the same, because the frequency of the collision is less at lower densities. Count Iblis (talk) 17:09, 16 October 2012 (UTC)[reply]

Additional complexity arises because these speeds are in the transsonic and supersonic regime; in other words, an air molecule traveling at 725 m/s at stratospheric densities is not in kinetic equilibrium. Each molecule is, in fact, in shock. I don't believe you can apply conventional gas temperature calculated from velocity according to the equipartition theorem, in this regime. Our article has information onattached shock, for molecules impacting the falling person; and detached shock, for the aerodynamic phenomenon that will inevitably form. Aerothermal heating in the transsonic and supersonic regime is widely studied. For the very enthusiastic thermodynamicist, NASA makes available an excellent free text, Facing the Heat Barrier: A History of Hypersonics - about the physics at much faster than super-sonic speeds. Aerodynamics of "hypersonic" flight are defined by the gas regime when thermal effects aremore important than any other physical aerodynamic effect. This is usually around Mach 5 or higher. This is one reason that spacecraft reentry systems use reaction control systems, and not flaps or parachutes - because conventional aerodynamic control surfaces do not behave as expected at these speeds. This book mostly applies to hypersonic aircraft; but I also found the more conventional book, Entry heating and thermal protection, which seems to provide a lot more practical approach, covering material more applicable to jumping out of a balloon at 120,000 feet. Nimur (talk) 19:38, 16 October 2012 (UTC)[reply]

Let's expand this a little. 80 kg * 9.8 m/s^2 = 784 newtons (kg m/s^2) of weight. The negative work (energy released) done by this weight is force x distance = 784 kg joules (m^2/s^2) per meter fallen. We know he falls 0.447(m/s/(mi/hr) conversion factor)*(800 mile/hour) = 357 m/s, so the energy released is his weight 784 kg m/s^2 * his velocity (357 m/s) = 280000 kg m^2/s^3 = 280000 watts. If I didn't foul up, that's a lot of power -at least twice what an automobile engine produces, according to the article. But for each second of this power output he passes through 357 meters of the surrounding air, releasing the original 784 joules per meter. So the question is, how much of a temperature differential will it require to get the surrounding air to take up 784 joules as it rushes past? Hmmm.... Wnt (talk) 18:02, 16 October 2012 (UTC)[reply]

Too bad they couldn't have fit one more thing, a temperature sensor, into his chest pack, if only for my personal curiosity's sake :). An ounce of actual measurement is worth a ton of theory. But from the mission's own site regarding hispressure suit: "The suit is designed to provide protection from temperatures of +100°F to -90°F." I guess there's my ballpark figure from the people closest to the real and true answer20.137.2.50 (talk) 20:50, 16 October 2012 (UTC)[reply]

So, from above, we have that he must be dumping 0.2 MW into the air at his peak velocity. Assuming an air density of 0.03 kg/m^3 at that height and an effective cross-section of say 0.8 m^2, that implies about 9 kg / s of perturbed air. With a specific heat of around 1 J / gram / K, that ballparks the heating of the air at around 25 Kelvin. So, it would seem that his problem is likely to be cooling due to low ambient temperatures rather than overheating due to friction, at least during the fastest part of his trip. Dragons flight (talk) 00:52, 17 October 2012 (UTC)[reply]