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October 25

Speed of the average water molecule at room temperature?

Can anyone give the speed in miles per hour of an average H2O molecule in a body of pure water at 1 Atmosphere and 68° Fahrenheit? The proper equation would be nice. μηδείς (talk) 02:23, 25 October 2011 (UTC)[reply]

v_ave = √ [8k_BT ÷ π ÷ m]

Plasmic Physics (talk) 02:34, 25 October 2011 (UTC)[reply]

You need to apply the equipartition theorem, because a water molecule is not a point-particle. The equipartition theorem states that the energy will be equal in all modes (on the average); so, you need to estimate the vibrational and rotational modes of energy; then count all the degrees of freedom, and divide the average total energy per molecule amongst each mode. Then, compute the velocity that corresponds to the translational motion modes. Nimur (talk) 02:42, 25 October 2011 (UTC)[reply]

Here it is: Stowe's Introduction to Statistical Mechanics and Thermodynamics. I have a hardcover red copy that I refer to as my little red book of thermal physics (they changed the cover in the 2nd edition, apparently). Stowe's textbook is essentially the only physics books you will ever need (because everything else can be derived from these principles - at least in statistical ensemble - and what else matters, besides ensemble behavior?) Unfortunately, reading this text does require at least a little preparation and familiarity with some more elementary physics and mathematics concepts. Nimur (talk) 02:53, 25 October 2011 (UTC)[reply]

Well, I would actually be happy with an approximation for ideal point particles. I just want something accurate on an order of magnitude. And I can figure out the equations with the ideal gas laws, or at least follow the derivation. But I figure there are complications for liquids, given they are not gasses. And I was hoping there must be something more obvious for water than an entire book, based on something llike pv=nRt. μηδείς (talk) 03:39, 25 October 2011 (UTC)[reply]

Okay, well here is a summary from HyperPhysics, Molecular Kinetic Energy. Unfortunately, to compute this molecular translational velocity correctly, you have to dive deep a little bit - the physics of molecular motion is nontrivial, even treated in a purely classical non-quantized way. There's no practical utility in incorrectly approximating molecular motion. Be sure you understand the difference between average speed, most probable speed, expected-value of velocity in one dimension, and so on. Ideal gases follow the Maxwellian distribution, elaborated here. Water is not an ideal gas, because it has viscosity, its molecules are asymmetric, hydrogen bonding and van der Waals forces are nonnegligible, and so on. Nimur (talk) 05:56, 25 October 2011 (UTC)[reply]

The redlink above is probably in reference to the Maxwell–Boltzmann distribution (sometimes Maxwell distribution) Grandiose (me, talk, contribs) 21:20, 25 October 2011 (UTC)[reply]

Is k_B the Boltzmann constant? Dualus (talk) 04:36, 25 October 2011 (UTC)[reply]

Yes. Plasmic Physics (talk) 04:37, 25 October 2011 (UTC)[reply]

Is there such thing as an ideal liquid, or would that just be the same thing as an ideal gas at extreme conditions? (I am wondering what would happen to the p variable.) Yes, I understand the difference between mean, median and mode. Yes, I understand that water molecules are polar, and that the angle between the hydrogen atoms varies. I can even figure out that unless a body of water is moving, the net average velocity of its particles will have to be zero. But since the atoms are indeed moving it seems reasonable to me to ask for some average speed at a given temperature and pressure. All I can remember from the gas laws is that the answer is calculable and on the order of 1000mph (or was it kph?) for air molecules, with a mean free path of ~10cm, if I remember correctly. Surely there is some sort of analog for water molecules. μηδείς (talk) 11:27, 25 October 2011 (UTC)[reply]

The assumptions of an ideal gas are dilute concentration which, a liquid is not, and very low temperatures. Plasmic Physics (talk) 11:38, 25 October 2011 (UTC)[reply]

The Van der Waals equation gives a fairly good description for a liquid while still keeping the math reasonably simple. Dauto (talk) 12:34, 25 October 2011 (UTC)[reply]

The system not being an ideal gas is irrelevant here . The translational part of the partition function factors out, so barring quantum effects, you have the Maxwell distribution for the velocity. Also that H2O is not a point particle is also irrelevant, because, again, the partition function for the translational part factors out of the configurational part rotational, vibrational parts (note that these vibrational, rotational and other internal parts may not factorize as they are coupled, but the translational part will factor out due to physics being invariant under translations).

The only issue with H2O being a liquid as far as the velocity distribution is concerned, is that there will be strong correlations between nearby molecules. Count Iblis (talk) 15:23, 25 October 2011 (UTC)[reply]

In other words, the only issue is that the actual molecular velocity depends on the non-ideal parts? I agree whole-heartedly. Except for the non-ideal parts, water is an ideal gas. The ideal velocity can ideally be calculated by ignoring any non-ideal effects. PV = nRT and such, for n spherical non-interacting water-molecules and cows. Count Iblis, while you are correct in stating that translational motion is not coupled to the rest of the molecular motion, your assertion implies that these other energy modes do not contribute to the temperature of the system: I think that assertion is incorrect.

Temperature is comprised by all the energetic degrees of freedom in a system - not just the translational kinetic energy; this is the definition of "temperature." You can't ignore the other energetic modes, especially in a molecule as complicated as the water molecule. You must define all the degrees of freedom and apply equipartition theorem. Unlike, say, Helium, these other degrees of freedom are non-negligible for water - which is why we call it a non-ideal gas (or fluid). Nimur (talk) 21:08, 25 October 2011 (UTC)[reply]

Count Iblis is right (as usual). Note that the question is not about the total energy of the molecules at a given temperature in which case you would be right and the equipartition theorem would have to be invoked. The question is about the speed of the molecules and that will be given by the Maxwell-Boltzmann distribution. Note that Count Iblis never said that the ideal gas is a good description for the liquid water. That would have been a ridiculous statement. What he said was that the Maxwell-Boltzmann distribution is a good description for the liquid water, and in that he is right. Dauto (talk) 03:38, 26 October 2011 (UTC)[reply]

Correct me if I'm wrong, but my impression from Temperature is that E_k = 1/2 kT for each degree of freedom; thus, there should never be any need to divide by the number of degrees of freedom to figure out the speed of any given particle at a given temperature. If you have a liquid made up of a mix of molecules that can stretch and rotate in a hundred different ways, and individual atoms floating free, the individual atoms don't have any way to know they're supposed to go faster than the others when they hit something (unless they have less mass, that is). Of course, those hundred different extra rotations do carry energy and are in equilibrium, so if you cool such a liquid I suppose it should release quite a bit more energy over time / larger heat capacity. Wnt (talk) 23:19, 26 October 2011 (UTC)[reply]

Lightsabers

Would plasma or lasers be better suited for a lightsaber, and which would be more practical? →Σ τ c. 05:59, 25 October 2011 (UTC)[reply]

Neither is practical. Light sabers are impractical; they make for exciting fiction, but they defy any attempt to explain their behavior according to known physics. The real behaviors of both plasmas and lasers are very dissimilar from the special-effects that were used in Star Wars. Nimur (talk) 06:05, 25 October 2011 (UTC)[reply]

Could you explain further? →Σ τ c. 06:14, 25 October 2011 (UTC)[reply]

Sure. Let's start with plasma. Think of a plasma as a very hot gas. It has other special properties, but your starting point for intuition should be "very hot gas." In what way do you think very hot gases behave like a sword?

Now, let's move on to a laser. Think of a laser as a very bright light. It has other special properties too, but your starting point for intuition about lasers should be "very bright light." In what way do you think very bright lights behave like a sword?

Ultimately, I'm sorry that you were exposed to fictional ideas about lasers and plasmas before you were ever exposed to nonfictional facts about them. Rest assured - both lasers and plasmas are fascinating for all sorts of reasons - but they have nothing to do with light sabers.

In fact, there are cold plasmas and dim lasers, but this doesn't change the fact that you can't build swords from either.

If you have some prior scientific knowledge, I recommend a better intuitive starting point for both concepts - albeit this requires more than laymans' terminology. A plasma is an ionized gas whose ensemble behaviors are determined by electromagnetic effects, rather than just by kinetic effects. A laser is a machine designed to amplify light, typically resulting in a very coherent, monochromatic, high power output signal. These more technical descriptions are much more accurate than the ones I gave above, but they are harder to "really" understand for a lot of people - which is why we have sci-fi stories about laser/plasma/swords. Nimur (talk) 06:43, 25 October 2011 (UTC)[reply]

The clearest way I know to explain the problems with a lightsaber in a layman's terms is to ask the following: if it is a laser, a beam of light, what caps it so it doesn't shoot off into infinity like any other beam of light, and if there is some physical cap, what holds it in place and why wouldn't a cap at the tip get caught up on things? If, on the other hand it is a plasma, what, like the glass of a neon light, holds the excited gas in place? I always figured it has to be some field-bound mono-molecular filament in some excited state. Surely there is a book on the physics of Star Wars? μηδείς (talk) 11:41, 25 October 2011 (UTC)[reply]

Better than that - there's a Wikipedia article: Physics and Star Wars. Alansplodge (talk) 12:52, 25 October 2011 (UTC)[reply]

The force bends the light. That's why you have to have the power of the force to use a light saber. Otherwise, it is just a clunky flashlight. -- kainaw ™ 12:51, 25 October 2011 (UTC)[reply]

This blog has some conjecture about light sabers. Alansplodge (talk) 12:58, 25 October 2011 (UTC)[reply]

You can purchase a lightsaber in most toy shops, or you could. They are basically a translucent plastic tube with a flashlight at one end. Have fun.--Shantavira|^{feed me} 14:02, 25 October 2011 (UTC)[reply]

I wouldn't utterly rule out a kind of deliberately shaped ball lightning, but ... it's a really big stretch. Also I remember reading a news report recently (<1 month) about transmission of light in packets which somehow shape themselves, even in vacuum, to keep a bit of light in a tight bunch, but I can't find it now; you might be able to devise some handwavy explanation about how you put "english" on the original composition of such clumps of light so they boomerang back. Still, until someone turns up with a bona fide light saber it's all mist and vapors. Wnt (talk) 14:27, 25 October 2011 (UTC)[reply]

The most practical light sabers are made from computer graphics. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:35, 25 October 2011 (UTC)[reply]

How about magnets and mirrors? →Σ τ c. 22:14, 25 October 2011 (UTC)[reply]

It is well-established that a lightsaber consists of plasma encased within a force-field. See the Wookieepedia article on lightsabers for more information. --130.216.172.199 (talk) 22:18, 25 October 2011 (UTC)[reply]

Ok, but since "force-fields" are complete make-believe that just brings us back to where we started. APL (talk) 03:02, 26 October 2011 (UTC)[reply]

Oh, well that's reassuring. Since the plasma is kept safely contained within a force field, you can't accidentally cut your fingers on it ... ? Wnt (talk) 23:06, 26 October 2011 (UTC)[reply]

Watch the following video series, http://www.youtube.com/watch?v=xSNubaa7n9o This is about as close as you will get to a working lightsaber, but obviously there are more practical uses of the technology used in Michio Kaku's design. ScienceApe (talk) 15:43, 26 October 2011 (UTC)[reply]

Hello chaps - digging way back into the Lightsaber article history gives us this choice info :-

"Lightsabers, made of immaterial beams of light, collide when they should pass right through one another without a sound. Moreover, laser beams propagate in a straight line as long as they do not meet an obstacle, therefore scientifically correct lightsaber blades would cut through a starship hull. This would mean that dueling with lightsabers would be like dueling with flashlights because the blades would pass through each other and continue to travel until they hit an object, not reflecting back onto themselves.

Instead of a laser-based device, the most believable design for a lightsaber-like device would use plasma confined by a magnetic field. Plasma is a super-heated gas and is also the fourth state of matter, the color and luminosity of which varies depending on the temperature and composition. Plasma could be ionized by a particle beam from a compact particle accelerator; at relativistic energies, the beam would produce its own blue glow along its axis from Cherenkov radiation.

Keeping a gas in the plasma-state requires considerable energy: 40 kW are necessary for a 10 centimeter (4 in) saber at 10,000°C (18,000°F). It would be difficult to fit the required generator into the saber's hilt. To control and increase the length of the blade, the plasma would need to be confined within a magnetic field. Although this design would behave like a lightsaber from the Star Wars movies, it is considered foolhardy to confine plasma magnetically. A handful of magnets would disrupt the confinement field, and plasma would spill onto the saber's wielder. Furthermore, the magnetic field would prevent the plasma from performing any cutting action because it would always be shielded from whatever the blade struck by the magnetic field.

To make a semi-solid beam of energy which could interact with both matter and energy would require containing a quasar and quantum singularity inside the hilt. The gravitational field would pull all the quasar's expelled plasma back moments after the quasar releases them. The speed of the returning plasma would form a chainsaw effect allowing it to cut through the matter with ease, while still being stopped by an opposing beam. A modulated gravity field would bounce, allowing for the reflection of energy beams. So far, no known substance exists which is able to contain a quantum singularity or to contain and/or focus a gravitational field.

A lightsaber in reality would be difficult to wield, since its center of gravity would be in the hilt next to the hand. A real sword has its center of gravity further from the hand, which allows the user more control and power in the swings and thrusts. With a lightsaber, a user would need to have much more control and wrist strength in order to accurately use a lightsaber with strength and power. However, because the blade is supposed to be made of light, it would have no weight at all. In that instance, the blade would move easily while the hilt moved slowly. According to commentaries on the prequel trilogy DVDs, the lightsaber props were indeed built so that their center of gravity was centered in the hilt, as if there was no blade at all."

Quintessential British Gentleman (talk) 19:56, 26 October 2011 (UTC)[reply]

" Furthermore, the magnetic field would prevent the plasma from performing any cutting action because it would always be shielded from whatever the blade struck by the magnetic field." -- Only if the "whatever the blade struck" was magnetic in its own right. 67.169.177.176 (talk) 06:01, 27 October 2011 (UTC)[reply]

I see Luke Skywalker pulling and pushing and kicking and hammering and cursing at a light saber stuck firmly to (or in...) the deck of the Millennium Falcon, while the ghost of Yoda softly whispers, "Use the force, Luke!" Wnt (talk) 17:48, 27 October 2011 (UTC)[reply]

Just watched an interesting program where in they will use a new type of plasma engine; basically the plasma is accelerated out the back of the engine using magnets that are kept at extremely cold temps and only acts as magnets when cold, so if you can now effect the shape of plasma and the speed it comes out of on a miniature scale say ""hand held"" scale then a very hot jet of plasma being brought across the enemy from a hand held device would certainly give you a bad day and would equate some what to a weapon or Sword...

... the drawback, of course, being that the radiant heat from this "plasma sword" would give the person wielding it flashburns as well. (BTW, why don't you sign your posts w/ four ~'s?) 67.169.177.176 (talk) 23:48, 29 October 2011 (UTC)[reply]

" dissolving " water in other solvents

How can I calculate the amount of water I can " dissolve " into another solvent. For instance how can I calculate the amount of water that can dissolve in octanol before one can percieve two different phases? — Preceding unsigned comment added by 137.224.252.10 (talk) 09:48, 25 October 2011 (UTC)[reply]

You can't, you have to find it through experimentation. Plasmic Physics (talk) 12:33, 25 October 2011 (UTC)[reply]

Are there tables where you can look up this kind of thing? — Preceding unsigned comment added by 137.224.252.10 (talk) 12:48, 25 October 2011 (UTC)[reply]

See solubility chart and solubility table. μηδείς (talk) 17:42, 25 October 2011 (UTC)[reply]

Octanol doesn't mix well with water, so you'll only be able to dissolve a small amount of water in it before you get phase separation. (Best way to find out how small is to actually try it.) 67.169.177.176 (talk) 01:29, 26 October 2011 (UTC)[reply]

Just checked the solubility chart: the solubility product is negligible, so you might not be able to dissolve any water in octanol at all. 67.169.177.176 (talk) 01:32, 26 October 2011 (UTC)[reply]

Are you sure you looked under octanol, and not octane? "At 25 °C, the solubility of water in octanol is quite large, approximately 0.275 mole fraction, but the solubility of octanol in water is just 7.5 × 10 −5 mole fraction." [[1]]. See the Ropel reference. Dominus Vobisdu (talk) 02:07, 26 October 2011 (UTC)[reply]

Yes, I checked under octanol, C₈H₁₇OH. However, it does appear to me that I was looking at the octanol-in-water number rather than water-in-octanol. I stand corrected. 67.169.177.176 (talk) 02:47, 26 October 2011 (UTC)[reply]

Everything disolves, a concentration measureable in nanomoles or picomoles per cubic decimentre is still a concentration. Plasmic Physics (talk) 01:42, 26 October 2011 (UTC)[reply]

Thanks to everyone for their replies. They have been very helpful! — Preceding unsigned comment added by 137.224.252.10 (talk) 11:12, 26 October 2011 (UTC)[reply]

Price of airplane ticket: weight vs. volume

First or business class tickets are much more expensive than tourist class. However, you weigh the same, no matter what class you fly, but occupy more space, depending of class. Does that means that for airplanes, the volume you occupy if more important for determining the cost than your weight? Or is it simply a market dynamic: they charge more because they can. Strangely, when you send a package they normally charge you by weight. Quest09 (talk) 13:20, 25 October 2011 (UTC)[reply]

Well, skimming the article, a Boeing 747 has a maximum takeoff weight of 735,000 to 970,000 pounds, and seats up to 550 passengers. (There are all kinds of variations painstakingly documented there, which I basically ignored; a more careful reading may get you more accurate figures) At anywhere from 1330 to 1760 pounds per passenger, the passenger weight is a pretty small factor. Wnt (talk) 14:17, 25 October 2011 (UTC)[reply]

Don't forget fuel weight, it makes a huge difference. Our 747 article doesn't contain enough info for a proper calculation, but for the Boeing 727 one can work out that fully loaded and fully fueled, there is capacity for somewhere in the neighborhood of 250 pounds per passenger, including luggage. Looie496 (talk) 20:17, 25 October 2011 (UTC)[reply]

Well, I suppose that's how they define "fully" fueled. But jetliners don't usually fly with full fuel. Wnt (talk) 16:49, 26 October 2011 (UTC)[reply]

Also, the US postal service has recently been ggressively advertising a Flat Rate Priority Mail program with presized boxes and a slogan "if it fits it ships", so I think this applies to some extent to air shipping also. Wnt (talk) 14:19, 25 October 2011 (UTC)[reply]

The smallest USPS Flat Rate envelope comes with a 70lb domestic and 4lb international weight limit.μηδείς (talk) 16:00, 25 October 2011 (UTC)[reply]

Drat, I thought I had found a cheaper way to mail plutonium to people I dislike. :-) StuRat (talk) 03:04, 26 October 2011 (UTC)[reply]

The cost of flying a plane from A to B is basically fixed. You then want to maximise the amount of money you make from the flight. That's determined by business concerns, not engineering ones. First class doesn't really exist to make money by being sold. I don't have any figures, but I don't believe many first class tickets are sold. The seats are often either empty or filled with people that got upgrades (either free or with airmiles). They exist to serve as a reward for their frequent flier programs (which exist to get people to buy more economy and business tickets), and also to make business class look cheap by comparison. The improvements from business to first are minimal, at least in the air (you do get some special treatment on the ground with a lot of airlines), so very few people would pay the enormous increase in price just to get them. You travel first class either because it's a free/airmiles upgrade, someone else is paying, or you want to do so for the sake of it (eg. to show off how rich you are, or because it makes your honeymoon more special). Once you are rich enough for the extra price not to be significant to you, you're probably chartering private jets rather than travelling on commercial flights. Business class is a significant improvement on economy, so is worth the money to a lot of people - I'm not sure airlines actually make more from a business class seat than they would by filling the same space with economy seats, but they would struggle to get businesses to name them are their preferred airline if they didn't offer it, even if the business usually flies people economy. (Apologies for the complete lack of references - I'll try and find some later and will come back!) --Tango (talk) 18:17, 25 October 2011 (UTC)[reply]

Minor comment: the cost of operating an aircraft does vary as a result of fuel consumption; and fuel efficiency does vary as a result of gross takeoff weight; so it's not a totally fixed cost to fly a plane from A to B. Otherwise, Tango's pretty spot-on; most of the short-term decisions in commercial aviation are made on the basis of marketing and business choices, not on engineering limitations of the aircraft. Over the much longer term, major airlines choose how to strategically invest in their fleets in order to keep their operations in line with their business objectives: for example, purchasing large aircraft versus small aircraft; regional- vs. transcontinental- aircraft; and so on. Southwest Airlines is often cited for operating an all-737 fleet - excellent, versatile, large aircraft. (No first-class on SouthWest, either). This consolidates planning, maintenance, training, and other costs, but also consolidates risk. Two business analyses, both from Forbes: Southwest Gives Fleet A Facelift, And Saves Some Dough Too; and following a 737-related aircraft safety incident, Southwest shares drop.... Corporate strategy is a finely-tuned game of highly-quantitative guess-work. Nimur (talk) 19:20, 25 October 2011 (UTC)[reply]

Boas vs. pythons

After reading this summary of the similarities and differences between boas and pythons, I had the following insight: without the anatomic and physiologic distinctions between the two families, the matter of inhabiting different locations would be irrelevant. Is difference in worldly habitat really, then, a difference between the two families? As a contrast, I offer the distinction between polar bears and giant pandas, which not only possess anatomic and physiologic differences, but a uniqueness in their areas of distribution that is, I think, somewhat necessary to their survival -- pandas could not exist in the arctic circle, although polar bears could probably live in China, as they do in zoos all over the world. DRosenbach ^{(Talk | Contribs)} 13:36, 25 October 2011 (UTC)[reply]

Well, I think you're missing a key concept here: There is a relationship, an interrelatedness between the environment an organism lives in and the particular physiological and genetic traits that produce them. Two important things: 1) None of the three factors is absolutely responsible: organisms change their environment, environmental pressures cause genetic changes, genetic changes cause changes in form and behavior, which cause further environmental changes. It is a series of complex feedbacks all affecting each other. 2) Nothing is static. The system of relationships changes continuously and never settles into any static set of relationships. There are periods of precarious equilibrium where some locations and some times show periods of stability, but nothing is ever permanent. --Jayron32 15:25, 25 October 2011 (UTC)[reply]

Your points are well taken, Jayron32, but my question revolves around the notion that, despite the loosely-fitting old-world/new-world classification of pythons and boas, respectively, introducing boas into Australia or Africa or pythons into Central America (or South Florida) doesn't seem to affect the species, unlike moving a panda into the arctic. If a difference is that boas don't naturally inhabit Australia and pythons don't naturally inhabit, but that difference is really only consequential as a side point (as can be demonstrated by invasive species of feral pythons in S. FL) -- is it really accurate to refer to it as a difference? DRosenbach ^{(Talk | Contribs)} 16:40, 25 October 2011 (UTC)[reply]

The fact that a species can survive in a place doesn't mean it isn't different from the species native to a place; invasive species can thrive and actually greatly alter an environment; it doesn't mean that they are somehow indistinct from native species. C.f. rabits in Austrialia. --Jayron32 20:38, 25 October 2011 (UTC)[reply]

Modern taxonomy uses cladistics which groups species by their evolutionary history. One cause of populations drifting apart evolutionarily into separate species is that two groups become geographically separated so they no longer interbreed. The differences in the geographic ranges of two families can say a lot about evolutionary distance. Rckrone (talk) 03:54, 26 October 2011 (UTC)[reply]

EMP effects on artificial pacemakers

What would be the effects of an exposure to EMP from a short distance on artificial pacemakers (will they stop working? for a limited period of time? permanently? and why?), and the person who owns it? (please correct my English)

It is difficult to find precise information, but the general conclusion seems to be that artificial pacemakers are susceptible to damage from a sufficiently strong EMP. Looie496 (talk) 18:04, 25 October 2011 (UTC)[reply]

I'd think it would have to be a very strong and/or close EMP, since the main susceptibility to EMP is due to long wires which build up charge over miles (such as electrical transmission wires on high tension towers). The wires on a pacemaker are relatively short. StuRat (talk) 03:09, 26 October 2011 (UTC)[reply]

...and the voltage required to screw it up is relatively small. Spinning Spark 21:46, 26 October 2011 (UTC)[reply]

Decapitation

Could any of you give me a brief description of what ligaments, muscles and bones would be cut through by a killer using an axe to sever a human head please? — Preceding unsigned comment added by IsonomicJedi (talk • contribs) 16:37, 25 October 2011 (UTC)[reply]

You can check google images for an "axial section of the neck," but this diagram looks like it answers your question. If you include the term 'Netter' (he's a very famous and accomplished anatomy illustrator) you'll likely get better results. DRosenbach ^{(Talk | Contribs)} 16:43, 25 October 2011 (UTC)[reply]

Note though that using an ordinary axe to sever a human head is essentially impossible. Even professional executioners, tremendously strong and well-trained and using a much broader and much sharper axe on a victim lying stretched out motionless in front of them, frequently failed on the first try. With an ordinary axe all the killer would do it make a gory mess. Looie496 (talk) 17:55, 25 October 2011 (UTC)[reply]

Hence the guillotine, leading edge technology of its day. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:28, 25 October 2011 (UTC)[reply]

Cutting edge technology, surely? Greglocock (talk) 00:28, 26 October 2011 (UTC)[reply]

Or bleeding edge technology, perhaps ? StuRat (talk) 03:17, 26 October 2011 (UTC) [reply]

D'oh! I missed that obvious punning situation. And don't call me... you know. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:02, 26 October 2011 (UTC)[reply]

...sharp? Pfly (talk) 03:26, 27 October 2011 (UTC) [reply]

Did you have to be so blunt? Nil Einne (talk) 09:34, 27 October 2011 (UTC)[reply]

Earthquakes and Brick Buildings

In terms of safety, would brick buildings be a good choice in an earthquakeprone area? Compared to, for example, timber construction or steel frame buildings. Thanks, Wanderer57 (talk) 18:20, 25 October 2011 (UTC)[reply]

Brick buildings are the absolute worst. In earthquake-prone parts of California they are forbidden. Steel is best, timber next. Concrete is bad, brick just disintegrates. Looie496 (talk) 18:34, 25 October 2011 (UTC)[reply]

Around these parts of California, you hear the terminology "un-reinforced masonry construction" - it's not merely that the bricks are present, its that structural integrity of the building is entirely dependent on the "glue" (the mason's mortar) that holds each brick together. An unreinforced brick building will "disintegrate," if the shaking gets bad enough. If the building had rebar steel inside it, and brick on the outside, the problem is less severe: the building may remain standing in an earthquake even if the mortar cracks and disintegrates, and bricks fall off. However, you still have the serious problem of falling bricks coming off the side walls: this can be very hazardous. Even if the building doesn't collapse, it only takes one loosened brick falling from just a few feet up on a wall to land on your head, causing a serious casualty or fatality. (I used to think - how unlikely is that? What's the probability of one brick falling and landing on your head? ... But after surviving a few earthquakes, I realized ... the entire region is shaking - somewhere in a fifty mile radius, somebody is standing near a brick wall that ~~could~~ will collapse). It's safer to avoid such construction altogether. Here's more information from the State of California: Typical Unreinforced Masonry Building Damage from our multi-county regional planning commission in the San Francisco Bay area; and here's Putting Down Roots in Earthquake Country from the Federal Government (the USGS). Lots of good photos and statistical information in that one; you can order a free paper-copy from the US Geological Survey if you want one. Nimur (talk) 19:05, 25 October 2011 (UTC)[reply]

Generally there are requirements to retrofit such buildings in earthquake-prone areas by putting in cross-braces, so that there is something that actually holds the building together should a quake occur. Mikenorton (talk) 19:20, 25 October 2011 (UTC)[reply]

Instead of bricks, if you use cinder blocks, and run rebar through the hole in each, you can come up with a reasonably safe construction. StuRat (talk) 03:15, 26 October 2011 (UTC)[reply]

Wood frame construction is surprisingly durable in an earthquake, provided it's tied to its foundation properly. As noted above, unreinforced masonry performs poorly: it can just come apart. While it's possible to engineer such structures to withstand earthquakes, that work involves inserting what amounts to a parallel structure into the masonry building to deal with the forces that masonry can't withstand. Examples of unreinforced masonry include the Washington Monument and Washington National Cathedral, which suffered significant damage in a rather moderate earthquake. Acroterion (talk) 17:16, 26 October 2011 (UTC)[reply]

Thank you all. My question was prompted by the situation of Bangladesh. I'm led to believe that Bangladesh is in an active seismic area, is substantially a delta susceptible to shaking, and uses bricks as a major construction material because the soil can easily be used to make bricks. Comments? Wanderer57 (talk) 05:05, 27 October 2011 (UTC)[reply]

It lies close to the plate boundary between the Indian Plate and the Eurasian Plate and the earthquake hazard will increase in the northern and eastern parts of the country where the boundary lies, although it is significant everywhere. This suggests that the country is relatively unprepared for a major earthquake, such as the 1897 Assam earthquake, which severely affected the area that is now Bangladesh, with most of the brick buildings in Dhaka being destroyed (according to this). Mikenorton (talk) 07:22, 27 October 2011 (UTC)[reply]

You can also have a look at the Modified Mercalli scale for earthquake damage to various types of masonry. Soft soils are especially prone to soil liquefaction. ~AH1 ^(discuss!) 20:28, 29 October 2011 (UTC)[reply]