Rawlemon’s Spherical Solar Energy Generation From Moonlight - Page 1
Forum section: electronics / general

author: EEVblog
author_rank: Unknown
timestamp: September 19, 2016, 10:13:11 am
content: I predict about half a bdW (Bee's Dick Watt) of power from the moonlight.http://inhabitat.com/rawlemon%E2%80%99s-spherical-solar-energy-generating-globes-can-even-harvest-energy-from-moonlight/

author: 2N3055
author_rank: Super Contributor
timestamp: September 19, 2016, 11:56:16 am
content: Quote from: EEVblog on September 19, 2016, 10:13:11 ambdW

author: RGB255_0_0
author_rank: Frequent Contributor
timestamp: September 19, 2016, 12:01:40 pm
content: It hurts the green movement when these lunatics are given the limelight (could have made a joke here but this is no joke). I wonder how long it'd take for the energy gathered from the "moon's" energy to recoup the cost to manufacture such monstrosity.

author: rob77
author_rank: Super Contributor
timestamp: September 19, 2016, 12:16:46 pm
content: wow ! what a stupid design !probably it's 35% better on concentrating sunlight/moonlight but it must weight a few tons .... who in his right mind would design a solar system which can't be installed on roofs because of it's sheer weight ?and i would also have a look at losses in that giant marble... probably the losses in the glass will be higher than the 35% gain

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 20, 2016, 10:24:56 am
content: https://what-if.xkcd.com/145/

author: Someone
author_rank: Super Contributor
timestamp: September 21, 2016, 07:47:14 am
content: Quote from: T3sl4co1l on September 20, 2016, 10:24:56 amExcept thats still an open ended question without any consensus on an answer, etendue is not preventing it from being possible.

author: Red Squirrel
author_rank: Super Contributor
timestamp: September 21, 2016, 08:17:27 pm
content: I jokingly mentioned here what if you could use a bunch of mirrors to concentrate moon light to a solar panel and get usable power. Someone ran the numbers and you need a ridiculous amount that may as well simply be said to be impossible. This design would be better served as a solar thermal generator.

author: Towz
author_rank: Contributor
timestamp: September 24, 2016, 08:04:28 pm
content: It's so ridiculous!! I doubt it would ever even match the energy wasted manufacturing it... but of course! We've got an elegant way to power a solar calculator at night! A bdW should suffice

author: TheWelly888
author_rank: Frequent Contributor
timestamp: September 25, 2016, 03:23:54 pm
content: If Wallace and Gromit can make a man and a rabbit nuts by concentrating moonlight then this is a perfectly viable idea.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 27, 2016, 09:28:39 pm
content: There is an even more ridiculous / confusing version: one could point a kind of telescope to a really dark part of sky. Here the radiation temperature is really low, well below ambient and thus very little IR light comes from there. So an IR sensitive photodiode would show a reverse voltage (other polarity than with normal light), as there is IR radiation send to the sky with no return. So you could generate power from sending out radiation to a cold place.I have not done the numbers but I would expect less than a bee's ...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 27, 2016, 11:36:28 pm
content: Quote from: Kleinstein on September 27, 2016, 09:28:39 pmThis isn't as bad as it sounds!But it will be thermal in nature.You could build a moderately effective solar-thermal array by doing...1. Build a radiator. Assuming clear skies overall, then: during the day, it will absorb sunlight and heat up. During the night, it will cool down.2. During the day, pump the heat into a holding tank. Preferably phase change (paraffin or molten salt) storage.3. At night, run a heat engine from the tank, using the radiator as the cold side.This boosts efficiency because the radiator goes down quite cool at night: perhaps 40 degrees below ambient.The heat transfer rate is fairly low, so you will need quite a lot of radiator, of course. But it will still perform better (higher Carnot efficiency) even if the radiator isn't much below ambient temperatures -- because, if you used convection cooling to ambient air instead, you can only ever have the heatsink somewhat hotter than ambient, never equal or below.Probably, the solar array should be a concentrator type, like the trench or tower kind. The Sun is highly directional, so it's a big net win to thermally isolate your hot side from anything that's not line-of-sight with the Sun. That way you can get really hot fluid coming out. (As opposed to a flat thermal field, like say, a blacktop parking lot with thermal pipes underneath, which loses heat as radiation in all directions, plus convection off the huge area of air it's in contact with.)Conversely, the cooling radiator should be exactly the opposite kind: because, at night (cloudless), the whole sky is largely quite cool, and you need to maximize area because the temperature is low and radiation has crappy conductivity at low temperatures.Okay so I lied, it's not as exciting as when I started typing... The reason the collector and radiator must be asymmetrical is because the radiation source and sink are asymmetrical in nature. So, unfortunately, there's no easy two-for-one sale on that.Tim

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 01:42:13 am
content: Quote from: T3sl4co1l on September 20, 2016, 10:24:56 amTotal nonsense! Is this a new tactic - fighting bad science with even worse science?The site is right in saying that lenses and mirrors are bi-directional and so that an equilibrium is reached, but it mixed up power and temperature. It claims that as the Suns temperature is 5000 deg, then if all the light from the Sun was directed to, say, a 1M square black body, it would only reach 5000 degrees maximum. As radiated power is approximately proportional to the fourth power of the temperature for a Sun and the Sun's surface area is over 6 x 1018square meters, the What-If page is possibly incorrect by a factor of over 100000000000000000000000000000000000000000000000000000000000000000000000000000000%.That is what I call the error of all errors!I have to admit I cannot say precisely how a 1m2black body behaves when receiving all the Sun's energy - I haven't done the test myself.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 28, 2016, 02:02:32 am
content: Well, I can assure you with great confidence that the 1 m^2 area would, very quickly at that, heat up to 5000 K. Assuming the material somehow remains within that 1 m^2 region at such temperatures, it would simply stay there in equilibrium.Global (so to speak) equilibrium would certainly be off by a bit, though -- all the power being radiated by the Sun would be reflected back into itself, creating the heaven's grandest greenhouse of all. It is, of course, identical to coating the inside of your Dyson sphere with a mirror finish -- why you put the bling on theinside, I have no idea. You must not be a very smart Dyson-sphere-constructor. Which kind of begs the question...But anyway, the Sun's surface temperature would rise, and the photosphere, puff up.The temperature, of our somehow-still-stationary 1 m^2 plasma cloud, would track nearly perfectly with the Sun's surface temperature (give or take propagation delay).Pretty soon, the temperature will rise to that which is normally seen at much deeper layers, and hard UV and even some soft x-ray radiation will dominate. Those better be some damned good mirrors!In any case, focusing that much radiation, from an astronomically sized body, doesn't sound like a good idea...Tim

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 02:11:37 am
content: Quote from: T3sl4co1l on September 28, 2016, 02:02:32 amTim, have you read what you have written? A 1m2black body at 5000 degrees would cause the Sun to heat? You realise that to heat the Sun significantly, the power has to be radiated from the 1m2black body. A 1m2black body at 5000 deg may radiate enough to warm a decent sized building, but not the Sun.To radiate the same power back to the Sun that the whole Sun is emitting, a 1m2black body would have to be at a nonsensical temperature - something like 3 x 1069degrees.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 28, 2016, 02:55:41 am
content: So what you're saying is, a cup of water, at room temperature, can't possibly be in thermal, radiative equilibrium with the Earth, because the Earth has a surface area of (whatever) and the cup doesn't, and so the total radiative powers of both can't possibly be reconciled?Tim

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 03:14:08 am
content: Quote from: T3sl4co1l on September 28, 2016, 02:55:41 amNo I am not saying that.The subject here is two objects of different sizes connected by lenses so there are two completely different power densities on each side of the lens, and I am saying that in this case they have to reach a radiated power equilibrium regardless of the temperature that that implies. If you put a cup of water in space with a huge lens focusing the Earth's emitted light, the water in the cup would boil. It has to because if you are putting megawatts of power into a cup, it has to get very hot. It cannot re-emit energy to return it back to the Earth without getting very hot. You seem to think that somehow megawatts of power can be emitted from the cup back to the Earth via the huge lens without the cup getting hot. That is not physically possible. Emitted power is driven by temperature. If there are megawatts of emitted power, then the cup must be very, very, very hot.

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 03:35:13 am
content: Just another example, places like the Livermore Labs have been using lasers to initiate Nuclear fusion. Now fusion needs temperatures like 14 billion degrees, and the lasers are not at 14 billion degrees. They use Neodymium glass lasers, so there is no way temperatures in the lasers can be very high at all.If you focus the power of simultaneous pulses from 192 x 2.5 Terrawatt Lasers onto a 2.3mm diameter target, you get 14 billion degrees along with an inward pressure on the core of the target of 300 Billion atmospheres. It is conservation of Energy not conservation of Temperature.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 28, 2016, 04:50:21 am
content: Quote from: amspire on September 28, 2016, 03:14:08 amThe power densities are equal -- as you stipulated, they are equal temperatures!QuoteFascinating...So, how far does the lens have to be from the Earth? Can it be on the Earth? Is the atmosphere refractive enough (given suitable conditions) to meet your requirement?So why does my cup not boil, sitting here?Perhaps another riddle will prove fruitful? Suppose you point an antenna at the Sun. (It should be a large enough parabolic dish type, so the beam is reasonably pointy, and can be said to be fully receiving solar noise.) For as-yet poorly explained reasons, the corona is microwave-active, with a temperature in the millions K in that frequency range. This is reflected at the antenna terminals as a few nV/rtHz noise (the same temperature, expressed into 50 ohms, or whatever impedance the antenna is matched to).You agree that this is all consistent, right? There's a resistance, there are two degrees of freedom (Gaussian noise on a resistor), and there's a temperature, which can be no other temperature than that of the plasma which is radio-opaque in the relevant range, correct?Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 28, 2016, 04:57:04 am
content: Quote from: amspire on September 28, 2016, 03:35:13 amNonequilibrium, and lasers are superluminous; in a certain sense, they have an exceedingly high temperature, over a very narrow passband. (But a mere thermal bandpass wouldn't exhibit the spacial and spectral coherency that is characteristic.)Tim

author: IanB
author_rank: Super Contributor
timestamp: September 28, 2016, 05:20:27 am
content: Quote from: amspire on September 28, 2016, 01:42:13 amThis is why science is a useful tool, and why we can use science to make fun of people who believe in free energy and other impossible schemes.I have to believe the xkcd argument because of science. There is a law of thermodynamics that says you cannot transfer heat from a colder body to a hotter body without doing work on the system. This law has been tested to exhaustion and is effectively immutable. It is as impossible to violate this law as it is to construct a perpetual motion machine. Under no circumstances can heat flow spontaneously from a colder body to a hotter body. You could spend your whole life trying and it would be as vain as the work of alchemists past trying to turn lead into gold.

author: Someone
author_rank: Super Contributor
timestamp: September 28, 2016, 05:43:40 am
content: Quote from: IanB on September 28, 2016, 05:20:27 amThe falling down of that explanation of the XKCD comic is assuming that the light coming from the moon is emitted by its temperature, when we all know that you can look at a fire with a mirror and the mirror is not too hot to hold in your hand. Sun -> Moon (inefficient mirror) -> Super Lens -> Deathray

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 05:44:56 am
content: Quote from: T3sl4co1l on September 28, 2016, 04:57:04 amThe output of these lasers is just UV photons. Photons have no temperature, so once the photons leave the laser, temperatures are totally irrelevant. The only thing that matters is the amount energy in the photon stream.If the photons hit a black body, all the photon energy will be absorbed by the black body raising the temperature. As the temperature rises, the black body will radiate energy that can go back to the source but the amount of energy radiated is totally a function of temperature. For a small object to radiate a huge energy, it has to be much hotter then the large emitter on the other side of the lens.To have a Black body release a large amount of UV, you would be looking at 10,000 deg to 20,000 deg but that is a very long way to the 14 billion degrees that can cause fusion in the case of the Livermore lasers.All radiation is photons and so it has no temperature. Temperature is a statistical measurement of particle motion and the old form of the 2nd law of thermodynamics was "Heat cannot of itself pass from a colder to a hotter body". The moment you add external heat pumps, lenses, lasers to push away hot atoms (in cooling helium down to near absolute zero), you can get heat passing from a cooler body to a hotter body as it is no longer a closed system. The way the 2nd law is worded nowadays, if you close the system (ie include the heat pumps, lenses, etc, power sources), then the total entropy of the closed system cannot decrease. That does not say that photons from the Moon cannot heat a small target in a vacuum to 10,000 degrees or more. As long as the total entropy is increasing, it is fine. It is merely a matter of how many photons you can get from the Moon and how small the target is. You could calculate the number of photons needed to get 10,000 degrees in a specified object.

author: IanB
author_rank: Super Contributor
timestamp: September 28, 2016, 06:04:30 am
content: Quote from: amspire on September 28, 2016, 05:44:56 amQuoteI rather think photons do have a temperature characteristic. That is how astronomers can measure the temperature of the cosmic microwave background radiation and compare it with predictions from the big bang.The important distinction that you mention in your post is the difference between an active system and a passive system. With an active system you can put work into it and can achieve any temperature you wish. A passive system with no external inputs cannot concentrate the photons from the moon to achieve any temperature greater than the surface of the moon. If it were possible it would violate the second law, no matter which way you look at it.

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 06:19:22 am
content: Quote from: IanB on September 28, 2016, 06:04:30 amYou could say a lens is passive but it definitely allows a colder body to heat a hotter body. As long as you look at the total entropy in a closed system, the 2nd Law is correct. If you get hung up on the colder body-hotter body thing, you can get yourself into trouble.

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 06:41:41 am
content: Just did some quick calculations. Based on an emission of about 26 lux, the amount of light radiated by a full moon would be about 1013W. This excludes IR radiation which would be much higher - this is just visible light.The amount of this power the Earth would receive during a full moon would be about about 1000th of this or about 10 Gigawatts.So to get a useful amount of power from Moonlight - say 1Kw would keep life's barest essentials going - light, TV. modem and computer - we would need a sphere that had a diameter that was about 3200 times smaller then the earth.That would be 8km in diameter and weigh 8 x 1016kg.A 1m diameter sphere would probably produce something like 16 uW during a full moon on a clear night. Over the month with clouds, you would be lucky to get 4uW. A 1m glass sphere would weigh about 1300 kg and if you assume a 25% efficient solar cell, you would get about 1uW at night on average.If you got a sphere the day you were born, and accumulated the moonlight energy every night, then on your 90th birthday, you could have the 1kW of power for a full 4 hours!

author: EEVblog
author_rank: Unknown
timestamp: September 28, 2016, 06:58:24 am
content: I want to do a video on this but am having a hard time getting a ballpark figure for insolation of the full moon. (and I guess insolation isn't the right term any more)Got some papers that mention solar flux so maybe I can covert that or something, hmm.

author: EEVblog
author_rank: Unknown
timestamp: September 28, 2016, 07:00:54 am
content: http://oceancolor.gsfc.nasa.gov/SeaWiFS/On_Orbit/lcal/docs/kieffer_stone_irradiance_moon_aj.pdfandhttp://nvlpubs.nist.gov/nistpubs/jres/118/jres.118.020.pdf

author: 2N3055
author_rank: Super Contributor
timestamp: September 28, 2016, 08:17:47 am
content: I dabble in photography.. White object in full moonlight is about -2 EV and full sunlight is about 16-17 EV... The scale is LOG, means 10 to 18th difference..Extrapolating from that, Sun gives cca 1e3 W per sqm, moonlight is cca 1e-18 down from that... so you need 1e15 sq meters to get 1kW....In that NASA paper they talk microwats...So yeah, won't work...

author: tszaboo
author_rank: Super Contributor
timestamp: September 28, 2016, 08:25:44 am
content: Quote from: 2N3055 on September 28, 2016, 08:17:47 amSo, you can store more energy in a 100uF, 12V capacitor, than get from the moonarpanel (tm) in a day.

author: 2N3055
author_rank: Super Contributor
timestamp: September 28, 2016, 08:34:09 am
content: After all what are we waiting... I'm sure someone has a solar panel (I was wondering why no one makes Moon panels) , a data logging meter, and a full moonlight... Should measure current... If you get milliamps you're good to go... Then all you need is 2 kilometers glass ball and you can charge your phone....I think you could harvest much more energy if you put bunch of antennas in different frequencies and rectifying and upconverting radio stations electromagnetic waves....Get close to large cell tower and I bet you could get a watt or two from it with proper antenna ..Not that I suggest stealing electricity....

author: EEVblog
author_rank: Unknown
timestamp: September 28, 2016, 08:38:49 am
content: Quote from: 2N3055 on September 28, 2016, 08:34:09 amYou'll get microamps at bugger all voltage, so microwatts.

author: 2N3055
author_rank: Super Contributor
timestamp: September 28, 2016, 08:49:18 am
content: Quote from: EEVblog on September 28, 2016, 08:38:49 amYou'll get microamps at ampermeter burden voltage, that's what you'll got...But hey, it looks like a giant crystal ball...

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 09:16:28 am
content: The figure I found somewhere was 26 lux at the Moon's surface, but I now think this number was wrong. I suspect this value is over 100 times too low.Lux and Lumens are weighted for the human eye, so it is not a good number to use for measuring energy.If you go back to the basics, this NASA fact sheet is not bad:http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.htmlThe W/m2from the Sun is about the same as the Earth: 1361.0 W/m2The Bond Albedo (Reflected light ratio) is 0.11 so that means the Moon is reflecting about 150W/m2The Moon diameter is 1737 km and the distance to the Earth is 384400 km. That is a ratio of 0.00451. Since the area of a sphere is proportional to the radius squared, then 1 m2on the Earth corresponds to 0.004512= 20 x 10-6m2on the Moons surface.That means the total Moon reflected light on Earth for a full Moon would be 150W/m2x 20 x 10-6= 0.003W/m2.This page:https://www.reddit.com/r/askscience/comments/15o5im/does_a_full_moon_provide_any_noticible_reflected/?st=itmma42x&sh=a298b82fcalculated about 0.0031W/m2.Now according to this pagehttps://en.wikipedia.org/wiki/Moonlight,light from a full Moon is typically 0.1 Lux but is about 0.26 Lux at high altitude near the equator, so there is probably a significant attenuation of Moonlight at low altitudes away from the equator.I think if you used 3mW/m2at high altitudes at the equator for a full moon, it would represent the upper limit on the possible amount of reflected energy from the Moon. IR black body radiation from the Moon would have an Upper Limit of 24 mW/m2.The reddit page also calculates the radiated IR radiation and got about 1182 W/m2so the IR radiation energy from the Moon is about 7.9 times the reflected sunlight.Richard

author: IanB
author_rank: Super Contributor
timestamp: September 28, 2016, 01:47:17 pm
content: Quote from: amspire on September 28, 2016, 06:19:22 amIt is possible to do an experiment to demonstrate this. For example, take an electric bar fire where the heating element has a certain temperature--let's say it is 1000°C. Now according to your hypothesis you could use a system of mirrors, lenses and other passive elements to concentrate the heat from the fire and achieve a temperature higher than 1000°C. For instance you could focus the heat on a thermometer element and make the thermometer read higher than the source temperature. I don't think you would be able to do so.

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 02:38:05 pm
content: Quote from: IanB on September 28, 2016, 01:47:17 pmOf course you can do this. If you are saying that no matter how much energy you send to a target, it stays at the same temperature, that is defying all the laws of physics.A simple thought experiment. You have set up one heater with a convex mirror and a target at the focal point heats up to 1000 deg. Now you add 9 more heaters with mirrors pointing to the same target. If the target is a black body, all the photons from the other 9 heaters will get absorbed by the target. A black body by definition absorbs all photons as it cannot reflect any photons. It is now absorbing 10 times the power that it was absorbing with one heater. If it stayed at 1000 deg, that would mean it can only radiate the same heat as it could with one heater. The only way it can radiate 10 times the heat it was radiating with one heater is for the temperature to go up to well over 1000 deg. Conservation of energy requires the temperature to increase to whatever is needed to increase the radiated heat by a factor of 10. You cannot have the heat from the 9 extra heaters disappearing into a void.It seems like you have the view that a a black body at 1000 deg can only receive photons from and object that is over 1000 deg and that is totally untrue since photons have no temperature. A black body at 1000 deg can receive photons emitted by a body at 1 deg absolute. A red photon emitted by some cryogenic LED at -200 deg and a red photon emitted by the Sun are absolutely identical and so the receiving black body absorbs both exactly equally.As I said, if you think the 2nd Law of Thermodynamics says that heat cannot pass from a cooler object to a hotter object, you will get yourself into trouble. That is why the 2nd Law does not say that. It never has and it never will. If you shine your LED torch at the Sun, you are sending heat to the Sun.

author: IanB
author_rank: Super Contributor
timestamp: September 28, 2016, 04:40:35 pm
content: I think you are overlooking that the black body is also a radiator. If the black body is absorbing photons from the heaters, it is also radiating photons back towards the heaters along the same path. If the black body is hotter than the heaters then the net flow of energy will be in the reverse direction away from the black body.We can analyze this in terms of thermodynamics. Let us suppose there is to be a net transfer of a quantity of heat \$ \delta q \$ from a colder body to a hotter body in a passive system (a passive system is one where no work takes place).The change in entropy is therefore given by:$$\Delta S = - {\delta q \over T_c} + {\delta q \over T_h}$$If$$T_c < T_h$$Then $${\delta q \over T_c} > {\delta q \over T_h}$$Which means that$$\Delta S < 0$$If the entropy decreases then the second law is violated.If I shine my torch at the Sun, then the torch will absorb photons from the Sun and get heated up. If I manage to aim it precisely the light from the Sun will be focused by the reflector onto the LED emitter and the LED emitter will get fried. The net transfer of heat will absolutely be from the hotter body to the cooler body.

author: amspire
author_rank: Super Contributor
timestamp: September 28, 2016, 06:46:08 pm
content: I think you have the wrong equations. They are for two systems in thermal contact. In two systems linked only by photons, there is no thermal contact. Conduction and photons behave differently. You can passively magnify the energy density of photon streams with a lens, but you cannot do the same with conduction. Once a photon is emitted, it will travel until it hits something regardless of the targets temperature. Conduction does depend on temperature.If you look back through this discussion, you will see an absurdity. If net photon energy transfer is driven by temperature difference, then a 1 meter square plate at 5000 deg can radiate the full power or the sun. So if someone takes a plate and heats it to 5000 deg, the Earth isimmediately is destroyed. When your equations are telling you things that cannot possibly be true, it is time to stop and see where you have made a mistake.

author: IanB
author_rank: Super Contributor
timestamp: September 28, 2016, 07:12:41 pm
content: Quote from: amspire on September 28, 2016, 06:46:08 pmNo, I disagree. Limiting the mechanism to thermal conduction is an arbitrary and unnecessary restriction. Thermodynamics deals only with state transitions and outcomes, regardless of how those transitions occur. The laws of thermodynamics apply to any system in the universe, without exception.This is the absolute beauty of thermodynamics. It can tell you whether something is possible or not before you attempt to design any kind of machine or device to achieve your proposed outcome. If you propose to decrease the total entropy of system plus surroundings by some state transition, then you know it cannot be achieved, by radiation or otherwise.QuoteHeat transfer by radiation can be treated with the appropriate transfer laws just like conduction. In industry furnaces are designed using radiation laws considering surface areas, temperatures, emissivities, view windows, and lines of sight and all of this is fully understood and successful. No part of the furnace is ever hotter than the hottest part of the flame, and nor can it be if only passive heat transfer by any mechanism is considered.QuoteThis is an absurdity because nowhere in this thread has such a thing been suggested. A square plate heated to 5000 deg can of course radiate with the sameintensityas the surface of the Sun. But saying this in no way suggests the total radiated power is equal to that of the Sun.Perhaps you could actually try the experiment with the bar fire and a thermocouple probe? Try to use lenses and mirrors to focus the energy of the fire down into the tiny volume of the thermocouple tip and make it hotter than the fire. You can surround the thermocouple with firebrick insulation to prevent heat escaping and try any other passive device you can contrive. I guarantee you cannot do it.

author: Someone
author_rank: Super Contributor
timestamp: September 28, 2016, 10:17:48 pm
content: Quote from: IanB on September 28, 2016, 06:04:30 amPhotos only contain quantised energy, its the spectrum (relative amounts of energy across the wavelengths) that defines the temperature of a black body radiator through Planck's law.You're way off the mark and overextending your understanding of physics, and have wandered off on a tangent ignoring the gaping hole in your original argument which started this.

author: amspire
author_rank: Super Contributor
timestamp: September 29, 2016, 01:54:48 am
content: Quote from: IanB on September 28, 2016, 07:12:41 pmIt could be that my original premise about sending all the Sun's radiation to a 1m2panel was the problem. It seems there may be laws of optics that state that if you capture all the radiation of the sun, you cannot focus it to a spot smaller then the Sun. I didn't know that. This would mean that to have a black body receive all of the Sun's radiation, it has to be at minimum the Sun's size.I cannot see why you cannot have a gigantic lens that focuses the sun's energy to a focal point the size of the Sun, and then have billions of mirrors in the Sun size focal point bouncing light to a much smaller target. If you are right, then there has to be some reason that means that doesn't work. I will have to think on this.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: September 29, 2016, 02:07:45 am
content: Quote from: IanB on September 28, 2016, 06:04:30 amAll good until the last couple of sentences. Those are true for thermally emitted photons from the moon. Much of the radiation from the moon is sourced by a much hotter surface, the sun.Things like this are why scam science succeeds. There are a lot of really bright people on this forum and even in this illustrious group it is hard for everyone to keep track of the pertinent facts. A quick skim misses key points that make a free energy project either impossible, or fails to kill the faith in said project because the pungeant argument has flaws. Less informed groups have little chance in comparison to the readers of this forum.Another example is the laser fusion comments. Those experiments do not depend only on radiative energy transfer. A significant part of the process is inertial confinement, transfering momentum from the laser beam to the target. Which then transfers the energy into a smaller volume with non radiative processes. And at no point in the process is any kind of equilibrium reached. The devil is in the details.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: September 29, 2016, 02:21:49 am
content: Quote from: amspire on September 29, 2016, 01:54:48 amOf course you can focus the entire surface of the sun on a spot smaller than the sun. Every solar telescope does this, and so does the pinhole in a piece of paper used to view eclipses. But that system doesn't capture all of the energy emitted from the sun. The sun radiates almost equally in all directions so most of the energy misses your telescope or pinhole. So you start making your mirror bigger. It ends up needing to be solar scale to reach all sides, and then when you do the simple geometric optics you run into the image size problem.Thermodynamics is powerful, but leaves mousetraps everywhere to catch the unwary. Think of the simple firebow firestarter. This in no way violates thermodynamics. Entropy increases. But the hot point of friction rises to temperatures far above anything else in the local system. The concentration of heat occurs through a non radiative process, and equilibrium is not achieved over the time period of interest.

author: IanB
author_rank: Super Contributor
timestamp: September 29, 2016, 02:55:01 am
content: Quote from: amspire on September 29, 2016, 01:54:48 amThis is a tricky problem to reason about, I agree.One possible approach is to consider beam lines. Suppose you had a giant lens focusing the sun down onto a small target, and you allowed the system to reach equilibrium so that nothing is changing with time.Now every point on the surface of the target is connected by a beam line to some point on the surface of the Sun. Since the system is in equilibrium the intensity of the absorbed radiation along every beam line is equal to the intensity of the emitted radiation. It follows that if the absorbed and emitted radiation intensities are equal at each end of each beam line, then the temperatures at each end of the beam line must be equal. If this argument applies to every beam line individually then it must apply to the system as a whole when integrating over all beam lines.I believe that if one object is not a black body then the argument still works, since the emissivity applies symmetrically to absorption and emission and therefore cancels out at equilibrium.Another thought experiment is to imagine a small ball suspended in a vacuum inside a much larger hollow sphere. If the interior walls of the hollow sphere start out hotter than the ball then the ball will gradually heat up by radiant heat transfer until it equals the temperature of the outer wall and thermal equilibrium is reached. This will happen even though the total inner surface of the sphere facing the ball is much larger than the surface area of the ball. A justification for this argument is that the interior of the hollow sphere will become filled with black body radiation at the temperature of the system. Being bathed in this radiation it is inevitable that the small ball must equalize its temperature with the surrounding radiation temperature.

author: IanB
author_rank: Super Contributor
timestamp: September 29, 2016, 03:02:56 am
content: Quote from: CatalinaWOW on September 29, 2016, 02:07:45 amI do agree with this and it is a definite loophole in the "temperature of the Moon" argument.However, what makes the exact calculations more complicated is that the radiation incident on the Moon from the Sun is the entire black body spectrum. The sunlight reflected from the surface is only a tiny fraction of the incident spectrum. Most of the incident radiation causes the Moon to heat up until it radiates the same energy out into space at thermal equilibrium. The absorbed and re-radiated energy is probably by far the larger part of the total lunar radiation (I don't have any facts to hand on this).

author: Someone
author_rank: Super Contributor
timestamp: September 29, 2016, 04:41:37 am
content: Quote from: IanB on September 29, 2016, 03:02:56 amIt isn't, the links from dave above show the distributions. Or consider that the moon reaches a maximum surface temperature of around 400 degrees Kelvin, which emits almost no visible light. Here you can see that emission is irrelevant:http://www.asterism.org/old/tutorials/tut26-1.htmThis is entirely an optics problem, there is no need to confuse it with thermodynamics.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: September 29, 2016, 04:43:00 am
content: The explanation in xkcd is given for passive lenses, not mirrors. Seems to me that a mirror isnota passive device. Any mirror that's going to concentrate the sun's radiated energy to one location is going to experience a net reaction force. So I would think you can put a parabolic reflector around the sun and then focus the result down, but that mirror is going to be pushing to do it!

author: IanB
author_rank: Super Contributor
timestamp: September 29, 2016, 08:23:52 am
content: Quote from: Someone on September 29, 2016, 04:41:37 amTheBond albedoof the Moon isreported by NASAto be 0.11. This means that of all the solar radiation landing on the Moon, 89% of it is absorbed and 11% is reflected back out into space.Since the Moon has been around for a long time we are safe to assume it has reached thermal equilibrium with the Sun. It follows that the Moon is emitting 9x more energy into space as thermal radiation compared to reflected light in the visible spectrum.

author: IanB
author_rank: Super Contributor
timestamp: September 29, 2016, 08:28:45 am
content: Quote from: Galaxyrise on September 29, 2016, 04:43:00 amIf a lens changes the direction of a ray of light then it will also experience a reaction force normal to the plane of the refracted angle. In other words, a lens that focuses a parallel beam of light down to a point is going to experience a net reaction force back towards the light source.However, in either case work is the product of a force and a distance. If a force acts on a mirror but it doesn't move then no work is done and the mirror is passive. Similarly, a table holding up a heavy weight is applying a reaction force to prevent the weight from falling, but the table is not doing any work.

author: Someone
author_rank: Super Contributor
timestamp: September 29, 2016, 10:24:42 pm
content: Looks like I'll have to take back the previous statement that etendue was not limiting the solution. The worked example is as follows:Burning paper requires at least 100kW/m2 flux (100mW/mm2), interesting but not needed for the proof.The sun is providing a flux of 1kW/m2 to the earths surface, and moonlight is at least 100,000 less intense in the visible spectrum, a maximum of 0.01W/m2 (https://en.wikipedia.org/wiki/Solar_irradianceandhttps://en.wikipedia.org/wiki/Daylight)Worlds best concentrating optics can achieve 100,000:1 gain in flux (https://eng.ucmerced.edu/sett/presentations-1/4-Nonimaging%20Optics.pdf)Sunlight to moonlight ratio 100,000:1Concentration peak 100,000:1So the best we can hope for is a radiant flux (irradiance) as bright as direct sunlight which we know doesn't ignite paper.Even trying to take a lucky segment and timing of the moons image that is much brighter than the average wouldn't get the missing factor of 100. Energy storage in capacitors from solar panels and then releasing that energy in small bursts however ....

author: stm
author_rank: Newbie
timestamp: January 06, 2019, 01:30:08 pm
content: Despite the warning, this thread was dead for over 120 days... Dave, do you still plan to make a video about the rawlemon? A colleague just pointed me to this and it smelled fishy somehow.. So i ended up here again., Would really appreciate it, to see your calculations on this 😊 Greetings, Stephan

author: fsr
author_rank: Regular Contributor
timestamp: January 06, 2019, 03:58:13 pm
content: That's one needlessly big, heavy and expensive loupe... What's the advantage? That the sphere doesn't needs to track the sun, and the loupe does? The PV cells do need to track the sun, however.About the difference in brightness of the sun and moon, that's data that astronomers have since a very long time, an this wikipedia article even calculates it:What is the ratio in brightness between the Sun and the full Moon?The apparent magnitude of the Sun is −26.74 (brighter), and the mean apparent magnitude of the full moon is −12.74 (dimmer).[calculations removed, because they don't display good here, look at the wikipedia article]The Sun appears about 400000 times brighter than the full moon.https://en.wikipedia.org/wiki/Apparent_magnitude#Example:_Sun_and_MoonThat probably means that at most, with a full moon, you can get 1/400000 of the energy than with the sun.

author: Zero999
author_rank: Super Contributor
timestamp: January 14, 2019, 01:03:23 pm
content: Quote from: amspire on September 28, 2016, 06:19:22 amNo it does not.Heat will always travel from a hotter body to cooler body. If you have two objects, one huge and one much smaller, with a lens between them and no external source of energy, they will reach the same temperature eventually. Once both objects are the same temperature, they will sit there for all eternity.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 14, 2019, 04:48:16 pm
content: It's non-obvious why this is; it has to do with another property of light -- which happens to be thermodynamically relevant -- conservation of étendue.https://what-if.xkcd.com/145/Tim

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: January 17, 2019, 08:58:47 am
content: Quote from: Zero999 on January 14, 2019, 01:03:23 pmOnly strictly true for conduction. Radiated heat is exactly the same thing as radio signals and light, except different wavelength. Think about it a minute; if it were true, your radio in a warm room would get nothing from a cold outdoor transmitting aerial.The reality is that all bodies with non-zero absolute tempexchangeradiant energy. Consider a universe in which there is just one star, at say 6000K surface temp. This will radiate EM radiation in all directions, and since there are no other bodies in the vicinity it will be simple case of all that energy being lost.Now consider that the star has just one planet, surface temp 300K. The planet has no internal heat source, all heating is passive, from the star. Obviously the planet absorbs thermal radiation, however the interesting point is that it also radiates infrared back in all directions. Most of that is lost to the vastness of empty space, but a small amount impinges on the star. The fact that the star's surface is hotter does NOT alter the fact that the radiation from the planet warms the planet-facing surface of the star, albeit just a little."You can't use lenses and mirrors to make something hotter than the surface of the light source itself."From the above you should be able to see that this is complete BS. For a real world example, a semiconductor laser can burn a hole in steel. The melting point of gallium arsenide (or whatever chip material) is a lot less than that of steel. The steel melts. The chip doesn't.HTH in understanding a subject which is counterintuitive in some respects.

author: Marco
author_rank: Super Contributor
timestamp: January 17, 2019, 11:47:43 am
content: QuoteQuote"So yeah, I'm going to pretend the moon is a black body any way"He never addressed the fact that most of the moonlight is reflection. You're not imaging the moon, you're imaging the sun, reflected by an imperfect mirror.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: January 17, 2019, 04:17:57 pm
content: -What is a black body? Well, an ideal black body is one which has 100% absorbtion, zero reflection, and zero transparency to a particular EM radiation wavelength. By Kirchhoff's Law of Thermodynamics (Which essentially says there are no thermal diodes) such a body will also be a perfect emitter of the same wavelength of radiative heat.Note that this is 'to a particular wavelength' and an object which behaves as a black body to one wavelength need not necessarily do so for others.Real substances are bound to fall short of this ideal, having some reflectance, and/or transparency. It also follows from Kirchhoff's Law that an object with high reflectance (albedo) or transparency will also be a poor emitter of radiation.The Moon's surface is only about a quarter as reflective as the Earth's, which partially explains why the sunward side gets so hot and the dark side so cold. The radiation we receive from it though, is mostly reflected, the Sun's rays predominating over black body radiation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 17, 2019, 06:37:04 pm
content: Quote from: IanMacdonald on January 17, 2019, 08:58:47 amYou're very nearly correct in your understanding of this subject, but it is imperfect --The radio example works because the radio's front end has an effective temperature below that of the signals it's intended to receive. Which depends on what frequency band it's designed for, because background noise varies, but let's say VHF or UHF where we can ignore atmospheric noise and consider the background being faint (except for the Sun) astronomic noise sources.The radio consumes power, and consequently is able to produce nonreciprocal elements. One consequence of this is, the effective noise temperature of the front end can be much smaller than the ambient temperature of the instrument as a whole. Indeed, adequate to resolve those distant astronomic noise sources, for some radios (although it's noteworthy that some of those radios are, in fact, cryogenically cooled, for even better performance; I don't know when that's required as far as effective noise temp).It's noteworthy that all the elements up to the radio front end (transmission medium, antenna, transmission lines) are (usually / preferably) low loss. That is, they do not contribute much of their own temperature to the noise level received. The optical equivalent of this is simply a mirror: a metal mirror can be quite hot, incandescent even, but it is still shiny. Its emission only dominates over the reflected signal when the reflected signal is below the emissivity of the mirror.Or for transmissive media, same thing, the bulk glow doesn't necessarily dominate. Hot glass looks really freaking cool: it does glow, but not much, because its emissivity in visible light is very low; meanwhile, it remains clear and colorless, so you can see through it, but also see its own glow through it.So too, the antenna needs to be low loss, so that it faithfully couples between free field and transmission line, without adding its own noise to the system. In effect, the antenna is a lens, and the transmission line is a fiber optic, shiny and reflective and clear, without adding its own noise to the system. (Of course, a sufficiently bad, and long, cable will have appreciable losses, and introduce its noise and temperature to the system. Such is the case with any medium.)Conversely, the laser is a special, active device. I'm not too certain of the V-I vs. illumination characteristics of a solid state laser, but at the very least, as a PN junction, it must act like a solar cell. Which responds to incoherent, wideband radiation, not just sharply tuned laser light. So already we can see there is some nonreciprocity here. (Well, anything inside a resonator will respond mainly to signals passed by that resonator -- but the point is, the response will still be wider than the laser emission line(s) are.)The laser emission is extremely intense (not just narrowband and coherent, but very parallel as well), and this occurs due to a feedback process. The equivalent temperature is very high indeed: to have such high emission at a narrow band, you'd need an incredibly hot source (like, Sun's core hot), sent through an incredibly narrow bandpass filter.Why does it melt metal, but not itself? Well, reciprocity is one thing; lasers can indeed be damaged by very reflective targets (poorly matched loads -- so too can radio transmitters!). As long as most of that energy is leaving, it's fine.You might just as well ask why an induction heater can melt metal, when its energy is very low (fractional MHz, so, whatever that is, photons of nanoelectronvolts?). Clearly its effective temperature must be incredible (and the bandpass on that noise, equally strong), and to get visible light emission somehow or another implies one heck of a multi-photon mixing/upconversion process. Well, something like that. Physics gonna physic. Suffice it to say, a quantum description of a macroscopic, classical system is a rather impractical approach, and since it happens, we can conclude that yes, somehow or another, such a mechanism exists.So, suffice it to say: classical physics is adequate here, and active devices not in thermodynamic equilibrium, need not obey the laws of thermodynamics on a local basis?Tim

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 17, 2019, 06:40:17 pm
content: Quote from: IanMacdonald on January 17, 2019, 08:58:47 amSemantics and details are why this conversation and subject are always difficult. The radiation from the planet does not warm the suns surface, it just slightly reduces the cooling. The net flow of heat will always be from the warmer body to the cooler body for thermal radiation.The laser welder example is not a case of thermal radiation. To solve the thermodynamics you have to consider the whole system. You can locally violate entropy in all sorts of ways. Every time you type an email, solve a jigsaw puzzle you are violating decay of local entropy. But in every case we have discovered to date, and as far as we understand the universe there are no non local exceptions.

author: TerraHertz
author_rank: Super Contributor
timestamp: January 25, 2019, 09:29:05 am
content: Quote from: amspire on September 28, 2016, 01:42:13 amExactly.I lost most of my respect for XKCD when he presented a cartoon-ized version of the AGW 'hockey stick graph' as fact. Thus demonstrating that he's quite gullible and doesn't look into things beyond a superficial level.