Wikipedia:Reference desk/Archives/Science/2009 October 3

= October 3 =

Is there really meaning to the speed of light?
This shoots off from the most recent comment from the last speed of light question a couple posts above: When God chose the speed of light, did he have a choice? Anyway, given that essentially every measurable quantity is dependent to an extent on the speed of light, is there meaning to its value? For example, all elementary forces are transmitted at the speed of light; the speed of light is the maximum speed of transfer of information; a change in the speed of light would influence the energies of atomic orbitals. So my question really is, if the speed of light were to be changed, universe-wide, how would all of these things it influences scale to eachother? Basically, would everything get smaller at the same proportion, or would the size of things like different atoms, or different bond lengths, scale disproportionately, as I suspect? Someguy1221 (talk) 05:56, 3 October 2009 (UTC)
 * Unfortunately there's no well defined answer to this question; it depends on arbitrary notational choices. For example, the energy of a photon is E = hν = hc/λ where ν is the frequency and λ is the wavelength, so changing c won't change the energy-frequency relationship but it will change the energy-wavelength relationship. But Planck could just as well have chosen a constant k with the property that E = k/λ = kν/c, and in that case it would be the other way around, even though the physics is the same. As another example, particle physicists quote masses in kg but also in MeV/c². If you change c then those no longer agree, so the theory isn't even internally consistent without further changes. It is possible to choose new values of the constants and get a new theory that works, but it's not well defined whether you've changed "just one" or "several". It's also possible to change the numeric value of c without changing anything of physical importance. For example, you can just redefine the meter such that c = 123456789 m/s and express all other constants in terms of the new meter. -- BenRG (talk) 09:34, 3 October 2009 (UTC)
 * For an interesting take on the potential effects of varying c (and other physical constants) you might like to read the "Mr Tomkins" books by George Gamow. --Phil Holmes (talk) 10:57, 3 October 2009 (UTC)
 * Those books aren't entirely accurate, though. --Tango (talk) 18:09, 3 October 2009 (UTC)
 * You are right in some cases. All the physical constants are related to each other. You can't change one without changing some of the others, but you can choose which ones to change and how. For example, $$c=\frac{1}{\sqrt{\mu_0\epsilon_0}}$$. So if you double the speed of light, you need to change one or both of $$\mu_0$$ and $$\epsilon_0$$, but you can choose the details. There are ways to do it where everything would change in such a way that the physics is identical and you wouldn't be able to tell. --Tango (talk) 18:09, 3 October 2009 (UTC)
 * If you use Planck units, all the universal constants are one, but all the masses, charges, etc. are arbitrary numbers. In a sense, this means that all the universal constants are meaningless, and it's just the attributes of elementary particles that matter. On the other hand, if you make all your definitions based on elementary particles, it's the universal constants that matter. I think the best you could say is that any given measurement is meaningless, and what matters is when you compare different measurements to each other. — DanielLC 18:31, 3 October 2009 (UTC)
 * I like to concentrate on dimensionless constants. It doesn't matter what the masses of electrons and protons are in kilograms, what matters is the ratio between the two, which is dimensionless and thus not arbitrary. --Tango (talk) 19:45, 3 October 2009 (UTC)


 * See absurdism. The long and the short of it is that nothing has concrete meaning. Vranak (talk) 17:43, 8 October 2009 (UTC)

Magnetic saturation
I would like to know what will happen if we try to magnetise a magnetic material beyond its saturation,and what behavioral change we get if the materaial back to below saturation.Roon93 (talk) 06:45, 3 October 2009 (UTC) Thanks
 * Your first question is answered in Saturation (magnetic). Your second question is answered in Magnetic hysteresis.  Red Act (talk) 09:16, 3 October 2009 (UTC)

solid mechanics
I just need to know what is stress and how can we realize a stress in a body? what is the physical interpretation of it?210.212.244.133 (talk) 09:32, 3 October 2009 (UTC)


 * In the simplest, where a body is stretched (in tension) or squeezed (in compression) by a single force and an equal and opposite reaction, the (average) stress is simply the force divided by the cross-sectional area perpendicular to the direction of the force. In more complex cases, where shear stresses are involved, stress has to be treated as a tensor. See stress (mechanics) for more derails. Gandalf61 (talk) 10:23, 3 October 2009 (UTC)


 * That article is rather math-heavy, unfortunately. The physical reaction is called the strain or deformation (mechanics).  In addition to tension and compression, there's also bending and twisting (torsion).  Heat and noise can be generated, too, and eventually a phase change or fracture may occur.  StuRat (talk) 00:32, 5 October 2009 (UTC)


 * The easiest physical interpretation of a stress that I can think of is that if I push with a force F directly on a sheet of metal with area A, then I am exerting a normal stress of magnitude F/A. If I push parallel to the surface of the metal plate instead of directly into it (e.g., placing your hands on a wall and pushing upwards), then I am exerting a shear stress (or shear traction) of the same magnitude. You'll have to clarify before I can give a meaningful answer to, "How can we realize stress in a body?", Awickert (talk) 16:59, 5 October 2009 (UTC)

Speed of light between Casimir plates?
When light moves through a dense medium such as glass or water it slows down. By the same logic, since the gap between Casimir plates is less "dense" in terms of energy than the normal space outside, would for example, a pulse of laser light shone between such plates arrive slightly faster than it would in "normal" space, and (assuming this was the case) could such a phenomenon be exploited for intelligible superluminal signaling?Trevor Loughlin (talk) 10:47, 3 October 2009 (UTC)
 * If you read further down in the casimir effect article, under "wormholes", you'll find reference to a paper in the journal Physical Review, which discuss what you propose! :-D The discussion is rather complicated I'm sure, as the speed of light enters as a constant in Casimir's own derivation of the effect. Not to mention that c is related to other constants like the Fine-structure constant and the planck constant. How these would change with a changing c is not obvious. EverGreg (talk) 11:04, 3 October 2009 (UTC)


 * There's an article on the Scharnhorst effect. I find the whole thing implausible because light speed is built into relativistic QFT at such a fundamental level. If you have Lorentz invariance and you can solve the initial value problem then you don't have faster-than-light signaling. So I suspect this is just a miscalculation, such as the first term of a series approximation that ends up getting canceled out by later terms. Or the photon might really acquire an effective tachyonic mass, but that doesn't imply superluminal propagation (as explained here). Most speculation about the Casimir effect is fueled by the idea that it has some special relationship with vacuum energy, which is not true, as explained in this paper. There's nothing wrong with invoking vacuum fluctuations in deriving the Casimir effect, but you can derive anything in QFT with or without reference to vacuum fluctuations. -- BenRG (talk) 12:36, 3 October 2009 (UTC)

Could a strong electric charge applied to the plates enhance any such (assumed) effect?Trevor Loughlin (talk) 10:09, 4 October 2009 (UTC)
 * I don't know, sorry. In the standard setup there's no net charge on the plates. -- BenRG (talk) 21:34, 4 October 2009 (UTC)

Surviving a Pyroclastic Flow
I have heard through talk, and read in Wiseman's SAS Survival Handbook, that the only remote possibility of a poor hapless soul surviving an oncoming wall of superhot gas and rock is to get into water and dive down for as deep and as long as possible. Considering that one may hold one's breath for about two minutes, and also taking into account that the water probably will get very very hot, would this method actually see you live? Lady BlahDeBlah (talk) 13:39, 3 October 2009 (UTC)
 * See: Time (min), Distance (Max), and Shielding (Max) for the protection from threats, real or percieved. These proven parameter adjustments can be applied to Items as varied as Mothers-in-law to hand grenades, and yes even to 'pyroclastic' flows. 68.245.155.69 (talk) 16:09, 3 October 2009 (UTC)
 * That's not really the answer to my question. I'm trying to be specific. Lady BlahDeBlah (talk) 18:15, 3 October 2009 (UTC)
 * You're asking for an absolute, "will it". It might or it might not, but you'll have a better chance of survival than if you stand there and let the hot gas and rocks cover you. →Baseball Bugs What's up, Doc? carrots 18:28, 3 October 2009 (UTC)
 * Yeah -- one in a billion is better than zero in a billion. Looie496 (talk) 23:20, 3 October 2009 (UTC)
 * Agreed. Considering that your other realistic options in this situation are:
 * a) freeze like the elsewhere-mentioned deer in headlights
 * b) scream for Jesus
 * c) briefly and bitterly reflect upon the fact that you've wasted your life
 * d) soil your britches
 * e) some combination of the above
 * --Kurt Shaped Box (talk) 00:13, 4 October 2009 (UTC)


 * f) reflect upon your best Wikipedia edits... StuRat (talk) 00:25, 5 October 2009 (UTC)


 * If you could dive down just a few feet in a body of water, the top foot might boil off, and thus not contribute to heating the rest, while the next foot might get up to near boiling, but then that heat would be mixed with the coolth (my word) of the rest of the water to make a warm, but not deadly hot, body of water. I would think all the ash would be the real threat, as a several foot thick layer of ash could bury you in the bottom of the water. StuRat (talk) 00:25, 5 October 2009 (UTC)


 * The water is probably the best option. The speed at which the clasts in the flow are moving is enough to cause significant injury, the temperature will certainly kill you, and the velocity of the flow makes running away rather implausible. But I don't think that the flow would pass before one would need a breath of air, so the true survivalist needs a mini airtank :). Another option I just thought of would be to try to run to the top of a nearby high point. The density-driven pyroclastic flow will converge towards topographic lows. Awickert (talk) 17:03, 5 October 2009 (UTC)

Visualizing magnetic fields in 3d
It's easy to visualize magnetic field lines in two dimensions by putting iron filings on a flat surface, as at right. Is there any such way to visualize the third dimension? I'm thinking of something like a flow of ferrous mist or an electron gun and special glasses or something. Thanks. --Sean 14:47, 3 October 2009 (UTC)
 * See Fluidized bed. Lab versions (glass) should suffice for your viewing pleasure. 68.245.155.69 (talk) 15:59, 3 October 2009 (UTC)
 * I don't see how that helps. SteveBaker (talk) 16:01, 3 October 2009 (UTC)
 * a Fluidized bed would allow you to suspend fine magnetic particles in a 3 dimensional arrangement. An electromagnet held in place and energized, would, conditions taylored, allow you to have a 3d view similar to the 2d example shown. 72.58.186.50 (talk) 16:14, 3 October 2009 (UTC)


 * Well, you could mount your bar magnet on some kind of non-magnetic axle that would allow it to rotate about its long axis. Then you could use the paper-and-iron-filings trick to visualise the field - take a photo - rotate the magnet (say) 10 degrees - tap the paper to let the iron filings rearrange themselves - take another photo - repeat 18 times so you have a photograph of the field lines every ten degrees over the entire volume around the magnet.  Then you could load the pictures into a computer - use GIMP or Photoshop to turn the white parts of the paper transparent then apply the photos as texture maps on a 3D model using Blender, 3DStudio or Maya or something.  My slight concern is that you can't use the paper-and-iron-filings approach to get a proper cross-section of the field because it's off to the side of the centerline of the magnet - but so long as the 3D model you build in the computer correctly reflects the position of the paper when each 'slice' of the field was imaged - that shouldn't matter...but for complicated shapes of magnet - you might not be able to get the paper close enough to image the field in detail close-up to the magnet. SteveBaker (talk) 16:00, 3 October 2009 (UTC)


 * Sometimes iron filings are suspended in some sort of clear viscous liquid or gel. Here's an example that showed up on google .  Electron guns are used to visualize how magnetic fields affect moving charged particles, but they don't show the shape of the magnetic field lines the way iron filings do.  The electrons' path is made visible by firing them through a gas that's ionized by the electrons and glows. Rckrone (talk) 16:03, 3 October 2009 (UTC)
 * Ooh, neat! This is exactly the kind of thing I was thinking of.  Thanks!  --Sean 21:11, 3 October 2009 (UTC)


 * See Magnetosphere, Plasmoid and Heliospheric current sheet (extra cool) where it can be seen (or rendered) in nature. 173.103.133.181 (talk) 16:33, 3 October 2009 (UTC)


 * Iron filings arrange themselves into only an approximation to the magnetic Field lines. The density of true field lines at any point is proportional to the field strength at that point while the lines of iron filings are equispaced. They also do not show the direction of the field vector. Cuddlyable3 (talk) 00:03, 4 October 2009 (UTC)

Mobile phone charging disabling vibration function
Is there a reason why when I charge my mobile phone (and all others I've had in the past) that the phone won't vibrate as it normally does when someone rings? Smartse (talk) 15:33, 3 October 2009 (UTC)
 * See this. as far as a workaround to the issue (I am assuming that you would not need to feel the vibration, but only need to hear the phone 'vibrate'). As to why the feature is disabled, I would need to know the model to research it.  Speculation on my part includes:


 * 1) Charging/power consumption circuit limitations
 * 2) Developer seeing no need to vibrate when unit is (presumably off the body) and charging.
 * 3) Developer's lack of consideration for the mildly obsessed.
 * 4) Developer's passive-aggressive desire to screw with the mildly obsessed. 68.244.39.0 (talk) 15:57, 3 October 2009 (UTC)
 * It's not really a problem but the three phones I've had: Philips Savvy, Sharp gx10i and Nokia 6300 have all done it. Considering this it seems reasonable to guess that there is a specific reason why beyond all three manufacturers wanting to make me confused! Smartse (talk) 16:25, 3 October 2009 (UTC)


 * I've only had two mobile phones. I always kept them on vibrate and rarely removed them from the charger.  Both vibrated while charging.  Currently, I have an LG Voyager that vibrates while charging without a problem.  Therefore, this is not an industry standard.  It just happens that the phones you've selected contain this feature. --  k a i n a w &trade; 21:04, 3 October 2009 (UTC)
 * Go with #2 above, Developer may have intended for a charging phone on the bathroom sink not to vibrate into the toilet, but plain omission of feature is most likely. Oh and if your phone will load the Vibrating wave file linked above...Go for it, and set the phone for Vibe & Ring. It will stay out of the crapper while charging and still make that buzz sound you NEED to here. 68.244.20.23 (talk) 21:39, 3 October 2009 (UTC)


 * I've always assumed it was so you could charge it in a docking station kind of charger (for example, a hands-free set in a car) without it breaking. --Tango (talk) 21:42, 3 October 2009 (UTC)


 * Perhaps they seek to avoid damaging the charging connector - so they avoid imposing vibration stress on it? Seems the most likely explanation to me. SteveBaker (talk) 23:14, 3 October 2009 (UTC)


 * I think Tango and SB have points here. Also remember depending on the type of connection it may actually come off enough to stop charging which many would likely find annoying Nil Einne (talk) 23:34, 3 October 2009 (UTC)


 * With all the phone's I've owned (all Motorola) the phone stores separate settings for both on and off the charger, and you cannot edit the other settings while not in that mode. Thus, to have it on vibrate at all times, you need to unplug it, set it to vibrate, then plug it in (says switching to loud mode or something of the sort) and you need to set it to vibrate yet again.  The difficulty is that you can't feel it vibrate when it's three rooms away, but your call. 76.228.199.201 (talk) 08:32, 6 October 2009 (UTC)Ehryk

That "deer caught in headlights" reaction - what is it?
I was reading the fight-or-flight response article to see if there was some way of explaining the reactin when a deer that freee when it's caught in headlights - or any creature (including humans) freezes. I'm thinking it's too much informtion causing an overload in the neural network. but, I'm not really sure. I suppose a severe enough situation might even be considered a transient ischemic attack, but I doubt it; just because blood vessels constrict and muscles tense doens't mean a clot is actually forming, which is what happens ina TIA, right? (Of course, I'm sure not many vets have tried to diagnose TIAs in deer :-) )So, is it just too much information having to be processed at once, as I suspect? What articl4e would point me in the right direction? Thanks in advance.4.68.248.130 (talk) 16:16, 3 October 2009 (UTC)


 * "Too much information causing an overload" is a reasonable first pass at this. What you're seeing is probably a combination of three things:  (1) a very strong form of the orienting response that causes an animal to direct full attention to anything that is intense and surprising, (2) a freezing response that many camouflaged prey species are "programmed" to emit in response to threat, (3) conflict between multiple possible motor responses (go forward, go back, etc) that results in none of them being executed.  The intense activation can sometimes cause a TIA (i.e., fainting) in humans, but the basic freezing-in-the-headlights reaction doesn't depend on that. Looie496 (talk) 17:05, 3 October 2009 (UTC)


 * See Collision. From a different frame of reference: It is the last thing that a person may see before the impact.  It will be the rut season here soon, and many a driver at the body repair shop may reflect on that image as they are having the front end damage appraised. 68.244.20.23 (talk) 21:30, 3 October 2009 (UTC)


 * As far as I was aware, animals freeze in headlights purely because in the wild they'd do the same when they saw a predator, in order to stay camouflaged and be less noticeable. I'm not a biologist, but I didn't think "overload" came into it. Chase me ladies, I&#39;m the Cavalry (talk) 23:27, 3 October 2009 (UTC)


 * Yeah, the old 'stand still - they can't see you if you don't move' thing (which failed so spectacularly in Jurassic Park). --Kurt Shaped Box (talk) 23:54, 3 October 2009 (UTC)


 * To clarify something in the original question, a TIA is generally caused by a constriction of blood vessels or by a cardiovascular problem that causes blood pressure to drop, not by a clot. In most cases a clot that lodges in the brain will cause a stroke, which is like a TIA except that it isn't transient -- clots don't go away quickly. Looie496 (talk) 00:44, 4 October 2009 (UTC)


 * I agree that it's the "freeze so they don't see me" reaction (they don't know that our headlights make them glow brightly against the black night). They have judged that the car is faster than them, so running from it will do no good.  Thus, they think their only hope is if they aren't spotted.  In their mind anything as large as a car always wants to eat them, they just can't imagine that we are driving down a road and want to avoid them as much as they want to avoid us.  I wonder if African deer (or whatever they call them there) react more like we'd expect, by running out of the road, due to being familiar with elephants and such.  Millions of years of evolution there may have taught them that large moving objects aren't hunting them, but still need to be avoided. StuRat (talk) 00:15, 5 October 2009 (UTC)


 * Although actually the deer has no idea how large the car/predator is. For that matter, neither do you late at night, you only make assumptions based on your prior experience with vehicle lighting, which the deer has none of. It could be a child on a bicycle with balsawood light mountings for all you know - but your human instinct/knowledge will cause you to flee anyway (cue Nelson saying "ha ha"). I'd also agree that it's a freeze-response - which is why hunting deer by night using intense lights is forbidden in many areas. (I'd link nightlighting, but it doesn't dab to a relevant page). Franamax (talk) 00:36, 5 October 2009 (UTC)
 * Round these 'ere parts, we'm callin' it Lampin', what seems to redirect tolerable well. 87.81.230.195 (talk) 03:06, 5 October 2009 (UTC)


 * You're just talking about the headlights, which is all you detect from a distance when in another car. But when outside and standing on the ground, the sound and especially bass sound and vibrations give a clue that the approaching object is large.  Deer, being particularly keen on observing for large moving objects they would think of as predators, would be well able to read all these signs to determine that the object approaching is much larger than them. StuRat (talk) 01:32, 5 October 2009 (UTC)


 * As a former resident of semi-rural Scotland and more recently as a regular driver through the New Forest at night, I can say from first hand experience that deer don't just freeze in one's headlights, they are totally unable to judge a car's speed and predict its (straight line) direction, and are liable to dash from out of the darkness off the road into one's path at the last second, particularly if one or a few are separated by said road from the bulk of their herd on the other side. This is unnerving, because apart from one's not wanting to hurt any animal unnecessarily, a collision with an animal in the New Forest is (1) an offense and (2) automatically deemed the driver's fault. The numerous free-roaming cattle and ponies are actually less of a hazard because they rarely startle or move quickly, having learned that traffic gives them the right of way. 87.81.230.195 (talk) 02:55, 5 October 2009 (UTC)
 * As a regular (ex-)driver of rural and wilderness Ontario, I can tell you for sure that the minute you see deer anywhere even close to the road, you slow down as quickly and smoothly as reasonably possible. Deer might decide the best way out is right through your windscreen. As I alluded in my reference to nightlighting (which is hunters driving slowly and quietly along back roads using a spotlight to illuminate the forest edge where deer come out to graze, so no Stu, sound has nothing to do with it, it's the dazzling light), deer freeze then suddenly make an entirely unpredictable escape response. The second rule of seeing deer near the road you're driving on (after slowing down right away) is to keep driving slow, 'cause where there's one deer, there are more somewhere in the darkness. They will be making sudden bolting movements too, and they can leap onto the roadway in a matter of seconds. Sorry, no links for this, just original and quite thorough research! Franamax (talk) 03:16, 5 October 2009 (UTC)


 * When your car gets within a certain distance, the deer may well figure it has been spotted by a predator, and dart off in a random direction. The goal here is to confuse and outmaneuver the predator.  It has no concept that the car is headed in a fixed direction, as this is alien to it's experience (or the experience of it's ancestors when they evolved their reactions). StuRat (talk) 02:02, 6 October 2009 (UTC)

The traditional explanation I grew up with was that the shadows adjacent to the road moved in frightening ways as the car approached and deterred the animals from leaving the road. Polypipe Wrangler (talk) 11:53, 6 October 2009 (UTC)

Atomic cannon
I'm not talking about this, M65 Atomic Cannon, but actually using a nuclear device as the propellant to drive a solid slug (tungsten probably) at high velocity. Lets say we use a nuclear device of about 50 kilotons or so, and a tungsten slug of 50 kg. We put the cannon in orbit, aim it at the ground, and fire it. Assuming the cannon can actually survive this, would the slug impact the ground with energy greater than the 50 kt nuclear charge? What about if the cannon was aimed at a target in space, and didn't have gravity to assist the slug? ScienceApe (talk) 17:17, 3 October 2009 (UTC)


 * I'd have to do some calculations to determine if the slug would vaporize, melt, or merely splinter into fragments. But even in the ideal case of merely splintering, the fragments heading toward the earth would be limited by terminal velocity in the atmosphere.  So probably more energy would actually be imparted to a target in space (assuming the target is at rest relative to the cannon) than to a target on the ground.  Ironically, the slug would impact the ground with greater energy if it were just pushed out of orbit at relatively low speed rather than being fired by a nuclear cannon, since the air resistance is less if the slug remains in one piece.  Red Act (talk) 18:11, 3 October 2009 (UTC)
 * Well terminal velocity only applies if the object is in freefall. Furthermore, it actually doesn't matter if the remains in one piece or not since the conservation of momentum states that the fragments will still impact at the same velocity. But Tungsten is very hard so I think it would survive atmospheric entry. There are real-life examples of kinetic bombardment using just tungsten rods. See, Kinetic bombardment. ScienceApe (talk) 18:37, 3 October 2009 (UTC)
 * An object 66 tons (Hoba meteorite) survived speeds of 11-72 km/s, far above current kinetic energy rounds, with only a thin melted crust, even though iron-nickel is less strong and tungsten is 2.5 times denser. It is supposed to be almost the largest meteorite that the atmosphere can slow down from space to terminal velocity, so even though tungsten's extra heft over iron will help 50 kg won't do it. All guns deflagrate inside instead of explode like a bomb, would it even be possible to make something like that that could survive being fired? Sagittarian Milky Way (talk) 19:07, 3 October 2009 (UTC)
 * I'm not really sure what your point is. Our article on kinetic bombardment indicates that tungsten projectiles would be able to survive atmospheric entry. The US Government has had projects (Project Thor) which involved firing tungsten projectiles from orbit to targets on the surface of our planet. In any case, I'm mostly just interested in the kinetic energy of the projectile, and how it compares to the energy of the nuclear charge. ScienceApe (talk) 19:25, 3 October 2009 (UTC)
 * I didn't reread the article but didn't remember the tungsten telephone pole/crowbar.
 * Fine. Proof of concept. Just use more usual means for acceleration. I doubt you'd would want to use a nuclear weapon, and just for fun, an equal mass of powder shoots a bullet about bullet speeds, so a 50 kiloton nuclear weapon should shoot a 50kg slug.. insanely fast. Large fractions of c fast.. Gravity would barely touch it. A projectile more on the order of the kilotonnage of the bomb in mass would be needed. Sagittarian Milky Way (talk) 20:12, 3 October 2009 (UTC)
 * The object is in freefall after it leaves the cannon. After it leave the cannon, there is no further accelleration except for gravity and air resistance.


 * Staying in one piece does matter, because air resistance is higher on a lot of little pieces spread about than on one big piece. When thinking about conservation of momentum in this problem, it's important to not neglect momentum that gets transferred to air molecules due to air resistance.


 * At this point, any way, kinetic bombardment is largely a science fiction concept. The article only references one rumor that it might actually have been developed.  Red Act (talk) 19:27, 3 October 2009 (UTC)
 * But the terminal velocity is different for an object fired from a cannon. Considering the transit time from orbit to the surface of the planet could be as low as just a few seconds, the projectile would not encounter air resistance long enough for it to be a significant factor.


 * Not at the extreme velocities that would be involved in this example though. The amount of energy absorbed by the atmosphere might just be a tiny fraction of the total kinetic energy of the fragments. But, I don't think tungsten would break up.


 * No, it's a real concept. Do a search on Project Thor, or Rods from God. In any case, I feel like this discussion is more about whether orbital bombardment is possible, rather than if the projectile's kinetic energy would surpass the energy released by the nuclear fuse used to propel the projectile. ScienceApe (talk) 19:43, 3 October 2009 (UTC)

I wonder how much of the energy of the nuke would go into the kinetic energy of the cannon and escape as heat/whatever? My elementary physics tell that should be less than what gravity adds, ie 50 kg * 10 m / s^2 * (some thousands of kilometers? idk) = much less than a gigajoule (this is an absolute limit as gravity is already weaker than 10 m / s^2 at sea level no matter where you are and air resistance is ignored). A kiloton is ~ 4 terajoules. The answer to the original question is no, but it would be nice to have some numbers to support. --194.197.235.240 (talk) 20:54, 3 October 2009 (UTC)
 * Kinetic energy is given by $$ E = \frac{1}{2}mv^2$$. Thus, velocity $$ v = \sqrt{\frac{2E}{m}}$$.
 * According to our article about TNT equivalent, 1 kiloton of TNT is equal to 4.184 x 1012 joules.
 * So, for 50 kilotons of TNT with a 50 kg slug, and assuming that 10% of the energy from the explosion is converted to kinetic energy in the projectile, numbers become
 * $$ v = \sqrt{\frac{2 * 50(.1)(4.184 * 10^{12} joules)}{50 kg}}$$.
 * This works out to a muzzle velocity of approximately 914000 meters per second.
 * The troposphere contains approximately 75% of the Earth's atmospheric mass, and is 15km deep. To determine how long the projectile will be in contact with the troposphere, the relevant equation is
 * $$ x = x_0 + v_0t + \frac{1}{2}at^2$$
 * where $$x$$ is the height of the target above the ground, $$x_0$$ is the initial height of the projectile (in this case, zero), $$v_0$$ is the initial velocity of the projectile, $$a$$ is the acceleration due to gravity, and $$t$$ is the time required for the projectile to reach the target.
 * Plugging in our numbers, the equation becomes $$ 15000 meters = 0 meters + (914000\frac{meters}{second})t + \frac{1}{2}(9.81\frac{meters}{second^2})t^2$$
 * Digging out my TI-89....
 * $$ t = 0.0164 seconds$$
 * I do not remember what equations are used to determine the amount of energy frictional heating can impart into an object, but considering that the projectile will be in contact with troposphere (and hence 75% of the mass available to heat it up) for just 1.5 hundredths of a second, I cannot imagine that any reasonably aerodynamically-shaped projectile would break up in flight. It simply would have no time to heat up and start to ablate. J.delanoy gabs adds  22:04, 3 October 2009 (UTC)
 * Just a note on the math here: it looks like you've got the sign wrong for acceleration due to gravity, it will be opposite to that of the initial velocity. By my Casio fx-350D with the keys that don't always work anymore, that would actually be 0.224 sec in the troposphere. None of which really matters, 'cause at that velocity the result would likely be (as we all watched the object leave the solar system) "oops, let's try aiming at the boss's car in the next test". :) Franamax (talk) 01:00, 5 October 2009 (UTC)
 * Air resistance is proportional to velocity squared, so would be enormous at those velocities. The time being so short means there wouldn't be time to dissipate any heat, so I can imagine the object would get very hot. I can't see why the time being short would stop it having time to heat up - that's like the old joke "I have to write this letter quickly before my pen runs out." For future reference, units in TeX need to be written inside a \mathrm{...} command, otherwise you've actually written the product of m, e, t, e, r and s. --Tango (talk) 23:59, 3 October 2009 (UTC)


 * Energy lost due to air resistance is of order:


 * $${1 \over 2} \rho_{air} v^3 (\Delta t) A C_d$$


 * Taking A as 200 cm^2 (for a 50 kg slug of tungsten), Cd as 0.3 (somewhat rounded), and density of air as 0.5 kg / m^3 and your numbers for velocity and time gives: 1.9 J or roughly 90% of the energy you presumed it started with. It wouldn't so much get hot as it would explode on impact with the atmosphere.  And that's neglecting the hypersonic aspects of the problem.  Air resistance gets worse, not better, when one tries to travel at very high velocities.  Dragons flight (talk) 01:35, 4 October 2009 (UTC)
 * If it's losing that much energy then you can't assume constant velocity, as you seem to be doing. That means the energy loss will actually be less than you say. It will still be a very significant amount, though - I would expect enough to vaporise it before it hit the ground (I can't be bothered to calculate it...). --Tango (talk) 02:16, 4 October 2009 (UTC)
 * Well since I expect it to explode / totally disintegrate in the atmosphere, I'd say the energy loss would actually be 100%. Dragons flight (talk) 02:31, 4 October 2009 (UTC)


 * The maximum amount of speed that can be imparted by Earth's gravity = 11 km/s, Earth's escape velocity. Gravity trails off quickly with distance so you could impart almost as much without having to go too far from Earth. You could also use much of your already existing orbital speed (8 km/s) by bleeding off a little velocity, so coming in at an angle and do this from low earth orbit. The effects of nuclear explosions article says 40-50% of the energy is commonly in the blast wave. If you want Earth to be hit by more projectile energy than the bomb put in simply make it heavy enough that it can't go too fast. This would go hand in hand with the whole not destroy cannon thing but remember the gravitational part of the energy was put in my you so isn't free. Sagittarian Milky Way (talk) 22:16, 3 October 2009 (UTC)
 * What do you consider to be "too far"? Gravity in Low Earth Orbit is only a few percent lower than on the Earth's surface. While gravity does drop off quite quickly (with the square of distance), the altitude of LEO is very small compared to the radius of the Earth. If you go to Geostationary Orbit, then gravity is quite a lot lower (about 36 times lower, I think). If I've done the calculation correctly (somebody please check me!) if you fall from 200km straight down (so that's LEO but without the horizontal velocity) you will hit the ground at only 2km/s, quite a lot less than escape velocity. --Tango (talk) 23:59, 3 October 2009 (UTC)
 * Okay, subjective. But compared to the infinity actual distance.. And if measured in time and delta-v it isn't much more than what we do to reach objects very near us now (ISS, 2 days, which is routine). If dropping straight down is discarded you could just redirect your LEO which is already there so it intersects with your target, getting much speed without having to give much speed or start up high. (of course you gave that speed when you launched it) It'd come in at an angle, which has air resistance implications. Sagittarian Milky Way (talk) 03:04, 4 October 2009 (UTC)


 * The specific heat capacity of tungsten actually varies considerably with phase and temperature, but for just a rough order-of-magnitude calculation, I'll use a constant value of the 24 J mol-1 K-1 that's listed in the tungsten article. Given also the atomic weight, boiling point, heat of fusion and heat of vaporization listed in that article, I calculate that 50Kg of tungsten at 0 K can absorb on the order of 272 MJ of heat before it vaporizes.  In comparison, a 50 kiloton bomb releases close to a million times as much energy.  So it's very hard to see how you could reasonably propel a 50 Kg tungsten slug with a canon powered by a 50 kiloton atomic blast without the slug vaporizing.  Red Act (talk) 23:11, 3 October 2009 (UTC)


 * Your calculations aren't necessarily correct depending on the way the device is designed. Certain designs of Project Orion (nuclear propulsion) allow for a pusher plate that doesn't ablate much if at all, which shows that specific heat of a substance doesn't dictate whether it will "vaporize" or not if propelled by a nuclear explosion. And also, just because it vaporizes doesn't mean the matter that makes it up disappears. Conservation of mass applies here, so the particles that make it up will still be given tremendous kinetic energy. In any case, you aren't really understanding the point of the question. I'm simply asking if the kinetic energy of the mass that the cannon propels will be equal or greater than the energy released by the nuclear charge used to propel the mass. ScienceApe (talk) 15:41, 4 October 2009 (UTC)


 * In both cases, the energy the slug has would be far less than the energy of the nuclear blast. This is because most of the energy released by the blast would be in the form of heat and radiation, with only a small portion being kinetic energy, and only a small portion of that would then go into the slug.  As for the case of the slug fired toward Earth, the acceleration due to gravity would be totally insignificant, while the air resistance would be overwhelming.  Much like the Tunguska Event, the object would vaporize in the high atmosphere, creating heat, light, sound, possibly an EMP, and an atmospheric shock wave aimed at the ground.  Since it would be spread out over a very large area, though, the shock wave might not even be noticeable, from such a small slug, when it hit the ground.


 * You don't seem to be getting the way air resistance works, it's the total air you are displacing and the speed with which it is moved that determines how much energy is robbed from the projectile. A faster object moves just as much air, and accelerates it to an even faster speed, so loses even more energy.  As a thought experiment, imagine making the same ground trip in your car at 60 MPH or at 120 MPH.  Which would use more energy (gasoline) ?  The 120 MPH trip, of course, even though the time the trip takes is half as much.  The reason, just as in your Q, is all the air the car moves along with it, at either 60 or 120 MPH. StuRat (talk) 23:57, 4 October 2009 (UTC)


 * No, if the blast is in space, most of the energy would be in the form of X-rays. If it were a pure fusion bomb catalyzed by perhaps antimatter as is discussed in this report, http://arxiv.org/PS_cache/physics/pdf/0510/0510071v5.pdf most of the energy would be in highly energetic neutrons.


 * I'm not really interested in debating whether or not the slug would vaporize in an atmosphere, I simply used the example of the weapon being fired from space because I thought it might be slightly more realistic to use such a weapon in orbit, rather than in a terrestrial setting. But if you are still curious about the feasibility of kinetic bombardment, I suggest you do some searches on "Project Thor" and "Rods from God", both of which involve using tungsten slugs fired from orbit to targets on the ground. ScienceApe (talk) 01:52, 5 October 2009 (UTC)


 * The idea behind those projects is to use gravity to accelerate the slug. You would fire some kind of rocket to knock it out of orbit and it would then hit the ground at 11km/s (minus a bit for air resistance, which wouldn't be anywhere near as significant as for your design since the speed is about 100 times less and velocity is squared and cubed in the relevant formulae, and minus a bit since you aren't dropping from infinity, but you would probably drop from pretty high since the orbital velocity you have to shed would be less). A 50kg tungsten slug hitting the ground at 5km/s (a number I've made up, but it should be in the right order of magnitude) has an energy of 625 MJ or about 150kg of TNT. That's not on the scale of a nuclear explosion, but it's a big bang. --Tango (talk) 02:11, 5 October 2009 (UTC)


 * "No, if the blast is in space, most of the energy would be in the form of X-rays." I said radiation, which includes X-rays.  And you keep assuming that faster travel through the atmosphere means less heating, when everyone here has told you it's just the opposite.  At the speeds you're talking about, the slug would be vaporized in the high atmosphere.  Yes, if just dropped, it would survive, but that's due to the much lower speeds and therefore air resistance. StuRat (talk) 14:24, 5 October 2009 (UTC)


 * StuRat, as I already stated, I am not interested in debating whether or not the slug would vaporize in an atmosphere, but in any case I never said that faster travel through an atmosphere means less heating. Tango, what I'm interested in is the kinetic energy of a tungsten slug fired from an atomic cannon that I described in my first comment. Whether for the slug or the sum of its fragments if it doesn't survive the explosion. ScienceApe (talk) 20:09, 5 October 2009 (UTC)
 * I'm not sure exactly what question about kinetic energy hasn't already been answered. The original question was "would the slug impact the ground with energy greater than the 50 kt nuclear charge?"  The answer to that question is definitely not, regardless of whether or not the slug vaporizes.  If the slug vaporizes, the vast majority of whatever kinetic energy the nuclear explosion imparted to the slug will ultimately go into heating the air, over a large area around where the slug exploded.  Very little of the kinetic energy the slug had will make it all the way to the ground.  And the effect of gravity on the tungsten vapor will be negligible.  If the slug doesn't vaporize in the atmosphere, that implies that as of when the slug left the cannon, it was so slow and so cool that only a tiny fraction of the energy of the nuclear explosion ever got transferred to the slug.  The slug will gain some additional kinetic energy due to gravity, but nowhere near enough to compensate for all the energy of the nuclear explosion that didn't get transferred to the slug.


 * Later on, your question became "if the kinetic energy of the mass that the cannon propels will be equal or greater than the energy released by the nuclear charge used to propel the mass". Correct me if i'm wrong, but it sounds like at that point you're just asking about events in the near vicinity of the canon, correct?  If that's the question, the answer again is definitely not.  The nuclear explosion will be considerably less than 100% efficient in transferring its energy to the slug.  And if you're only asking about events in the near vicinity of the canon, the effect of gravity is negligible.


 * If the above doesn't answer your question, could you please be more precise about exactly what it is that you're asking? Like it sounds like maybe you want us to solve the problem while pretending that the atmosphere didn't exist, and pretending that tungsten has an infinite melting point?  Pretend problems like that are fine, but if that's the case, what efficiency would you like us to pretend the cannon has?  Red Act (talk) 22:00, 5 October 2009 (UTC)


 * Actually, the problem can basically be answered "definitely not", even under the assumptions that the earth doesn't have an atmosphere, and the melting point of tungsten is infinite. Even without any atmosphere, the energy gain due to gravity would be less than the energy of a 50kg mass at the escape velocity of 11km/s, which is 3GJ or 723kg of TNT.  So the impact energy would have to be less than the 50kt nuclear charge, unless the canon was considerably more than 98.5% efficient, which it realistically can't be.  It'd definitely be more efficient to just drop the nuke.  Red Act (talk) 23:20, 5 October 2009 (UTC)

I'm interested in knowing what the kinetic energy of the slug would be (a rough estimate is fine), and/or the muzzle velocity of the slug. For comparison sake, I would also like to know the kinetic energy of a bullet fired from an M-16 compared to the energy released from igniting the powder used to propel the bullet. How efficient is the powder at transferring kinetic energy to the bullet? Assume all takes place in a vacuum. ScienceApe (talk) 19:40, 6 October 2009 (UTC)

light
does light accelerate near black holes?117.197.208.7 (talk) 17:55, 3 October 2009 (UTC)
 * In a sense it does. It doesn't move any faster, but it gains energy, I think. Looie496 (talk) 18:11, 3 October 2009 (UTC)
 * Not really. Light always travels at the speed of light, a constant. It does gain or lose energy, but that results in a change of wavelength rather than a change of speed. See gravitational redshift. --Tango (talk) 18:12, 3 October 2009 (UTC)


 * Yes. You don't even need a black hole. Any gravity field will accelerate light. Amazingly light speed won't change when it is accelerated though its velocity, momentum, and energy will change. @tango: remember that speed and velocity are two different things Dauto (talk) 18:16, 3 October 2009 (UTC)
 * Good point. I interpreted the question as referring to a change of speed, and I expect that is how it was meant, but it isn't actually what was said. --Tango (talk) 18:25, 3 October 2009 (UTC)
 * Actually I'm going to raise some issues with that point. Your interpretation of lensing in a gravitational field is only valid for wave-treatment of light. Since we don't have a theory of quantum gravity we can't comment of the behavior of a photon in a gravitational field, other than to say in the classical limit, the path of a beam of photons will be bent. Elocute (talk) 19:30, 3 October 2009 (UTC)
 * We know empirically what light does in a gravitational field. It makes no difference how you interpret light, we know what we see. --Tango (talk) 19:39, 3 October 2009 (UTC)
 * Ah it does make a difference, because quantum particles don't accelerate per se, their vectors just in an exchange of what in this case would be the hypothetical graviton. I suppose if you consider an instantaneous change of velocity acceleration then they do accelerate, but they spend zero time actually accelerating. I did chuckle a bit at "We know what we see" as an approach to an inherently quantum problem. Elocute (talk) 19:48, 3 October 2009 (UTC)


 * That's such an inane point. I define the bending of the path as an acceleration. Dauto (talk) 20:04, 3 October 2009 (UTC)
 * Well I am sorry to have offended but all I was trying to add was that when using the more traditional definition of acceleration as dv/dt you will find that a=0 for all regions where a is well defined. Elocute (talk) 20:15, 3 October 2009 (UTC)
 * Obviously you need to average the change in velocity over some reasonable length of time. This isn't an inherently quantum problem. What you say holds just as well for billiard balls. They don't accelerate, the atoms sub-atomic particles (I suppose I should be precise, since you seem to want to be excessively pedantic) in them just exchange virtual photons, but obviously that is not a useful interpretation when playing snooker. It also isn't a useful interpretation in astrophysics. --Tango (talk) 20:17, 3 October 2009 (UTC)
 * The difficulty here is not quantum but general-relativistic. Dauto says "I define the bending of the path as an acceleration".  Fine, except in what sense does the path "bend"?  Locally, it does not bend; the light follows a geodesic in spacetime, which is a straight path, as best a straight path can be defined on a manifold of nonzero curvature.
 * If you refer everything to asymptotically flat spacetime removed from the gravity well, you can talk about the path "bending" in the sense that if you take a vector representing the photon going in to the well, and subject it to parallel transport through flat spacetime to where the photon comes out, you'll find they're no longer parallel. It may be possible to make a reasonable definition of "acceleration" from that; I'm not sure.  But the conceptual and definitional problems are not trivial. --Trovatore (talk) 20:25, 3 October 2009 (UTC)


 * You're making things more complicated than they have to be. The geodesic path followed by the light beam will only be perceived as a straight line by an observer in free fall. But why should we use an observer in free fall? The reasonable choice of observer is one that is inside of a rocket which has its engines on and poiting downwards in such a way that it is hovering above the black hole at a constant distance. That observer will see the light beam follow a curved path from which he can inferer that the light is indeed subjected to an acceleration. Dauto (talk) 21:52, 3 October 2009 (UTC)

I would assume the frequency would change, and become blue-shifted. ScienceApe (talk) 19:47, 3 October 2009 (UTC)

The answer is "not really", in that the light doesn't undergo any proper acceleration. It could be said to undergo a "coordinate acceleration", but that really just reflects an arbitrary choice of coordinate system. The light's proper velocity does not change, and the light travels along a perfectly straight line (a geodesic). Whether the light's energy and momentum appear to change depends on what choice of coordinate systems you choose to use at the beginning and at the end, which is arbitrary. Red Act (talk) 20:07, 3 October 2009 (UTC)
 * That's a property of gravity, rather than light, though. Nothing in free-fall experiences any proper acceleration. I think it is clear that we are talking about acceleration in some observer's frame, not the light's own frame. --Tango (talk) 20:14, 3 October 2009 (UTC)
 * I was about to make that point as well. Thank you Tango. In other words, massive particles also follow geodesics (though a different family of geodesics) and that does not keep us from saying that they do accelerate under gravity. Dauto (talk) 20:17, 3 October 2009 (UTC)
 * For simple problems, it can be convenient to think of a massive object as accelerating due to gravity. But that’s really treating the problem loosely.  In a more rigorous sense, what’s really happening is that the massive object is moving at a constant velocity.  It just appears to be accelerating, when viewed in an accelerating coordinate system, such as an accelerating coordinate system that’s stationary with respect to the surface of the earth.  Red Act (talk) 20:47, 3 October 2009 (UTC)
 * What do you mean by "some observer's frame"? There is no inertial frame of reference that encompasses the entire spacetime involved.  "Some observer's frame" isn't a well-defined concept in a gravitational problem like this.  Red Act (talk) 20:32, 3 October 2009 (UTC)
 * Of course it isn't well defined, I haven't specified the observer, but that doesn't really matter. "Inertial reference frame" is a concept from Special Relativity, it doesn't really apply to General Relativity, where all frames are equivalent. Obviously we can't encompass the entire spacetime since there is a singularity in it, but we can choose a useful co-ordinate system that works away from the black hole. --Tango (talk) 20:43, 3 October 2009 (UTC)
 * It’s not just the singularity of the black hole that causes a problem with “acceleration in some observer's frame, not the light's own frame”. There would still be a problem with that idea even if it was, say, a star that was affecting the light’s path.  There is no global coordinate system that behaves “nicely”, i.e., has spatial and temporal intervals that correspond to the coordinates in all directions, except in the region of spacetime that’s far enough from the gravitating object that it’s essentially flat.  In other words, the “nice” coordinate chart has to leave out the interesting part of spacetime, in which light is affected by the gravitating object.  Sure, if you only look at the light’s path before it enters the interesting part of spacetime, and after it leaves the interesting part of spacetime, the angles and positions are as if the interesting part of spacetime was flat, and the light’s path was bent, i.e., the light was accelerated.  But that’s not what really happened.  What really happened was that the light travelled in a straight line, on a curved spacetime.


 * If the coordinate system doesn’t ignore the interesting part of spacetime, then it’s necessary to make rather arbitrary choices as to what coordinate system you use, since due to the curvature of spacetime there is no coordinate system that has all of the features one might hope for. Schwarzschild coordinates are fairly popular, but there are some disadvantages to those coordinates, and various advantages to other coordinate choices, such as isotropic coordinates, Eddington coordinates, Gaussian polar coordinates, Kruskal–Szekeres coordinates, or Eddington–Finkelstein coordinates.  Since you’re forced to make an arbitrary choice of coordinates anyway, you might as well choose a coordinate system in which along the light’s path in question, only real (i.e., proper) acceleration appears as an acceleration, instead of dealing with a coordinate system in which there is an additional, unreal “acceleration” that’s really just an artifact of the choice of coordinates.  Red Act (talk) 04:05, 4 October 2009 (UTC)


 * Red Act, what you are saying is true, but the exactly same points are also true for the trajectory of a massive particle and that does not keep us from describing that as an acceleration. For consistency we should also describe what's happening to the light beam as a acceleration, even though the proper acceleration vanishes. Dauto (talk) 20:54, 3 October 2009 (UTC)
 * It’s more important to bring up the distinction between proper acceleration and mere coordinate acceleration in this problem than in a problem in which only massive objects are involved, because the constancy of the speed of light is such an important principle. I suppose if the OP is just wondering whether or not light is affected by gravity at all, then bringing up the distinction is just unnecessary pedantry.  But my impression is that the OP might be wondering why acceleration of light near a gravitating object doesn’t wind up violating light always traveling at exactly the speed of light.  If that’s what the OP is really wondering, then pointing out that light doesn’t really accelerate at all is the most accurate way of pointing out why the speed of light doesn’t get violated. Red Act (talk) 04:05, 4 October 2009 (UTC)

about photosynthesis,
I'm a grade 11 student from sri lanka. i faced a question in a paper and i wasnt able to find the answer for it. it asked whats the source for O2 which is produced in the process of photosynthesis. the answers were H2O CO2 NO3- CO3- please help me to solve this question. —Preceding unsigned comment added by 112.135.199.4 (talk) 18:50, 3 October 2009 (UTC)
 * Wikipedia isn't a school help place. Abce2 |  This is  not a test  18:51, 3 October 2009 (UTC)
 * Right, but more useful would be to point out out article on photosynthesis. --Stephan Schulz (talk) 19:04, 3 October 2009 (UTC)
 * Actually, the help desk is a "school help place". The fact that a question arises from something studied in school in no way invalidates the question. Our "we don't do homework" policy must be confusing someone. Asking for clarification of something you've read or been tested on in school is a good use of the reference desk and should be encouraged, not discouraged. That said, if our original questioner will look at the photosynthesis article, under the section on Light reactions, he (or she) will see that the overall equation for the portion of photosynthesis that produces oxygen is:
 * 2 H2O + 2 NADP+ + 2 ADP + 2 Pi + light → 2 NADPH + 2 H+ + 2 ATP + O2
 * The source of the O2 is H2O.
 * Regarding the other choices, the oxygen in CO2 (carbon dioxide) is incorporated into the sugar produced by photosynthesis in the light-independent or dark reaction of photosynthesis, but this reaction does not produce oxygen. While NO3 (nitrate) metabolism is important in plants, it has nothing to do with what the question is asking. And carbon trioxide (CO3) is an unstable molecule; even if the bicarbonate ion was meant, it also has nothing to do with photosynthesis. - Nunh-huh 19:44, 3 October 2009 (UTC)

Quantifying Color Blindness
Has their ever been a method devised for ascertaining the degree of a subject's color blindness?—so, for example, the color blind individual could then correct the saturation of certain hues in an image they've generated, allowing people with full color vision to see it correctly?

Alfonse Stompanato (talk) 20:17, 3 October 2009 (UTC)


 * There are varying degrees of Color blindness, as noted in the article. There are various tests for color blindness. One I've seen involves a field of dots with some of them colored differently to form the shape of a numeric digit. If you can't see the number, you have that type of color blindness. →Baseball Bugs What's up, Doc? carrots 20:36, 3 October 2009 (UTC)


 * See Color blindness and Ishihara color test for a start. 68.244.20.23 (talk) 21:20, 3 October 2009 (UTC)


 * The question is about the degree of colorblindness. The Ishihara test doesn't really do that with much precision.  I don't know of a test that does this precisely. SteveBaker (talk) 23:11, 3 October 2009 (UTC)


 * A color blind artist creates an image that looks ok to them. That means it resembles what they see in the world around them. If part of their system of 3 types of color sensitive cone cells is weak or anomalous they do not see, and can therefore hardly be expected to reproduce, some colored features. However there is no way to process the image to generate missing detail, nor any certain way to make it "more real" than the artist was able to make it. Compare this with the difficulty of reproducing for normal-sighted people what a color blind person actually perceives. Cuddlyable3 (talk) 23:24, 3 October 2009 (UTC)


 * er...the 2nd link in the .23 reply describes how the severity can be determined as the test progresses. It should be a good start for the OP with a couple of good internal and external links if they need to delve in further.  Its about the journey, not the destination for me.  So for the OP, I just point in the direction of TYBR.  173.100.133.44 (talk) 00:25, 4 October 2009 (UTC)

Why can't you treat carbon monoxide poisoning with a blood transfusion?
As far as I understand part of the problem during CO poisoning is that the CO binds to the hemoglobin preventing oxygen from doing so. I have often wondered whether this could be partially alleviated by replacing a proportion of the patient's blood with tranfused blood, thereby replacing the compromised hemoglobin with fresh hemoglobin. Having read Carbon_monoxide_poisoning there is no mention of any attempt to do this. I wonder, is it the case that doing so would not be useful because CO causes other problems (including binding to myoglobin) so that solving the hemoglobin issue alone would not be beneficial. Or is it that it could, in theory, be useful but has practical problems preventing its use. Thanks Illegal Cat (talk) 21:00, 3 October 2009 (UTC)
 * See Emergency! as Wikipedia cannot give medical advice. As far as the TV show goes, both Early and Brackett would issue out the perscription for 100% Oxygen (along with the generic: administer D5W (or Ringers'), monitor the patient and transport immediately). 72.58.9.205 (talk) 21:15, 3 October 2009 (UTC)
 * This clearly does not fall under Reference desk/Guidelines/Medical advice as I would hope that it is obvious that I am not asking for medical advice. As I have indicated above, I have read about the treatments (including 100% oxygen) that are given for CO poisoning and am asking why blood transfusion is not one of them. Illegal Cat (talk) 22:36, 3 October 2009 (UTC)
 * The practical problem, to which you alluded, is nothing more than "blood transfusion" is not a particularly speedy process. Factor that into your thinking?  --DaHorsesMouth (talk) 22:38, 3 October 2009 (UTC)
 * OK, but would it even be effective? It might be slow, but if it was in some way effective, then perhaps it would be used in bad cases, even if it was slow. After all, it seems as though the half-life of the bound CO is 80 minutes on 100% oxygen, so that is hardly speedy. It seems unlikely that speed is the only issue to not doing a transfusion. Perhaps the problem with clotting factor when too much blood is replaced would be a more substantial issue. Illegal Cat (talk) 23:04, 3 October 2009 (UTC)
 * No (to the 1st responder). Wikipedia does not give medical advice directly nor indirectly by endorsing a TV show. Proposals for improving the CO poisoning article are normally raised and discussed on the article Talk page. Cuddlyable3 (talk) 22:46, 3 October 2009 (UTC)
 * Sorry, you've lost me. Are you saying that you feel my question does fall under the "no medical advice" guidelines? As for the article's talk page, I am not suggesting an article improvement. I have, it appears, made up a treatment and I want to understand why it wouldn't work. Is that not a reasonable question to ask here, and one that does not fall it the realm of asking for medical advice? Illegal Cat (talk) 23:04, 3 October 2009 (UTC)
 * I just did a bit of research -- apparently the story is that although this idea has been suggested many times dating back to the nineteenth century, it doesn't actually work very well, because transfusing a patient with severe CO intoxication leads to a set of pathological reactions known collectively as "reperfusion syndrome" (see ), which are greatly reduced by hyperbaric oxygen. Looie496 (talk) 23:16, 3 October 2009 (UTC)
 * Thanks, I will read up on this effect. I suspected that there would be some kind of side effect, well either that or the fact that fixing the hemoglobin alone had little benefit. Illegal Cat (talk) 23:19, 3 October 2009 (UTC)
 * C3, the .205 answer was to what happened on the TV show when a CO case came up. I think I made that clear that I was reciting the plot not passing out triage chits.  I'm not a Doctor, but I have watched an actor play one on TV.  The episode (from Emergency!) I remember most was a knucklehead who wandered into the ER with his son after installing a new furnace.  I think it was Early on the horn with Roy(Back at the subjects house, with the rest of the family) on that one.  But it is interesting that he was spot on 35 years back w/ the current research.  Never did see them bring the poor dog out of the house, Hope the little guy made it.   70.4.152.122 (talk) 00:06, 4 October 2009 (UTC)
 * The procedure you are suggesting is called an exchange transfusion. Your question is good, and has been asked before, the latest I found was this letter to the editor of Chest, which advocates for further studies. At the time the letter was written (2002), the author was able to find only one published case report. --NorwegianBluetalk 08:01, 4 October 2009 (UTC)


 * Because carbon monoxide doesn't just inhibit hemoglobin oxygen binding, but also oxygen binding in cytochrome c oxidase and other enzymes, leading to cellular asphyxiation. It's not just the oxygen transport that gets interrupted. --Pykk (talk) 17:34, 4 October 2009 (UTC)


 * I suspect that the timing is the critical factor. That is, if the patient had bad enough CO poisoning to need a transfusion, they would be dead before it could be set up.  On the other hand, if the carbon monoxide poisoning is at a low enough level that there's time to set up a transfusion, then giving them pure oxygen is enough to save them, and transfusions, with all their added risks, are not needed.  StuRat (talk) 23:33, 4 October 2009 (UTC)