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January 7

Walking from the Earth to the Sun

If one built a long enough and strong enough ladder, would it be possible to walk from the Earth to the Sun? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:10, 7 January 2012 (UTC)[reply]

The distance between the moon and the Earth varies from around 356,400 km to 406,700 km. How fast do you walk? 5km/h? How would you transport the supplies needed for such a long trip? Von Restorff (talk) 03:18, 7 January 2012 (UTC)[reply]

(edit conflict) Walking the distance that would be between the earth and sun? Not possible in a lifetime, given that the earth is 150,000,000 kilometers from the sun, and a person can walk, what, 8 minutes per kilometer? That's a speed of like 7.5 kilometers per hour, so it would take someone 20,000,000 hours to walk there. The human lifespan is about a century at the outside, if we're generous, and at 24*365 = 8,760 hours in a year, that means you could get as far as 87,600 hours in a lifetime. 87,600*7.5 = 650,250 kilometers. You'd not even make it to Venus, assuming you did nothing but walk (no stopping to sleep, eat, shit, etc) for 100 years, and that's at a pretty decent clip. --Jayron32 03:20, 7 January 2012 (UTC)[reply]

You lost an order of magnitude when you computed the number of hours in a 100 year lifetime (and then you transposed the 7 and the 6 before multiplying against the speed), but it hardly makes a difference as even at 6.5 million km a century, the trip takes over two millennia. -- ToE 13:01, 7 January 2012 (UTC)[reply]

150,000,000??? Was the moonlanding a hoax? Von Restorff (talk) 03:23, 7 January 2012 (UTC)[reply]

What does the moon have to do with the OPs question? --Jayron32 03:32, 7 January 2012 (UTC)[reply]

Oops, meth is a hell of a drug. Von Restorff (talk) 03:35, 7 January 2012 (UTC)[reply]

"And for that, by the gods, if I say the moon be the sun then to you, good wife, it shall be so" -- ToE 03:51, 7 January 2012 (UTC)[reply]

"Good wife I am in name only, good husband, and thus it is the moon and 'tis the moon no matter what thee says!" Von Restorff (talk) 04:12, 7 January 2012 (UTC)[reply]

(edit conflict) Slower than that with the proper attire. Clarityfiend (talk) 03:22, 7 January 2012 (UTC)[reply]

I am guessing climbing a ladder is slower than walking on a flat surface. But there is a bit less friction in space so that would improve the speed a bit. Von Restorff (talk) 03:26, 7 January 2012 (UTC)[reply]

Yes. The ladder would also need to be elastic, given that the Earth's orbit is elliptical - you'd also need to anchor it at both one of the Earth's poles, and one of the Sun's, on some sort of bearing to allow for their rotation. Assuming you had a spacesuit that could support you indefinitely, and ignoring the fact that it would take rather a long time at a walking pace, you'd initially be 'climbing' the ladder to escape Earth's gravity, only to find yourself under increasing gravitational pull towards the Sun as you approached. I'll leave it for someone else to speculate how close you could get before losing your grip - and then there is the problem of increasing heat etc... I think that anyone with the sort of technology to approach that close to the Sun would use something more sophisticated than a ladder. AndyTheGrump (talk) 03:24, 7 January 2012 (UTC)[reply]

I agree. A space elevator is a bit more sophisticated than a ladder. Von Restorff (talk) 03:30, 7 January 2012 (UTC)[reply]

I meant theoretically if you could "walk" for as long as it took to get from the Earth to the Sun. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:36, 7 January 2012 (UTC)[reply]

Well Jayron answers that pretty well. I guess, yes, theoretically if you could "walk" for as long as it took without dying you could do it, but it would take many normal lifetimes to do so. Additionally in all that time, and as others have pointed out, you'd need a massive amount of supplies of food, drink, and oxygen, and some sort of surface to "walk" on. Then somehow you've to counter the varying gravity, from escaping the Earth, to essentially zero out in space, to massively high as you approach the sun. And frankly it would be a sod of a boring walk, there wouldn't be that much variation in what you were seeing for periods of 50yrs or more, so somehow you've got to deal with that. And then as you approach the sun you've somehow got to figure out how to deal with the heat (reminds me of an old Goon Show skit where they wanted to fly to the sun in a wooden rocket, when asked how they were going to prevent the rocket from burning up as they got near the sun they said they were going to go at night). So the answer is probably theoretically "yes" in the sense that almost anything is theoretically possible, but in any sense of reality, then "no". --jjron (talk) 04:18, 7 January 2012 (UTC)[reply]

Everyone is ignoring the fact that it is not possible to walk in a weightless environment, period. Walking depends intimately on gravity pulling the body down towards a surface. Without that, only floating is possible, not walking. --99.237.252.228 (talk) 06:58, 7 January 2012 (UTC)[reply]

I guess that if you have a ladder you could probably "walk" on it by pushing your heels against the rungs. Maybe a bit of velcro makes it even more comfortable. Von Restorff (talk) 07:05, 7 January 2012 (UTC)[reply]

You could also do it if the walking path were iron-based and you were wearing magnetic shoes. ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:06, 7 January 2012 (UTC)[reply]

Maybe even something like an inverted jetpack is an option. Von Restorff (talk) 07:09, 7 January 2012 (UTC)[reply]

I was expecting some type of elastic based attachment to the walking path. But since we're just making all this up anyway, I also thought that you may just use some type of artificial gravity device. --jjron (talk) 08:01, 7 January 2012 (UTC)[reply]

I think we're getting into the realms of fantasy now. Caesar's Daddy (talk) 08:49, 7 January 2012 (UTC)[reply]

All we can usefully say is if you had the technology to continuously travel for as long as you liked at a constant speed equal to a walking pace of 7.5 km/hr it would take you about 20 million hours or about 2,300 years to travel a distance equivalent to the distance from the Earth to the Sun. Gandalf61 (talk) 13:10, 7 January 2012 (UTC)[reply]

Several people have mentioned gravity, but you need to account for it together with the fact that the Earth is orbiting the Sun. Once you've climbed out of the Earth's gravitational field, you will be pretty much weightless because you are in free-fall (you, the ladder and the Earth are all going around the sun at orbital velocity). As you get closer, your orbital period stays constant meaning you are now orbiting too slowly and are not in free-fall any more. The net effect of all the various forces and your rotation is quite difficult to work out and can be very counter-intuitive, but I expect you would end up having more difficulty holding onto the ladder against the sideways forces caused by you trying to orbit at orbital velocity and the ladder orbiting much slower (these same forces would rip the ladder apart, but let's ignore that) than you would holding on against the Sun's gravity (some of the sun's gravity is, loosely speaking, cancelled out by centrifugal force, which ought to help, although I'm not sure how far you could get because not enough is being cancelled out and you get pulled off the ladder). Can someone that is better are celestial mechanics than me work out exactly what would happen? --Tango (talk) 16:38, 7 January 2012 (UTC)[reply]

The ladder would be going around the Sun at Earth's orbital period, so the Sun's gravity would soon be felt. The point of zero G would be at the L1 Lagrange point. Wnt (talk) 18:02, 7 January 2012 (UTC)[reply]

Once you were far enough from the Earth you could just slide down the ladder like it was a fireman's pole. With that technique you could probably make the journey in a single lifetime. APL (talk) 06:12, 9 January 2012 (UTC)[reply]

This reminds me of the science fiction short story (maybe "The Long Way Home" by Fred Saberhagen?) about the disabled spaceship being laboriously dragged slowwwly back to Earth by its crew's descendants by rope (attached to a "space anchor"). Clarityfiend (talk) 03:02, 8 January 2012 (UTC)[reply]

Yes, that's Saberhagen's The Long Way Home, which is collected in his Of Berserkers, Swords and Vampires anthology. It just happens that that short story is one of the free sample chapters visible at Baen Ebooks, linked from the table of contents here. -- ToE 08:39, 8 January 2012 (UTC)[reply]

Several compound names

There are about vitamin B12 total synthesis, I am not sure about their meanings:

α-corrnor sterone(corr = corrin ?)
thiodextrolin(dextro- = dextrorotation ? lin = ?)
cyanocorrigenolide(corrigen = oleandrin ?)

Bold parts of these words is most confusing. I have linked them to some pages, and I want to confirm if they are right. --Makecat _Talk 05:30, 7 January 2012 (UTC)[reply]

Thiodextolin doesn't seem to exist outside of this article. Oddly, Google Scholar finds zero cites for that term: [1] which is a red flag. The only places a regular google search finds it is in the Wikipedia article you cite and mirrors thereof. I'm working on the other two... --Jayron32 05:38, 7 January 2012 (UTC)[reply]

The article made a spelling mistake...--Makecat _Talk 05:56, 7 January 2012 (UTC)[reply]

I dug this up: [2] Skip the Wikipedia article and work from there instead. --Jayron32 05:40, 7 January 2012 (UTC)[reply]

Thanks. And I want to understand the meaning of the prefix corr- and dextrolin. Could explain them?--Makecat _Talk 05:56, 7 January 2012 (UTC)[reply]

this was all I can find using Google Scholar on "dextrolin". It isn't much. --Jayron32 06:01, 7 January 2012 (UTC)[reply]

My best guess is that corr is from corrin, given the presence of the corrin structure in B-12. You may also be interested in norsteroid. --Jayron32 06:06, 7 January 2012 (UTC)[reply]

And my guess is that "corrigen" is "corrin" and "-gen-" as meaning "forming" like in oxygen, if you look at structure 57, it is clearly about to form the corrin ring structure. --Jayron32 06:24, 7 January 2012 (UTC)[reply]

I got lucky with Google (Hmmm, that makes it sound like a porno search...) and found [3], which says "It has been a loose but fairly consistent practice of the Zürich group to refer to the B/C moiety of the vitamin B₁₂ molecule as the 'eastern' half, and the A/D area as the 'western' part. By contrast, Cambridge custom has tended to utilize the terms 'right' and 'left'. In such trivial affairs consistency is hardly a matter of moment, and we need not cavil at the action of the Zürich group in adopting the name 'dextrolin' for a B/C building block which they first succeeded in constructing..." The same source describes [4] "the compound LIII, which we have more or less jocularly dubbed 'α-corrnorsterone'. The 'corr' in this appellation represents our hope that the substance is destined one day to be transformed into a corrin; the 'norsterone' devolves from the fact that LIII is a ketone whose skeleton is that of a norsteroid if the nitrogens be ignored; and finally, if the name be pronounced in Slurvian, it becomes 'cornerstone'!" The Zürich group is introduced at the beginning of the section [5] as the laboratory of Albert Eschenmoser. Wnt (talk) 18:20, 7 January 2012 (UTC)[reply]

porno search...Does it mean that "dextrolin" is just a symbol without any special meaning? --Makecat _Talk 05:55, 8 January 2012 (UTC)[reply]

Well, he doesn't analyze their motives, but at least the "dextro" part just means "right". Besides being on the right hand as (arbitrarily) drawn on a chalkboard, it has nothing in common with dextran sulfate or dextrose or dextrorphan or manual dexterity. (Wow. We don't have an article for dextran sulfate yet...?) Wnt (talk) 15:34, 8 January 2012 (UTC)[reply]

Accidents/Injuries jumping into sea from a height

How deep a person will go if he jumps into the sea from a height of 12 meters/14 meters? — Preceding unsigned comment added by 175.100.188.30 (talk) 05:33, 7 January 2012 (UTC)[reply]

Ourdiving article states that Olympic divers use a 10m high board, and the pool is at least 5m deep for safety. I think it would be reasonable to extrapolate that another few metres of drop would likewise require a little more - but there is more to it than this. From a greater height, the instantaneous deceleration as one hits the water will have more effect, and one may be more at risk from this than with possible impact with the bottom of the pool. There is an increasing risk from getting the initial entry wrong - a belly flop from a few metres may result in little more than immediate pain, whereas a misjudged dive from higher might have severe consequences: to quote from our article "hitting the water flat from 10m brings the diver to rest in about 1 ft. The extreme deceleration causes severe bruising both internal and external, strains to connective tissue securing the organs and possible minor hemorrhage to lungs and other tissue. This is very painful and distressing, but not life-threatening". AndyTheGrump (talk) 05:58, 7 January 2012 (UTC)[reply]

That suggests another question or two: What is the greatest height anyone has fallen into (sufficiently deep) water and survived? And how deeply into the water did they go? And is it better to go in feet-first or head-first? All of these questions assuming you hit the water vertically. ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:02, 7 January 2012 (UTC)[reply]

Read this and look at this chart. Windresistance is a factor, and the weight because of the momentum. You do not even need water to survive falling from a great height: Vesna Vulović, Ivan Chisov, Nicholas Alkemade, Alan Magee. Von Restorff (talk) 07:15, 7 January 2012 (UTC)[reply]

See also shallow diving. Oh, and @Bugs, just in case anyone is reading this post and considering testing this out I'll put out a safety warning as a disclaimer. The Lifesaving Society guidelines for entering water are to only jump into water of known (and safe) depth. If it's unknown you should walk in, or use some type of 'slide entry' if you can't walk. And if you have to jump in, then you always go feet first (better to break your leg than break your neck). There is a technique, the which name eludes me, where you can jump into deep water feet first from a reasonable height without your head going under water - you have to be able to do this to get even a basic lifesaving qualification such as the pool Bronze Medallion (which, for the record, I have). --jjron (talk) 07:51, 7 January 2012 (UTC)[reply]

Actually, the evidence seems to suggest that when falling from a great height, your chances of survival (small as they are) are less than when falling on land, where you have a slight chance of coming down through trees, into snowdrifts etc. Water is water, and impact is likely to be predictable - over land, you have at least a theoretical chance of landing on a duvet factory's surplus output repository, or the like... AndyTheGrump (talk) 07:59, 7 January 2012 (UTC)[reply]

If you want to break the Guinness book of records world record you need to defeat Harry Froboess who fell 361 feet (110 m) into Lake Constance from the airship Graf Zeppelin on June 22, 1936; that sounds easier than doing what Vesna Vulović did (if the story is true and not just propaganda). Von Restorff (talk) 08:02, 7 January 2012 (UTC)[reply]

If which story is true and not just propaganda? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:11, 7 January 2012 (UTC)[reply]

The story about Vesna Vulović. According to the section called "2009 report", there is a possibility the story about her surviving the highest fall without a parachute ever: 10,160 metres (33,330 ft) is made up to cover up a serious mistake made by the Czechoslovak Air Force. Von Restorff (talk) 15:25, 7 January 2012 (UTC)[reply]

All those people would have reached terminal velocity. Once you're at terminal velocity it makes no difference how far you've fallen from, only how you hit the ground. Rckrone (talk) 20:11, 7 January 2012 (UTC)[reply]

That's not necessarily true. If you've only only just reached terminal velocity (after about 15 seconds), you're probably still breathing, conscious and reasonably warm. If you fell from 10 km, you might be hypoxic, unconscious and half-frozen, because the air is quite thin and cold for the first part of your fall. This could affect you chances of surviving the "landing". Mitch Ames (talk) 23:58, 7 January 2012 (UTC)[reply]

Being unconscious might increase your chances of surviving the landing. It would mean your body was completely relaxed. Compare landing feet first with your legs held out straight and landing with them relaxed. The latter is much less painful. --Tango (talk) 14:10, 8 January 2012 (UTC)[reply]

Yes, which means that less of the energy of impact is absorbed by your legs and feet, and more is transmitted through the torso and the vital organs, thus, ironically, increasing the probability of death. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:44, 9 January 2012 (UTC)[reply]

Imaging of potato growth

I'm trying to find a method which would allow me to visualise potato tubers as they grow in a field, but aren't sure what the best technology to use would be. X-ray tomography can be used to visualise plant roots [6], but all the studies I can find grow plants in pots, rather than the field e.g. [7]. Is there a reason that it can't be used in the field? X-ray tomography may not be necessary anyway, since the resolution is extremely high (μm) whereas potato tubers are obviously a lot larger. Ground-penetrating radar seems like another option, but the only relevant thing I could find is this thesis and the resolution isn't great. Could 3D ultrasound be used or does anyone have any other ideas? SmartSE (talk) 15:09, 7 January 2012 (UTC)[reply]

Typically scientist don't go quite so high tech, for cost reasons, among others. A rhizotron can be dug relatively cheaply, though mini and micro rhizotron rigs can get more costly, see paper here: [8]. SemanticMantis (talk) 15:39, 7 January 2012 (UTC)[reply]

That might work, but it sounds as if you can only see structures that grow around a tube placed in the soil, meaning that you wouldn't be able to see all the tubers on a plant without seriously disturbing the soil around it. SmartSE (talk) 16:12, 8 January 2012 (UTC)[reply]

From a strictly mechanical standpoint, x-ray tomography requires the x-ray source and detector to be on directly opposite sides of the volume being scanned. To get an accurate and unbiased three-dimensional reconstruction, you need to be able to rotate source and detector a full 180 degrees. If you've seen clinical CT scanners, you know that they're doughnut shaped, and that you slide the patient through the doughnut hole to collect sequential image slices. That's easy to do with a plant in a pot; that's very difficult to do with a field full of soil. TenOfAllTrades(talk) 19:19, 7 January 2012 (UTC)[reply]

Ah that explains why I can only find papers using pots or soil columns. Thanks for that. SmartSE (talk) 16:12, 8 January 2012 (UTC)[reply]

Transparent soil? {The poster formerly known as 87.81.230.195} 90.197.66.116 (talk) 09:00, 8 January 2012 (UTC)[reply]

Concur. I don't really know the name of this stuff, but at least over here you can get a transparent soil substitute (in the form of crystals or somesuch) which apparently works fairly well (haven't tried it myself though). --Ouro (blah blah) 13:45, 8 January 2012 (UTC)[reply]

You mean products like that sold here: [9]? These would be very expensive to fill a field with, and I think would be too wet for potatoes. Also, it seem the OP might be interested in 'natural conditions', which these 'crystals' would not replicate. 'Hydroponic potato' does yield some google hits, but those seem to be mostly pots filled with perlite, with small amounts of nutrients dripped through. SemanticMantis (talk) 14:34, 8 January 2012 (UTC)[reply]

That's an intriguing idea, but SM is right that I'm interested in natural conditions. Even growing plants in pots causes enough problems, so artificial media aren't much use. Also, if you used that with potatoes, you'd end up with green tubers. SmartSE (talk) 16:12, 8 January 2012 (UTC)[reply]

Exactly, that's something like that. I know it's not natural but it would satisfy the criterion of you being able to inspect the underground development easily. Sidenote, I for one have always been a fan of planting plants in weird things, so I used all manners of disused teapots and even cut enormous glass jars in half (had them cut, okay, I didn't do it myself) to plant cacti in. --Ouro (blah blah) 20:10, 8 January 2012 (UTC)[reply]

Shooting down a falling bomb

Is it possible to destroy a falling bomb or artillery shell in midair? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:06, 7 January 2012 (UTC)[reply]

Possible, yes. Easy, no. But see close-in weapon system. --Stephan Schulz (talk) 16:17, 7 January 2012 (UTC)[reply]

In the 1982 Falklands War, the Argentine garrison defending Stanley airfield claimed a number of "kills" with the Roland missile system, which did not relate to lost British aircraft. Our article on the missile says: " This system fired 8 out of the 10 missiles it was deployed with and is credited with shooting down one Harrier Jump Jet and two 454kg (1,000lb) bombs." Alansplodge (talk) 17:24, 7 January 2012 (UTC)[reply]

The Iron Dome system aims to intercept artillery shells. In the short term at least, such a system is useful only in the unusual environment of Israel, or perhaps in defending small vital areas. As a general defence against artillery, it's much too expensive - artillery rounds and pieces are cheap and readily available, whereas these interceptors and their related launch and control infrastructure are super expensive. -- Finlay McWalterჷTalk 17:37, 7 January 2012 (UTC)[reply]

To make this more entertaining - is it possible for a skilled military sniper, seeing the bomb falling down toward his position on a flat plain, to use his rifle to detonate or deactivate the bomb in mid-air and thus somehow survive? Wnt (talk) 17:53, 7 January 2012 (UTC)[reply]

Theoretically possible, but practically impossible. — Lomn 18:01, 7 January 2012 (UTC)[reply]

Shrapnel shell gives the terminal velocity of a WW-1 era 3" artillery shell (presumably modern, longer range round like NATO 155mm, which travel much further, have much higher terminal velocities). That gives a very conservative speed of 400 ft/sec, which is about 120 m/s. So even if the sniper knew exactly where the projectile was, and could see it (and it's essentially impossible to see a 3" black circle at a range of a km or more, against the bright daytime sky) and given 10 seconds to aim and fire (in practice snipers need much longer than that, for long distance shots) he'd be shooting a 3" target 1.2km up, which is the about the limit of accuracy of a conventional sniper bullet. This all discounts the practical impossibility of knowing where to point his gun to begin aiming, of seeing the shell, of needing much more time to properly aim, of shooting upward with a heavy rifle, and the unpredictable layers of high-speed winds above him - as Lomn said, in practice it's impossible for a person. Gravity bombs have a terminal velocity of around 300m/s (reportedly; I can't find a very reliable source) and missiles and high-performance bombs like Tallboy get to over 1000m/s. Things like Goalkeeper only have a chance of working because they have very accurate long-range radar, fully automated operation, and spam the sky with bullets. -- Finlay McWalterჷTalk 18:30, 7 January 2012 (UTC)[reply]

It is extremely likely that incidents of fratricide amongst munitions have happened many times in war, especially in 20th and 21st century war. If you drop a stick of bombs, or salvos of artillery, fused to detonate at a given height above ground, the fragments could hit other munitions and detonate them. This would not be much of a boon to the targets below. A bomb or shell detonating when it struck a hard target might throw back fragments which detonated a bomb or shell which had not yet reached the target, which would be an advantage to the target. Could a shell be detonated by a piece of shrapnel or a bullet? There have been numerous news reports of people setting off "war souvenir" shells by firing small calibre bullets at them, but there is little chance of intentionally shooting a bullet which hits a falling bomb or shell. Fratricide is probably more likely than someone on the ground firing flak or bullets and detonating a bomb or shell. While a bomb is still in the airplane, a fragment from an antiaircraft shell might hit and detonate a bomb. If the bomb is falling with its descent slowed by a parachute, like the atom bombs dropped on Japan in 1945, hitting it with fragments from high velocity antiaircraft shells would be easier. Edison (talk) 21:19, 7 January 2012 (UTC)[reply]

It certainly is. I am aware that in early (maybe 30 years ago) trials of the Seawolf missile, it destroyed an incoming 4inch shell (Although I dont think I can provide a ref for that)--92.25.100.174 (talk) 11:16, 8 January 2012 (UTC)[reply]

Just found it in Sea Wolf (missile) in History section.--92.25.100.174 (talk) 11:22, 8 January 2012 (UTC)[reply]

I want a Goalkeeper CIWS for my birthday. Von Restorff (talk) 14:53, 8 January 2012 (UTC)[reply]

This page on the British Ordnance Collectors Network forum says that special "radar enhanced" 4.5 inch shells were produced and used in the Sea Wolf trials. Shooting down a normal shell might be more difficult. Alansplodge (talk) 18:37, 8 January 2012 (UTC)[reply]

How do you radar enhance a shiny metal cylinder?--92.25.99.162 (talk) 20:12, 8 January 2012 (UTC)[reply]

It is an experimental 4.5" Mk8 radar enhanced round; the top half is made of opaque fibre with an aluminium cross section insert which acts in the same way as a radar cone on a boat ie it reflects and so can be seen by the radar. Click the link. Von Restorff (talk) 20:34, 8 January 2012 (UTC)[reply]

They're made for missiles and aircraft, but a Close-in weapon system could theoretically also hit bombs or shells. 68.156.149.62 (talk) 23:18, 9 January 2012 (UTC)[reply]

There's also SDI: Strategic_Defense_Initiative. --Sean 20:22, 10 January 2012 (UTC)[reply]

Thick, dripping liquids

When thick liquids like saliva or semen drip from a surface, why do they form a column of liquid with a rounded drop at the tip, which can be pulled back up into the air in its entirety? Wouldn't such a structure be unstable and break apart into separate drops very quickly? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:13, 7 January 2012 (UTC)[reply]

Gravity and fluid mechanics. Yes. Von Restorff (talk) 16:29, 7 January 2012 (UTC)[reply]

Surface tension may be important too. Breaking into drops increases the surface area, which surface tension acts against. If the surface tension is high enough, then it will overcome gravity and cause the liquid to form a single, spherical shape. --Tango (talk) 16:44, 7 January 2012 (UTC)[reply]

Surface tension works to change liquid into spherical drops. It takes less energy to change a narrow stream of liquid into many small drops than into one large drop. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:19, 7 January 2012 (UTC)[reply]

Many small drops have a larger surface area than few large drops or a single column. Surface tension acts to minimise surface area. You are wrong if you think it encourages water to break up into drops; rather, it encourages liquid to stay in a single mass. --Colapeninsula (talk) 19:19, 7 January 2012 (UTC)[reply]

In any case, wouldn't a long, thin column of liquid with a round drop at the tip be at or near a maximum-energy state? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:25, 7 January 2012 (UTC)[reply]

Perhaps those thick fluids aren't just liquids, but gels. In particular, they may start out as liquids, but, as the water evaporates, the reminder forms a gel. The proteins, in particular, seem to play a role in gel formation. StuRat (talk) 21:24, 7 January 2012 (UTC)[reply]

Both saliva and semen exhibit this behavior even without significant water evaporation, and neither one starts out as a gel. You try dripping those liquids from an object - you will end up with exactly the same behavior that I described. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:29, 7 January 2012 (UTC)[reply]

Note that evaporation occurs very quickly from a strand of saliva, etc., presumably at body temperature to start, due to the extreme surface area to volume ratio. StuRat (talk) 22:08, 7 January 2012 (UTC)[reply]

Do these liquids contain proteins that might affect their characteristics as a liquid? Richard Avery (talk) 22:30, 7 January 2012 (UTC)[reply]

They definitely contain proteins. Whether those particular proteins tend to form the weak cross-linking characteristic of a gel, I cannot say. StuRat (talk) 22:48, 7 January 2012 (UTC)[reply]

What does the weak field do?

I think I understand some of what the other fields do:

Gravity field: attracts (or repels?) mass, bends space-time.
Electric field: attracts or repels charged particles.
Magnetic field: attracts or repels other magnetic fields and by extension causes moving electrical charges to circle around. (Interchanges with the electric field in other ways, including light polarization.)
Strong field: attracts or repels quarks and other particles with color charge.

I can visualize all of these.

Now suppose I had a device of some sort that generated a weak field. What exactly would it act on? What sort of particles would it attract, repel or make travel in circles? Hcobb (talk) 20:43, 7 January 2012 (UTC)[reply]

The strong force holds neutrons together with protons in the atomic nucleus and also holds quarks together. The weak force is responsible for beta decay. StuRat (talk) 20:58, 7 January 2012 (UTC)[reply]

The weak force does the flavor changing, but what if you had a device that generated a dipole (or tripole?, monopole?) in the weak force. What would that do? Hcobb (talk) 23:50, 7 January 2012 (UTC)[reply]

It could deflect all weakly-interacting particles (which is basically all particles except gluons and right-handed fermions, I think?), albeit very weakly and the "device" would be necessarily be extremely short-ranged. (Unlike gravity and the electromagnetic field, which are long-ranged because their force carriers are massless, weak field has an effective range that is determined by the mass of its carrier bosons and it is smaller than the typical size of a nucleus.)--Itinerant1 (talk) 22:33, 8 January 2012 (UTC)[reply]