Can you use a magnifying glass and moonlight to light a fire? (2016) - <AUTHOR> https://what-if.xkcd.com/145/ <comment><USER3> Interesting article. Right in many places. Wrong (possibly) in main conclusion.

Entropy argument - correct in the sense that using radiation from black body we cannot use lenses to heat another body to the temperature higher than original. Easy to understand why - the first body has a temperature, radiation has the same temperature, if we apply the radiation to another object it will not heat up more than the radiation's temperature.

Also the argument about impossibility of concentrating light into a dot is correct (although even if it were possible we still would not be able to get higher temperature - light would not be energetic enough for that). The important part is - we could concentrate light into a dot only if it consist of parallel rays - i.e. only for an object that is infinitely far away.

Moon surface temperature argument is incorrect. A body at 100 degrees Celsius does not radiate in visible spectrum, so the light we see is not produced by Moon's temperature. It is reflected Sun light. So Moon's temperature doesn't matter. Moon surface does absorbs some light, changing spectral composition from about 5.7kK (Sun's surface temperature) to about 4kK. So we should consider moon to be a part of optics not emitter.

Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)? I don't have answer for that - need to do calculations. What can be a deal breaker? Remember that Moon is much closer than Sun, so rays come to us even less parallel, so the area into which we can concentrate light reflected from the Moon is even larger than the Sun's, so together with lower intensity of light from Moon we might have trouble achieving the necessary intensity for combustion. However big enough lens probably will work.

And yes - I'm a physicist by training.

<comment><USER4> The conclusion is correct. It doesn't rely on the moon being a black body radiator, but on etendue, which shows that the most you can do with a system of lenses and mirrors is to create an environment (on earth) as bright as the environment on the moon. The fact that rocks on the moon's surface only reach 100°C shows that an environment like that is not bright enough to light a fire.

<comment><USER3> Trying to follow your argument. Suppose we put absolutely reflective mirror on the orbit. Since its absolutely reflective its temperature will be 0 Kelvin (in theory, very low in practice - there is a reason solar bound spacecrafts are covered with reflective surfaces). So we would not be able to use the light it reflects to heat anything?

<comment><USER5> Specular reflector preserves the etendue of the light rays that fall on it. Diffuse reflector (such as the moon) does not; it becomes a new light source instead, resulting in higher etendue.

In short - perfect reflector preserves etendue, but imperfect does not.

<comment><USER6> The question is to what extend the moon surface is imperfect reflector. For example, if the mirror has a hole, it will be an imperfect reflector, but you can compensate for it using a bigger lense.

<comment><USER7> By imperfect it's sufficient to be diffuse. You can verify the Moon is diffusive to a good approximation because of the approximately uniform appearance of the full Moon.

The idea is that you can "organize" or "revert" any ray bundle from a system of non-absorbing lenses and specular reflectors, but if your reflector has billions of tiny irregularities it's not viable to build such a system (it's equivalent to an ideal diffuser, in which light is isotropically reflected). The ideal diffusion process is clearly not reversible by itself: if you shine a beam onto a diffuser it spreads the light; if you expose it to the same light (with reversed directions), it again diffuses it instead of reverting to the original beam. In theory again the physical laws of electromagnetism are time reversible, but in practice the effort to revert some systems might be too demanding (you can even do better -- see Maxwell's demon); manipulation of physical apparatus and information acquisition/manipulation itself has a cost that surpasses any gains.

</comment></comment></comment></comment></comment></comment><comment><USER8> > You can't use lenses and mirrors to make something hotter than the surface > of the light source itself.

This is an interesting argument. Can I not reflect some sunlight off a mirror, then do the magnifying-glass-to-start-a-fire trick in daytime? Doesn't the mirror stay cool? Isn't the moon just a (poor) mirror for the sun's light?

<comment><USER9> You've found the right question to ask.

Your mirror in sunlight works because the reflectivity or albedo of the mirror is very high relative to whatever target you're lighting on fire.

In a magical closed system where radiative heat transfer was the only factor, objects of differing reflectivity would eventually reach temperature equilibrium through black body radiation.

We aren't interested in closed systems though. The moon's temperature is set by the equilibrium between incident solar radiation and black body emission, most of which flies off into deep space making the system very open. Just like your mirror's temperature is the equilibrium of incident radiation, black body cooling, and convective cooling in Earth's atmosphere.

If the moon had a high reflectivity and/or a powerful cooling mechanism like convection, its equilibrium temperature would be far lower than the temperature of its emitted + reflected light. Unfortunately the moon's albedo is just 0.12 and black body radiation is all it's got, so the modest difference between its temperature and that of its light isn't enough to start a fire.

<comment><USER10> Really?

According to the Stefan-Boltzmann Law, the energy put out in blackbody radiation is proportional to temperature to the 4th power. Therefore something that is 5000 degrees reflected off of an object with albedo 0.12 is putting out 0.12 times the energy it originally did, while something that is 2500 degrees only puts out (1/2)^4 = .0625 times as much. So the "temperature of the Moon's light" should be more than hot enough to light something on fire if it is focused right.

As a sanity check, compare how much light the moon puts out as a black body in shadow with what it reflects from the Sun. As another sanity check, compare how bright the Moon is versus a fire.

What am I missing here?

<comment><USER11> The angular concentration of the light.

The moon is diffuse, so an incoming ray of sunshine is spread by the optically rough surface of the moon from an incident solid angle of 6 * 10^-5 steradians out into 2 pi steradians of the night sky, or a reduction in angular concentration by a factor of about 100,000 (totaling ~1 million after the albedo is accounted for).

It is like you are looking at the sun through a mirror so rough that the image of the sun is blurred over literally half the sky. Because this process does not create new photons, the blurred image must be far, far dimmer. This circumstance corresponds to the "most we could do" with lenses and mirrors focusing the moon, which is to fill the sky with an image of the moon/"blurred sun".

Unconcentrated moonlight corresponds to the same picture, except we only see a "cutout" disk of this blurred sun-image which is the size of the moon in the sky. Our crappy moon-mirror does not fill our vision, it is a porthole letting through only a tiny fraction of the blurry sun-image.

And of course if you imagine yourself as the ant under the magnifying glass, with your entire sky filled with moon, there is no way you could spontaneously become hotter than your moon-y surroundings.

<comment><USER12> This is the primary reason. You can no more start a fire with sunlight reflected off the moon at night than with sunlight reflected off a sheet of paper during the day. (In fact you can do better with the sheet of paper, because you could in theory surround it with lenses to recapture the diffuse light.)

</comment></comment></comment></comment></comment><comment><USER13> Wow... separately I had no idea the surface of the moon reaches (and goes above) 100°C... that's _hot_! Literally boiling hot. Turns out at night it goes down to almost –200°C. That's _insane_.

Quick searching of how the astronauts survived this, turns out it seems they timed landings to the lunar dawn for an in-between temperature, that the lunar surface doesn't conduct heat well (all dust?), of course there's no atomsphere to conduct heat, and that their boots were extremely well-insulated.

<comment><USER14> Also, since the moon is tidally locked to the earth, it rotates at the same rate it revolves, which means that sunrise and sunset only happen once a month, which makes dawn a pretty long time.

There's a pretty good song about how the moon's day/night cycle would affect a lunar mining colony: [https://www.youtube.com/watch?v=GDPUdUGJpjc](https://www.youtube.com/watch?v=GDPUdUGJpjc)

<comment><USER13> Ah, that makes so much more sense then... two weeks of straight unfiltered sunlight is gonna heat things up... and two weeks of total darkness gives it time to freeze, freeze, and freeze some more.

</comment></comment></comment><comment><USER15> Thermodynamics argument seems iffy to me. I can cover the entire surface of the earth with solar panels to harvest the moon light and use the combined electricity to melt iron. If this is ok with thermodynamics then so is using lenses.

The rest of the argument seems to be that light cannot be optically condensed to a single point as there will always be some dispersion due to diffraction, and the size and shape of the dispersion is dictated by that of the source. That is you can't make the target denser then source, hence the temperature must be less.

EDIT: Several commenters submitted that lenses are reversible while solar panels are not, and this makes all the difference. My retort is that I can make non-reversible lenses by covering them in a thin layer of dust. Since the lense system is now non-reversible can I use these sub-par lenses to create higher temperature than I could with clear lenses?

<comment> 08-15 > I can cover the entire surface of the earth with solar panels to harvest the > moon light

Are you sure about that? First off, take an off-the-shelf solar panel, point it at the moon in the middle of the night. You get a grand total of nothing. Okay, it was a cheap panel, that might not generalize to anything.

But more importantly, by the argument laid out in the article, your solar panels _cannot work_ in moon light. (Or maybe they work at horrible efficiency, because the moon is a bit warmer than the earth.) I'm not sure I buy that argument; maybe you should run the experiment.

<comment> 08-15 I'll repeat for the reflexive downvoters:

Solar panels will work in moonlight _if and only if_ you can make fire from moonlight with a magnifying glass.

The answer depends on how good a mirror the moon is. It calls for a real experiment, not a thought experiment. I don't really know which way it will go.

<comment><USER15> You're not addressing the essential part of my comment. Let me clarify:

I put in place 1000 solar panels and aim them at the sun thus procuring 200kw of power. Next I use it to power an arc welder, producing 10,000 C of heat. Therefore I have used a 6,000C sun to produce 10,000C on earth.

Would a similar setup work with the moon? I don't know. That was not my point, my point was that it certainly is possible to produce higher temperature at the target than it was at source.

<comment><USER16> Monroe's explanation only applies to optical systems, ie reflecting and focusing light, which is a thermodynamically reversible process (yes, even with dust. Dust doesn't make the lens thermodynamically irreversible, it just scatters and absorbs some of the light so it doesn't reach the lens. The irreversibility is a property of the lens). Solar panels are converting light energy into electrical energy, which is not an irreversible process - <USER2> creates entropy, and therefore the rules are different.

</comment></comment></comment></comment></comment><comment><USER17> So thinking about this some more, the argument here is that if you use "just" a magnifying glass of arbitrary size and shape you can't do this. I can buy that based on the arguments presented. Basically it says that the moon emits (really reflects but that's immaterial) F photons per second per square meter, and while you can concentrate that into a very small area, all of those photons will not be enough to raise the temperature (that is input enough energy into the system) to a sufficiently high level. This is partially because you can't make a small enough point with a single lens, and partially because there just aren't enough photons. The sun doesn't care because it has such a high flux that the concentration ends up being high enough for the area.

The area argument is more important here because the lens cannot increase the number of photons per second, but it can decrease the area. If F = N / (t * A) where N is the number of photons, t is time, and A is area, the lens can change the area, but not to 0. And if you need a sufficiently high F to get to the right temperature, the only way to get there with limited N is to bring A sufficiently close to 0.

If you have multiple magnifying glasses and mirrors I am fairly certain that you can. That is the equivalent of using a set of solar panels that power a laser. But that was not what was postulated in the original thought experiment, so it does not apply.

I am still fuzzy on the thermodynamic argument, but I was never good at intuiting thermodynamics. The argument presented is that if you have one body at 100 degrees C, and you put another body next to it, you cannot make the second body hotter than the first. That makes sense. But if the first body is constantly generating and transferring heat to the second with at most 100 degrees C temperature, _and_ the second body has some way to store heat energy, then it is possible to heat a local area of the second body to higher than 100 degrees C. The storage of energy here is what I think counts.

<comment><USER16> The thing is, radiative processes (like light) can't store energy. To do that, you need some kind of engine, either electrical or mechanical.

Of course you could put a solar panel connected to a battery in moonlight for a few months and build up enough stored energy to power a laser for long enough to fry something. But that's not "burning something with moonlight using a magnifying glass."

<comment><USER17> Right. That's what I am saying. Using the original setup of the problem, you cannot use a device to store energy.

</comment></comment></comment><comment><USER18> Oh god, can we please get a real physicist in here? This entire thread is a mess of computer programmers "well actually"ing other computer programmers and everyone being wrong.

<comment><USER19> The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). A lot of the counter-arguments/speculation here in the comments is wrong.

Reading this is is equivalent to reading a thread on a physics forum where with people arguing about an article saying that O(n*lgn) is really the best possible runtime complexity for a comparison-based sorting algorithm, and trying to disprove it.

It's also worth pointing out that Randall Munroe is a physicist (to the extent an undergrad degree counts anyway).

<comment><USER20> _The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive)._

It's interesting what bothers different people. While many of the statements in this thread are probably wrong, not many of them bother me. But I find the lack-of-self-doubt and appeal-to-authority in your message to be genuinely offensive. Where does your certainty come from?

With that out of the way, could you give some links to well known arguments that you refer to? Specifically, I feel certain that one can start a fire with sunlight reflected from a room temperature mirror, and don't understand the difference between a mirror and the moon within Munroe's argument.

His conclusion might be correct (in practice, you may not be able to concentrate moonlight enough to start a fire) but I don't think the details of his argument can be. I currently don't believe that the temperature of the reflecting surface can be the limiting factor, and I think this is central his argument.

<comment><USER5> >Specifically, I feel certain that one can start a fire with sunlight reflected from a room temperature mirror, and don't understand the difference between a mirror and the moon within Munroe's argument.

A mirror does specular reflection and thus conserves the etendue of the sunlight. You're concentrating the image of the sun in the mirror, not light from the mirror itself.

The moon in contrast is mostly a diffuse reflector - it scatters most of the light that falls on it (and absorbs and re-emits most of the rest), so it is effectively a new light source.

<comment><USER20> Yes, I can agree with that. The problem for me is fitting it into Munro's argument. He says (in boldface): "You can't use lenses and mirrors to make something hotter than the surface of the light source itself".

This is true for a black body, but is it always true for an object being illuminated by another? I'm don't know that we can consider a diffuse reflector with a surface temperature of 100C as being equivalent to a black body with temperature of 100C. I think his conclusion is likely true (the moon is too dim to start a fire even with a really big magnifying glass) but I don't think he's right to point to the surface temperature of the moon as being the evidence of this conclusion.

Assume the sun was much brighter, so that ignition on earth is possible with a sufficiently large magnifier. Presumably if the moon was the same, this would mean the surface temperature of the moon was much higher. Now change the moon to be more heat conductive (causing the surface temperature to drop due to more heat loss on the dark side), and more reflective (causing the surface temperature to drop further due to less absorption). I'd guess that if you tweak the parameters sufficiently, you could end up with a surface temperature low enough that Munroe's argument would say that ignition is impossible, even though we've increased the intensity of the moonlight over our baseline.

How does Munroe know that we aren't in that second regime? I don't think there's enough information in his argument to distinguish. Alternatively stated, we know that there is some current temperature to which we can heat an object using concentrated moonlight. We also know that if we can change the shape and composition of the moon, we can reduce the surface temperature without reducing the intensity of moonlight. Unless there is some limit to the effectiveness of the heatsink that we can put on the moon, I think this means there is some possible arrangement that violates the assumption that the surface temperature must always exceed the temperature achievable with a magnifying glass.

(Thanks for helping me to puzzle this out)

<comment><USER5> An excellent point!

 _> Now change the moon to be more heat conductive (causing the surface temperature to drop due to more heat loss on the dark side),_

Yeah, although this isn't too big in the case of the moon (unlike the Earth, it doesn't have an atmosphere and doesn't rotate rapidly, so there isn't too much redistribution of heat across its surface), it is definitely something that would confound the calculations. We'd still be able to just look at the effective temperature of light falling onto the moon and that would limit the temperature that we could light the object up to. But we wouldn't be able to use a direct measurement of the temperature of the surface of the moon.

 _> and more reflective (causing the surface temperature to drop further due to less absorption)._

To the extent that it's a gray body (and most objects are approximately graybodies), this wouldn't actually lower the temperature. Absorptivity < 1 causes it to absorb less energy from the light, but for a gray body emmisivity equals absorptivity so it also radiates out less light too, and you actually end up reaching the same equillibrium temperature as fully absorptive black body.

<comment><USER20> Thanks for the kind response. I now see that the final average core temperature should remain the same, but I'm less sure about the surface temperature. There are a lot of complex processes involved, and I'm not sure that one can conclude that everything cancels out to keep the answer constant.

I respond late to offer a link (that's hidden on the second page of this thread) that I think presents that argument I was trying to make better than I managed to: [https://physics.stackexchange.com/questions/370446/is- randal...](https://physics.stackexchange.com/questions/370446/is-randall- munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-th/370600). I thought the comments on the answer were a helpful reframing of the problem.

</comment></comment></comment></comment></comment></comment></comment><comment><USER21> Physicists here.

He's trying to explain everything in terms of a simple blackbody in thermal equilibrium, peacefully radiating its energy away only via thermal photons. That's not the reality of the radiation from the sun or moon. Solar physics is an entire branch of physics, and such simple toy models are not even wrong.

Sun doesn't just radiate away its existing energy via thermal photons.

First, it keeps burning its fuel via a series of nuclear reactions, which by the way keeps pumping energy into the system, essentially acting like a battery (so there's no perpetual motion here).

Second, sun emits photons that are much more energetic than the thermal photons from the surface. Some of the radiation is not thermal, and comes directly from different types of nuclear reactions (which provides signatures regarding the kind of reactions happening in the sun) and various other processes.

</comment><comment><USER22> Discussed at the time: [https://news.ycombinator.com/item?id=11211454](https://news.ycombinator.com/item?id=11211454)

</comment><comment><USER23> So what about non-linear optics and optical frequency doubling?

[https://en.wikipedia.org/wiki/Second- harmonic_generation](https://en.wikipedia.org/wiki/Second-harmonic_generation)

It was just an option on my MSc in LASERs but I thought it was cool (with the potential to be very warm).

Although my frequency has been halved and I've been working in software for decades

<comment><USER24> I thought of that as well. Once you have a different frequency you can also do more bizarre fun things. [http://optics.org/news/7/1/11](http://optics.org/news/7/1/11)

</comment></comment><comment><USER25> I don't buy the thermodynamic argument. Here's a version I would believe: if you have a gadget that, exposed only to the sun and to empty space, heats some target hotter than the sun, then that gadget must _not_ work if you take away the empty space part. This is because your gadget could be used to drive a heat engine, which is impossible without a temperature difference, and the sun is more or less a blackbody emitter. Lenses and mirrors aren't magically taking advantage of the cold parts of the sky, so there you go.

But the moon is not blackbody, and I think the whole argument falls apart. Here's a thought experiment: go stand on the moon, and assume the moon is made of rock that diffusely reflects, say, half of the indicent 500nm light. Stand somewhere that's in shadow, so you can't see the sun. Wrap a piece of paper and some air in perfectly insulating, perfectly reflecting material, except that the material lets 100% of 499-501nm light through, but only on the moon side. The target will be in a bath of 499-501nm light at 1/2 the intensity (energy density per unit volume) of the sun, which is far more than half the temperature of the sun. It'll catch fire after a while.

Now do the same experiment on the Earth, at night, with lenses to bathe it in moonlight from all sides. Fire! So I claim that lenses+mirrors+filters can start a fire with moonlight.

Another interesting question: can you use a luminescent solar concentrator or other fluorescent material to pull this off without taking such egregious advantage of the spectrum of moonlight? These types of materials can violate conservation of étendue.

<comment><USER26> > The target will be in a bath of 499-501nm light at 1/2 the intensity (energy > density per unit volume) of the sun, which is far more than half the > temperature of the sun. It'll catch fire after a while.

Unconcentrated sunlight is slightly under 1400 watts per square meter. It's equivalent to a temperature of 122C.

You can concentrate sunlight coming from the sun, because the sun only fills five millionths of the sky. With a simple lens you can focus hundreds of megawatts per square meter onto a surface.

But once you bounce that light off a diffuse surface, whatever concentration you had becomes the new maximum.

In your experiment, bathing something in moonlight would max out at 700 watts per square meter.

700 watts per square meter doesn't set things on fire. It can only heat a blackbody to 60 degrees C.

Even the full brunt of unaltered sunlight can only bring a blackbody up to 122C.

-

Treating the moon as a blackbody or not doesn't actually change the equations. The important property is that it _diffuses light_. It resets your maximum concentration of light, because light that comes evenly from every direction can't be concentrated.

(I'm ignoring the part about wavelength filtering because it's confusing and would only make your piece of paper heat up less.)

<comment><USER25> > 700 watts per square meter doesn't set things on fire. It can only heat a > blackbody to 60 degrees C.

> (I'm ignoring the part about wavelength filtering because it's confusing and > would only make your piece of paper heat up less.)

I'm afraid you're ignoring the interesting bit. You say that 700 W/m^2 doesn't set things on fire. This is not true. Sure, 700 W/m^2 applied to some target that is allowed to radiate its own blackbody light out to the sky won't get it very hot, but that's not what I'm suggesting. I'm suggesting that you insulate the target very well so that its blackbody emissions don't escape, but you let in the short-wavelength moonlight. Thermodynamics requires that you also let _out_ the short wavelength blackbody emissions, but those are negligible until the target gets very, very hot.

This effect isn't science fiction — it's just the greenhouse effect, amplified. Greenhouses (the glass ones and the atmospheric ones) exploit the fact that sunlight doesn't match the _Earth's_ blackbody spectrum, so a filter (glass or gaseous) can allow incoming radiation in but trap most outgoing radiation.

In effect, I'm suggesting that a very good greenhouse plus some lenses could get hot enough to start a fire.

<comment><USER26> I see. I couldn't quite follow which parts you were saying to insulate, and your mention of "half the temperature of the sun" lead me to misinterpret. That's fine then, I think. Not an expert on wavelength filters.

</comment></comment></comment></comment><comment><USER6> The article reasoning can be shortened to observation that passive optical system does not change the wavelength of photons and to trigger a fire the wavelength has to be short enough.

But the conclusion of the article is wrong. The surface temperature of the Moon has very little to do with the wavelength of the reflected photons.

Consider a surface covered with ideal tiny mirrors each pointing to random direction. Only a tiny proportion of these mirrors will reflect light from the Sun towards observer. Now consider that 90% of those mirrors are painted black reducing the reflected enrrgy flux by further factor of ten. The Moon is like that.

Still the reflected light has original wavelength of the light of the Sun. Collect enough of it and that triggers fire.

<comment><USER4> That's incorrect. It has nothing to do with the wavelength of the light.

> Still the reflected light has original wavelength of the light of the Sun. > Collect enough of it and that triggers fire.

You can't collect enough of it into one place with a passive optical system, because it's been irreversibly scattered by the moon's surface (Read: irreversible increase of [https://en.wikipedia.org/wiki/Etendue](https://en.wikipedia.org/wiki/Etendue)).

<comment><USER6> Yes, I stand corrected.

Essentially an optical system will bring Moon's surface closer, but even if it brings the surface within 1 cm from the wood, the defused Sun light scattered from that surface is not enough to ignite the fire.

</comment></comment></comment><comment><USER27> Well, this doesn't look right to me. Imagine that the moon is actually a filter at the path of the sunlight.

Sun's temperature is 6000 K. Moon's surface is pretty black: it reflects only 12% of the light.

So, effective temperature of the Sun reflected by Moon, considering that thermal radiation is proportional to T^4, is 6000 * .12 ^ (1/4) ~= 3500 K. That's quite enough to light up some fire! Of course, the spectral composition of the light will be not thermal etc, but the estimate should be close enough.

Why doesn't the Moon itself heat up like that? Well, the rocks on Earth don't heat up to 6000 K either... I think, it's partly that they are "not surrounded by sun", partly that the Moon is a giant cold heatsink

</comment><comment><USER28> I'm a bit skeptical of the conclusion and the reasoning process used to arrive there.

For example, the article states "In other words, all a lens system can do is make every line of sight end on the surface of a light source, which is equivalent to making the light source surround the target."

If you forget about optics for a second, imagine that the outer surface of the sun were wrapped around a point (think of the image shown in the article). If you consider conservation of energy for the energy flux from the surface of the sun being entirely directed to a single body of matter that absorbs this heat (assume it's a penny), the steady-state blackbody emission of the penny would have to equal the energy flux from the entire surface of the sun. I think this situation would end up making the 'temperature' of the penny much higher than the surface of the sun for the same reason that the center of the sun is hotter than the surface: There is energy expended, and it comes from the fusion of light elements inside the sun, so there is no violation of entropy as stated in the article: "you'd be making heat flow from a colder place to a hotter place without expending energy."

<comment><USER29> That doesn't work because when the penny is hotter than the sun it radiates an equal amount of energy back to the sun.

The sun is a blackbody, it radiates light because it's hot. Once the penny is the same temperature it will radiate back at the sun until the two are in equilibrium.

<comment><USER28> If the size of the penny was reduced to just a few atoms, wouldn't changing the number of atoms in the penny affect the rate at which the penny itself could emit radiation through blackbody emission? If so, I would expect there would be a size at which the penny could be small enough that it's temperature could grow arbitrarily large once the heat flux in = heat flux out from the penny.

<comment><USER30> I don't know very much thermodynamics, but I'm not sure that thermodynamic temperature is well-defined for a system consisting of "just a few atoms".

</comment></comment></comment></comment><comment><USER31> Isn't the bigger problem that sunlight is nearly parallel and moonlight is reflected off of a spherical surface?

How are you violating conservation of energy if you're taking all of the light that would hit a square mile of the earth and concentrating it down to the size of a penny?

If you can't concentrate light that way then how do focusing lenses on cutting lasers function? Makes no sense.

<comment><USER5> It's not because the moon is a spherical surface. If that were the issue, there would be a bright spot on the Moon where you can see the reflection of the Sun (you can see such specular reflection spots on e.g. many cars), and you could use that spot to light a fire. The issue is that the Moon barely reflects any of the light that falls on it. Most of the light is scattered, and most of the rest is absorbed and re-radiated.

<comment><USER32> The _scattering_ seems it should cancel out. As far as total energy reaching a point on earth (or the circular disk of a lens) just as much should get to you only due to scattering as failed to get to you due to scattering.

I don't think this changes the ultimate answer to the question.

</comment></comment></comment><comment><USER17> I don't buy the argument that you can't concentrate two beams on the same spot. Sure you might not be able to do that with one lens, but concentrating on a small area is as good an approximation. But if you require that it really be a point, why can't I do that with N lenses and mirrors that are fully reversible but all align to aim at the same point from different angles?

<comment><USER26> The argument is that you can't concentrate two beams to hit the same spot _from the same direction_.

So yes, you can have a bunch of lenses each focusing from a different direction.

And if you do a good job of aligning them, each lens will look as bright as the sun/moon from the target.

But that's your limit.

</comment></comment><comment><USER33> I'm not so sure about this.

Imagine you had a large cloud of planar mirrors, each, specifically can be aimed at any given point --even points that overlap.

While I agree that you cannot focus a whole image to a smaller area than the diffraction limit allows for a continuous lens surface, if you omit diffraction, mirrors could certainly do it.

<comment><USER34> You are describing a parabolic concentrator.

</comment></comment><comment><USER35> > Lenses and mirrors work for free; they don't take any energy to operate.

Wait a minute. Does that mean that I could get a tiny solar panel and light it up with a lens, instead of getting big panels? Energy output of a panel is proportional to the amount of light that hits it, right?

I guess that lenses are 'free' only if they have no impurities, but even then, assuming solar panels are costlier than plastic lenses, I could save money.

Come to think of it, how is it that I haven't seen or heard about parabolic reflectors with solar panels in the focal point? Right now I've found an article about parabolic troughs that are apparently used to heat old-school fluids instead: [https://en.wikipedia.org/wiki/Parabolic_trough](https://en.wikipedia.org/wiki/Parabolic_trough)

<comment><USER36> Making solar cells that effectively can handle concentrated sun light is difficult (if they heat up efficiency goes down, ...), so it's easier to just fill an area with solar cells instead of with mirrors pointing at a smaller cell area.

<comment><USER35> > if they heat up efficiency goes down

So you're saying I should cool them with water and use the steam to move turbines, then I'm golden!

\\($ ∇ $ )/

<comment><USER37> I'm not sure if you are joking or not, but something like this is still used for solar power generation.

[https://en.wikipedia.org/wiki/Solar_power_tower](https://en.wikipedia.org/wiki/Solar_power_tower)

</comment></comment></comment></comment><comment><USER38> Randall might be correct for conventional optics, but what about metamaterial lenses that break the diffraction limit? [https://en.m.wikipedia.org/wiki/Superlens](https://en.m.wikipedia.org/wiki/Superlens)

</comment><comment><USER39> I think They missed something. If the max temp you can get from moon light is 100c that doesn't mean you can't start a fire. You just need something that has a very low ignition temp. There must exist something that can start a fire at this temp.

<comment><USER40> Maybe: [https://en.m.wikipedia.org/wiki/Carbon_disulfide](https://en.m.wikipedia.org/wiki/Carbon_disulfide)

Doesn't appear that many substances have auto ignition as low as 100c.

If you're hunting for easy to ignite stuff, it might be better to go for low flash point stuff, and strike a spark?

Eg gasoline will work in pretty cold environments with a flashpoint of - 43c.

</comment></comment><comment><USER41> There is a very interesting image in the article. It shows a bunch of light coming into a block and emerging as a beam. It also has a caption saying this is impossible.

It struck me as very similar to the setup in this video. [https://youtu.be/awADEuv5vWY](https://youtu.be/awADEuv5vWY)

At about 4 minutes in, a nearly identical setup is shown with a beam of light emerging from a block of opaque material using holographic techniques.

It seems plausible to me that a specially designed anti-moon hologram could allow reconstruction of the incident light from the sun, thus allowing a fire to be started without violating any law of thermodynamics.

<comment><USER41> I think I finally realised what rubs me the wrong way about these xkcd musings, and the related discussions.

[Start rant]

They take the situation to its absurd conclusion, then quit with a full finality that people take as truth. Further absurd conclusions are ignored, and their word is law.

Some more absurd arguments for why you will be able to light a fire with a magnifier and moonlight are as follows:

1: use a pre-magnifier to create a spot on the moon with the same temperature as the sun, then magnify the light from that spot to start your fire.

2: wait a long time... eventually a meteor will hit the moon creating a spot bright enough to focus.

3: wait even longer... Eventually the random mollecular collisions between the wood and the (presumably oxegen rich) air around it will convert it into carbon dioxide and water while serendipitous individual high energy blackbody photons help break it down.

4... insert absurd^4 answer that brings in tunneling effects, or moving mirrors, or some other "impossible" reason that is only impossible because they didn't think of it for you. [End rant]

</comment></comment><comment><USER42> I'm not buying this. The Moon is only a reflector, not a producer of light. The temperature of the light is that of the reflected source. Moonlight is sunlight, more or less. The fact that the moon reflects poorly is compensated by the size of the gathering lens or mirror.

Suppose we build a 100 foot mirror which reflects 10% of sunlight, such that the spectrum remains the same. We could still make a fire with the reflected light, if we just gather 10 times more of it with a larger lens. We could do this in the Arctic, with the mirror's temperature at below zero; the mirror's temperature is irrelevant.

<comment><USER43> What matters is that we have intensity with quadratic falloff. Light from the sun has quadratic falloff based on the distance from the sun and hence can't be bent to become more intense than at the suns surface. Similarly light from the moon has quadratic falloff based on the distance from the moon and hence can't be bent to become more intense than at the moons surface. If you put a mirror on the moon, then the light from it will have quadratic falloff from the reflected sun and not the moon, which is why it can be used to heat to sun level temperatures.

<comment><USER42> Yes, since the moon is a scattering reflector, in fact the inverse square dropoff of moonlight is based on the distance from the moon.

Note that the article claims that no matter how much moonlight we are able to gather (i.e. we are allowed to overcome the inverse square law however much we want) we cannot create a temperature that will ignite paper.

</comment></comment></comment><comment><USER44> You can, if you remember "the scientific principles of the convergence and refraction of light."

"The scientific principles of the convergence and refraction of light are very confusing, and quite frankly I can't make head or tail of them, even when my friend Dr. Lorenz explains them to me. But they made perfect sense to Violet."

Violet Baudelaire goes on to use the scientific principles of the convergence and refraction of light to set fire to a piece of sail cloth using only moonlight and the lens from a spying glass in "The Wide Window" by Lemony Snicket.

It's possible that this book is a work of fiction.

</comment><comment><USER45> This explanation does not "click." I can't say whether it's technically wrong; I just note that it lacks some of the features of a successful explanation. First time I've seen a dud from Randall.

</comment><comment><USER46> Here's a thought experiment to consider. Imagine creating a lens to focus the blackbody radiation from a stack of bricks onto a single brick, and heating up the brick.

Now focus the light back onto just a small region of the brick pile, and heating up that region, which in turn heats the rest of the bricks by conduction.

This in turn increases the amount of heat collected from the pile, ad infinitum or until the brick pile melts.

Short of letting the brick pile melt itself, imagine tapping into the excess heat and using it to power an electric motor for a useful purpose.

<comment><USER47> The problem is that some of the radiation from the moon is not blackbody radiation.

</comment></comment><comment><USER48> And what of materials that burn at a temperature below the temperature of the Moon? If 100C is the limit, there are materials that burn at much lower temperatures such as Phosphorous (34C).

<comment><USER5> I mean you can just take white phosphorous to a warm place and let it spontaneously combust.

The use case would be materials that don't burn at the Earth's surface temperature, but do burn at the Moon's peak surface temperature. But you could probably get those hot enough just by rubbing them or something.

</comment></comment><comment><USER49> One has to coat the material to be set on fire with something that has low infrared absorbance (and thus low cooling through heat radiation) but high absorbance for low wavelength (blue/ultraviolet). This is called selective coating.

[https://en.m.wikipedia.org/wiki/Solar_thermal_collector](https://en.m.wikipedia.org/wiki/Solar_thermal_collector)

Combined with the lens, this might work.

</comment><comment><USER50> I don't really follow this argument, and I would like to.

I think one thing which would help me develop an intuition for it would be to see the calculation of the lens size for heating a one square-centimeter area on the earth to as high a temperature as possible by the light of the moon, and what that optimal temperature is.

Anyone reading for whom this is straightforward? Even a description of how to do the calculation would go a long way.

<comment><USER26> You get the highest temperature by reflecting moonlight from every angle.

For a single lens you just want something really big. Make it take up 90+% of the sky from your target.

For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get.

The calculation way goes as follows: Near earth you get 1400 watts per square meter of sunlight, so if that bounces perfectly off the center of a full moon and gets through the atmosphere with no losses, your target will get 1400 watts per square meter. That's equal to a black body at 122C. After taking into account the spherical shape and atmospheric losses you might get less than half of that, so ambient heat might drown out your results.

<comment><USER50> > You get the highest temperature by reflecting moonlight from every angle. I > see the thermodynamic principle at work, here, but I don't really understand > how it's operating at a mechanistic level. Is it possible to demonstrate > that assertion optically?

<comment><USER26> Take your target and trace a ray in every direction away from every point on its surface. The more of these that hit the energy source, the more energy you're getting. And 100% is obviously the best you can do.

<comment><USER50> From an optical perspective, how does the argument dismissed at the start of the OP break down? Is there some reason the moon's light, gathered from a lens covering hundreds of acres, carries insufficient energy to light a fire in concentrated form?

<comment><USER26> The optical argument goes like this: Measure the brightness (in energy per square meter) right at the moon's surface. No matter what you do with lenses, you can't concentrate moonlight beyond this level.

You can make an enormous lens where all the energy coming off a particular acre of moon goes through it. But you can't focus all of it onto a spot smaller than an acre, no matter what you do.

<comment><USER50> Thanks for the explanations.

The current top comment here seems to agree with my original intuition, though.

[https://news.ycombinator.com/item?id=18739120](https://news.ycombinator.com/item?id=18739120)

<comment><USER26> As a few people pointed out, the reflection off the moon ruins the étendue, so there's no concentrating it beyond what all the rocks on the surface already experience.

<comment><USER50> Thanks again for your help. I think that led me in the right direction.

This is the optical argument for maximum concentration given conservation of étendue (the same page has an optical argument for the conservation):

[https://en.wikipedia.org/wiki/Etendue#Maximum_concentration](https://en.wikipedia.org/wiki/Etendue#Maximum_concentration)

The angle subtended by the moon is approximately 0.54 degrees, or about 0.01 radians, so the maximum concentration factor is about 10,000. The moon provides about 0.1 lux of illumination, so the maximum illumination you can achieve by optics is about 1000 lux. The sun subtends about the same angle, and we receive 30,000-100,000 lux from it, and the maximum illumination you can achieve from concentrating it optically is about 1B lux.

I'm willing to believe that you can't light a fire from the moon, if the intensity's a million times lower.

</comment></comment></comment></comment></comment></comment></comment></comment></comment><comment><USER51> Reading the comments I found something I don't understand. What is the difference between 1.black body photons and 2.laser photons An object will heat only up to the original temperature of the black body source in the first case, but to an arbitrary high temperature in the laser case...

<comment><USER3> Black body photons are coming at random times. Laser photons are coherent, so are timed. Think of swing - if you try to push it randomly you'll get it swing as far as hardest push. But if you push periodically, you can swing it very far with small pushes.

<comment><USER7> That's not important at all; what's important is that black body radiation has a fixed maximum flux -- its spectrum or lack of coherence isn't why you can't heat another body to a greater temperature. It comes back to etendue, or if you prefer the 2nd law.

You could reproduce any fixed black body spectrum (to arbitrary accuracy) from a set of thermal sources and filters (or a set of lasers, LEDs, etc. with random phases) to arbitrary fluxes just like a laser has, and use this light to heat objects to arbitrary temperature. But if the original emission is of black-body type, you cannot -- the flux is given by the quantum mechanical process and a function of local temperature only. From then it follows from etendue conservation you cannot achieve higher temperatures.

<comment><USER51> Interesting, this makes more sense. Now I have to read more on the conservation of entendue.

</comment></comment></comment></comment><comment><USER52> I had the presumption that you could make a mega hot spot with a large enough lens back in the 90s. Fortunately usenet set me straight. It was a long and facinating journey to understand all the whys about it.

</comment><comment><USER53> It is a truism: no matter how many times this is explained, some engineer will come up with a complicated system that violates the laws of thermodynamics and refuse to admit that their idea is extremely unlikely. The laws of thermo are some of the best understood and most well-supported physical systems that humans have yet invented. Every time somebody comes up with a perpetual motion machine, it gets shot down because the person who invented it literally ignored all the really well-understood math and physical theory in thermo.

it's like there's a brain bug where engineers think they can outsmart 200+ years of scientific progress with a clever arrangement of mirrors.

<comment><USER20> _It is a truism: no matter how many times this is explained_

What exactly is the "this" that you refer to? I don't think the issue is that the "engineers" disagree with the physical principles, rather they tend to disagree that the physical principles apply in quite the way that the author claims. Many of the engineers probably believe is that Munroe is correct in claiming that on cannot start a fire with a low temperature blackbody radiation source regardless of the size of one's magnifying glass, but disagree that it is reasonable to consider sunlight reflected by the moon as fitting this model.

Presumably, you agree that it's possible to start a fire using a magnifying glass using sunlight on earth. I'd guess you also believe that it's possible reflect the light from small handheld mirror into the magnifying glass, and still start a match, even though the mirror is much lower than the temperature of the sun? While the specular reflection from the mirror is different than the diffuse reflection from the moon, one might note that the words "specular" and "diffuse" don't appear in Munroe's exposition. Would you agree that Munroe's argument would appear to prohibit this behavior?

Now assume that the moon was replaced by an equally sized parabolic mirror aimed to be focused on the earth. Would it be possible to light a match using this light if one's magnifying glass was large enough? Which thermodynamical principle am I violating in thinking that it might be possible? And which part of Munroe's argument do I invalidate by making these modifications? Again, my point isn't that Munroe's conclusion is wrong, just that there might be something flawed about the argument he uses to reach that conclusion. This might be a "brain bug", but from the inside it just feels like an attempt to understand truth.

<comment><USER53> The this I refer to is "the principle of etendue": [https://en.wikipedia.org/wiki/Etendue](https://en.wikipedia.org/wiki/Etendue)

and I agree it's counterintuitive and it's OK for people to come up with ideas, but when you hit the point of "hey, the thermo people have a nice collection of proofs demonstrating this, and it fits very well with the underlying theory, oh, and if you do manage to violate etendue, you could probably build a perpetual motion machine", if you willingly continue to argue and get shot down, it's time to go back and re-read the books.

BTW, what's your obsession with Monroe? What he's referring to is a scientific phenomenon, Monroe is just a science popularizer, and if he got the details wrong- well, the point of xkcds like that is more to inspire people with ideas, than get the exact details right.

<comment><USER20> _The this I refer to is "the principle of etendue"_

Great principle. I agree with it, and think that most of the people offering objections here do as well. Our question is whether it's it's being applied correctly in this case. For me, the sticking point is whether surface temperature can be used as a proxy for the brightness we care about.

 _BTW, what 's your obsession with Munroe?_

I have none. He's a great explainer, probably occasionally gets details wrong, and (so far as I can tell) is a positive force for spreading scientific understanding. My "obsession" seems to be that I have poor tolerance for overly broad smackdowns of critics. If you are going to tell someone else that they are wrong (as opposed to Munroe who is trying to explain what he believes is true) I think you have a higher obligation to get all the details right. I presume I'm sensitive to it because I'm often on the receiving end.

How about you? Why does it bother you so much that some people say that Munroe's argument is logically flawed?

<comment><USER53> I don't have any problem with people complaing about Monroe's argument. I am fairly certain he wrote what he did in consultation with optics engineers, and then modified it so it was intellectually comprehensible by a nerd-but-not- optical engineer audience.

<comment><USER20> Thanks for the response. If you are still interested in this, there's an useful link on the second page of this thread to a discussion of the same question: [https://physics.stackexchange.com/questions/370446/is- randal...](https://physics.stackexchange.com/questions/370446/is-randall- munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-th/370600). If you expand the comments on the answer, the back and forth presents most of the argument I would make, in a clearer form than I could muster.

</comment></comment></comment></comment></comment></comment><comment><USER54> I've often wondered how big of a lens you would need to grow a plant using a light source from outside of the solar system, if it's even possible.

</comment><comment><USER55> Is it possible to light a fire using sunlight during an eclipse?

If not, at what percent totality does it become impossible?

<comment><USER56> Should be possible. Your hot spot under the lens is just as bright as always, it's just smaller, and crescent shaped. As to percent totality, that would depend on how fast your wood/fuel was dissipating heat.

</comment></comment><comment><USER57> One thing that makes me uncertain about this is the fact that the sun is generating the light from a fusion reaction, thus expending energy. In other words, the system entropy does not decrease because the fusion reaction makes up for any lost entropy by the cold-to-hot temperature flow. This is the same reason why a system consisting of a battery and a fridge would work.

</comment><comment><USER58> I'm pretty sure if I were to go pick up a number of magnifying glasses and focus them on the same point using moonlight, the temperature at that point would increase with every additional magnifying glass.

Am I to accept that the additional magnifying glasses would cease increasing the temperature once the temperature matched that of the moon's surface?

<comment><USER59> I don't know enough about physics to judge whether Randall's argument is correct, but in general it is possible for a sequence to be strictly increasing yet bounded.

For example, 1/2, 3/4, 7/8, 15/16, ...

</comment></comment><comment><USER60> this article seems to refute some aspects of xkcd's answer:

[https://physics.stackexchange.com/questions/370446/is- randal...](https://physics.stackexchange.com/questions/370446/is-randall- munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get- th/370600#370600)

<comment><USER20> This is indeed a great link (expand the comments on the answer), and shouldn't be languishing at the bottom. Thanks for digging it up!

</comment></comment><comment><USER61> I love xkcd, but this is completely wrong.

It is well known that the spectral temperature of the moon is about 4000K. See for example : [http://www.lumec.com/newsletter/architect_06-10/the_sun_the_...](http://www.lumec.com/newsletter/architect_06-10/the_sun_the_moon.html) or [https://physics.stackexchange.com/questions/244922/why- does-...](https://physics.stackexchange.com/questions/244922/why-does- moonlight-have-a-lower-color-temperature) . That is the maximum temperature that you can achieve with light from the moon, no matter how concentrated. 4000 K is plenty hot enough to start a fire.

<comment><USER62> Randall has a response to this argument in the article.

<comment><USER63> Does he though? Here's all I see -

> "But wait," you might say. "The Moon's light isn't like the Sun's! The Sun > is a blackbody—its light output is related to its high temperature. The Moon > shines with reflected sunlight, which has a "temperature" of thousands of > degrees—that argument doesn't work!"

> It turns out it does work, for reasons we'll get to later.

But the rest of the article is about etendue, and I don't see how the issue of reflected light is addressed (though possibly the answer is implied with an etendue argument I missed).

I'm very curious now - it seems to me that if the Moon was a perfect mirror, you should be able to start a fire with it. Maybe the Moon's low albedo is the reason? Fun fact about the Moon: its albedo (roughly the fraction of incident light it scatters) is about the same as asphalt - not very reflective at all! [0] It just looks bright to us because the Sun is so tremendously bright.

[0] - [https://www.lcas- astronomy.org/articles/display.php?filename...](https://www.lcas- astronomy.org/articles/display.php?filename=albedo_effects&category=observing) , also learned this in an astronomy class

</comment></comment></comment><comment><USER36> EDIT: I'm not sure about this anymore.

<comment><USER64> I think TFA refutes your argument, but I'm not smart enough to know for sure. Can you please expand on this?

<comment><USER65> What if the sun was a bunch of parallel laser beams? Then you could focus them all into a single point.

<comment><USER64> Right, but that point would be exactly as hot as the surface that the light was coming from, and no hotter.

(Is the argument as I understand it).

<comment><USER66> If this is true, how do laser cutters work? The focused beam can melt through steel, but surely the temperature of the gain medium is much lower than the melting point of steel?

But the thermodynamics argument seems like it makes sense too...

<comment><USER29> Lasers have negative temperature. The light emitted from the sun is nothing like laser light.

[https://en.wikipedia.org/wiki/Negative_temperature#Lasers](https://en.wikipedia.org/wiki/Negative_temperature#Lasers)

</comment></comment></comment></comment></comment></comment><comment><USER67> This is great, I always love the xkcd "What if" explanations. I even have the book sitting on my desk next to me.

</comment><comment><USER68> You can with magnifying glass bigger 2.3 million times</comment>


Nvidia Unveils First Mobile Supercomputer for Embedded Systems - <AUTHOR> http://nvidianews.nvidia.com/Releases/NVIDIA-Unveils-First-Mobile-Supercomputer-for-Embedded-Systems-ad8.aspx <comment><USER2> If they would only stop hating OpenCL (read: Actually support it) I'd purchase one in a heartbeat as a testing rig and experimentation rig.

Having a board which works only with a certain vendor specific API is not really that useful unless I'd specifically want to develop a full fledged product using exactly that particular board.

<comment><USER3> This SDK will have fully support of OpenCL 2.0

<comment><USER4> Do you have a link for that? I would be super excited to see them get back on top of their OpenCL support. They have been a 1.1 for a long while now.

</comment></comment></comment><comment><USER5> So many posters here do not have a clue. Can I use it as a media server? So its similar to parellela! Xyz is better, cheaper, faster... People - This is the beginning of an epic shift in focus. We are at beginning of an exciting new period for computing. The focus is shifting to function - out there in the real world. The goal behind this nvidia drive is no less than the creation of useful real world robotics. This marks the very beginning of a curve that will take us all the way to west world. As long as the semiconductor industry keeps cranking out the production nodes, this story will remove humanity from the process of production -forever. You are here at the beginning, you are here to bear witness, don't let the gravity of this moment escape you.

<comment><USER6> I totally agree. This is so long overdue!

My only gripe is that I decided to look at OpenCL and CUDA today and felt kind of.. underwhelmed. I have a lot of OpenGL experience and I guess it's not surprising that they are so similar to framebuffers and shaders. But I was really hoping for true general purpose computing, more like a hybrid between VHDL and say MATLAB. Ideally it would work like Go, where you would send a C-style function off to an execution unit or refer to other units by id, and use something like fork/join to tabulate the results, and maybe just give up on the notion of a global memory space. Instead it looked more like image processing kernels for doing convolutions and stuff, and I mean that's great, but didn't exactly wow me.

Maybe I'm missing something fundamental? Does anyone know of a site with more of a computer science approach, say like [http://golang.org](http://golang.org), instead of so much emphasis on graphics? Or possibly a compiler that would convert Go/MATLAB/Python to OpenCL/CUDA? Am I alone in feeling a little mystified here?

</comment></comment><comment><USER7> I hope Nvidia throws more support behind this than they did with past Tegra developer kits. I bought a Tegra devkit and perpetually got the feeling that the driver support / dev relations was being done by one person in their spare time.

</comment><comment><USER8> This looks pretty similar to the parallela, with better price, nicer software support and better manufacturing horizon.

Oh well, that was somewhat predictable: parallela competed with nvidia in something that nvidia cared about and could execute.

<comment><USER9> no, they are very different.

NVIDIA is SIMD, a 32 thread warp doing the same instruction in lockstep (in 4 cycles usually on 8 SPs, the 192 SP chip can run 24 warps simultaneously, though specific numbers may change a bit from version to version) and heavily penalized for in-warp divergence.

Parallela - i.e. Adapteva ([http://www.adapteva.com/epiphanyiv/](http://www.adapteva.com/epiphanyiv/)) is 64 independent (execution-wise) RISC cores.

</comment></comment><comment><USER10> this is very cool. It operates in the "sub-10 watt space", had pretty decent io options for a device of that size, no apparent moving parts, decent graphics capabilities, and a reasonable price point. Seems like a great little device for a media machine or a graphics installation; somewhere between a raspberry pi and a mac mini.

</comment><comment><USER11> 192 kepler cores: don't bother. this is just a desperate attempt to get someone to take TK1 chips off their hands. coincidentally, I'm sure, but AMD is shipping their AM1 platform.

</comment><comment><USER12> I wonder if something like this could be the start of a decent key-value appliance. Hardware designed to perform the role of memcached on the network or similar.

<comment><USER13> Hmm. At first I thought this would be a waste, because you couldn't use all of its fast compute capabilities on doing key-value lookups. But these things also have much, much more memory bandwidth than regular CPUs (I can't remember if it's 10x or 100x, but that ballpark), and that might work well for these applications.

But I don't know how the GPU's memory caching infrastructure works. If the bandwidth is only for serial reads, that could be a problem.

<comment><USER14> K1 has about the same memory bandwidth as every other mobile SoC, which is to say not all that much. Don't compare it to discrete GPUs.

<comment><USER12> True, but in my experience if you can avoid hitting the memory the amount of computation you can do per thing is absolutely terrifying.

The main thing I had in mind here was hashing for bloom filters. i.e. do the different algos in parallel on a GPU, then pass those values for the main lookup to be done by the CPU.

</comment></comment></comment></comment><comment><USER15> I'm impressed by the price point ($192) and pretty curious to see what I can do with it besides computer vision. Bioinformatics co-processor that works with my laptop? I've wanted that for a long while now.

</comment><comment><USER16> Is this for processing USB webcam data? I'd like to use industrial or broadcast-level cameras but the board seems to be lacking support for these kinds of inputs. Also, in both of those domains cameras can deliver true 10 to 12 bits per color channel uncompressed performance. Webcams don't come close, as far as I know.

Am I wrong?

<comment><USER14> Tegra 4 has a MIPI-CSI interface for image sensors and I would assume K1 is similar. Supposedly you can combine the ISP and GPU for image/video processing.

</comment></comment><comment><AUTHOR> I am surprised that no hackernews users has responded to this news yet. It sounds an exciting platform for all kinds of innovations.

<comment><USER17> The people who were interested probably bought the AMD-powered Gizmo Sphere already? (www.gizmosphere.org)

<comment><USER18> Well, this one has about ~6x the computing power (if their numbers are to be believed), and it has an ARM core rather than x86. Some differences at the least.

<comment><USER19> Have we got non-NVidia benchmarks yet?

</comment></comment></comment></comment><comment><USER20> What about using this for bit/alt coin mining? Do you guys think it is a good ratio of performance / power consumption?

<comment><USER21> Notice they use the GigaFLOPs metric, which is a (maximum) floating point measure. Bitcoin hashing (it's been a while for me) is compute intensive on integer math. For a good ratio of performance/power you'd want to look at architectures like the GreenArray devices. But the world of bitcoin hashing is now firmly positioned in ASICs. You can't cut architecture fat beyond that so you'd have to look at (general purpose) architectures built on a faster silicon technology. Wouldn't count on it.

[http://www.greenarraychips.com](http://www.greenarraychips.com)

</comment></comment><comment><USER22> Could be a great platform for WPA2 key cracking and encryption bruteforcing when paired with tools like Pyrit or HashCat

<comment><USER23> I wonder how many pmks it could get through per second in Pyrit. I imagine it's a bit more power efficient than a bunch of GPUs going through pmks.

</comment></comment><comment><USER24> Anybody know if the FPGA's branching model has been improved?

<comment><USER25> What FPGA? Do you mean GPU? If so, then - it's improved a bit over the years, but it's still much different than CPU programming, and branching remains relatively expensive.

<comment><USER24> I have my terms mixed up. I thought GPUs used FPGAs, but I was wrong. Thanks for clearing that up.

</comment></comment></comment><comment><USER26> Could something like this work for a home server?

<comment><USER27> Why would you need a powerful GPU in a home server?

<comment><USER28> I've been looking for something to replace my Ouya as an XBMC box, but with enough power to support the incoming 4K content. I'm not sure this does it.

<comment><USER29> You'd probably be better off waiting for a 4K codec to settle out, and for hardware decoding support in new chipsets. As it stands, HDMI is barely capable of pushing 4K to the handful of displays that will accept it.</comment></comment></comment></comment>


A Brief History of Welfare for Middle-Class Americans - <AUTHOR> http://www.theatlantic.com/national/archive/2011/01/a-brief-history-of-welfare-for-middle-class-americans/68991/

<comment><USER2> Wait; let me get this straight... The government not taking as much of my own money away from me is welfare?!

<comment><AUTHOR> The point is broader than that. For instance, many people (excepting children) who go on food stamps, disability, or welfare are often people who have paid into the system over the course of their life. In that sense, the social safety net is an example of the government simply giving back their tax money.

So in that same sense, there exist massive programs of subsidization of the middle class and even the wealthy, just as there is of those who take in welfare or unemployment benefits.

Ultimately, the point the article is making is that society is an interlocking set of dependencies, and that demonizing one group for receiving government payouts ignores the complex social dependencies that all economic classes have in some form or another.

<comment><USER2> Are there people that are on food stamps, disability, or welfare that haven't paid sufficiently into 'the system' to off set the cost of them consuming said services?

<comment><AUTHOR> Of course there are, but determining someone's access to basic survival based on how much they had paid into the system would be problematic and Dickensian, to say the least.

And most importantly, societies don't benefit from denying basic survival and dignity to people. Crime rates rise, social stratification increases, the net effect of social problems due to economic entropy of a class of people can be detrimental to the health and well-being of a society as a whole. Especially for the middle class, which lacks the wealth to isolate itself from said social problems the way the upper classes can.

<comment><USER2> That's not what I was getting at.

You said '[people on] welfare are _often_ people who have paid into the system' (emphasis mine). Right? What would you guess the ratio was of what the _majority_ of people on welfare paid in as opposed to what they get in pay outs? (I am getting to an important point; stick with me; please and thanks.)

<comment><AUTHOR> I don't have access to that information, and haven't known any studies to have been done.

Assuming the statistics would fit your expectations, what would your point be?

<comment><USER2> I want to see _your_ expectations.

I started by saying that not taking more of _my_ money does _not_ constitute welfare. You countered my point by saying that _many_ people that later get paid out first paid in. So I am asking -- how many people? What ratio for paying in and taking out?

You're saying that me keeping more of my own money (you know, the money I worked for) should be called welfare because I _might_ take some of the money -- taken from me in the first place -- back.

In effect: we all put in to a kitty. One day Cardboard holding, Camel smoking, Corner lollygagging, 'Carl' takes some out -- next day I take some out. Right?

Well I am asking you; how much does 'Carl' put in and how much does he take out? It's easy for me to say I have tax withholdings and I know how much welfare I've used.

What's _your_ guess for the average 'Carl'?

<comment><AUTHOR> My guess is that a specific number for who "deserves" what social investment is a chaotic system which cannot be quantified properly. And I would consider it an exercise in futility to determine a social theory based on something so chaotic.

Conversely, I would suggest that my payment into a welfare system benefits me, since I am merely middle class, and cannot escape the detrimental social effects of abject poverty nearly as easily as the wealthy.

<comment><USER2> for the sake of argument let's say I agree with everything you said. Further, let's say I am so moved by your argument that, even though it doesn't seem like it, I now see that it is in fact better for all men for me to have even more of my money pilfered.

what I still don't understand is why we should call pilfering my money from my hands should be called me receiving welfare?

<comment><AUTHOR> Is it possible that you find the word "welfare" to be an emotionally and politically loaded term?

For instance, libraries exist as a form of social welfare.

<comment><USER2> so all this time you where primarally talking about roads and libraries?

<comment><AUTHOR> Nowhere near that narrow. I'm talking about the totality of the social contract.

<comment><USER2> for the sake of clarity: could you tell me what you think should be the upper bound as a percentage of my income in this social contract should entail?

<comment><AUTHOR> Since I agree with Thomas Paine's view on the necessity of progressive taxation, what you specifically are taxed would be dependent on your income, but given the extraordinary circumstances (two major wars, numerous conflicts, near economic collapse, etc), I would support a temporary 90% tax rate on the upper income brackets similar to what Eisenhower enacted in order to pay down the debts and make major necessary social investments (education, infrastructure, etc).

That said, I favor the Nordic social democracies to be good economic models, as they have been proven to support greater class mobility than the current U.S. system. Norway, for instance, ranges from 0-49% tax rates.

However, as this may be relevant to the important point that you would like to make, I don't think my personal views on the proper rate of taxation take away from the need to recognize the importance of the existence of economic social relationships.

<comment><USER2> wait; let me guess... and I would pay 49% and Carl would pay 0%, right? So his ratio would be 0:1 and mine would be 1:0.09 (for roads and libraries, of course. Can't forget my welfare benefits now can I?). Thank you for finally answering that question.

And when I forked it over I should say, 'thank you for my welfare' and Carl will say, 'thank you for my wel-fairer'!

Got it!

.

Well, either way 90%!? My friend I think we'll just have to agree to disagree.

<comment><AUTHOR> 49% assumes you are in a very, very high income bracket. And 0% assumes that your mythical Carl would not be able to survive if he were taxes on what little income he had. You would benefit greatly from a society that does not allow Carl to fall through the cracks, this has been proven. Carl may use public infrastructure and a social safety net to pull himself out of poverty and set himself on a path of entrepreneurship. Conversely, you may find yourself without your high income, your savings eliminated, and in need of the social welfare that was used to help others out of a cycle of poverty.

I'm not clear on what your important point was, but I think the point you try and make here is relatively short-sighted.

<comment><USER2> Well the point was that I started by saying that you should not be calling the government _not taking as much_ of my own money welfare.

You turned it into the government _taking more_ of my own money as being welfare.

So in the end my initial statement still stands. You cannot call the government letting me keep my own money welfare 'cause no body's building libraries for me with that money, are they?!

<comment><AUTHOR> I think the term "welfare" fully applies, because it pertains to the welfare of the society you live in. Again, I believe you're reacting based on an emotional response to what is often termed "welfare" by detractors of social programs

All classes benefit from social welfare programs, albeit in different ways. If your tax dollars are not paid towards society you benefit from, but instead returned as a tax credit, that is as much a form of welfare as the person who pays into the system for 30 years and then receives disability, TANF, or unemployment benefits.

<comment><USER2> Maybe this will help; are the following two statements contradictory in your mind?

      1.) The government taking *more* money from me should be called welfare.

      2.) The government taking *less* money from me should be called welfare.

You see, I was trying to get you to argue both sides and lo-and-behold you did it. The next step was to have you either explain how both of those (to me at least) mutually exclusive statements can be true or you have to accept one or the other as true and reject the other. (Or you can also say that the Aristotelian law of non-contradiction is phooey...)

You might be tempted to change the sense in which you are using the term welfare (I have noticed sacre quotes showing up in your last posts) but that's what i would called a fallacy of the fourth term. (And since you keep qualifying the term with 'social programs' I think it would be disingenuous to switch to the general second term and then back to the specific first without making it abundantly clear that you are doing so.)

In my mind (and in most common dictionaries) welfare means a. _governmental_ provision of economic assistance to _persons in need_ ; _or_ b. absolutely anything that aids or promotes anyone's well-being regardless of their prior need.

(So at best you could say all of the first sense are of the second but you could not then say that all of the second sense are of the first. i.e. it would be a a concentric venn.)

I thought that the article, you, and I where all using the term in the first sense but, once it became expedient to you, you seemed to switch to the second.

The only other option as I see it is that you think it all belongs to the government and I should just be happy that they let me keep any of their money at all. And surely if I had it I wouldn't be as enlightened in my use of it anyway so it's really best for me if I just hand it over.

But, and this is a very important 'but', the fact that you have declared that a 90% income tax rate is fine in your mind I now see that all hope of us becoming of one mind is utterly hopeless. I have no category that allows for that detestable state of affairs regardless of any circumstances. So, really, I think the best we can do is agree to disagree...

<comment><AUTHOR> _I was trying to get you to argue both sides and lo-and-behold you did it._

Simply because, and I don't this this is difficult, but social welfare is not what is taken, but the result of what is done.

I would hope at least you've gotten some sense from this the idea of considering your role as a small part in a greater social tapestry, than simply an argument about what percentage of your income should be taxed.

<comment><USER2> So, as you see it, my roll in society is to work hard so that I will have money that the government can then take by force?

How 'bout the charity work I do, or the two homeless guys I got off the street and got jobs, or the orphanage on Kat Island I supported with groceries, or the goat and 2 chickens I sent this Christmas, or the tithes I donate that go to benevolence, or the family that lived with me for six months (a mother and her _four_ daughters in our three bedroom house)? (And I mean it all and could share more!)

Do you have any idea how much more I could help if the government would stop giving me 'welfare'?

And since no matter whether they take my money or let me keep my money you'll still call it 'welfare' I guess everyone is on 'welfare' too, huh?

But the thing that gets me is that I am supposed to _thank_ my government for letting me keep what is _my personal property_ to begin with. Or do you not believe in personal property either? (Or are we only supposed to call 10% of our money personal and the rest is the government's and that's why I should call them letting me keep it 'welfare'?)

<comment><AUTHOR> Who you decide to thank is immaterial. Public investment in infrastructure, education, etc., all public goods, is how you built your wealth. Your wealth is as dependent on society as anyone else. And while your charity is commendable, it is utterly undemocratic.

<comment><USER2> Right; and so that's why all my wealth belongs to the public. I got it. Thanks for clearing that up.

</comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment></comment><comment><USER3> Just because gov't programs for the middle class cost a lot of money doesn't mean they are delivering actual benefit.

Given the example of medicare vs. insurance tax deductions for employers, I think the economy would be much better off with insurance tax deductions for everyone and no medicare. Or no insurance tax deductions and no medicare. There are plenty of ways to make healthcare cheaper and less bureaucratic.

If the middle class put the squeeze on gov't expenditures that won't mean that middle class benefits will be the first to be cut, they'll likely be last.</comment>