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October 9

Mucokinetics

The article on Mucokinetics says these drugs make mucus much easier for a patient to cough up (or I assume sneeze out). But what is done for patients who, due to injury or disease, are unable to cough up or sneeze ut mucus? Are there drugs that dissolve it so much they can easily swallow it?--178.167.189.6 (talk) 00:31, 9 October 2011 (UTC)[reply]

Normally, when you cough it up you do swallow it. You can't swallow it until you've got it out of your throat. The trachea (breathing tube) and oesophagus (eating tube) are only connected at the mouth. There are special massage techniques that can be used to move mucus up to the mouth. They are often used for people with cystic fibrosis. --Tango (talk) 01:17, 9 October 2011 (UTC)[reply]

Agree with Tango; I just want to add a link to mucociliary clearance, which I was taught to call the "mucociliary escalator". We normally swallow quite a bit of mucus, one of the reasons public spitting seems so unnecessary and selfish (no benefit, and some risk to others). -- Scray (talk) 03:20, 9 October 2011 (UTC)[reply]

Immune system strength

Is there any way or scale to measure immune system's strength (overall or against specific infections)? --178.182.107.62 (talk) 11:31, 9 October 2011 (UTC)[reply]

I am in no way an expert old chap, but I imagine there are two key criteria - one is the count of antibodies that are specific to the infective agent in question and second is the overall white blood cell count. The more antibodies the quicker they will latch on the infective agent and the more white blood cells the faster the agent will be attacked and destroyed. Quintessential British Gentleman (talk) 13:45, 9 October 2011 (UTC)[reply]

To measure reactivity against a specific antigen, the level of antibody activity is the most oftenly used parameter. If the response is primarily of the IgM isotype, it indicates that the infection may have taken place recently, if it is predominantly of the IgG isotype, it indicates that the response may be caused by an infection in the past. Some tests also measure cellular responses, such as the Mantoux test, which depends on T cell activity.

There is no simple way of presenting someone's "immune system's strength" as a number or anything like that. The immune system is complex, many things can go wrong, and it is when a patient has an abnormal frequency of infections, and especially of infections that are rarely seen in healthy people, that doctors will suspect that the individual has an immune deficiency. An incomple list of things to check:

Immunoglobulin (IgG, IgA, IgM and IgE) levels. Some doctors would also like to test the IgG subtypes at this point (IgG1, IgG2, IgG3, IgG4).
A blood count, to see whether the leves of lymphocytes, monocytes and granulocytes are normal.
A test of specific lymphocyte subsets (CD3, CD19, CD4, CD8, various combinations with other markers).
Measurement of complement levels and functional activity in both the classical, alternative, and Lectin complement activation pathway.
Depending on symptoms, various functional defects might be suspected. In chronic granulomatous disease there is a functional defect in the granulocytes, which can be measured.
Some lab's measure proliferative responses (mainly T cells) to antigens and mitogens, other's don't, and it is not universally agreed that this is a useful parameter.

I'm sure I've left something out. Check out the article Primary immunodeficiency, and its "diagnosis" section. --NorwegianBlue^talk 21:31, 9 October 2011 (UTC)[reply]

Weight-guessing

This is supposedly a skill displayed at county fairs and so on, though I've never actually witnessed it. How does one go about guessing a person's weight based solely on visual clues? I think it would be an interesting skill to have. 198.228.193.194 (talk) 14:44, 9 October 2011 (UTC)[reply]

I don't see much use, unless at places like county fairs. For describing people, categories like round figure, skinny, etc would be more than enough. Quest09 (talk) 15:19, 9 October 2011 (UTC)[reply]

I've not heard of guessing the weight of people at fairs - usually it's guessing the weight of cake. If you wanted to guess the weight of a person, you could do pretty well by guessing their height and then making a rough estimate of their build and comparing those against memorised numbers (perhaps based on Body Mass Index - if you can get a reasonably close guess of someone's height and BMI, you can calculate their weight). One thing to be careful of is that there are two ways to be heavy for your height - lots of fat or lots of muscle (or, I suppose, lots of both, but that's rare). You need to allow for both. --Tango (talk) 15:24, 9 October 2011 (UTC)[reply]

"Weight guessers" basically give out cheap prizes if they are wrong <g> The idea is that they actually try to be accurate on men, but always guess low on women -- who rarely are unhappy. They are entertainers, not more. That said, they can be within 10% on men just by looking at height and face ... Cheers. Collect (talk) 15:54, 9 October 2011 (UTC)[reply]

Weight guessing seems like a very fundamental capability which would be of great importance for some animals. I bet if you ran fMRI you'd find a special part of the brain assigned to do it, which I bet is much enlarged in raptors that snatch up their prey in flight. But it's hard to search for this in the literature... Wnt (talk) 16:45, 9 October 2011 (UTC)[reply]

It's clearly important, but I don't think they are all that accurate at it. One of the standard pieces of advice if you are threatened by a large predator is to try to make yourself look bigger. If animals were good at estimating weight, that wouldn't work. Looie496 (talk) 18:36, 9 October 2011 (UTC)[reply]

Hmmm, but is the animal estimating weight there? I think an animal that projects power further into the environment - for example, a porcupine or a boxer with a long reach - is more dangerous despite equal mass. Wnt (talk) 01:30, 10 October 2011 (UTC)[reply]

Even if there were an evolutionary advantage, you can be sure that evolution wasn't using pounds (or kilograms) when it was invented. A weight guessing person would still need to train themselves to express their judgments in the units we use. Dragons flight (talk) 18:49, 9 October 2011 (UTC)[reply]

Of course I'm just speculating, but the amygdala contributes to the verbal or written expression of emotions in humans, not (in a narrow sense) an option in other animals. Wnt (talk) 01:30, 10 October 2011 (UTC)[reply]

Here's a brief NPR segment on weight guessers, It doesn't really go into specifics, but it does touch on the idea that the prizes are worth less than the price to play the game. [1]

APL (talk) 04:53, 10 October 2011 (UTC)[reply]

Integrity of the Cryogenically Preserved Brain

When a rich person has one of the various companies in existence nowadays cryogenically preserve their brain, is there any evidence that the currently most-widely-used procedures succeed at maintaining the brain's network of synapses and their associated activation thresholds? Peter Michner (talk) 17:14, 9 October 2011 (UTC)[reply]

Has anything subject to that type of "preservation" ever been reanimated? ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:24, 9 October 2011 (UTC)[reply]

Yeah, frogs thaw out all the time naturally, but apparently large mammalian cells such as some connective tissue, spindle neurons and ova are subject to ice damage. Thawed mice remember how to run mazes, but the females are sterile and they all have substantially reduced lifespans. 64.134.157.164 (talk) 19:59, 9 October 2011 (UTC)[reply]

I don't see how there could be evidence, because currently we can't even come close to reactivating a brain that has been frozen -- far too much tissue damage. But in principle, nearly all the information ought still to be in there -- synapses are pretty large structures in molecular terms, and the receptor molecules that determine synapse activation thrshold ought to survive freezing pretty well. Looie496 (talk) 18:32, 9 October 2011 (UTC)[reply]

I suspect we can't even come close to reactivating the brain of a dead mammal in any case, whether or not it's been frozen (pace Mary Shelley). Tonywalton ^Talk 22:30, 9 October 2011 (UTC)[reply]

That would depend on your definition of death. If we use the old "heart has stopped" def, then plenty of people have been revived from death, brains included. StuRat (talk) 01:59, 10 October 2011 (UTC) [reply]

The question "does the technology to reactivate the brain currently exist?" is a different question than "is the information that is contained in the synapse network and the activation energy levels in each synapse preserved?" I don't pretend to think that the two factors alone in the latter question (which was the only question I was asking) are sufficient for retaining the information necessary to preserve a human mind, but I'd bet that those two things are necessary if it is at all possible. Peter Michner (talk) 22:58, 9 October 2011 (UTC)[reply]

I'm not sure that cryogenically preserved brains are any better than chemically preserved brains, as far as retaining the information in the brain. Has anyone ever compared the two ? I'd think that the risk of power failures (perhaps from the bankrupt company no longer paying the bill) over centuries would also be a major concern for cryogenically preserved brains. StuRat (talk) 02:03, 10 October 2011 (UTC)[reply]

What do you mean by chemical preservation? All humans who have been revived after hypothermic comas were not likely to have more than a little brain tissue frozen, and there was some brain damage for those who were partly frozen but the extent was varied and the effects weren't all permanent. I would have to review for information like long term memory loss. For extremities, we have frostbite which gives you some idea of the kind of damage to expect. 67.6.175.132 (talk) 19:14, 11 October 2011 (UTC)[reply]

I mean preserved in chemicals like formaldehyde. Obviously that will kill the cells, but if the goal is just to read the pattern of the brain cells and their connections, with some type of future scan, for the prupose of programming a computer brain to match, there's no reason the cells need to be alive. As for hypothermia, that's a long way from having your brain actually frozen solid. StuRat (talk) 17:19, 13 October 2011 (UTC)[reply]

arthritis

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

→Σ τ c. 19:33, 9 October 2011 (UTC)[reply]

Although, I don't think a healthcare professional would advice someone living in Florida to climb the Appalachian Mountains to get rid of his arthritis :) .Count Iblis (talk) 21:05, 9 October 2011 (UTC)[reply]

coin drop

if, say, we can dig a hole from north pole up to south pole, and drop a coin (or anything you would like to drop) in that hole, ignoring whats in earth's core, what will happen to the coin? — Preceding unsigned comment added by 203.112.82.128 (talk) 22:13, 9 October 2011 (UTC)[reply]

If you also ignore air drag, the coin will accelerate towards the center of the earth, overshoot and decelerate all the way to the oposite pole where it will poke it head (or tail) up and start falling again back to the original position. If you further assume a uniform density earth, the coin will be observed to follow a simple harmonic motion. --Dauto (talk) 22:29, 9 October 2011 (UTC)[reply]

(Edit conflict) If conditions were absolutely perfect, also assume there is a vacuum in the hole, the coin should accelerate to the centre of the earth. Whereupon reach the centre, it should undergo dampened occilation, before coming to a rest levitating at the centre of mass. Although conditions aren't perfect, which is to say that there is a variety of possible outcomes of how the experiment would fail. Like the centre of mass for the Earth does not remain in the same position over time, due to random fluctuations in density. Plasmic Physics (talk) 22:32, 9 October 2011 (UTC)[reply]

If you are assuming a vacuum, what causes the dampening? If the hole is full of air, then the coin will quickly reach terminal velocity, which for a coin is about 65 mph (according to Mythbusters). As it gets closer to the centre of the Earth, gravity will reduce (see shell theorem) and the density of the air will increase. That means the terminal velocity will reduce until it reaches zero at the centre. I'm not sure if the coin would slow down fast enough to avoid overshooting, but I wouldn't expect it to overshoot by much. --Tango (talk) 23:36, 9 October 2011 (UTC)[reply]

Dampening is caused is caused by a loss of momentum. Kinetic energy of the coin is converted to gravitational potential energy and as it overshoots the centre of mass. It is the same reason why a rubber ball bounces lower on each rebound. Plasmic Physics (talk) 00:02, 10 October 2011 (UTC)[reply]

Dampening occurs with your rubber ball due to inelastic collisions. Without losses to air resistance (or eddy currents) the gravity train acts more like a (magical) perfectly elastic superball which would keep on bouncing for ever. -- 49.230.105.185 (talk) 00:30, 10 October 2011 (UTC)[reply]

Other factors might cause dampening to occur: electromagnetically-induced currents in the surrounding material, transfer of gravitational momentum, vacuum energy pressures, etc. ~AH1 ^(discuss!) 01:08, 10 October 2011 (UTC)[reply]

Why would the air density increase? Tonywalton ^Talk 00:03, 10 October 2011 (UTC)[reply]

The immense gravity at that depth would squeeze any air to a very high pressure, and additionally heat it to above the boiling point of water, unless a near-vacuum is created. ~AH1 ^(discuss!) 01:08, 10 October 2011 (UTC)[reply]

Immense pressures, yes, but note that the greatest gravity occurs at the surface. -- 49.230.105.185 (talk) 01:38, 10 October 2011 (UTC)[reply]

(E.C.) Air pressure varies with elevation. The radius of the earth is significantly greater than the height of the substantial atmosphere, so it would seem that the air deep in the tunnel should liquify, but the critical point of nitrogen is 3.3978 MPa or only 34 bar (with T_cr = 126.19 K). With atmospheric pressure increasing by 0.5 bar in the last 6 km above sea level, 34 bar would be reached well before a depth of 34 bar * 0.5 bar / 6 km = 408 km (with the effect of the ever increasing heavier layers of air overhead greatly overwhelming the slight decrease in gravitation acceleration). From there on down the nitrogen would be a supercritical fluid. -- 49.230.105.185 (talk) 01:31, 10 October 2011 (UTC)[reply]

You might find the Gravity train article interesting. Tonywalton ^Talk 22:34, 9 October 2011 (UTC)[reply]

That article is missing an important piece of information: the maximum gravitational acceleration (it is different when the accelerated object is inside another object of mass that is exerting the gravity) assuming an infinite terminal velocity. For humans, determine whether an average body can survive the acceleration. ~AH1 ^(discuss!) 01:08, 10 October 2011 (UTC)[reply]

I'm not sure what you are implying here, AH1, but per the shell theorem (and square / cube considerations), the maximum gravitational acceleration will occur at the surface, and will decrease linearly to zero at the center. -- 49.230.105.185 (talk) 04:49, 10 October 2011 (UTC)[reply]

what would happen if i threw my mother-in-law in the hole? =D — Preceding unsigned comment added by 203.112.82.1 (talk) 23:51, 9 October 2011 (UTC)[reply]

You are describing a "gravity train" scenario. However, heat and atmospheric pressure/liquidity considerations are important. ~AH1 ^(discuss!) 01:08, 10 October 2011 (UTC)[reply]

In case there is air in that tunnel, the immense heat and pressures would likely slow down the coin as it spun around its own axis in the air, and very quickly vaporize. If vacuum engineering and other considerations fail to anticipate the launched object (or person!) arriving short of the surface target, retrieving the launchee may be a problem.

A question - in the case of a magnetically-levitated train, would any effects occur upon interaction with the Earth's polar magnet, and even considering a non-magnetic vacuum train: how does one ensure that the projectile does not hit the side of the tube, whatever material it is constructed of, and instantly vaporize? ~AH1 ^(discuss!) 01:08, 10 October 2011 (UTC)[reply]

Dampening is occuring to momentum being converted to waste heat. Work is done on the coin to draw it towards the centre of mass. As we all know, no thermodynamic process is perfect. Plasmic Physics (talk) 01:43, 10 October 2011 (UTC)[reply]

The whole atmosphere will be sucked into that tunnel. The pressure as a function of height is given by:

P(x)=P(0)\exp \left(-{\frac {mMG}{2kTR^{3}}}x^{2}\right)

where M is the mass of the Earth, m the average mass of an air molecule and R is the radius of the Earth. This is assuming constant temperature and validity of the ideal gas law. It follows from this that the total mass of air in the tunnel is given by:

M_{t}={\sqrt {\frac {2\pi mR^{3}}{MGkT}}}\exp \left({\frac {mMG}{2kTR}}\right)P_{\text{at}}A_{t}

where $P_{\text{at}}$ is the atmospheric pressure at the surface and $A_{t}$ is the cross sectional area of the tunnel. The total mass of the atmosphere times g equals the atmospheric pressure times 4 pi R^2. Before we build the tunnel, the atmospheric pressure is 1 bar, and that gives a total mass of about

M_{\text{at}}=5.28*10^{18}{\text{ kg}}

Then after the tunnel is completed, air will flow into the tunnel until the above formula for $M_{t}$ becomes consistent with an atmospheric pressure of $(M_{\text{at}}-M_{t})g/(4\pi R^{2})$ For a cross section of 1 m^2, this leads to an atmospheric pressure of about 10^(-151) bar. Count Iblis (talk) 04:12, 10 October 2011 (UTC)[reply]

Neither of your assumptions (isothermal and ideal gas law) seem reasonable to me. Dauto (talk) 04:39, 10 October 2011 (UTC)[reply]

They're not wrong by 150 orders of magnitude, though. You need to take into account what happens to all the matter you remove in order to create the hole, though - that will have the same volume as the air that replaces it. If you place it all on the surface, it will expand now it isn't under pressure and will take up quite a lot of space. I'm not quite sure what that would do to the atmospheric pressure, though. --Tango (talk) 11:37, 10 October 2011 (UTC)[reply]

Actually it is, once the pressure reaches the point that gas liquifies you aren't going to increase the density more than about 10 times beyond that. At that density less than 1 millionth of the atmosphere will actually fit in your 1 m^2 tunnel. Dragons flight (talk) 18:46, 10 October 2011 (UTC)[reply]

Yes, I agree that the estimate based on the ideal gas approximation is not appropriate. Count Iblis (talk) 14:49, 11 October 2011 (UTC)[reply]

Where can I see a pressure / density graph of super critical fluid nitrogen. Supercritical fluid#Phase diagram includes such a graph for CO₂ -- would one for N₂ be similarly shapes, albeit with different values? -- 110.49.225.244 (talk) 05:15, 11 October 2011 (UTC)[reply]

In any case, the coin should melt or even vaporize since the earth's core is thousands of degrees. Googlemeister (talk) 13:18, 10 October 2011 (UTC)[reply]

Well obviously the evacuated tunnel of unobtanium is cryogenically cooled. Dragons flight (talk) 18:35, 10 October 2011 (UTC)[reply]

How about Coriolis effect? I expect it will have an important role if the hole were through equator.--Almuhammedi (talk) 12:42, 12 October 2011 (UTC)[reply]

Sure. You're shooting an object through a tube that is also turning. If the tunnel is straight, the coin will hit the side. I'm not sure what path the coin would take if you cut the tunnel so that it wouldn't hit the side. The coin would be starting with pretty substantial angular momentum, so it would follow a sort of orbit, but the gravitational field inside the Earth is different than the familiar orbit setup. Rckrone (talk) 15:14, 12 October 2011 (UTC)[reply]

If we give the coin a fairly big (in two dimensions) cavity in which to maneuver, I think it should still describe an ellipse, based on the "simple harmonic motion" mentioned above. If r is the distance from the coin to the center, the force of gravity should be proportional to the mass within r (r^3), but diminish by r^2, thus gravity proportional to r. (Ignoring that the core is denser than the mantle) The difference is only that this ellipse has the center of the Earth at the center, whereas a Keplerian orbit has the center of the Earth at a focus! Seems like a lovely coincidence that surely must not be... Wnt (talk) 16:32, 12 October 2011 (UTC)[reply]

I once read a science fiction novel where this sort of thing was proposed, called a "bouncing orbit". It involved two masses joined above the north pole, falling straight down. At some point the two masses would electromagnetically repel each other, go around the Earth, and meet on the other side opposite the south pole (using magnetic repulsion to slow their approach and recharge the energy storage for the next separation). The center of mass would bounce up and down through the planet just as the picture of the coin above. The story's antagonist would hold the Earth hostage on every orbit, threatening to disable the separation phase if certain demands weren't met. I wish I could remember the title of that book. ~Amatulić (talk) 22:29, 13 October 2011 (UTC)[reply]