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October 23

Killed so quickly pain is not registered

Is is really possible to kill someone so quickly their pain receptors will not signal? I assume lasers would do as such; but what about something more conventional?— Preceding unsigned comment added by CensoredScribe (talk • contribs) 01:16, 23 October 2013‎

Yes, destroy their nerves faster than the rate of pain. Plasmic Physics (talk) 01:27, 23 October 2013 (UTC)[reply]

I don't know if you'd consider nuclear weapons to be "more conventional", but they can do the trick very well -- someone caught in the actual fireball would be vaporized before even the ion channels can open in the pain receptors, let alone the signal traveling to the brain! 24.23.196.85 (talk) 01:38, 23 October 2013 (UTC)[reply]

I have added a link to ion channel to your post. 220 of ^Borg 05:22, 23 October 2013 (UTC)[reply]

I’ve thought about this in the past and I suspect if a shipping container fell on your head without you “seeing it coming”, you’d be dead before you knew it, or felt any pain.. I think it takes about a second for your pain receptors to really kick in to full swing in your brain. Watch people who break a limb playing a sport, like soccer, it’s not uncommon that they don’t even realize it for a short while. Vespine (talk) 02:59, 23 October 2013 (UTC)[reply]

Isn't that just adrenaline? Sagittarian Milky Way (talk) 12:25, 23 October 2013 (UTC)[reply]

(edit conflict)The sensation of pain is 'created' in the brain in response to impulses from nociceptors. The brain itself though feels no pain, if I recall my anatomy & physiology classes, though the meninges covering it do.[1]

Therefore any action that destroys the brains' function quickly enough should cause no pain sensation.

For example, a gunshot in an appropriate spot (or large enough caliber) that causes brain death should work. There are though examples of people with grave head wounds involving large portions of the brain that survive with surprisingly little disability.[2], 43% brain lost

Vespines unexpected "shipping container" on the head, & being on top of a nuke when it explodes as our IP editor 24.23... suggests, indeed anything that near 'instantly' totally destroys the entire head & brain would seem to be good candidates too. Hand grenades or other explosive held to the head, or a rocket propelled grenade head-shot too perhaps.

This seems to be a rather hard thing to prove (volunteers would seem hard to come by), but I wonder if any tests on animals have been conducted? 220 of ^Borg 05:22, 23 October 2013 (UTC)[reply]

Yes. It is of utmost importance to well run abbatoirs. For hooved animals at any rate, three methods have been shown to work well, provided the operator does what he is supposed to do: a) a "bolt" of the right diameter impact driven into the head to just the right depth and at the right spot. The animal may well suffer though if the operator does not do it at the right point. b) an large electric current into the brain. The device should contain automatic circuitry so that the animal gets instant lights out or nothing at all. c) carbon dioxide asphixiation (used on pigs in some places). It takes up to 10 to 15 minutes or more to kill, but the animal is rendered painlesslessly unconscious long brefore death. The downside is operators may not wait long enough, and pigs recover consciousness with brain damage during the butchering process. That's not good. See the Temple Grandin lectures on YouTube. 124.178.152.227 (talk) 08:14, 23 October 2013 (UTC)[reply]

I have my doubts about that last. CO2 poisoning (not "asphyxiation" but really poisoning; CO2 in high enough concentrations is lethal even if there's plenty of oxygen) sounds fairly nasty to me. Our hypercapnia article says "symptomatology progresses to disorientation, panic, hyperventilation, convulsions, unconsciousness, and eventually death". (They seem to have left out headache — is that a myth?) --Trovatore (talk) 09:01, 23 October 2013 (UTC)[reply]

I suggest you read up on the ways of dispatching pigs. The effect of carbon dioxide is more complex that the article suggest. But the article does include headaches. Small amounts of CO2 just make you breath faster, without any ill effects. For that reason, gas fire suppressant systems, such as Inergen, which work by driving oxygen out of the building by displacemnt with intert gasses, also include CO2 to stimulate deaper breathing so humans can better utilise what oxygen partical pressure remains. At higher concentrations than are used in fire suppression, it can produce the symptoms described in the article. I have experienced a full Inergen dump in a computer room and felt no discomfort whatsover. At really high concentrations, CO2 asphyxiates as the blood looses its capacity to carry oxygen, which the lungs can't get enough off anyway, due to displacement - unconsciousness follows quickly with little or no distress. Note however what I said before - if the animal is not held in a CO2 chamber long enough, the oxygen starvation of the brain WLL produce nasty symptoms if and when it wakes up. The article also does not make that clear. The table of effects in the article only covers low CO2 concentrations experienced for long durations as might be encountered in human activity and you want the person to live. Pigs are dispatched with high concentrations as you want them to die as quick as possible. 124.178.152.227 (talk) 10:24, 23 October 2013 (UTC)[reply]

I should note that the experience of breathing carbon dioxide is readily available to anyone who has just finished consuming a bottle of soda pop, by inhaling from it. What's surprising is that it seems painless in this form, or when breathing from above a container of dry ice, but I've found it can definitely give a soda-pop-up-the-nose discomfort if it is humid (cloudy looking) enough. (I was curious what I was putting the mice through...) Wnt (talk) 15:48, 23 October 2013 (UTC)[reply]

@ IP124.178 I was thinking more of having sensors or electrodes attached to the animal (likely in their brain) to detect if they feel any pain at the instant they are despatched, if that is possible. The CO² thing would seem to fall outside the original question posed by CensoredScribe. I think it's implied that a fully conscious lifeform is suddenly killed without warning. -Δ-220 of ^Borg 09:39, 23 October 2013 (UTC)[reply]

I don't think that would be allowed, unless they use that kind of medicine that works by simply slowing down nerve signal speeds to bearable levels (instead of blocking brain consciousness or getting you high (opiates)), and then figuring out the full no-drug speed. Or for full scientificness, crush the animal's foot and then destroy it's brain before the signal could get there. And time it's nerve speed. Myelinated signals travel at up to 0.1 to 0.15 km/s, so over 340 mph should be unfeelable, even if it started at your toes. Sagittarian Milky Way (talk) 12:25, 23 October 2013 (UTC)[reply]

The Guillotine was introduced for this very purpose but it's still a matter of debate whether it is painless or not see: [3]. Of course, knowing that you are going to be executed is obviously a 'painful' experience in itself. Richerman (talk) 12:51, 23 October 2013 (UTC)[reply]

Supposedly, decapitation was the preferred method for condemned royals, over the other option, hanging, as decapitation was assumed to be as quick and painless as it could get. Common criminals had no say in the matter - hanging was it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:15, 23 October 2013 (UTC)[reply]

Of course it's possible if you have something like a nuclear explosion, but trying to answer this question in a comprehensive way gets into surprisingly deep philosophical problems. Daniel Dennett has devoted a lot of effort to explaining those problems -- see our article on the Multiple drafts model for a brief overview, or http://ase.tufts.edu/cogstud/papers/time&obs.htm for a more extended account, or even better, chapter 5 of Dennett's book Consciousness Explained. The basic point is that if you don't believe in dualism, then there can't be a specific point in space and time at which experience originates. Looie496 (talk) 15:55, 25 October 2013 (UTC)[reply]

which I refute with the example of a safe dropped on yours (or Dennett's, if you prefer) head. μηδείς (talk) 02:58, 26 October 2013 (UTC)[reply]

Cryogenic neutrons

How low must the temperature become, before free neutrons are electrostatically bound? (Are we talking nanoKelvin, picoKelvin, etc.?) At standard pressure, would a neutron gas convert to a solid or liquid below this temperature? What sort of packing would the solid assume, HPC? Plasmic Physics (talk) 01:21, 23 October 2013 (UTC)[reply]

There are no bound states at zero temperature and standard pressure. Compare to Helium where you also don't have a solid state because the zero point motion would provide enough energy for the Helium atoms to escape. Count Iblis (talk) 01:28, 23 October 2013 (UTC)[reply]

Then what is the minimum pressure required for bound state. Plasmic Physics (talk) 01:39, 23 October 2013 (UTC)[reply]

You'll probably have to be careful defining "bound state". Liquids exist because of inter-particle attractive forces, which could lead to a "bound state": where particles remain in close proximity, even if it is en masse rather than individually. Liquid helium should be regarded as a bound state under this definition, especially as this state occurs at zero pressure and temperature. The context can then be transposed to neutrons (which, for all we know, could exhibit a miniscule van der Waal's force due to the neutron's internal structure (or other slight attractive force). The best approach might be to try to determine the phase diagram, or pressure–volume diagram of a neutron gas at zero temperature. Chances are such information would not be easily determined, though. And as Plasmic Physics subtly suggests, and overall pressure may lead to T–P–V points where two phases are in equilibrium, where one of these phases could be considered to be a bound state – maybe even for neutrons. — Quondum 02:55, 23 October 2013 (UTC)[reply]

Is it possible to predict a lowest pressure boiling point and/or melting point using the highest predicted dipole moment? Plasmic Physics (talk) 22:05, 24 October 2013 (UTC)[reply]

You don't seem to be getting much response from those with the knowledge to answer this, and I'm no expert. I have no idea what you mean by "highest predicted dipole moment". Each neutron will have the same magnetic dipole moment. How these will align I have no idea. A fluid of neutrons at 0 K will probably be a form of degenerate matter, suggesting an inherent pressure related to their density. However, the dipole interactions may be a significant contributor to (positive or negative) pressure, if the neutrons do not simply pair up in pairwise antiparallel states. The answer is probably yes to the question of predicting phase diagrams, but only by an expert. — Quondum 07:16, 26 October 2013 (UTC)[reply]

At the size of neutrons, nuclear forces should dominate whatever electrical forces are created by temporary dipole moments. Also, quantum mechanics comes into play at such small scales; Bose-Einstein condensate may give some insight into how neutral particles coalesce at low temperatures.--Wikimedes (talk) 20:51, 26 October 2013 (UTC)[reply]

This would depend on the density. At low density, nuclear forces might be small, but nevertheless could serve as a small short-range attractive force (only over the limited range 0.7 fm to 2.5 fm), helping towards a "bound state". I'd think Fermionic condensate might be more appropriate, since neutrons are fermions, but without some long-range interaction, it is not clear that the neutrons would pair up. Nevertheless, they should be superfluid, since this is a characteristic of degenerate matter. At high density (around that of atomic nuclei), the nuclear force would dominate, and should be attractive. The high density scenario occurs in neutron starts, where there is a large degeneracy pressure, independent of nuclear forces. It would be interesting to see whether the nuclear forces could dominate the degeneracy pressure at 0 K at high density (thus forming a liquid at minimum pressure). If it could, cold neutrons might bind as a liquid at low pressure. In any event, neutrons are superfluid at these densities, even at the elevated temperatures of neutron stars. — Quondum 23:22, 26 October 2013 (UTC)[reply]

How would the liquid/solid phases be affected by the Pauli exclusion principle, given that the dominant force as you say is a nuclear force? Plasmic Physics (talk) 01:23, 27 October 2013 (UTC)[reply]

I meant dominant particle-particle force (which does not include degeneracy pressure). The actual mechanism though which the degeneracy pressure arises is the momentum of the neutrons. This pressure is proportional to density to the power 5/3, until the energy levels become very high: very similar to adiabatic compression of a monatomic ideal gas. This will be true throughout the low-density range, starting to deviate as the density approaches that of a nucleus. At some low density, other small forces (primarily magnetic moment coupling, less so electric dipole/polarizability, though this was what your original question seemed to aim at) may modify this curve. Enough to produce a phase change (ferromagnetic neutron liquid/gas?)? I rather doubt it. An externally applied magnetic field could modify this. At high density (around that of a nucleus), neutron–neutron attraction would probably be substantial, possibly comparable to the degeneracy pressure. The detail of how each changed with density would determine whether a classic gas/liquid phase change will occur. — Quondum 23:31, 27 October 2013 (UTC)[reply]

According to the article (Neutron), the dipole moment of a neutron in general is currently predicted to exist, with an established theoretical maximum limit(s). I thought that 'degenerate matter' was a nonholonomic term. Plasmic Physics (talk) 01:23, 27 October 2013 (UTC)[reply]

I have no idea what you mean by a "nonholonomic term". Whatever term you want to use, I'm referring to a state in which the pressure is dominated by particle momenta implied by the Pauli exclusion principle, as opposed to due to thermal energy. It would also be characterized by being a superfluid and the associated heat conduction. — Quondum 23:31, 27 October 2013 (UTC)[reply]

Thank you for the interesting answers. Plasmic Physics (talk) 23:53, 27 October 2013 (UTC)[reply]

Pin tumbler lock

Europrofile Cylinder set with 2 double cylinders and one single cylinder (left) and the screw to secure the cylinder

Today's main page picture is from pin tumbler lock. I understand that picture and the mechanism well. However I have never understood how locks where you can enter the key from either side of the door work - indeed you can fit two keys at once, although the second key often won't turn regardless.

How does the two sided mechanism work especially given that the key isn't symmetrical. -- SGBailey (talk) 08:20, 23 October 2013 (UTC)[reply]

It's a double cylinder, simply 2 identical tumbler mechanisms, one on each side. See pic, left a single, right a double one. Ssscienccce (talk) 08:52, 23 October 2013 (UTC)[reply]

So if you wanted to, then you could make a double lock with key A for one side and key B for the other side? I had visions of complex levers actuating the pins. -- SGBailey (talk) 11:21, 23 October 2013 (UTC)[reply]

Yes, you could. A locksmith would be able to re-key the one side if you wished by replacing the pins. — Quondum 23:26, 26 October 2013 (UTC)[reply]

Escape velocity

Hi,

In this week's "what if? - xkcd" Randall says the following: "However, the weird thing about escape velocity is that it doesn't matter which direction you're going.". I'm inclined to believe him as he's rarely wrong. But how is this? Surely you are more likely to escape from a planet if you are moving directly away from the centre than if you are moving tangent from the surface? Can anyone explain this to me? Thanks! 80.254.147.164 (talk) 10:01, 23 October 2013 (UTC)[reply]

Ditto, have you ever thought about moving towards the planet, and expecting to escape from it? Good luck. The escape velocity must have a vector which is greater then tangential. If the escape velocity is equal to the tangential, then you would have a simulated orbit. Plasmic Physics (talk) 10:21, 23 October 2013 (UTC)[reply]

If the planet has no atmosphere and is not rotating and the escaping object is moving ballistically and its trajectory does not intersect the planet's surface, then its escape velocity would be independent of direction, and is more correctly an "escape speed" - it is simply the speed at which the projectile's kinetic energy plus its (negative) gravitational potential energy is zero. The xkcd article is discussing a hypothetical small non-rotating asteroid, and it says "If you go faster than the escape speed, as long as you don't actually go toward the planet, you'll escape" - which is correct in that context. In practice, planets have atmospheres and rotate and rockets are not ballistic projectiles, so confusion arises when people try to apply the "escape velocity" concept to rocket launches. It makes more sense when applied to orbital mechanics in space. Gandalf61 (talk) 10:31, 23 October 2013 (UTC)[reply]

Even if you throw the body towards the planet, but make a tunnel in the planet for the body to go through, it will escape from the other end of the tunnel - WikiCheng | Talk 12:01, 23 October 2013 (UTC)[reply]

The part about the chaotic orbits of elongated objects is interesting, but I think the article fails when it says you'd have to worry about going into a tumble if you run too fast. Provided you were otherwise accustomed to the weird gravity, staying vertical rather than having periods of fast spinning, etc., your "near orbit" run would not be in a chaotic domain, I think. Wnt (talk) 15:55, 23 October 2013 (UTC)[reply]

A counterfactual example might be helpful here. Suppose it was harder to scape if you started your orbit horizontally. Then you would have to reach a point of maximum altitude and start falling back. at that point you would have to move slower than an object following a circular orbit at that altitude - That must be the case because your orbit would have a smaller radius of curvature even though you were under the same gravitational acceleration. But an object in circular orbit has a speed smaller than the scape speed contradicting our original assumption that your speed was equal to the scape speed. Dauto (talk) 16:03, 23 October 2013 (UTC)[reply]

Here's another way to think of it: The planet creates a well of potential energy. Think of it like a depression in a surface. To "escape" means to have more kinetic energy than the difference in potential energy where you are, and the limit at infinity. Kinetic energy is a scalar; it doesn't matter what direction you're moving (unless, as others have noted, you hit atmosphere, or the planet itself.

Things do get more complicated if there's more than one object you're trying to escape, especially if some of them have mass comparable to yours -- see n-body problem -- because you could lose energy to the system, or even steal energy from it. If the planet is non-homogeneous and rotating, this might possibly have some (small) effect on what's required to escape, and that could depend on the direction.

Responding to Gandalf, I would back off from the "velocity is a vector, speed is a scalar" shibboleth. I don't think actual physicists pay much attention to that; it's just a convenience for high-school physics teachers, who have a better chance of keeping the distinction fixed in the minds of their students if they can use a different word. Certainly there's no etymological basis for it (velox just means "fast", not "fast in a particular direction"), and terms like "high velocity" or the "velocity of light" are used regularly and I would argue correctly. --Trovatore (talk) 04:06, 24 October 2013 (UTC)[reply]

How do mirrors work ?

I've been reflecting on a problem in my head, the details of how an optical mirror reflects light. I'm using a particle model to try to understand. In that model, I believe the basic idea is that photons bounce randomly off a rough surface, and you get white, or they bounce in parallel off a smooth surface, and you get a mirror reflection. However, photons are subatomic particles, and surely all surfaces must be rough at that scale. That seems like bouncing tennis balls off a pile of boulders and expecting them not to go in a random directions.

So, do I need to completely abandon the idea of a photon as a particle here ? Even using a wave model, it's hard to see how a field of boulders doesn't scatter the wave packet randomly. Or does some quantum weirdness apply where the photon behaves as if it was a much larger particle, like in one variation of the double-slit experiment, where each electron seems able to go through both slits at the same time. StuRat (talk) 14:04, 23 October 2013 (UTC)[reply]

Actually, metallic surfaces are rather smooth even at that level due to the sort of metallic bonding (i.e. "sea of electrons") that minimizes the distinguishability between atoms. It's why metals make good mirrors and reflect so well. --Jayron32 14:40, 23 October 2013 (UTC)[reply]

That's not right. Seas of electrons, atoms, and metallic bonds are on a far small scale and have nothing to do with it, as StuRat has come to realise. Metals make good mirrors because they are ductile and easy to polish to a low surface roughness. Glass is not a metal. Everyone knows that in a domestic mirror, a greater amount of light is reflected off the silvering on the back surface, but you get quite a bit reflected of the front air/glass surface too. In vacuum tube colour TV cameras, unsilvered sheets of glass were used to split the light from the lens into 4 paths. It worked because each glass sheet reflected some light as a mirror, and let the rest straight through to the next piece of glass. Coloured film was used to block cyan from the red pickup tube, a magneta blocking film for the green pickup tube, etc. The fourth tube got unfiltered light and provided the luminance signal. As the eye cannot see detail in colour the luminence tube was the best possible, the other three were smaller cheaper versions. If you polish ceramic to a surface roughness less than 30 namometres, it will work nicely as amirror too. I've worked with ceramic pole insulators that you can see your face in. Not all ceramics can be polished that good though - it depends on what's in the mix before firing. We've all seen good reflections in water when there is no wind. Still water makes a good mirror becasue it is a liquid and thus self levelling.124.178.152.227 (talk) 14:54, 23 October 2013 (UTC)[reply]

This is where you need to look at it on the basis that light is a electromagnetic wave. Sometimes you need to consider it as photons, sometimes you need to consider it as waves. The wavelength of visible light is in the range of 400 to 700 nanometres, or 0.0004 to 0.0007 mm. To function as an optically smooth mirror, the surface variation needs to be sensibly small compared to the wavelength or scattering and destructive interference will occur. If the surface roughness is sensibly small compared to the wavelength, there is little time of arrival diffrence between waves and they will reinforce, only in the direction of propagation. Now what is a typical surface roughness of a reasonably good mirror? It's in the order of 30 nanometers or less. Check the wiki article on mirrors. Quite ordinary float glasses have a surface roughness of around 50 nanometers, which can be signifantly improved by polishing. 124.178.152.227 (talk) 14:42, 23 October 2013 (UTC)[reply]

Here you have to take into account the spread of the wavefunction in the perpendicular direction. Suppose you have a particle that moves in the z-direction. Then initially it will be within some area in the xy-plane, so it starts out as a beam with some cross section. By the uncertainty relation, the smaller that initial beam cross section is the larger the uncertainty in the momentum in the perpendicular direction will be, therefore the beam will diverge faster.

When light reflects off a mirror, you have to consider the area of the mirror that is illuminated. If e.g. the light from the source is collimated to a narrow beam then only a small area of the mirror will reflect the light. Each photon starting out from the light source that makes it into the beam will be spread out in the perpendicular direction when it arrives at the mirror over that area. Suppose the mirror is perfectly smooth. Then the reason why the photon would reflect off the mirror in the epxected way is due to intererence of all the possibilities that you have within that reflecting area. Perpendicular to the incident beam, the phase is constant. This means that at the reflecting area oin the mirror you have phase shifts. In the outgoing beam, the phase is again constant accross the beam, and that requires that the angle of the reflection is equal to that of the incident beam but opposite w.r.t. the perpendicular direction.

The narrower the incident beam on the perfect mirror is, the less well defined will the reflecting beam's direction be. You always get an inteference pattern where the central peak correpsond to the "correct direction" but the spread in that peak becomes larger then smaller the beam width is. Then what happens for incoherent light sources is that the interference pattern gets washed out (each photon will have a different interfence pattern), so you won't see inteference fringes, you will see that the reflecting beam diverges. Instead of a small incident beam, you can also use a tiny mirror here. If the mirror is not perfect, you will have the effect of introducing additional phase shifts accross the reflecting area which has the effect of distorting the reflecting beam.

The classical picture of the moving and reflecting particle is only obtained when the beam width is much larger than the wavelength, the so-called geometric optics limit. For perticles like electrons, you have such small wavelengths that you can pretend that the wavelength is zero. But for light you will easily see the effects of it havng a finite wavvelength, and then you get the paradoxical result that classical behavior only arises in the limit that the beam width is infinite, which is actually not classical at all. This is due to the uncertainty realation, if you want to have a well defined beam moving in a particular direction, the uncertainty in the momentum in the perpendicular direction has to be small, but that implies that it must have a finite width.

Example. Take camera with a perfect lens of 5 cm diameter. Photons of 500 nm from a certain direction will have an angular spread of the order of 500 nm/5 cm = 10^(-5) radians. This means that in a picture of an object at 1 km distance you can't resolve detals smaller than 1 cm. So, evewn the most expensive lens with that aperture on a gigapixel camera won't give you perfectly sharp pictures. Count Iblis (talk) 15:35, 23 October 2013 (UTC)[reply]

One thing that can throw you off is thinking in terms of "bouncing" at the atomic/photon level like billiard balls. Rather photons are absorbed and re-emitted by the electrons in the mirror's metal surface. The re-emitted photons interfere to give a coherent image. It gets into quantum electrodynamics I'm afraid. There is always QED: The Strange Theory of Light and Matter which talks a lot about mirrors. 88.112.41.6 (talk) 15:58, 23 October 2013 (UTC)[reply]

The article on Surface Plasmon may be of interest to StuRat and others trying to answer this question.--86.157.138.182 (talk) 14:52, 24 October 2013 (UTC)[reply]

Thanks for all the answers. StuRat (talk) 04:41, 26 October 2013 (UTC)[reply]

Ammonia to remove mildew from clothes?

Having established above that my clothes and towels are riddled with mildew, I'd like to kill off what I can. A gamma irradiator would be perfect for this, but the one I have access too is way too small and we're not allowed to clean our clothes with it :P Can I use ammonia indiscriminately or would I need to check each items material composition and avoid certain or all colours? I'd like to avoid damaging my clothes and towels. I don't really have the opportunity to hang my clothes outside (I live in an apartment block; I'm guessing that hung outside would UV-treat them, but I guess that also fades the colours). I don't mind using chemicals so long as they don't damage the fabric or my health. --129.215.47.59 (talk) 14:05, 23 October 2013 (UTC)[reply]

Have you tried either bleach or bleach alternative? Ammonia may damage the fabric. --Jayron32 14:36, 23 October 2013 (UTC)[reply]

Before doing either (bleach can damage fabric as well), I'd give it a run through the washer again with hot water and a good detergent to see what comes of it. But yeah, bleach would be the next step; not ammonia. Matt Deres (talk) 15:55, 23 October 2013 (UTC)[reply]

Some bleach alternatives are safer on fabric, and can still have some deodorizing effect. Clorox 2 and OxiClean are two different name brands with different formulations. These are sometimes called "color safe bleach", though I believe that the new term "bleach alternative" is currently more in use. --Jayron32 16:42, 23 October 2013 (UTC)[reply]

...and never use ammonia and bleach at the same time. Mixing the two can produce chloramine and other toxic gasses.--Srleffler (talk) 16:53, 23 October 2013 (UTC)[reply]

I forgot my gym t-shirt for for four days, leaving it (soaked in sweat) for four days in the pocket of my gym bag. On discovering it, it was extremely mildew-ey; I thought it was a gone-er. At a last ditch, I left it in neat white vinegar (the cheapest generic kind the supermarket has) overnight, and then laundered it with other clothes. It's fine, free of nasty odours and with neither mildew marks nor any bleaching. Vinegar is a great fungicide, and supermarkets sell it almost as cheaply as bleach (which lightens, weakens, and damages many things like fabrics which vinegar does not). -- Finlay McWalterჷTalk 18:14, 23 October 2013 (UTC)[reply]

Since vinegar is just weak acetic acid, couldn't I just fill the bathtub with a weak acid and clean a load of clothes at the same time? Since acetic acid is just a weak acid, couldn't I just as well use an equivalent solution of hydrochloric acid? 2.97.26.56 (talk) 04:45, 24 October 2013 (UTC)[reply]

I've found that real bleach can be used without damaging most clothes:

1) Wear junk clothes when doing laundry, in case you get some on your clothes.

2) Run washing machine a bit first, to get all the clothes soaked, then add bleach.

3) Don't use super-concentrated bleach.

The only thing I found that still fades when bleach is used is my black socks. I wash those separately. Also, don't use with "rubber backed" drapes, rugs, etc. Those melt in bleach. StuRat (talk) 19:16, 23 October 2013 (UTC)[reply]

Repotting a Marigold and Inducing Flowering

I visited my parents in September, we had Chinese takeout, the restaurant parking lot was full of brilliant orange marigolds run riot, sprouting from the gardened soil and having spread to the sidewalk and blacktop. I pulled out a single small plant from the blacktop and put it in a vase. The next time I visited it was still in full bloom and had developed copious new roots, so I planted it in a small pot of topsoil, very lightly fertilized it, and dead-headed the blooms. I am quite certain it is a variety of Tagetes, almost identical to the one pictured, just a pumpkin orange color instead. It was doing so well the squirrels were eating it. Since it will frost by Halloween I put it inside in a southern-facing window that gets about six hours of direct sun. I cannot find any advice on how to get it to bloom all winter. (I have a red "geranium" that does so.) Where can I get advice on care and how to induce blooming? Thanks. μηδείς (talk) 21:17, 23 October 2013 (UTC)[reply]

It seems that photoperiodism plays a large part in flowering see:[4], also temperature and access to nutrients [5]. There's also some interesting info at [6]. Incidentally, you seem to be American (sidewalk, blacktop) but talk about 'soil' and not 'dirt'. Do most Americans use both terms? Richerman (talk) 12:22, 24 October 2013 (UTC)[reply]

Yes, Americans say both "soil" and "dirt". Some of us might even make some distinctions in meaning :) SemanticMantis (talk) 13:44, 24 October 2013 (UTC)[reply]

It's a question of attitude. I said soil because the concrete paved area had four square areas left unpaved, filled with nice topsoil, and intentionally planted. Had my nephew started playing in one of them, I'd have told him to get out of the dirt. (One of my undergrad majors was biology with a focus on plant ecology, and I worked for two years in the crop science lab, so I do know all about soil and photoperiodism (I keep poinsettias year-round, and have grown crops of marijuana). I just don't know the facts in this case.) μηδείς (talk) 18:22, 24 October 2013 (UTC)[reply]

I only asked about usage of terms because a lot of times on TV I hear Americans talking about 'dirt' where in the UK we would say 'soil'. We use them pretty much as as in the example you gave. My American cousins (from Illinois) found it confusing at times when they visited England because a lot of times we use the same terms but with a different meaning e.g. back yard, potato chips etc.and as for that line that goes "with kisses on the bottom, I'm so glad I got 'em" - we won't even go there :) Richerman (talk) 21:15, 24 October 2013 (UTC)[reply]

You will definitely hear people who do not know the distinction between dirt and soil, and who will use only dirt, and even people who say "put it on the floor" when they mean on the ground. (THis is very common in NYC, of course.) But for most educated people, if you said "They were doing construction and left the roadway covered in dirt" they will picture a smear of dry earth that needs to be washed away, while if you said, "...covered in soil" they would imagine a pile of fresh earth that will presumably go back in place when the work is done. Or at least I would. μηδείς (talk) 01:43, 25 October 2013 (UTC)[reply]

Using floor instead of ground is common here too - one of my pet hates. Richerman (talk) 04:54, 25 October 2013 (UTC)[reply]

I suspect that it will not work well to keep a marigold ever-blooming indoors. The general natural history of Pelargonium is to be evergreen perennial, while marigolds are (mostly) annuals. The Tagetes article says there are some perennial varieties, but I don't think they are commonly used as plantings in restaurants... Probably the one you got from the blacktop was a volunteer from the previous season's seeds, and they will senesce no matter what you do. Still, the best bet would be to religiously dead head the plant, because it is usually seed completion that sends senescence hormones in temperate annuals. Worth a try though, and little to lose! SemanticMantis (talk) 13:44, 24 October 2013 (UTC)[reply]

Actually, I don't mind if it truly is an annual, I want it to bloom. I read that almost all marigolds except for some specialized hybrids actually are perennials, but they are allowed to die of cold and dry over the winter, rather than being brought indoors in harsh areas. Looking at the way the plants were growing by the restaurant (size and spacing) I got the impression the ones in the soil, which were much larger, and which had obviously not just been planted that season given their wild growth habit, were survived perennials and the smaller plants growing all over were from the previous year's seeds. μηδείς (talk) 18:22, 24 October 2013 (UTC)[reply]

This article [7] seems to say that you can force marigolds to bloom indoors, but it might be hard to access. Most of the stuff I'm finding seems to be about starting them from seeds in the winter, to have them blooming indoors by e.g. Feb or March. For better googling, include the term "force" or "forcing" which is the horticultural term for making plants bloom when then otherwise wouldn't ([8]), along with /marigold indoor bloom/ etc. Finally, you can probably get better answers by asking the same question at e.g. GardenWeb [9]. SemanticMantis (talk) 14:29, 24 October 2013 (UTC)[reply]

Yes, I am familiar with forcing, I did it to my marijuana. Unfortunately the ebsco link you gave simply says you can bring them indoors and says how to pot them from the soil. μηδείς (talk) 20:05, 24 October 2013 (UTC)[reply]

The last link I gave above says about marigolds "Flowers are available round the year. It is a day neutral plant which takes 60 - 70 days from seeding to harvest." I suppose it would be easy enough to keep a succession of them going all year but that's not really what you were asking was it? Richerman (talk) 21:27, 24 October 2013 (UTC)[reply]

Yes, I really would like to keep this plant blooming, it was a really striking color. I believed it was budding again about 10 days after repotting from the vase, when the dastardly squirrels ate some buds off. (Seing it, I banged on the window, but the squirrel only stared at me in contempt, so I ran outside hissing and clicking squirrel obscenities, and brought the plant in.) I am tempted to go get some seeds from the garden shop. Tomorrow I will bring in the geranium I kept in the window last winter. I'll have to post some pictures to wikimedia of my prize Poinsettia. Frankly, I have a very green thumb. I am just curious whether I can expect this marigold to bloom before next spring, and if so, how to encourage (force) it. μηδείς (talk) 22:02, 24 October 2013 (UTC)[reply]

I just found out that 'African' marigolds come from Mexico - how did that happen? Usually when I plant them quite a few get their leaves stripped by slugs and snails. That's our damp climate for you! Richerman (talk) 05:05, 25 October 2013 (UTC)[reply]

You might want to import some Virginia opossums and Raccoons, as well as Skunks and woodchucks. When I was in grade school, these animals were quite rare (they were hunted for food until well after the Depression. Six inch slugs were so common it was suicidal to go barefoot after dark, or lean on a tree without looking first. Nowadays those mammals are very common, and I haven't seen a slug or a slime track in almost 30 years. μηδείς (talk) 05:38, 25 October 2013 (UTC)[reply]

Hmmm, we have enough trouble already with American grey squirrels and mink. I blame the Romans for the snails - they introduced them to Britain as a delicacy. Well, actually, we did have some already but you have to blame someone :). I remember years ago there was a hedgehog in the garden so I started putting cat food out to try and keep around to eat the slugs. Then one night I saw it run straight past a big fat juicy slug to get to the cat food. Richerman (talk) 13:36, 25 October 2013 (UTC)[reply]

So, the "African" marigold is from India in French. (Rose d'Inde, Oeillet d'Inde). Now, snails. The Romans tried to introduce the large "Burgundy" snail, that the French call "gros blanc". Failed. We only have the "petit gris". Which is not all that petit, unfortunately. Itsmejudith (talk) 20:37, 25 October 2013 (UTC)[reply]

Ah, snails? Just ye wait till they rewild The Highlands with mammoths and saberteeth. μηδείς (talk) 02:55, 26 October 2013 (UTC)[reply]

Bring 'em on! Itsmejudith (talk) 18:27, 26 October 2013 (UTC)[reply]

cancer evolution

Are cancer cells evolving? They are so successful like adaptive microorganisms. Is that like they have other proteins as accomplices to overlook the mutation? Even when around 95% gene (junk) is evolved to sacrifice bearing mutations, some proteins can't just overlook the mutation. Are proteins that allow the cancer cells enjoy any evolutionary benefits? — Preceding unsigned comment added by Anandh chennai (talk • contribs) 23:23, 23 October 2013 (UTC)[reply]

I believe individual cancer do evolve, or at least adapt, in that they develop a tolerance to specific drugs, because only the least susceptible cells survive and spread. However, it's important to note that Bob's cancer doesn't then jump to Tom, in this evolved form. Tom's cancer starts over again from scratch. So, in that sense, they aren't able to evolve and pass down those mutations, beyond the current host.

Of course, some cancers are caused by viruses, like HPV, and the viruses can evolve. StuRat (talk) 03:36, 24 October 2013 (UTC)[reply]

StuRat is correct. Another difference is that species evolution gives rise to new characteristics and capabilities. This has never been reported as happening with cancer. What happens is that cancer cells turn on metabolic processes that should be turned off. When a small cancer gets going, the cancer cells divide and multiply, increasing the size of the tumour until the distance of inner cells from the nearest blood capillary becomes too great. Then the tumour stops growing because the cells are starved of oxygen and nutrients, and can't adequately get rid of CO2. At this stable starvation phase, the tumour is in no way a threat to the host animal. However, after some time, a tiny fraction of cancer cells in a fraction of tumours somehow turn on something that is necessary in the animal's growth phase, especially in the embryo stage. These cells begin to secrete a substance that causes blood capillaries to grow toward them. So the nearest capilliaries do what they are programmed to do, and penetrate the tumour. The tumour then can begin another growth spurt. It's now this tiny fraction of cancer cells that get the closest blood supply, so they divide and multiply faster than the other cancer cells. 120.145.195.215 (talk) 05:17, 24 October 2013 (UTC)[reply]

It's quite rare, but there do exist cancer cells which have the opportunity to evolve beyond the extent to which they have the time to evolve within one animal, namely, the clonally transmissible cancers. These include the devil facial tumour disease, in which the cancer cells are spread by devils biting each other, Canine transmissible venereal tumor, in which the cancer cells are sexually transmitted, and contagious reticulum cell sarcoma of the Syrian hamster, in which the cancer cells are spread via mosquitos. Red Act (talk) 08:19, 24 October 2013 (UTC)[reply]

WP:WHAAOE: Somatic evolution in cancer. From my brief skim of it actually looks like a damn good article (well, it's well referenced at least, which puts it in a better state than many), although it reads more like a scientific review article than a wikipedia article at present. Equisetum (talk | contributions) 11:07, 24 October 2013 (UTC)[reply]

Just remember that the basic concept of evolution is that certain adaptations enable the adapted organism to survive and prosper better than the unadapted ones. As StuRat notes, Bob's cancer can't keep on going; either it's unsuccessful and gets killed by chemo/radiation/whatever, or it's successful and kills Bob — and thus it too dies. Barring weirdnesses such as HPV and the non-human devil facial tumour disease, cancer can't keep going after it kills its "parent" organism, so there's no way for cancer itself to evolve. Nyttend (talk) 21:05, 27 October 2013 (UTC)[reply]

Yes, that is true, each person's cancer is an independently "evolving" collection of cells and the evolution doesn't continue beyond the course of that single person's disease. However, evolution (specifically natural selection) is blind to the future - just because an evolutionary change is "doomed" to failure doesn't mean it won't occur. If a change allows a cell lineage to propagate better, it will, well, propagate (the trouble with evolution is that the concept is so simple it is almost a tautology, which makes it difficult to talk about or explain, since human language doesn't seem particularly good at handling tautologies). Note that there are multiple types of selection going on here - genetic alterations that suppress cancers propagate because they stop organisms dying of cancer and thus increase reproductive success, whereas genetic alterations that promote cancer propagate (ultimately fruitlessly) because they allow a given population of cells containing that alteration to grow beyond normal limits. Each of these is initially successful for the gene in question, it is only with hindsight and overview that we can identify one as fruitful and the other as fruitless. Equisetum (talk | contributions) 18:39, 2 November 2013 (UTC)[reply]

Evolution is not related to survival alone. An organism A wants to kill B, another organism. But surviving after killing B is not a concern for A, all it wants is to kill B and survive in the interim. So, A originates and invades B and surviving in the mean time. Finally, it kills B and succeeds in its intention. But in the process, A is also killed. A has achieved what it wanted and so it becomes 'selected' in evolutionary terms. "A" is cancer cell and "B" is the host. Cancer is like a different species, originating and leading its life till the host dies. Here is an article that i got to see: http://newscenter.berkeley.edu/2011/07/26/are-cancers-newly-evolved-species/. I questioned if these camouflaged species are helped by host proteins (and, is there a similar host-parasite relationship in any organisms?). anandh, chennai