Wikipedia:Reference desk/Archives/Science/2011 April 19

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April 19

A few questions about the Japan earthquake...

1. How much radiation is in the sea water?
2. How did the radiation even get out of the nuclear power plant?
3. What types of radioactive material was in the water? (e.g. Was it radioactive metals, gases, or liquids? Did it have gamma rays, alpha rays, or that other kind of rays?)

PLEASE answer. — Preceding unsigned comment added by MickWithoutGlasses (talk • contribs) 00:45, 19 April 2011 (UTC)

Please read our article on the Fukushima I nuclear accidents.--Shantavira|^{feed me} 06:40, 19 April 2011 (UTC)

I thought you asked about the earthquake. The damage to the reactors was caused by the tsunami to the local infrastructure at the reactor sites. The data from the media is all we have available and that data is at best questionable and certainly not reliable. They don't cite the type of detector used the more sensitive the detector the scarier the report, which is of course exactly what they want ... fear. My own Geiger counter gives a result of around 16 counts per minute of background radiation localy. My1 1/2x1 1/2x3" NaI(Tl) Gamma Scintillator which is much more sensitive gives a background reading of over 20000 counts per minute. There is no more radiation but the scintillator is several orders of magnitude more sensitive. While there may be some risk to the inhabitants of Fukushima and its immediate locality there is no appreciable risk to the rest of the world unlike the Chernoble incident which was much more serious and that killed less than 35 people (proved) and they were the workers directly irradiated by the exposed core. Any increases in the global background radiation as a consequence of the Fukushima incident are so small as to be insignificant and are barely measurable. The Granite worktop in your kitchen or the bricks and concrete in your building are more radioactive than anything you will see from fallout from Fukushima.

Gamma's are high energy photons and pass through pretty much anything. Alphas are essentially helium nuclei and cannot penetrate skin or a sheet of paper, Beta are electrons or positrons and can travel only a few feet in air at best and are completely blocked by 3mm of aluminium (aluminum). Radiation is around us everywhere everyday. It is unavoidable and a natural part of the world we live in. Dont Panic !!! — Preceding unsigned comment added by Brian Thwaites (talk • contribs) 02:58, 20 April 2011 (UTC)

"System of Care" health care

Using Google I discovered that System of Care is a health care thing...www.systemofcare.org, which is exactly what I wanted to know...but I was hoping hat Wikipedia could have a page about System of Care. Trust me, Im not the guy to write the Wikipedia page. Thanks and I love Wikipedia! 99.188.102.172 (talk) 02:28, 19 April 2011 (UTC)

One definition given for system of care is "a framework within which health care is provided, comprising health care professionals; recipients, consumers, or patients; energy resources or dynamics; organizational and political contexts or frameworks; and processes or procedures. Current theory recognizes that an analysis of the provision of health care requires knowledge of the systems of care."

It looks like this is probably well covered by our article health care system --- Medical geneticist (talk) 16:05, 19 April 2011 (UTC)

Atomic number and qualities of the element

Perhaps I am misunderstanding this concept completely, but how can adding one proton to an atom completely change the qualities of the element? For example, Hydrogen has one proton and is extremely flammable and explosive while Helium has two protons and is completely inert. I do not understand how adding or subtracting a single proton can change so much about an element when it is seemingly such a small change. It seems very random that an atom with three protons would be a silvery metal while an atom with 29 protons would be a reddish-orange metal, and so on through all the elements. Is there a reason why the qualities of the elements change based on atomic number or is is just a mystery of nature? Any help understanding this would be greatly appreciated, thanks! Zipidiedoda (talk) 02:54, 19 April 2011 (UTC)

My basic understanding: if you change the atomic mass, you're actually going to also change the electron configuration. The things you're talking about (reactivity, etc.) are chemical functions that have to do with the electrons, not the nucleus. Things like color, the way in which the atoms fit together in a lattice, how they bond with other atoms (and release energy, etc.) are chemical reactions that depend on the electrons, not the nucleus. --Mr.98 (talk) 02:59, 19 April 2011 (UTC)

It's true that adding protons requires adding electrons, because the positive charge pulls them in. But what is mysterious is how orbitals work, which is tied to the strange science of quantum mechanics. It turns out that if you have one electron "circling close around" a nucleus, it can readily accept another that is spinning "in the opposite direction". So hydrogen will strive to pair up. Yet throw in a third electron, and there's no way for it to spin around in the same way - it gets stuck in the upper bunk, so to speak, and you have lithium, which is eager to get rid of it wherever it can. Now why can't three or four electrons circle just like two - it's complicated, and hard for me to understand in an intuitive way. It has to do with eigenvalues and the Schroedinger equation and the Pauli exclusion principle and all sorts of nasty math, which is derived not because it seemed like the logical way things ought to happen but because it fit the experimental data. Wnt (talk) 03:08, 19 April 2011 (UTC)

The problem is that the model of electrons existing as little balls orbiting a nucleus in various ways, and physically "spinning" actually doesn't work. The actual model is based a lot more on the wave-like properties of the electrons. There are two perspectives on the actual electron, both of which work perfectly well with the quantum mechanical model, and so we tend to think of them both working simultaneously. First, you can think of an electron as a complex standing wave (like the wave formed on a plucked guitar string as it vibrates). The orbitals around an atom represent harmonics of the fundemental wave. Secondly, you can think of electrons as existing as three-dimensional Probability distribution around the nucleus, indentifying the relative probability of finding a particular electron in a particular location after an abritrarily long time. After you make a scatter graph of all of the possible locations one could find that electron in the space around the nucleus, you basically get the shape of that 3D standing wave I just described above. This probability/wave definition of an electron is what the wavefunction and the Schroedinger equation is all about. The deal with the Pauli exclusion principle is that, if we define an electron in these terms, then we cannot have two electrons that have identical wave functions; two wave functions can only co-exist in the same space if they are orthogonal to each other; that is if the two waves cannot interfere in any way; much as two sine waves at right angles cannot interfere. This is where "electron spin" comes it; this is the property that introduces orthogonality to the wave function. It's not a real "spin" in that there is no little ball rotating on its axis; nor is there two "balls" orbiting in different directions. Its just a (somewhat arbitrary) name given to the part of the wavefunction that allows two electrons to coexist in the same physical space. --Jayron32 04:15, 19 April 2011 (UTC)

Hmmm. You say electron spin isn't really spin, but... when electron spin flips (relative to one spatial dimension), it is transmitted away by means of a photon with an angular momentum of hbar. The spin of this photon is a real thing, isn't it? A circular polarization that can be measured by passing it through filters. Am I wrong? Wnt (talk) 05:38, 19 April 2011 (UTC)

Sort of. Particle spin in physics is a real physical property of particles; it's just not actually spin. When you tease the property out of the particles, it behaves like angular momentum, and it has directionality (the terms "spin up" and "spin down" are real aspects of particle spin, relative to the z-axis). However, with point particles, its not clear how such a particle can physically spin. The name is an artifact of treating electrons like little balls; this is explained well at Spin_(physics)#Elementary_particles. So, it's called spin because mathematically it behaves like classical angular momentum, but there's nothing to spin in fundemental particles, since they have no internal structure to spin about. Spin is likely more related to some sort of fundemental directionality; the other aspects of a particle's fundemental properties (like charge and mass) make no directional distinction, spin uniquely does. It is through spin that we can establish a non-arbitrary set of axis to define the regions around an atom, for example. Still, though it is tempting, one must resist using this property of spin to treat electrons like little balls; they clearly don't behave that way. After all, quarks aren't really colored, and fundemental particles don't really taste like anything. --Jayron32 05:58, 19 April 2011 (UTC)

No—color has nothing to do with color, and flavor has nothing to do with flavor, but spin really is spin. Just as you can develop a theory of instantaneous velocity by taking the limit of the velocity over successively smaller time intervals, so you can develop a theory of "angular momentum at a point" by taking a limit of successively smaller spinning objects. The theory you get when you do that is the theory of particle spin. That doesn't mean that particles are spinning points. They might be very small spinning objects, or they might be Kerr–Newman black holes, or something else entirely. It's a mistake to take any aspect of the standard model of particle physics too literally. All we really know is that particles behave like they're spinning. -- BenRG (talk) 07:01, 19 April 2011 (UTC)

When I look at the figures in circularly polarized light, it certainly looks like something is really spinning. A photon should look like a brief segment of such a wave, trailing off at either end, moving forward like a corkscrew. Now I am still slightly hazy on the spin number of light - apparently it is hbar in two dimensions, adding up to sqrt(2) hbar - but my impression is that the spin of the light in the dimension of its travel should match this classically perceptible rotation. Question: can you actually look at the E and B fields of a circularly polarized photon and derive hbar as the angular momentum of the energy it carries?? Wnt (talk) 07:40, 19 April 2011 (UTC)

The deal with particle spin, which I guess by BenRGs reasoning is that it is OK to call it a spin (by a certain perspective) is that it doesn't interact with other, macroscopic angular momenta, outside of the "particle spin" mode. Think about a molecule like triplet oxygen. Triplet oxygen has "particle spin". This is because, as a diradical, it has unpaired electrons. This particle spin generates a magnetic field (as all spinning charges, point particle or macroscopic) should. This creates the property of paramagnetism in groundstate dioxygen. However, the oxygen molecule itself also has a rotational axis (two in fact) perpendicular to the bonding axis of the molecule. Being a non-spherical molecule, that means it has an axis about which it can rotate. However, I am not aware of any way that the particle spin has any effect on the rotation of the molecule; I am willing to accept that I am wrong on this, but it is my understanding that the "particle spin" which accounts for dioxygen's paramagnetic nature does not increase the molecule's rotational energy, which is pretty much exactly what spin is. If we take the situation to a classical analogy; we could look at dioxygen like a helicopter. The existance of angular momentum of the helicopter blades tends to introduce a torque into the helicopter, that's why there needs to be some counteracting force in the form of some stabilizing counterrotation of some sort. Likewise, if the "spin" of oxygen's electrons were the same sort of thing, it would only stand to reason that this would cause the molecule itself to rotate. The same should happen in ANY radical molecule, such as Nitrogen dioxide. I am not aware of such an effect, though I am open to an explanation of either a) why my reasoning is wrong or b) where such an effect actually occurs. --Jayron32 13:24, 19 April 2011 (UTC)

I'm really on the wrong side of the bar here, but to quote [1], "The energy of a molecule can be approximated as the sum of the contributions from its different modes of motion (translation, rotation, and vibration), the distribution of electrons, and the electronic and nuclear spin. Given that the energy is the sum of independent contributions, ... the partition function is a product of contributions: q = q^T q^R q^V q^E q^S where T denotes translation, R rotation, V vibration, E the electronic contribution, and S the spin contribution. The contribution from electronic spin is important in atoms or molecules containing unpaired electrons. For example... the Cs atom, which has one unpaired electron ... contributes a factor of two to the molecular partition function."

Now to what degree I understand it, this means that the amount of entropy, at least, is increased by the spin (it doesn't really increase the molecule's energy in the sense that the spin can't drop to zero, though pairing spins is welcomed in chemical bonding). The molecules with electron spins are free to move energy back and forth between those and rotational spins, and since angular momentum is always conserved I assume that means that the direction of the electron vs. molecular spin matters. So I really would think of electron spin as a "real" spin, contributing to the overall motions going on within the molecule. Wnt (talk) 19:38, 19 April 2011 (UTC)

That quantum spin can couple arbitrarily to other quantum numbers is unsurprising. This sort of thing happens between all of those various contributions. More telling in your description is that the rotation of the molecule and the spin of the molecule are treated differently; that is the quantum rotation (that of the entire molecule spinning) is an independent value from the quantum spin value (that calculating by summing all of the individual electron spins). To me this tells me that there's no special physical relationship between those energies, over that one would expect from any coupled modes of energy, such as vibration and electronic. In other words, spin and rotation are unique properties, and though each is classified in terms of "angular momentum" they actually measure different things about a molecules. --Jayron32 19:54, 19 April 2011 (UTC)

Well of course the rotation and spin should be different - in rotation the two nuclei are moving end over end, after all. Though it would be interesting to consider the rotational mode of a helium atom, or more to the point, monatomic hydrogen! What I did find is the spin isomers of hydrogen article which says that only para hydrogen can fall into the lowest energy symmetrical rotation state. Wnt (talk) 22:40, 19 April 2011 (UTC)

Basic answer (as oultined above) is that chemical properties are determined by electrons, not by nuclei and nuclear processes. So it is not adding or subtracting protons (or neutrons) that changes the chemical behaviour of an element - it is the associated addition or subtraction of electrons. Example - sodium chloride has chemical properties very different from either sodium or chlorine, yet its atomic nuclei are standard sodium and chlorine nuclei; its chemical properties are due to the transfer of electrons and the resulting creation of ionic bonds. Gandalf61 (talk) 11:03, 19 April 2011 (UTC)

Chemical properties are primarily determined by electrons (especially in covalent compounds), but those electrons are certainly affected by the protons (especially in ionic compounds). For example, they are what actually causes the ionic charge once the electrons have been added or subtracted to get a closed shell. "Why is it +2?" Because it has 2 more protons than electrons. Why two? Because that's how many electrons it lost to get to a stable electronic configuration. But why did it have "two more than a stable electronic configuration" when it was neutral? Because that's how many protons it had. And isotope effects are real chemical-reactivity differences due solely to mass not atomic number or charge. DMacks (talk) 12:50, 19 April 2011 (UTC)

Why only 92

Since we know of only 92 naturally occuring elements, the other 17 transuranics (maybe more now) being man made, and since we know about the enormous pressures, temperatures and densities that exist in many types of stars, it seems inconceivable that mother cosmos has not been able to make any elements with a greater atomic mass than U238. questions. (1) Has there ever been any observed indications of natural elements heavier than U238

(2) If the answer to (1)is no, then does this indicate some naturally limiting factor. If yes, then what is it.

(3) If "no" again then what might that limiting factor be.Phalcor (talk) 03:02, 19 April 2011 (UTC)

Neptunium (239, I assume) is actually produced in trace amounts in natural uranium ores. (On the other hand, technetium was synthesized). See nuclear shell model - nay, rather Magic number (physics) - for a diagram of how the largest known nuclear shell marks the end of the stable sequence at bismuth. Some believe that an island of stability of super heavy elements exists, but the number of neutrons for the best isotopes keeps going up, making it very difficult even for supernovae to make them out of smaller nuclei with fewer neutrons. Wnt (talk) 03:13, 19 April 2011 (UTC)

(edit conflict) (1): Yes, there have been observed natural elements heavier than U238: Plutonium is the heaviest Primordial nuclide, and is found in extremely small quantities in nature left over from the Earth's formation. It has also been formed more recently in the Natural nuclear fission reactor of Gabon, though again, has almost completely decayed by now. Other, heavier elements were (and are) created in stars, but are not found on Earth only because they've decayed away in the 4.5 billion years since Earth's formation. Buddy431 (talk) 03:16, 19 April 2011 (UTC)

But neptunium has amu of237.0482 less than U238 with amu 238.03 and plutonium exists in nature only as an isotope of U238. the Element plutonium is greater amu 244.0 but is man made as you said at gabon and other places and was used in (fat boy) over Nagasaki. If other heavier elements have existed and decayed then there would be a non-radioactive element left in it's place, such as uranium to lead. If I understand it correctly they do not simply disappesr. correct my if I'm wrong but all the elements 93 to 109 are all man made I'm asking about natural elements above 92 —Preceding unsigned comment added by 190.56.115.96 (talk) 04:32, 19 April 2011 (UTC)

Plutonium-244 is created by natural processes, and has an atomic number greater than 92. --Jayron32 04:40, 19 April 2011 (UTC)

(edit conflict with Jayron) Stars produce many elements above 92, but uranium is the heaviest element that decays slowly enough that it's still here on Earth after 4.5 billion years. The reason heavier elements decay faster is well answered by Wnt. In short, there's nothing special about U-238 is terms of being produced by nature, it's the decay times that determine whether it still occurs naturally on Earth. U-238 decays slowly enough that there's still a lot of it after 4 billion years. Pu-244 decays slowly enough that there's a tiny amount of it after 4 billion years. Buddy431 (talk) 04:44, 19 April 2011 (UTC)

Thanks guys I just checked out "island of stability" Guess I should have done that earlier Huh. Boy do I feel updated, or dated or something. Yeh I'm gonna have to throw some books away and graduate to the computor.190.56.14.159 (talk) 05:32, 19 April 2011 (UTC)

is there a formula from wavelength in nanometres to RGB?

Suppose I have a sharp band of 632.4 nm light. What RGB value should I use to represent it? John Riemann Soong (talk) 06:03, 19 April 2011 (UTC)

Depends how accurate you need it, if it's just rough, my guess is about 255,50,0. I'm seeing a few conflicting sources, some saying it's roughly red, some saying it's more orange. Vespine (talk) 06:38, 19 April 2011 (UTC)

Actually just found a site that does it, it says 255, 70, 0.. Hey! My guess was pretty good:) lol. Vespine (talk) 06:41, 19 April 2011 (UTC)

Because the RGB receptors in the retina don't match the RGB mapping of your screen (they actually use a different system, and vary from person to person), you will never get a "perfect match" for everyone. Dbfirs 07:14, 19 April 2011 (UTC)

Human vision is not an RGB system. There are three color receptors, but they are not simple red, green, and blue detectors.--Srleffler (talk) 17:22, 20 April 2011 (UTC)

I think that applet is somewhat inaccurate, as it claims to use this code, which links to this page, which displays this chromaticity diagram generated using the algorithm, which is rather different from this chromaticity diagram, which I trust more. You could use an eyedropper tool on the latter diagram to get a better answer. There is actually no correct sRGB value for any monochromatic color; you have to desaturate the color (mix it with white) to get something in the sRGB range. And color perception does vary somewhat from person to person. -- BenRG (talk) 07:34, 19 April 2011 (UTC)

I agree with previous replies; there is no single "correct" relationship. A lot of thought has been put into finding good approximations, see e.g. this article. (I had once seen a nicer one with the pictures directly in the article, but I can't find that one just now.) — Sebastian 17:56, 19 April 2011 (UTC)

Electrodynamics

what is the physical significance of displacement current? — Preceding unsigned comment added by Karan khajuria (talk • contribs) 11:03, 19 April 2011 (UTC)

It produces a magnetic field just like an actual alternating current would produce. Dauto (talk) 15:17, 19 April 2011 (UTC)

But its not a real current in that it cannot be measured directly.--2.98.149.24 (talk) 21:35, 20 April 2011 (UTC)

Chickens in a container

I recently heard something on TV which has me puzzled. It was talking about a sealed shipping container containing chickens on a weighbridge. The show's presenter said that if you could get all the chickens to fly into the air at the same monent, it would not affect the measured weight of the container on the weighbridge. I sometimes fleetingly think I understand it, but can anyone explain, please? Si1965 (talk) 12:19, 19 April 2011 (UTC)

Mythbusters tackled this one, but I can't remember what they found. HiLo48 (talk) 12:27, 19 April 2011 (UTC)

See here. --Mr.98 (talk) 17:27, 19 April 2011 (UTC)

Flight works by increasing air pressure beneath the bird (and decreasing it above), which spreads out as a roughly cone-shaped region above and below it. This applies forces to the floor and roof of the container which add up to the weight of the birds. So, the overall weight of the container doesn't change. Now, if it was open at the top and bottom (say with gratings), then the air pressure would go down below the container, and apply to whatever was below it. If the bridge also had a grating form, then the air pressure change would continue to spread down and out, eventually being supported by the ground. The same is true, in reverse, for the decrease in air pressure above the birds. So, there could be a tiny difference when the birds were all in flight. However, note that not very many flying chicken could fit in the container, as collisions and the air pressure from those birds flying near the roof would both force lower chickens to fall back down. StuRat (talk) 12:44, 19 April 2011 (UTC)

Interesting. I could find no article dealing with chickens flying in shipping containers. Your answer may well be definitive Sturat, but it occurs to me that the kinetic energy transfered to the air by the wings would be completely or mostly used up by the work of compressing the air locally beneath the wing which would then completely decompressed in all directions, thereby the weight of the chicken would not be transfered to the bottom of the container.190.148.134.75 (talk) 14:33, 19 April 2011 (UTC)

Mythbusters used pigeons. They fly better than chickens. HiLo48 (talk) 17:34, 19 April 2011 (UTC)

The force does dissipate in all directions, but the downward component of the force eventually works on the ground.

Perhaps it would help to imagine something bigger and more powerful. Imagine a helicopter in a box. Clearly, the helicopter is pushing down on the bottom of the box.

The higher the helicopter is, the harder it is to notice that downward force because it's spread out over a wider area. APL (talk) 14:53, 19 April 2011 (UTC)

Unless said helicopter is in a box where the air disturbances can not dissipate. Googlemeister (talk) 15:04, 19 April 2011 (UTC)

Right. APL (talk) 15:10, 19 April 2011 (UTC)

In any case, chickens don't really fly all that well. The longest I ever saw one stay in the air was less then 10 seconds. Googlemeister (talk) 16:02, 19 April 2011 (UTC)

Also discussed in episode 8 of QI (H series).--Shantavira|^{feed me} 16:39, 19 April 2011 (UTC)

How sealed? If it is air tight, there should be literally zero mass difference. You are not weighing the bottom of the container, you're weighing the whole container unit, air, chickens, and all. --Mr.98 (talk) 17:27, 19 April 2011 (UTC)

Best of all, if it's air tight the chickens eventually stop moving so you can get an accurate measurement on your scale. APL (talk) 18:46, 19 April 2011 (UTC)

This is true - until the chickens stop moving up and down, the weight on the scale will keep changing. It will only average the same as when they're not flying. Wnt (talk) 19:14, 19 April 2011 (UTC)

Having sex after working out

Since body testosterone plays an important part in muscle growth, does having sex shortly after exercising intensely improve the results of the work-out? I am assuming the person in question assumes a more passive role in sex, in order to favour recuperation. Leptictidium (mt) 13:20, 19 April 2011 (UTC)

Assuming there is an effect, it is not clear to me why you would need to do this "shortly after exercising". What is known is that exercising an hour or so after the main intense exercise does help recuperation. Count Iblis (talk) 16:47, 19 April 2011 (UTC)

Excessive caffeine and hair loss

Is there any correlation between the two? I read that caffeine can cause numerous nutritional deficiencies (iron, vitamin b, etc) which can lead to hair loss; or that it could lead to an increase of DHT, which in turn causes hair to fall off at a faster rate. However, I didn't find any reliable sources concerning this.

24.202.44.252 (talk) 13:46, 19 April 2011 (UTC)

You are probably stretching things too far. A -> B and B -> C both being true in different studies does not always equate to A --> C in real life. There are so many different variables involved that you can't always put 2 and 2 together. This article suggests that caffeine might actually stimulate hair follicle growth at least in vitro. However, a little bit of personal WP:OR suggests that this is not working in vivo. Perhaps I should be using a caffeine shampoo for better effect. --- Medical geneticist (talk) 15:57, 19 April 2011 (UTC)

If someone is losing his hair due to genetic causes, I doubt coffee will change the picture. 212.169.181.129 (talk) 22:09, 19 April 2011 (UTC)

Caffeine is a Featured Article -- it's a pretty good source for reliable information. Looie496 (talk) 02:21, 20 April 2011 (UTC)

vitamins

I heard somewhere that a lot of the benefits of vitamin pills are counteracted by the body not absorbing the vitamins well. Is this because all the vitamins show up in a large bolus? In that case, wouldn't it make sense to divide the vitamins into say 4 parts and take one part every 6 hours or so?

Googlemeister (talk) 16:01, 19 April 2011 (UTC)

The bolus isn't really the issue, normally. There are many reasons why they may not be absorbed properly:

1) Some are fat soluble, so need to be taken when eating fats.

2) Some can't be absorbed in the presence of certain other foods.

3) Sometimes they give you a different form of the vitamin than you get from food.

4) And, of course, you may not have a deficiency in the first place.

One case where the bolus does seem to be a problem is when they give dialysis patients a week's worth of iron in a single ferritin injection. This causes local problems in the area of the injection. StuRat (talk) 16:25, 19 April 2011 (UTC)

materials treatment

Could waste materials be heated up to a high enough temperature leaving all the elements of the waste by themselves then could these be filtered by their density? —Preceding unsigned comment added by 82.38.96.241 (talk) 16:12, 19 April 2011 (UTC)

Possibly, but that would take a huge amount of energy. StuRat (talk) 16:31, 19 April 2011 (UTC)

At microscopic/atomic levels, this is (partly) how Mass spectrometry works, but it is difficult to envisage this being scaled up to macroscopic quantities economically. {The poster formerly known as 87.81.230.195} 90.197.66.111 (talk) 16:47, 19 April 2011 (UTC)

Methods based on this principle might be used extra terrestrially at some point in the future (possible by the Chinese at the speed they are advancing). http://www.permanent.com/i-distil.htm --Aspro (talk) 18:54, 19 April 2011 (UTC)

Great website aspro —Preceding unsigned comment added by 82.38.96.241 (talk) 12:47, 20 April 2011 (UTC)

Yes, it exists, it's called: Plasma arc waste disposal. Ariel. (talk) 01:26, 21 April 2011 (UTC)

Absolute

I've been fixing the links to the disambiguation page absolute. I can't seem to find a Wikipedia article about absolute as the opposite concept of relative in a scientific context. Does anyone know of such an article? Thanks, P. D. Cook ^{Talk to me!} 17:39, 19 April 2011 (UTC)

There's Absolute time and space. Does that help? --Tango (talk) 18:11, 19 April 2011 (UTC)

There's also Absolute temperature (redirecting to Thermodynamic temperature), but I suspect that these are too specific for the OP. I'd say that is because "absolute" is rather a case for Wiktionary; it's hard to imagine a whole Wikipedia article about so general a term. — Sebastian 18:36, 19 April 2011 (UTC)

Well the concept of an absolute measurement (versus a relative one) comes up many times in math and science (temperature scales, coordinate systems, scoring systems, etc.), so I could see there being an article about it or maybe a section in a broader one. I had been considering just linking the articles in question to Wiktionary, and now that you mention it as well, I might do that. Thanks for your input. P. D. Cook ^{Talk to me!} 19:15, 19 April 2011 (UTC)

You are raising a good idea, maybe it would make sense as a subsection of Measurement. Still, at least for me, the absolute/relative dichotomy often isn't so clear cut. Take the height of Mount Everest. You'd probably say it's ~9km absolute, and some lesser value relative to the base camp. However, one could just as well argue that 9km is the value relative to sea level, and measure the absolute value from the center of the earth. I see this as a mere situational convention, rather than an underlying physical dichotomy. — Sebastian 21:34, 19 April 2011 (UTC)

Testosterone Levels

In a recent article in the New York Observer, the author writes:

"Indeed, it's no longer so easy to be male. If you're a frog, this is literally true—modern environmental toxins can actually turn you into a female. If you're a human, they've merely halved your sperm count since the 1940s and zapped 15 percent of your testosterone since the 1980s."

We'll let the frogs be for now – are the claims he makes about men really scientifically documented? Wouldn't a 15% change in a testosterone levels have a profound effect on anyone? And, what types of environmental toxins, if any, could alter testosterone levels like this?

Alfonse Stompanato (talk) 18:37, 19 April 2011 (UTC)

Sounds very dubious. As usual, the impetus is on the originator of the claim to provide evidence; it's very hard to publish a scholarly article that provides counter-evidence to every inane claim ever made. Do you need help finding epidemiological studies of hormone levels? Nimur (talk) 18:49, 19 April 2011 (UTC)

The lower sperm count may have more to do with modern fashions, where tight pants overheat the testicles. Scotsmen had the right idea with kilts, apparently. StuRat (talk) 18:52, 19 April 2011 (UTC)

The article on Endocrine disruptors might interest you.--Aspro (talk) 19:08, 19 April 2011 (UTC)

FPGAs in space applications

Where can I find extensive information about that topic? Thanks. --Belchman (talk) 19:18, 19 April 2011 (UTC)

I previously worked on these: Rad Hard FPGAs from Actel; but I think Actel has seen better days. I'm not sure if that datasheet is up to date. You probably cannot find "extensive" information about them, because they are subject to export control and other regulation. Nimur (talk) 21:37, 19 April 2011 (UTC)

That's useful, thanks. --Belchman (talk) 22:32, 19 April 2011 (UTC)

ChemDraw diagrams in articles

how can I place into body article chemical structures, like polysaccharides, written by Chemdraw programBermanel (talk) 19:53, 19 April 2011 (UTC)

You have to save the ChemDraw drawing as a graphics file format (png, gif, jpeg, etc.) rather than as a ChemDraw document (cdx, etc.). Use the ACS template when making your diagram, and save the image as a PNG format with a resolution-option of 720dpi. Then, upload that image file. An image is its own item on wikipedia, not "a component of another page". Finally, edit the page where you want the image to appear and enter the wiki-code for it. Wikipedia:Creation and usage of media files has lots of specific details about how to upload files and insert them in pages. WT:CHEMISTRY is a place to discuss chemistry issues related to wikipedia articles (formatting and style, etc). DMacks (talk) 20:20, 19 April 2011 (UTC)

Actually for something like that SVG will almost definitely be preferred if they render properly. If you only create a PNG it's likely you're creating future work for someone to recreate the image in the future. According to Wikipedia:Manual of Style (chemistry)/Structure drawing#Generating SVG files ChemDraw 12 should be able to generate SVG files. Nil Einne (talk) 10:33, 21 April 2011 (UTC)

Until the very latest release and highest-end product, I don't think the ChemDraw suite supports SVG export:( Of more concern, XfD has extensive trail of SVG images that are "correct" but render incorrectly on WP. Not sure if ChemDraw does them viably, or if they require extensive post-processing to render corectly. It's clearly a better format in many respects and may be the way forward, but CHEMMOS does not state a preference either way at this time. DMacks (talk) 14:18, 21 April 2011 (UTC)

Um the page I linked to earlier is part of the MOS and says "SVG files are preferred, because they are smaller in filesize, can generate high quality images for printing and are easier to translate if text is added to mechanism." In any case, if CHEMMOS didn't say anything it wouldn't overide the general wikipedia (let alone commons) consensus SVG is preferred for anything suitable for vectors images (provided there's no reason it can't be used like the content doesn't render properly which I earlier specified as an exception well aware it can sometimes be a problem). The plenty of images I'm guessing including some chemical structures tagged for replacement atest to that hence I think a good reason to encourage people to start with SVG in the first place, and only use PNG if they can't (note I didn't ask the OP to jump thru hoops to try and make SVG, simply pointed SVG is prefered and any PNGs are likely to require replacement in the future and ChemDraw 12 from our own MOS is allegedly able to generate SVG). BTW the MOS I quoted earlier does say ChemDraw 12 hence why I specified that but doesn't say anything about needing a specific variant/highest-end version hence why I didn't specify that, if it's deficient in this regard it should be rectified. P.S. I'm aware the smaller in file size isn't always true, but I didn't add that text and this isn't the place for such discussion. Nil Einne (talk) 23:27, 21 April 2011 (UTC)

Last I looked (and it has been a year or so), svg did not appear to be present in the student-/entry-level chemdraw packages (vs full chemoffice or chembiodraw suites) and/or only on Windows platforms. The page you linked does agree with windows-only, and the only hits for I'm finding now via google for "svg" on cambridgesoft.com talk about the full suites. It may well be there, I'm just not seeing it:( I'd love to see svg widely used, I hate that it involves hoops and poor results when it should be no harder than other formats and give good results (the "may not be rendered correctly on Wikipedia" note on the same MOS page you link). DMacks (talk) 01:21, 22 April 2011 (UTC)

I see Wikipedia advocates .svg, but I'm not clear on why. There's one good freeware program to draw the files, Inkscape, but anyone using Inkscape is frowned upon - in part because svg files often don't look the same in Inkscape as on your browser! (Those isotope and decay maps we were looking at recently, for example) All the svg policy means is that most people can't/shouldn't bother trying to make images. Oh, and if we ever do get good tools for making svgs, shouldn't they include something that can automagically convert pixelated images into svg images assuming that what look to the eye like straight lines are meant to be straight and so forth? Wnt (talk) 06:59, 22 April 2011 (UTC)

Windward Islands and trade winds

Trade winds (yellow arrows) come from the northeast in most places, including at the Antilles.

There is a great comparison table at Windward Islands#Terminology, which shows that (in English) the "Windward Islands" lie to the south and southwest of the "Leeward Islands". How can this be, given that the trade winds come from the northeast? — Sebastian 21:23, 19 April 2011 (UTC)

The Windward Islands lie to the southeast of the Leeward Islands. See File:LesserAntillesIslands.png. Most ships probably approached the Carribean from the coast of South America, moving northwestward. The first islands they would hit in this path would be the Windward Islands, exactly as the name implies. --Jayron32 04:53, 20 April 2011 (UTC)

Clearly, Aruba lies southwest of any of the Leeward islands. Or are you saying our table at Windward Islands#Terminology is wrong? If we leave out the portion from Isla de Margarita to Aruba, that would take away some of the worst contradictions.

Still, even then the name doesn't fit the wind. Since the remaining islands are not arranged in a straight line, it's a bit moot to quibble over whether they are arranged in a south-north fashion or in a southeast-northwest one. Instead, let's actually measure concrete angles:

• First, let's look at the wind direction. The yellow arrow in the above map indicating the trade wind pointing at Guadeloupe, has an angle of 255°. (We probably have a measurement error of about 10° for this and the other measurements.)
• Next, let's look at the course from Dominica to Martinique, right at the border between "Windward" and "Leeward" islands. That is about 155°. If it were 165°, it would be exactly perpendicular to the wind. It's a bit less than that, so that one can say that Martinique is a little bit upwind from Dominica, but that difference is less than the size of the islands; the downwind side of Martinique is still upwind of the upwind side of Dominica.
• Now let's look at the other extreme, the furthest downwind of each island chain: The course from St Croix to Trinidad is about 160°, which is also about at a right angle to the wind.
• Lastly, let's look at the most upwind islands: Barbuda and Barbados. There, too, the angle is about perpendicular to the wind.

I have to conclude that the island groups are not up- and downwind from each other, but almost exactly abreast. How could Britannia, who ruled the waves, get the direction of the wind wrong by 90°? — Sebastian 07:38, 20 April 2011 (UTC), corrected 08:02, 20 April 2011 (UTC)

The term Leeward Antilles may be of some interest to this discussion, vis-a-vis Aruba; the specific use of "windward" vs. "leeward" in relevence to the islands off the coast of Venuzuela is somewhat in flux; so lets throw those out for the purpose of the discussion. Secondly, the terms "windward" and "leeward" apply not merely to the wind, but how the wind carries your ship to get there. The wind gets you to the Windward islands first on the standard trip from Europe, hence the name. Thirdly, lets look at your supposition: You claim that terms invented by people who had a practical, working, everyday knowledge of seafaring are wrong at their own job, and that you can determine this by looking at a tiny, low-resolution, amateur-created map of the entire world with some yellow arrows on it, yellow arrows of no confirmed reliability, and which are on a map that gives no indication of how literally and acurately we are supposed to understand their direction, and using a protractor on the yellow arrows of dubious reliability and applicability, claim that the entirety of the British Navy, whose job description is largely "know how to get a boat where you want it to do" didn't know how to get their boats where they wanted them to go? Really? --Jayron32 12:05, 20 April 2011 (UTC)

We seem to agree on most of the details. The Leeward Antilles are just the portion from Isla de Margarita to Aruba which I mentioned above, and we agree to throw them out for now. We agree that the British were very good sailors and that it is unlikely that they would choose a name that doesn't fit reality. We also agree that there is some measurement error, but I see no reason to assume an error of 90°. Trade winds are known to have a northeasterly direction, and they're pretty reliable; you can't twist them to become southeast winds just to fit the name.

Where we differ is in the conclusions we draw. You seem to be saying: "Because the British Navy said so, all other evidence must be dismissed." I, OTOH, say: "This is odd. Let's investigate this further. Let's ask at RD/S, there are probably people who like this sort of riddle."

So now, let's investigate this further. You suggest looking at how the wind carries your ship to get there. That is an interesting suggestion which hadn't occurred to me; If I may paraphrase, you translate "windward" as "that which you reach first". You go on to state that "The wind gets you to the Windward islands first". At first glance, the opposite seems to be true, as the Leeward Islands are closer to Britain. (Moreover, the wind direction should not make a difference in this case, because the general direction of the wind is favorable, and ships don't just drift like leaves in the wind, but have great liberty as to the course. In fact, for square rigged ships "best speed is made in a quartering wind"[2].) Well, maybe we need to look at how the current carries your ship to get there. (The Antilles Current, to be exact.) Alas, at about 0.7 kn (about 1/20 of the ship's speed) and with a direction similar to that of the wind, the effect should be too small to counter the bigger distance.[3].

But this gives me another idea: Maybe it is wrong to assume that ships were coming from Britain. Britain was engaged in the triangular trade, so the ships that came from Africa would reach the southeast before the northwest. Could that explain why the English nomenclature is different from the German, as Germans did not engage in the triangular trade, and would therefore come directly from Europe? This sounds like a bold hypothesis; what do other editors think? — Sebastian 06:45, 21 April 2011 (UTC)

As my last question is not a science question, but rather a history question, I posted it at WP:RD/H. Please post any history related replies there. — Sebastian 07:09, 21 April 2011 (UTC)

It appears this was never answered and the paragraphs describing the terminology on each island group's pages is as confusing as ever. We need to answer two very basic questions 1.) Were this islands named in relation to one another or from the relation of England to both groups of islands (it seems it can't be the latter, as it doesn't make sense), and 2.) exactly which way does the wind blow in relation to both sets of island groups? This has been incredibly hard for me to nail down. I literally found a source yesterday describing the winds coming from "the south" which is completely different from most sources saying that the winds come from the northeast and southeast (basically meeting at the islands). So, were most English ships (or at least the ships that gave these islands their names) coming from the southeast (Africa) or the northeast (Europe)? For me, if the winds are coming from the south and/or southeast as were the ships, then the names of the island makes since to me as the leewards would be downwind of the windwards. The direction would be, then, ships traveling the curve from the south to the northwest. If the winds and ships are coming from Europe, I guess it still makes sense, as I guess you'd eventually hit where the winds collide so you'd go from sailing with the windward (leeward thus the leeward islands would be landed upon first) to be sailing into the wind (windward) to hit up the islands to the south/southeast?

I guess what I'm saying is that I want to see some detailed wind maps, because that makes all of the difference in the naming of the island groups. --Criticalthinker (talk) 11:41, 14 September 2017 (UTC)

IBM Watson's size in 10 years.

According to Moore's Law, how small would Watson get by his 10th anniversary? Would his entire hardware finally fit behind a game show podium? What would be his estimated dimensions? --70.179.169.115 (talk) 21:30, 19 April 2011 (UTC)

You're making a few strong assertions: that Moore's law applies to Watson's technology (which is dubious); and that Moore's law implies volumetric scaling of the computer-system (when in fact Moore's law was originally an empirical observation of a trend in transistor feature size). We can't make a conclusive speculation about what future computer architectures and parameters will be like, but you may be interested to read about emerging technologies at IBM AlphaWorks, the official distribution source for IBM research and technology. Nimur (talk) 21:36, 19 April 2011 (UTC)

Our article says Watson runs at about 80 TeraFLOPS, which puts it 94th on the most recent TOP500 List. The number 94 supercomputer on the list 10 years ago was 208 GigaFLOPS. These days, that's slightly faster than a top-end PC (I'm not citing a reference for that because I can't find a good one, but I have found several bad ones that all agree, so I'm fairly confident about it!). If you wanted to, you could easily get 208 GigaFLOPS in a game show podium. So, if we assume past trends continue into the future (they've continued for a while now, so it's a reasonable assumption) there should be no difficulty getting Watson into a game show podium in 10 years. (And that's just considering hardware - if you allow for software improvements (it's the software that is impressive with Watson) then it should be able to get it that small significantly quicker. I've also ignored memory requirements - I don't know what proportion of Watson's volume is memory.) --Tango (talk) 22:03, 19 April 2011 (UTC)

It's very likely that Watson doesn't use floating-point arithmetic in its operation, given its purpose is to answer questions given in a natural language. Therefore, I don't think it's possible to work out when Watson can fit inside a game show podium by estimating when a PC can match or exceed Watson's LINPACK score. I think a better approach would be to look at what hardware attributes enable Watson (my guess is memory capacity, memory bandwidth, and integer performance) and when a PC or podium-sized computer is expected to match or exceed Watson in these areas. Rilak (talk) 06:09, 20 April 2011 (UTC)

Whatever metric you use, you see the same basic trends. I think the fact that I neglected improvements in software will have a greater impact than using floating-point rather than integers. --Tango (talk) 19:40, 20 April 2011 (UTC)

If cars followed Moore's Law from the 1925 Model T

At 6.9 HP per liter for a 1925 Model T, if cars started following Moore's Law from that year, how many HP would an economy car's 1.0 liter engine contain? What devices have that much power today, and what would that kind of power do?

Or if we only needed 1,000 HP, how big would the engine need to be? What other object(s) would it then be the size of? --70.179.169.115 (talk) 21:30, 19 April 2011 (UTC)

Well, if HP per liter doubled every two years, starting in 1925, the current value would be 2⁴² times the 1925 value.

That 1.0 liter engine would output more power than is consumed by every piece of equipment on earth. APL (talk) 21:39, 19 April 2011 (UTC)

Wow. Okay then. So if we decided to leave more room for passengers and cargo by shrinking engines, how small would an engine need to be to output 1,000 horsepower if it followed Moore's Law like so? (To give you a starting reference point, 1,000 HPs would need to fit in a 145-liter engine in 1925, strictly calculating from the Model T's stats.) --70.179.169.115 (talk) 22:31, 19 April 2011 (UTC)

Automobile locomotion is not governed by the same physical principles as microelectronics; so your attempt to apply Moore's Law to engine scalability is flawed. As the joke goes, "if GM were like Microsoft..." Nimur (talk) 22:43, 19 April 2011 (UTC)

Hence the "If" in the question. StuRat (talk) 00:28, 20 April 2011 (UTC)

Yes, and to be clear, Moore's 'law' isn't any sort of physical law, just a descriptive pattern for the past few decades. No reasonable scientist thinks it will hold true indefinitely. Even our article says this pattern is only "expected to continue until 2015 or 2020." SemanticMantis (talk) 00:38, 20 April 2011 (UTC)

Actually, the article says "...until 2015 or 2020 or later" (emphasis mine). That sentence is referring to the earliest that Moore's law might end, not the latest. And Moore's law has always been something that's unclear as to whether it'll be possible to sustain for more than about another ten years. Gordon Moore's original 1965 paper just said that the exponential increase was expected to continue "for at least ten years". Moore's law can't be sustained forever, but it won't be possible to predict the timing of the law's demise much in advance. Red Act (talk) 04:48, 20 April 2011 (UTC)

Also note that the article's reference for the 2015 to 2020 dates refers to a news article that was published six years ago, so the referenced article was looking 10 to 15 years ahead with those dates, not 4 to 9 years ahead. And the cited article was actually only predicting that that will be about the time range when the use of silicon for computing will start being replaced by one or more of a large number of newer technologies that are being worked on, not when the exponential growth of computing power might end. That lead paragraph really needs to cite a more current reference about the future of Moore's law; six years ago is ancient history in the computing world. Red Act (talk) 07:05, 20 April 2011 (UTC)

Since you're assuming that these HP/liter would double every two years, and that HP/liter does not depend on total engine displacement, we could also phrase this hypothetical 'law' as saying that the displacement necessary to produce 1,000 HP halves every two years. To get the current displacement for take your 145 L estimate and divide by APL's figure of 2^42. SemanticMantis (talk) 00:44, 20 April 2011 (UTC)

That is 32,970 cubic micrometers of engine displacement that would hold 1,000 HP if it followed Moore's Law. What other objects can you think of that has 32,970 cubic micrometers of volume? (I doubt Google could help me on that.) --98.190.13.3 (talk) 03:40, 20 April 2011 (UTC)

In other words, 32.97 picoliters. I have Wolfram Alpha to thank for that. --98.190.13.3 (talk) 04:15, 20 April 2011 (UTC)

At 3 x 10^-11 L (or 3 x 10^-14 m³), that's about the volume of a grain of sand. Perhaps that should be "grain of salt"; it seems silly to apply Moore's law this way. -- Scray (talk) 04:05, 20 April 2011 (UTC)

Is there an article about wheel-mounted engines? I've read that the cars of the future would have an engine embedded on each wheel. I guess if there ever comes a sandgrain-sized engine, then we could fit many of them to a wheel, or even the lugnuts of each wheel. --98.190.13.3 (talk) 04:15, 20 April 2011 (UTC)

The Toyota Hybrid Highlander and some other high-end electric and hybrid gasoline/electric vehicles have assist motors mounted on two wheels, in addition to the Continuously Variable Transmission drive train. You can read more about this: Highlander Drive-Train specs; the rear motor on 4-wheel-drive variants provides an extra 50 kilowatts and nearly 100 lb-ft. of torque to supplement the Hybrid Synergy Drive. But it is hardly "small as a sand grain."

You may also be interested in the Apollo LRV, which had independent wheel-mounted motors.

To be honest, both of those vehicles represent the fore-front or "frontier" of wheeled vehicle propulsion; the original poster would learn a lot about automotive technology by reading up on such contraptions. Extrapolating trends in digital electronics technology is totally inapplicable to this topic; instead, the OP could read about trends in motor design and vehicle dynamics, to construct a meaningful extrapolation for future technology. Idle speculation without any grounding in actual physical principle, or engineering practice, is never going to help us solve the actual problems of transportation. Nimur (talk) 18:03, 20 April 2011 (UTC)

I assumed that he didn't want to learn about automobiles, he simply wanted a metaphor to enhance his understanding of Moore's Law as it applies to microprocessors. APL (talk) 02:14, 21 April 2011 (UTC)

Craps probability question

Hello scientists. I wanted to post this over at the maths desk, but the page seems to be protected against anons at the moment.

Anyway, I have devised an extremely stupid craps "strategy" where I basically hope against hope that a 7 will never come up on a roll of a pair of dice. I make "place" bets on every number, and as long as 7 never comes up I keep making money. Inevitably, 7 is always rolled and always much sooner than I expected.

I think I read on a gambling website that, even though 7 is the most common die roll, the chances of rolling a 7 on any given roll is only about 16%. But what is the probability that a 7 will come up at least once in two successive rolls? Three rolls?

Could you show me how this is calculated, all the way up to 20 rolls? I want to see mathematically how stupid my strategy is.

Cheers, --24.188.235.80 (talk) 22:51, 19 April 2011 (UTC)

Your win condition is the equivalent of a series of independent Bernoulli trials, where the probability of success (not rolling a 7) is 5/6. Since the probability of winning is 5/6 in each trial, the probability of n successive wins is (5/6)ⁿ. That's 5²/6² = 25/36 (about 70%) for two rolls, 5³/6³ = 125/216 (about 58%) for three rolls, and you can go from there. TenOfAllTrades(talk) 23:08, 19 April 2011 (UTC)

(Expanding slightly on my previous response, there are 6 possible outcomes for each die, and 6*6 possible outcomes with a pair of dice. Six of those combinations give the losing total of 7, so the chance of losing on any given roll of the dice is 6/36, or 1/6: about 16.7%. That leaves the probability of not losing at 5/6.) TenOfAllTrades(talk) 23:12, 19 April 2011 (UTC)

(ec) And, if you are wondering why the chance of rolling a 7 is 1/6 (or 16.6667%), I will describe why. Since there are two dice, with 6 numbers each, this allows for 36 possible rolls (6×6). Of these, 6 rolls add up to 7: (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). So, the chance of rolling a 7 is 6/36 or 1/6. StuRat (talk) 23:16, 19 April 2011 (UTC)

Thank you, those answers make it very clear. And if it were truly a 5/6 success/failure for each throw, it would indeed be a great system. However in craps, you can only make place bets on 4,5,6,8,9 and 10. 7 means you lose; 3, 11, 12 are a push in my system... so rather than looking at it as binary W/L scenario, could we organize odds into win/lose/tie probabilities?

• LOSE: The probability that I will get a 7 and lose all the money on the table is 16.67% on every throw
• WIN: what is the probability that I will win (i.e, get 4,5,6,8,9 and 10)
• TIE: what is the probability for a push (2,3 or 12)

--24.188.235.80 (talk) 23:26, 19 April 2011 (UTC)

Sure, let's start with the odds for each possible roll:

  ROLL  ODDS
  ====  ====
    2   1/36
    3   2/36
    4   3/36
    5   4/36
    6   5/36
    7   6/36
    8   5/36
    9   4/36
   10   3/36
   11   2/36
   12   1/36

You seem inconsistent on what is a push, I will assume that 2, 3, 11, and 12 are. So, then, the odds are:

• LOSE = 6/36 = 1/6

• WIN = 24/36 (3/36 + 4/36 + 5/36 + 5/36+ 4/36 + 3/36) = 2/3

• TIE = 6/36 (1/36 + 2/36 + 2/36 + 1/36) = 1/6

StuRat (talk) 00:09, 20 April 2011 (UTC)

You might also want to know the probability of an eventual win or loss. We can determine this by removing the possibility of a tie, which leaves us with 5/6 of the cases. To "unitize" this (my own term), multiply all odds by the reciprocal, or 6/5:

• LOSE = 1/6 × (6/5) = 6/30 = 1/5 = 20%

• WIN = 2/3 × (6/5) = 12/15 = 4/5 = 80%

StuRat (talk) 00:19, 20 April 2011 (UTC)

Thanks for the detailed answer. Because we're discussing gambling, it's important to distinguish odds, from 'chances' or probability. Odds of 2:1 against means the probability of success is 1/3. I think you mean 'probability' or 'chance of occurrence' in your table above. SemanticMantis (talk) 01:04, 20 April 2011 (UTC)

Yes. (And, true to your name, you attacked based on semantics.) :-) StuRat (talk) 01:51, 20 April 2011 (UTC)