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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. -
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...
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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..
So, you can store more energy in a 100uF, 12V capacitor, than get from the moonarpanel (tm) in a day.
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... -
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.... -
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...
You'll get microamps at bugger all voltage, so microwatts. -
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...
You'll get microamps at bugger all voltage, so microwatts.
You'll get microamps at ampermeter burden voltage, that's what you'll got...
But hey, it looks like a giant crystal ball... 
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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.html
The W/m2 from the Sun is about the same as the Earth: 1361.0 W/m2
The Bond Albedo (Reflected light ratio) is 0.11 so that means the Moon is reflecting about 150W/m2
The 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 m2 on the Earth corresponds to 0.004512 = 20 x 10-6 m2 on the Moons surface.
That means the total Moon reflected light on Earth for a full Moon would be 150W/m2 x 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=a298b82f calculated about 0.0031W/m2.
Now according to this page https://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/m2 at 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/m2 so the IR radiation energy from the Moon is about 7.9 times the reflected sunlight.
Richard
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You 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.
It 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. -
It 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.
Of 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. -
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.
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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 is
immediately 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. -
I think you have the wrong equations. They are for two systems in thermal contact.
No, 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.QuoteIn 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.
Heat 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.QuoteIf 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 is immediately 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.
This 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 same intensity as 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.
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Photos 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.Photons have no temperature
QuoteAll radiation is photons and so it has no temperature.
I 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.
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.
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It could be that my original premise about sending all the Sun's radiation to a 1m2 panel 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 think you have the wrong equations. They are for two systems in thermal contact.
No, 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.
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. -
Photons have no temperature
QuoteAll radiation is photons and so it has no temperature.
I 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.
All 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. -
It could be that my original premise about sending all the Sun's radiation to a 1m2 panel 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 think you have the wrong equations. They are for two systems in thermal contact.
No, 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.
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.
Of 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. -
It could be that my original premise about sending all the Sun's radiation to a 1m2 panel 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.
This 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. -
All 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.
I 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). -
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).
It 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.htm
This is entirely an optics problem, there is no need to confuse it with thermodynamics. -
The explanation in xkcd is given for passive lenses, not mirrors. Seems to me that a mirror is not a 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!
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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).
It 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.htm
This is entirely an optics problem, there is no need to confuse it with thermodynamics.
The Bond albedo of the Moon is reported by NASA to 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. -
The explanation in xkcd is given for passive lenses, not mirrors. Seems to me that a mirror is not a 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!
If 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. -
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_irradiance and https://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 100,000:1
Concentration peak 100,000:1
So 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 .... -
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