Wikipedia:Reference desk/Archives/Science/2011 March 22

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March 22

Earth's tilt and space radiaton.

Greetings!

I have a rather complicated question concerning the tilt of the Earth, the way the atmosphere protects us from radiation, and the effects—for a limited time—of exposure to said radiation.

Right now, our planet's orbit is tilted 23-1/2 degrees to the ecliptic. (This is why the Sun rises 23-1/2 degrees north and south of due east on the solstices.) In the (northern) summer solstice, the Earth approaches perihelion—the closest it will get to the Sun—and approaches aphelion—the farthest away it will get—at the (northern) winter solstice.

In spite of this, however, the southern hemisphere gets just as hot a summer (and just as cold a winter) as the northern hemisphere. This suggests to me that it is not distance per se, but a property of the atmosphere that gives us our tellurian seasons. I'm not an expert at all in the sciences, but I do know that the atmosphere absorbs virtually all Gamma Rays, X-Rays, Ultraviolet Rays, and most Infrared Waves from space. (Only Visible Light, Radio Waves, and some Infrared Waves make it all the way down to the surface). And I also know that around the polar regions, there are "holes"—for lack of a more proper term—where space radiation penetrates far more freely.

Now, suppose that the Earth were tilted on its axis as, or more severely than, Uranus. For simplicity's sake, let's say at an even 90 degrees to the ecliptic. (Also, for all intents and purposes, let's say that all other factors remained unchanged: orbital speed, distance from the Sun, etc.) How would that change the way that space radiation affects us?

Over the course of one Earth day (23 hours and 56 minutes) the Sun would pass through both poles on each of the solstices.

Here are my questions:

As the Sun's radiation penetrated to the surface of the Earth through the "ozone hole" and other "holes" that exist at high lattitudes, what would change about life on Earth?

—Would we see an abundance of natural disasters? (floods, volcanic erruptions, cyclones, earthquakes, etc.)

—Would anatomy and physiology of Earthly lifeforms change? (We know for a fact, the photosynthesis in plants depends on the Sun, and even we humans synthesize Vitamin D from it.)

—Would geology change in any way; to wit, would rocks become lighter or more porous, or some way different. Would the magnetic field change direction or become stronger (or less strong)?

—Would the seasons change in length or demeanor, since the atmosphere would factor differently?

And finally:

—Can such a state be replicated today, under laboratory conditions, in a small, self-contained experiment?

--Thank You! Pine (talk) 00:31, 22 March 2011 (UTC)

So, assuming that you mean the North Pole would point directly at the Sun during one season, and the South Pole would point directly at it in the opposite season, with things being about how they are now, when halfway between the seasons, here's what I would expect to be different:

1) Far more extreme changes between the seasons. During winter at each pole, temperatures would drop low enough for carbon dioxide to form dry ice, and possibly low enough to produce liquid nitrogen from the air. During summer at the poles, water might boil. This would cause massive storms and currents in the air and water in-between. I doubt if (multi-cellular) animals could survive the extremes at the poles. Bacteria and some other simple life forms might, though. Migrations would become far more extreme, with animals moving to the summer side continuously. Unlike now, this would likely need to include humans. Life might be far more concentrated around the equator than it is now.

2) As for radiation at the poles, yes, there would be more, but this would be a minor concern compared to the temperature extremes.

3) I don't see how this would affect geology or magnetic fields.

4) You would see far more extreme weather (cyclones, hurricanes, tornadoes, etc.), but not earthquakes or volcanoes, because they don't get their energy from the Sun.

As for creating a physical model, I can't see how you would replicate the oceans and atmosphere in such a model. Therefore, a computer model (using computational fluid dynamics) would be the way to go. StuRat (talk) 02:08, 22 March 2011 (UTC)

Just to note that you've got the perihelion/aphelion–solstice business backwards, Twinpinesmall. At the summer solstice, Earth is approaching its aphelion; and at the winter solstice, its perihelion. See Apsis#The perihelion and aphelion of the Earth. This shows that it's the planet's tilt, far more than its distance from the sun, that determines seasonal variations in temperature (I'm not sure what you mean by your reference to a "property of the atmosphere"). Deor (talk) 02:29, 22 March 2011 (UTC)

The statement that the northern and southern hemispheres get equally hot summers is false, I believe. Even if they received equal insolation it would be hard to make valid comparisons, though, because (1) a much higher fraction of the southern hemisphere is ocean, and (2) the southern polar region is an ice-covered continent whereas the northern polar region is ocean. The statement that if the tilt were 90 degress the sun would pass over both poles at the solstice is also false -- in fact each pole would experience permanent day for 6 months and permanent night for 6 months. Looie496 (talk) 02:41, 22 March 2011 (UTC)

tidal energy backup

Would a tidal energy turbine be a good backup energy supply for nuclear power plants? especially in Japan where the plants are mostly on the coasts and they have alot of earthquakes. —Preceding unsigned comment added by 98.221.254.154 (talk) 03:30, 22 March 2011 (UTC)

No:

1) Being on the coast, they probably would have also been destroyed by the quake and tsunami.

2) The wires used to deliver electricity from them would have also been destroyed.

3) A backup power supply should be something with very little cost when not in use. However, tidal generators are exposed to the sea and weather, and, as such, would require constant maintenance. This could only be justified if they were producing constant energy.

4) The cost of building a tidal energy turbine is quite high, again meaning it only makes sense if it can be run at capacity for many years to pay off the costs.

Now, if you used the turbine to produce energy along with the nuclear power plant, normally, then issues 3 and 4 would go away. Issue 2 could possibly be addressed by putting the wiring underground, although quakes could still be an issue. Issue 1 is the toughest to overcome. StuRat (talk) 04:11, 22 March 2011 (UTC)

I think this sounds like an excellent backup system. The next time a company wants to build a nuclear power plant, they should build the tidal power system first... then scrap the rest of the plan. Wnt (talk) 04:24, 22 March 2011 (UTC)

I like to hijack that question: why don't they use stirling engines for backup cooling? As long as there is a need for cooling there is also a heat source to drive the engines. 93.132.164.231 (talk) 19:12, 22 March 2011 (UTC)

Interesting idea. Is a stirling engine quake-proof and tsunami-proof ? I suspect not, as it sounds a lot like the nuclear reactor's own cooling system. StuRat (talk) 03:24, 23 March 2011 (UTC)

Given that an earthquake can have arbitrarily high strength, nothing can ever be really quake-proof. But unlike a diesel a stirling engine won't mind being under water for some time and could be built very, very robust. 95.112.197.146 (talk) 08:58, 23 March 2011 (UTC)

I'd argue that there is an upper limit to quakes that can be expected on Earth (say 10.0 on the Richter scale), and a lower limit than this for most locations. However, I'd even settle for a design which would survive the latest Japan quake/tsunami. Is there some reason why a sterling engine can be built more robustly than the nuclear plant's own cooling system/steam engine ? StuRat (talk) 15:13, 23 March 2011 (UTC)

(ec) I guess that even the type of diesel engine used at Fukushima could have been built more robust than it really was. In addition, stirling engine does not need external air and can survive (and maybe to some extend even operate) being submerged in water. Besides, as far as I heard, the engines were not damaged by the earthquake itself but by the tsunami. 95.112.197.146 (talk) 15:43, 23 March 2011 (UTC)

A so-called "energy Island" to store electricity, see e.g. here does seem to be a good idea. Apart from the cooling problems, in Japan you now have a shortage of electricity. Also, you can think of applications that need many Terrawatts of power for a limited time (e.g. launching satellites using an electromagnetic rail gun). You can then store the needed energy into such a facility over a long period of time and then release it in a matter of seconds. Count Iblis (talk) 15:35, 23 March 2011 (UTC)

On first sight I thought the island would work be being filled with water some meters above sea level, but in truth it is emptied below sea level, and energy is gained by letting sea water in, correct? So the tsunami would have filled it, as would have even a medium sized storm. Or do I have some concept wrong? 95.112.197.146 (talk) 16:41, 23 March 2011 (UTC)

It could be done either way. StuRat (talk) 17:04, 23 March 2011 (UTC)

If it is filled above sea level, how much energy could be stored? 95.112.197.146 (talk) 17:49, 23 March 2011 (UTC)

Whether above sea-level or below, that would depend on the volume of water and the elevation difference. Creating a tank, like a water-tower, wouldn't be practical because of the high cost relative to the volume and height. You'd need to have a natural elevation difference already. (I don't know if there are hills around the nuclear plants or not.) StuRat (talk) 17:53, 23 March 2011 (UTC)

From the pictures on google maps there are no hills near the factories, none worth mentioning. From the pictures shown after the tsunami I have got the impression that a good part of the land is even below sea level. The link given by user Count Iblis talks of an area of 10km x 60km (you can't spare such an area on a densely populated country, inland) and the picture indicates that the dam cannot be much more than perhaps 10m. I'm out of practice doing calculations and if I do there are frequently grotesque mistakes, that is why I asked instead of presenting the numbers I came up. If I'm correct, such a energy island can buffer the output of one reactor of only one hour if it works by storing water above sea level, and the output of 4 hours if it works by pumping water out, 40m deep. 95.112.197.146 (talk) 18:16, 23 March 2011 (UTC)

Sorry, I was lying about that 60km, it's only 6km. With 60km it would last 10 or 40 hours. 95.112.197.146 (talk) 18:27, 23 March 2011 (UTC)

If it's 1/10th the size you thought, shouldn't it last 1/10th as long as you thought ? Note that it wouldn't need to provide anywhere near the full output of a reactor, just enough to pump in cooling water. However, this approach still sounds totally impractical here. I'm back to my idea of building the reactor core below sea level so that gravity can provide water, even with no electricity. StuRat (talk) 21:41, 23 March 2011 (UTC)

No, I actually calculated with 6km, not with 60km. But I found another error in that the height goes in quadratic, not liner. Building the reactor below water level would have other backdraws. But what I'm really wondering about is why they had humans to do the dangerous tasks. Sony has a robot that plays the violin, so I don't understand why there are no robots for this kind of task that could at least be remote controlled from a slightly bigger distance and possibly some shielding. 95.112.197.146 (talk) 22:05, 23 March 2011 (UTC)

Well, if nobody forces them to do so, they will go with the (immediately) cheapest option, which is not to prepare for a disaster. StuRat (talk) 22:23, 23 March 2011 (UTC)

As for my idea of using gravity to deliver water, so no electricity is required, the AP1000 reactor does this, although they use an above-ground fresh-water tank rather than ocean water: [1]. This prevents the fuel from being contaminated and rendered useless, but above-ground tanks can crack and quickly empty in earthquakes, so it doesn't seem quite as reliable. Perhaps a combo of the two designs could be made with the fresh-water water tank below ground, with the reactor core below that. StuRat (talk) 22:41, 23 March 2011 (UTC)

Escape velocity - vector or scalar?

Physicists have obviously been confused all these years. Escape velocity, contrary to what the name might suggest, is not velocity at all, but simply speed, and should be renamed as soon as possible. Well, at least, so a number of contributors (?) to the article would have readers believe. I'm amazed that this article could be allowed to spread these weird notions - WP obviously and desperately needs review panels to vet some of the rubbish perpetrated in its name. Androstachys (talk) 05:14, 22 March 2011 (UTC)

Escape velocity is infact a vector quantity, as it is relative to the centre of gravity. Plasmic Physics (talk) 05:49, 22 March 2011 (UTC)

If a craft travels at an accute angle relative to the tangent of centre of gravity, and its speed is slightly over the escape speed, then it can not escape the gravitational well. However, if the same craft travels radially from the centre of gravity, and at escape speed, then it will escape. Thus, a esccape velocity is a correct term that specifies the direction of the speed. Plasmic Physics (talk) 05:59, 22 March 2011 (UTC

(edit conflict) No, escape velocity is a scalar. It must be measured in the center of mass reference frame, but that alone does not make it a vector, as it has no direction. If the speed of the object exceeds the escape velocity at its location, then it will escape regardless of its direction of travel (provided it doesn't hit anything along the way). Similarly, exceeding the escape velocity implies an orbital eccentricity > 1, and a hyperbolic trajectory. Incidentally, the escape velocity article also stated plainly that it is a scalar before Androstachys started edit warring over it. Dragons flight (talk) 06:58, 22 March 2011 (UTC)

My edit warring your mumpsimus... Androstachys (talk) 07:30, 22 March 2011 (UTC)

Escape velocity is a scalar. It is the solution to ½v² = Φ, where Φ is the gravitational potential (relative to infinity). Since the velocity only shows up as v² (that is, v·v), the solution is directionless. Plasmic Physics is incorrect (as is the original poster). -- BenRG (talk) 06:47, 22 March 2011 (UTC)

Perhaps someone should tell NASA that the angle of launch of space probes is not important!! Androstachys (talk) 07:03, 22 March 2011 (UTC)

NASA launches have to fly through an atmosphere that adds drag and is best avoided. The definition of escape velocity assumes there is no drag or other additional forces. Dragons flight (talk) 07:57, 22 March 2011 (UTC)

Not to mention that NASA launches are rocket-powered, so escape velocity, with or without accounting for drag, is hardly relevant. The rockets leave the launch pad at a lot less than 11 km/s (which is the escape velocity at Earth's surface). It is relevant for probes like Pioneer after the rocket propulsion and gravitational slingshotting is over with. -- BenRG (talk) 08:21, 22 March 2011 (UTC)

Correct - the term for this particular type of continuously-powered flight is "gravity burn", indicating that the rocket expends some work against the gravity well that does not contribute to the kinetic energy required to reach escape velocity. One objective of a rocket engineer is often to minimize the amount of "wasted" energy that does not convert to kinetic energy of the rocket payload. On the other hand, if the objective is a lander rocket (such as the Apollo Lunar Module descent stage), the rocket engineer must optimize to minimize kinetic energy of the descent - that is, to land with a low impact-velocity and avoid "crashing." In that case, the spacecraft approaches the target gravity well, gains potential energy from gravity; and the rocket motor must supply enough energy to overcome the kinetic energy gained from losing the "escape velocity", plus additional energy to be "wasted" in gravity burn to control the descent speed through the entire time-integral of the trajectory. Nimur (talk) 16:39, 22 March 2011 (UTC)

You could fly parallel to the earth's surface at 11 km h^-1, and you wouldn't escape the gravity well. Plasmic Physics (talk) 09:56, 22 March 2011 (UTC)

Well, assuming you meant 11 km s^-1 not 11 km h^-1, then it all depends what you mean by "fly parallel to the earth's surface". If you took away the atmosphere, removed any inconvenient mountains, stopped the Earth from rotating and launched a projectile in a horizontal direction at 11.2 km s^-1 then yes, it would escape the Earth's gravity well. However, if you apply continuous thrust to keep its flight path parallel to the Earth's surface (i.e. to keep its height constant) then it is in a powered orbit, and the fact that its speed is equal to escape velocity is irrelevant because it is not on a ballistic trajectory. Gandalf61 (talk) 10:08, 22 March 2011 (UTC)

Yes you would (I assume you mean s not h). Low earth orbit is about 9 km/s, any faster and you will orbit higher (slowing down in the process). At 11 km/s you won't slow down enough to stay in orbit. Ariel. (talk) 10:18, 22 March 2011 (UTC)

As it said in the article's intro: Escape velocity is the minimum initial velocity imparted to an object that will enable it to escape a gravitational field without any further boosting. This is a vector as the imparted velocity is directly away from the body's centre of gravity or normal to its surface. Any other angle of launch at the same speed will not escape the gravity well. Anyone here done even a basic course in physics? Androstachys (talk) 11:03, 22 March 2011 (UTC)

Not so. If an angle of 90 works, and 180 works (the two extremes), then obviously anything in between will work too. The only think left is down toward the ground, but that works too (ignoring crashing into the ground, which has nothing to do with the mathematics of escape velocity). Stop thinking of the earth as a giant sphere, and instead think of it as a tiny dot. No matter which initial direction you go, at some point you will be heading away from the dot. Also, I think you are expecting escape velocity to mean the object is no longer under the influence of the mass - i.e. it will travel in a straight line. That's not necessarily the case, the object might travel in an infinite spiral around the mass, and still have escaped from it because the spiral will never bring it back to the mass - it will circle around it, at greater and greater distance, but forever making circles. Ariel. (talk) 11:09, 22 March 2011 (UTC)

Minor correction - a ballistic trajectory has to be part of a conic section, so this rules out a spiral path. A ballistic escape trajectory is either a parabola if speed = escape velocity, or a hyperbola if speed > escape velocity. "Straight up" can be regarded as a degenerate parabola or hyperbola that happens to go through the centre of the Earth. Gandalf61 (talk) 11:32, 22 March 2011 (UTC)

Reply to Androstachys - I am sure that everyone who has replied to you has done at least a basic course in physics, and I expect some have much higher qualifications. I have an A-level in physics and a degree in mathematics. It is you who are clearly out of step on this one. When you are in a hole, it is best to stop digging. Gandalf61 (talk) 11:43, 22 March 2011 (UTC)

"If you took away the atmosphere, removed any inconvenient mountains, stopped the Earth from rotating and launched a projectile in a horizontal direction at 11.2 km s-1 then yes, it would escape the Earth's gravity well." Well, no. Discounting gravitational effects from the Sun and the planets, it would go into a highly eccentric elliptical orbit around Earth. I would suggest you ask for your money back from whichever institute doled out the A-level in physics. Androstachys (talk) 13:07, 22 March 2011 (UTC)

Androstachys, you are incorrect and Gandalf is correct. If an A-level wasn't good enough for you, than may be my PhD in Physics will impress you more. A ballistic trajectory at a speed higher than the scape velocity will lead to a hyperbolic trajectory and object will scape if a collision can be avoided. No highly eccentric elliptical orbit will come out of it. So, indeed, Scape velocity is a scalar despite its name. Dauto (talk) 15:26, 22 March 2011 (UTC)

Please read Gandalf's statement properly: "If you took away the atmosphere, removed any inconvenient mountains, stopped the Earth from rotating and launched a projectile in a horizontal direction at 11.2 km s-1 then yes, it would escape the Earth's gravity well." Now read your own:"A ballistic trajectory at a speed higher than the scape velocity will lead to a hyperbolic trajectory". Small wonder the article is in such a mess. Androstachys (talk) 06:55, 23 March 2011 (UTC)

Androstachys - it would be fascinating to hear you explain just what exactly you think is inconsistent between my responses and Dauto's. I am saying that a projectile launched horizontally at a speed equal to escape velocity follows a parabolic trajectory; Dauto is saying that a projectile launched at a speed greater than escape velocity follows a hyperbolic trajectory. Both of these are escape trajectories. Gandalf61 (talk) 11:05, 23 March 2011 (UTC)

Androstachys, I think you have already amply demonstrated that this simple topic goes over your head. As Gandalf said, it would be best if you stopped digging. Gandalf's statement above and mine are both correct. Dauto (talk) 18:51, 23 March 2011 (UTC)

Indeed. Androstachys - stop digging and see if you can follow the demonstration from first principles given below. Gandalf61 (talk) 15:30, 22 March 2011 (UTC)

Everyone seems to be ignoring the effective potential well introduced by conservation of angular momentum. This is deeper the more angular momentum you have, and so your necessary escape speed will increase at greater angles from the radial. —Preceding unsigned comment added by 92.20.215.104 (talk) 12:04, 22 March 2011 (UTC)

No. A different angular momentum just means a different angle with respect to the radius vector at a given distance. Escape velocity is independent of the direction in Newtonian physics. In General Relativity, it isn't, but the deviations from Newtonian physics are tiny when speaking about planets. In the case of a non-rotating black hole however, even light won't escape if the direction of motion is tangential and the distance from the center (or more correctly: the radial Schwarzschild coordinate) is less than or equal to 3/2 times the Schwarzschild radius. Icek (talk) 12:19, 22 March 2011 (UTC)

Okay, gloves off, let's do this from first principles. From conservation of energy we know that the speed s and radial distance r for an object in a ballistic trajectory must satisfy

$s^{2}-{\frac {2GM}{r}}={\text{constant}}=E$

and from conservation of angular momentum we know that

$r^{2}{\dot {\theta }}={\text{constant}}=h$

Suppose that at some point in its trajectory, when r = r₀, the object's speed is greater than or equal to the escape velocity i.e.

$s\geq v_{e}={\sqrt {\frac {2GM}{r_{0}}}}$

then we know that

$E\geq 0$

The speed of the object is given by

$s={\sqrt {{\dot {r}}^{2}+r^{2}{\dot {\theta }}^{2}}}$

so if ${\dot {r}}=0$ we have

$s=r{\dot {\theta }}={\frac {h}{r}}$

$\Rightarrow {\frac {h^{2}}{r^{2}}}=s^{2}={\frac {2GM}{r}}+E$

$\Rightarrow Er^{2}+2GMr-h^{2}=0$

If E = 0 there is only one value of r that satisfies this quadratic equation, and if E > 0, there are two roots, but only one of them is positive, and we know r must be positive. So if E ≥ 0 there is only one value of r for which ${\dot {r}}=0$ i.e. distance from Earth is a maximum or a minimum. We know the object is not in a circular orbit - it is travelling too fast for that - and it cannot be in an elliptical orbit either, because then there would be two values of r for which ${\dot {r}}=0$ (at apogee and perigee). Therefore it is on an escape trajectory. QED. Gandalf61 (talk) 15:30, 22 March 2011 (UTC)

Another way to see this is to look at the orbit equation.

$r={{h^{2}} \over {\mu }}{{1} \over {1+e\cos \theta }}$

Where r is the distance from the center of mass, μ is the standard gravitational parameter, h is the specific relative angular momentum, $\theta \,$ is the direction to the orbiting body (true anomaly), and e is the eccentricity.

Note that the eccentricity is a constant such that

$e={\sqrt {1+{\frac {2\epsilon h^{2}}{\mu ^{2}}}}}$

Where $\epsilon$ is the specific orbital energy.

Since $\epsilon \geq 0$ if and only if $v\geq v_{escape}$, this implies $e\geq 1$ whenever $v\geq v_{escape}$.

If $e\geq 1$, then there exists $\theta$ such that $1+e\cos \theta =0$, which from the orbit equation implies that r goes to infinity. The angular momentum affects the eccentricity, and thus the direction of travel at infinity, but as long as $\epsilon \geq 0$ there is no value of the angular momentum (expressed in h) such that the eccentricity could be less than 1. Dragons flight (talk) 20:43, 22 March 2011 (UTC)

Refering to my earlier statement, I was of the impression that a vector description was not limited to linear paths, but also curved paths as in some matrices. Plasmic Physics (talk) 08:12, 23 March 2011 (UTC)

I understand the reasoning behind linear vectors, I was simply going by the definition of a vector. Plasmic Physics (talk) 08:15, 23 March 2011 (UTC)

I think the problem here is that the question itself assumes the speed-is-scalar-velocity-is-vector shibboleth drilled in in high-school physics classes. In high school, one of the hard questions to answer is "how can the thing be accelerating when it never speeds up?" and for this purpose it is very convenient to have two different words and insist on the distinction.

I don't think physicists, though, really pay much attention to it. It's pretty much always obvious whether you are referring to the velocity vector or to its magnitude, and so you don't need to disambiguate in the choice of the word. So we have escape velocity, phase velocity, the velocity of light, all used quite blandly and with no sense that anything is wrong. And why should it be? Velocity is just the Latinate word for "speed"; velox simply means "fast", not "fast in a particular direction". I suspect that using the Latinate word for the vector quantity and the Anglo-Saxon one for the scalar was a quite arbitrary choice at some point, made just for convenience. If it's not convenient for us, there's no reason to follow it. --Trovatore (talk) 08:56, 23 March 2011 (UTC)

That sounds like a solution to me - leave the article as it is. Plasmic Physics (talk) 09:19, 23 March 2011 (UTC)

What is the most efficient wind turbine design, in terms of dollars-per-watt?

Or hopefully, watts-per-dollar?

I'm sure the 3-bladed design that we're all familiar with is getting a little old; we already have eggbeater wind turbines and other designs. Now of all the available designs, which would be the most efficient? (Please provide stats and links to more info about that design of turbine.)

And if it is, why isn't it as popular yet? --70.179.169.115 (talk) 07:49, 22 March 2011 (UTC)

Does Wind turbine design help? The section on blade count concludes that 3 blades is best (more doesn't help much, is heavier, and has vibration issues). Ariel. (talk) 08:17, 22 March 2011 (UTC)

"Watts-per-dollar" alone doesn't define the ideal design:

1) Since construction cost is a major factor, this makes "watts-per-dollar" better, the longer it stays in operation. Maintenance cost also figures in.

2) Whether you have high or low wind speeds, and constant versus intermittent winds, would also effect the ideal design.

3) Some consideration also needs to be made for how the electricity is used. If it can be sold to a utility company at a constant rate versus being used immediately at a home, for example, which leaves open the possibility of "wasting" electricity which can't be used. In the second case, you'd want a design that produces a lower, constant rate of electricity over one that varies dramatically. StuRat (talk) 14:03, 22 March 2011 (UTC)

Reading through Wikipedia articles, I found that the "eggbeater" design was patented in 1931 but the three-blade commercial design only came about in 1957. You might also find Unconventional wind turbines interesting. 75.41.110.200 (talk) 15:05, 22 March 2011 (UTC)

1. What are some tougher game shows for Watson to tackle? 2. ...

1. What are some tougher game shows for Watson to tackle, and how will they be tougher for the A.I.?

2. What would it take to build an android named "Watson Jr." that'll have enough computing power crammed in its body to actually go to Kindergarten with warm-blooded classmates, get its height readjusted by its scientists every time it advances a grade, pass a high school with high honors, and enroll in a college?

Since a decade and a half is too long for IBM to conduct that experiment, how about one grade level every 2 weeks? What could enable a "Watson Jr." android to become feasible this way? --70.179.169.115 (talk) 07:55, 22 March 2011 (UTC)

1: Game shows are specifically designed not to be hard - they want the viewers at home to play along. 2: No one knows. Every prediction of the amount of computing necessary has come and gone, so no one will make predictions anymore. Besides the hardware (which everyone assumes will eventually arrive) there is also the software, and no one has a clue about how to do that part. I think most people assume the software will not be written, but rather the machine will learn or evolve on it's own, but both of those require a fitness function, and we don't have that either. Ariel. (talk) 08:12, 22 March 2011 (UTC)

Watson is a first step toward human A.I. the way climbing a tree is a first step toward reaching the moon (standard analogy). We just don't have the faintest idea how to do human A.I., and Watson doesn't help. IBM doesn't do A.I., anyway; it creates enterprise data management solutions for big corporations with deep pockets. That's what Watson is meant to advertise.

There's a lot of variety in game shows. "Who Wants to be a Millionaire?" would probably be much easier than "Jeopardy" since it's multiple choice. "Wheel of Fortune" is basically mindless and a computer could probably beat any human player, especially if you let it calculate the precise amount of torque to apply to the wheel. "The Newlywed Game" and "Survivor" are too human-centric to be playable. "The Price is Right" might be an interesting target for future research. -- BenRG (talk) 08:53, 22 March 2011 (UTC)

The Price is Right would be simple for Watson. The data for average prices could easily be loaded into Watson and then it's just a matter of probabilities for most of the games. I wouldn't be surprised if Watson hit every price within ~1%. Dismas|^(talk) 09:16, 22 March 2011 (UTC)

Actually, Price would be an interesting challenge for Watson, for a couple of reasons.

1. Even if accurate to 1%, guessing high (even by a dollar) knocks you out of contention -- so Watson would have to lowball the price. This is algorithmically easy, but it's not something you find in Jeopardy (in case we're talking "Watson-as-is" vs "Watson-with-minor-changes").
2. Unless Watson is the last player in the round, it would be easy for the human contestants to use the "Watson +$1" strategy. Watson's only actual defense against this would be to guess the price exactly, but that conflicts with problem #1.

That said, assuming Watson got into the final round, it would hold a major advantage with problem #2 removed. — Lomn 13:37, 22 March 2011 (UTC)

Watson is a specialist machine. It is not a generalist. It could probably be tweaked to do similar trivia shows. But it's not like Watson, as is currently programmed, would be able to play Wheel of Fortune. The inputs and outputs would just not make any sense to its program. --Mr.98 (talk) 12:54, 22 March 2011 (UTC)

Off-topic slightly, I'll bet a piece of software to play Wheel of Fotune would be a lot simpler than Watson. APL (talk) 15:54, 22 March 2011 (UTC)

Indeed, and not off topic at all! The point is that Watson is a specialist. Even very easy computational tasks are probably well beyond the specific algorithm that plays Jeopardy. It is like the classic (perhaps true? I don't know) anecdote about frogs that only know how to eat flies when they are flying or moving around, and will starve if put into a basin of freshly-killed flies. --Mr.98 (talk) 16:26, 22 March 2011 (UTC)

But will it be Randy Marsh? Nil Einne (talk) 17:03, 24 March 2011 (UTC)

2) Attending each grade for only 2 weeks wouldn't work. Watson learns by trial-and-error, like humans. As such, it would need the same number of trials and errors that children need (or perhaps slightly fewer, since it has perfect memory). Therefore, the only way to get it to learn quicker would be to provide input quicker. Since a regular school can't run at hyper-speeds, this accelerated input would need to be provided in another manner. StuRat (talk) 13:52, 22 March 2011 (UTC)

If you allow it to keep its current database, Watson could probably pass most highschool and college standardized tests now. If you skip the essay questions, anyway.

Erasing its brain and Learning purely from class would be problematic. For us that kind of classroom instruction depends on a huge amount of knowledge we just pick up from observing the world around us. Without background general knowledge even the simplest textbooks would be incomprehensible.

Imagine if you read in a text book "All animals with four legs are quadrupeds.", and then on an exam you were asked "Is a cat a quadruped?" you'd have no problem. You learned this in the book! But you didn't. You only learned part of it from the book. You had outside knowledge that cats have four legs.

IBM gave that kind of general knowledge to Watson by having it read and remember millions of pages of written material. (It wouldn't surprise me if Watson has Wikipedia in his brain somewhere.) Of course, no human could learn that way.

Even solving issues like speech recognition in the classroom, I don't think that Watson is compatible with human methods of learning. It was designed to use massive data-dumps in text format. (Even if the exam had a photograph of a cat it wouldn't help Watson.) APL (talk) 15:54, 22 March 2011 (UTC)

I always wonder why thinking like a human is still the gold standard for AI. (I know it was for Alan Turing, but a) he was thinking speculatively and b) it was a long time ago. I don't expect a car to get around as well as I do; I expect it to get around much more quickly. I don't expect my computer to sort a file of a million records in the way I would; I expect it to sort the file in minutes. From what's said above, it sounds like Watson solves problems in a very different way from a human. It should be quite good at Countdown, both the anagram games and the arithmetic. Itsmejudith (talk) 16:17, 22 March 2011 (UTC)

Even your filing algorithm is "like a human," it's just faster (and probably better optimized than most humans, but humans manually could do a insertion sort, too, if they thought to do it). Speed is not the interesting issue; even the most primitive computers and calculators can beat humans at speed. The reason humans are the gold standard is because as far as we can tell, the algorithms are well beyond what we can approximate with machines, even with all of their speed. --Mr.98 (talk) 16:29, 22 March 2011 (UTC)

Both ways of "thinking" have their uses. One use for a computer that "thinks like humans" is in search engines. The current methods often give results that any human can tell don't match the (intent behind the) search criteria. StuRat (talk) 16:30, 22 March 2011 (UTC)

But it's not even a way of "thinking". One can marvel at the speed in which computers do things, but their ability to do anything complex is very limited. They usually do just very simple things very fast. I don't consider that "thinking" in any meaningful sense. Just because a computer chip can shuffle cards faster than I can doesn't mean that shuffling cards is a form of "thinking". --Mr.98 (talk) 21:31, 22 March 2011 (UTC)

That's why I put "thinking" in quotation marks. Would you prefer "processing", to collectively refer to human thinking and machine calculations ? StuRat (talk) 03:20, 23 March 2011 (UTC)

First question of 1): University Challenge. Probably the toughest tv quiz in the world. 92.15.23.133 (talk) 17:58, 22 March 2011 (UTC)

By the way, when I said the Watson could "probably" pass an exam now, I mean its underlying technology. Some custom software would be needed to create an interface between the Watson engine and the test. (Like whatever software they wrote to handle the "game" aspects of Jeopardy!) APL (talk) 18:46, 22 March 2011 (UTC)

University Challenge is an interesting suggestion - of course Watson could know all the others, but the buzzer element is very interesting. If I said "Lima is" you could probably shout out "Peru" and be right, or something like but harder in UC. I'm not sure an AI would be able to play the probabilities like that. Grandiose (me, talk, contribs) 19:09, 22 March 2011 (UTC)

High altitude effects indoors

I was recently watching a basketball game on television, broadcast from Denver, Colorado. The announcers made reference to how the play of the game would be affected by the high altitude of the area. Because the game was being played indoors, would the effect be significantly less than if it were to be played outdoors? Kansan (talk) 13:27, 22 March 2011 (UTC)

The game being indoors vs outdoors won't change the altitude effects, assuming the basketball arena is unpressurized (I don't know of any that are). Naturally, playing indoors vs outdoors would affect many aspects of basketball, but those effects wouldn't be related to altitude. — Lomn 13:32, 22 March 2011 (UTC)

So all other things being equal, the air inside a building would have the same air pressure of the surrounding area, right? Kansan (talk) 13:34, 22 March 2011 (UTC)

Yes. There are stadiums with soft covers that are partially supported by increasing the air pressure inside, but I believe that pressure difference is tiny compared with that due to the elevation difference. StuRat (talk) 13:42, 22 March 2011 (UTC)

Certainly within the tolerance of "has a measurable effect on athletic performance". Any building is going to have slight positive or negative pressure relative to its environment, but that's orders of magnitude smaller than the overall change in air pressure from sea level to Denver. Sea level to Denver is a difference of about 17 kPa, which is also about a 17% change. We note that the differential of a positive pressure enclosure is only about 0.05 kPa, a change of 0.06%. — Lomn 13:47, 22 March 2011 (UTC)

Thanks for the help. Kansan (talk) 14:19, 22 March 2011 (UTC)

Some pressurized buildings seem to go up to 1kPa, or 1% of atmosphere, but that effect would still be negligible. Googlemeister (talk) 19:56, 23 March 2011 (UTC)

A good way to visualize the relative air pressures is to think about ear popping. You ears probably pop several times as move from sea level to Denver's elevation and back down. How often do they pop when entering or exiting a building ? Not often. So, this shows the pressure differential is far less. (If it was as high, you'd probably burst your eardrums when entering or exiting such a building, as that much pressure change in that little time would be too much for them to handle.) StuRat (talk) 08:08, 25 March 2011 (UTC)

What is the maximum recommended amount of salt in water for dairy cattle?

What is the maximum recommended amount of salt in water for dairy cattle?

Back ground if your interested: My firm recently drilled a water well for a gentleman who in setting up a rather small dairy farm. The water contains 1520 ppm of sodium, ideally of course there would be none. We are trying to seal the lower parts of the well with a concrete mix, but the question remains. What is the maximum recommended amount of salt in drinking water for dairy cattle? Any help would be greatly appreciated. JohnQposter (talk) 14:00, 22 March 2011 (UTC)

This site says less than 3000 ppm of "total dissolved solids", which includes sodium, is "Usually satisfactory for most livestock". This site says (specifically about salinity) < 1000 is good, and for 1000-3000 ppm, "Generally no problems; possible temporary diarrhea to animals not accustomed to this water". --Sean 15:00, 22 March 2011 (UTC)

Please note that Wikipedia does not give medical advice, and I assume that means veterinary advice also. This may be an academic response in good faith, but don't trust us when it comes to the cows' (and farmer's) well-being. Wnt (talk) 02:58, 24 March 2011 (UTC)

Time dilation

I remember reading some time ago an explanation of time dilation in special relativity based on transfer of information and the constancy of the speed of light. So if A is stationery and B travels some distance away they will observe each other's clock to tick slower than their own.

The explanation involved A (using his own clock) sending B one signal each second but due to c being constant for all observers, B receives the signals at more than one second apart (based on B's clock). Does anyone know of this explanation? Thanks. Zain Ebrahim (talk) 15:48, 22 March 2011 (UTC)

That sounds like the Doppler shift (which exists even in Newtonian physics). At relative rest, signals sent at a constant rate are received at the same rate, but with recessional motion, the received rate is slower:

         A/ / B         A/ / / B
         A / /B         A / / /B
         A/ / B         A/ / /B
         A / /B         A / / B
         A/ / B         A/ / B
         A / /B         A / /B

You can derive all of special relativity from the rule that the ratio of the sent and received rates depends only on the relative speed. Time dilation is not especially easy to derive, though, because it requires synchronized clocks. The twin paradox is easier. -- BenRG (talk) 18:53, 22 March 2011 (UTC)

Ah yes, but if you assert the signal velocity is still c in the observors rest frame you get time dilation, as explained by the OP, whereas in CM doppler shift, the signal speed would be different in the two frames. —Preceding unsigned comment added by 92.21.86.36 (talk) 15:22, 23 March 2011 (UTC)

Thanks guys - I was actually looking for this (which I still don't understand - see below). I didn't remember it properly. 163.202.48.108 (talk) 10:14, 24 March 2011 (UTC)

Another question about neurons

How do axons and dendrites meet up? What is their physical method of motility? Layman's terms would be appreciated, even if simplification makes the answer less than perfect. 20.137.18.50 (talk) 19:16, 22 March 2011 (UTC)

They meet at specialized junctions called synapses. Dendrites don't extend very far from the cell body, and they grow in intricate tree-like patterns whose rules are not all that well understood. Axons, in contrast, can extend for enormous distances. The tip of a growing axon is a structure called a growth cone, which extends numerous tiny fingers of protoplasm called filopodia, which interact chemically with things that they touch, sticking to some and being repelled by others. These attraction-repulsion processes cause the growth cone to travel through the brain or body, sometimes by very complex routes, extending the axon behind it as it goes. Here is a link to a youtube video that explains the process and shows it in action. Looie496 (talk) 19:51, 22 March 2011 (UTC)

Real-time glacier data

Hi. Is there any reliable source on the Internet where I can select any monitored glacier from a worldwide database and get real-time or recent data on parameters such as flow rate, temperatures, height changes, precipitation, status of retreat or advance, density, any moulins or glacial lakes and dams, calving events, etc? Thanks. ~AH1^(TCU) 20:27, 22 March 2011 (UTC)

What questions are you trying to answer? Most glacier monitoring looks at things like mass balance (e.g. thickness) and edge location, which are variables that change slowly and hence can be recorded infrequently. The number of intensively monitored glaciers is probably very low. Dragons flight (talk) 00:13, 23 March 2011 (UTC)

Agreed. There's no need for real-time data on things which only change at a glacial pace. StuRat (talk) 03:16, 23 March 2011 (UTC)

Wikipedia has a stub article about the World Glacier Monitoring Service that collects standardised observations on changes in mass, volume, area and length of glaciers with time. Here are real-time data from the Hubbard Glacier (Alaska)] and the Tweedsmuir Glacier (Canada). You may also find glacier webcams. Cuddlyable3 (talk) 10:30, 23 March 2011 (UTC)

Poppy seeds

Could I grow poppies from the seeds on poppy seed bagels? --70.244.234.128 (talk) 23:14, 22 March 2011 (UTC)

Maybe if you bought some seeds directly, but the seeds on a bagel are baked, and thus dead. If you buy a bottle of seeds make sure they are not toasted. Ariel. (talk) 23:22, 22 March 2011 (UTC)

Or Irradiated. It wouldn't surprise me to find that poppy seeds are routinely irradiated (it's not uncommon for spices), which would kill parasites, but also the seed. Ariel. (talk) 05:24, 23 March 2011 (UTC)

You can grow the seeds from many other fresh fruit and vegetables. Experiment and see. Searching for growing pips (no ""s) on Google gives a lot of results. 92.15.14.45 (talk) 12:13, 23 March 2011 (UTC)

I think this is the "definitive source" for a certain type of home enthusiast. The type who often regrets his efforts, that is. Still, any transgression against the cartel's supply side is a blow against Islamic terrorism... Wnt (talk) 04:39, 24 March 2011 (UTC)

It never occurred to me that the OP might be interested in that aspect. I understand that only opium poppy specie(s) contain enough opium to have any effect. In my country poppies are a common weed and garden flower. The best book for growing seeds from many different types of fresh vegetables and fruit is "The Pip Book", once published by Penguin but now out of print. 92.28.242.170 (talk) 19:59, 25 March 2011 (UTC)

Forgive me if I'm wrong, but so far as I know poppy seeds come from Papaver somniferum. Wnt (talk) 21:53, 26 March 2011 (UTC)