Wikipedia:Reference desk/Archives/Science/2010 October 21

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October 21

Griswold - knitting industry machine

Just watching Michael Wood's Story of England, on the BBC. He was looking at the frame-knitting industry in Kibworth, and mentioned a machine called a "Griswold", which people would buy on tick and use at home. Do we have anything on the machine or its inventor? Couldn't see anything on the page Griswold. DuncanHill (talk) 00:02, 21 October 2010 (UTC)

Henry Josiah Griswold gets a brief mention in Timeline of clothing and textiles technology in 1878. Otherwise, the best description I can find is here, photos here, and if you happen to be near Nottinghamshire, you can watch a demonstration of their use at the Ruddington Frameworks Knitting Museum. Clarityfiend (talk) 01:36, 21 October 2010 (UTC)

Thank you! DuncanHill (talk) 10:37, 21 October 2010 (UTC)

There are several videos of circular knitting machines at work on youtube. Here's a Gearhart machine in action, which is similar. I've had a one for a couple of years now and they are quite easy to use. I can make a pair of socks in an evening while watching the TV. --TrogWoolley (talk) 22:52, 21 October 2010 (UTC)

Thank you for the link, very interesting. DuncanHill (talk) 23:17, 21 October 2010 (UTC)

Explanation

http://www.youtube.com/watch?v=foZlciP6gUQ&

How does this work? I know that the bullet will be spinning after leaving the gun, but now, it's spinning in a completely different direction, so its angular momentum is in a completely different direction. I guess the ground must have produced the torque, but when I try to do force diagrams, the torque due to the normal force seems to be pointing in the wrong direction. 76.68.247.3 (talk) 00:29, 21 October 2010 (UTC)

It looks like the bullets are spinning along their axis in every case, just as the rifling originally imposed. However, the ease of changing the direction of spin can be seen very close to the end of the video clip, where the bullet wanders in many directions on the ice. I'm actually rather surprised that it didn't shift from spinning along the axis to tumbling end over end, since I'd think that the ends would be heavier and would tend to push outward, but perhaps the many small forces from the ice didn't provide a push in that direction? Wnt (talk) 01:32, 21 October 2010 (UTC)

Those rounds seem VERY underpowered to me. My guess is the bullet was spinning from the rifling. It hit the ice/snow, bounced, and happened to land nose side down in a depression (or made one in snow), and spun. Why do you say "spinning in a completely different direction"? Ariel. (talk) 03:58, 21 October 2010 (UTC)

Well, they're still spinning along their axis, but the angular momentum vector has changed direction, no (now in the vertical direction instead of the horizontal direction) ? 76.68.247.3 (talk) 11:24, 21 October 2010 (UTC)

Right, the angular momentum vector has changed. As you pointed out this is due to torque resulting from the first impact. I think Ariel explains it as well as a first approximation. As for getting the normal force vector from a simple diagram, this is futile. Fracture_mechanics are amazingly complex, and resultant forces are probably best described statistically. Notice that it does take several tries to get it to work. --SemanticMantis (talk) 16:35, 21 October 2010 (UTC)

Hitting the ice and bouncing probably caused it to tumble, this changed the direction of the spin. The direction is "captured" if it happens to land in a small depression at the moment when the random tumble has the nose pointed down. The depression prevents it from continuing the tumble. Ariel. (talk) 18:16, 21 October 2010 (UTC)

Thanks a bunch. 76.68.247.3 (talk) 20:55, 21 October 2010 (UTC)

Visible, or not visible?

The Atlantic Wind Connection article talks about 80m towers 20 miles out to sea being "almost invisible from the shore". Could someone run the numbers with a little more precision? For instance,

• If I'm standing on the shore (my altitude = 0 above sea level), can I see a turbine? OR
• How far off the ground must I be (2-story bldg, 20-story bldg, etc) in order to see a turbine?

Either one would be helpful to me. THANKS! DaHorsesMouth (talk) 00:38, 21 October 2010 (UTC)

According to my calculations, the top of the 80m tower would theoretically be just visible from fractionally above sea level 20 miles away (just a few millimetres in theory). The amount of the top of the tower that's visible then increases with height. For example, from 1m above sea level you'd theoretically see about the top 16m of tower, and then from 2m above sea level about 22m. The eyes of an average person standing with feet at sea level would be somewhere between the two. Even where the line of sight theoretically allows it, whether in practice you could see this sliver of the top of a thin tower from 20 miles away would depend a lot on perfect lighting and weather conditions I imagine. 86.135.26.218 (talk) 01:31, 21 October 2010 (UTC).

Excellent, thanks much. DaHorsesMouth (talk) 01:55, 21 October 2010 (UTC)

Looking at the Horizon page, it mentions "that the actual visual horizon is slightly farther away than the calculated visual horizon, due to the atmospheric refraction of light rays." Therefore, at an average eye-level height of 1.70m, you could possibly see as much as the top 32m of the towers. Cheers, Mark 150.49.180.199 (talk) 02:01, 21 October 2010 (UTC)

An equally important question, though, is how big would that 32m appear? I'm guesstimating it would look like something an inch high at arms length. Does that sound right? Someguy1221 (talk) 07:33, 21 October 2010 (UTC)

No, way off! By simple trigonometry, I get an answer of about one millimetre (1⁄16 of 1⁄16 of an inch)) at arms-length. That's visible, but hardly a "blot on the seascape". Physchim62 (talk) 12:25, 21 October 2010 (UTC)

1 inch (25 mm): 1mm is not (1⁄16 of 1⁄16 of an inch)). Bazza (talk) 12:45, 21 October 2010 (UTC)

Oops! indeed! I was think 254 mm, not 25.4 mm, my mistake. Physchim62 (talk) 12:47, 21 October 2010 (UTC)

Thank you both, I found my math error. Someguy1221 (talk) 20:44, 21 October 2010 (UTC)

The atmospheric visibility toward the horizon, the presence of any fog and large waves would be important factors affecting the visibility of the turbines at head height. ~AH1^(TCU) 20:10, 23 October 2010 (UTC)

Age of the universe by isotope ratios

Some years ago, in an earth science class, I encountered a clever estimate for the age of the universe based on measuring the ratio of certain isotopes. One or more of these was radioactive (I don't recall exactly). The calculation was based on the assumption that they would be created in known ratio(s) relative to each other but decay at different rates. So measuring their ratio today and knowing the rate of decay would somehow allow one to know how long the universe had existed. The end result had a large uncertainty (maybe a factor of two?) but was an important constraint on cosmology decades ago when more direct observations of cosmological parameters were also very limited. Obviously I don't remember the details very well, but I remember being impressed by the clever approach. Does anyone know what calculation I am thinking of? I'd like to look it up again if I can find it. Dragons flight (talk) 05:35, 21 October 2010 (UTC)

Radioactive_decay#Occurrence_and_applications briefly deals with this. It states that only radioactive isotopes of the first five elements came along with the big bang, and that none of these isotopes could still exist. Beach drifter (talk) 05:53, 21 October 2010 (UTC)

Big_Bang_nucleosynthesis some more technical information. Beach drifter (talk) 05:56, 21 October 2010 (UTC)

No, no, this is about heavier elements and evolution after big bang nucleosynthesis. There was some trick to it. Even though, one didn't know how much material was formed via stellar nucleosynthesis over time, one could predict the ratio of isotopes and use that to form a constraint. Strictly speaking it probably didn't apply to the very first generation of stars, but within factors of two that's probably not an important detail. Dragons flight (talk) 06:01, 21 October 2010 (UTC)

Is it in the Radiometric dating article? If not I will have a look later.--06:33, 21 October 2010 (UTC)

Perhaps the ²³²Th & ²³⁸U chronometers, using the assumption of a constant nucleosynthesis rate [1]. Mikenorton (talk) 10:05, 21 October 2010 (UTC)

There's a section in Peebles' book on Cosmology from 1993 here (click on page 104). It's been a long time since I last heard of that method in connection with the age of the Universe. --Wrongfilter (talk) 10:37, 21 October 2010 (UTC)

This looks pretty easy to google. Here's one source: [2] detailing how the r-process and others that are active during a supernova explosion, can be used to estimate the age of our galaxy, and by that give a lower bound on the age of the universe. A more readable introduction, using U-235 and U-238, can be found here: [3] EverGreg (talk) 10:37, 21 October 2010 (UTC)

Turning flight

I’m a flying instructor (part-time.) Think about an airplane making a steady turn while maintaining altitude with no change in power setting or bank angle. There is a good wind blowing. The airplane’s ground speed reaches maximum when it’s heading downwind, and minimum when it’s heading upwind. While the ground speed is reducing from maximum to minimum the airplane is obviously decelerating relative to the Earth’s surface, and while the ground speed is increasing from minimum to maximum the airplane is obviously accelerating.

One of my students asked me this question: Using the Earth’s surface as the reference frame -

• what force causes deceleration while the airplane’s heading is changing from downwind to upwind?
• what force causes acceleration while the airplane’s heading is changing from upwind to downwind?

At first I said that both the deceleration and acceleration are caused by the force of the wind but our textbook says there are only four forces acting on an airplane – weight, lift, thrust and drag. Next I said that the deceleration would be caused by drag and the acceleration would be caused by thrust but now I’m not so sure. Is there a good scientific explanation? 203.110.136.172 (talk) 06:01, 21 October 2010 (UTC)

I don't think it matters where the wind blows. I think the weight of the airplane has the biggest factor here. (Although I'm not so sure, I haven't done physics for a while) Minimac (talk) 06:08, 21 October 2010 (UTC)

It's drag both times. When the wind is speeding up the plane it's negative drag, but still drag. Thrust is only what the engines produce. And yes, you can have negative thrust too (thrust reversers). Negative lift would be adjusting the flaps to force the plane down fast. Negative weight doesn't really happen (although I suppose if you have some upward momentum that's negative weight). Ariel. (talk) 06:38, 21 October 2010 (UTC)

The question confuses physics language with everyday language. In everyday language "acceleration" means to an increase in speed, and "deceleration" means a decrease in speed. Speed is a scalar quantity. But in physics, when we say a force causes acceleration, by "acceleration" we mean a change in velocity, which is a vector quantity. When a plane is flying in a circle at constant speed, it is continuously accelerating toward the center of the circle. The force causing this acceleration is lift, or specifically the horizontal component of lift. (Banking the aircraft causes the lift to be directed at an angle, the vertical component keeping the plane aloft while the horizontal component makes it turn.)

In the situation asked about, you have four forces acting horizontally. The thrust offsets the drag caused by the plane's motion, so those can be ignored. The other two are the horizontal component of lift as in the previous paragraph, and the drag caused by wind, as Ariel said above. The resultant of these two forces varies in magnitude as well as direction, since the wind drag always acts in the downwind direction and so the angle between them varies.

If you want to think about the plane's speed rather than its velocity, you then need to resolve this resultant force into one component parallel to the direction of the plane's motion (i.e. acting forward or back), and another at right angles to the plane's motion. You will see that the "parallel" component is sometimes forward and sometimes backward, and that's what accounts for the change in speed.

--Anonymous, 07:12 UTC, October 21, 2010.

The textbook is simplifying statement about effective forces, or net forces - and only in the 2D "flatland" plane that is parallel to the aircraft's motion. In reality, there are lots of individual forces, and this simplified 2D view of "lift/weight/thrust/drag" may be oversimplified in this case - we have a crosswind, which is acting out-of-the-2D-plane. Really, we're worried about a "3D" effect here - forces acting in the X/Y (left/right) direction. Particularly with something complicated, like wind, the true force acting on the airframe is going to be a complicated vector-field that varies in intensity at every point on the airframe. (Fluid dynamics here would be particularly nasty). The effective, net force of wind will be a single force vector, that acts on the center of mass of the aircraft. The simplified 2D model boils this down to "net forces" - so why four, instead of two (net vertical and net horizontal)? Well, because there may also be a net torque (but you know this, because you can apply control surfaces to pitch the aircraft). Torque can't be described by a single net-force; so for the simplified model, they need at least one additional parameter; and as I explain below, they have kept a bit of redundancy in order to "fully" describe the aircraft in 2D while staying "intuitive."

So let's simplify this situation without loss of generality. (And of course, when I say "plane" I mean the geometric concept; when I say "aircraft", you know what I mean). An aircraft is a six-degree-of-freedom object. It can move in three dimensions (x,y,z) and can rotate in three dimensions (roll, pitch, yaw). It has a position and a velocity and an acceleration along each of those dimensions. (See angular acceleration if you're unfamiliar with the way a torque can induce an angular acceleration). In each dimension (including the angular orientation), positions and velocities are determined by integrating the accelerations; the accelerations are determined by Newton's law (F = m a for linear motion, and τ = I α for angular acceleration).

The net force in each dimension is the (scalar) sum of all the forces. And the net torque is determined by the vector sum of the parallel components, applied to the center of mass. This is why, in the simplified 2D case, it is necessary to keep lift/weight (to parameterize both rate of climb and pitch rate) and thrust/drag (to parameterize both forward acceleration and also to contribute to pitch-rate). Surprise surprise! We have some redundancy in the 2-D, 4-force model. We could get away with only 3 numbers here in 2D Flatland, if we reformulated the pitch-rate to only depend on one vector (a torque) instead of being determined by the combination above.

But a few pieces of the 6-DOF puzzle are missing! Where is roll? Where is yaw? These are determined by out-of-plane vectors - forces acting to the left and right. Two effective force vectors will determine the roll rate and the acceleration left or right (think of a crosswind correction maneuver). Two more effective force vectors will determine the yaw rate, and we again have 4 numbers to specify three parameters (a bit of redundancy that could be eliminated by converting to an angular domain).

So - in conclusion - yes, if you have a cross-wind, the 4-force textbook model breaks down. It is a 2D-only model and does not account for side-acting forces. Somewhere in the FAA guide-book (or your textbook) should be a whole chapter about crosswind compensation - in there they will introduce a simple and practical model that pilots can use to compensate for crosswind without breaking it down into full-blown physics. An autopilot (if properly designed) can independently control all six degrees of freedom - which is no easy task. It does this by relating the aircraft controls (i.e. control surface motions, thrust/throttle, spoilers) into an orthogonalized 6-DOF coordinate space. This mathematical task is not very intuitive, because we like to think that elevators only affect pitch and rudder only affects yaw, even though we know they also affect drag and lift and roll by a nonzero amount. But a computer can do this very well - it can be calibrated for a specific aircraft under specific conditions; i.e., given a 6 knot left-to-right crosswind, by what non-zero amount will applying the ailerons affect pitch rate? And it can use a control algorithm to lock each degree of freedom (rate-of-climb, pitch-rate, and so on) to the "dialed-in" rate - compensating for the weird side-effects you didn't expect. This process allows you to control each parameter "independently." Nimur (talk) 12:52, 21 October 2010 (UTC)

Minor pedantic quibble: The net force in each dimension is not the (scalar) sum of all the forces, it is either the vector sum projected onto that axis or the scalar sum of the components in that dimension ;) Physchim62 (talk) 13:25, 21 October 2010 (UTC)

In retrospect, I should probably refrain from posting physics analysis (even elementary mechanics) before 6AM... any reader should independently validate my claims before flying aircraft. Nimur (talk) 14:38, 21 October 2010 (UTC)

If you're just after a qualitative understanding, think like this: Assume the wind blows from the west and you're circling clockwise. Consider the moment when your heading is due north. Drag cancels thrust, so we can ignore them. Gravity points downwards. Since the aircraft is banked, lift points upwards and to the east. The upwards part cancels with gravity (since you're maintaining altitude). What is left is the horizontal part of the lift, pointing due east, which will be changing your velocity.

Relative to the air, you're moving the same way as your heading, i.e. due north. Since your acceleration is due west, it will change the direction of your travel (so you need rudder), but not your airspeed.

Relative to the ground, you'll be moving with the wind in the west-east direction while still flying northwards through it, so you're tracking north-northeast. But your acceleration vector still points due east (it is the same in all reference frames), so viewed from the ground your acceleration is not perpendicular to your velocity. The lift has a component in the direction you're tracking, which works to increase your ground speed. In the other side of the turn, when you're heading south, it's opposite: tracking SSE and accelerating due W will retard your ground speed. –Henning Makholm (talk) 04:07, 22 October 2010 (UTC)

Thanks gents for your time and expertise.

Special thanks to Anonymous! You wrote “When a plane is flying in a circle at constant speed, it is continuously accelerating toward the center of the circle. The force causing this acceleration is lift, or specifically the horizontal component of lift.” When I read this, I had one of those light bulb moments!

A few of us have been discussing this round the whiteboard, including lots of vector diagrams with the horizontal component of lift. As you said, in circular motion (such as using the atmosphere as the reference frame) the horizontal component of lift is always perpendicular to the velocity vector so velocity constantly changes direction but not magnitude. In cycloidal motion (such as using the Earth’s surface as the reference frame) the horizontal component of lift is not exactly perpendicular to the velocity vector so velocity constantly changes both direction and magnitude.

Also special thanks to Henning Makholm! Your explanation is very clear and well suited to pilots. I agree with every word of it (except that I’m not sure rudder is required just because of changing direction.)

Thanks Wikipedia! 203.110.136.172 (talk) 05:24, 22 October 2010 (UTC)

You're welcome. Yes, my rudder comment was oversimplified. –Henning Makholm (talk) 17:43, 22 October 2010 (UTC)

isolated system

please help me...for an isolated system when delu=0 what would be value of dels,and how did we arrive at result —Preceding unsigned comment added by Sameerdubey.sbp (talk • contribs) 06:54, 21 October 2010 (UTC)

I'm sorry, but I can't do anything with this without some explanation of what u and s are and what is being calculated. I doubt anyone can (but please do, if you can guess). Wnt (talk) 09:05, 21 October 2010 (UTC)

This is a thermodynamics question, where U is the internal energy, and S is the entropy. Red Act (talk) 09:41, 21 October 2010 (UTC)

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our policy here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Red Act (talk) 09:46, 21 October 2010 (UTC)

I think we can safely point the OP to our article on the second law of thermodynamics. Physchim62 (talk) 10:04, 21 October 2010 (UTC)

wireless transfer of elecrticity for long distance transmisan

i want knowabout it., i know that it is possible for short distance(by mutual indactance) but this wireless system able for long distance like radio wave if it happen than we trasfer energy withoutloss; it may be done by leaser/photodiode system also,,, but here also big problem any other option —Preceding unsigned comment added by Vikash dhillon (talk • contribs) 09:27, 21 October 2010 (UTC)

The big problem with any long distance wireless transfer of energy is something getting in the way of the energy beam. Transferring useful amounts of energy by wireless means would fry anything which got in the way of the beam. Physchim62 (talk) 10:07, 21 October 2010 (UTC)

Have you read our wireless energy transfer article? It contains quite a few links that you may find useful. Smartse (talk) 10:56, 21 October 2010 (UTC)

Wireless transfer of electric power happens inside a Waveguide. Cuddlyable3 (talk) 20:17, 21 October 2010 (UTC)

Tesla fanciers believe the eccentric inventor could transfer useful energy long distances without wires, to actually power things, as opposed to signalling. Others think he was delusional. Edison (talk) 04:23, 22 October 2010 (UTC)

Water in the equilibrium expression?

Oxalic acid and ethanol form an equilibrium reaction as follows:

$(COOH)_{2}+2C_{2}H_{5}OH\rightleftharpoons (COOC_{2}H_{5})_{2}+2H_{2}O$

The equilibrium constant given is $K={\frac {[(COOC_{2}H_{5})_{2}][H_{2}O]^{2}}{[(COOH)_{2}][C_{2}H_{5}OH]^{2}}}$

Why is the water included in the equilibrium expression? I thought only gases and aqueous solutions, and not liquids, were included. --220.253.253.75 (talk) 09:41, 21 October 2010 (UTC)

I'm no expert chemist, but in that equation, water is a reagant so needs to be included in the equilibrium, even though it will also be the solvent that the reaction is occurring in. Having looked at chemical equilibrium though, you might be right that it shouldn't be included. I'm sure someone will be able to enlighten you further. Smartse (talk) 10:54, 21 October 2010 (UTC)

Both ethanol and diethyl oxalate are liquids at room temperature, so why should they be included and water not? When a reaction is conducted in aqueous solution, the water is often omitted from the statement of the equilibrium constant (see acid dissociation constant for a well-known and widely used example) because the concentration of water doesn't change during the reaction. In this reaction, it would not be conducted in aqueous solution but in ethanolic solution, so the concentration of water will change and be relevant to the position of the equilibrium. The various different conventions for expressing equilibrium constants are certainly confusing for students but, unfortunately, the powers that be have yet to issue any clear guidelines so textbooks have to present all the possible ways of doing the calculations. Physchim62 (talk) 11:19, 21 October 2010 (UTC)

The short answer is that the equilibrium expression is set up such that the absolute concentration of the substances in the expression affects the Activity of the substance. If the activity does not change during the reaction, then that substance is left out of the equilibrium expression. --Jayron32 23:40, 21 October 2010 (UTC)

white light fringes in the Michelson interferometer

Why are they only visible with zero path-length difference? Qualitatively I sort of see why, but I can't find an equation examining this quantitatively, i.e. maybe expressing the width or sharpness of the fringes. John Riemann Soong (talk) 12:41, 21 October 2010 (UTC)

The interferometer shows fringes clearest with a monochromatic i.e. single wavelength light source. If the two paths differ by a whole number (e.g. 0, 1, 2 ...) of wavelengths, there is constructive interference and a strong signal at the detector; if they differ by a whole number and a half wavelengths (e.g., 0.5, 1.5, 2.5 ...) there is destructive interference and a weak signal. Note the ellipsis "..." that mean (with ideal source) the fringe pattern is infinitely periodic. However, a white light source emits wavelegths in the visible range 390 to 750 nm (about 32% bandwidth) which causes the contrast of the fringes to decrease with increasing order of difference. Cuddlyable3 (talk) 20:07, 21 October 2010 (UTC)

Improved Brownian ratchet would overcome the second law of thermodynamics?

The Brownian ratchet thought experiment, in which a nanoscopic paddle moving randomly in both directions due to impacts from molecules (Brownian motion) connected via an axle to a ratchet fails to violate the second law of thermodynamics because the ratchet, being nanoscopic itself, undergoes random Brownian motion, which makes it fail. If the ratchet is at a lower temperature, the Brownian motion decreases, the ratchet works better, and rotation occurs-but heat conduction through the axle will eventually equalise the temperature and motion will stop. But supposing the paddle was in a high temperature thermally insulated chamber, and the ratchet was in a low temperature thermally insulated chamber, and the random paddle rotation was transmitted to the ratchet using nanoscopic permanent magnets, meaning no physical contact to conduct to equalise the heat? Would this be a true violation of the second law of thermodynamics, resulting in perpetual motion? Alternatively, what if the nanoscopic paddle used a series of initially nanoscopic but ultimately microscopic gears to scale up the rotation to a microscopic level, and then use a microscopic ratchet (not subject to Brownian motion due to its size) to extract the energy? Would this also be a loophole in the second law? [Trevor Loughlin]80.1.88.15 (talk) 13:39, 21 October 2010 (UTC)

(For reference, see Brownian ratchet.)

As an ironclad rule, if you find yourself asking the question "Is this a loophole in the second law of thermodynamics?" the answer is always "No; something was overlooked in the thought experiment". In this particular case, the problem is that joining components together using permanent magnets does not actually thermally isolate the two chambers for the purposes of this experiment. The Brownian ratchet fails to violate the second law not because the two paddles are connected by a thermally conductive shaft, but because the motion of the ratchet (imparted by the rotation of the shaft) transfers energy to the lower-temperature reservoir, heating it up directly. In that system, you're still building a simple heat engine, fully in compliance with thermodynamic laws.

In the second case, consider what happens to the amount of force available to turn a gear or lift a pawl as you make the driven gear larger and larger. Remember that you can't extract any more work out of the gear train than you put in at that nanoscopic paddle. Bear in mind that there's no magical threshold where Brownian motion disappears completely; it just has a much smaller relative effect on macroscopic objects. A macroscopic (or even microscopic) ratchet and pawl finely balanced enough to be turned by your nanoscopic gear train would be just as readily affected by Brownian effects as a nanoscopic ratchet, while a macroscopic ratchet resistant to Brownian motion would be too stiff for your nanoscopic paddle to turn. TenOfAllTrades(talk) 14:18, 21 October 2010 (UTC)

(edit conflict with below) That seems a pretty strong statement, that "there can never be a loophole in the second law of thermodynamics". The second law is not like the first law. The first law is a pretty well tested "law of nature"; no one has ever seen energy created or destroyed, and so we feel pretty confident saying that it won't be. The second law is "merely" a statistical law. I put the scare quotes on "merely" because it is supported by overwhelming statistics; the number of particles involved in any macroscopic measurement is huge, and we can be confident that entropy will increase in any macroscopic process. I have yet to see a really convincing argument, though, that the second law must hold true, in all circumstances, at all times. The second law (like other laws of nature) began as a descriptive law based on observations, about how efficient steam engines could be. We now understand more about why the second law is true, but realizing that it's a statistical law, we can look for loopholes. Yes, most proposed methods of circumventing the second law have ended up still obeying it, but I don't think it's possible to say that no loophole will ever be found. Buddy431 (talk) 14:42, 21 October 2010 (UTC)

Given that it is a statistical law, it is posible that it does not hold true in a experiment involving a small nnumber of particles but it can not be predicted to fail in any particular experiment and the probability of failure goes towards 0 as the number of experiment goes towards infinity. This does not prove that the law is correct but the fact that it is a statistical law does not tell us that it is not valid in all cases, just that the predictions are the average in the limit. --Gr8xoz (talk) 15:34, 21 October 2010 (UTC)

You won't find a loophole allowing one to convert heat to work, generally. If we just forget about the second law and focus on the fact that the fundamental laws of physics conserve information, then that's easy to see. Under a time evolution, two different initial states of an isolated system will always evolve to two different final states. This means that from the final state you can always, in principle, deduce what the intitial state had been, so the information contained in the initial state never gets erased.

The entropy of a system quantifies the amount of information that you lack about its precise state. Consider a device that is able to violate the second law systematically. Suppose the device is always in the same initial state. Under a time evolution, the combined system comprising of the device plus systems it acts on, can be in a large number of initial states. The number of final states for the system must be the same because each intitial state evolves to a unique final state. Since the number of final states of the systems the device acts on becomes less, it follows that the number of possible final states the device can end up in, becomes larger to compensate for this.

So, what is not possible is for the device's final state to be always some fixed state. If the device's initial state is not fixed, this doesn't affect the conclusion (that the entropy of the device increases), unless the initial state of the device is correlated with the precise state the systems it will act on are in. Using prior information about the precise state the systems are in, the device can lower the total entropy. But then the entropy of the systems wasn't actually what we thought it was, because entropy is defined as the amount of information we lack (or don't have access to) about the system. Count Iblis (talk) 16:07, 21 October 2010 (UTC)

But in the second case, why not use millions of nanoscopic brownian ratchets to power the one large gear? And in the first case, lets assume frictionless magnetic bearings, so no heating of the isolated ratchet. —Preceding unsigned comment added by 80.1.88.8 (talk) 14:30, 21 October 2010 (UTC)

In the second case, the nanoscopic brownian ratchets does not work so they can not power a large gear. The argument was to power a large rachet by a smal paddle, to connect milions of smal paddles afected by random forces from brownian motion makes the forces cancel out each other. In the first case, remember that heat is the same thing as unorderd vibrations or random motions, the rachet must be exposed to this random motions and it will make th internal particles in the rachet move relative to each other, the rachet will be heated. --Gr8xoz (talk) 15:34, 21 October 2010 (UTC)

Is sushi healthy?

I am talking about your standard fare, rice, nori, raw fish, avacado, cucumber, carrot etc. I have never seen a nutrition label for the stuff, and it seems like it would be healthy food, low fat, low sugar, low salt, but don't really know. Googlemeister (talk) 14:31, 21 October 2010 (UTC)

Yes, for the reasons you listed. However, one must be careful of contamination (bacterial or... um... worm-ial) of the fish. In addition, there are health risks when eating too much large predator fish; tuna especially can have a high mercury content. → ROUX ₪ 14:39, 21 October 2010 (UTC)

Many sashimi cuts are actually very fatty. Toro is even called "fatty" tuna. Nimur (talk) 14:40, 21 October 2010 (UTC)

POV. Healthy for you; not so healthy for the victim of course.--Shantavira|^{feed me} 15:10, 21 October 2010 (UTC)

Is the tuna a victim if it gets eaten by a shark instead? Googlemeister (talk) 15:42, 21 October 2010 (UTC)

The oil in fish generally contains massive amounts of omega 3 fatty acids, which are widely thought to be beneficial. Sushi isn't necessarily low in salt, though -- it is typically dipped in a soy sauce mix, which is very salty. Looie496 (talk) 17:36, 21 October 2010 (UTC)

Tuna’s End, a recent NY Times article, discusses the relation of ōtoro sushi and the conservation status of bluefin tuna. -- 124.157.218.53 (talk) 18:19, 21 October 2010 (UTC)

Are you sure sharks can catch tuna? I'd expect the tuna to have a massive speed advantage. Smartse (talk) 20:16, 21 October 2010 (UTC)

healthy food: low fat, low sugar, low salt This is a short form which makes me very suspicious. Food should be viewed as a whole, single compounds are most of the time not what anybody should look at. --Stone (talk) 21:07, 21 October 2010 (UTC)

If every food was low in calories, you would starve! --Chemicalinterest (talk) 22:46, 21 October 2010 (UTC)

Who said anything about low calories? Nil Einne (talk) 02:44, 22 October 2010 (UTC)

There are downsides to having a diet heavy in fish these days (e.g. high mercury levels and other unpleasant things in the modern fish stock). In general though, if the fish were all "good" fish (which actually excludes tuna, if I have read correctly), my understanding is that a raw fish diet is pretty good for you when combined with other things. Unfortunately we've managed to completely foul up fish over the last few decades, and much of what is popular now either poses health threats or major environmental threats. Sigh. --Mr.98 (talk) 00:37, 22 October 2010 (UTC)

Yeah, eating tuna sushi every day would be a very bad idea. My young niece got mercury poisoning quite badly from eating nothing but tuna fish and blueberries for six months (she basically had an eating disorder). Some of her teeth fell out, the hypotonia was so bad that she went from being a very active jock to missing a lot of school because she had to spend her days in bed, and the neuropsychiatric symptoms were so bad that she's actually currently institutionalized. She's perhaps ruined her entire life, just due to eating too much tuna fish. Red Act (talk) 08:26, 22 October 2010 (UTC)

Coincidentally I just read this news story on Wednesday: I had a tuna meltdown - 'Mercury poison' suit. Doesn't sound as serious, and I certainly don't think he should be suing the supermarket, but interesting nonetheless. A tuna company spokeswoman said that to their knowledge there's "never been a case of mercury toxicity from eating commercial seafood in the US". Perhaps you should try and set the record straight with them. Best wishes to your niece. the wub "?!" 09:11, 22 October 2010 (UTC)

I was initially rather startled to hear the tuna company spokeswoman make that statement, but I don't know why I would be. After all, tobacco companies were publicly claiming that there was no solid evidence that their product caused any health problems for decades after studies started showing a strong link between smoking and lung cancer. I guess the moral of the story is to listen to independent third parties like the FDA or something, rather than to companies that have a strong vested interest in a particular outlook.

Admittedly, my niece is a rather unusual case. Very few people make tuna as big a portion of their diet as she did. I don't think my sister-in-law and her husband considered suing the tuna company or anything. I'll have to ask them when I see them at Thanksgiving. Red Act (talk) 10:42, 22 October 2010 (UTC)

See The Straight Dope Raw fish can contain parasites (worms) which get passed to the consumer, unless the restaurant is zealous about only buying the best fish and the chef is zealous in the preparation. That seems like leaving a lot to the capabilities and intentions of random food preparers. They may get to your stomach and merely cause intense pain until the stomach acid kills them, or they may take up residence in the intestine. Cut up raw fish is a fine ingredient for thoroughly cooked dishes, and it is also useful bait for fishing. Edison (talk) 20:13, 22 October 2010 (UTC)

In severe cases of mercury poisoning, Minamata disease can result. However I'd think that sushi is generally a healthy food due to the variety of combinations. ~AH1^(TCU) 19:25, 23 October 2010 (UTC)

Focusing on a blurred image

I am nearsighted and have a small amount of astigmatism and normally wear glasses. At night, when I look across the room at a glowing LED, like a clock, the image is very blurred and unreadable. The blur can become barely readable when I squint my eyes hard. In the past few years, I have noticed something very interesting that is hard to describe: If I concentrate while looking at the blur and shift my depth of focus a little bit (not squinting), I can focus upon the blur itself, and it does not look like a blur anymore — it looks like a hemisphere of a net of glowing lines connected to circles. All in sharp focus, as though I were looking at an image of a glowing tree. What is occurring here? Could this be a trick of the mind (I hadn't thought so, because the image stays stock still as I focus on it), or is something really occurring optically? (The usual disclaimer: I am not seeking medical advice, just information on the optics. For all I know, people with 20/20 vision may experience this too when looking at something particularly far off and hence blurry.) Green Bark (talk) 18:15, 21 October 2010 (UTC)

Yeah, I've noticed those too. I think you can actually see the surface of the eye this way. Not sure though. 206.248.178.11 (talk) 18:32, 21 October 2010 (UTC)

I have also noticed them. I've always assumed that they were some sort of reflection of part of my eye. Possibly the veins on the retina. But I've never had anything to back that theory up. APL (talk) 18:41, 21 October 2010 (UTC)

Floaters? Nimur (talk) 18:53, 21 October 2010 (UTC)

They are not floaters. It is a reflection or diffraction. I saw it before I had eye surgery, but I don't see it anymore. -- kainaw™ 19:01, 21 October 2010 (UTC)

If what you see resembles Newton's rings then it is a diffraction pattern. That is possible when the light is monochromatic, and I guess it could arise from internal reflection in your cornea. Cuddlyable3 (talk) 19:31, 21 October 2010 (UTC)

Maybe you are looking at the "canals" of Mars or Vanus [4], I do not know if this is posible without a telescope. --Gr8xoz (talk) 21:28, 21 October 2010 (UTC)

What I see is not floaters or Newton's rings, but is closer to the canals of Mars and Venus linked above. But of course they glow against a black background. Green Bark (talk) 22:41, 21 October 2010 (UTC)

It's simply an out of focus image - which is to say, the surface of the cornea maps to the shape of the blur. Partially closing eyelids affects part of it, for example. Whether that's the upper or the lower part should be determined whether it's near- or far-sightedness, I believe. But there's a confounding factor also: any minor variation in the angle of the lens surface has a major effect on where the image projects. Inhomogeneous fluids and oils on the corneal surface therefore will redirect it, converting a simple blur into an apparent set of "canals". Wnt (talk) 23:22, 21 October 2010 (UTC)

I think you can sometimes see veins. See [5] which while concerning damage, the second post suggests even normal eyes may be able to. Also [6] and [7]. However as per these discussions I think it's only common when you shine a light on a dark adapted eye (not that I'm recommending you try that). It goes without saying if you have any concerns about your eyesight you should see a professional. Nil Einne (talk) 11:42, 22 October 2010 (UTC)

If you are looking at a small, not dazzlingly bright light in a relatively dark room, your pupil may be dilated. This cause some of the light to pass through the outer portions of the eye's lens. A nearsighted persons image is different from, say, a camera image which is out of focus, because the human eye's lens is not as high quality a lens as a camera's. See "Eye aberrations". Your focus adjustments may cause changes in the size of the pupil to a small extent, and the shape of the lens to a large extent. There is likely to be more distortion and aberration when light is not restricted to the center of the lens by a constricted pupil. Irregularities in the lens are likely what you see when a very nearsighted person looks straight at something like a distant streetlight with a dark sky and ground. These optical aberrations are not normally seen in daylight because they require a dark background to be seen. The distortion includes chromatic and spherical aberration and diffraction, as well as structural flaws in the outer lens. A different effect is obtained when an eye doctor examines your retina with an opthalmoscope, or when you are in a dark room and a small bright point source of light hits the eye from the side. The arteries on the retina may then become clearly visible, even if you have perfect vision. They look like a roadmap of wavy lines radiating out from the center.See Fundus (eye). Edison (talk) 20:10, 22 October 2010 (UTC)

I'm pretty sure the OP was talking about a blur that maps to the front of the eye. Veins in the retina can be seen easily enough, but naturally they cover the entire field of vision. Wnt (talk) 01:38, 23 October 2010 (UTC)

Tritan

whats Tritan plastic —Preceding unsigned comment added by Kj650 (talk • contribs) 18:19, 21 October 2010 (UTC)

A brand of BPA-free unbreakable plastic produced by Eastman Chemical Company. -- kainaw™ 18:24, 21 October 2010 (UTC)

is it polyethylene based —Preceding unsigned comment added by Kj650 (talk • contribs) 18:35, 21 October 2010 (UTC)

It is a copolyester. They use polyethylene terephthalate in production, but claiming it is "polyethylene based" is not as accurate as "copolyester based". -- kainaw™ 18:52, 21 October 2010 (UTC)

I'd go further: I'd say that "polyethylene-based" was wrong (or, at the very least, unnecessarily misleading) in polymer terminology. A polyester is very different from a "polythene" as the latter tem is usually understood. Physchim62 (talk) 19:20, 21 October 2010 (UTC)

MCH and MCHC units

Why did whoever it was decide to use picograms to measure mean haemoglobin content in cells, but then decide to use g/dL for the mean haemoglobin concentration? Could you not use something like pg/fL or something like that? Seems hideously inconsistent but I'm sure there's a reason for doing it. Is it simply to do with sensible powers? Regards, --—Cyclonenim | Chat  21:52, 21 October 2010 (UTC)

Standard Units for nearly all measurements are usually chosen to make the numerals fall somewhere between 1-1000 for the scale of the measurement. Generally 2-3 digit numbers are easier to work with, especially cognitively, for most people. If you notice, the numerals in BOTH measurements are 2 digit numbers. This is likely the driving force behind chosing the units. Its also why units such as Angstrom (for atomic sizes) and millibar (for airpressure) are chosen. The Angstrom gives atomic sizes in nice 2-3 digit numerals. --Jayron32 23:36, 21 October 2010 (UTC)

Haemoglobin concentration in the blood is typically quoted as g/dl or g/l. MCHC is about twice or three times the blood concentration and it is convenient to use the same units. Axl ¤ [Talk] 09:14, 25 October 2010 (UTC)

Coplanar forces

Resolved

 – ʄɭoʏɗiaɲ ^τ _¢

Just as a heads up, no this isn't my homework. We had a test, got the results back today, and I'm positive the teacher made a mistake somewhere. The question involves a semicircle (from 180° to 0°) with a radius of 2 ft. At the origin is point D. A force of 50 lb acts on a point along the circumference, pulling on it at a 240° angle. The force is acting on the circumference at 210° (based on D). The moment (torque) at point D needs to be calculated. Hopefully this crude diagram does the trick.

Anyways, my calculation was to bring the 50 lb force down along its line of action and find the perpendicular distance. Based on the angles given, the distance leg would fall at an angle of 150° from point D to meet the force at a perpendicular angle, creating a triangle with point D, B, and a new point (C)

Triangle BCD has a right angle at C. B is 30° and D is 60°. Knowing B is 30°, the length of D is cosine B times the length of the hypotenuse, in this case 2 ft. This totals 1.732 ft.

Up until this point, I am marked correctly on the question.

My understanding is that in such a triangle, the moment is equal to the force (50 lb) times the distance from D. 50 lb * 1.732 ft = 87 ft*lb in a counter-clockwise direction.

Given all this; am I wrong? Or perhaps, is the teacher wrong? Cheers, ʄɭoʏɗiaɲ ^τ _¢ 23:58, 21 October 2010 (UTC)

You need to use a dot-product (in other words, only the projection onto the relevant axis) - in this case, tangent to the circle. It seems today is a good day to forget to apply the dot-product; I made a similar mistake a few posts up (airplane). So your calculation should include 50 lb * 1.732 ft * cos(θ), where θ is the angle between the tangent and the direction of force. The remainder of the force is won't produce a moment - it will be counteracted by a normal force if the object is resting on the ground. Nimur (talk) 00:14, 22 October 2010 (UTC)

This is pre-equalibrium. Basically I should have broken the force into components and calculated them individually based on x-bar and y-bar? - ʄɭoʏɗiaɲ ^τ _¢ 00:20, 22 October 2010 (UTC)

I apologize; my explanation was both unclear and not exactly correct. Properly, to compute torque (moment) you need a cross product. My maths do work out if you selected the right θ but this is not the conventional way to "do it". A much more simple, elegant, and easy-to-remember approach is to cross-multiply the force by the moment-arm to yield the torque. The moment is (by most definitions) the magnitude of that torque. I really should get some more sleep, my proper use of vector multiplication is way off today. Nimur (talk) 00:25, 22 October 2010 (UTC)

So the good ol' M=Fd would apply, with F as the 50 lb force and D as the length of the moment-arm? Given what you know, and the diagram, what would you calculate the moment as? The teacher gave the answer as 50 ft*lb, but I am coming to either 87 ft*lb or 75 ft*lb (the latter when I use your earlier approach and calculate omega as 60°) - ʄɭoʏɗiaɲ ^τ _¢ 00:31, 22 October 2010 (UTC)

Oh I see what the teacher is doing. 50 ft-lb is the moment relative to the bottom - in other words, the object is pivoting around the lowest part of the hemisphere - where it makes contact with the ground. It was a "trick question" to make us think the torque should be calculated with respect to the center of the circle - but there's no physical reason to pick that geometric point as the reference. As always, a torque or moment is always relative to some chosen reference-point - which does not have to be the origin of the coordinate system. In this case, it's the torque relative to the point of contact (where the object will pivot). Nimur (talk) 00:34, 22 October 2010 (UTC) I am mistaken, the teacher asked for the torque at point D. I am sorry to continue posting errors here, let me rescind previous comments and hold off until I am sure I have the right answer. Nimur (talk) 00:38, 22 October 2010 (UTC)

You got it! That would make sense... if we were that far along. This is my first year of statics for civil engineering. We began with trig review, moved to concurrent coplanar, then non-concurrent. This test covered up to that point, and now we are moving on to equalibrium with respect to ground points and pivots. The actual question reads "Calculate the moment about point D for the force acting at point B". The solution sheet shows the components of the force, a right triangle with BD as the hypotenuse to calculate the x and y distances between B and D, and the calculation -50 cos 60° * 2 sin 30° + 50 sin 60° * 2 cos 30° = 50 ft*lb - ʄɭoʏɗiaɲ ^τ _¢ 00:44, 22 October 2010 (UTC)

Okay. I unstruck the comment that was correct. Nimur (talk) 00:47, 22 October 2010 (UTC)

That can't be right... it's pivoting about D, not the bottom of the hemisphere. - ʄɭoʏɗiaɲ ^τ _¢ 00:55, 22 October 2010 (UTC)

Going by the diagram you posted, the perpendicular distance to D from the line along which the force is acting is 1 ft, not √3 ft. Check your trig there. Rckrone (talk) 03:05, 22 October 2010 (UTC)

Hmmm... You've made me realize that the mistake I made was using cosine instead of sine. If I had, the moment arm would be 1 ft and the moment 50 ft lb. I may be able to get a better mark knowing that it was one mistake. - ʄɭoʏɗiaɲ ^τ _¢ 03:46, 22 October 2010 (UTC)

Yes, your method (when you get is right) is shorter than the teacher's method of resolving then adding the components of the moment. I would have used a third method using the distance from D to the point 2 ft above D where the line of action crosses the vertical through D, but it's all the same result and really a matter of choice. Sometimes one method is easier and sometimes another, and sometimes it is just a matter of preference. Dbfirs 08:32, 22 October 2010 (UTC)

So long as you perform that method correctly :p. The teach' gives partial marks, so I'm hoping if I go and explain that my calculations were correct, but I simply mistakenly used cosine instead of since, I may get an extra mark or two (which is the difference between my current 77.5 and 90). - ʄɭoʏɗiaɲ ^τ _¢ 13:08, 22 October 2010 (UTC)

Yes, if you had explained your method clearly, showing that you understood the principle, I would have given most of the marks, but not if you just wrote down an incorrect answer without explaining what you were doing. The advantage of your teacher's method is that mistakes in calculating distances are less likely. Dbfirs 08:07, 25 October 2010 (UTC)