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July 3

How are microbial toxins broken down?

The advice for dealing with spoiled food is to throw it away, because the microbial toxins won't be destroyed by cooking it. The temperatures needed are so high that your food would be carbonized. But since microbes in Nature are constantly busy breaking down biological materials and producing the same sort of toxins, these toxins would accumulate without limit in Nature unless they do get broken down at room temperature by interacting with other chemicals. So, why can't we use such processes to deal with spoiled food or to slow down food spoilage? Count Iblis (talk) 00:42, 3 July 2017 (UTC)[reply]

Basically, you are looking at using a fungus to decompose another fungus or bacterium. That may not be all that appetizing. There is no universal solvent that can break down all toxins. Indeed, part of being a good toxin is not to be easily degradable. There's also the option of feeding slop to pigs, and eating the pigs, or fermenting the mash with yeast or bacteria, and using the ethanol for the usual purposes and solid wastes as compost or food, depending on the end products. The issue is too complicated to say that everything can be made into Soylent Green.

μηδείς (talk) 16:37, 3 July 2017 (UTC)[reply]

Microbial toxin is not all toxic or deadly. In yeast cells, ethanol is produced through fermentation when the oxygen is exhausted. This kills yeast cells, but it surely makes wine tastes like wine! 50.4.236.254 (talk) 01:05, 3 July 2017 (UTC)[reply]

There are several things to discuss here. First, there's no "universal toxin". Chemicals that are toxic to one organism can be completely harmless to another. Indeed, that's pretty much what antibiotics are; they're toxic to bacteria (and sometimes other microbes) but not to you. Caffeine and nicotine are deadly poisons to insects; to humans they're stimulants. Also, not all decomposition is done by organisms. Complex organic molecules left alone will slowly degrade over time from heat, light, oxidation, etc. For instance, fats rancidify. So yes, you could leave spoiled food alone to slowly let toxins degrade, but everything else in the food will degrade as well, and you'll have a disgusting soup of organic molecules. We slow down food spoilage by preventing or inhibiting the aforementioned processes. --47.138.161.183 (talk) 19:34, 3 July 2017 (UTC)[reply]

Social norms and Asperger's

Is willfully breaking written social rules ("Say excuse me after you burp", "Don't chew with your mouth open", "Use your inside voice", "Take your hat off inside a building", "Don't go naked in public", "Don't talk about sex/feces/genitalia in public", "Don't cross-dress", "Don't use the wrong gender restroom", etc.) symptomatic of Asperger's syndrome? Barney the Dinah Shore (talk) 03:16, 3 July 2017 (UTC)[reply]

Not necessarily. Some of those things can be symptoms of mental illness. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:16, 3 July 2017 (UTC)[reply]

Hmmm. I could believe that a perceived need to say something after burping or to take a hat off when entering a building could be viewed as a form of obsessive-compulsive disorder... Wnt (talk) 11:08, 3 July 2017 (UTC)[reply]

By the way... which one is the 'wrong' gender? - Nunh-huh 04:46, 3 July 2017 (UTC)[reply]

One other than the one you routinely use. — Preceding unsigned comment added by Baseball Bugs (talk • contribs) 07:34, 3 July 2017 (UTC)[reply]

Symptoms are felt or experienced by the patient, signs are noted by the clinician or other person. The activities in the question may also be present in someone who is confused or disinhibited by intoxication from alcohol or other substance. Richard Avery (talk) 07:16, 3 July 2017 (UTC)[reply]

Sure, and it's situational too. Going nude in public is not only OK but is encouraged... on a clothing-optional beach. ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:39, 3 July 2017 (UTC)[reply]

What does "inside voice" mean? ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:36, 3 July 2017 (UTC)[reply]

Volume (and perhaps pitch) appropriate in a confined space. —Tamfang (talk) 08:24, 3 July 2017 (UTC)[reply]

Also, there is a school of thought that you should not apologize after burping, because that merely draws further attention to it, and hence is more rude. ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:38, 3 July 2017 (UTC)[reply]

Id say in most cases its primarily symptomatic for being a Punk. --Kharon (talk) 19:53, 3 July 2017 (UTC)[reply]

From personal experience, few people with Aspergers wilfully break social norms. They simply, but genuinely, don't intuitively pick up the norms and cues. "Ordinary" people pick up such cues intuitively. (Explicit and formal lessons on Etiquette are not the general norm is most western countries). A wilful, deliberate, and conscious departure from accepted social norms would not be "the norm", for the most part, for an individual with Aspergers. They are often genuinely oblivious to the non-conforming nature of their behaviour. 2001:8003:5333:8C00:DD9A:DA00:429F:8475 (talk) 02:37, 5 July 2017 (UTC)[reply]

BJT voltage divider bias

I made an small experiment (in a simulator) with different combinations of resistors. In particular, the combination of 10k/5.49k, 100k/59k and 330k/240k resistors (the collector resistor and the emitter resistor were 2.2k and 1k respectively in all 3 cases) all biased the transistor to a collector voltage of 2.5V (5Vcc supply.) The base current, too, was roughly the same at 2.48uA, so was the collector current (roughly 1.18mA-1.36mA.) What, then, is the principal difference and is there a rule how to choose bias resistors? I know voltage divider bias calculations are done using equivalent voltage (the actual voltage at the junction between the two resistors and disconnected base) and equivalent resistance (R1||R2), but these are all different (1.77V/3.54k, 1.8V/37k, 2.1V/138k respectively) 78.53.24.21 (talk) 10:09, 3 July 2017 (UTC)[reply]

PS BJTs are capital-W weird. Actual EEs, as opposed to hobbyists like myself, I wonder, what is their ratio of intutive understanding vs "shut up and do the math"? And what is the math? 78.53.24.21 (talk) 10:30, 3 July 2017 (UTC)[reply]

The base current is I_collector/h_FE and very small, just a few µA (microamps). So when the current in the base bias resistors is much larger (in your examples 323, 31, 8.8 µA respectively) the voltage drop in the equivalent resistance (0.009, 0.09, 0.34 V) is negligible. The conclusion is that using low value resistors in the divider gives no advantage. Using higher value resistors saves wasted current, which could translate into slightly longer battery life in portable equipment. However using extremely high resistances would make the biasing more dependant on h_FE which varies with temperature and is not exactly specified in ordinary transistors. See the article Bipolar junction transistor, particularly Bipolar junction transistor#h-parameter model. Blooteuth (talk) 12:17, 3 July 2017 (UTC)[reply]

Thorns on opium poppy engineered or natural?

I can understand why man would have bred or engineered the thorns on an opium poppy plant. Is their evidence for this? thank you — Preceding unsigned comment added by Rextrent2 (talk • contribs) 11:27, 3 July 2017 (UTC)[reply]

Use of Papaver somniferum, the opium poppy was known as far back as Sumer 4000 BC and the ancient Minoans. The stem and leaves are sparsely covered with coarse hairs. I can't understand why any poppy farmer would want to breed thorns on his crop. Blooteuth (talk) 12:28, 3 July 2017 (UTC)[reply]

There are poppies with thorns but not the opium poppy. Dmcq (talk) 12:35, 3 July 2017 (UTC)[reply]

The Sacramento prickly poppy for example. Alansplodge (talk) 19:18, 3 July 2017 (UTC)[reply]

Big cylinder in the gravity well of a black hole

Suppose one had a magically strong 400 mile long cylinder, made up of over 1000 decks each 600 metres high, just outside the event horizon of a black hole, trying to reverse its way out, but making only very very slow progress, so effectively almost stationary.

(Any resemblence to events in the most recent two episodes of a major television franchise entirely non-coincidental).

For those not in the UK, this question was almost certainly triggered by a recently broadcast Doctor Who episode. Bazza (talk) 14:40, 3 July 2017 (UTC)[reply]

Events on the flight deck (level 0, nearest the black hole) appear time-dilated by a factor of 150,000 when viewed on closed-circuit tv from floor 1056, four hundred miles away.

(i) is it possible to calculate (or guesstimate) the relative time dilation between floor 507 and floor 1056 ? I'm guessing the relationship is not linear with distance, so rather a lot less ?

(ii) how would signals from floor 1056 (eg transmitted by a remote life-signs monitor on that floor) appear when viewed at floor 0 ? Would they appear sped-up by a factor of 150,000 ? (Related question: if an observer in free fall crosses the event-horizon, do they see the universe behind them become incredibly blue-shifted -- or is the transition for them apparently imperceptible, with clock-signals from outside still reaching them at a steady rate; since it is only for the outside observer that they will have appeared to have become red-shifted to eternity as they approach the horizon?)

ALSO: What other bodies are buried in this scenario?

-- eg: What is the local gravitational acceleration that the ship is resisting, and that any bodies on its decks would be subjected to if it wasn't for magical artificial gravity ?

-- if our cylindrical object were rotating (a primitive way to make artificial gravity, here replaced by the more advanced magical method), would the difference in time dilation mean enormous twisting forces on the structure -- one end having rotated appropriately for 2 1/2 days, the other for 10,000 years?

-- lifts can be operated between floors. If they run at a steady local velocity, how much more quickly than usual does the journey from 400 miles away appear to take, when viewed from floor 0 ? How magical do they have to be, to extract mass from so close to the event horizon to 400 miles away?

Thanks for any thoughts, Jheald (talk) 11:34, 3 July 2017 (UTC)[reply]

From Gravitational time dilation there is this simplified formula:

t_{0}=t_{f}{\sqrt {1-{\frac {r_{s}}{r}}}}

where

$t_{0}$ is the proper time between events A and B for a slow-ticking observer within the gravitational field,
$t_{f}$ is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate),
$r$ is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate),
$c$ is the speed of light, and
$r_{s}=2GM/c^{2}$ is the Schwarzschild radius of $M$ .

So you can calculate different t0 values for the different r values. With such a big time dilation, the level 0 is really close to the event horizon, and the energy to lift anything from that up is close to the rest mass of the object*c². Graeme Bartlett (talk) 11:52, 3 July 2017 (UTC)[reply]

@Graeme Bartlett: Thanks.

So, to plug in some numbers:

If we approximate time on floor 1056 as approximately undilated, then r₀ ≈ r_s (1 + 1 / 2.25 x 10¹⁰) -- so, as you say, very very close to the event horizon; with the slower that time is running on floor 1056, the even nearer that floor zero must be.

From Schwarzschild radius, r_s for a one solar-mass black hole is about 3km, and floor zero would be within about one tenth of a micron of the event horizon (which would be a bit tight), with time running at 99.7% of normal on floor 1056, and 99.5% of normal on floor 507.

At the other end of the spectrum, from the same article, the supermassive black hole at galactic centre has r_s = 13.3 million km. 640 km beyond this, time would be running at 0.7% of its speed in free space, while on floor 507 it would be about 0.5% -- the most you can achieve in halving the distance to the event horizon is a reduction of about 70% (ie 1/√2) in the rate of time.

For this mass, for a time dilation of 0.007 / 150000, floor zero needs to be within a fraction of 2 x 10¹⁵ of the event horizon -- about 7 microns.

So the maths doesn't entirely work out.

A couple more numbers: a time dilation of γ = 150,000 corresponds to an escape velocity β = 0.99999999998 of the speed of light; any free photons would have to be aimed very directly away from the black hole to get out.

I'm not sure whether the formula for gravitational acceleration g = 0.5 r_s (c / r)² still applies in such a strongly relativisitic environment; but plugging numbers into it gives 1.5 x 10¹² earth gravity for the event horizon of the one solar-mass black hole; and still 3 million g for the super-massive black hole -- so the magic cylinder (and magic artificial gravity system) do indeed have to be quite strong. Jheald (talk) 15:32, 3 July 2017 (UTC)[reply]

Event horizon says that the local gravitational acceleration for an object in fact tends to infinity (presumably as a consequence of now including relativistic corrections), the nearer an apparently stationary object is to the horizon. My thanks to User:Gem fr for the pointer. Jheald (talk) 15:52, 3 July 2017 (UTC)[reply]

I am still curious as to what the rest of the universe looks like to a free-fall observer crossing the event horizon. From the outside it seems the observer never crosses the horizon -- their time slows down to nothing, and they appear to remain there forever (though red-shifted to oblivion), along with every clock-pulse sent in from outside. But from their own perspective, this is surely not right. They (classically) would continue to fall on the far side of the horizon, and clock-pulses from outside would continue to reach them at a steady (finite) rate. Jheald (talk) 13:11, 3 July 2017 (UTC)[reply]

When I crunched some numbers I got 0.01mm from a 1000 km radius black hole. So just consider this a work of fiction, as the acceleration would vary too much over level 0. Most of the dilation variation would just be close to level 0. You may be interested in Ehrenfest paradox involving rotation in relativity. I can assure you that no material will support the strength required of your cylinder. No need to worry about it twisting either! Graeme Bartlett (talk) 13:26, 3 July 2017 (UTC)[reply]

event horizon is a so tricky thing, i confess not understanding it. So I just read event horizon#Interacting with an event horizon, which may be not perfect, but clear enough. Gem fr (talk) 15:29, 3 July 2017 (UTC)[reply]

The way you put things, you suppose some sort of fixed, absolute event horizon of the black hole, and some way to measure some absolute distance from this fixed boundary. This is not the way relativity works. There is no such thing as a single "event horizon of the black hole", it is all relative to the observer, and is not the same for a man A on the deck 0 as it is for a man B on deck 1. When vertically aligned, man A and man B won't find each other at the same distance and their clocks won't run at the same speed. They are engaged in some sort of twin paradox experiment

redshift is not absolute, either, and its' reverse is not a "blue shift", but ... another red shift ! Objects appear redshifted when they seem accelerated away, so if A see B redshifted, so does B see A redshifted, too. (I am not 100% sure, i confess)

Gem fr (talk) 15:55, 3 July 2017 (UTC)[reply]

See Schwarzschild radius. It is not dependent on the speed or location of the observer. It is dependent on the mass of the object. 209.149.113.5 (talk) 16:47, 3 July 2017 (UTC)[reply]

except that, Mass in relativity depends on speed Gem fr (talk) 08:30, 4 July 2017 (UTC)[reply]

Assume you are correct. The speed of an observer alters the mass of the black hole. So, the faster the observer, the less mass the black hole has, and the Shwarzschild radius shrinks. Light travels so fast that black holes would lose so much mass that the Shwarzschild radius would shrink to the point that light would have no trouble escaping the black hole. Where do we observe that happening? I can't find any reference to it happening. Instead, I see that Shwarzschild radius is not dependent on the observer. It is dependent on the mass of the object. In relation to the question here, the ship outside the black hole is the observer. The black hole is the object and, in relation to the observers at opposite ends of the ship, it is sitting still. 209.149.113.5 (talk) 18:02, 6 July 2017 (UTC)[reply]

Suppose the black hole is moving at 0.5c relative to you. Everything around the black hole seems flattened in the direction of motion by Lorentz contraction. Does the Schwarzschild radius also seem flattened? If so, then there are probably some details that article glosses over ... if not, double. Wnt (talk) 18:05, 3 July 2017 (UTC)[reply]

Thanks @Gem fr:, anon, and @Wnt:. Like Gem, my instincts -- very much formed from Special Relativity and the twin paradox -- were to feel that the opposite of redshift would not be straightforward blueshift, but something else. However, SR is very different from GR, so I'm not sure how much I trust those first instincts any more; I'm certainly more aware now that I don't have much feel (yet) for GR.

There's also a clear difference between the twin paradox, which is symmetric between the two observers; and the gravitational time dilation, which arises from their having very different proper accelerations (i.e. accelerations with respect to free fall). The gravitational case also differs because it (might) be symmetric with respect to time-reversal, which the SR one isn't -- the signals the twins see moving away aren't the same as what they see re-approaching each other.

But I think the link event horizon#Interacting with an event horizon, that you gave above, is very useful, and it smells right to me, at least so far as it goes.

It says that "observers stationary with respect to a distant object will all agree on where the horizon is"; and that the proper distance to the horizon is finite. (The proper distance might even be the distance between r and r_s -- though I suspect the definitions of the radiuses r may be a bit more subtle than that).

On the other hand, if there's time dilation, then the apparent distance to somewhere near the horizon (eg measured by the time taken for a photon to get there and come back) will appear to tend to infinity. This is not a contradiction, but does appear to indicate that the apparent height of each level (as observed from eg level 1056 -- but not as observed from a lift going slowly up the ship) will appear to get higher and higher.

I *think* photons going down the gravity well get increasingly blue-shifted, as seen by stationary observers on each deck they pass, corresponding to energy they have gained, falling down the gravity well. The rate of them also increases, as more seconds of emissions from a lower deck are being mapped to a single second on the higher deck. So there's a twofold effect, subjecting the higher decks nearer the horizon to an increasing number of increasingly hard X-ray particles. As seen from the lower deck, I think the wavelength appears the same, but in this perspective an increased number of particles fit into the apparently distended space between each deck. The two perspectives would be equivalent.

This, as I understand it, is ~~what leads to the controversial idea that there should be a "black hole firewall"~~ - a very very high temperature region adjacent to the event horizon.

No: BH firewall is something else, (and, on the face of it, very very ad-hoc). Jheald (talk) 20:22, 3 July 2017 (UTC)[reply]

But this would only be perceived by an apparently static observer, hovering close to the event horizon. As I now understand it, the perception of an observer in free-fall, not experiencing the enormous proper acceleration of the static observer, would be very very different -- and so this is how, per my original post above, they continue to hear clock pulses from outside reaching them at a steady rate, with no essential difference made by crossing the horizon. Jheald (talk) 19:32, 3 July 2017 (UTC)[reply]

Twin_paradox#What_it_looks_like:_the_relativistic_Doppler_shift is of interest here; of course there is no "return trip", but the difference of pattern seen from the static to the traveling, and from the traveling to the static, illustrate how tricky things are. Also, observe how long it take to explains just a part of it.

Bottom line: i think we should close the question, we already gave the most relevant links, going further is above what can be expected of reference desk (and, my own expertise, AFAIAC)

Gem fr (talk) 10:15, 4 July 2017 (UTC)[reply]

Regarding the Lorentz contraction issue, I found [1][2] which sort of address it. The event horizon is fixed in space, apparently, because it is determined by the stress-energy tensor that is a property of space, not observer motion. The Lorentz contraction doesn't affect what is in or out of the hole... so it must be able to affect the shape of the event horizon. I think. Wnt (talk) 15:37, 6 July 2017 (UTC)[reply]

If something were fixed in space, it would be a perfect absolute frame of reference, while the whole purpose of relativity is to cope with the absence of such. And of course something fixed just cannot change shape, so what do you mean?

Gem fr (talk) 09:30, 7 July 2017 (UTC)[reply]

See absolute horizon and its companion article apparent horizon. Gandalf61 (talk) 12:51, 7 July 2017 (UTC)[reply]

Recent weather trend

This indicates that the recent hot weather pattern in southern Europe is also man-made. Does this mean that the recent thunderstorm/hail and cloudburst pattern in Europe was caused by heavy evaporation from that heat which formed rain clouds that traveled north? This trend covers a vast area roughly from Germany to Russia. In this month snow also reportedly fell in Murmansk and some other Russian cities. Brandmeister^talk 12:14, 3 July 2017 (UTC)[reply]

Weather is to complex or chaotic to prove a direct correlation between longterm influence and specific local events but its trivial to conclude in general that more heat implies more energy, which implies more extreme weather events. --Kharon (talk) 19:39, 3 July 2017 (UTC)[reply]

it's trivial but it is ... utterly wrong xD

heat and temperature are very different things. Actually, in real life more air temperature usually means LESS atmosphere energy: The air above Sahara as less energy than the air above London, even though it is at quite higher temperature (London air as much higher latent energy of water condensation, than more than make up for the minute energy difference caused by difference in temperature). Foehn wind increase temperature while losing energy. A storm is always much lower temperature than calm region around it. Energy is also a function of pressure, so, for a same temperature and water content, anticyclone have more energy than cyclone, however anticyclone are not associated with extreme weather event, cyclone are. Etc.

and of course, we are talking about an average 1% energy change of every day energy change. That is to say, if everyday change is 0 to 1 and back to 0 again, instead of averaging 0.5, it would average 0.51. No statistical tool is able to link a so minute change to rare events in a so badly known, so naturally noisy, so chaotic, process as weather.

Hence the cautious formulation of IPCC, that can be sum up as "we think it is probably/likely/verylikely so and so (but we cannot prove it)"

Gem fr (talk) 09:47, 4 July 2017 (UTC)[reply]

In meteorology and climatology we have the concepts of dry static energy and moist static energy to quantify the contributions of temperature, elevation, and moisture content to the total energy of an air parcel. So it is more correct to say that the energy content of air depends on both temperature and moisture (and on elevation).

Returning to the example given by Gem fr, presently the surface the air in London (Northolt) has a temperature of 19.1 C and relative humidity of 62%. This yields a moist static energy value of 315 kJ/kg. At Jeddah the temperature is 31.2 C and the relative humidity is 40%, for MSE of 334 kJ/kg. Shock Brigade Harvester Boris (talk) 01:14, 5 July 2017 (UTC)[reply]

The same trend of bad things can be observed in media about just anything: economics, terror and war, disease, etc. This is called sensationalism. I don't see anything notable in snow in Murmansk#climate, i guess the average 0.5 snow day in June is like some year without any and a few days of snow a year in a while ; this happened this year, not previous year and probably not the next (but media won't talk about it, will they?).

Humans always need scapegoat for bad things, they just don't accept that Nature/gods are lunatics, when something bad happens someone must have offended them. That's specially true of weather. Plagues of Egypt includes meteorologic events, caused by Pharaoh's unwillingness to obey God. You'll find many instance of human sacrifice such like Iphigenia's to appease gods and get back "normal" weather.

(the above was added by someone else)

This kind of question is not very useful, as it provokes a straw man asking whether lower emissions would've led to a lower temperatures in Germany, higher temperatures in Murmansk, etc., which is essentially asking to backtrack and remodel the very thing that inspired the "butterfly effect", when the base question of whether the Earth is warmer now than it was a few decades/centuries ago, to which the answer is a unanimous "yes"[3]. 93.136.60.225 (talk) 00:43, 5 July 2017 (UTC)[reply]

There are some relevant issues like whether warming affects the size of the Hadley cell or causes the jet stream to become more variable. But if it increases the potential variety of weather slightly, that doesn't prove that any one event is "due" to it. Causality is something of a religious concept, and assigning cause can get iffy. I mean, if someone asks the dealer to give the deck one more shuffle and you lose the card game, did that person cause you to lose? Even if the deck really was better shuffled? Wnt (talk) 15:41, 6 July 2017 (UTC)[reply]

I find it quite strange to read that "Causality is something of a religious concept". Causality has its own article.

Obviously we are not talking about a cause, but about a causal factor, among many other. Did a shuffle more or less change your probability to win or lose? We are talking about something more similar to adding a 10 in a deck of blackjack: it obviously change your chance to blackjack, but if blackjack occurs you just cannot say "it's because the 10 was added"

Gem fr (talk) 09:21, 7 July 2017 (UTC)[reply]

Death dreams (an experience of the dying brain not a normal dream about death)

I watched Jacob's Ladder a long time ago and became aware of the concept of a 'death dream' wherein a dying brain depolarises in a less-controlled way than usual, resulting in a surreal and possibly unpleasant experience for the person dying. It seems like Wikipedia doesn't have an article on the concept of a death dream. Is there any credible information about the concept or is it just a vague idea referred to in some movies? --145.255.244.249 (talk) 12:48, 3 July 2017 (UTC)[reply]

A possibly related concept might be effects experienced by those who lived a near death experience, but that may not be enough damage to really be equivalent... There's of course also "meat's dream"

. —Paleo Neonate - 12:56, 3 July 2017 (UTC)[reply]

I wouldn't call it a "concept." I would call it an 80s movie trope. 209.149.113.5 (talk) 14:32, 3 July 2017 (UTC)[reply]

the concept is much older. See An Occurrence at Owl Creek Bridge. -Arch dude (talk) 18:49, 3 July 2017 (UTC)[reply]

Since there is never recovery from brain death, there can't really be any credible information. As PaleoNeonate said, the closest thing you get is a near death experience. Looie496 (talk) 16:04, 3 July 2017 (UTC)[reply]

That article actually claims you can have a flat EEG from deep anaesthesia or cardiac arrest. I should look into this... obviously there is no logically essential reason why all the neurons couldn't just stop at once for a bit without dying, so it's just a matter of when and to what degree it can be observed in practice. Wnt (talk) 18:10, 3 July 2017 (UTC)[reply]

Which reminds me of Flatliners... —Paleo Neonate - 07:01, 4 July 2017 (UTC)[reply]

Do elephants get boogers in their trunks?

Topic says it all. ScienceApe (talk) 17:30, 3 July 2017 (UTC)[reply]

I would suspect that when elephants are dehydrated the mucus in their trunks would thicken, yes. Note that when they find a body of water, they suck water up through their trunks and spray it back out, and this may serve to clean it out. StuRat (talk) 17:44, 3 July 2017 (UTC)[reply]

But they are probably very, very thin, given the length. --Kharon (talk) 19:26, 3 July 2017 (UTC)[reply]

An example. 209.149.113.5 (talk) 19:31, 3 July 2017 (UTC)[reply]

Pressure on the ground, speed of vehicle

How does the speed of a vehicle correlates to the pressure of its tires on the ground? --Hofhof (talk) 17:46, 3 July 2017 (UTC)[reply]

No direct correlation. See Friction#Dry_friction. The normal force, which is approximately the same as the weight of the vehicle on level ground, under most conditions, does relate to the frictional force, which limits the acceleration of a wheel-driven vehicle and also can limit the max speed, as the normal force needs to be sufficient for the wheels to exert enough frictional force to overcome the aerodynamic drag of wind resistance. The pressure on the area of tires in contact with the ground is roughly the normal force divided by the contact area. So, there is distant relationship between the two, but nothing direct. StuRat (talk) 17:57, 3 July 2017 (UTC)[reply]

As above, no direct correlation in most situations and under most reasonable assumptions. However, some vehicles (including cars with a functional spoiler instead of a decorative spoiler) have an aerodynamic shape that uses airflow to push the vehicle toward the ground, resulting in more pressure as the speed increases. Also, a vehicle going over a hill will see increases and decreases in the vertical component of acceleration, These vary with speed and result in varying pressure. At the extreme, if you pop over a hill you will leave the ground completely, resulting in no pressure. -Arch dude (talk) 18:38, 3 July 2017 (UTC)[reply]
In theory, if your speed is high enough and you are on an astronomical body with a small enough radius and low enough gravity, an increase in speed will reduce your effective weight and therefore the pressure as centripetal acceleration counteracts gravity. This effect is probably not observable on Earth or even on the Moon, but would be appreciable on small bodies such as the moons of Mars or small asteroids. -Arch dude (talk) 18:44, 3 July 2017 (UTC)[reply]

Short: The answer depends on how the car is build. If it has wings the pressure will fall as the vehicle gains speed. If it has spoilers pressure will rise with speed. --Kharon (talk) 19:23, 3 July 2017 (UTC)[reply]

I beg to differ. Above considerations only apply to a perfect road, where ... no tire would be needed! In real life, you need tires, with a correct pressure, and you need them all the more than you speedy drive. When you drive on a speed bump slowly, the pressure barely change in the tire; if you do it with speed, a dramatic spike in pressure will occur.

since pressure is a direct measure of energy, and energy correlates with square of speed, i think that pressure variation in tires correlates with square of the speed of the vehicle. Variation, of course

Also, it is well known that pressure is increased by heat, and that heat is increased by speed. I.e., pressure is greater when you drove for a length of time at, says, 30m/s (108km/h), than if you did at 10m/s (36km/h)

I also expect this effect to be of the square kind (but different)

Now, a proper answer would need to be a tire designer

Gem fr (talk) 10:55, 4 July 2017 (UTC)[reply]

In a way this seems a very simple question: aside from the aforementioned aerofoil effects, and presuming the car's orbital velocity is also negligible, the car has to rest on its tires. The pressure of the tires on the ground x the area of the tires will have to equal its weight. Even if the car is bouncing up and down, the pressure on the tires would have to match its weight plus its current upward/downward acceleration (caveat being that the actual center of mass must be considered, which includes parts in the suspension that are meant to move so that you don't). Of course, the pressure of the tires on the ground is not the same as the pressure in the tires: you can touch a firmly inflated tire gently to the ground, while a run-flat tire might not have pressure and it still puts pressure on the ground to hold the car up. All that is left is the small but significant change in area as the tires rotate faster - which admittedly is not so simple to figure out. Wnt (talk) 15:01, 6 July 2017 (UTC)[reply]

Well, in a gas in a closed envelope (such as a not flat tire), pressure is pretty much the same everywhere (variations are acoustic waves, out of scope here), so the pressure of the tires on the ground IS the same as the pressure in the tires, provided of course there is pressure -- making your flat tire example irrelevant

A typical tire has pressure in the 2+ bar range, very easy to exert for a man and even a child. That's 200 000 kg/m², meaning a 2t car only need 1/100 m² ground contact to be supported ; that's 25 cm² per tire, if equally spread on 4 tires.

I must add that any force variation, for any reason, as effect both on the pressure on/in the tire and its contact area with ground, and, actually, more on area than on pressure because tires are round and deformable. Hence, pressure doesn't change much in any event. In fact, this is what makes tires useful.

Gem fr (talk) 08:57, 7 July 2017 (UTC)[reply]