Wikipedia:Reference desk/Archives/Science/2008 November 9

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 9

do electric blankets save energy vs. an electric heater?

Is an electric blanke energy-efficient versus my electric heater, because I'm right next to it, and it heats into an insulated place? It seems it would only need to get a fraction of as hot as a normal heater, and there's no such thing as more or less efficient heat generation via electricity -- it's all 100% efficient, isnt' it? (though maybe not directed optimally, but the amount of total heat generation from every x-watt electric blanket pulling y amps is the same, isn't it? —Preceding unsigned comment added by 82.124.214.224 (talk) 02:47, 9 November 2008 (UTC)

if you operate an electric blanket at night and turn the thermostat down, you will probably save considerable energy. You can be cozy under the electric blanket while the house is allowed to become quite cool. When the trade-off is turning up the electric heater (as opposed to a gas furnace) the savings are even clearer. The downside is that some people worry about the small electric field from an electric blanket as an example of EMF Electromagnetic field which is hypothesized to be harmful. An ad for an electric blanket [1] says you can save 10% on your heating bill by turning the thermostat down while you sleep under the electric blanket. Edison (talk) 02:58, 9 November 2008 (UTC)

You appear to be presuming the OP is referring to central heating but I don't see any reason from the question to presume he/she is. The IP looks up to France, where I believe central heating is the norm, so you may be right but there is still the possibility the OP happens to live in a house without central heating. Definitely here in Auckland (or heck NZ) where central heating is relatively rare, it's likely you'd only heat your bedroom when your sleeping. While an electric blanket is still likely to be quite a bit cheaper, the costs savings are probably going to be less. Of course, if your me, a good feather-down duvet is enough for most winter nights in Auckland anyway. Nil Einne (talk) 11:14, 9 November 2008 (UTC)

The answer isn't a "yes" or "no". It is an "it depends". What is the watt rating on your electric blanket? What is the watt rating on your heater? I have a ceramic heater that is 12 watts and an electric blanket that is 35 watts. So, I figure I can run the little heater twice as long as the blanket and get the same energy usage result. So, you want watts to calculate kwatt hours used for some number of hours running each device. -- kainaw™ 03:05, 9 November 2008 (UTC)

I do not believe that you have an electric space heater which draws only 12 watts. Please check the nameplate more closely. I would believe 12 amperes. Electric blankets use far less power than any appliance calling itself an electric space heater. 03:07, 9 November 2008 (UTC)Edison (talk)

I agree with Edison that it is very unlikely that you would have a space heater of any kind that drew only 12 watts. (I would believe 1200 watts, or 12 amps, or somesuch.) In any case, just the rating on the packaging isn't necessarily going to tell you the whole story. Both blanket and heater will likely contain thermostats which regulate their output; neither device is likely to be operating at full power all the time. TenOfAllTrades(talk) 16:59, 9 November 2008 (UTC)

I'm sorry. I didn't mean to imply that I was referring to a space heater. It is a foot heater under my desk. I don't own a space heater. So, all I had to go on in my example is my electric blanket and my foot heater. -- kainaw™ 21:14, 9 November 2008 (UTC)

Unfortunately, heating your feet and heating your bed are rather different tasks, so the example isn't very useful. Space heaters use *enormous* amounts of electricity. --Tango (talk) 21:29, 9 November 2008 (UTC)

I can't even believe a small space heater under the desk only draws 12 watts. That is like a very small light bulb. Edison (talk) 22:00, 9 November 2008 (UTC)

Kainaw called it a "ceramic heater", perhaps it's like a heated ceramic tile that you rest your feet on? That could be quite efficient. --Tango (talk) 22:25, 9 November 2008 (UTC)

That is exactly correct. As I said - I don't have a space heater, so I just used what I did have for the example. My point was that when talking about power consumption, you don't have to guess. You can look at the power rating on the device and calculate rather accurately how much power it will consume over a period of time. -- kainaw™ 00:28, 10 November 2008 (UTC)

I don't think calculating it's that easy though. As you stated, you need to calculate how many kilowatt hours you use, but most heaters aren't on constantly so you'd need to work out how long the heater is actually on, e.g. by timing how long it's on for. I have used several oil filled heaters, either 1kW or 2.4kW. However even if the thermostat's at maximum, they don't usually remain on constantly. (And I personally have never felt the need to turn then on maximum except sometimes when I turn it off after I let it get very hot.) Indeed the 2.4kW one has 3 levels and then a variable thermostat. I'm not sure what the 3 different level setting does, it may adjust the power consumption (as opposed to simply affecting at what temperature the heater turns on and off). The simplest thing would be to get a power consumption metre you can plug the heater into. What is clear is that the electric blanket will win although by how much will vary. (Here in NZ again, our houses tend to be rather poorly insulated which would increase consumption of space heaters a fair amount Nil Einne (talk) 09:30, 10 November 2008 (UTC)

obviously, if the purpose is to heat your body from ambient temp to 65 or something, the less additional mass you have to heat, the more efficient. putting the heater in contact with your body is obviously a step in that direction, compared to heating the mass of air up in the room and wafting it over towards yourself.

on the other topic, yeah, 12 watts sounds about right for something you contact with your body/feet to keep warm, but inadequate for a space heater in anything but a small box (see above) if you think about the heat from a 12 watt bulb. Just for fun; the added heat from AGW currently is estimated about 1.7 watts per square meter, so imagine keeping your tootsies on that little warm tile. it's not insignificant.Gzuckier (talk) 20:23, 11 November 2008 (UTC)

You just reminded me that it is time for our local Darwin award nominees. Every year, about this time, people start setting their rooms on fire - usually killing themselves, by sticking space heaters under their blankets. -- kainaw™ 02:13, 12 November 2008 (UTC)

Speciation in the lab

What new species, if any, have been created in labs? NeonMerlin 03:24, 9 November 2008 (UTC)

See Mycoplasma laboratorium. --Arcadian (talk) 03:48, 9 November 2008 (UTC)

Not seeing species creation, at least not in that article... --98.217.8.46 (talk) 04:14, 9 November 2008 (UTC)

See Speciation#Artificial_speciation. --98.217.8.46 (talk) 04:14, 9 November 2008 (UTC)

spacetime deformation

If you have a supermassive object a great distance away from a planet or something.. The planet will gradually "slide down" the gentle (at that distance) slope of the object's gravity well, right? But a ball falling from a tower on the planet will accelerate far more rapidly than the planet is accelerating. So doesn't that mean that the gravity gradient is steeper locally than relative to the object? In other words, won't the planet just stay slumped in its own deformation of spacetime? Or is this just a trick of relativistic geometry? I know it's common sense that the planet doesn't have to labor up its own gravity well to move anywhere, but does relativity actually explain WHY it doesn't? This is what I mean: http://img204.imageshack.us/img204/2894/wellsbp1.jpg Thanks 71.176.179.91 (talk) 04:32, 9 November 2008 (UTC)

First, gravity wells are a Newtonian thing, not a relativistic thing—see gravity well. In Newtonian gravity the acceleration of an object depends on the net force acting on it, and the net force is just the sum of the individual forces. An object can't exert a net force on itself. So the forces acting on the planet are a bunch of internal forces, which are large but add up to zero (because they point in different directions), plus the force from the distant supermassive object, which is small but nonzero. The net result is an acceleration toward the distant object, and it's the same acceleration as if the internal forces hadn't been there at all. If you like, the object doesn't have to climb out of its gravity well because the gravity well is always instantaneously aligned with the object. Unlike the object itself, the gravity well doesn't have inertia. Gravity acts instantaneously and without any apparent mechanism.

In a relativistic field theory, like Maxwell's electromagnetism or general relativity, the field of an object can't be in constant "communication" with the object because of the light speed limitation. If the object's motion changes, the field won't know about it until some time later. So relativistic fields do have a life of their own, and they do have inertia. If you change the motion of a charged particle it experiences a force due to the delay in updating its own field, and I think this could be described as "laboring up its own potential well". (This force is called the "self-reaction" or "back-reaction", and Wikipedia doesn't seem to have a good article about it.) But that doesn't matter when the particle is being accelerated by gravity, because everything gravitates equally, including fields. The gravity of the distant supermassive object pulls on the planet and the planet's gravity well in equal measure, so the planet doesn't have to fight against the well.

(I should add that I don't actually know if there's a gravitational version of the electromagnetic back-reaction. It's tricky to discuss such things in gravity's case because everything gravitates. You can do electromagnetic experiments by pushing charged objects around with uncharged sticks, but you can't do gravitational experiments by pushing massive objects around with massless sticks. Even light has a gravitational field. It may not even make sense to say that there's a gravitational back-reaction, because there's no way to change an object's motion that the gravitational field won't know about ahead of time.) -- BenRG (talk) 14:02, 9 November 2008 (UTC)

The article self-force looks fine to me; did you not find it? --Tardis (talk) 01:36, 10 November 2008 (UTC)

mechanical reactionless drive?

File:BalloonsReaction.jpg

Supposing you have two spherical balloons full of water connected by a pipe full of water all the way though. if you squeeze one balloon equally in all directions, the pressure change in the pipe will make the other balloon expand equally in all directions. If they were in outer space doing this would not move the system backwards of forewords since all forces cancel out. But hold on-the center of mass has been moved without reaction in violation of Newtons third law of motion. Rotate the deflated balloon around 180 degrees and reverse the process and you have a reactionless drive. Or am I missing something?

--Trevor Loughlin (talk) 06:20, 9 November 2008 (UTC)

You aren't squeezing "equally in all directions" if you have a hole for water to flow through. Dragons flight (talk) 07:14, 9 November 2008 (UTC)

And the apparatus would probably move away from you... Due to viscosity and friction concerns, forces YOU apply will not be 100% returned via the opposing balloon, and there will be a net force away from you that will be unopposed... --Jayron32.talk.contribs 11:30, 9 November 2008 (UTC)

I thought someone would bring up the "hole the water flows through" but since this "hole" is connected to a solid metal tube with water all the way through it (no air bubbles to be displaced) connecting to the other balloon, there is only a pressure change and I assume this would not cause a backwards reaction. As for friction and viscosity, what if you used liquid helium? An easier to understand thought experiment equivalent to the two balloons (in fact more practical) would be two bicycle chains with a j shaped weighted section on two spindles oppositely opposing each other in a reflection at the top of the j to prevent rotation. If both were synchronized and the weighted section was pulled up on the bottom j chain and down on the top, I am sure that the center of mass would move without any reaction in the opposing direction. I will of course try my experiment unless you can convince me I am wasting time and money, but (as with EM drive) a space based test would be the only convincing evidence.

It is not as if I haven't broken fundamental laws of physics before-I have already built and tested in (lucrative) real world applications a Quantum Superluminal Communication Device which violates causality by transmitting USEFUL binary data from the future to the past. But I am not putting the plans for that here because I intend to rule the world using it! —Preceding unsigned comment added by Trevor Loughlin (talk • contribs) 15:34, 9 November 2008 (UTC)

I think you'll find that you can move the center of mass around, but the balloons won't actually move anywhere. And rotating the balloons around is cheating- it requires energy from outside the system 71.176.166.28 (talk) 17:31, 9 November 2008 (UTC)

No, the hole implies a pressure imbalance, which implies a net force, and is what causes water to flow from one side to the other. Dragons flight (talk) 23:06, 9 November 2008 (UTC)

An electric motor with or without a gyro will spin the system. Without a gyro it will move in the opposite direction, but as long as its mass is similar to the reactionless chain assembly it WILL easily rotate the centre of mass.

The change in PRESSURE causing a reaction to spoil this system IS a more serious objection. But this applies to the balloon experiment. What about the chain based experiment in the image above? It also has four linked units to counteract all rotational forces. I will build this unless you can find a flaw.—Preceding unsigned comment added by Trevor Loughlin (talk • contribs)

Regarding the balloon drive, I think the fly in the ointment is the need to "Rotate the deflated balloon around 180 degrees". Is this rotating one balloon or rotating the entire system around the centre of mass? CBHA (talk) 06:12, 10 November 2008 (UTC)

The whole system is rotated by a motor, which will rotate in the opposite direction at the same speed if it is the same weight, but an alternative might be to have two separate motors on each balloon and alternately switch them on when each balloon inflates. The balloons would have to be connected by a U shaped solid tube or they WOULD impart a reaction and spoil the plan, because the balloon would push against the tube as it expanded outwards pushing the whole system backwards. But with a U shaped tube the expanding balloon would push DOWNWARDS forcing the system to ROTATE UPWARDS (not the required sideways movement from the top motor needed when the appropriate balloon has expanded. This problem is eliminated by pairing an opposing (upside down U) reactionless drive as a mirror image,so that each rotation cancels out, as with the chain drive version's. —Preceding unsigned comment added by Trevor Loughlin (talk • contribs) 13:50, 10 November 2008 (UTC)

This is an easy one: The center of gravity (strictly, the center of mass) of the system composed of the two balloons, the water and the pipe would not move as you squeezed the balloon. So the whole machine would move as you squeezed - but then stop when you stopped squeezing. When you release the balloon and the water sloshes back again, the center of gravity still doesn't move - although the machine moves back the other way - and you're back exactly where you were. Rotating the system about it's center of gravity leaves that center of gravity in the exact same place - so there is no net motion and this doesn't buy you anything. To rotate it other than around the center of gravity requires reaction mass...so the only way to propel yourself forwards is to expel reaction mass...pretty much as you'd expect. SteveBaker (talk) 14:56, 10 November 2008 (UTC)

Rotating around a point other than the center of gravity could be accomplished by attaching two such devices at a common rotation point, and have them rotate in opposing directions. Of course, the center of mass would still not be changing position.

The problem of a reactionless drive has existed for ages. It's a staple of science fiction. The Dean drive is the most infamous example of a "real" reactionless drive. ~Amatulić (talk) 22:24, 10 November 2008 (UTC)

AHA-but if the balloons are connected by a U-SHAPED TUBE then this "sloshing back" will create a ROTATIONAL rather than a LINEAR reaction. This "rotational reaction" could be CANCELED OUT by a second version of the U+balloons linked in a mirror image, since the two rotations would be in opposing directions. (I am not talking about the desirable sideways rotation of the whole system created by the motor to spin the smaller balloon about the bigger one, but an undesirable top to bottom spin of the single U which force the centre of mass back to where it started.) —Preceding unsigned comment added by Trevor Loughlin (talk • contribs) 02:30, 11 November 2008 (UTC)

Look - you can make this as complicated as you like - it's NEVER going to work. The whole "every action has an equal and opposite reaction" thing means that you're screwed no matter how complicated you make it. The best you can hope for is to exhaust our patience in breaking down these increasingly bizarre (and useless) contraptions to find the 100% inevitable flaw in them. That doesn't mean that you succeeded in inventing a reactionless drive - it just means that you became boring and we gave up trying to explain one of the most fundamental laws of physics to you. Feel free to consider that this may already have happened. SteveBaker (talk) 04:02, 11 November 2008 (UTC)

I am fully aware of the implictions of Newtons third law of motion, and am skeptical of reactionless drives, otherwise I would have built and patented this design rather than put it in the public domain. However, rather than dismissing it out of hand, look at the diagrams and tell me how it will not work in detail-for example where do the forces NOT counteract, leading to the system moving back to where it started? —Preceding unsigned comment added by Trevor Loughlin (talk • contribs) 05:14, 11 November 2008 (UTC)

Why bother though? Either Newton's laws work and the exercise is meaningless. Or (and this is important) Newton's laws don't work, and the whole thing is even more meaningless, because then we would literally have no way of predicting what your device (or any other device) will do in any untried situation.

If you have outside, physical evidence that Newton might be wrong about something, then that's a whole different story. But this question is simply asking us to use Newton's laws to prove Newton's laws. It's just an exercise in futility. APL (talk) 05:46, 11 November 2008 (UTC)

I agree with SteveBaker. My interest ran out when you suggested a U-shaped tube instead of a straight one, thereby changing the entire configuration.

But if you think the chain drive mechanism has potential, please do not be discouraged by negative comments here. A lot of good ideas have been subjected to negative comments. Go ahead and build one. If it works, I expect NASA or some other space agency will be interested to see the results. CBHA (talk) 05:59, 11 November 2008 (UTC)

I will. If it is going to pass the "swing" test it will need several blocks in series of the four unit drives illustrated so that there is always some momentum, like a four stroke engine (assuming it works) and if it can constantly pull a swing at an angle this might work. —Preceding unsigned comment added by Trevor Loughlin (talk • contribs) 14:21, 11 November 2008 (UTC)

Come on guys, this one is easy. SteveBaker had the answer already. Let's assume frictionless pipes and zero viscosity water just for fun. Look at the first diagram, top left at the top of the section. When you compress the right balloon, it exerts a force against the balloon in every direction, all of which cancel out - except the upward/downward direction, since the water is free to flow down the pipe. It reaches the end of the "down" section and exerts a force downward on the down side of the horizontal pipe section, cancelling out the until-now-uncancelled up force it exerted on the top of the balloon during compression (thus this balloon will move slightly up, keeping the centre of mass fixed as the water travels down). Since it is stuck in the rightmost elbow joint and has stopped going down, it exerts forces on that joint in all the lateral directions too. These all cancel out except for the left/right forces, as the water flows to the left along the pipe. Eventually it reaches the end of the horizontal bit of pipe and exerts a force on the left wall of the pipe, finally cancelling the force it exerted to the right (so the pipe will move slightly right, keeping the centre of mass fixed as the water travels left). More forces in every direction, but the water can flow up into the left balloon. There's a delay between the force on the down side of the left end of the pipe and the force on the top of the left balloon, so this balloon will move slightly downward, keepign the centre of mass fixed at all times. So as you do this you will set the device rotating (in rotation that will be cancelled once you stop pumping water through as the forces cease in reverse order), but its centre of mass remains fixed at all times. Finding the inevitable flaws in "free energy" machines like this is a mildly entertaining but ultimately pointless endeavour. Maelin (Talk | Contribs) 14:41, 11 November 2008 (UTC)

Indeed.

I could come here and describe a Heath Robinson (Rube Goldberg for you Americans) contraption with hundreds of swinging, spinning, spring loaded parts with hydraulic and pneumatic connections using exotic mixtures of non-Newtonian fluids, clockwork parts, gyroscopes, magnets, complicated chemical and biochemical reactions, active electrical parts and photonic circuits, with complex computer software driving them using neural networks and evolutionary algorithms - being driven by a trained parrot...and then ask the people at the science desk to tell me why it doesn't work as a reactionless drive. The result is likely to be exceedingly difficult to analyse at a detailed level - it would result in days of complicated discussions and arrive at the same conclusion. You can't expect us to do that. It's utterly unreasonable - BUT it's also utterly unnecessary.

To analyse your machine (or mine), we simply place an imaginary black box around whatever bizarre contraption we are presented with and boldly assert (per Newton's first law of motion) that the center of gravity of "the system" (the black box) doesn't move no matter what happens inside because "the system" is in a state of motion that it will continue in unless some external force is applied to it.

In the case of a rocket engine - the exhaust from the engine remains inside the box - so the center of gravity of the rocket and all of it's exhaust gasses doesn't move. In the case of a light-sail, the light source, the craft and all of the photons are inside the box - and the box doesn't move. So for the craft to move bodily one way - something else has to go off in the opposite direction to keep the center of gravity of "the system" where it is. Hence there are no "reactionless" drives - per Newton's first law.

If you want to assert that this is not the case then you are explicitly denying Newton's laws - and we're going to laugh at you. If Newtons' laws are somehow incorrect at "human scales" of speed, mass, distance and time - then we're in a much weirder universe than we believe. It's utterly inconceivable that any device you could think up like this could break those laws without doing something relativistic or quantum-level or close to a black hole or in some other way radically 'pushing the envelope' of the realms of experimentation we've done as a civilisation for the past 300 years. Mundane stuff like pipes and liquids and spinning things are just too well tested for the laws to be incorrect at those scales.

The whole beauty of these kinds of fundamental law is that they allow simplification. We don't NEED to calculate the energy in each part of the system, analyse the torque and the tensors and the inertia and all of that stuff. We know for an absolute certainty that if we went to all of that trouble - the answer would be "NO!". We have a simple law of nature that's always proven correct that says that we can shortcut all of that analysis and treat the system as a black box. It's the same deal with perpetual motion machines - the laws of thermodynamics say "NO!". So there you go: "NO!" - and the explanation is very simple indeed. SteveBaker (talk) 15:05, 11 November 2008 (UTC)

I admit that this idea is so simple that it is highly probable that someone must have tested it at some time and failed, though I have made a search. It will not be my highest priority to build! So what do you think the chances of an idea that really does "push the envelope" such as EM drive, has of actually moving a craft in a straight line in space without rockets?

Certainly the device IN THE FIRST DIAGRAM will ROTATE about its centre of mass and not move anywhere when transfering the fluid. Worse still, without any air resistance or gravity it would end up on its SIDE by the time all the fluid is transfered, so instead of the motor being able to swing the small balloon around the big one, and then reverse the process, the motion is now upwards instead of sideways, and another cycle will put the device the wrong way round. But what about the second diagram, when two mirror image units counter each others rotation in the Z-axis? —Preceding unsigned comment added by Trevor Loughlin (talk • contribs) 02:54, 12 November 2008 (UTC)

It's not necessary that someone tested something just like this. You're missing the "big message" here. Newton's first law says that if we draw a line around some kind of 'system' (your machine in this case) then unless some external force operates on it - it's just gonna sit there with it's center of gravity not moving by so much as the diameter of an atom no matter what weird-assed motion it's going through internally.

We've tested all of the parts of standard machines and we know how such things behave. So:

• The chances of a reactionless drive working with 'ordinary stuff' such as you describe is zero - not gonna happen.
• The chances of making a reactionless drive that uses some aspect of relativity or quantum theory (or preferably, both) seem exceedingly slim - the science is pretty well known.
• If I'm pushed into imagining some kind of experiment (likely, just a thought experiement) where there might be some doubt as to whether Newton's law might break unexpectedly - then I guess it's remotely possible that some insanely 'extreme' design that has the mass of three galaxies - or which was just over a 'planck length' long or only works when cooled within a billionth of a degree above absolute zero - or which would have worked within a picosecond of the big bang but cannot work at any time since - or which only works in 17-dimensional space....something like that might maybe expose a new aspect of the laws of motion that could perhaps require a small adjustment to Newton's first law that might just open a chink of hope for a ridiculously impractical reactionless drive.

But I really don't imagine any such thing could ever be remotely practical - simply because everything within the realms of "possible" have been so well researched.

SteveBaker (talk) 05:10, 12 November 2008 (UTC)

One other idea. Supposing you had two spindles with a heavy chain inside a rectangular box, and a motor at each end wound/unwound the chain from one spindle to the other. The box would rotate about its center of gravity due to the spin of the motor at each end. But if we have TWO of these boxes back to back what would happen? Would the box move linearly backwards as the in the opposite direction to the winding of the chain instead of spinning? If it did not, then mass would have been transered without reaction,so it probably would, but I am not sure what would happen. It certainly WOULD move backwards if the chain was pulled in without being still wound on the other spindle. But as long as the chain is attached to the two spindles, it can't move backwards(?), otherwise with large enough spindle and rope the device would move further than its length in any case,violating Newtons third law in any case(?)

LISTEN TO ME. NO! It's Newtons' first phreaking law. You will NEVER come even close to breaking it with any crazy contraption you can come up with. Give it up and go back to squaring the circle, performing a trisection with ruler and compass, making a perpetual motion machine, calculating the last prime number (and the last digit of PI)...those are all just as likely as you coming up with a reactionless drive. SteveBaker (talk) 01:18, 13 November 2008 (UTC)

Look. Whenever you have closed system consisting of one object exerting a force on another object and things moving each other around however you like, all the forces involved will end up balancing each other out perfectly, and the centre of mass of the whole system will continue with the exact same momentum as it had initially. If you have some complicated construction made of pipes and water and balloons and motors to spin bits around and pumps to move the water, no matter how you rig it up, if it's a closed system its centre of mass will stay exactly where it is the whole time. This is unavoidable.

Every time someone proposes a reactionless drive, or perpetual motion machine, or whatever else that will violate conservation of momentum (or conservation of energy, or entropy, or whatever), they have always taken a very basic machine and tacked on bits and altered it until it is sufficiently complicated that they can no longer understand how the forces will cancel each other out, then they say, "lo! a perpetual motion machine!". But it isn't. You can take one thing that isn't a reactionless drive, and staple it to another thing that isn't a reactionless drive, and guess what: you will still not have a reactionless drive. No matter where you move the water, the centre of mass stays the same. No matter how you rotate bits, the centre of mass stays the same. You can't escape conservation laws just by being complicated. Maelin (Talk | Contribs) 11:24, 13 November 2008 (UTC)

If the chain is continuous spiral winding/unwinding should not cause back reaction (I think) and the mirror image device in the same box will prevent rotation.

If the chain was not continuous it would be subject to Newtons third law and not work;

The statement above is about a reactionless drive finding a loophole in Newtons third law, not a perpetual motion machine breaking the first law of motion-that would be a tall order. This machine would use electric motors with either a solar panel or RTG battery as the power supply.

Human brain -> computing power

Hi,

One thing I've wondered is how much memory space does the human brain harbour, in terms of computing power (eg Gigabyte, terabyte, etc.). I'd be interested to know the answer to this! Thelb⁴ 13:07, 9 November 2008 (UTC)

The best answer on this Yahoo! answer is pretty good. —Cyclonenim (talk · contribs · email) 13:55, 9 November 2008 (UTC)

Thanks, that's very interesting! 81.151.36.130 (talk) 14:23, 9 November 2008 (UTC)

But we don't store information like a computer. That explanation vaguely hints at "compression" but I think this misrepresents what we are storing. When a computer stores a video file, it stores every single pixel of every single frame, and every single sample of every audio signal, at some fixed sampling rate (30 frames x 320x240 pixels)/sec, + (44100 audio samples per second). Then, effective compression locates redundancy in these stored data, and removes that redundancy (with some "loss" in the exact representations, sufficiently engineered to be minimally noticeable to a human).

But when a human "remembers" a movie, it doesn't do anything at all like that! If you were asked to recall Pixel (122,61) at Second # 4051 in your favorite movie (that you've watched a dozen times and memorized all the dialog for), you would not be able to do ANYTHING like recall that information. You wouldn't even be able to get a glossy "key-frame", or the blurred-pixel-square representation from Second #4050 at Pixel (128,64) which you would presume is probably close to the request. You might not even be able to remember the color of the main character's shirt. At the same time, if a clever psychology-experimenter made you watch the same movie twice, with the only changed-detail being a different shirt-color on a minor character, there's a significant chance that you would have noticed that detail. How did your brain selectively notice something which you might not be able to recall if directly questioned? Clearly information-storage is selective and complex.

It's really hard to estimate the capacity for human brain's information storage system when we know so little about how it actually works, so I would be very reluctant to assign any numeric estimate of capacity - you're comparing serial, precise machine storage to a mash-up of biological/analog distributed fuzzy storage, where the concept of "one bit" probably does not apply. Nimur (talk) 17:23, 9 November 2008 (UTC)

It doesn't really make sense to discuss the information stored in the brain in terms of bits and bytes, so estimates of storage capacity in terms comparable to those of a computer are pretty misleading. But, your question can be answered, and the answer is surprising: the human capacity to store factual memory appears to be near limitless. A Russian neurologist named Dr. Luria studied a man named Solomon Shereshevskii who apparently had limitless memory. Since then it was taken for granted that the Luria case was an extreme outlier, but in 2006 researchers at UCI discovered first one, then almost immediately afterwards 2 more people with seeming unlimited memory, a condition they termed Hyperthymesia. The speed with which new cases were discovered suggests that this capacity may actually be somewhat common, but like synesthesia, affected individuals may take for granted that they are no different from the rest of the population. --Shaggorama (talk) 22:59, 9 November 2008 (UTC)

I don't believe in this limitless-capacity idea. A poem, even an epic poem, doesn't take much RAM to store. Moby Dick is only about 10⁶ characters long, and state-of-the-art compression algorithms can get it down to 100-200 kilobytes. The pi-memorization world record seems to be 67890 digits, which is only 28190 bytes (10⁶⁷⁸⁹⁰ ≈ 256²⁸¹⁹⁰). That's the record among everyone who's ever participated in these contents; I have a hard time believing that Luria's mnemonist would have done better.

There's a famous paper by Thomas K. Landauer, How Much Do People Remember? Some Estimates of the Quantity of Learned Information in Long-term Memory (Cognitive Science 10, 477–493 (1986), online here), which estimates the total memory capacity at about 10⁹ bits. There's no consensus that he's right, but at least his measurements are based on actual empirical tests of what people remember. The neuron-counting numbers are an attempt (though not a very principled one) to estimate something else, namely the number of bits necessary to describe "the current structure of the brain". But even in an analog computer the number of states is astronomically larger than the number of computational states. The raw physical states don't behave predictably enough for reliable computation. -- BenRG (talk) 13:36, 10 November 2008 (UTC)

Obviously, it's not limitless - but the idea that memory can become fuzzier and less detailed as the memory becomes less and less important does allow it to SEEM limitless because we're able to recall the important details but merely have a broad-brush fuzziness about the less relevant parts. This does indeed suggest that some pretty sophisticated 'lossy compression' is going on - and it's happening on-the-fly. So a new memory pretty much has to displace some older information - but it's not like a computer where you delete an old file to make room for a new one...learning calculus doesn't make you forget your wedding day...what it does is to make you forget the color of your best man's shirt and whether you gave your bride an orchid or a rose. This is more like turning down the JPEG or MP3 'quality' dial on older files to make them take less space so that there is room for new files. SteveBaker (talk) 17:48, 10 November 2008 (UTC)

I mean, I don't really know what to tell you guys. Whether you 'believe' it or not, the cases I described exist. BenRG, the paper you cite is from 1986, which is a pretty old citation for neurology paper. The neuron counting attempts look nice and produce impressive numbers, but they assume an antiquated, folk-scientific principle known as the "Grandmother Neuron," a theory which states that individual objects of memory are stored in their own respective neurons. How memories are encoded is still very, very unclear, but current community consensus is leaning more towards distributed neural network implementation, an idea which makes it very difficult to give an estimate of bits and bytes. If the neural net theory is right, then the net is signficantly greater than the sum of its parts (i.e. can retain more information than the number of involved neurons may suggest). Furthermore, related ideas may be stored in mapped areas, further compressing data. When it comes down to it, the fact remains: the architecture of the brain does not resemble that of contemporary computers, either in hardware or software, so these bitwise comparisons are meaningless to begin with. Obviously the human capacity for memory is not actually limitless: physical constraints on the brain obviously must put a cap on the possible amount of data we can store. But, since we just don't know how the data is stored, there's no good reason to just assume that the memory capacity of the brain necessarily must be exceeded in a human lifetime. Although they are outliers, the 4 cases I described (in particular the latter 3) strongly support this. As far as steveabakers "lossy compression" goes, current theory seems to suggest that its more likely lossy retrieval: memories are stored intact, but are cached in a way that permits for more rapid access of the most pertinent information (and consequently, normal people lose access to alot of the details). If it was just lossy compression, repressed memories wouldn't be so problematic for traumatized individuals: the bad parts just would have been erased/stored poorly. Instead, it seems that the bad parts just often aren't accessed consciously. --Shaggorama (talk) 16:29, 11 November 2008 (UTC)

Bell inequality

I get the idea behind it and I get the experiment and scoring. But I don't really understand the actual implications of the Bell inequality being violated. This is sort of how I understand it.. is one of these correct, or am I totally off? The article is tough!

1. After observing a particle's spin, you can guess its entangled partner's spin (even measuring orthogonally to the first measurement) with better-than-random success by guessing the same as the first particle's spin.
2. After observing a particle's spin, you can guess its entangled partner's spin (even measuring orthogonally to the first measurement) with better-than-random success by guessing the opposite of the first particle's spin.

71.176.166.28 (talk) 14:38, 9 November 2008 (UTC)

Mmm... you've kind of got the implications wrong. If we lived in a world without Bell inequality, you'd have 100% chance of guessing the spins. Then, as Einstein wanted it to be (see EPR paradox), quantum mechanics would allow you to have "complete" knowledge of a given particle despite the Uncertainty Principle, which would mean that Uncertainty Principle was based primarily on issues regarding observation. But lo! Because of Bell inequality you can't just guess the spin—you will be wrong a significant part of the time, and your knowledge of the universe is incomplete, per Uncertainty Principle, and thus Einstein was wrong. Does that clarify things a bit? It actually resolves a fundamental question about QM and the Uncertainty Principle, one that most people thought could never really be resolved (most thought that one's take on whether particles contained all information—but that some of it was just hidden due to UP—was a matter of epistemological opinion, and couldn't be actually tested). --98.217.8.46 (talk) 15:03, 9 November 2008 (UTC)

No, the original poster has it the right way around—the nonclassical correlations in quantum mechanics are stronger than classical correlations at somealmost all angles, meaning that you have a better chance of guessing the spin. E.g. if the two spin measurements are separated by 60° then given the outcome of one you can predict the outcome of the other with 3/4 accuracy in quantum mechanics, while you can't do better than 2/3 in any local classical theory. (That's the specific case used in the gambling-game example I linked below.) Newtonian physics is deterministic, but Bell's theorem applies to any theory that follows the rules of classical probability, even theories that have true randomness or theories in which measurement has an unavoidable effect on the thing being measured. -- BenRG (talk) 17:34, 9 November 2008 (UTC)

Same vs. opposite doesn't matter—it's just a question of how you set up the spins. If you started with the state |↑↑〉 + |↓↓〉 then they'll be the same, if you started with the state |↑↓〉 + |↓↑〉 then they'll be opposite. In practice the easiest way to get a Bell state is from the decay of a spin-0 system, which (by conservation of angular momentum) has to lead to antialigned spins. That's why you'll more commonly hear "opposite" in descriptions. But there's no deep difference—you can turn one into the other by rotating one of the particles 180 degrees, or by simply redefining your coordinate axes. In the remainder of this reply I'll assume equal (not opposite) spins.

I'm not completely sure what you mean by "orthogonal" since orthogonality of wave functions is different from orthogonality in space. (For example the spin states |↑〉 and |↓〉 are orthogonal as wave functions but parallel in space.) But if the spin measurements of the two particles are made at angles differing by 90° in space, then your same-spin guess will be right 50% of the time, i.e. it's no better than flipping a coin. If the measurements are made in the same direction (0°) then you will be right 100% of the time. If the measurements are made in opposite directions (180°) then you will be right 0% of the time. So far there's nothing very surprising here. But quantum mechanics says that, in general, if the separation angle is θ then you will be right cos² (θ/2) · 100% of the time, whereas there's an argument from fairly general classical assumptions that you can't be right more than (1 - θ/180°) · 100% of the time for those other angles, no matter what internal algorithm the two particles are using to decide the outcome of the measurement. (edit to add: sorry, silly mistake here. Obviously if the measurements are separated by more than 90% you can guess the opposite of what you saw and be right more often than that. The formulas should be max(cos² (θ/2), 1 − cos² (θ/2)) and max(1 − θ/180°, θ/180°) (for 0 ≤ θ ≤ 180°). The quantum formula is at least as large as the classical maximum for every θ and it's strictly larger except when θ is a multiple of 90°.)

But I wouldn't worry about it. There are lots of ways of showing that quantum mechanics is nonclassical, and Bell's is not especially clear. He proved too much—it's not necessary to get a result valid for any angle, you just have to show one example of quantum mechanics violating classical assumptions. I like this gambling-game version (which isn't original to me, though that explanation is mine).

All of the above assumes spin ½ particles (like electrons). If you're using spin 1 particles (like photons) then you should divide all of the above angles by two, and your Bell state will look more like |↕↕〉 + |↔↔〉 (or |↕↔〉 + |↔↕〉). -- BenRG (talk) 17:08, 9 November 2008 (UTC)

So the .707 figure means that if you guess "same" for the second particle you'll be correct 71% of the time, instead of the 50% that is the non-QM answer? I guess that makes sense, especially in light of your point about same vs opposite just being a matter of how you set up your axes 71.176.166.28 (talk) 17:41, 9 November 2008 (UTC)

(see my edit above) The probability depends on the angle between the measurements, and I don't think there's any simple angle for which it equals .707 (i.e. there's no simple θ for which cos² (θ/2) = √2/2). The non-QM maximum is 50% at an angle of 90°, and the QM prediction at that angle is also 50%. You need to use some other angle to get a difference. Note that the QM formula is a prediction of quantum mechanics, while the classical formula is not a prediction, it's a maximum over all possible physical theories that satisfy certain assumptions. -- BenRG (talk) 11:33, 10 November 2008 (UTC)

That link is great. So it's saying that if each player had a sheet of paper that says something like:

• If the number you're given is a 1, say YES
• If the number you're given is a 2, say NO
• If the number you're given is a 3, say YES

then they make on average (1 & 2 is +1) + (1 & 3 is -2) + (2 and 3 is +1) = ZERO. But they can actually make more money using entangled particles than with a predetermined answer key? That's amazing! thanks a ton 71.176.166.28 (talk) 18:36, 9 November 2008 (UTC)

Yes, but moreover I'm arguing that they didn't have any better option than to use a crib sheet like that, i.e. it's not just the best crib-sheet-based strategy but also the best strategy of any kind (subject to certain assumptions about the way the world works). -- BenRG (talk) 11:33, 10 November 2008 (UTC)

Is the space shuttle capable of breaking Earth's orbit?

Assuming that they wanted to, is NASA's space shuttle physically capable of breaking out of Earth's orbit? 67.184.14.87 (talk) 15:15, 9 November 2008 (UTC)

I've just done some research into that and I think the answer is "no". The main engines can get the shuttle into Low Earth orbit, they would then need about 3.2km/s of delta-v (according to that article) to reach an escape orbit. As far as I can tell, the only engine they could use for that is the Orbital Maneuvering System, which only carries enough fuel for 0.3km/s of delta-v, far short of what they would need. They also have Reaction control system engines, but I can't find the maximum delta-v for those, I very much doubt it's enough to make up the extra 2.9km/s they would need, though. --Tango (talk) 16:20, 9 November 2008 (UTC)

The main cargo bay should have ample capacity (volume and mass) to store enough extra fuel for escape. As far as I know nobody ever bothered to design a tank to be fit in the cargo bay, and it might also be a major problem to get the fuel from there to the main engines. —Preceding unsigned comment added by 84.187.126.130 (talk) 16:45, 9 November 2008 (UTC)

That's assuming the engines can handle a significantly longer burn. Even if they can, you're talking about a massive retrofit. --Tango (talk) 17:09, 9 November 2008 (UTC)

Maybe things like low energy transfers, Interplanetary Transport Network and gravity assist could eventually reach the needed speed. It would be too slow to be suited for a manned mission. PrimeHunter (talk) 17:00, 9 November 2008 (UTC)

There are dozens of other considerations, such as whether the radios and communications equipment has sufficient range to operate in other orbits; whether the life support systems can sustain longer voyages; if the crew compartment is sufficiently protected from radiation in higher orbits; whether it would be safe or structurally feasible to store fuel in the cargo bay; in short, the space shuttle was designed for its current orbit. With significant modifications, it could probably be retrofitted to do a lot of different things, but a sufficiently re-engineered vehicle would no longer be the space shuttle as we know it. Nimur (talk) 17:03, 9 November 2008 (UTC)

I don't think any of those things would help with getting from LEO to an escape orbit. Low energy transfers, etc, would require reaching a Lagrangian point, I think, which are all far above LEO. A gravity assist isn't going to help, that requires you to pass near a massive object that is in orbit around the object you're orbiting (usually the Sun), the only relevant massive object when you're in LEO is the Earth itself, and you can't get a gravity assist from the object you're orbiting (the moon does, in a sense, via tidal interactions, but it is many orders of magnitude bigger than a space shuttle and even it only gains altitude at a rate of centimetres a century). --Tango (talk) 17:09, 9 November 2008 (UTC)

In space aerodynamics are irrelevant, so they could add on whatever structure and supplies were needed for an extended mission. The extra food, fuel, life support assets, and even extra solid fuel rockets could be taken up in one mission and used in another. I do not think the low earth orbits of the shuttle are significantly less exposed to radiation than places farther from earth. The big problem I see is that the shuttle is designed to be strong enough and aerodynamic enough for reentry,with wings, control surfaces, parachutes, and landing gear, and that makes it more massive than a craft would have to be whose only goal was to get beyond earth orbit with a human crew. Or did someone say the craft should land on another planet, orbit another planet, or return the crew to earth? The devil is in the details of the mission requirements. If you want to send a shuttle out of earth orbit, it is certainly possible with today's technology, as long as you do not expect it to land a crew somewhere and come back. Edison (talk) 21:54, 9 November 2008 (UTC)

Surely when Apollo 11 orbited the moon it could have performed a gravitational slingshot out into space. It would only have taken a small tweak. It's not exactly the space shuttle, but it is ancient technology. Plasticup ^T/C 15:56, 10 November 2008 (UTC)

In fact, the maximum Δv capability of the Saturn V would have caused the Apollo capsules to approach the Moon much faster than they actually did, reducing the trip from three days to one day. The slight problem with that is that the capsules would have been on a hyperbolic lunar orbit, and wouldn't have been able to reduce their speed enough to enter a stable elliptical orbit, so they would have zipped past it.

Going back to the original question, the Shuttle doesn't have enough fuel to get out of Earth orbit right now, but it could have if an extra fuel tank were added inside the orbiter (with the tradeoff of a significant safety risk, orbiter redesign, and a drastically reduced payload capacity), and/or larger SRBs were used. The SSME's should handle it, as they are [supposedly] designed for multiple uses. Titoxd^{(?!? - cool stuff)} 19:12, 10 November 2008 (UTC)

Apollo 11 was launched with much a much bigger rocket than Space Shuttles are. Making bigger rockets isn't that difficult, it's just more expensive. --Tango (talk) 21:49, 10 November 2008 (UTC)

Even with a cargo bay full of fuel - there could still be significant problems: The cargo bay doors MUST open because without them the orbiter doesn't have enough electricity to keep going for more than a few orbits - and it doesn't cool properly without them opened to increase the surface area of the orbiter. I very much doubt they are strong enough to stay open while doing a burn to a higher orbit...so maybe this is still impossible...I'm not sure. Certainly there would be MANY other obstacles...not least, on reentry, the orbiter would be going much faster than it's designed to come in at...so right there is a big problem. Certainly many of the onboard computer systems would have to be reprogrammed (a not inconsiderable task because every flight-critical software function was written independently by several different teams of programmers and the results compared so that software bugs may be eliminated (mostly!). It's not a simple matter. But the shuttle is already being phased out - this could never happen. SteveBaker (talk) 21:56, 10 November 2008 (UTC)

Please source the bit about the bay doors having to be open for electric generation. Solar cells? radiators from fuel cells? The rockets to boost out of earth orbit need not be as powerful as the ones to lift off from Cape Kennedy, so they can be sized to as low a force as the structure can withstand. 1 G of acceleration is certainly within the structural capabilities of every part of the shuttle, since it endures it on earth, and it would allow an amazing velocity over a little time. Reentry? Where? The question was LEAVING earth, not landing safely back at earth or on Mars or elsewhere. The biggest problem, as I stated, is all the extra mass the reentry equipment represents. A craft to leave earth orbit and go BEYOND need not be aerodynamic at all, and need not have wings, landing gear, parachutes, etc, unless a landing stage for Mars or somewhere is desired, and that does not need to be the whole interplanetary craft. Add enough booster rockets, or enough fuel for the present shuttle, and we could get the thing out of earth orbit with today's technology. The program modifications are relatively trivial. Edison (talk) 03:23, 11 November 2008 (UTC)

Inside the cargo bay.

You shouldn't really need a source for that - just take a look at any photo of the shuttle in orbit and note that the inside of the shuttle bay doors are completely covered with thermal radiators and solar cell panels. Those have to be on the inside because they'd never survive launch and reentry if they were on the outside. Here is a (google cache) explanation of how those radiators matter so much in orbit: [2].

I guess, technically, the OP's question doesn't require the successful return of the craft - but in practical terms, it's bloody stupid to send a reusable spaceplane on a one-way trip. It's not unreasonable to assume that the OP was really talking about a return capability - and for that the re-entry issues are surely significant (and, I believe, insurmountable). SteveBaker (talk) 03:50, 11 November 2008 (UTC)

So we get rid of the wings, the landing gear, parachutes, possibly add a landing craft, add whole new rockets and fuel tanks, etc. In what way is this still the Space Shuttle? You've just constructed a whole new interplanetary craft, and yes, we know we have the technology to get to Mars, we've sent numerous probes there. The only difficult bit is life support (and man rating everything), we've done everything else. --Tango (talk) 11:37, 11 November 2008 (UTC)

Yes - exactly. You wouldn't be able to take people to Mars in the shuttle anyway - there isn't remotely the space or facilities you need to allow people to exercise - or to generate spin gravity - and for sure you couldn't carry enough oxygen, food, water, recycling gear, etc. (Particularly because you filled the entire cargo bay with fuel tank already!)...you'd also find so many other little subsystems that you'd have to replace: Are the radios powerful enough to reach earth from that distance? Does the inflight software cope with radio transmission delays greater than a second when communicating with earth-based systems without 'timing-out' and giving up? Can the antenna be pointed with enough precision to aim at a more distant earth?...and that's considering just one of dozens of critical subsystems. SteveBaker (talk) 14:33, 11 November 2008 (UTC)

If the point was simply to leave orbit for the sake of doing it, perhaps a large solar sail could be used. It probably wouldn't require major structural changes to the shuttle. Given enough time, and a large enough solar sail that can be furled and unfurled strategically you could get pretty far away. The shuttle could carry an enormous sail in it's cargo bay and still have room for extra oxygen and water supplies. The longer time frame of the mission could (in part) be offset by a smaller crew.

The problem with this idea, of course, is that there's no good design for a solar sail of that size yet. The obvious designs don't hold up in computer simulations. APL (talk) 16:18, 11 November 2008 (UTC)

There's another potential problem - there is still a significant amount of atmosphere in LEO, atmospheric drag from the sail would probably be greater than the light pressure. You could try having it only open when it would be angled in line with the direction of travel, but I think the best time to have it open is when the light would push the shuttle forwards, rather than up (see the discussion somewhere above about the opposite case of pushing backwards being a better way to de-orbit that pushing down), so it would be extremely inefficient. If it worked at all, it would probably take months to get up to escape velocity, far longer than even a small crew could survive on even a modified shuttle (you could probably do it with major modifications, but then you're back into the realms of it not really be a space shuttle any more). --Tango (talk) 16:33, 11 November 2008 (UTC)

I think that we have established that the answer to the bare question is "Yes, the shuttle could be rocketed out of earth orbit." Other answers are "No, it could not do it as presently configured and equipped, due to a lack of fuel. And the crew could not survive indefinitely, although the habitability could be extended considerably by carrying extra food etc. And since it is built with much of its mass devoted to reentry into the earth's atmosphere, it would be silly to use the energy required to send all that useless mass out of earth orbit." Many of the objections could be answered by a little creative engineering. Do you recall that a Saturn stage was adapted to be a Skylab, and that was a far greater re-purposing than just accelerating the shuttle with sufficient thrust for sufficient time to increase its velocity to that needed for escape. The shuttle has been called the most complicated machine made by humans, and maybe the idea would be just to push it out into the solar system on its own as a monument. The radios aren't powerful enough? Give me a break. More powerful transmitters and more sensitive receivers are not very high technology, and 1970's technology allows communication with probes at the edge of the solar system. The radiators inside the doors must be exposed to space? Then open the doors. There is no wind resistance. The real problem is the lack of a mission or purpose or goal in the project. How much would the target mass be for a manned mission to Mars, compared to the shuttle? Wikipedia articles on space exploration are very short of such numbers. Space shuttle says the mass of the craft is 2,029 metric tons, apparently including the external fuel tanks, loaded, and the solid fuel rockets. The space shuttle orbiter(Endeavor) had an empty weight of 68.6 metric tons, per Space Shuttle orbiter. The proposed Russian Martian Piloted Complex of 1975 would have a mass of 1630 metric tons, apparently the mass when it was assembled in low earth orbit before blasting off for Mars. Edison (talk) 17:06, 11 November 2008 (UTC)

Yes, Skylab was partially adapted from a Saturn V but no-one would describe it as one. You could certainly use the Space Shuttle as a starting point to design and build an interplanetary craft (it probably wouldn't be very wise, though), but it wouldn't be a Space Shuttle afterwards, they would share very little in common. Adapting the Space Shuttle would be far harder than you make out. The crew surviving longer than the week or two Shuttle missions are designed to last would require far more than just carrying extra consumables. You need to consider radiation shielding, room for exercise, room so they don't go stir crazy cramped together in a tiny Shuttle for months, etc. You basically have one cargo bay to work with, it can probably be used to solve each of the problems, but not all of them together. It can't serve as a fuel tank, an extra engine, more pressurised space for the crew and storage for additional consumables all at the same time. Your only option would be to attach more things to the outside which means launching them separately, which means you're even further away from it actually being a Space Shuttle, it would be a larger craft which happens to contain a modified Space Shuttle as one component. --Tango (talk) 17:42, 11 November 2008 (UTC)

I'm reminded of the (terrible) film "Deep Impact". In that film the passenger compartment of the space ship looked like a hastily retrofitted space shuttle, essentially just bolted onto a set of much larger engines (in orbit, apparently.) The design of that ship was my favorite part of that film. APL (talk) 17:57, 11 November 2008 (UTC)

i would so no, a priori, because 1) if it were, they would have done it at some point and 2) conversely, if it were capable of doing that, it would be heavier than if it weren't, and adding excess unneeded weight onto the space shuttle so it could do things it never actually does, and thereby requiring all the extra fuel to hoist the extra mass up into orbit, is a big nono. Gzuckier (talk) 20:12, 11 November 2008 (UTC)

Total ship water displacement

Dear Wikipedians, what might be an educated guess towards how much water the world's shipping displaces? And, I seek to know, how much has the ocean risen from this displacement? Curiously and humourously yours, 80.202.246.253 (talk) 16:55, 9 November 2008 (UTC)

From Displacement (ship), "A floating ship always displaces an amount of water of the same mass as the ship." So, you are essentially asking for an estimate of the total tonnage of shipping (transport/cargo, military, recreational, etc). I would estimate that cargo ships make up the lion's share of worldwide tonnage, with other contributions probably being negligible. If I don't mention it first, Steve Baker will probably have to remind me that this is the Science Desk and not the Speculation Desk, so I would focus my efforts on locating actual reasonable estimates for international shipping before excluding any quantities. Nimur (talk) 17:07, 9 November 2008 (UTC)

Doing the calculation the other way around (working out how much shipping would be required to raise sea levels by a certain amount) is somewhat easier, so I'll do that. According to ocean, the oceans cover approximately 361 million square kilometers. If we ignore curvature (which I think we can for very small increases in sea levels) and assume all coastlines are vertical cliffs (again, reasonable for small increases), we find that a 1mm increase in sea levels would increase the volume by 361 billion metres cubed, corresponding to a mass (ignoring salinity) of 361 trillion kilograms. I'm struggling to find numbers for the mass of container ships, but it seems the biggest has a deadweight tonnage of 300,000, that's not a measure of displacement, but I reckon it gives us a rough order of magnitude at least, so let's say a container ship weights 1 million tonnes, or 1 billion kg. That means we would need 361,000 of the biggest container ships to raise see levels by 1mm. I can't find estimates of how many ships there are, but even with the most generous estimates I can't see the sea level rise caused by all the shipping we have being more than a few millimetres. --Tango (talk) 17:34, 9 November 2008 (UTC)

This says that in 2005, the total deadweight tonnage of the world's bulk cargo ships was 960 million tons - and that, fully laden, that goes up to 7100 million tons...which is 6x10¹²kg - or 6 billion cubic meters of water (OK - that's pure water, not salt) - which falls well short of the 360 billion that Tango estimates we need for a 1mm raise. So roughly 1/60th of a millimeter - which gets us to (as the popular press would be SURE to say) "about the thickness of a human hair". But since global warming has been raising the sea level by about 1.8mm per year for the last century (surely much faster that over the past few years)...I don't think we have to be too concerned about the displacement of our shipping fleet! SteveBaker (talk) 18:29, 9 November 2008 (UTC)

Here is our discussion of this topic from June. — Lomn 15:37, 10 November 2008 (UTC)

Input resistance of nerve fibres

Why is the input resistance of nerve fibres (accroding to cable theory): rinput = 0.5*sqrt(rm*ri) where rm is the membrane resistance and ri is the internal resistance ? 131.111.8.104 (talk) 17:57, 9 November 2008 (UTC)

Nerve fibers conduct nerve impulses electrochemically, not by electrical conduction like a wire or coax cable. The question makes an unwarranted assumption. Edison (talk) 21:44, 9 November 2008 (UTC)

low-cholesterol heart attacks

I was reading this, which starts

“ People with low cholesterol and no big risk for heart disease dramatically lowered their chances of dying or having a heart attack if they took the cholesterol pill Crestor, a large study found. ”

But goes on to say

“ [the drug] gave clear benefit in the study, but so few heart attacks and deaths occurred among these low-risk people that treating everyone like them in the United States could cost up to $9 billion a year — "a difficult sell," one expert said. ”

I don't get it: the group is already low-risk. Wouldn't they prefer to receive even $1 billion as life insurance payoff to the families of the dead among them than to force the whole group to both take an unneeded pill and pay out $9 billion for the privilege? Or is my logic off... 82.124.214.224

Hi, 82.124.214.224, please sign to avoid confusion. Julia Rossi (talk) 22:36, 9 November 2008 (UTC)

Well, I would prefer to live than give my surviving relatives a million Euro each, but that's just me (a selfish fecking bastard :-) ). But other that that, you are correct. What the article says is that a study proved that there is a benefit. This is proved by many studies very often. But cost-benefit analysis comes in at this point. There are numerous things that we could do to save a few extra lives each year, but the costs are just too high, and the money better spent elsewhere. This appears to be one of those cases. Fribbler (talk) 00:11, 10 November 2008 (UTC)

To the OP; that's the idea. Drug companies are looking for a way to make money off of people that don't need to take their drugs by convincing them that if they don't, they will DIE. Turns out that most statin drugs, while they lower cholesterol, don't actually increase your lifespan, life quality, or otherwise prevent heart attacks in any measurable way. It's the old "correlation is not causation" or post hoc ergo propter hoc problem. People with high cholesterol have more heart attacks; however, there is no mechanistic connection. Turns out, that you can lower cholesterol levels via these drugs and have no appreciable effect on heart attacks. It could very well be that high cholesterol levels are likely a symptom of the underlying health problems, and not the actual cause of them. However, methods that lower cholesterol via lifestyle changes do lower chances of having a heart attack. So, rather than taking drugs to cure a symptom, it's better to change your lifestyle to fix the underlying symptom. However, you will never hear this from the health system because that message doesn't make drug companies any cash... --Jayron32.talk.contribs 03:22, 10 November 2008 (UTC)

Just for the record, check out Cholesterol#Clinical_significance. The connection isn't well understood, and there are some skeptics, but the mainstream opinion is that high cholesterol increases the risk for a heart attack. Saying that there is no connection isn't a "flat earth" opinion, but it's definitely an unusual stance. SDY (talk) 16:08, 10 November 2008 (UTC)

I should note that the linked quote doesn't support your claim. It says that people who take Crestor have a dramatically lower chance of having a heart attack, but these people already have such a low chance that it isn't a big drop in the number of people dying from heart attacks. However our article statins does support your claim Nil Einne (talk) 09:08, 10 November 2008 (UTC)

Except there were TWO problems with the Crestor study cited above (and I am familiar with that 1 study). 1) Its primary sponsor was Astra Zeneca, who, um, makes Crestor. Creates a major conflict of interest problem. 2) The study was stopped early, according to the company because, and I am paraphrasing here, "this drug is so fucking awesome, it would be a crime to stop people from taking it by waiting for the study to finish". However, as noted the number of people effected in a positive way by the drug was so statistically insignificant that the shorter study likely only prevented more reliable data from being obtained. They basically stopped the study early because the results, though likely a statistical abberation, were in their favor. The company who made the drug had nothing to gain by keeping the study going, since increasing the participants would only have averaged out the outliers, and would have reduced the "good results" they got. There have been DOZENS of truly independent studies (i.e. those NOT done and controled by the drug companies) that support the idea that Statins like Crestor don't actually improve quality or length of life. --Jayron32.talk.contribs 15:14, 10 November 2008 (UTC)

Please don't use the RefDesk to spread non-sense like this. The test was ended prematurely because the independent review board concluded the results were so obviously significant (>99.9% chance of reducing major cardiovascular events) that further study was unnecessary and delaying the release of the results was contrary to the interests of public health. Yes, you can criticize them by saying that most of the people in the study would have been healthy anyway, but among that fraction who wouldn't be, it did show a clear benefit in this study. Saying the result were statistically insignificant is simply wrong. Dragons flight (talk) 17:12, 10 November 2008 (UTC)

So according to data extracted from our articles on statins, heart attack and stroke - and this claimed $9 billion cost to treat 300 million people (which seems a little low to me - but let's go with that): $9 billion for 300 million people is just $30 per person per year. They estimate a 60% reduction in risk of a heart attack across the board, a 17% reduction in stroke risk and a possible 50% reduction in colorectal cancer risk - balanced against a new 0.00044% per year risk of the rather nasty (and sometimes fatal) muscle issues that are the major side-effect and an even smaller (and presumably non-fatal) risk of some increased production of liver enzymes. So to put numbers on this: if we dosed every American with this stuff - we'd expect to get around 13,000 new cases of the nasty muscular problems per year - versus maybe 600,000 fewer heart attacks and (since 40% of those are fatal) 240,000 less deaths as a result. The cost of treating a heart attack averages to around $40,000 in the first 90 days - so if the treatment reduces your personal risk of a heart attack by only one part per thousand each year - then it's worth the $30 just in the savings in the cost of treatment. To paint the "big picture": the US spends $110 billion annually on treatment for heart attacks - so the $9 billion cost only has to reduce the number of heart attacks by 10% to be cost-effective...since it's claimed to reduce them by 60% then regardless of the humanitarian issues - the cold cash benefits are hard to deny (IF you believe in the results of the study). Free provision of these drugs could be funded by adding a 50c tax on every burger McDonalds sells...or considerably less if we put the tax on every fat-rich fast food item. SteveBaker (talk) 14:41, 10 November 2008 (UTC)

Or we could use a cheaper generic statin with less of a chance of side effects. Crestor is a bit of a loose cannon, and some groups have even called for yanking the approval on it because there are drugs without the problems. See Crestor#Debate_.26_criticisms. SDY (talk) 16:08, 10 November 2008 (UTC)

Was it $9b to treat everyone, or $9b to treat all the low risk people (with the high risk people already being treated since they're high risk)? If it's just the low risk people then that increases the cost per person (I'm not sure how much by, but I'd guess significantly - how many Americans actually have low cholesterol?). Also, are they recommending statins for children or is it just everyone over a certain age? --Tango (talk) 17:28, 10 November 2008 (UTC)

Thank you very much for your detailed analysis, it is greatly appreciated! (just out of personal interest). Will you tell me if you really mean "600,000 fewer heart attacks", this doesn't at all mesh with my understanding that heart attacks are nearly unheard-of among people with extremely low cholesterol (to begin with, independent of any drug). Will you explain how you arrived at that figure? —Preceding unsigned comment added by 82.124.214.224 (talk) 17:31, 10 November 2008 (UTC)

I haven't checked Steve's arithmetic, but it's important to remember that there are lot of people in the US so even if the risk of an individual having a heart attack is very low you still end up with a fairly large number of heart attacks in total. --Tango (talk) 17:40, 10 November 2008 (UTC)

Ok, I've done my own research and checked Steve's numbers and his look right but they are numbers for the population as a whole, not for low risk groups. So, Steve's conclusion is actually that, if everyone's risk was the same it would be worth treating everyone, but that's a false assumption. Generally speaking, you get better value for money treating high risk people than low risk people, which may well mean it doesn't make good financial sense to treat the lowest risk people. --Tango (talk) 17:55, 10 November 2008 (UTC)

Certainly you get better bang for buck by only treating the needy - but I was attempting to demonstrate that there was essentially nothing to lose (and a heck of a lot to gain) by simply treating everyone regardless of whether they need it or not (a similar argument to putting vitamins in milk or flouride in the water supply). Remember - there is a 'cost' to testing everyone too. Could you give everyone in the USA a cholesterol-level test every year - and pay for that by the savings you get by not treating those who pass it? Would the deaths due to preventable heart attacks amongst the false-negatives be less than the additional deaths due to side-effects from the stattin drug? I don't have numbers - but I strongly doubt it. I doubt you could administer the cholesterol test for less than $30 (it's a blood test - so you need a professional and a clean needle - and you've probably broken the $30 barrier right there) - and because heart attacks are SO common, I bet more people would die from the occasional unnecessary heart attack death following a screwed up test that said "don't take the drug" than would die from the side-effects of taking the drug unnecessarily. So I believe that you're STILL better off treating everyone, regardless. SteveBaker (talk) 21:47, 10 November 2008 (UTC)

don't forget, the study was done, published, and presumably the press releases sent out by the manufacturers of crestor. so, guess what they would answer to the question "wouldn't it be better to just give the money to the people at low risk rather than buy them all crestor?" Gzuckier (talk) 20:09, 11 November 2008 (UTC)