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November 22

Big Xbox is watching you?

Trying to ask this question without sounding like a loon, so here goes:-

Is it technically possible for Them (CIA/MI5/whoever your local Them are) to use Xbox One's controversial new Kinect faciliy to watch and record you in your own home wihthout your knowledge?

Just curious about the technological side of it - not trying to advance any conspiracy theory here. Needless to say, They will probably find some way to watch you anyway, if They are interested in your activities, wheteher or not you happen to have an Xbox One in your home. Thanks. --146.90.108.78 (talk) 00:01, 22 November 2013 (UTC)

• While there are plenty of conspiracy folks speculating on this topic, there are some legit sources that express concerns (e.g.: [1]).

You might have better luck regarding specific tech details here: Wikipedia:Reference desk/Computing; but, I am certain that if it could be done, it will be done. See also: PRISM (surveillance program) ~E:71.20.250.51 (talk) 01:07, 22 November 2013 (UTC)

A web-connected camera and microphone can be accessed by third parties without any real trouble by people in the know for these kinds of thing. It is absolutely possible. Is it happening? I don't think They care that much about you, but it is technically a thing that could happen. Norton explains possibilities for webcams here, and I have no reason to think Xbox is not vulnerable. Mingmingla (talk) 01:32, 22 November 2013 (UTC)

The companies aren't even trying to hide the spying any more. Just look at any set of Terms and Conditions, any company - for example, Kinect. [2] You don't own it, you can't examine it, you can't export it anywhere that allows people to look at it, you don't have any claim to privacy, you don't have any access to the courts, and they can do whatever they want to you at any time. The bottom line is that the only security you have with any computer equipment (so long as you are allowed that) is to physically disconnect things, preferably by never taking them into your home. We are far, far beyond the level of technology that our social mores are able to sustain, and there will be a reconciliation of the overall tech level on one side or the other. Wnt (talk) 01:58, 22 November 2013 (UTC)

"They" have always had the technical ability to spy on you through any web-connected camera (see Webcam#Privacy), provided they are able to inject code into your machine. My own privacy concerns (over stupid trolls, rather than the government) has always led me to disable the webcams on any computers I buy, or receive from my employer. Someguy1221 (talk) 02:02, 22 November 2013 (UTC)

• Anyone who has a camera or microphone permanently enabled on an internet-connected device is completely crazy in my opinion. I'm not even particularly paranoid, but it scares the sh*t out of me. Why would you even take the remote chance that some pervert or lunatic could spy on you in your home? I don't understand how people can be so blasé about it. I'm not happy unless the camera is physically disconnected or covered when not in legitimate use. Unfortunately, in-built microphones are harder to physically disable because even covering them won't completely block the sound. 86.169.185.129 (talk) 04:04, 22 November 2013 (UTC)

What's the difference between a Kinect and every laptop and cell phone sold in the past 10 years? It seems like there are hundreds of millions more of those devices compared to a gaming console that just launched. Yet no one seems particularly paranoid about them. Why is a Kinect any different? --209.203.125.162 (talk) 23:49, 22 November 2013 (UTC)

What you want is a white noise machine and an earbud. Still, better keep up with the happy thoughts, just in case. InedibleHulk (talk) 04:35, 22 November 2013 (UTC)

Those who have nothing to hide have nothing to fear. This is the high-tech equivalent of Will Rogers' axiom, "Live your life so's you wouldn't be ashamed to sell your parrot to the town gossip." Just in case, though, you could disable the camera by simply placing a piece of electrical tape over it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:32, 22 November 2013 (UTC)

A quick question, what's your banking account and password? Oh and a list of the numbers, CVVs, expiry dates, full name, addresses, sample signatures and PIN numbers for any credit or debit cards you own would be great too. Nil Einne (talk) 13:24, 22 November 2013 (UTC)

Well you're at it, I want to know what your sexual habits are, and what you look/sound like when having sex. Creeped out? You should be. --Bowlhover (talk) 15:24, 22 November 2013 (UTC)

I don't do bank accounts or credit/debit cards - I operate strictly on a cash basis. As for the other thing, just picture the typical rabbit during mating season. Creeped out? You should be. >:) ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:11, 22 November 2013 (UTC)

That's fine then. You don't sound like you're extremely poor, so a quick question where you store your spare cash (specific locations, e.g. if in your house under your mattress specify that level of detail as well as a full address preferably including GPS coordinations to a 10 metre or so range)? Can you also let me know where you store any keys etc required for access for any of said locations as well as any passwords, pin numbers etc for safes, combinations locks, secured guards, whatever. Oh and don't worry if my friend visits these locations sometime soon, they're just checking this stuff out, ignore them since you have nothing to fear. Nil Einne (talk) 15:46, 23 November 2013 (UTC)

The "you have nothing to hide" thing isn't true. To take a real example - my local neighborhood has a problem with people letting their dogs poop on the sidewalk. I proposed to our home owners association that we provide free poop bags at stations around our neighborhood and needed to research the cost. So I Googled for people selling those bags. Unfortunately, I accidentally clicked a "sponsored link" instead of just a regular search result. Consequence: For the last month, I'm plagued with adverts for dog-poop cleanup devices on just about every freaking web page I visit! Even to the point that I'm trying to find a recipe for a dish at thanksgiving - and when I print it out for my wife, it has TWO dog-poop-related adverts on the printout! Aaaaarrrrgggggh!

Privacy is a huge issue. Big businesses want to know as much as possible about you so that they can target you with adverts and other things that will entice you to spend more money with them. There is strong evidence that people like Amazon dynamically vary their prices depending on what they know about you. There is a very strong possibility that something you do online (thinking you're doing it in private) will result in them jacking their prices up when next you go to buy a book or something from them. Dunno about you - but I'd rather that didn't happen!

SteveBaker (talk) 15:51, 22 November 2013 (UTC)

While your response is correct, I don't think you have looked far enough forward to grasp the full problem. A merchant might say that his knowledge of you might entitle you to a discount instead; that's the first step. What about when they start saying that your Facebook/Twitter status determines the attractiveness of your online persona? What about when you're told that you have to have a degree in Microsoftology if you expect to be hired anywhere? What about when failing to be filmed for a few hours of your life is treated the same way as a ten-year gap on your resume is treated now, and they treat you like a terrorist? It is possible that the Revelation of John provides as good an insight as is to be had on such society, and on the fate of those who fail to comply with its dictates. Wnt (talk) 18:20, 22 November 2013 (UTC)

Steve, you might be able to fix the flood of dog poop adverts by clearing the cookies from your browser; or am I being naieve to the reach of the dog poop or big brother? Astronaut (talk) 20:05, 22 November 2013 (UTC)

Snakebite

If you get bitten by a venomous snake, is it possible to suffer permanent damage? This is NOT a request for medical advice. 24.23.196.85 (talk) 01:42, 22 November 2013 (UTC)

Well, yes, such as death (I assume the science reference would consider death permanent). Check out the article snakebite for some other long term effects. (In that article there is a picture of a particular injury that may be upsetting. Proceed with caution.) 88.112.41.6 (talk) 01:52, 22 November 2013 (UTC)

Definitely. The Hemotoxin bites in particular can cause huge wounds that may never fully heal. Google and you'll see lots of awful pictures. Shadowjams (talk) 03:13, 22 November 2013 (UTC)

Power required for hovering

Hi, please see the question here which I asked a quite a while ago. Returning to this problem, I could not remember how I had worked it out, so I tried to do it again. However, this time I get the answer Power = (M g)^(3/2)/(2 sqrt (rho A)), which is different by a factor of sqrt(2). Although the original answer was said to be correct by one respondent, I now doubt whether it was. Can anyone adjudicate as to which (if either, I suppose) is correct? 86.160.221.246 (talk) —Preceding undated comment added 03:52, 22 November 2013 (UTC)

The problem is under-constrained. There are literally an infinite number of correct answers. I gave a very technical answer, with citations and specific examples, when this was asked again in "Hover," in November 2012. "Helicopter flight is characterized by aerodynamic work against a non-conservative force. What this means in practice is that the amount of energy or power required for a maneuver depends on the helicopter's configuration. A helicopter pilot can independently control the collective, the cyclic, and also throttle the engine rpm. The pilot can also change the attitude of the helicopter to maintain a hover at many different engine settings and aircraft attitudes."

If your "ducted fan" has fixed blade pitch, your problem becomes a little bit better constrained; but you still need to completely define the angle of attack and the profile of the "fan" blade airfoil shape. Nimur (talk) 04:32, 22 November 2013 (UTC)

In a real-life helicopter there are masses of other considerations beyond what I am trying to calculate in my highly simplistic model. Forget all those. I believe that what I am trying to calculate is well-defined. It is just based on the amount of kinetic energy that must be imparted to the air in order to generate the right force to keep the mass hovering. 86.129.18.115 (talk) 12:15, 22 November 2013 (UTC)

Correct, but because of the way that physics works in the real world, there are many ways to put different amounts of kinetic energy into air, and achieve the same effective net force. Ultimately, the problem with your approach is that you are trying to apply conservation of energy, but you want to "conveniently ignore" the numerous ways that energy can manifest in air: as turbulence, as cyclic flow, as heat, as convective flow, as laminar flow, as a compression or pressure perturbation, and so on. Because the engine performs no useful work (when hovering, the force is exerted over no net distance upward), you must use a model that considers the energy lost to lossy terms like aerodynamic drag and turbulent flow and heat. These are not "optional" parts of the equation.

The real work - force exerted over a distance - is performed horizontally, against the air in the form of the rotating fan-blades pushing against the air as it turns. Work is performed against the drag term. How much work? Depends on the drag - which depends on the airfoil shape, the angle of attack, an| the parameters of the air, like density and viscosity; and on non-ideal terms that characterize turbulent flow.

If you can't quite grok how there can be an infinite number of correct solutions to a physical equation, it's time to review the theories covered in advanced algebra or ordinary differential equations. There are an infinite number of correct answers, but that is not identical to saying any solution is correct. Nimur (talk) 12:41, 22 November 2013 (UTC)

Yes, I do want to conveniently ignore all the other complexities. I know that the answer obtained will have little to do with the amount of power needed by a real-life helicopter. All I want to do is confirm the answer to the question "how much power to impart sufficient kinetic energy to the air to keep hovering", ignoring absolutely everything else, no matter how unrealistic it is in the real world. 86.129.18.115 (talk) 12:51, 22 November 2013 (UTC)

Let's replace air with a solid material, then, since you are choosing not to treat it a fluid - you are intentionally ignoring the most fundamental and critical properties of air ("it can flow"). So, the amount of energy required is zero. The hovering machine rests on a solid material, exerting a contact force equal to, and exactly countering, its own weight. No work is performed, because no force is exerted through any distance. This model is only accurate as long as we can ignore the properties of air as a fluid. Nimur (talk) 13:00, 22 November 2013 (UTC)

Or, let's take the other hypothetical extreme, treating air as a fluid that is so perfect that it never exerts any aerodynamic drag, nor ever exhibits any turbulent motion. The fan blades begin to spin, and lose no energy to the air! But there's a problem... the blades pass effortlessly through the air. They just ... go right through the air molecules, in total contraindication to all prior physical intuition! Because the fan blade cannot even feel the air, it cannot exert a force to push any air downward. (That would require drag and shear force to direct the rotational flow downwards). The fan keeps spinning, and keeps spending energy against its own internal friction, but air never flows, and the vehicle cannot hover.

Maybe you're starting to get the intuition that these "idealizations" are not only unrealistic - they're useless, ...and they aren't even consistent with any theory we could contrive if we started with first principles of physics, solving the kinetics for one single gas molecule, and then extrapolating statistically to billions and billions of other gas molecules. We need to deal with terms like shear force and viscosity. Those terms describe the way air really behaves, whether we observe it in a lab or if we model it with the kinetic theory of gases. If you choose to ignore those terms, you aren't writing "simplified" equations about your problem: you're just describing something else entirely.

We use drag coefficients and similar aggregate parameters so that we don't have to solve the harder, full-form analytic equations that describe fluid flow as an absolutely immense n-body problem with n=6.022x10²³. It's already in a "simplified form!" But if you just leave out these terms, you're going to get answers that are so incredibly wrong that they predict perpetual motion, or free energy, or zero energy, or something else totally ridiculous. Nimur (talk) 13:18, 22 November 2013 (UTC)

I see that there is a complete failure of communication here, in terms of me conveying the problem that I am actually trying to solve. If it helps, forget air. Forget fans. Forget all the complexities that are plaguing you. Imagine a device being fed sand. It has to fire a stream of sand downwards at sufficient rate to keep hovering using some imaginary mechanism that wastes no energy. The density parameter is the density of the stream of sand emanating. 86.129.18.115 (talk) 13:42, 22 November 2013 (UTC)

Even then, the answer is still underconstrained. A smaller amount of sand at higher velocity can accomplish the same task - providing identical impulse - as a larger amount of sand at lower velocity. The power expenditure depends on how you feed sand into the machine: are you constrained by mass flow rate? Are you constrained by maximum sand-pump power? None of these? Then, if you have a control system to ensure the craft hovers, you still have one unconstrained variable, so the engine can run at any power setting; a different amount of sand needs to flow at a different velocity depending on the power setting. That has to be controlled independently of the power setting. Your control system must calculate and match power- and flow-rate, (maybe it "opens the sand valve" to the desired level, at the same time it throttles the engine power to the desired level; alternately, the engine power can be set, and the "sand valve" is opened or shut); or else the craft won't actually hover. Again, because you're choosing to ignore energy loss terms, you've left open the prospect that we can, for example, take a 200 horsepower engine and use it to impel one single grain of sand at some preposterous hypersonic velocity every second, thus levitating the vehicle. A real engine cannot actually apply its entire power to drive a single grain of sand. Nimur (talk) 14:12, 22 November 2013 (UTC)

The power expenditure does indeed depend on the flow rate of sand. That is accounted for by the cross-sectional area and density parameters. You are correct, the resulting equation would in theory work with a tiny flow rate and extreme speed. That is as expected. 86.129.18.115 (talk) 14:17, 22 November 2013 (UTC)

...so, power is force times velocity, which is just "weight times sand-velocity" in this case. Interestingly, that's the gravity-burn equation I linked for you... nearly the first answer you got, back in 2011! But, explain again why cross sectional area limits the sand velocity? Perhaps you mean to say ... the flow is choked, because it is non-ideal? Fascinating, that we would need to model non-ideal parameters for a simplified problem like this! Nimur (talk) 14:20, 22 November 2013 (UTC)

(edit conflict × 2) In the first round Dragons flight spoke of the air being incompressible; note that your sand idea is at odds with this (because obviously sand grains can be brought closer together with no energy). That said, the equations I can see here are $F=Mg={\dot {m}}\Delta v$ (conservation of momentum, and a sand supply separately suspended at your altitude), ${\dot {m}}=A\rho v'$ (note that this depends delicately on the issue of compressibility), and $P={\dot {m}}(v'^{2}-v_{0}^{2})/2$ (rate of kinetic energy deposition). For sand, we can suppose that $v_{0}=0$ (a horizontal conveyor feed), so $v:=\Delta v=v'$ and we can simplify to $Mg=A\rho v^{2}$ and $P=A\rho v^{3}/2={\frac {(Mg)^{3/2}}{2{\sqrt {A\rho }}}}$, which is what you have this time around. At least three of these four equations are separately wrong for a fluid: a fluid exerts pressure forces in addition to reaction forces, has a compression-dependent flow rate, and is moving when it arrives at the impeller. (Of course, if Dragons flight has a derivation of the older form, let's hear it; as Nimur says, this is treacherous ground.) --Tardis (talk) 14:28, 22 November 2013 (UTC)

Thank you so much for understanding what I was asking! It looks like last time I was off by a factor of sqrt(2), as I suspected. 14:39, 22 November 2013 (UTC) — Preceding unsigned comment added by 86.129.18.115 (talk)

Tardis, you didn't apply the product rule on your derivative to compute "rate of deposition of kinetic energy":

${\frac {d}{dt}}{\frac {1}{2}}mv^{2}={\frac {1}{2}}(v^{2}{\frac {dm}{dt}}+2mv{\frac {dv}{dt}})$... surely you don't think we can idealize away that velocity term? And that's why your answer for power totally lacks a term proportional to exhaust velocity. I'm not so very good at following equations, but I can spot an error a mile away if it forgets a term that has physical significance... and in this case, you left off v_E. Not that it's important, it's just where the majority of the energy is getting put! It also makes for a catchy song lyric... udv +vdu, you left off vE... something wasn't quite right about it... Nimur (talk) 15:11, 22 November 2013 (UTC)

I don't know whether this answers your question, but in my working-out the exhaust velocity gets accounted for in the form v = sqrt(F/(A rho)). 86.129.18.115 (talk) 18:05, 22 November 2013 (UTC)

I'm not quite sure which of those two terms I'm omitting: the first is appropriate if you consider the "object" to be the slug of falling sand, which (neglecting gravity on it since we don't have to pay for that) is moving at a constant velocity but has a growing mass (yielding $P={\frac {1}{2}}v^{2}{\dot {m}}$: here v is a constant, so is your $v_{E}$), while the second is appropriate if you consider an infinitesimal mass — but then you need to integrate over the time during which the impulse is applied, giving $dE=dm\int _{0}^{t}v{\frac {dv}{dt}}dt={\frac {dmv_{E}^{2}}{2}}$ so $P={\frac {dE}{dt}}={\frac {v^{2}}{2}}{\frac {dm}{dt}}={\frac {1}{2}}v^{2}{\dot {m}}$ (where I have again taken $v:=v_{E}$). (There are, I'm sure, some very technical arguments that could be made about integrating and then differentiating with respect to t in two different senses, but the rigorous approach of considering the simultaneous distribution of energy parcels of mass that are at different points in their acceleration will yield the same result because of linearity.)

Anyway, I didn't think of it with that equation at all. I simply said that each parcel of material changes from 0 velocity to $v:=v_{E}$, so its specific kinetic energy change is just the final value ${\frac {v^{2}}{2}}$. Since it is at a rate ${\dot {m}}$ that we dole out that specific kinetic energy, the power is just the same product from before.

You're quite right that specific impulse is relevant. However, here it is constrained by the known $\rho$ and A, whose product is a linear density. It is an entirely fair objection that $\rho$ at least is not known a priori, but must be found by equation of state considerations from the flow and varies non-trivially over space and time. However, our trivial "sand conveyor" case is reduced to one parameter in a physically plausible way (though the engineering is absurd) by the specified density. A different question would be "how can we minimize the power required by changing the output density?", to which the trivial answer is of course "make the density as large as possible to mimic a solid support where the power is 0". --Tardis (talk) 01:39, 23 November 2013 (UTC)

I believe that the solid support case is qualitatively different, and cannot be approximated or approached by making density large. 86.128.4.176 (talk) 14:29, 23 November 2013 (UTC)

I think the best "qualitative" interpretation I have is that the first term reflects energy added to the slug of sand to move it through the impeller, and the second term reflects the energy added to each slug of sand to accelerate it through the impeller. It's not really physically meaningful to distinguish between these components of the equation, except in the infinitesimal sense: there is an instantaneous change in energy associated with the velocity, and another term associated with the acceleration, for each infinitesimal unit of mass. Nimur (talk) 09:01, 24 November 2013 (UTC)

Your question has no reasonable answer. The energy required is entirely dependent on the nature of the fluid you're pushing downwards.

• An object can sit on top of a block of concrete indefinitely using no energy whatever - the power is zero.
• I have a desk ornament of Dr Who's Tardis that floats effortlessly in the air using magnetic levitation - again, zero power consumed.
• Work done is force through a distance. If the object is perfectly still - with a perfectly efficient mechanism, the distance travelled is zero and the power is zero. Ergo, any power that is needed is "losses" due to a less than perfectly efficient levitation scheme...so the amount of power required depends only on the relative efficiency of the machine.
• A hovercraft needs some power to keep it's soft skirt inflated - but otherwise is sitting on a block of air.
• Consider a helicopter, which needs much more power than a hovercraft because the air isn't contained and is continually leaking out of the sides.
• Even in a helicopter, the numbers are drastically different between a 'ground-effect' hover and a high altitude hover or a hover beneath an overhang (eg hovering underneath a bridge).
• Those water jet packs keep someone hovering very nicely with very little power because their working fluid is water - but regular rocket packs need huge amounts of power because they are exhausting hot gasses.
• A rocket hovering in a vacuum needs different amounts of power than one hovering in air - which is different again to a jet motor doing the same job.

Anyone who claims to have a simple formula for such a general question is talking nonsense. I think this was adequately explained to you the last time you asked this. One person gave you some equation or other - but it was incorrect, please don't latch onto that as "The One Good Answer That I Got" - it wasn't...it was the one complete bullshit answer you got!

If you can narrow things down more specifically to precisely how this "levitation" is being performed, we'd have a chance at getting a better answer for you.

SteveBaker (talk) 13:55, 22 November 2013 (UTC)

Please read my last message. The question that I am posing is well-defined and has an exact answer. 86.129.18.115 (talk) 14:11, 22 November 2013 (UTC) Also, ignore ground effects. If any other real-life complexity occurs to you, IGNORE IT. 86.129.18.115 (talk) 14:13, 22 November 2013 (UTC)

There's some out-of-order posting here, (which is okay, just a little confusing). I've responded above. Your new problem restatement is still underdefined. Nimur (talk) 14:16, 22 November 2013 (UTC)

I completely understand that you're asking for a spherical cow kind of answer - and I can tell you that the equation for that is "P=0". Gravity is a "force", not an energy source - so in the abstract "spherical cow" universe, producing an equal and opposite "force" requires no energy whatever - which is why magnetic levitation consumes no energy and this little TARDIS on my desk hovers so effortlessly - or why we could place an object at one of the "Lagrange points" and it would just sit there without falling.

Power equals force times distance moved...if the distance moved is zero, then the power/energy-consumption is zero too. The problem is that 100% of the energy consumed in a practical machine is in producing that force in a situation where there is nothing solid to push against. The details and losses that you're trying to ignore are 100% of the resulting power requirement - so you don't get a simple abstract answer. Many people here are trying to tell you that - they know the answer and you don't (which is why you asked them!) - so stop arguing with them and listen! I'm sorry, life sucks, there isn't always a simple answer to even the simplest question. SteveBaker (talk) 14:35, 22 November 2013 (UTC)

The question that I am asking is well-defined and has an exact answer that is not zero. It is not difficult to solve for anyone with a knowledge of physics. For me, however, it is on the margins of what I can work out. The biggest difficulty appears to be explaining to other people what the question actually is. 86.129.18.115 (talk) 14:42, 22 November 2013 (UTC)

I'm sorry - but you say that you don't know what the answer is - but you somehow do know that it's not difficult and you know that it's not zero and you know that it's well-defined? Well, if you know so much, how come you don't know the answer? The truth is that you know none of those things - really - you don't. Everyone here is telling you that. Ask yourself what information you have that tells you that there it's "not difficult to solve"? How do you actually know that? The answer is that you don't...you're guessing that it's easy to solve, that it's well-defined and so forth - and we're telling you that your guess is incorrect. So, you asked us a question - we've told you the answer. Stop arguing and listen! SteveBaker (talk) 15:36, 22 November 2013 (UTC)

No, you listen. "We" have done nothing. At least one person understands what I am asking. The fact that you can't grasp it is your problem, not mine. 86.129.18.115 (talk) 18:00, 22 November 2013 (UTC)

See the article on thrust, specifically https://en.wikipedia.org/wiki/Thrust#Thrust_to_power. --Modocc (talk) 15:02, 22 November 2013 (UTC)

Hey, is that the same as my equation??!! 86.129.18.115 (talk) 18:34, 22 November 2013 (UTC)

Or see this pdf: www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf‎ (notice the square root of 2, so your old answer was correct). This is a typical physics introduction handbook problem, one that a highschool student should be able to answer. Judging by the reactions you got, most people here would have flunked their physics exam. Ssscienccce (talk) 18:58, 22 November 2013 (UTC)

Here's another one: FTM106: U.S. Naval test Pilot School Flight Test Manual chapter 5 Hover performance Power required assuming an ideal rotor: equation 5.24 Ssscienccce (talk) 19:26, 22 November 2013 (UTC)

And we even have an article on it: momentum theory Quote:

For a stationary rotor, such as a helicopter in hover, the power required to produce a given thrust is:

$P={\sqrt {\frac {T^{3}}{2\rho A}}}$

Ssscienccce (talk) 19:32, 22 November 2013 (UTC)

Do you have any idea why the formulas here and here seem to differ by the same factor of sqrt(2)? My working-out was based on my pretty sketchy knowledge of some basic physics equations, and I did not consult any of these references, yet spookily I seem to have produced two differing equations that match the two differing equations in the articles. Unfortunately, I cannot now remember how I arrived at my original equation, so I cannot see where the reasoning diverges. 86.129.18.115 (talk) 20:05, 22 November 2013 (UTC)

The formula in the thrust article would be valid for a rocket, not for a helicopter (you best ignore the comment in that section about incoming air). To quote A review of rotor induced velocity field theory (Office of Naval Research, 1954) "this was one of the most important features of momentum theory. It showed that the induced velocity at the disk was one-half its value in the ultimate wake." It can be derived with the Bernouilli equation, but to put it simply, you're not just pushing air away from the rotor, you're also pulling air towards the rotor. Look at it this way: the rotor creates a pressure difference between top and bottom, and the upward force is equal to (Pb-Pt)A . Pb will be higher than ambient pressure, Pt lower. Pb-Pa = Pa-Pt = (Pb-Pt)/2, so the upward force is 2(Pb-Pa)A. With a rocket, the upward force is only (Pb-Pa)A. Ssscienccce (talk) 21:57, 22 November 2013 (UTC)

Note that real rotors will require more power to achieve the required trust. The calculated power will only produce T*FM thrust, with FM being the figure of merit of the rotor. Values between 0.7 and 0.8 represent good hovering performance, while state-of-the-art rotors may reach 0.82. The actual required power is P/FM, this is the power that the rotor must receive, doesn't include transmission losses between engine and rotor. Maybe add 10% for those losses and you get a value that, assuming good design and optimal conditions should suffice to hover without ground effect. Ssscienccce (talk) 05:05, 23 November 2013 (UTC)

A very similar factor explains the difference between the OP's two answers: since the fluid must already be moving when it reaches the impeller, its change in kinetic energy is larger ($\Delta (v^{2})=(v_{0}+\Delta v)^{2}-v_{0}^{2}=(\Delta v)^{2}+2v_{0}\Delta v$). --Tardis (talk) 01:39, 23 November 2013 (UTC)

A bit of searching turned up this power point presentation on propeller thrust: [3]. --Modocc (talk) 03:14, 23 November 2013 (UTC)

I would like to momentarily go back to the "sand conveyor" simplification, which eliminates all the issues of fluid behaviour. Assume that the sand is made available to the device at zero velocity and zero cost. My method of working out the power requirement is as follows. (I know that this probably should be done with calculus but I want to stick with exactly what I have if possible, to avoid getting lost, as my grip on this is pretty tenuous.)

F is the force (thrust) required to be produced
v is the ejection velocity of the sand
m is the mass of sand ejected in each timestep of Δt
P is the power needed
A is the cross sectional area of the ejected stream of sand
ρ is the density of the ejected stream of sand

From basic principles:
(1) m/Δt = vAρ
(2) F = mv/Δt ⇒ m/Δt = F/v

From (1) and (2):
(3) F/v = vAρ ⇒ v = sqrt(F/Aρ)

From kinetic energy equation:
(4) P = (mv^2/2)/Δt

Substitute (2) and (3) into (4):
(5) P^2 = F^3/(4Aρ)

Now, what I would like to know is whether it is possible to adapt this method with just a small tweak in order to produce the forumula at Momentum theory, or whether a different method altogether is needed. I believe that in my original result from several years ago I may have had v = sqrt(2F/Aρ), but I can't figure out now how I arrived at that. 86.128.4.176 (talk) 00:12, 24 November 2013 (UTC)

As you have set up your equation, power is the dependent variable. You have set up an equation to calculate the power as a function of thrust. You specify the thrust (e.g., set it equal to vehicle weight), and you assume that some valid configuration exists for which you can deliver that thrust. You have (intentionally) ignored that some thrust values are impossible to achieve; and some power values are impossible to achieve; and that thrust and power are coupled in a real engine. Those are acceptable simplifications. It's all legal, but it's a little weird, because usually power is the variable we can actually control, and thrust is a result, not an input. You're sort of solving an inverse problem.

But there are still some flaws in your approach. Equation 1 assumes an incompressible fluid; that's also fine. You are in good company when you make that assumption.

Equation 2 is incorrect. It's just flat out wrong. Force is not "mv/Δt". Force is equal to the total change in momentum with respect to time. So, F = d(m v) /dt = (m dv/dt + v dm/dt). This is called the product rule for differentiation. In your approach, you are entirely neglecting one of these terms, which means you are failing to account for a significant amount of power. In order to calculate this term, you must provide some constraint equation. How do you accelerate each element of mass? Typically we could use the properties of a fluid (or sand) to constrain this term. But, I already know you don't want to bring those details into play, and you're happy to sacrifice physical accuracy. That's fine... you can constrain this term by some other method - by decree, if you like: dv/dt = 1. But you can't just leave it out of the equation. That would be mathematically incorrect.

Equations 3 and beyond are invalid because they depend on Equation 2. Equation 4 makes a similar mathematical error.

If you want to solve physical dynamics problems like this one - even if you choose to abstract some details away - you must become better acquainted with the correct applications of differential and integral calculus. These tools are not optional. Nimur (talk) 10:47, 24 November 2013 (UTC)

First, I disagree that this "inverse problem" is "a little weird". It is entirely reasonable to ask how much power is needed to hover a certain mass. Second, the basis of F = mv/Δt is simply F = ma. Over a time interval Δt, a mass m (of sand) is accelerated smoothly from 0 to v. Therefore a = v/Δt and the associated reaction force is mv/Δt. I would like further opinions on the validity of this please. I don't understand why it is wrong. 86.169.36.80 (talk) 12:30, 24 November 2013 (UTC)

It's legal to solve inverse problems; they are a central consideration of control theory and engineering. But they are frequently ill-conditioned, which means that you must use absolutely flawless mathematical rigor when calculating solutions. In your specific problem, you are considering power and force to be totally independent. Is this valid? Well, sure, if you drop the terms that relate the two! And then you are left with an equation that is wrong. Surely by now you've seen those mathematical jokes that prove 2+2 = 5 by cleverly hiding a division by zero somewhere in a simple-looking equation? Those are the consequences of incorrectly solving ill-conditioned problems. Nimur (talk) 16:17, 24 November 2013 (UTC)

(Disclaimer. I'm just another guy on the web that likes to seek relevant references too). From our Impulse (physics) article we have $d(m\mathbf {v} )\,=(dt)\mathbf {F} \,=d\mathbf {J} \,=(dm)\mathbf {v} _{e}\,$ which justifies the derivation given in the article on thrust which is essentially the same as yours. These derivations glosses over the some of the finer detail that Nimur is trying to constrain/define when he takes the derivative, but I see nothing inherently incorrect with your approach either (I may not have had enough sleep, and my maths can suffer when that happens). The power required is proportional to the jet velocity, see [[4]] but that means its better to propel larger masses when using lower exhaust velocities. The difference I see between what you have now and before is the difference between the forces produced by the rockets and helicopters that others pointed out above. Propellers have an incoming velocity added in addition to the exhaust velocity because the force is doubled (its a push pull system). A similar result (with a doubling of an upward force on the hand) can be obtained by dribbling a basketball. --Modocc (talk) 13:30, 24 November 2013 (UTC)

Let's be sure there is no confusion about the meaning of m. In my derivation, m is, as explained above, the mass of a chunk of sand propelled out over Δt. It is not the mass of the vehicle. In my concept dm is zero (or is not a relevant measurement). The sand is available in the "milieu" at zero cost and does not need to be carried by the device (this is obviously unrealistic in the case of a rocket which must carry its own fuel, but is intended to be analogous to the air in the case of a propeller). That being said, do you feel, then, that my derivation above can yield the equation at Momentum theory just by applying a factor of 2 at the relevant point? This is the part I really can't get my head around. 86.169.36.80 (talk) 13:48, 24 November 2013 (UTC)

The equation at Momentum theory can be obtained from the powerpoint presentation's equations that I provided earlier (you can simplify those by first setting the plane velocity to zero and then combine the equations for thrust and power). When you do that, you will see that for a given air velocity the thrust is twice that of the rocket. To understand that, envision a car with two props, one underneath expelling air from its interior and another one on top pulling it in, so you have left out half the force. --Modocc (talk) 14:12, 24 November 2013 (UTC)

Yeah, thanks, I looked at that before but I can't understand it unfortunately. That is why I just wanted to stick with what I had worked out and make a small change, if at all possible. I'm puzzled that my formula, P^2 = F^3/(4Aρ), actually needs less power for given thrust than the Momentum theory one, P^2 = F^3/(2Aρ). If I am somehow "missing half the force" then shouldn't it be the other way around? 86.169.36.80 (talk) 14:32, 24 November 2013 (UTC)

The problem is that your entire equation is wrong: it states an incorrect relationship between power and force; and you are still trying to draw conceptual conclusions from it. If you make unphysical equations, you get unphysical results: 2 = 5, perpetual motion, faster-than-light sand conveyors. The least of your worries is a scale factor of 2. Your continued attempts to make these problems go away by substituting sand for air belies that you are not understanding the fundamental problem. Flow equations work equally well for grains of sand and for microscopic gas molecules. You still need to define shear terms, viscous interaction terms, ... it's almost as if the Navier-Stokes equation was written without loss of generality!

Hey, what's this? An entire textbook chapter of correct physical equations, available at no cost, from a reliable source? It's almost stunning that nobody linked to this already! Except, that I did, twice before. You can download the entire textbook, at no charge, from the FAA; or you can buy a paper copy for about $15. I can recommend some good physics and calculus books if you need those, but the Helicopter Handbook uses engineering-charts instead of calculus-based derivations, so the math is quite simplified.

Have a look at chart 2-32. See how many parameters go into the power equation? The power required depends on the interaction between the impeller, and the fluid that is being impelled. The very same applies, whether we consider a rotor in air, or a conveyor belt pushing sand, or a ship propellor in seawater. The parameters are different, but you cannot ignore that required power is coupled intricately with the messy details of the motion of the fluid (or sand). Nimur (talk) 17:49, 24 November 2013 (UTC)

An actuator disk accelerating a fluid flow from right to left

Woeps, I should have paid more attention, wasn't as simple as straight forward as I thought, ignore what I said previously about rocket versus helicopter. What's missing is the continuity equation: we assume the air is incompressible, so the volumes above and below the rotor have to be the same. Since the velocity of the air under the rotor is higher than above it, the diameter of the air flow must contract to comply with the continuity equation. That's what you see in the figure: the rotor accelerates the flow to the left, the rotor is at point S, and you see that the air flow contracts, and the velocity increases up to the "ultimate wake". So the air doesn't exit the rotor at maximum speed, the pressure differential between the rotor and point S1 accelerates the air, and only at S1 will it reach the maximum velocity. On the right of the rotor, the same thing happens. Instead of using the velocity of the air exiting the rotor to calculate the kinetic energy, you have to take the ultimate wake velocity, which is larger, that's the reason for the higher power requirement. The ultimate wake velocity comes from momentum theory, that's why one handbook said: "this was one of the most important features of momentum theory. It showed that the induced velocity at the disk was one-half its value in the ultimate wake."

This also explains why the sand analogy can't be used, because pressure difference play an essential role.

I'll post a more decent overview later Ssscienccce (talk) 18:14, 24 November 2013 (UTC)

Here goes: Air far above the rotor is at ambient pressure p_a and zero velocity, far downstream it's at wake velocity V_w and at ambient pressure p_a again (we assume inviscid and frictionless fluid, so the wake and the surrounding air don't mix). Directly under the rotor, the velocity is V_d and the pressure is p_d.

The thrust delivered is equal to the change in momentum with time, so (with dotted m meaning the mass flow rate):

$T={\dot {m}}V_{w}$ (why V_w and not V_d? Because at V_d there's an additional pressure component)

The power delivered by the rotor is

$P={\tfrac {1}{2}}{\dot {m}}V_{w}^{2}\quad or\quad P={\tfrac {1}{2}}TV_{w}$

But at the location of the propeller disc, the power is equal to the work done by the thrust:

$P=TV_{d}$

So

${\tfrac {1}{2}}TV_{w}=TV_{d}\quad or\quad V_{d}={\tfrac {1}{2}}V_{w}$, the velocity under the rotor is half of the ultimate wake velocity.

The mass flow rate is:

${\dot {m}}=\rho V_{d}A$

We can write the thrust as:

$T={\dot {m}}V_{w}=\rho V_{d}AV_{w}={\tfrac {1}{2}}\rho AV_{w}^{2}$

This gives us Vw in function of the thrust

$V_{w}={\sqrt {2T \over \rho A}}$

So the power becomes:

$P={\tfrac {1}{2}}TV_{w}=T{\sqrt {T \over 2\rho A}}$

That should be it, I think... Ssscienccce (talk) 13:53, 25 November 2013 (UTC)

So the essential thing to remember is: the air leaves the rotor at speed V_d, but the pressure difference with the ambient air continues to accelerate the air, so the thrust is actually mV_d + p_d (p_d being the overpressure under the rotor) Ssscienccce (talk) 14:16, 25 November 2013 (UTC)

It's all very clever to point out that there is a ${\frac {dm}{dt}}$ term that arises from the product rule. You can even use that term (in my "slug of sand" case earlier). But there's a reason that students are taught $F=ma=m\Delta v/\Delta t=mv/\Delta t$ (further expressions appropriate in the case of constant acceleration (or average force) and zero initial velocity, respectively); you can always work merely with that expression. (The common cases where people use the ${\frac {dm}{dt}}$ analysis, like pushing a cart that's simultaneously being filled with material, can instead be interpreted using the normal second law if you consider that you are applying the force (perhaps indirectly) to the added material: its instantaneous velocity change times its mass flow rate yields the correct force.) (In reply to your comment higher up in response to me, to simplify things:) Acceleration requires work because it changes kinetic energy. Mere movement does not, unless it is in a potential field. Here, we can ignore the effects of gravity on the sand, because we can take it to be at zero velocity (and otherwise supported) until disbursed and to be irrelevant afterwards. --Tardis (talk) 21:44, 24 November 2013 (UTC)

About the only thing the derived thrust equations for the helicopters and rocket (which is nothing more than a souped up version of a sand blaster :-)) have in common exactly is their mass flow, and the change in volume elements with the helicopter's flow does not effect that. In addition to what I interpret to be a force $mv/\Delta t$ being applied twice (once during the inflow and secondly during the outflow), the induced power Tv that is dumped into the helicopter's wake is also twice that of the rocket Tv/2, thus the rocket is more efficient, producing an equivalent thrust with less power as the OP pointed out and, in any case, the difference between the relations turns out not to be very large and is simply the cube root of 2. -Modocc (talk) 00:11, 25 November 2013 (UTC)

Let me try to rephrase: the quantity of energy conveyed away by the exhaust, during each unit of time, is a conceptually distinct amount of power from the power necessary to generate said exhaust. The former is purely derivable from a kinetics calculation, while the latter depends on things like engine efficiency. Even idealized engines have maximum efficiencies. Consider, for example, the Carnot cycle. An ideal rocket in a no-net-motion gravity burn is one limiting case, where all the momentum is carried away by the exhaust, and all the useful power is conveyed to the exhaust gas. But that is not the only power we need to consider: the original ask was for the power required to hover - not the subset of that power that gets wasted in exhaust gases; nor the subset of power that corresponds to only the kinetic energy of the rotor wash; nor the subset of power that corresponds to the kinetic energy of moving sand. When you fail to account for the lossy process by which an engine converts potential energy into kinetic energy, you are describing a perpetual motion machine. At that point, we may as well "further idealize the problem," connect the exhaust nozzle to the engine intake, and use perpetual motion to hover with zero energy input. Lots of power goes through, but no new energy gets put in by the engine. Everybody who keeps "simplifying" these equations is performing exactly this flawed process.

The ask was for how much power is needed to hover. That is asking how much new energy the engine must add, per each unit of time. The ask was not how much power denotes the rate of energy exchange in the airflow. The ask was not how many ways can we play with dimensional analysis to yield a term whose units are measured in terms of energy per unit time. The ask was "how much power is needed to hover." To answer that, we need to know how efficiently the engine converts input energy into useful work. Whether the engine is a rocket impulse engine, or a helicopter rotor, or a sand chute, the laws of thermodynamics still apply. If you want to calculate the energy conveyance for each granule of sand, and set up an n-body problem, you can do so, but what you cannot do is assume a magical sand conveyor belt that is perfectly efficient, has no friction, yet still performs useful work accelerating sand.

Think of it conceptually! If the conveyor belt has no friction, it cannot move the sand, because there is no traction. This is exactly analogous to a viscous drag term in a conventional fluid. You can not evade thermodynamics, no matter how many simplifications you make. You must account for how the conveyor belt exerts force on each sand granule. And you must integrate each differential unit of work over all n-gazillion sand grains. And did you remember to conserve angular momentum for each granule of sand with respect to (n-gazillion minus one) other grains? Oh, woops, you forgot? No problem, it's only that angular momentum is a fundamental property of the universe, whose conservation is more fundamental than the conservation of energy... not like anyone would notice if we totally use wrong physics to constrain the motion! We can just throw out one conservation principle so that it's easier to calculate the other one, right? Besides, each grain is so small that it has almost no momentum... how much different would our answer be? There's only n-gazillion very small errors... Or, you could use a bulk approximation that is physically accurate in the statistical ensemble, ergo, applying regular fluid flow equations. If you don't use the correct physically-guided equations, your answer is wrong. Period. Nimur (talk) 18:19, 25 November 2013 (UTC)

I see you are trying to teach the OP something he didn't even ask about. How do you know he hasn't already figured that some processes are more losy than others? Did he ask for the power needed to hover with a balloon? No. He asked about some specific derivations regarding thrust and you don't seem to have a problem with the P(T) derivations nor their utility. -Modocc (talk) 18:57, 25 November 2013 (UTC)

I understand perfectly well that all real machines are lossy. None of Nimur's obfuscations seem to be any help in deriving or explaining the formulas at Momentum theory and Thrust#Thrust to power, and they may be just a smoke-screen to disguise his early failure to recognise what seems to be a fairly standard equation. 86.161.61.128 (talk) 17:51, 26 November 2013 (UTC)

• @Ssscienccce, thank you for your derivation above. I need to spend time working through it to see if I can get my head round it. I wonder, if you have time, whether you could cast your eye over my "sand conveyor" workings above (piece starting 'I would like to momentarily go back to the "sand conveyor" simplification') and give your opinion on whether this is correct within its own ambitions. Nimur keeps insisting that it is flat wrong but I am not so sure. What do you think? 86.161.61.128 (talk) 18:11, 26 November 2013 (UTC)

You are working with an equation so non-standard that it cannot be cited in any textbook or paper! But you believe your equation anyway, ...because some other anonymous Wikipedia contributor wrote a bunch more equations you barely understand! So tell me, when you read the article on thrust and power, which type of power do you think it is referring to? The Pilot's Handbook of Aeronautical Knowledge (yet another free physics textbook for you) doesn't seem to have your "standard" equation in it. Actually, it states this, about comparing power and thrust:

“ It is possible to compare the performance of a reciprocating powerplant and different types of turbine engines. For the comparison to be accurate, thrust horsepower (usable horsepower) for the reciprocating powerplant must be used rather than brake horsepower, and net thrust must be used for the turbine-powered engines. In addition, aircraft design configuration and size must be approximately the same. When comparing performance, the following definitions are useful:

• Brake horsepower (BHP)—the horsepower actually delivered to the output shaft. Brake horsepower is the actual usable horsepower.
• Net thrust—the thrust produced by a turbojet or turbofan engine.
• Thrust horsepower (THP)—the horsepower equivalent of the thrust produced by a turbojet or turbofan engine.
• Equivalent shaft horsepower (ESHP)—with respect to turboprop engines, the sum of the shaft horsepower (SHP) delivered to the propeller and THP produced by the exhaust gases.

”

That comes from... what's this? A whole chapter on comparing different types of power between different types of engines. Oh! How relevant! It looks as if a group of experts on aerodynamics got together, wrote a physics book, and explained your problem very thoroughly! Were I to actually trust my life to an aircraft designer, I would sure hope they use that procedure, instead of "some equation from Wikipedia!" But, I'm sure you guys will be able to work through the small bug in your math. While you and your fellow contributors work through some more equations and conduct original derivations in total contravention of Wikipedia's standard policy, and in complete defiance of standard, published, approaches to solving this problem, I'll be out flying; the weather is great today. Nimur (talk) 19:17, 26 November 2013 (UTC)

If you believe that strongly that the P(T) derivation at thrust is incorrect and needs a citation, tag it, it was added over a year ago by some other guy... and if you didn't realize that we have been talking about two different equations (neither of which were the OPs although he has derived both independently apparently), then please reread this thread to see what you missed. --Modocc (talk) 19:44, 26 November 2013 (UTC)

Nimur, I am getting very tired of your worthless and irrelevant rants. I would prefer it if you made no more contributions to this thread. 86.179.114.69 (talk) 19:57, 26 November 2013 (UTC)

Unless his contributions are disruptive or violates WP:NPA he is still welcome, its the wiki-way, so please bear with it. --Modocc (talk) 20:17, 26 November 2013 (UTC)

Arbitrary break

From the article on thrust we have $\mathbf {P} ^{2}={\frac {\mathbf {T} ^{3}}{4\rho A}}$. From momentum theory we have $P={\sqrt {\frac {T^{3}}{2\rho A}}}$ or $\mathbf {P} ^{2}={\frac {\mathbf {T} ^{3}}{2\rho A}}$.

The OP's equation using sand is exactly the same as the thrust article's equation. Thus, whether or not the OP used a rigorous derivation, if his thrust equation happens to be incorrect then the article's equation is also incorrect and would need to be removed. Apparently, Nimur snubbed a poorly written article, unless he was under the false impression that the two articles' equations were identical (and he would have misinterpreted much of what I and others have written about it). When Nimur says it "...doesn't seem to have your "standard" equation in it." emphasis mine, I'm not sure if he is referring to the article's equation, the OP's equation or both. -Modocc (talk) 20:53, 26 November 2013 (UTC)

It may be correct for sand (as far as I can see it is), it's not correct for air: sand will leave the accretion disk at it's ultimate speed because there's no force acting on it (ignoring gravity). Air that leaves the propeller is still accelerated until the pressure drops to ambient. So the velocity at the disk is correct for the mass flow (density * area * velocity), but for the thrust and power, the wake velocity must be used. That's the main result from momentum theory, and every source repeats it:

• http://www.aerostudents.com/files/aircraftPerformance2/helicopters.pdf‎ page 1: The ideal power of the helicopter now becomes equation (2.3)
• FTM106: U.S. Naval test Pilot School Flight Test Manual chapter 5 Hover performance page 5.16: The power required to accelerate the air mass through the disc, induced power, assuming T ≅ W at a hover, is equation (5.24) (T is thrust, W is weight of the helicopter)
• "Rotorcraft Aeromechanics" By Wayne Johnson, page 41: equation (3.5)

Every time the same equation, $P={\sqrt {\frac {T^{3}}{2\rho A}}}$ Ssscienccce (talk) 22:05, 26 November 2013 (UTC)

Instead of moving air downward with rotors, the model used with momentum theory, one can build rockets with pressurized air, so same fluid but different equations. The OP was correct the first time for rotors, and has simply gotten our article's rocket equation the second time. But we've already pointed this out to the OP though, but I'm not sure if Nimur will agree that the P(T) equation at thrust is correct, so we will likely need to quote or locate a reliable source for it. -Modocc (talk) 22:33, 26 November 2013 (UTC)

I agree, it would be good to find a source for that. It says here that P = Fv/2, which spookily enough is an equation I have written down on the pad in front of me where I did the workings-out. If we could also validate that F = Aρv^2 then that would seem to nail it. 86.179.114.69 (talk) 00:14, 27 November 2013 (UTC)

fluid mechanics

how to know flow analysis of fluid — Preceding unsigned comment added by 213.55.110.123 (talk) 13:37, 22 November 2013 (UTC)

After you've exhausted the information in our article on fluid mechanics, you might be able to ask a more specific question, or more clearly define your request.

Do you need help finding information in a different language? We have thorough articles in German, Italian (it:Fluidodinamica), and Arabic, and even a stub in a stub article in Oromo. There are related articles in many other languages. Nimur (talk) 13:40, 22 November 2013 (UTC)

Fluid mechanics is one of the tougher subjects out there (and one of the least well-understood ones). There is simply no way to contain all that you need to know in a response here - or even in a full encyclopedia article. You're going to need at least a good book on the subject - and possibly some college-level courses to properly get to grips with it.

The mathematics behind it is complicated - and even with the best math available, we need gigantic wind-tunnels or wave tanks to properly confirm the results of mathematical/computer estimations.

I can perhaps sketch out one approach to this in computers that may prove helpful to you:

• Divide the volume of fluid we're interested in into tiny imaginary cubes - producing a three-dimensional grid.
• Divide time into tiny steps.
• Within each cube we'll approximate the flow as a linear motion through the cube and the temperature and pressure as being constant throughout - clearly that's not true in the real world - but if we make the cubes small enough, it may be good enough.
• Now, we can formulate a relatively simple equation using the temperatures, pressures and kinetic energies of all of the adjacent cubes to figure out how they influence this cube during one tiny time-step. We figure out heat transport and fluid flowing into or out of this one - which gives us a new temperature, pressure and flow rate through this cube at the end of one tiny time step.
• Some cubes have boundaries that face the object we're interested in testing - so their equations have to take that into account.
• We start with all of the cubes being identical with the fluid flowing through them being stationary and the temperature and pressure being uniform throughout them all.
• Armed with those equations, we can have a computer run that equation for each of the cubes of fluid for one tiny time step - then move onto the next time step and do it again. You can then tell the cubes at one end of this massive grid that there is fluid being injected into them and gradually, after many more time-steps, the system starts to look more like how fluid flows in the real world.

Unfortunately, this misses a lot of real-world effects - but for some applications, it produces enough insight into the fluid flow to be useful. Sadly, to model something like an airplane flying, you'd need a mesh of cubes maybe a centimeter or two on a side - and there would be perhaps billions of them. The time step might need to be a fraction of a second - so even the fastest computers would take days to calculate anything useful. In many cases, these systems have to run on giant super-computers in order to get useful and realistic results.

Of course my description misses many subtleties of how this is really done - and there are certainly other ways to approach it.

SteveBaker (talk) 14:22, 22 November 2013 (UTC)

For reference, Steve is basically explaining the idea behind a finite element method to solve the PDEs governing fluid flow. More at Finite_volume_method_for_unsteady_flow. He's right that this is a difficult subject. Though some engineers can learn how to manage some simulations in undergrad, the real nuts and bolts of the theory are usually reserved for graduate level mathematics and physics instruction. On the flip side, we do have a system of equations that can answer most of our questions, the Navier–Stokes_equations, but no general solutions are available (as of yet), and so we tend to approximate solutions with numerical methods, as outlined above. SemanticMantis (talk) 14:44, 22 November 2013 (UTC)

Don't forget about the finite volume method, which is also popular for CFD. --Tardis (talk) 02:24, 23 November 2013 (UTC)

For how many percents stress can raise the glucose in the blood?

I'm looking for an accurate information about that. It's interesting me because last week I saw someone that the EMTs cheeked her glucose and her levels was high (320) and the EMT said that it because the stress. I doubt if that true too much (because I know that the normal levels are between 80 -140 14:07, 22 November 2013 (UTC) — Preceding unsigned comment added by 5.28.171.111 (talk)

You're not going to get a good answer here because there are some people doggedly enforcing policies against "medical advice" in a very broad sense. Note that an EMT may not give local busybodies a full medical history of people they are working on. We're hard pressed (and not supposed) to diagnose a patient who knows his history, so you're pretty much SOL for an "accurate" answer. If you're interested in the topic I encourage you to explore blood glucose, information from health organizations, etc., but that won't tell you what was going on in a specific situation. Wnt (talk) 18:12, 22 November 2013 (UTC)

Ok, let's make order, I wouldn't want a medical edvice or somthing, so let's be relax... I would only like to understand the topic of the influences of the stress on the glucose in the blood, that all. I'm not talking about specific case, what I told you, it was only as an example. I came here because this issue of "stress influences on the level of the glucose" is a new for me and I'm looking for that some information. I think that we exaggerate sometimes when we take the warning about the medical advice - too much even to the place that we shouldn't need that at all and then people like me lose information. Although, I definitely understand you so Thank you. 19:45, 22 November 2013 (UTC) — Preceding unsigned comment added by 5.28.171.111 (talk)

This is not medical advice, this is a list of references that address the general topic of stress and blood glucose. Here are two links, to authoritative sources that confirm that stress can increase glucose levels (though they don't say by how much). Mayo clinic ([5]), and the American Diabetes Association ([6]). I see that WnT has supplied one of these links already. If you want quantitative information, you might have luck searching google scholar for /stress blood glucose/, like so [7]. If you cannot access these articles, you might try the same search at your local library. SemanticMantis (talk) 21:56, 22 November 2013 (UTC)

Also, since people react differently to stress, I'd also expect their blood glucose to respond differently. StuRat (talk) 22:05, 22 November 2013 (UTC)

As per our article on stress hyperglycemia, "the glucose [due to the stress of illness] is typically in the range of 140–300 mg/dl (7.8-16.7 mM) but occasionally can exceed 500 mg/dl (28 mM)." - Nunh-huh 22:28, 22 November 2013 (UTC)

Projectile penetration

If a projectile hit the same target at 400 m/s and 800 m/s, the penetration doubles or became four times higher (i.e. the penetration of projectile is directly proportional to momentum or kinetic energy?)?95.246.215.102 (talk) 21:21, 22 November 2013 (UTC)

Impact depth is the relevant article. If the speed of the projectile is great enough for elastic deformation in the target to be neglected, the penetration depth is directly proportional to the momentum, so doubling the speed will double the depth. Tevildo (talk) 21:53, 22 November 2013 (UTC)

(ec) I doubt if it's as simple as either. A faster projectile has the ability to concentrate the pressure more on the point of impact, while a slower object gives the material which is struck time to respond, and distribute the force over a wider area. The also depends on other factors, like the material and temperature. If you look at video of a high speed bullet striking a rubber sheet, for example, it seems to shatter, not bend, just as it would if frozen of it made of a brittle material. (The difference between my answer and the previous one is that I did not assume that we can neglect deformation.) StuRat (talk) 22:00, 22 November 2013 (UTC)

Penetration depth depends on the shape of the projectile, whether it's subsonic, supersonic or hypersonic (relative to the speed of sound in the target material, not in the air), the velocity range when it's subsonic, and lots of parameters like unconfined compressive strength. A range of penetration depth formulas exist, most of them similar to one of these forms (V is velocity, a, b, c, d, e are various parameters or constants):

• a.log(1+bV²)
• a.(b.V-c.log(1+e.V))
• a.V^1.8+b
• a.log(1+bV²) for V < 200fps, c.(V-d) for V > 200fps

An in-depth overview of penetration depth formulas can be found here (master thesis, 85 pages). Too bad that they only did five actual impact tests: velocity in m/s, corresponding depth between brackets in m: 277(0.173) 410(0.31) 431(0.411) 499(0.48) 567(0.525), so three times the depth at twice the speed.

A U.S. Army Engineers report on penetration in rock gives some data from tests, between 300 and 800 m/s. In that range, twice the velocity results in 2.4 to 2.9 times the penetration depth.

Experiments on porous rock with projectile speeds between 18 and 42 m/s showed "penetration depth quadratically related to the initial projectile kinetic energy" ( "quadratically related"? is that a convoluted way of saying it was proportional to the speed?) with dry rock, but proportional to the kinetic energy when the rock was saturated with kerosine (so double the speed, four times the depth).

See also a blog about slingshot tests here

All the test results mentioned are subsonic, at supersonic speeds penetration depth should be proportional to kinetic energy, so double the speed, four times the depth. Ssscienccce (talk) 00:04, 23 November 2013 (UTC)

I would say that the penetration, if the material is homogenous and doesn't have any particular properties, should be directly proportional to the square of speed: this because the material resistance can be regarded as a strong form of friction, and if an object double the speed, appling the same force, the drake distance became four times higher. 95.246.216.236 (talk) 18:23, 23 November 2013 (UTC)

Yes, your model is based on fibrous materials or other situations where friction is the main stopping force. It is very different from Newton's model (linked above by Tevildo). In practice, I suspect that reality will lie between the two models for most materials. Dbfirs 21:38, 23 November 2013 (UTC)

Yet another room temperature superconductor prediction notable?

Is this wild guesstimate:

http://www.sciencedaily.com/releases/2013/11/131121135635.htm

Notable enough to include under High-temperature superconductivity? Hcobb (talk) 22:27, 22 November 2013 (UTC)

This question would probably be better on the article talk page, but Physical Review Letters is an impeccably reliable source, and I see no reason not to include the reference. Tevildo (talk) 23:12, 22 November 2013 (UTC)

Topological insulator might be a better place. They are related though. --DHeyward (talk) 08:58, 23 November 2013 (UTC)

November 23

Genetically engineering hair color and eye color

Is it true that hair and [[eye color are coded for by a single gene, and therefore would be the easiest to genetically modify (compared to other traits)? ScienceApe (talk) 02:55, 23 November 2013 (UTC)

No. I have linkified the relevant articles if you want to read them. μηδείς (talk) 03:05, 23 November 2013 (UTC)

Redshift experiments

The existence of dark energy AFAIK was deduced from differences between type-1 supernova derived distance and redshift-derived velocity of high-z galaxies. Surely all alternative hypotheses explaining these differences have been ruled out? So, for example, nonlinearity of redshift in the optical wave lengths at high-z velocities in general can be rejected how? Direct experiments? Which direct experiments measuring redshifts at the relevant velocities have been done? Any hint is welcome. --SCIdude (talk) 11:16, 23 November 2013 (UTC)

The origin of these hypotheses is Einstein's cosmological constant, which was explicitly introduced as a fudge factor rather than as part of a more "elegant" theoretical equation. The astronomical data you mention establish that the cosmological constant is, in fact, non-zero and has a definite measurable value. Dark energy is an hypothesis that attempts to explain why the cosmological constant is non-zero based on known existing physical phenomena - on metaphysical grounds, the answer "it just is, it's an inherent property of the vacuum that we can only measure rather than predict" isn't (apparently) acceptable. Tevildo (talk) 11:37, 23 November 2013 (UTC)

No, all alternate hypothesis cannot ever be ruled out. Your un-orthodox nonlinear redshift would require some huge revolution of physics though and is considered to have a negligible a priori probability. Dauto (talk) 14:12, 23 November 2013 (UTC)

So the Nobel prize is rather for the SN1A observation, like the one for CMB discovery by Penzias/Wilson. I misunderstood that, thanks. - SCIdude (talk) 18:31, 23 November 2013 (UTC)

To answer the other part of your question, about experiments that directly measure redshift at high velocities: Novotny et al. have done spectroscopy on ion beams at 0.33c, not too different from the recessional velocity of supernovae at z=1 or so [8]. This confirms the expected shifting of spectral lines. Moreover, synchrotron light sources and free electron lasers use ultrarelativistic electrons with Lorentz factor much higher than anything that comes into play for supernovae. If you try to somehow tweak the Lorentz transformations to give a different result for cosmological redshift, you will almost certainly break SSRL and FELs in a way that would be immediately noticeable. People do still perform experimental searches for Lorentz violation, but they're looking for tiny effects that would make no difference at all for the original data that provided evidence for accelerating expansion back in the 1990s. --Amble (talk) 22:35, 23 November 2013 (UTC)

Very good. Many thanks. Could you please augment the redshift article to include the relevant parts of your answer? - SCIdude (talk) 05:24, 24 November 2013 (UTC)

I'm afraid that might fall afoul of WP:SYNTH without some secondary sources to make the connection. --Amble (talk) 16:40, 24 November 2013 (UTC)

Structural elements

Are there any good diagrams which show which part of a cantilever structure are trusses, supports and cantilever members? And which ones are beams?Clover345 (talk) 12:20, 23 November 2013 (UTC)

This page goes through the terminology at a very basic level. Tevildo (talk) 12:53, 23 November 2013 (UTC)

I've looked for such sources a few times myself. Questions like what's the difference between beam and girder (http://www.eng-tips.com/viewthread.cfm?qid=164292).

Cantilever seems to be the easiest to identify: it's an overhanging part of a construction. A beam, a truss or any other construction can be supported at both ends; but you can also support it for example at the left end and somewhere in the middle, in which case the part of the construction to the right of that point is considered a cantilever.

A truss is a structural "unit" made up of straight elements connected as triangles. List of truss types should give you a good idea. You could consider it a metal plate where much of it is cut out to save weight but still keep the stiffness. The term is mostly used for a two dimensional structure, not sure if for example an antenna tower is considered a truss or a construction made of several trusses. The Allen truss bridge shown here has three trusses (or six, when you cross a bridge you have a truss to your left and a truss to you right) not one, because the parts are connected at only one point, so the bridge as a whole isn't stiff.

A support can mean different things. For example Truss#King_post_truss calls the top chords of the truss "supports". Cantilever uses the word for the points where the structure is "supported", the "external" connections you could say, which is the most common meaning I think. It can also mean the support column a truss rests on.

A beam is usually an element that has to bear a bending load: for example, a bridge can consist of two trusses, in between them is the deck that carries the traffic, this deck rests on top of beams that are connected to the two trusses. While the elements that make up the trusses only have to withstand either tensile or compressive forces (and a minor bending force due to their own weight), the beams are only or mainly subjected to bending forces.

Vocabulary of trusses explains some of the terminology and has diagrams. Keep in mind that the terms aren't always used consistently. Ssscienccce (talk) 17:28, 23 November 2013 (UTC)

How (non-rad) hard would it be to turn every cellphone into a radiation detector?

https://medium.com/war-is-boring/21dc0b023f1d Homeland Security Agency Wants to Turn People Into Nuclear Tripwires

Why not just mandate that this be used:

http://www.researchgate.net/publication/253023219_Detection_mechanisms_employing_single_event_upsets_in_dynamic_random_access_memories_used_as_radiation_sensors

To have every cellphone track the Single event upsets in its own chips and report to the cellphone network provider (and hence of course the NSA), whenever the rad counts hop over the background level? Hcobb (talk) 20:08, 23 November 2013 (UTC)

That would lead to too many false positives (from radon exposure, cosmic rays, even thunderstorms). Cell phones CANNOT reliably detect radiation that way -- this is a job for a dedicated instrument (such as that which DHS wants to use). 24.23.196.85 (talk) 21:06, 23 November 2013 (UTC)

I have to agree with 24.23.196.85 in part on this. At the current sophistication of cell phones they may be able to detect ionizing radiation but to be useful the Signal-to-noise ratio would need to be clear. However, having said that, if a thousand cell phones all reported an indication of a higher level of ionizing radiation above background, then yes, they could be used as trip wires.--Aspro (talk) 21:46, 23 November 2013 (UTC)

The memory chip area available for use as a detector in a cell phone would, for a start, be far too small, meaning no detection events for months at least, even with significant increases in radiation levels. Because the chip is enclosed in the cellphone case and hidden by internal structures, it would essentially only detect cosmic rays - not what Homeland Security would be most interested in. The electronics industry / electronics hobbyist magazine Elektor carried a series of articles last year or ealy this year on making a home made radiation detector using a standard commercially available semiconductor device as a detector, instead of a gieger tube. Read that article, and you'll see that a detector chip has to have sufficient area (of the order 1 x 1 mm - vastly bigger than even a large number of DRAM memory cells), and be enclosed in special pakaging that blocks light completely while letting low energy particles through. 60.228.240.47 (talk) 00:12, 24 November 2013 (UTC)

Memory uses error correction and detection. It used to be parity detection, but they grew the word line and sophistication so a 256 bit line can add 10-12 bits to correct either single bit or multiple bits. Usually this goes unseen to the user as the 270 or so bits still return a 256 bit corrected line. The ECC is flagged on access, not when the event happened. It cannot identify the cause of failure. An unreliable cell or noise could cause the error. As long as the device can correct it, it continues. Multi-bit failures that exceed the ECC capacity generate the parity type of halt/BSOD hardware error. The next question is how separate is neutron radiation from alpha particle radiation? Alpha particles dominate SEU from ionizing radiation in semiconductors not the least of which is the use of lead in packaging (although being phased out). I do not know if neutron radiation will cause enough ionization failures over background alpha radiation to be reliably detected. Feature size also makes cumulative damage difficult to measure in a single device. The mission for memory is to correct, not accumulate errors. To be honest, I think the best starting points are more like the CMOS digital camera arrays with methods to filter and isolate the radiation being sought. --DHeyward (talk) 04:44, 24 November 2013 (UTC)

Note that most phones do not have ECC. I'm not aware if there is even a LPDDR3 (or any LPDDR) standard for ECC, our article Mobile DDR doesn't mention one nor do I find anything from a quick search.

And it's perhaps worth remembering that most desktops and laptops don't have ECC memory either. (Remember you will generally need a Xeon on the Intel side. Although on the AMD side there is limited support with some/many?/most?/all? of the AM3+ CPUs but many of the motherboard manufacturers still don't bother to official support it.)

The CPU cache may or may not have ECC I'm not totally sure.

Either way, most desktops and laptops, let alone mobile devices don't even have the capability to clearly detect many errors. This isn't surprising, as many above have indicated, in normal circumsances these errors are so rare, they just aren't worth worrying about for non mission critical systems. (Some people even use servers an workstations without ECC.)

Sure you can do some sort of test with a known result like people do for stress testing to detect errors, but that won't do wonders for your battery life.

(Just to emphasise the earlier point, if you have done stress testing, you would know whether one of the memtest variants or whatever, with a stable system with nonECC RAM you can run for over a week or more with no errors on any test. Incidentally as far as I know from such tests, you can generally obtain reports of ECC failures from the CPU if you want to.)

Now I guess you could mandate ECC, some may even be happy for it for the obvious other benefits. But to force a feature which is not going to work very well for what you're want to achieve, when you could at least force something which is designed for the purpose, just seems dumb. Although I share concerns with our resident cranky Perth engineer who likes to pretend to be multiple people, that you still won't be able to achieve much given the size and design limitations and perhaps cost ones as well (edit: for clarification I mean even if you require chip purposely designed to monitor radiation levels).

Nil Einne (talk) 13:15, 24 November 2013 (UTC)

ECC is not just DRAM. SRAM on the microprocessor incorporates ECC. It's inherent and invisible to the system or user. ARM and Intel both have ECC on L2 Caches. --DHeyward (talk) 05:10, 25 November 2013 (UTC)

Not to mention that cellphone manufacturers really want to MINIMIZE the number of RAM errors - not MAXIMIZE them! What would be needed would be some kind of additional detector chip - and the cost of that makes this proposal impractical unless somehow a law were passed to require it - which seems overwhelmingly unlikely. SteveBaker (talk) 16:48, 24 November 2013 (UTC)

Oil for plastics

I think petroleum products are used in the manufacture of certain polymers. When viable oil runs out, will we see certain types of plastic that we still have now disappear? Are there already alternatives in place? — Preceding unsigned comment added by 78.148.107.181 (talk) 23:04, 23 November 2013 (UTC)

Bioplastic. -- Finlay McWalterჷTalk 23:44, 23 November 2013 (UTC)

Hey, thanks. I was aware that there were already polymers that could be manufactured using non-crude-oil-derived plastics but I was wondering whether there were certain applications not yet covered. — Preceding unsigned comment added by 78.148.107.181 (talk) 23:58, 23 November 2013 (UTC)

The earliest plastics were made from cellulose; see: Parkesine. -- Also: History of Plastic ...And from milk (casein); see: Casein / The Plastics Historical Society. ~E:71.20.250.51 (talk) 05:57, 24 November 2013 (UTC)

See also Fischer–Tropsch process. In addition to carbon monoxide, it is also possible to turn methane or carbon dioxide into synthetic hydrocarbon compounds, which could then be used to make traditional polymers. So running out of fossil fuels will not make traditional polymers disappear, but it would make them more expensive (digging something out of the ground is often far cheaper than making it from scratch). Due to this increased expense, you may still wind up seeing bioplastics instead. Someguy1221 (talk) 06:13, 24 November 2013 (UTC)

Indeed, we can make them, doesn't mean we will. Hard to find any Bakelite today. Unless you go to a chinese webshop. "Replacement Bakelite Saxophone Mouthpiece", two belgian inventions for the price of one! Ssscienccce (talk) 09:57, 24 November 2013 (UTC)

Firstly, I don't think we'll ever run out of oil. If we keep burning it at the rate we are, we'll cook the planet long before we run out of the stuff. However, if we somehow did, we can make perfectly good oil-substitutes from plants, bacterial mats, algea and so forth. The problem is that it takes a lot of energy to do that...often more than you get back from burning the resulting oil. However, for the purpose of making plastics, the energy consumption isn't likely to be the primary issue.

What might be a problem is that as we burn through the easy-to-dig-up oil we'll have to start being more agressive about mining oil shale and oil sands - which will certainly push up the cost of oil - possibly to the point where we don't want to use it for making plastics anymore.

Fortunately, there is no shortage of alternatives. We already make plastics like PLA (Polylactic acid) from corn starch or sugar cane...and PLA is a perfectly useful plastic for many applications. You can make sheets of the stuff, extrude it in a 3D printer, injection mold it and so forth. It's also biodegradable and recyclable - which is useful - and it's easy to adjust vrious chemical properties to vary the time before it degrades, the melting point and so forth.

There are many other useful plastics that can be made from plant material without going through the intermediate step of converting it to oil. Cellophane is made from wood pulp, cotton or hemp. Plastarch is like PLA but better for high temperature applications. Then there is PHB (Polyhydroxybutyrate) - which is made using micro-organisms. Then there are whole families of exotic materials like amorphous metal glasses which can be stronger than titanium and yet heat-formed just like a plastic.

Hence, I don't think we have anything to be concerned about here.

SteveBaker (talk) 16:43, 24 November 2013 (UTC)

I don't think he's concerned about running out of oil, he's asking if there are certain kinds of plastics that will disappear once the oil used to make them is no longer available. As you note, it's much more likely that we'll render oil too expensive to extract before we literally run out of it, but that doesn't alter the premise of the question. For example, polyvinyl chloride is the third most widely used plastic (according to our article) and it's made from the vinyl chloride monomer which is derived from petrochemicals. Once oil becomes too pricey, will we no longer have any PVC? I don't know, but that would seem to be the implication. Polyethylene is the most widely used plastic and our article notes that it's available in bioplastic form, so presumably, it's not one of the forms that would disappear (not that plastic really disappears, but you get the drift). Our article on polypropylene (the second most widely used plastic) is so full of jargon, I have no idea how it's actually made in real-world terms and so can't even register a guess. Matt Deres (talk) 18:35, 24 November 2013 (UTC)

PP, PE and PVC can also be made from natural gas or from coal -- so they won't "disappear" even if we run out of oil. 24.23.196.85 (talk) 03:09, 25 November 2013 (UTC)

Vinyl chloride is made from ethylene, which can be produced by steam cracking of fossil fuels, but the same process can also be applied to bio-ethanol. Basically, thermal cracking, steam cracking, hydrocracking apply high pressure and temperature to create a mixture of lots of chemicals, catalysts are added to increase the yield of useful products, the resulting mix is separated by distillation, unwanted chemicals are put through another cracking process. After all, heat and pressure is what produced the oil in the first place. It may be more costly when starting from biofuel, but that's no reason why it wouldn't be used. Look at gasoline production: light crude oil contains a lot of gasoline and diesel that can be extracted by simple fractional distillation, but we're also mining tar sands and using energy intensive processes to produce them. Ssscienccce (talk) 19:03, 25 November 2013 (UTC)

November 24

DNA to RNA to protein synthesis

When we say that a child resembles a parent, or that siblings resemble each other, other than things like hair color, I'd say we primarily focus on things like distance between the eyes, shape of the nose and chin and cheekbones, etc. Is it that biological design and construction of all of these physical features are directed by proteins? I mean, at first glance, I don't necessarily see how proteins are responsible for the structure of the human body -- I think of them more as directing function. But is that merely because science education focuses, perhaps, on physiology rather than anatomy, and the shape of bones and the relationship they share with other bones is also rigorously directed by proteins? DRosenbach ^{(Talk | Contribs)} 00:56, 24 November 2013 (UTC)

DNA is the basis for the creation/synthesis of proteins but for familial traits just look not further forward than DNA. When the fertilized ovum cell multiply, a chemical message it sent out. For instance. Some cell will be told “you are going to be a nose” (differentiation). Sandra Bullock and Barbra Streisand have very different honkers. That's not proteins; that is the arrangement of genetics. DNA/RNA doesn't come into it in this sense.--Aspro (talk) 02:00, 24 November 2013 (UTC)

Shape of organs is determined by development of the embryo which is both influenced by genetics and the environment. Genetics can do a lot of shaping via overlayed secretion of transcriptions factors, especially homeobox proteins; they are crucial for evolution of anatomy features. - SCIdude (talk) 05:38, 24 November 2013 (UTC)

It absolutely is the proteins, DRosen (honestly not sure what Aspro is talking about. Everything DNA does to the body is a result of its transcription). But understanding how proteins lead to body shapes is difficult to understand because you are talking about an interaction of massive scale - billions of cells each containing at least as many proteins of thousands of varieties. It is easier if you confine your self to looking at small effects in smaller animals. The entire DPY (dumpy) family of proteins in C. elegans, for instance, controls the length of the body, generally by controlling the properties of the extracellular matrix. And even in this case, researchers learned this by looking for extreme changes in body length/width. The philosophical equivalent would be trying to understand what controls height by looking at human dwarves, and indeed you can read about some of the genetic causes at Dwarfism#Causes (of course, a "genetic cause" exerts its effect through the presence or absence of a particular protein or RNA product). Understanding far more subtle differences in body shape is going to be far more difficult, since genetic causes may be hidden by natural variation. So I would not call this any sort of failure of education, but simply a result of the fact that emergent phenomena are very hard to predict, and can be equally difficult to explain. In case you are wondering why I am focused on explaining differences rather than explaining the phenomenon directly, this is just me thinking as a geneticist - the easiest way to learn about a machine is to break a piece and see what happens, or find a machine that's already broken and try to figure out how. There are researchers tackling this problem from the other direction, taking a minimal system and trying to use genes (and thus their protein products) to direct cells to assemble into particular shapes with particular dimensions. This research is slow-going and using it to explain the shape of an entire human body would be quite a leap. Someguy1221 (talk) 11:46, 24 November 2013 (UTC)

See morphogenesis. Gandalf61 (talk) 14:23, 24 November 2013 (UTC)

Kirchhoff's first law of spectroscopy

Kirchhoff's first law of spectroscopy says that "A hot solid object produces light with a continuous spectrum." I'm wondering why an hot object wouldn't emit a discrete spectrum. 74.15.137.253 (talk) 03:22, 24 November 2013 (UTC)

This is basically the behavior of a black body, which is a well-studied subject in physics. Kirchhoff is well known for studying black body radiation. --Jayron32 05:23, 24 November 2013 (UTC)

Because much more energy states are available. In a hot gas, it's individual atoms (or molecules) emitting, the electrons have a limited number of discrete energy levels they can occupy. In solids there are much more energy states possible. See density of states Ssscienccce (talk) 10:36, 24 November 2013 (UTC)

Thanks. I don't understand where these extra energy levels are coming from though... 74.15.137.253 (talk) 17:57, 24 November 2013 (UTC)

In two words: band theory. --Tardis (talk) 21:45, 24 November 2013 (UTC)

See also Spectral bands: molecules (like a diatomic gas) will have more energy states because the molecule can rotate and vibrate: the vibration or oscillation can be at once, twice, three times .. the harmonic frequency, and each transition between those can combine with an electron state transition, so instead of one line per possible electron transition you get a group of lines close together. The rotation can add more lines even closer together because the energies are smaller. The bigger the molecule the more combinations are possible, and in solids the spectrum will essentially be continuous (unless its a crystal where you may have band gaps) See also Energy_level#Energy_level_transitions and Rotational-vibrational spectroscopy Ssscienccce (talk) 21:26, 25 November 2013 (UTC)

Main Battle Tank

When you read the specifications of Challenger 2 and M1 Abrams you find that the speed of both vehicles off-road is 40 km/hr although they differ in their engines power , so why ? is it limit for any vehicle which drives off-road ? Tank Designer (talk) 12:32, 24 November 2013 (UTC)

There is a five ton weight difference between them. Rmhermen (talk) 13:52, 24 November 2013 (UTC)

But the M1 has higher power to weight ratio, and is faster on the road. Maybe it's the suspension, M1 uses torsion bar, Challenger has hydropneumatic suspension. You don't choose a more complex mechanism unless it offers some advantage, I guess... Ssscienccce (talk) 15:36, 24 November 2013 (UTC)

With something like a car, top speed is generally limited by air-resistance and small variations in the weight of the car or the size of the engine are less critical than how slippery the design of the body is. In a car, power-to-weight ratio has little to do with top speed - but everything to do with acceleration.

But tanks don't go fast enough for air-resistance to matter much and their complicated drive trains make frictional losses much more critical. Hence, any small variation in the way they are designed (and especially their weight) will have dramatic effects on their top speed for any given engine horsepower. Analyzing the reasons why one goes faster than another is likely to be a complicated business.

SteveBaker (talk) 16:16, 24 November 2013 (UTC)

Think you'll find this is a safety issue. A WW2 Sherman taking a bend too fast would over-steer and skid off sideways. The low loading per square foot of track gave it little grip. On soft ground the same manoeuvre would lose it one or both of its tracks (too much resistance to side forces). Not a good thing to happen in the middle of a battle. Then there is the issue of pitching up and down over rough terrain. In the heat of battle with the adrenalin coursing through the body, the bod driving, can go too fast and give himself whiplash. Again, this could occur at just the wrong moment in time. --Aspro (talk) 18:37, 24 November 2013 (UTC)

Also. At 40 km/h diving through rough scrub and into blind ground that one has not seen before and assuming the standard 1.5 second reaction time. Then the tank will have traveled (1.5 X 40 000) / 3600 = 16.6666667 metres be before the driver can even think “Oh Sh-one-T” and try to stop.--Aspro (talk) 18:56, 24 November 2013 (UTC)

I can't see that the "off road" speed is calculated in any scientific way. Is there a standard "off road" terrain? I can imagine that on some surfaces, either tank could almost reach their top road speed, while on others it would be a lot less than the 40 kph quoted. After all, the Land speed record is set "off road" isn't it? Alansplodge (talk) 20:55, 24 November 2013 (UTC)

One does not 'need' an "off road" speed calculated in any scientific way (other than injuries received). Jump into a 4x4 and head out off road at 25 mph (40 km/h). Notice, that even with seat belts on, if the ground is rough, you're being bounced up and down quite violently. That's on a modern vehicle with good suspension. Even though the modern tank cost a couple of million, the tank designers don't put a lot of effort into providing a smooth diver experience. So at 40 km/h the human body has a lot of kinetic energy that muscle power can not compensate for. The WP article also states that:The M1 Abrams' powertrain comprises a 1,500 shaft horsepower (1,100 kW) Honeywell AGT 1500 (originally made by Lycoming) multi-fuel gas turbine, and a six speed (four forward, two reverse) Allison X-1100-3B Hydro-Kinetic automatic transmission, giving it a governed top speed of 45 mph (72 km/h) on paved roads, and 30 mph (48 km/h) cross-country. With the engine governor removed, speeds of around 60 mph (97 km/h) are possible on an improved surface; however, damage to the drivetrain (especially to the tracks) and an increased risk of injuries to the crew can occur at speeds above 45 mph (72 km/h). Try driving across Dartmoor in a Landrover at just 25 mph and you will see what I mean. So tanks apparently are governed to keep the speeds to with in safe limits in all terrains. In the last Gulf War they may have well have had these limits raised, as the terrain was flat and the TV coverage showed them belting-a-long at fair rate on knots (well, the British built tanks did). But off the production line, they will only go so fast and no more.--Aspro (talk) 23:17, 24 November 2013 (UTC)

It's not just a matter of what's safe or 'comfortable' for the crews; the tanks themselves are more fragile than they look. Tanks require a lot of maintenance and repair even under ideal circumstances. There are a lot of heavy moving parts, working under (mechanically) stressful conditions. Consider the M1—there's more than an hour of maintenance required for every hour of operation. A set of tracks are good for 1000 (really) to 2000 (ideally) miles. They suffer a failure that affects combat ability every couple of hundred miles. Remove the governors and start really shaking up that equipment and the enemy won't have to blow up your tanks—you'll do it for him. TenOfAllTrades(talk) 14:55, 25 November 2013 (UTC)

That would seem to be appallingly bad design. A track life of 2000 miles at an average speed of 20 km/hr is a life of 100 hours. I owned a Caterpillar D4 bulldozer at one time for work on my hobby farm. It was second hand, originally owned by a Shire council for towing road compaction equipment, firefighting, and landfill rubbish work. When I bought it, it had 16,000 hours on the clock, and it still had the original tracks, albiet seriously worn though perfectly usable. Such eqipment is considered obsolete now. Skid-steer machines "Bobcat" style ie machines with wheels and no tracks) have taken over most of the market, but bulldozers are still available, with rubber tracks. It has been found that rubber tracks last even longer and require no maintenance. 121.221.118.84 (talk) 23:43, 25 November 2013 (UTC)

But you did not have to guarantee combat readiness for your equipment! There is a difference between having a piece of equipment that works, and having a piece of equipment that can be trusted to work when lives depend on it! If you quantitatively study combat engineering, you will find that vast over-expenditure on preventative maintenance is a statistically better choice. Here's another book from the same library: Developing the Armored Force..., which is essentially an interview transcript with Major General Robert Sunell, who oversaw the U.S. Army's armored technologies and operations. He has much to say about maintenance, (and much more to say about tank track design choices). Nimur (talk) 01:12, 26 November 2013 (UTC)

Valid points in your first para Nimur, but the difference in track life is so extreme (100 hours versus >16,000 hours), and Cat bulldozer tracks so trouble free, that we just have to suspect something wrong with the tank tracks. Last purchase prices nothwithstanding, the military does get junk from time to time. Not all the time, but some of the time. 121.221.118.84 (talk) 02:22, 26 November 2013 (UTC)

Have you been driving your bulldozer for 16,000 hours through Golan Heights-style rocky terrain, off cliffs, and over land mines, while scout snipers fire fifty-caliber anti-materiel rifle rounds at your treads, all while enduring a barrage from 300,000 artillery rounds per hour per square kilometer? Because the M60 can only do that for about 250 500 hours (provided it's equipped with steel track). But it's sort of a high-maintenance vehicle. Nimur (talk) 02:38, 26 November 2013 (UTC)

Furthermore, the Cat D4 weighs about 5 tons - the M60 weighs 50 tons. Right there is a large part of the problem! The Cat's tracks are relatively lightweight and the stress on the links between them is small - but the M60 is an altogether more massive engineering problem. Worse still, the Cat's top speed is nowhere near 40km/hr - and even if it were that high, it doesn't spend much of it's operational life moving anywhere near top speed. The tank, on the other hand drives that fast anytime the terrain allows it. The force on the track links is going to depend critically on weight and speed. Worse still, the Caterpillar is designed to be reliable enough to last for thousands of hours with minimal maintenance - and doubtless there are aspects of it's engineering where compromises were made in other aspects of it's design to get that kind of reliability. But the M60 is designed to get in, do a job and get out again - then get hours of careful maintenance. If they have a trade-off to make between (let's suppose) minefield-survivability and track life - then you can guess which one wins. So it would be no surprise whatever if the lifespan of the tanks tracks were 160 times less than the Caterpillar. SteveBaker (talk) 14:22, 26 November 2013 (UTC)

Actually, the top speed of my D4 (in top gear) was about 30 km/hr. D4's were often used to tow road compaction devices at up to around 20 km/hr - the same order of magnitude as the tank's average speed. A D4 can weigh a lot more than 5 tons. The basic tractor weight of mine with the particular tracks (D4's were made with 2 optional track types) fitted was about 4.5 tonnes. The blade, blade arms, and hydraulics etc weighed another 3 tonnes, and I had a tree lever on it as well - about another tonne - total 8.5 tonnes. (A tree lever on a little D4 is perhaps not much use - the D4 has only enough traction to push over quite small trees, no more than about 150 mm trunk diameter - but I had the tree lever option anyway.) In any case, if you like, you can compere the tank with a D9 (50 tonne nominal weight typical) - which has at least the same track durability & reliability as the D4. You have a point perhaps about trading off performance against reliability, but hey, the difference is so extreme, you just have to question whether the tank has a deffective track design. 121.221.118.84 (talk) 15:15, 26 November 2013 (UTC)

In the future tanks will be remote controlled drones and one won't have to worry about puny humans inside them. It'll cut down the required armor and ease a lot of problems but even so I'm not sure they'll be able to go much faster over rough ground. Dmcq (talk) 22:25, 25 November 2013 (UTC)

Once you have the concept of a drone - the entire concept of a tank may prove unimportant. Without human lives at risk, you can consider a vehicle that is little more than a gun on wheels - if dumping the armor makes it drastically cheaper (which I'm sure it does) then maybe you can have a lot more of them for the same $$$ cost. With more of them, and cheaper - you can more afford to lose them due to enemy action. These kinds of consideration will doubtless produce something that looks nothing like a tank in the end...just as a drone aircraft looks nothing like a conventional manned surveillance plane. Classic surveillance aircraft thinking pinnacled with the SR71 - one of the fastest aircraft in service...modern drone thinking resulted in the Predator - which is probably the slowest! Consequently, I think it's dangerous to speculate about how future drone "tanks" will be...my bet is that there won't be any tanks. SteveBaker (talk) 14:22, 26 November 2013 (UTC)

Just a question/observation - wouldn't the speed depend a lot on the track and groove depth? Thinkof of a tank like a train the lays down its own track, the potential throw the track off would depend mostly on speed and side stress and the depth of the guard to keep the wheels on the track. I think the speed of the tracked vehicle is determined by the track design and less by power and weight except as it relates to the stress on keeping the wheels on the track. --DHeyward (talk) 04:27, 26 November 2013 (UTC)

Gluten and Salt

I'm reading On Food and Cooking, and it states that salt increases the elasticity of the gluten. How does this happen at a molecular level? Is it something specific to the chemistry of sodium or chloride ions? Mostly, I'm trying to make puff pastry without adding salt, and I'm trying to understand what I could use in lieu of NaCl to produce the functional effect. I'd like to avoid KCl if used in any significant amount because of taste concerns, but if it's a function of binding available water instead of something specific to the chemistry of the chloride salts, I'm guessing I could try sucrose instead? 2001:558:600A:2F:45D:4092:30A6:49E3 (talk) 21:31, 24 November 2013 (UTC)

I don't know the answer specifically, but I do know that the elasticity of gluten results from two factors: the tendency of gluten (protein) molecules to curl up, and their tendency to stick to other gluten molecules. Salt solutions can affect both of those things, but not as a function of binding water -- dissolved salt doesn't do that. Instead the ions work their way into the protein structure and alter the force geometries. Sugar would not have the same effect. Also putting any substantial amount of sugar into a dough will radically alter the result you get when you bake it.

Integration of body temperature signals

When I'm cycling in the winter, my torso usually overheats while my face (and ears if exposed) and feet can get cold. I was wondering whether the body of humans or any other animal ever integrates temperature signals from different areas of the body to do something more sensible like delivering excess heat from the torso to the extremities? — Preceding unsigned comment added by 129.215.47.59 (talk) 22:13, 24 November 2013 (UTC)

The body does have pretty sophisticated mechanisms for thermoregulation. I would be cautious about "more sensible" - evolution would favor survival over comfort, and it seems likely that heat loss (the major lethal risk) would be greater with diversion of heat to the extremities, especially when the thermoregulation system cannot "know" how soon the body will find warmer shelter. So, I would start by defining "sensible" - or realizing that the body's system is quite good already. -- Scray (talk) 22:24, 24 November 2013 (UTC)

The body is already acting "sensible" by not moving more heat to an area that is loosing it quickly. Having your nose, fingers, ears or toes freeze is a lot better than having your core temperature drop. Also having those body parts freeze does not always mean you are going to loose them. Over the years I've frozen my nose and my fingers (those several times) and still have them all but my fingers do get cold very easily. See hypothermia and frostbite. CambridgeBayWeather (talk) 00:23, 25 November 2013 (UTC)

When the core temperature rises, the body will start sweating, hairs on the skin lie flat, preventing heat from being trapped by the layer of still air between the hairs. Arteriolar vasodilation occurs. This redirects blood into the superficial capillaries in the skin increasing heat loss by convection and conduction. But that alone is not enough to keep exposed skin warm, because you can't get enough heat to the skin to compensate the cooling by the cold air. Keep your hand in -20°C brine while you're in a sauna, and your hand will still suffer frost bite. Whether local cold affects the blood supply to that part of the body, I'm not sure, but blood viscosity increases at lower temperature, so that would decrease the flow in cold skin. Ssscienccce (talk) 14:39, 25 November 2013 (UTC)

• I've done quite a bit of winter cycling and experienced all these things. The mechanisms are a bit tricky. To begin with, when your core temperature begins to drop, the body reduces blood flow to the hands, feet, and face -- that's why they get cold. If you exercise enough to bring your core temperature above optimum, the peripheral blood flow will usually come back, especially to the hands. However, the switching is not immediate, and you might have to sweat pretty hard for a while to make it happen. I've done hard riding in 20 degree conditions with no gloves or face mask -- the usual effect is that for the first few minutes I feel a lot of pain, but then the "heat comes on" and my hands warm up and feel fine. For some reason, though, my feet usually get cold, even if I'm wearing good shoes and socks. Looie496 (talk) 17:01, 25 November 2013 (UTC)

Seems that blood flow in extremities is more sensitive to local temperature than ambient temperature according to this. It also mentions vasodilation when the temperature drops to 4°C (39°F), the Lewis hunting response, in which the alternating vasodilation and vasoconstriction occurs. At 10°C, constant vasoconstriction will occur, with blood flow dropping gradually for ten minutes and then staying constant at that low level (in Eskimos it takes 90 minutes). So your cold feet and warm hands may be because your feet aren't cold enough to initiate the Lewis hunting response. Ssscienccce (talk) 07:38, 26 November 2013 (UTC)

Snake bite article

The article on snake bites includes a section on first aid. Am I correct in thinking this equates to medical advice and that some of the details there should be removed? Bazza (talk) 22:39, 24 November 2013 (UTC)

Reference desk requests for urgent medical aid are different from discussion of medical matters in articles. Talk and articles are different things. I'm not going to specifics, just saying on a general level - which sort of reflects the underlying difference. 88.112.41.6 (talk) 22:53, 24 November 2013 (UTC)

It does make an interesting question. Where does one draw the line between medical information or knowledge and medical advise? Anyways, the applicable policy is WP:NOTHOWTO. While I do not think the problem is that it is medical advise, but that it is a "how to" section. The information on how snake bites are treated seems OK, but the step-by-step part should be removed. Help yourself if you like, Bazza, since you saw it. Richard-of-Earth (talk) 06:15, 25 November 2013 (UTC)

it's open to interpretation, it uses the imperative mood, but is it instructing the wikipedia reader or merely illustrating how most first aid guidelines instruct their readers? NOTHOWTO writes: "describing to the reader how other people or things use or do something is encyclopedic", so putting "they" in front of every sentence would fix the problem... WP:MEDICAL states: "Nothing on Wikipedia.org or included as part of any project of Wikimedia Foundation, Inc., should be construed as an attempt to offer or render a medical opinion or otherwise engage in the practice of medicine." Wile this disclaimer is compatible with providing general information on medical topics, it seems hard to reconcile with answering concrete requests for medical advice on the reference desk. Practicing medicine requires a patient (I think). There's no patient involved when writing an article, there may be when you answer questions from individuals. Ssscienccce (talk) 15:33, 25 November 2013 (UTC)

Permeability of rocks in the soda canyon

Are the rocks of the soda canyon in Colorado mostly permeable? — Preceding unsigned comment added by 99.146.126.108 (talk) 23:59, 24 November 2013 (UTC)

According to http://pubs.usgs.gov/sim/3224/SIM3224_pamphlet.pdf, the rocks of Soda Canyon are essentially all part of the Goodridge formation, which consists of sandy shale, sandstone, and cherty limestone. All of those should be pretty permeable. Looie496 (talk) 16:33, 25 November 2013 (UTC)

November 25

Identifying dinosaur species from tiny bone fragment

This query was prompted by seeing dinosaur-bone beads and jewelry at an art fair: given one of those items (which are tiny, spherical beads or cabochons and presumably retain no trace of their original shape), would it be possible to determine in a non-destructive manner what species (or even higher taxon) it came from? 69.111.191.53 (talk) 02:10, 25 November 2013 (UTC)

I am reminded of the Astragalus of Necrolemur, a paper I read several years ago. Its author made his entire career by identifying extinct primates almost exclusively from tiny ear-bone and wrist fragments. The scientific skeptic in me knows that there is at least a little hand-waving in the process; but the paper has many elements of rigorous palaeontological investigation. And its author has a certain literary flourish: announcing to a world who had been desperately holding its breath, "the previously unknown astragalus of the Eocene omomyid Necrolemur has been discovered." Finally! Nimur (talk) 15:35, 25 November 2013 (UTC)

Interesting... I don't blame him for sounding excited, though, because anything called the Astragalus of Necrolemur is clearly a magical McGuffin. (PS: I'm the OP, my IP just seems to have changed overnight.) 69.111.73.99 (talk) 16:58, 25 November 2013 (UTC)

It might be possible to say something on the basis of a bone-bead, because there are differences in bone density and perhaps one or two other aspects of structure that might survive in a fossil. But if the shape has been lost, it wouldn't be possible to identify the species or anything close to it. Looie496 (talk) 16:16, 25 November 2013 (UTC)

...all of which leads me to wonder whether the vendor of these items is really selling dinosaur bones in the first place. Since birds are now considered to be therapod dinosaurs, you might be getting an old chicken bone! SteveBaker (talk) 18:43, 25 November 2013 (UTC)

Except I don't think even the oldest of chicken bones would be fossilized. Unless there's a way to turn fresh bone into stone in a matter of days-- perhaps a Young Earth Creationist might know? 69.111.73.99 (talk) 21:27, 25 November 2013 (UTC)

Fossil says that fossils can be as recent as 10,000 years old - which (I presume) would mean that bird fossils might be much easier to obtain than a 66 million year old dinosaur. SteveBaker (talk) 23:42, 25 November 2013 (UTC)

Fair point, but according to Chicken#Origins domestic chickens probably don't go back quite that far (though presumably some related birds do). Also this page says that the avian fossil record is poor because birds have hollow bones. So I doubt that bird fossils would be more common than non-avian dinosaur ones (nor would their bone structure be as conducive to carving into beads, if they're hollow and more fragile). 69.111.73.99 (talk) 03:08, 26 November 2013 (UTC)

Electrochemistry question

Is the dropping mercury cathode still used for voltammetric analysis? I'm currently working on a design project for a multipurpose potentiostat which can make several different analyses using interchangeable electrodes and multimode software, but we can't agree on whether to include a polarography mode -- I'm in favor of at least making provision for it, while two of my teammates argue that it's not needed at all. Thanks in advance! 24.23.196.85 (talk) 03:20, 25 November 2013 (UTC)

Yes, there are several journal articles from the last decade discussing current DME applications, and potential applications (no pun intended). From what I can find, DMEs seems to be uncommon in polarography. Nonetheless, it seems appropriate to make provision for it as you suggest. Plasmic Physics (talk) 10:39, 25 November 2013 (UTC)

Electrolysis

Why is it so that electrowinned copper occasionally exhibits a latency when redissolving in concentrated hydrochloric acid? The electrolyte consists of mixed sulfates. Plasmic Physics (talk) 10:50, 25 November 2013 (UTC)

Lack of dissolved oxygen or oxides? Copper shouldn't dissolve in hydrochloric acid, should it? Maybe the answer by pisgahchemist here can shed some light... Ssscienccce (talk) 16:18, 25 November 2013 (UTC)

Copper will not dissolve in hydrochloric acid, but it will in nitric acid, due to the ability of nitric acid to oxidize the copper. It's the nitrate ion, not the acid part, that does the dissolving. See [9]. You get blue copper nitrate solution and brown nitrogen dioxide gas. But the chloride ion is a pretty crappy oxidizing agent, so HCl shouldn't do squat. --Jayron32 20:21, 25 November 2013 (UTC)

Does that mean I broke my copper, seeing as how it does dissolve? Plasmic Physics (talk) 21:26, 25 November 2013 (UTC)

Are you sure it was copper, and not already oxidized copper, like red Copper(I) oxide? Because dropping copper into concentrated HCl should do absolutely nothing interesting, per the Standard electrode potential (data page), which clearly shows that the E^o of Cu + 2H⁺ --> Cu²⁺ + H₂ to be -.340 V. Either you're providing some excess energy somewhere, or its simply not going to do anything. I can leave a penny in HCl overnight, and get a penny back the next day. It'll be a bit cleaner, but otherwise unchanged. --Jayron32 00:45, 26 November 2013 (UTC)

I'm positive it was copper. It does not seem likely that cuprous oxide would be deposited at the anode. Plasmic Physics (talk) 02:20, 26 November 2013 (UTC)

From this topic on Gold refining forum: "As I understand it, it seems that, once the HCl + oxygen dissolves a little copper, it becomes self-perpetuating, from the action of the copper chloride that is produced." Some tests another poster did seem to agree with that hypothesis.

There's a paper with the title "Dissolution of copper in hydrochloric acid solutions with dissolved molecular oxygen", available for $41.95. The abstract mentions this was in the presence of cupric chloride, and "the dissolution rate of copper exhibited first order kinetics with respect to CuCl2 concentrations above 0.01 mol dm⁻³" Ssscienccce (talk) 05:48, 26 November 2013 (UTC)

OK, so now that we've established that copper is attacked by oxygenated hydrochloric acid, let us get back to the original question - why does the reaction intermittently show a latency? Plasmic Physics (talk) 08:23, 26 November 2013 (UTC)

Oxygenated hydrochloric acid is likely forming chloroxy anions (hypochlorite, hypochlorite, chlorate, perchlorate, etc.) For example, ClO2(g) +  H+ +  e− <-> HClO2(aq) has a standard potential of +1.9 V, which is more than enough to oxidize copper. You would need enough dissolved oxygen to generate enough ClO2 in situe to actually do the reaction, but I suppose that it's possible. --Jayron32 18:40, 26 November 2013 (UTC)

Spectroscopic film ?

I saw on a science show where they held up a "special film" to a neon light, and the film then showed a series of lines, the spectrum of neon. Presumably this works because each line on the film is opaque to all but one narrow frequency of light. What is the name of this special film, where can I get some, and how much does it cost ? StuRat (talk) 14:08, 25 November 2013 (UTC)

Maybe it was a diffraction grating? DMacks (talk) 14:15, 25 November 2013 (UTC)

While I obviously can't definitively say that what StuRat saw was a diffraction grating, that demonstration certainly works with a diffraction grating. Assuming you are based in the US there are several formats of diffraction grating available from Edmund Scientific, such as these, which cost $7.95 for 15, or in fashionable cardboard glasses format for a dollar fifty. Equisetum (talk | contributions) 14:53, 25 November 2013 (UTC)

Some of those say they are "single-axis" and others say "double axis". I'm guessing that single axis is the one I saw ? With double axis, would I see two spectrographs overlapping, at right angles ? If so, what's the use in that ?StuRat (talk) 21:42, 25 November 2013 (UTC)

With christmas time coming up you can probably find those glasses locally for a couple of bucks, if you don't mind the spectrum ending up displayed as images of snowflakes or "happy holidays" instead of nice clear bars. :-) They also show up around the 4th of July for watching fireworks. Katie R (talk) 17:36, 25 November 2013 (UTC)

Sounds annoying, but I will look for them. I wonder how to determine which local stores would have them. StuRat (talk) 21:44, 25 November 2013 (UTC)

My mother-in-law in the Detroit area works at English Gardens, a garden center chain, and usually brings home a few pairs every year. If you happen to be near one it would be worth checking out. They work best with point sources - I think the idea is to use them to look at Christmas lights . Now that I think about it, they're probably a lot more interesting with LED lights than incadescent. I've pointed a laser through them, and it ends up projecting the image very clearly onto the wall. Katie R (talk) 12:52, 26 November 2013 (UTC)

Social inhibition of urination

Why do some people find themselves unable to urinate when standing next to someone else at a urinal? Does this phenomenon have a name? How prevalent is it? 129.215.47.59 (talk) 17:26, 25 November 2013 (UTC)

In laymen's terms, it is called piss shy; that link is a redirect to the real article Paruresis --Jayron32 17:44, 25 November 2013 (UTC)

It has also slightly more tastefully been called Shy bladder. Mingmingla (talk) 17:46, 25 November 2013 (UTC)

You've tasted what? --Jayron32 17:50, 25 November 2013 (UTC)

Drinking urine is blue. Well, the link is, whether or not methylene blue is used. DMacks (talk) 18:09, 25 November 2013 (UTC)

I notice that the article states "The codes and procedures for drug testing in sports are set by the World Anti-Doping Agency (WADA). Enquiries to WADA reveal that their doping codes do not cater for the condition at all, and they say they have never had any reports of problems with it." Yet a report on a WADA website mentions "Elite athletes have to undergo doping-controls. In over 50% of cases delays occur because the athlete experiences a sudden inability to urinate. 46% of these delays exceed one hour." Ssscienccce (talk) 21:50, 25 November 2013 (UTC)

Sounds like they need some cross-transfer of ideas. Those of us who are male, and old enough to have potential prostate problems, sometimes have doctors order a urine external flow test. For this you are requested by a (usually) pretty nurse to urinate while she uses a simple gadget to measure the flow rate in ml per second. When she is ready, she tells you to "let it go!" and then maybe says "is that the best you can do?". Nurses usually tell you to imagine having a shower while peeing - it works nearly every time apparently. 121.221.118.84 (talk) 02:40, 26 November 2013 (UTC)

It has also been called Bashful bladder syndrome. Richerman (talk) 18:13, 25 November 2013 (UTC)

I suspect that bashing someone with a full bladder in said area could be a way of overcoming the problem... DMacks (talk) 18:25, 25 November 2013 (UTC)

Nope. The feeling to go completely leaves. A similar condition is an overwhelming need to go when arriving home regardless of feeling before. Brain is a weird mind farking thing. --DHeyward (talk) 08:38, 26 November 2013 (UTC)

Funnily enough, the thing that's telling you that info about your brain is ... your brain. It seems to have a low opinion of itself. Maybe time for a brain transplant, and this time try to get one with a little bit more self-esteem.  :) -- Jack of Oz ^{[pleasantries]} 09:14, 26 November 2013 (UTC)

Platinum Daniell cell

If I were to make a Daniell cell with a platinum cathode instead of copper and platinum chloride instead of copper sulphate, would I need platinum(II) chloride or platinum(IV) chloride? I am asking this question on behalf of user:DPL bot who keeps bugging me for an answer, but I don't know. SpinningSpark 18:55, 25 November 2013 (UTC)

Liquids at zero pressure

Are there any substances that exist as a liquid at zero pressure? — Preceding unsigned comment added by 74.15.137.253 (talk) 19:56, 25 November 2013 (UTC)

Zero pressure (like absolute zero) is an idealized state that does not really exist, except as an asymptote you can approach, but never realize. A perfect vacuum cannot be realized. As soon as you put a liquid into a vacuum, it will start to evaporate instantly, and then you don't have a vacuum anymore. --Jayron32 20:18, 25 November 2013 (UTC)

True. Is there a substance that doesn't sublimate at zero pressure? 74.15.137.253 (talk) 21:09, 25 November 2013 (UTC)

That would imply a vapor pressure of zero at a temperature above 0° K, not sure if that is possible, wouldn't quantum effects, uncertainty principle statistical randomness prevent that? Ssscienccce (talk) 22:06, 25 November 2013 (UTC)

There is no such thing as zero pressure, and by extension, there is no such thing as a substances that exist as a liquid at zero pressure. Plasmic Physics (talk) 07:02, 26 November 2013 (UTC)

The fact that asteroids exist implies that those solids do not sublimate much at temperatures and pressures near 0, as found in deep space. StuRat (talk) 07:03, 26 November 2013 (UTC)

Could you please clarify this comment? Why doesn't zero pressure exist(at least as an idealization)? 74.15.137.253 (talk) 02:18, 27 November 2013 (UTC)

Sublimation of solids at near absolute vacuum become a significant factor when discussion the heat death scenario of the end of the universe. Plasmic Physics (talk) 07:54, 26 November 2013 (UTC)

What would happen if you started heating the asteroid, but kept it in a zero pressure environment? Would it ever melt? Is there any substance in deep space that wouldn't melt if you heated it in deep space? 74.15.137.253 (talk) 02:20, 27 November 2013 (UTC)

Entirely sterile rooms

Which facilities commonly have absolutely sterile (including air) rooms and do such compartments have some specific name? I assume that operation rooms and many scientific facilities don't have sterile air as well. Perhaps in such highly bioclean rooms food wouldn't mold and corpses wouldn't corrupt? 93.174.25.12 (talk) 22:56, 25 November 2013 (UTC)

I've heard of the 'clean room'. Plasmic Physics (talk) 23:05, 25 November 2013 (UTC)

That's a cleanroom. Note that they are not sterile, but have a "controlled level of contamination." SemanticMantis (talk) 23:24, 25 November 2013 (UTC)

Note: Rare New Microbe Found in Two Distant Clean Rooms; November 06, 2013, Jet Propulsion Laboratory ~E:71.20.250.51 (talk) 23:35, 25 November 2013 (UTC)

Manufacture of sterile active pharmaceutical ingredients is generally conducted within clean rooms; these can't be sterilised later, unlike sterile medical supplies like syringes, bandages, pads, swabsticks which are usually sterilised after production and packaging by applying ethylene oxide in a vacuum chamber. (thats's why the packaging is made of porous paper); others are the preparation of vaccines and parenteral products (food that is administered by infusion), an example of hands-on training of aseptic processing can be seen here. Air in operating rooms is filtered 20-25 times per hour to achieve low numbers of airborne pathogens (source), absolute sterility isn't really necessary for patients with a working immune system. Complete aseptic environment is most important when working with products that are a good growth medium for bacteria or molds, that have to be stored for a period of time and that would be damaged or destroyed when sterilized after production. Regarding specific names: "aseptic (filling, finishing, processing) core" is widely used, in job ads for example. Ssscienccce (talk) 01:38, 26 November 2013 (UTC)

Bear in mind that any food or corpse introduced into such an environment would bring their sources of corruption with them. I doubt rate of decay would be much different, unless they were irradiated or otherwise sterilized first. Rojomoke (talk) 05:32, 26 November 2013 (UTC)

Engineering definition

I recently heard a definition of engineering I've never heard before, coming from a former engineering director. He said that engineering is not the application of science to design, build and maintain new products, structures and services but the application of science to make money. Is this definition correct? Can't this definition be used for any discipline under a for profit company, which is most disciplines? — Preceding unsigned comment added by Clover345 (talk • contribs) 23:54, 25 November 2013 (UTC)

That sounds more like cynicism or sarcasm than a "definition". ~E:71.20.250.51 (talk) 00:02, 26 November 2013 (UTC)

A common engineering maxim is "engineers make for a penny what any fool can make for a pound". Cynisism it may be, but there is some truth in it. To me, engineering is about making things in a professional environment and very little about applying scientific principles. As a practising engineer 99% of what we did (I am retired now) was based on experience rather than hard science worked out from first principles. SpinningSpark 00:13, 26 November 2013 (UTC)

Do you think it's only a handful of detailed design teams which are concerned with hard science? Clover345 (talk) 00:13, 27 November 2013 (UTC)

There's lots of them but usually with a directed outcome. A lot of research grants are really hard science engineering. Room temperature super-conductors for example is an engineering problem. Super-string theory is not. --DHeyward (talk) 00:23, 27 November 2013 (UTC)

There is more than one way to say what is realy the same thing. Previous posters have nailed it. Another way is what I was taught in company induction training in my first job after campleting engineering at university: Engineering is about the Three M's - Where is the (specialised) Machinery comming from? Where is the (skilled) Manpower comming from? Where is the Money coming from (loans? other financing?) ? These three questions are what senior engineers, engineering managers, and the heads engineering companies are fundamentally concerned about. 121.221.118.84 (talk) 00:42, 26 November 2013 (UTC)

I think it defines the person who said it more than it defines engineering. He never heard of nonprofit organizations, open-source, volunteers, hobbies? Working on an irrigation project in the third world, building a three stage amateur rocket, designing nuclear weapons, it's all engineering... Ssscienccce (talk) 02:05, 26 November 2013 (UTC)

Here's one of a series of strips of Dilbert talking about engineering.[10] ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:40, 26 November 2013 (UTC)

Engineering is about producing products quickly, cheaply and with high quality. But you can only pick two of those. It runs the gambit of the three combinations of two choices. --DHeyward (talk) 00:18, 27 November 2013 (UTC)

November 26

Must an element's pth ionization energy be greater than its qth ionization energy if p>q?

I think the issue of 12th ionization energy of Aluminium in Molar ionization energies of the elements has been left there for some years. There is a note saying 12th ionization energy, which is 223366KJmol^-1 "Should be less than the 13th; perhaps 201266." 201266 is the value in Ionization energies of the elements (data page), while 223366 seems unreferenced. By the way, is it possible that 12th ionizaton energy is greater than 13th one?--chao xian de lun zi (talk) 01:45, 26 November 2013 (UTC)

Feel free to change it, I saw a graph of the ionization energies in a book, no exact values given but the 12th was in the region of 20000, and definitely lower than the 13th, so 201266 seems reasonable, certainly better than 223366. Ssscienccce (talk) 15:33, 26 November 2013 (UTC)

Real life transformer

In real life, do they match the primary coil turns with designed input voltage (so 220v means 220 primary coil turns)? Because in science questions I frequently encounter 120v for 240 turn coil 140.0.229.39 (talk) 13:28, 26 November 2013 (UTC)

It's strictly the ratio of the turns of the two coils that matters. The resulting voltage can vary very quickly if the input voltage changes. Hcobb (talk) 13:47, 26 November 2013 (UTC)

The number of turns per volt depends on the core area, the frequency and the maximum flux density of the core (limited by saturation losses) ; the rated power VA determines the core area needed, say for example 10 cm² for 100VA (at 50Hz). The Transformer universal emf equation gives you the E_rms that the maximum flux density will produce: E_rms=4.44.f.a.N.B_peak With a maximum flux density of 1.5 Tesla and a frequency of 50 Hz, you get E_rms/N=333a, a= 0.001m² so E_rms/N=0.333. So a core of 10 cm² will require 3 turns per volt. Ssscienccce (talk) 15:27, 26 November 2013 (UTC)

Numbers in exam/revision questions rarely have anything to do with real life examples. They are chosen more to make it easy for the writer to set, the student to calculate, and the teacher to check. It is frequntly the case that if your answer isn't a whole number or a simple fraction then you have done something wrong in a test question. Conversely, in real life answers are rarely simple or whole numbers. A mains transformer will usually have many more turns on the primary than the number of input volts. SpinningSpark 15:36, 26 November 2013 (UTC)

I have spent a fair amount of time working with transformers and offer an observation (which could be readily backed up by references). I really liked Ssscienccce's response with very specific design information. When the other windings of a transformer are open(unloaded), and you excite the primary winding with its rated voltage at rated frequency, the transformer and the winding should act as an inductor with sufficient inductive reactance to limit the exciting current to a reasonable low value. Thus there will be a minimum number of primary turns for satisfactory operation. If you had too few turns in the winding, excessive exciting current would flow, causing overheating of the winding or blowing a fuse, even if there were no load on other windings. This exciting current has both real (overcoming resistance of the wire) and reactive components. To the extent other windings are then "loaded" by drawing current to a load, the primary current will increase above the exciting current. A caveat: in a large utility transformer the initial inrush of exciting current will vary somewhat depending on the point in the cycle it is energized and the previous recent magnetization history of the transformer. Edison (talk) 17:39, 26 November 2013 (UTC)

Possibility of chromatophore bacteria using complex pigmentation for crypsis

Is there a chemical reason why there could not be chromatophore bacteria that use the right pigments to achieve the same level of crypsis as chameleons, if not better? I'm aware no such species of microorganism is known to exist, however would a living incredibly stealthy pigment goo be as scientifically impossible as something like faster than light speed time travel? I cannot think of a reason such a bacteria would have access to all those pigments in nature without being mobile like higher organisms such as chameleons; so I'm assuming this would have to be made by scientists or be from space. Even an example from science fiction of such a bacteria would be welcomed. CensoredScribe (talk) 22:11, 26 November 2013 (UTC)

Would visual camouflage help a bacterium hide? The only bacteriovores i can think of do not hunt by sight. Someguy1221 (talk) 01:03, 27 November 2013 (UTC)

I agree with you that visual camouflage is largely ineffective against most animals as they rely more on smell; however if the pigments it used were highly poisonous as well, like a lot of commercial paints, that could give the bacteria an additional level of protection against predators. Smell would also not be a factor in anaerobic conditions, like an asteroid; I don't know why it develop camouflage in a cave where there is little if any visible light. Nor would being able to survive on it's own be important where it artificially created in a laboratory environment; where that the case it's only means of survival could well be in forming a symbiotic relationship with humans.

Acid rusted tools

I accidentally exposed some of my sockets in my cheap socket set to very concentrated hydrochloric acid and now they've rusted up. Can they be saved? It's not a fancy set but I'd like to keep them if I can. I'm not sure how they were manufactured/what their composition is. --78.148.107.181 (talk) 22:12, 26 November 2013 (UTC)

A l-o-n-g time since I had a chemistry class, so chemistry experts please be kind. This is not advice, but Hydrochloric acid says the acid is used to remove rust from steel, in a "pickling" process which has as inputs the acid, surface rust, and the iron itself and , with results ferrous chloride and water. There would be some loss of the metal, and perhaps pitting. Perhaps the acid removed a protective chrome layer. An article about mild steel in hydrochloric acid notes pitting and evolution of hydrogen [11]. Have you tried washing it to remove acid residue, drying, wirebrushing to remove rust (while wearing suitable personal protection against eye injury or inhalation of particles and spraying with WD40 to discourage additional rust? If there is significant pitting and loss of metal I'd pitch the damaged sockets, since "cheap" socket sets can be had for a few dollars, and they will be even less useful than they started out, especially the works inside a ratcheting socket wrench. [12] says that while concentrated hydrochloric acid removes rust from steel, a dilution of the acid causes rapid rusting. Edison (talk) 23:57, 26 November 2013 (UTC)

The acid just removed the very thin coat of oil that is applied in the factory, allowing oxidization to precede rapidly. Get some more hydrochloric acid (an excellent rust remover) and diluted down to 5%. Adding acid-to-water. The original strength will be written on the container (use some eye protective goggles). Dunk all sockets in this, until the rust has been removed ( just a few minutes will do). Wash in boiled water and 'immediately' coat in oil. Anything will do, like engine oil but something really tenacious like Wynn’s engine oil additive or STP engine oil additive is even better. Wipe off surplus oil. --Aspro (talk) 00:17, 27 November 2013 (UTC)

Correct approach to deriving the shell theorem?

I know the easy approach to proving the shell theorem is to take a point particle, integrate all the force elements over an infinitesimally thin shell, integrate over all shells, and then integrate all the point particles. I now asked myself, why not do it all at once? The problem is that it arises in a nested volume integral.

Here, let V₁ and V₂ represent the regions occupied by my two spherical objects (assuming they do not overlap). Let the center of the first object be at the origin, and let d be a displacement vector between a mass element in the first object and another mass element in the other object, taken to point towards the second object. Let dM be a mass element of the first object and dm a mass element of the second. Let u, v, and w represent the position of a mass element in V₂. Let f(R) and g(r) be the densities of the two objects respectively (they are functions of the radius only, as required by the shell theorem), and finally, let the center of the second sphere be located at a displacement vector of X=<A,B,C>. G is the universal gravitation constant.

Then: $\mathbf {F} =-G\iiint \limits _{V_{1}}(\iiint \limits _{V_{2}}{\frac {dm*\mathbf {d} }{d^{3}}})dM=-G\iiint \limits _{V_{1}}f({\sqrt {x^{2}+y^{2}+z^{2}}})(\iiint \limits _{V_{2}}{\frac {\mathbf {d} *g(\Vert \mathbf {d} -\mathbf {X} \Vert )}{d^{3}}}dV_{2})dV_{1}$ $=-G\iiint \limits _{V_{1}}(\iiint \limits _{V_{2}}{\frac {f({\sqrt {x^{2}+y^{2}+z^{2}}})((u-x)\mathbf {\hat {i}} +(v-y)\mathbf {\hat {j}} +(w-z)\mathbf {\hat {k}} )g({\sqrt {(u-x-A)^{2}+(v-y-B)^{2}+(w-z-C)^{2}}})dV_{2}}{((u-x)^{2}+(v-y)^{2}+(w-z)^{2})^{\frac {3}{2}}}})dV_{1}$

In the integration, X is fixed but arbitrary, so the six variables of integration are u, v, w, x, y, and z. As you can see, I don't have many ideas on how this reduces to (using my notation) $\mathbf {F} ={\frac {-GMm\mathbf {X} }{X^{3}}}$, because f and g are arbitrary continuous functions of one variable that determine the densities of the two objects as functions of their radii, and entire integrand is very convoluted. If I were to evaluate the whole thing using one set of substitutions, then the Jacobian determinant would have to be of order 6 (!). Therefore, I have to do something to simplify the mathematics. I know this is a convoluted approach, but really, this is what is really done when we apply the shell theorem - summing up the forces contributed by all the possible pairs of mass elements. If the above notation is not correct, how would this "all-at-once" approach be executed?--Jasper Deng (talk) 01:39, 27 November 2013 (UTC)

I don't know. Possibly through Karma or some other "mythological" human idea formed at some time in our evolutionary history!. Dr. Madhattan (talk) 01:45, 27 November 2013 (UTC)

(edit conflict)At another glance, $-G\iiint \limits _{V_{1}}f({\sqrt {x^{2}+y^{2}+z^{2}}})dV_{1}$ reduces to -GM, but the remaining integrand is still overly convoluted.--Jasper Deng (talk) 01:46, 27 November 2013 (UTC)

Oh, wait, you can't do that because the second integral is a function of x,y, and z.--Jasper Deng (talk) 01:49, 27 November 2013 (UTC)