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April 9

Gravity in conditions of the planet Earth

Did in conditions of the planet Earth the gravity been different mathematical values in different physical environments?--83.237.207.119 (talk) 12:20, 9 April 2015 (UTC)[reply]

If you are asking about variation of gravity on earth, I suggest you read Gravity of Earth.--Shantavira|^{feed me} 12:58, 9 April 2015 (UTC)[reply]

Suggesting a gravity in water.--85.141.232.245 (talk) 14:20, 9 April 2015 (UTC)[reply]

No. --OuroborosCobra (talk) 19:40, 9 April 2015 (UTC)[reply]

Capacitor

An AC circuit contains a 0.01 microfarad capacitor and 0.06 micro farad capacitor that are connected in series. What is the total capacitance of this circuit — Preceding unsigned comment added by 64.56.106.62 (talk) 12:47, 9 April 2015 (UTC)[reply]

You might be interested to read Capacitors in series.--Shantavira|^{feed me} 13:00, 9 April 2015 (UTC)[reply]

Or you might find someone else to do your homework assignment for you. Short Brigade Harvester Boris (talk) 00:31, 10 April 2015 (UTC)[reply]

This is easy, you just add them.--79.97.222.210 (talk) 03:59, 10 April 2015 (UTC)[reply]

That is only true if they are in parallel. For capacitors in series you'd have to be talking about the inverse of capacitance for that to be right. Dmcq (talk) 20:57, 10 April 2015 (UTC)[reply]

Prototypes of computer

Did the code-making machine Enigma been prototype of computer?--85.141.232.245 (talk) 14:15, 9 April 2015 (UTC)[reply]

See the Wikipedia article titled History of computing hardware. --Jayron 32 14:23, 9 April 2015 (UTC)[reply]

I did not understand, did a binary calculation system machines been a prototypes of computer – the machines of program binary codes?--85.141.232.245 (talk) 15:22, 9 April 2015 (UTC)[reply]

No. Dmcq (talk) 16:06, 9 April 2015 (UTC)[reply]

So, it to become me to avow that the USSR did not invented a program language and computer, I’m suppose that the USSR had got to invented a binary calculation system machine which was been one of much from word, of course it did been electromechanical (radiotechnical) and electronical versions, but not microelectronical version, so the USSR did it more simplest than over members, because the economy of the USSR was been progressing for win. I don’t know, how binary calculation system was been relating with binary programming?--83.237.200.102 (talk) 17:30, 9 April 2015 (UTC)[reply]

The Enigma machine was not a prototype of the computer as we now use the word. As for your other comments, you may find it useful to learn the term popovism. Finally, you are strongly encouraged to ask questions in the Wikipedia of your native language. (If you have already tried that, and they have sent you here, you are encouraged to take your questions to another Wikipedia that may be able to understand you.) RomanSpa (talk) 18:09, 9 April 2015 (UTC)[reply]

Much thanks for you explanation. I’m should interesting, did been a different between calculation and programming?--83.237.200.102 (talk) 19:30, 9 April 2015 (UTC)[reply]

No. --OuroborosCobra (talk) 19:41, 9 April 2015 (UTC)[reply]

In Russian language: В чем состоит логическая посылка – что-то является частью чего-то, то есть что в ней общие а что частное, вычисления или программирование?--83.237.205.199 (talk) 22:03, 9 April 2015 (UTC)[reply]

OK, I think the "troll" flag has been well and truly raised for this account. I'm out. RomanSpa (talk) 01:48, 10 April 2015 (UTC)[reply]

Yes. Most programming includes some calculation, but I have written programs with no calculation; conversely, most programming involves actions that are not calculation, such as deciding (according to empirical conditions) what to do next. —Tamfang (talk) 18:30, 11 April 2015 (UTC)[reply]

Enigma wasn't by any stretch a "computer". The meaning of the word "computer" has changed quite a bit over the years - back in the second world war, a "computer" would be a person who was employed to do arithmetic calculations...later, the term "mechanical computer" came to mean "a machine that can do what a human computer can do" - which is to say, basic arithmetic. But these days, we don't generally consider a simple four-function pocket calculator to be "a computer" (although it might easily contain one). The more modern definition would be something that's Turing machine equivalent...which requires that it's able to take decisions and execute loops. That said, I don't think Enigma came remotely close to making any of those definitions.

What Enigma basically did was to have a bunch of disks with electrical contacts around the outside and a bunch of wires that connected every contact to a different contact. A stack of those wheels produced a scrambling of wires leading from the keyboard switches to a set of lights. Every time you pressed a key, the wheels would turn, re-wire the keys to a different set of lights and light up whichever light happened to be connected to the key you pressed. A machine like that can't do even simple addition or subtraction - so it's not a computer by any reasonable definition.

The simplest mechanical computer I know of is the Digi-Comp I (I own one of these - the second photo in the article is mine). It effectively has just two bits of memory...yet is (arguably) a horribly limited turing machine - and therefore should properly be called a "computer".

SteveBaker (talk) 06:12, 10 April 2015 (UTC)[reply]

Stop engaging him. Stop answering him. He's a troll. Steve, you put a fair amount of effort into writing that post; I assure you it is entirely wasted. This IP will likely write a response completely ignoring everything you wrote, possibly making some off the wall philosophical declaration that either has nothing to do with this topic or still declared Enigma to be a computer, and the USSR to not have some language something or other. It isn't worth it. --OuroborosCobra (talk) 11:36, 10 April 2015 (UTC)[reply]

Of course the USSR was made it's own choice for electronical versions of technics, but not microelectronical versions of technics, because it is been necessary for win, so because in electronical versions it been made the much more microtransistors on square decimeter of perimeter on electronical matrix, than on microelectronical matrix, that’s why the USSR had not got a microelectronical versions (technology) of technics, it may been only composite versions of technics.--83.237.192.41 (talk) 12:39, 10 April 2015 (UTC)[reply]

Of course the USSR had not do a complex calculations, because the USSR did not used the program languages or always been used a low levels program languages, so that's the microelectronical technology did not need to the USSR.--83.237.218.251 (talk) 14:50, 10 April 2015 (UTC)[reply]

It did’s all been depended on by technology, but of course the most simplest program languages are always been much faster that over program languages, if technology it did’s been could.--83.237.218.251 (talk) 16:28, 10 April 2015 (UTC)[reply]

In Russian language: Проблема состоит в том, что простейшие языки программирования всегда строятся на примитивной математике (на примитивном логическом уровне).--83.237.198.209 (talk) 17:42, 10 April 2015 (UTC)[reply]

But, in what is been basis of computer in calculation or programming?--83.237.192.41 (talk) 11:49, 10 April 2015 (UTC)[reply]

Нет --OuroborosCobra (talk) 16:17, 10 April 2015 (UTC)[reply]

Thanks. I’m should thinking that the calculation versions of computer and programming versions of computer are been two different versions of computers.--83.237.218.251 (talk) 16:28, 10 April 2015 (UTC)[reply]

I’m russian, and I’m understood that my nation is been very greedy!--83.237.218.251 (talk) 16:46, 10 April 2015 (UTC)[reply]

Russian Wikipedia can answer your questions in Russian at This link Please use them, as your poor English language skills makes it almost impossible for us to help you. --Jayron 32 17:18, 10 April 2015 (UTC)[reply]

It is impossible, because as I’m understood, I’m was been blocked in Russian version of Wiki.--83.237.198.209 (talk) 17:42, 10 April 2015 (UTC)[reply]

Blocked. Well, that explains a lot! Try Polish Wikipedia or some such.--Aspro (talk) 18:03, 10 April 2015 (UTC)[reply]

So, you've been blocked at Russian Wikipedia, and came here to cause problems? Right. Thanks. --Jayron 32 18:11, 10 April 2015 (UTC)[reply]

The Polish use is been not serious!

I see no problems in my membership in Wiki, because at first I’m did not answering for medical advanced problems, so I’m did not done the medical problems at all, and at second I’m did not discovered the secrets which nations law been defended.--83.237.198.209 (talk) 18:58, 10 April 2015 (UTC)[reply]

You are not a 'member' of anything. And your endless spamming of our reference desks with frequently incomprehensible questions followed by even more frequently incomprehensible commentary on any responses is a gross abuse of our facilities. That you haven't as yet been banned from posting has much more to do with internal Wikipedia disputes than it has with your postings themselves - and sooner or later, we are going to send you packing, since you are clearly under the delusion that this is some sort of combined forum and exercise in forensic linguistics. Though frankly I see little reason to believe that your convoluted grammar etc is actual evidence of poor English - it looks to me to be intentionally worded for maximum confusion. I suggest you find another location for your trolling, before you are obliged to. AndyTheGrump (talk) 19:15, 10 April 2015 (UTC)[reply]

I see no laws problems in I'm did. As I’m know, the nations laws always been defended from break and theft new ideas, but not from asking and taking idea.--83.237.198.209 (talk) 19:19, 10 April 2015 (UTC)[reply]

Fuck off and troll somewhere else. AndyTheGrump (talk) 19:20, 10 April 2015 (UTC)[reply]

I understood that break law not very difficult, as many people thinking.--83.237.198.209 (talk) 19:25, 10 April 2015 (UTC)[reply]

Thank you for confirming that you are a troll - nobody who understood even the most rudimentary English would ignore my last post, unless they were intent on continuing this façade... AndyTheGrump (talk) 19:48, 10 April 2015 (UTC)[reply]

Dura Lex, Sed Lex! — Preceding unsigned comment added by 83.237.221.202 (talk) 20:14, 10 April 2015 (UTC) --83.237.221.202 (talk) 20:15, 10 April 2015 (UTC)[reply]

Good bye 83.237.221.202. (Illegitimi non carborundum)--Aspro (talk) 20:45, 10 April 2015 (UTC)[reply]

Pressure reducing valve question

I have a vented immersion heater connected to a 90 gallon water tank in the loft. The tank supplies 0.5 bar of water pressure. There is a vent pipe from the immersion heater that allows any expansion of water to release safely outside. Obviously if I connected the immersion heater directly to the 3 bar mains water supply the water would overflow out of the vent pipe continuously.

However, could I hypothetically replace the loft tank with a pressure reducing valve set to 0.5 bar and connected directly to the mains? From the perspective of the immersion heater it would still receiving 0.5 bar of water pressure, so it shouldn't make any difference as far as I can see. Am I correct in that assumption or have I missed something vital?

Note; this is merely a hypothetical question and I will not be undertaking any work of this nature. I'm just curious if it would actually work. Pressure reducing valve (talk) 17:15, 9 April 2015 (UTC)[reply]

Yes, your talking about what is referred to as a unvented hot water system. However, you will also need some other bits -- all explained here: [1]--Aspro (talk) 18:46, 9 April 2015 (UTC)[reply]

No that's not right at all.

It would still be a vented system and work at low pressure, the only thing that would change is that the low pressure 0.5 bar of supply water to it comes via a "pressure reducing valve" instead of a loft head tank. If too much pressure builds up inside the immersion heater it would still overflow out the vent pipe.

Unvented systems are entirely different and typically deal with high pressure. I'm not asking about converting a vented system into an unvented system (which is not possible without replacing the entire immersion heater and boiler system) I am asking about changing only the method of water supply to a vented system while keeping everything else the same.

Generic answers won't help here because the system I am proposing is a complete bodge and probably doesn't even have a name. It's an exercise in inventive thinking. Pressure reducing valve (talk) 19:28, 9 April 2015 (UTC)[reply]

OK lets go back to basics. I'm assuming your immersion heater is upstairs (two story typical home) and 'immediately' above that you have loft with the tank. Your presuming that tank deliver water at 0.5 bar. Highly unlikely. Reason: measure the hight from the immersion heater top to the loft tank overflow. Divide that distance in feet by two and it will give the approx pressure in PSI. A pressure of 0.5 bar will raise water some 7½ feet – probably above your loft tanks over flow level. Resulting in all your hot water flowing into the garden (or where ever its vented). One would need an adjustable pressure relief valve to ensure this did not happen. Does that make more sense?--Aspro (talk) 20:24, 9 April 2015 (UTC)[reply]

Perhaps I've over-complicated the question and need to boil it down to something more basic. The immersion heater expects 0.5 bar of water pressure from a loft head tank. Any more than 0.5 bar will cause the water to escape out of the vent. We don't want this to happen. Can the loft head tank be replaced by something else that supplies water at 0.5 bar pressure without water escaping out the vent pipe, or is there something special about the way loft head tanks work that can't be replicated by simply replacing the head tank with a 0.5 bar water supply? Pressure reducing valve (talk) 19:43, 9 April 2015 (UTC)[reply]

Using a pressure reducing valve (set correctly) would have the same result as the tank only if the flow rate was the same as well.

I'd say that the flow rate is more important than the pressure with the heater. Rmvandijk (talk) 19:55, 9 April 2015 (UTC)[reply]

I don't think you understand the difference between pressure and flow rate. Whatever the flow rate (unless it is zero) if the pressure is more than 0.5 bar the vent will overflow. A lower flow rate just means it will overflow slower. Is there anyone on here who can give an actual answer instead of just making vague and inaccurate guesses? Pressure reducing valve (talk) 23:20, 9 April 2015 (UTC)[reply]

@ Pressure reducing valve. Understand pressure head and then you will realize that the 0.5 bar figure is not applicable outside a very tight conditions in this situation. You are asking a form of complex question]that rests on a questionable assumption. I.E. the 0.5 bar pressure head. We do are best to answer reasonable questions here. If you don't like the replies to your complex question then take it else where and waste somebodies else time but stop wasting ours.--Aspro (talk) 14:14, 10 April 2015 (UTC)[reply]

If you don't have an answer to the question then don't post. And especially don't post disparaging comments about "wasting peoples time". You are acting like you've gone above and beyond in trying to help when in fact all you did was post PDF file that wasn't relevant and then claimed the conditions of the question are too" tight" for you to answer. Well then, don't answer it! You are free to not "waste" your time by refraining from posting and let someone else answer, and I'd be very grateful if you did. Pressure reducing valve (talk) 19:32, 10 April 2015 (UTC)[reply]

@Pressure reducing valve: Replacing the cistern this way shouldn't be a problem. In practical terms: Assuming that the actual inlet pressure provided from the cistern is 0.5 bar at rest, you'll have the same pressure at the hot water outlet if on the same level or a minor difference if not. When you install a pressure valve to put away with the cistern, you need to compensate for fluctuations in pressure due to heat expansion, spikes when turning water on and off and finally fluctuations from the valve itself. With a cistern you have a "preset max. high pressure" which is at the highest fill level. Pressure valves might not be as precise at this critical point so setting the valve's pressure slightly lower and/or raising the vent pipe should do. Hope that helps.--TMCk (talk) 21:00, 10 April 2015 (UTC)[reply]

That's a very good point about heat expansion and other fluctuations, I didn't consider that. Thanks a lot that's really helpful! Pressure reducing valve (talk) 23:28, 10 April 2015 (UTC)[reply]

And the PDF that I linked to explained expansion. Why didn't you read it first before shooting it down?--Aspro (talk) 12:47, 12 April 2015 (UTC)[reply]

Reservoir storage

What information is needed to calculate storage volume of a reservoir. I have inflow, outflow and a value for 2s/dt-Q but I don't know what dt is so this is useless. I also have a value for max capacity of river downstream and my guess is this is what I need to use. I also have a storm hydrograph but as I'm only considering baseflow storage volume, I've ignored this. But now I'm not sure what to do. 90.201.190.183 (talk) 18:55, 9 April 2015 (UTC)[reply]

Building our own water-mill or hydro power station are we? Think you'll find that dt just means Δ T or in other words the difference in temperature in kelvins or centigrade. The density (mass per cubic volume) changes with temperature.--Aspro (talk) 19:08, 9 April 2015 (UTC)[reply]

I should think the expansion of the water due to temperature is small enough that it could be ignored (and it's not linear, anyway, so would require a look-up table to calculate accurately). Just taking the initial volume of water, then adding in the elapsed time since then, multiplied by the net flow (inflow-outflow) would seem to get the job done:

V_current = V_initial + (V_in - V_out)t

If that volume is less than zero, then the reservoir is empty (except for the small amount currently flowing through it). If that volume exceeds the max capacity, then the reservoir will be at max capacity (possibly with the surrounding area flooded). Evaporation (and precipitation directly into the reservoir) might be significant on an open reservoir, however, and that depends on not only temperature, but also humidity, wind, and sunlight. As a practical matter, you might just add fudge factors to V_in and V_out, which would vary by season, to account for precipitation and evaporation.

Also, if not obvious, using a formula to predict the future volume of the water in the reservoir makes sense, but if you want to know the current volume, just go a measure it. (I suppose this could get complicated, though, if the reservoir isn't a simple shape.) Without taking current measurements and updating V_initial to match, your end calculation will get farther and farther off, the larger t gets, due to inaccuracies in V_in and V_out. StuRat (talk) 19:18, 9 April 2015 (UTC)[reply]

Whoops: dt = time duration. Water Resources Engineering: Principles and Practice page 52. Sorry.--Aspro (talk) 19:50, 9 April 2015 (UTC)[reply]

What is need is either (a) a topographic map of the reservoir, including areas that are currently under water, or (b) a detailed history of inflow, outflow, direct rainfall, and surface elevation, encompassing the full range of depths that the reservoir supports. In practice you probably need a map, because it is very difficult to compute the amount of water added when it rains directly on the reservoir. Looie496 (talk) 14:26, 10 April 2015 (UTC)[reply]

That equation "2s/dt - Q" pulls up many hits on search engines (some with the sign of Q reversed) which call it a storage indication curve. I don't see enough to learn the derivation from scratch myself, but it's clear that there are differential equations being used to account for the change in outflow based on the amount of water stored, with various assumptions/approximations being made since real reservoirs have their own little wrinkles. An underlying formula is I - Q = dS/dT, i.e. the rate of change of the amount of stored water is equal to inflow - outflow; it looks like some simply take a numerical approach after this, though you can try to solve the integral (beware though that I and Q don't have to be independent of either S or T... diffy-q's are usually a huge pain in the rear!) Wnt (talk) 20:17, 12 April 2015 (UTC)[reply]

Radio question 1

WLW, Cincinnati, holds the record for the strongest ordinary US radio broadcast, at 500,000 watts for several years during the Depression. As noted by our article, it's well recorded that people living near the transmitter could hear broadcasts coming out of non-radio metal objects (fenceposts, mattress box springs, etc.), and there were even reports of people hearing the broadcast out of their metal tooth fillings. Fast forward to the present, and we have reports of electromagnetic hypersensitivity, which are generally discounted. Would radio-in-teeth be considered a form of electromagnetic hypersensitivity (and thus the stories would be considered mistaken folklore), or is electromagnetic hypersensitivity pretty much completely unrelated to the disturbances that would quite understandably result if your teeth started playing music? Nyttend (talk) 21:40, 9 April 2015 (UTC)[reply]

It's completely different. Radio playing from metal fillings, though I'm not sure if it's actually even been reliably recorded, is well within the laws of physics. The purported electromagnetic hypersisitivy on the other hand, has no basis in physics or biology, the energy levels of the average exposure are far too low to cause the sorts of physiological changes reported, tissue being far less affected by radio waves than metals. As stated in our excellent article, 'sufferers' can't even distinguish a placebo field from a real field. Psychosomatic disorders is where you want to be looking for them.... Fgf10 (talk) 00:28, 10 April 2015 (UTC)[reply]

Lucille Ball claimed that she picked up radio signals through her fillings. Our article says: Ball revealed in this interview that the strangest thing to ever happen to her was after she had some dental work completed and having lead fillings put in her teeth, she started hearing radio stations in her head. She explained that going home one night from the studio, as she passed one area, she heard what she thought was morse code or a "tapping". She stated that "as I backed up it got stronger. The next morning, I reported it to the authorities and upon investigation, they found a Japanese radio transmitter that had been buried and was actively transmitting codes back to the Japanese." - but these claims are really pretty preposterous and Snopes were unable to find confirming evidence. SteveBaker (talk) 05:54, 10 April 2015 (UTC)[reply]

You may be interested in this similar question I asked a while back [2]. Steve also brought up Lucille Ball at that time :) I did learn about Magnetostriction, which still seems relevant to this question. SemanticMantis (talk) 14:50, 10 April 2015 (UTC)[reply]

I think we should leave the much beloved Lucille Ball out of this because there is some history to that account that makes the account questionable but for all the right reasons. So I will leave that there. Back (before solid state diodes and mosfet detectors) – Radio Hams that didn't get their earthing sorted out properly found they where getting RF burn to their lips from the microphone. Their lead-amalgam fillings (as where popular in the day) also tingled. Whether this diode in the teeth actually demodulated AM to the point were they could 'hear' the broadcast like on the radio is debatable. I am going back 30- 40 plus years here but the impression I got is that they could only sense low fidelity music. Speech was unintelligible unless they had the gobs stuck up close against the mike an were sensing their own speech whilst transmitting. So Lucille Ball (or her entourage) would have been familiar with this commonly reported phenomena and used it, to deflect the political pressure she found herself under, by conjuring up this anecdote.. As to mattresses picking up RF, it was determined in the last few decades that the reason why so many people reported hearing sounds of a meteor despite it be in 60 to 40 miles up was that many conductive materials around them acted as half wave antenna to the low frequency radio wave generated and vibrated thus producing an audible sound.--Aspro (talk) 14:58, 10 April 2015 (UTC)[reply]

We had a metal coffee table frame that picked up a radio station. I also had a radio that picked up CB broadcasts, even when turned off. So, it's not all that rare to pick up radio frequency signals on devices not intended for that purpose (or frequency). But it seems less likely that any part of the human body could act as an antenna (other than implants, like fillings). StuRat (talk) 02:01, 11 April 2015 (UTC)[reply]

@Fgf10: So far as I know, a strong radio station imposes a quite significant variation in the local magnetic field. The biological capabilities of detecting magnetic field are fairly well documented: they include magnetotactic bacteria that literally align with the Earth's field, and magnetoception by animals such as pigeons that accurately estimates positions using a number of sensors, including cryptochromes and specialized structures in just a few neurons in the brain. [3] So the only piece that is lacking in biological detection of radio signals is the ability to detect rapid changes in the magnetic field (something we would doubt has been selected for, but who knows), and to convert the rate of change of magnitude of the maximum magnitudes of the changes (for AM) into a sound impression. I would say it sounds unlikely, but... nothing is impossible in biology. Wnt (talk) 20:06, 12 April 2015 (UTC)[reply]

Radio question 2

While reading on the above subject, I found this chat thread. I was amused by the idea from anomalous_cowherd's comment: I suppose it would work to use a super-powerful radio broadcast to charge batteries, but if you did that, would it have a discernable effect on local radio transmission, enough that neighbors would notice and engineers would be able to find you? I understand that the guy's batteries would be absorbing the signals, not reflecting them, but isn't that also the case for anyone using a normal radio receiver? Nyttend (talk) 21:40, 9 April 2015 (UTC)[reply]

It's a matter of degree. To make a significant dent in his electric bill he'd have to absorb on the order of a kilowatt. I can't find information on how much power is absorbed by a typical AM/FM radio, but I think it must be a tiny fraction of a watt.

Since the total radiated power of that transmitter is only ~1 MW, it's hard to believe that he could collect enough to be worth the trouble. And since the wavelengths are 250m and up, and his garage was presumably much smaller, it wouldn't cast a sharp shadow and I don't see how it could cause problems for anyone else. So I doubt the story is true, but I'm not very sure about my reasoning. -- BenRG (talk) 04:04, 10 April 2015 (UTC)[reply]

Crystal radio sets can be completely powered by the radio signal...I made one as a kid that was able to extract sufficient energy to drive an earpiece without a battery. Clearly it was extracting energy from the radio signal - so we can say with certainty that you can quite easily extract at least some energy from the signal alone. If you subtract energy from the signal by receiving it, then that has to attenuate the signal for people further from the transmitter. However, it's going to be a very, very tiny attenuation. SteveBaker (talk) 05:45, 10 April 2015 (UTC)[reply]

Correct, the power absorbed by a typical consumer AM/FM radio would be measured on the order of magnitude of millionths or billionths of a watt, or less. For example, see Signal strength, which lists some realistic electric field values at common distances from an ordinary AM or FM transmitter. You can convert from electric field strength (in units of volts per meter) into voltage by multiplying by antenna length; you can convert from voltage to received power by computing V²/R, and for lack of any better value of R, you can assume an antenna whose input impedance is perfectly matched to the impedance of free space (377 Ω). This is all sloppy back-of-the-envelope kinds of math: of course, if the answer matters, you should perform more sophisticated calculations and make less flimsy assumptions. You might want to pick up an advanced textbook on radio engineering (and what better one than Fred Terman's Radio Engineer's Handbook!? It's quite old, but it's got a very practical and pragmatic approach to calculating such quantities, although Terman still uses archaic nomenclature like "kcps" instead of kHz, and all his circuit examples are based on tubes that haven't been readily available for decades). Short summary: a real AM/FM radio is extracting very little power. Steve correctly points out that this is enough to power a crystal radio's earpiece - but not its amplifier - and as I recall, when I built an unamplified crystal radio, I used a massive antenna - the water pipe attached to the shower - to pick up the strongest possible signal! You can extract just enough voltage to drive a piezoelectric speaker, and hear sound, but it's quite faint unless you add a powered amplifier.

Radio enthusiasts might also read our article, dBm, a normalized logarithmic unit of true received radio power that radio engineers use. A WiFi base station might transmit at +20 dBm and the computer receiver might pick up +0 dBm. A common AM/FM or shortwave receiver may work around -75 dBm. People who work with spaceships - satellite television radio engineers ("SATCOM") and research scientists - commonly work in the world of -100 and -120 dBm. Keep in mind that these are logarithmic scales, and that the gain-bandwidth product (which is a material property, and is the same for any radio built out of common CMOS semiconductor materials) essentially dictates your total power/bandwidth/noise!

Nimur (talk) 17:50, 10 April 2015 (UTC)[reply]