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June 12

NASA Venus transit video artifacts

In the NASA SDO Venus transit video, [1] (APOD) and [2] (YouTube), there are a couple of points, such as at 0:33 and 0:43, where Venus appears to be fractionally translucent. As Venus crosses into the sun's field, the edge of the sun appears to be partially visible through the planet's disk. Is this just my imagination, an optical illusion, an artifact of the video encoding/decoding, or is this evidence that the video was manipulated and possibly fabricated? I'm not questioning the transit itself. I just think the video seems odd. -- Tom N (tcncv) talk/contrib 05:45, 12 June 2012 (UTC)[reply]

I had noticed this too; I instantly knew it would be a hit in the flat earth community. Ever take a digital picture of a really bright object (like the sun)? The over-exposed area will tend to bleed into nearby under-exposed areas. This is called Sensor blooming, and I suspect a similar effect is at play here. The image at right is an extreme example; not only does light bleed over the silhouette of the tower, but into the landscape itself. It can be mitigated by high-quality sensors, but I wouldn't be surprised if there was a slight effect left over, and the folks at NASA would probably find it scientifically dishonest to digitally remove it for aesthetic purposes.

Another related effect is called halation, but I think it's solely a phenomenon related to photographic film. That link doesn't show any examples, but this is a good one. -RunningOnBrains^(talk) 06:06, 12 June 2012 (UTC)[reply]

Have you seen the entire video? Sensor blooming cannot explain why Venus often appears transparent. There should be absolutely no way for the solar surface behind Venus to be visible, because it's blocked by Venus. --140.180.5.169 (talk) 06:45, 12 June 2012 (UTC)[reply]

It may appear transparent (really, translucent), but it's not. If you watch closely (I recommend full-screen), you'll see that there is never any anomaly over Venus's disk when it passes completely over a bright part of the Sun, but it does show an anomaly when passing near a bright spot. This is especially evident from a close watching of the sequence starting at 0:58. In response to the below points, I doubt any filtering in the traditional sense is to blame. Obviously some digital filtering had to be done to get a single wavelength from the raw signal, but this should be pixel-by-pixel, and not result in any translational anomalies. I am still convinced that it is a purely optical effect. -RunningOnBrains^(talk) 08:01, 12 June 2012 (UTC)[reply]

You don't isolate a wavelength with a digital filter. (Or an analog one for that matter.)--Srleffler (talk) 17:15, 12 June 2012 (UTC)[reply]

Hmm, I can't find any papers on the exact mechanism by which satellites isolate different frequency channels; I always just assumed it was a digital filter of some kind. Do you have any insight on exactly how this is done? -RunningOnBrains^(talk) 17:35, 12 June 2012 (UTC)[reply]

I would say that the video is "manipulated" for those sequences, and I doubt that NASA would deny it. There would need to be some pretty intense filtering to remove the background brilliance of the sun, and it's very difficult to do intense filtering without producing artifacts. Looie496 (talk) 06:38, 12 June 2012 (UTC)[reply]

Why does filtering need to produce artifacts? Have you ever looked through a neutral density filter or any other type of solar filter? Any solar filter that isn't total junk introduces absolutely no visible distortions or artifacts. --140.180.5.169 (talk) 06:45, 12 June 2012 (UTC)[reply]

The more I think about it, the more I think this might be the result of lossy video encoding and compression. If the encoder attempted to generate the darkened disk by defining some kind of transform to previously rendered data or , it could yield "close", but imperfect results. It would be interesting to see the set of original still images. -- Tom N (tcncv) talk/contrib 07:11, 12 June 2012 (UTC)[reply]

I've now found some still images at http://svs.gsfc.nasa.gov/vis/a010000/a010900/a010996/index.html, including a couple that match the video at about 0:45. The still images show a much more distinct black disk that lacks the haze which gives the appearance of translucency in the video. -- Tom N (tcncv) talk/contrib 07:42, 12 June 2012 (UTC)[reply]

I'm not sure which images you are referencing, but the still images look exactly the same to me. The bleed-through is especially evident in the 193 Angstrom image, just as in the video. -RunningOnBrains^(talk) 07:46, 12 June 2012 (UTC)[reply]

Here are some excerpts of both the video and the equivalent still image.

I think the difference in quality is fairly obvious, and supports the assertion that any apparent translucency in the video is an artifact of the video creation process. -- Tom N (tcncv) talk/contrib 08:27, 12 June 2012 (UTC)[reply]

It's unfortunate that the pictures aren't larger. But my thought as a non-expert is that the video is of lower resolution, which lends itself to being "fuzzier". Note that the texture of the sun is also less sharply defined in the video frame than in the still picture. ←Baseball Bugs ^{What's up, Doc?} carrots→ 09:18, 12 June 2012 (UTC)[reply]

Hmm, I see what you mean now, but if you look at the still composite I linked above the translucency is still apparent. This single image also features it. -RunningOnBrains^(talk) 15:32, 12 June 2012 (UTC)[reply]

Three explanations of light inside Venus' disk might apply:

Long decay in display. The effect is seen in an oscilloscope that continues to show weakly a trace after it has been switched off, and I have seen a full image persist many seconds after a TV is switched off in a darkened room. It would give a persistent ghost image of the edge of the Sun after it was in reality obscured by Venus
Long decay in image sensor.
Reflection by Venus. The surface of Venus' outer atrmosphere is neither a mirror nor totally black but something in between, and it may receive some light by backscatter within its atmosphere or by reflection from our own Earth, Moon or even interplanetary dust.

Case 1. would vary depending on the many different kinds of display that we each have. Case 2. probably involves a number of different sensors used by NASA etc. Cases 1. and 2. would give identical effects on video but 1. would not be seen on a still frame. The edge crossing at 0:43 in the NASA video comes closest to identifying them because there is visible activity on the Sun's edge. Is there edge activity visible "through" Venus or is the edge seen there just a static hangover from earlier frames? I can't decide. I do see low level light patterns on Venus at 0.42 before the transit even begins which support explanation 3. Added a small 2nd thought.DriveByWire (talk) 14:00, 12 June 2012 (UTC)[reply]

That last possibility is a really interesting one; I hadn't even considered it. I really want to get a NASA scientist on the phone. -RunningOnBrains^(talk) 15:32, 12 June 2012 (UTC)[reply]

Anybody who is looking for scientifically-valid data in the individual pixels of a compressed image or video frame is bound to be surprised. Here are some of the (many) practical engineering details that will derail your analysis, presented in roughly decreasing order of interference, ranging from "a standard engineering procedure that catastrophically destroys the scientific validity of individual pixels" and ending with "a standard design principle that only destroys the validity slightly, but in a recoverable way."

Start by reading about image compression.
Next, read about the optical design of practical lenses and telescopes, in our article on photographic lens design.
Read about digital signal processing, CMOS imagers, and digital signal processing in general.

Finally, if you're looking at SDO data, you can access Solar Dynamics Observatory raw data free of cost - so there's really no excuse for anyone to still be looking at summary preview images, and drawing conclusions about Venus' optical properties. JPEG and video-files are also available, because they "look neat" and help citizens and scientists appreciate the sort of "big picture" objectives; but if you actually want to analyze the observations, you have got to use the scientific data. Its error margins are well-known and well-documented, and all that information is available free-of-charge. Nimur (talk) 16:39, 12 June 2012 (UTC)[reply]

Thank you for the links - particularly the image download site. Unfortunately, the downloadable images appear to only be available as .jpg files, which may have already gone through some degree of lossy compression. I accessed a few leading transition images for AIA 171 gold (here), AIA 193 bronze (here), and AIA 304 red (here), and they all still show "translucent-like" artifacts. Additional higher resolution images may be retrieved from this page by selecting a date/time range of 2012-06-05 22:00:00 to 2012-06-05 22:30:00, but they show the same. Perhaps lossless versions of the images may exist but are just not publicly accessible. The download link seems to have sufficient bandwidth for such data. Anybody out there with insider connections? -- Tom N (tcncv) talk/contrib 20:55, 12 June 2012 (UTC)[reply]

To clarify the Q, I see two odd things:

1) The slight reddish color of Venus on the video. The is absent in the still.

2) Small red tendrils extending into the black disk of Venus on the still. This seems like the harder to explain issue, to me. StuRat (talk) 21:24, 12 June 2012 (UTC)[reply]

Assuming I'm seeing your "tendrils", they're ordinary imaging artifacts, with about the same level of reality as the Martian canals. --Carnildo (talk) 00:05, 13 June 2012 (UTC)[reply]

Fascinating video. I had originally considered the idea that the effect may have been a reflection by Venus' atmosphere, but I don't believe that's the case. Being somewhat familiar with the mechanics of digital video, I'd say those are primarily compression artifacts. The aforementioned halation may be playing somewhat of a role as well, though I doubt it is to any significant degree, given the still/video comparison posted by Tom. Evanh2008 ^{(talk|contribs)} 00:57, 13 June 2012 (UTC)[reply]

I tend to agree with Evanh and others regarding the likelihood that compression artifacts are the main culprit. Given that the image sampling rate is on the order one frame per minute and that NASA would be expected to be using high quality image sensors designed to handle stark contrasts, I would expect a better quality from the raw data. The Hinode image File:Hinode Views the 2012 Venus Transit.jpg appears much more crisp - with a very nice atmosphere glow. I'm hoping to find similar quality in the SDO images. I found some potentially better quality FITS and .jp2 images here and here, but I have not been able to view them. I tried GIMP, but it was unable too open either format. Does anyone have a recomendation for a free FITS viewer that can handle these filess and runs on a WIN7 machine? -- Tom N (tcncv) talk/contrib 03:43, 13 June 2012 (UTC)[reply]

Petrol engine horsepower rating?

The power rating of the petrol engine in an ordinary car. That is usually specified as "100 hp". Does that mean the chemical power of the fuel that goes in, or the power that comes out as mechanical energy? (ignoring heat) Electron9 (talk) 13:11, 12 June 2012 (UTC)[reply]

Horsepower of an engine indicates the power output. If it were the input, all engines that use the same fuel would be have the same rating, no? See Horsepower#Measurement and dynamometer. There are several different points where output can be measured, and where it is measured will depend on circumstances. I don't know where the standard place of measurement for cars is, but I'd guess the crankshaft or flywheel. SemanticMantis (talk) 13:45, 12 June 2012 (UTC)[reply]

You can generally get the power rated for a custom car either "at the fly wheel" or "at the road" (i.e. at the tyre using a dynamo). 203.27.72.5 (talk) 22:13, 12 June 2012 (UTC)[reply]

The power of petrol engines is given slightly different ways by different organisations, but never as the "chemical power" of the fuel consumed, as that has no practical value. The energy value of fuel is given as kilojoules per kilogram (metric countries) or calories per pound (USA). The most common way of rating an engine is what is known as the brake power, measured at the flywheel, and given in kilowatts (metric) or horsepower (USA). This is the actual mechanical power output of the engine in usuable or complete form. The term "brake" comes from the use of a brake in a dynamometer test to load up the engine. For accuracy, it is important to understand what is meant by usuable & complete. For example, for a car engine, apart from friction and thermodynamic losses inside to the engine, power is absorbed by the water pump, alternator, and radiator fan, thus reducing the power available at the flywheel to move the car along. Power is also aborbed by the power steering pump and airconditioner (if fitted), but these are not installed for a power output test or calculation. Power output varies slightly according to the energy value of the fuel, the ambient air temperature and humidity, and the altitude. Therefore, there are agreed standards in the USA and Europe on the fuel to be used when testing, and engines are rated at 25 C (Europe), and 64 F (USA, if I remember correctly) at sea level, 50% humidity. For a marine engine, as supplied it will not have a radiator fan (usually) and often not an alternator either. In such cases the power rating quoted will not allow for losses in the raditor fan and alternator. The SAE in USA, and the ISO in Europe publish standards on exactly how to do it, and how to apply corrections for unstandard condtions (eg high altitude) Ratbone58.170.167.209 (talk) 13:49, 12 June 2012 (UTC)[reply]

Car testers are most familiar with the chassis dynamometer or "rolling road" installed in many workshops. This measures rear wheel brake horsepower which is generally 15-20 percent less than the brake horsepower measured at the crankshaft or flywheel on an engine dynamometer. This video shows workshop measurement of a car's power. The measured power curve in kW is shown at 3:39 and comments below the video include one about the factory spec power by cuddlyable3, a former valued Wikipedia editor. DriveByWire (talk) 14:21, 12 June 2012 (UTC)[reply]

If one were to use a diesel engine for driving a electric generator then how these power ratings are done have significant impact. It was like I suspected, but I rather be sure. I added this information to the Petrol engine#Power measurement article. Electron9 (talk) 15:00, 12 June 2012 (UTC)[reply]

Motorcycle on an incline

A motorcycle mass $m$ drives on an incline $\theta$ its wheels separated by a distance $L$ . Its center of mass is between the wheels and height $h$ . What is the normal force at the wheel in the back? Apparently the answer is $mg({\frac {h}{L}}\sin \theta +{\frac {1}{2}}\cos \theta )$ ; how do you derive this? --150.203.114.37 (talk) 14:53, 12 June 2012 (UTC)[reply]

Draw a diagram of the motorcycle seen from the side that shows its center of mass and the two points where the wheels touch the incline; these 3 points when joined form an isosceles triangle. The base of the triangle is the length L and is at angle

\theta

to the horizontal. The gravitational force on the motorcycle is

mg

and its direction is vertical (down). You have to split this force into the following two components which do different things to the back wheel. 1) The component normal to the slope. Each wheel bears half of this force. (As a check: if

\theta

is zero, the motorcycle is on the level and this component is just the whole weight

mg

. Remember cos(0) = 1.) 2) The component parallel to the incline. Think of the front wheel contact point as a fulcrum and this force component tries to rotate the rear of the motorcycle into the road, resisted of course by additional force at the back (lower) wheel. Look for the lever relations that involve

h

and

L

to calculate this additional force. DriveByWire (talk) 19:59, 12 June 2012 (UTC)[reply]

We have to assume the motorcycle is stationary and not being driven because we lack data on its weight, speed and acceleration. DriveByWire (talk) 21:15, 12 June 2012 (UTC)[reply]

Mass is given by

m

. As long as the acceleration is zero I think the speed is irrelevant; the motorcycle need not be stationary. --150.203.114.37 (talk) 21:29, 12 June 2012 (UTC)[reply]

That is true for the normal force at the wheel. However for any driving movement except free-rolling downhill there must be additional tangential traction force at the wheel(s) for accelerating and braking. DriveByWire (talk) 23:55, 12 June 2012 (UTC)[reply]

Sleeping

You know when you wake up in the morning and feel all warm and comfortable and don't want to get up. Why is that? How can one get that same feeling at night instead of feeling uncomfortable and tossing and turning constantly? Sleeper4545454545 (talk) 15:11, 12 June 2012 (UTC)question replaced and comments re-hidden.203.27.72.5 (talk) 21:12, 13 June 2012 (UTC)[reply]

Waking by alarm clock, spouse, children, neighbor workshop etc.. or by yourself ? Electron9 (talk) 15:32, 12 June 2012 (UTC)[reply]

A classic question. Getting more exercise and more sleep might help, but for me going to sleep at night is rarely as enticing as staying in bed in the morning :) SemanticMantis (talk) 15:37, 12 June 2012 (UTC)[reply]

See Circadian rhythm. You may also be interested in Delayed sleep phase disorder, not that the phenomenon you describe sounds like a disorder.--Srleffler (talk) 17:19, 12 June 2012 (UTC)[reply]

Discuss melatonin with your doctor. Subjectively I find it induces yawning and that tingly "time to sleep" feeling if taken on an empty stomach in a darkened room. μηδείς (talk) 20:39, 12 June 2012 (UTC)[reply]

The warm part is likely due to body heat that built up overnight under the blanket. StuRat (talk) 21:15, 12 June 2012 (UTC)[reply]

No, the circadian clock cranks up metabolic activity shortly before dawn, when it functions normally. And it probably isn't the answer the OP is looking for, but the most straightforward way to get that warm and comfortable feeling at night is sex. Looie496 (talk) 22:43, 12 June 2012 (UTC)[reply]

Off-topic jokes and discussion

...followed quickly by a wet and sticky feeling. (In case the implication isn't obvious, many people would feel the need to shower afterword, which would delay sleep considerably.) StuRat (talk) 04:20, 13 June 2012 (UTC) [reply]

I do not want your stupid comment on my question but every attempt to either hide it or remove it has been rebuffed by people who apparently think comments about your semen have a place on the reference desk. So I have removed my OWN question. Thank you to the people who tried to help before this person decided to post their jokes here and others felt it a good use of their time to keep adding back the crude sex joke when the question author clearly does not like it. Sleeper4545454545 (talk) 19:12, 13 June 2012 (UTC)[reply]

I don't see what was wrong with the hide box and it was apparently removed without comment so I'm putting it back. 203.27.72.5 (talk) 21:12, 13 June 2012 (UTC)[reply]

Ok, the first time it was removed without comment. The second time it was removed with a comment to the effect of, "this will just draw more attention to it." The guidelines for anwering questions on the reference desk say that, "we never set out deliberately to offend, and we endeavor to quickly remove needlessly offensive material in questions or responses." Sleeper noted that they were offended (though no one is saying that it was intended by StuRat or Looie), so given that those particular comments do not serve to answer the question in anyway helpful to the asker, hiding them is appropriate. 203.27.72.5 (talk) 21:37, 13 June 2012 (UTC)[reply]

If you're that sensitive that you have to run and hide when you see such jokes, you might do better posting to a forum for kids or religious conservatives. Or, if you do post here, warn us how sensitive you are, and we will try to keep it clean. Of course, you will need to avoid reading the rest of the Ref Desk Q's and answers, too. StuRat (talk) 21:27, 13 June 2012 (UTC)[reply]

Conker tree (Aesculus hippocastanum)

I have a small Conker tree (Aesculus hippocastanum) growing in a medium-sized pot. The tree is ~10 years old and 6 foot tall. It has not flowered yet. It is doing wonderfully and continues to grow at a rate of about 1.5 feet per year. This however is becoming a slight problem as it is now dominating my tiny garden and starting to block access. As the constrains of the pot have not deterred its growth, I need to reduce its height or at least stop it from growing any taller somehow. Will it be able to take some pruning and cutting back, or will doing so likely kill it? I do not wish to plant it in the ground at this time as I would like to be able to take it with me when I move. Aesculus hippocastanum (talk) 15:15, 12 June 2012 (UTC)[reply]

I suggest you look into bonsai techniques, specifically pruning of both branches and roots. That is probably the simplest method for controlling size given your situation. 65.95.22.197 (talk) 15:35, 12 June 2012 (UTC)[reply]

Conkers recently came up on the ref desks, here: [3], and the consensus was that they can take a lot of pruning. Bonsai would be fun for this specimen, I also encourage you to look into it. You could knock it back to 3 feet tall, and fit it into a ~2 gallon pot (better if it's wider than deep). After a year or so, you can train it back to a nice shape. Note that it will still continue to grow, the idea is that you prune it back as necessary. You could try downsizing the pot only, and leaving the top intact, but that may just force the tree to self-thin its branches, i.e. cause some to die off. SemanticMantis (talk) 15:43, 12 June 2012 (UTC)[reply]

Do you know any free sources for beginners on bonsai? Itsmejudith (talk) 22:43, 12 June 2012 (UTC)[reply]

The Royal Horticultural Society has a page here, if that helps. Tonywalton ^Talk 23:52, 12 June 2012 (UTC)[reply]

Uses of gravitational acceleration

How much of today's technology depends on the discovery that gravitational acceleration is similar for all objects on different places on earth, neglecting friction? Did we have to discover this fact in order to build spaceships? in order to build airplanes? If so, how does it help us? Jobnikon (talk) 16:34, 12 June 2012 (UTC)[reply]

Well, if it differed, then a plane designed for one part of the world would be very difficult to control in another, with a much greater or less amount of relative lift.--Gilderien Chat|List of good deeds 16:48, 12 June 2012 (UTC)[reply]

Not much. Scales come to mind. Note that gravitational acceleration is not exactly the same everywhere on Earth. See Gravity of Earth and Gravimetry.--Srleffler (talk) 17:24, 12 June 2012 (UTC)[reply]

Are you talking about the fact that bowling balls and feather pillows accelerate at the same speed when they fall? I mean, it is pretty fundamental, in the sense that anything which uses the gravitational acceleration equations somewhere is going to be based on that understanding. But figuring out exactly where that would matter is a tough thing; if gravitational acceleration was correlated with mass, for example, then a whole lot of things would have to be wired up differently to take that into account. It would be easier (in a sense) to answer that question than to ask what technologies would be capable if people were simply ignorant of the acceleration equation. --Mr.98 (talk) 22:01, 12 June 2012 (UTC)[reply]

You couldn't understand orbits, and you couldn't predict the flight paths of spacecraft. That means no space ships, no weather satellites, no GPS, no satellite phones. You also couldn't predict all of the motion of celestial bodies, for example the motion of those colliding. As Mr.98 said, it's pretty fundamental though. If NASA blew up one rocket because they didn't know that gravity acts the same on all ojects, they'd have enough data from that mission to deduce that relationship. On earth, gravity is generally opposed by air resistance, which depends on the objects shape and composition. This effectively masks the fact that all objects fall at the same rate, so no realising the details and thinking that objects fall at different rates due to gravity can still lead to accurate predicitons.

I suppose the reason this is hard to answer definitively, is that if you imagine a world where we didn't realise that the earth's gravity provides the same acceleration to all objects, the people in that world also can't have yet understood air resistance. They possibly don't realise the air is there at all. Their level of understanding of the natural world is so far backward it's hard to believe that they could have much modern technology at all. 203.27.72.5 (talk) 22:28, 12 June 2012 (UTC)[reply]

None of today's technology depends on that discovery. All of our technology would work even if that discovery had never been made. People might not understand why it worked, but it would still work. Looie496 (talk) 22:38, 12 June 2012 (UTC)[reply]

Actually, some fairly fundamental technology does. If gravitational acceleration varied from place to place, or if it was not known to be constant, it would not be possible to sell goods measured by weight: you couldn't order a ton of wheat from Chicago, another ton from Denver, and count on having two equal-sized loads of wheat arrive at your bakery in New York. --Carnildo (talk) 00:12, 13 June 2012 (UTC)[reply]

I think the OP really wanted to ask, "what technologies required an understanding of the way gravity works in order for them to be developed?". 203.27.72.5 (talk) 00:56, 13 June 2012 (UTC)[reply]

Disagree, you'd either calibrate spring type scales centrally, or use 'comparison of mass' type scales. Greglocock (talk) 01:05, 13 June 2012 (UTC)[reply]

Calibrating the spring type ones centrally wouldn't work. They would need to be calibrated with a standard weight before use in any given location (many extemely accurate pan balances do this in real life automatically which accounts for the variation in acceleration due to gravity that exists). 203.27.72.5 (talk) 01:20, 13 June 2012 (UTC)[reply]

As a side note; my understanding is that the force of gravity is slightly different on different objects. If you think of a massive object, e.g. the moon, the gravitational acceleration between it and the earth should be greater than something small e.g me at the same distance from earth. Maybe this just depends upon your reference frame though; is the moon accelerating toward the earth at the same rate as a small object but it hits it earlier because it also accelerates the earth in its direction? 203.27.72.5 (talk) 01:03, 13 June 2012 (UTC)[reply]

You're mixing up two things: the force of gravity depends very strongly on the mass of the object. The acceleration due to that force does not. The force the Earth exerts on the moon is much greater than the force it exerts on a spaceship an equal distance away. The moon has more mass, though, so it takes more force to produce the same acceleration. The two effects exactly cancel.--Srleffler (talk) 17:43, 15 June 2012 (UTC)[reply]

Prediction reliability threshold

Is there some scientific threshold, which separates mere coincidences in the alleged prediction capabilities from actual prediction power, that is, a threshold that says that the number of correct prediction is unusually high for coincidences? How many correct prediction it equals? — Preceding unsigned comment added by 176.241.247.17 (talk) 17:10, 12 June 2012 (UTC)[reply]

Yes, but it's not a fixed number of predictions. How many predictions you need depends on how distinct the predictions of the competing hypotheses are and how precise your data is. If you have two hypotheses with very different predictions, and your data is very precise, one experiment may be enough. If the predictions of the theories are very similar and the data is not very precise, one may need to do an enormous number of experiments to decide which hypothesis to reject.

You may be interested in Statistical hypothesis testing and Confidence interval.--Srleffler (talk) 17:32, 12 June 2012 (UTC)[reply]

Or in statistical significance, which is the precise name for the concept being discussed here. Looie496 (talk) 17:46, 12 June 2012 (UTC)[reply]

If you're having trouble working out what the "other" hypothesis should be, see null hypothesis. 203.27.72.5 (talk) 22:38, 12 June 2012 (UTC)[reply]

Closed bottle at "room" temperature equilibrium?

If bottle is filled to half with a liquid like water and then closed with a cap. The pressure in the gas section will initially be the same as the surroundings. But when time goes to infinity the pressure in the gas portion will have an elevated pressure by the vapor pressure over the surroundings, ie 2.3 kPa in this case? Electron9 (talk) 17:20, 12 June 2012 (UTC)[reply]

That's right, the vapor pressure will reach its equilibrium value, and that means that the total pressure will have become larger. The fact that you go from equal pressure to unequal pressure with the environment means that you could extract work from this system, and that in turn derives from the fact that you started out with a non-equilibrium situation. In equilibrium, all the water should have been evaporated, because ambient relative huidity is typically a lot less than 100%. So, it's not completely nonsense to say that a car can in theory run on pure water... Count Iblis (talk) 17:39, 12 June 2012 (UTC)[reply]

...until the water completely evaporates, at which point you have to do more work to reliquefy it. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:39, 12 June 2012 (UTC)[reply]

At which point you open the valve for more tap water.. Electron9 (talk) 18:41, 12 June 2012 (UTC)[reply]

Just as a point of order, a back-of-the-envelope calculation (which may be wrong, I made a bunch of questionable late-afternoon assumptions), you could get about 2kJ of energy from a cubic meter tank using this method, assuming you could perfectly extract all the available energy (for reference, gasoline provides about 350 kJ per liter, or 1320 kJ per gallon). I'm loathe to believe that running a car on this puny amount of energy will ever be possible, and if you could, I think you'd be better off using solar power so that the car would never have to stop to refuel. Also, our article on Compressed air cars suggests that the efficiency of energy extraction would be less than half that of an internal combustion engine. -RunningOnBrains^(talk) 21:48, 12 June 2012 (UTC)[reply]

You're going to be limited by carnot efficiency. This is a heat engine because the water draws heat from its surroundings in order to evapourate and increase the pressure. The difference between

T_{H}

and

T_{C}

is going to be minute, so your efficiency will be terrible. To increase the efficiency, you need to heat the tank of water (and now you have a steam engine). 203.27.72.5 (talk) 02:45, 13 June 2012 (UTC)[reply]

The Carnot efficiency is irrelevant here, because the water will be evaporating at ambient temperature leading to entropy increase. If the evoporated water is to be used in a close circuit, then since the internal state will return to the starting point, and then the total entropy change is due to the heat that goes into the system and out of the system, in which case you recover the Carnot formula for the maximum work that can be done.

Now, our case is different, the water vapor is allowed to escape and you can refill the water tank. The maximum amount of work that is available for the car can be computed as follows. Take the initial state to be the car with an empty water tank, the final state is the same car again with an empty water tank but where a water reservoir contains a little less water and there is a bit more water vapor in the atmosphere. Also some work has been performed by the car engine. We assume that both the initial and final states are states of thermal equilibrium with a well defined teperatures and pressure equal to that of the environment.

Then since the car's state can be taken to be the same in the initial and final state, we only need to focus on the water reservoir and the water vapor in the atmosphere. The question is then simply how much work you can extract by bringing some amount of water from the water reservoir to the atmosphere. Water has a latent heat of L = 2450 kj/kg at room temperature, so if we evaporate 1 kg of water and also let it perform an amount of W of work, we would have extraced a amount of heat of Q = L + W. This leads to an entropy change of - (L+W)/T, where we are not yet taking into account the entropy increase due to the added evaporated water.

The increase in entropy due to the phase change of water from liquid to vapor would be L/T if the relative humidity would be 100%. Because the water will be added to the atmosphere at some lower relative humidity r, it will be expanding by a factor 1/r more and thus the entropy increase will be larger. If we treat the vapor as an ideal gas, then the entropy increases by an amount -N k Log(r). The total entropy increase due to entropy drop from extracting the heat Q from the environment and the evaporation of the water is thus -N k Log(r) - W/T. At best this is zero, so the maximum amount of work we can extract is

-N k T Log(r), which is about 69 kJ/kg at 20°C and relative humidity of r = 0.6. Count Iblis (talk) 18:25, 13 June 2012 (UTC)[reply]

I'm a bit confused by that. Are you saying that if the water is used in a closed circuit then Carnot efficiency applies, but if you release it then it doesn't? A regular petrol engine doesn't work in a closed circuit, i.e. it releases it's hot working fluid as an exhaust, but it's still a heat engine. Are you saying that steam engines are not limited by Carnot efficiency? 203.27.72.5 (talk) 21:52, 13 June 2012 (UTC)[reply]

Yes, but you can still apply Carnot's formula if you ask a different question. So, instead of asking what the maximum possible amount of work is you can generate from gasoline, you can ask what the maximum work is given a setting where it is burned in some way. Then tdue to the burning of the fuel, you have a de-facto high temperature reservoir. So, given that the temperature in some part of the engine is whatever it is, and you extract heat from there, you can put a limit on how much work you can extract using Carnot's formula. But if you lift this contraint and ask how much work you can possibly generate from one liter of gasoline, no matter what technology is used, then Carnot's formula becomes irrelevant. Compare e.g. the way your muscles burn glucose. The efficiency for converting energy to work here is about 25%, which is way more than what a Carnot engine could achieve if it were limited to operate between temperature differences that can occur in the human body. Count Iblis (talk) 22:30, 13 June 2012 (UTC)[reply]

I don't see how this arrangment is different to a steam engine, other than that you heat the steam engine's boiler. It seems to me like this is just a steam engine that waits for the water to evaporate by drawing energy from the surroundings, instead of supplying it with heat from a furnace. Is there something missing in my logic? 203.27.72.5 (talk) 01:50, 15 June 2012 (UTC)[reply]

With the vapor pressure of Acetone at 14.7 kPa and Ammonia at 800 kPa, I think they are quite interesting candidates for an experiment. The catch being that they are quite corrosive and toxic. Electron9 (talk) 18:33, 12 June 2012 (UTC)[reply]

And a bit more expensive than water. But with the both, you could extract energy in that way, then take the gas in the head space as a fuel for an ICE then wait for more to volatalise. 203.27.72.5 (talk) 22:43, 12 June 2012 (UTC)[reply]

Surface Charges on an open circuit

I have a circuit diagram with two 1.5 volt batteries in series, an 8 ohm resistor and a 32 ohm resistor. These would all be connected in a single loop, however there is a gap between the battery and the 32 ohm resistor. What do the surface charges look like on this open circuit? Would there be any net charge at any point? Widener (talk) 18:09, 12 June 2012 (UTC)[reply]

Yes. Let us consider the two 1.5 V cells in series to be a battery and ignore the internal node where the two cells connect to each other. There would be a net positive charge on the components and wires connected to the positive terminal of the battery and a net negative charge on the components and wires connected to the negative terminal of the battery. Jc3s5h (talk) 18:29, 12 June 2012 (UTC)[reply]

Wouldn't that cause a current to flow? Widener (talk) 18:34, 12 June 2012 (UTC)[reply]

No, because the battery's own polarity opposes any resulting current. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:38, 12 June 2012 (UTC)[reply]

I don't think that explanation clearly describes this situation. Polarity is the wrong word here; and voltage does not "oppose" current. Nimur (talk) 19:31, 12 June 2012 (UTC)[reply]

Okay. How do the surface charges differ in this circuit than if the circuit was closed in a loop? You would have charges on the nodes of the battery, but nowhere else? Whereas you would have surface charges at every point in the wire if the circuit were closed? Widener (talk) 18:40, 12 June 2012 (UTC)[reply]

The battery "pushes" electrons in one direction. This builds up a concentration gradient of electrons which eventually creates a high enough potential difference that it equals the battery's electron-pushing power and stops the current. In a closed circuit, the electrons would be pushed all the way around the circuit to the other side of the battery, which sucks them up. In an open circuit, this cannot happen, and the growing potential difference that the electrons must be pushed up shuts down the current flow. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:57, 12 June 2012 (UTC)[reply]

You get a surface charge gradient in a closed circuit don't you? I thought that was what made electrons flow and caused the current. Widener (talk) 19:03, 12 June 2012 (UTC)[reply]

Electric current is the movement of electric charge. It is perfectly valid to have a non-uniform charge distribution, and no current (no motion of charge) in the steady state. Of course, to get to steady-state, charge needed to move (if the initial distribution was uniform). That temporary flow of current is what we call a transient response and requires the tools of electrodynamics for full analysis. In basic physics, we usually learn about electrostatics (steady-state) first, because the math is simpler; but you are correctly intuiting that there's something "unphysical" about it. You cannot move from one static condition to another without a dynamic transient response in-between; but typically, you just ignore that bit until you develop the tools to fully study it. Nimur (talk) 19:26, 12 June 2012 (UTC)[reply]

So there's no difference in the surface charges between an open and closed circuit? Widener (talk) 19:40, 12 June 2012 (UTC)[reply]

An electric capacitance exists between any two physical objects - even between the two leads of a resistor, or between one portion of a conducting wire and another portion of the same wire. However, in your scenario, that capacitance is very small. If you needed to estimate the actual net charge that builds up, you would need to measure or model the parasitic capacitance of your system; you could add this to your schematic. In practice, you can directly measure this very small capacitance using an oscilloscope or certain multimeters; in principle, these work by measuring an R-C time-constant). Then, you can trivially calculate the net charge, using the capacitor equation, charge Q = C V (capacitance times voltage). Nimur (talk) 18:45, 12 June 2012 (UTC)[reply]

The initial current flow that charges the capacitance of the gap in the circuit, which Nimur calls the transient response, is what Maxwell called Displacement current. Good luck measuring what is probably an extremely small (brief) time constant. DriveByWire (talk) 21:07, 12 June 2012 (UTC)[reply]

No, that is not correct. In fact, displacement current is a different other physical parameter with a different definition. A system that is not in equilibrium may undergo a transient response, which may include both a displacement-current and a true current. Nimur (talk) 00:37, 13 June 2012 (UTC)[reply]

Rebuttal to Nimur. The only current that ever flows through a capacitance is displacement current. It is no less a true current with consequent magnetic field and heating effect. DriveByWire (talk) 16:49, 14 June 2012 (UTC)[reply]

So just to clarify - In the steady state, the surface charges between the open and closed circuits are exactly the same? And what do they look like around the resistors? Widener (talk) 23:09, 12 June 2012 (UTC)[reply]

The short answer is no. To clarify the question, we have (a) a closed circuit comprising two 1.5 V batteries plus a 32 ohm resistor and an 8 ohm resistor (total 40 ohms) all in a loop, and (b) the same thing but a gap between the 32 ohm resistor and the battery. In cases like this it is best to define a reference point, considered to be zero voltage and zero charge (after all the whole world has a voltage and charge with respect to some point in space). Let the ref point be the junction of the battery and the 8 ohm resistor. In case (a) we have 3 x 8/40 ie 0.6V at the junction of the two resistors, and of course 3V at the opposite end of the 32 ohm resistor (2.4 V across the 32 ohms). This means there is a certain charge at the junction of the two resistors, and a larger charge at the opposite end of the 32 ohm resistor. Just what the charge is numerically (in coulombs) we can't say - as Nimur has implied, you need to know the (stray) capacitance across each part. In case (b) no current is flowing due to the break, so there is no voltage across either resistor. Therefore, with respect to the reference point, there is no charge at any terminal of either resistor, but there is still a charge across the battery. The 2 situations are different, becasue in (a) we have 3 V across the 2 resistors, and in case (b) with have 3 V across the gap instead. Keit60.228.241.58 (talk) 03:03, 13 June 2012 (UTC)[reply]

Polarization?

A large thin plastic disk with radius R = 1.5 m carries a uniformly distributed charge of -Q = -3 x 10^-5 C. A neutral circular piece of aluminum foil is placed d = 3.0 m from the disk, parallel to the disk. The foil has a radius of r = 2.0 cm and a thickness of t = 1.0 mm.
(i) Show the charge distribution on the foil
(ii) Calculate the magnitude and direction of the electric field at the center of the aluminum foil, inside the foil
(iii) Calculate the magnitude q of the charge on face of the foil facing the charged disk.
For the first part, due to polarizability, I assume we have negative charges on the side facing away from the disk and positive charges on the side facing the disk. What about the other two parts? I would have thought you would need to know the dielectric constant of the foil. Widener (talk) 18:33, 12 June 2012 (UTC)[reply]

Is this homework? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:44, 12 June 2012 (UTC)[reply]

No, it comes from here: [4] Widener (talk) 18:52, 12 June 2012 (UTC)[reply]

Part (iii) is easy once you have the answer to part (ii), of course. --130.56.77.205 (talk) 23:10, 12 June 2012 (UTC)[reply]

"herpautochory" -- seed dispersal

At Seed_dispersal#Types_of_dispersal, we have "herpautochory (the seed crawls by means of trichomes and moist changes)." Being interested, I tried to find more information -- with no avail. Googling "herpautochory" returns an echo chamber sourced from our page, there are zero hits in google scholar as well as google books. I suspect there is a nugget of truth here, so what is it? Perhaps a simple typo? Any help is appreciated. SemanticMantis (talk) 18:42, 12 June 2012 (UTC)[reply]

I agree with you that the term cannot be sourced, and neither can the term "boleochory" that appeared in the same sentence. I have removed that sentence. It was added on 10 January of this year by an IP editor, 79.158.109.74 (talk · contribs), without any explanation. Looie496 (talk) 19:06, 12 June 2012 (UTC)[reply]

What a lesson on the influence of WP! ~250 pages have pulled that content since then... Still, you've helped me quite a bit, because you read "boleoautochory" and wrote "boleochory"-- effectively undoing the IP's mistake ;) Searching boleochory led me to this book: [5], it is a fairly authoritative textbook, and defines boleochory, as well as herpochory, so it was more of a typo/idiosyncratic terminology situation after all-- "herpautochory" should have been "herpochory". I'll put the article back together with the right words, and cited the book. SemanticMantis (talk) 19:40, 12 June 2012 (UTC)[reply]

Hah, that's pretty funny. I swear I've never seen any of this terminology before. Looie496 (talk) 22:13, 12 June 2012 (UTC)[reply]

Resolved

Speed of Response of Interconnected Ganglia

Hello. Why can interconnected ganglia in the autonomic nervous system respond faster than ganglia not interconnected? Thanks in advance. --Mayfare (talk) 19:13, 12 June 2012 (UTC)[reply]

Might be the diff between serial and parallel computing. That is, if there are a certain number of calculations to perform, then performing them all at once is quicker than one at a time. StuRat (talk) 19:45, 12 June 2012 (UTC)[reply]

I don't understand the question. All of the ganglia of the autonomic nervous system are interconnected -- at least, they are all connected to the central nervous system in some way or other. How did this question arise? Looie496 (talk) 22:11, 12 June 2012 (UTC)[reply]

I assume they are talking about a variable degree of interconnectivity, as some ganglia have many more connections than others. StuRat (talk) 00:20, 13 June 2012 (UTC)[reply]

It is to my understanding that: Sympathetic ganglia connect to each other to form the sympathetic trunk. Whereas parasympathetic ganglia run from the brain stem or the sacral region and synapse in or near the affected organ but do not connect to each other. --Mayfare (talk) 01:53, 13 June 2012 (UTC)[reply]

Okay, that makes sense. But why do you think the connection pattern determines the speed of response? It is natural that the "fight or flight" responses of the sympathetic system take place on a faster time scale than the "rest and digest" responses of the parasympathetic system, but there are lots of differences between the systems beyond their levels of connectedness. Looie496 (talk) 03:57, 13 June 2012 (UTC)[reply]