| ▲ | lopsotronic 10 hours ago |
| For real. It's an enormous problem solved only with 1) sheer scale, and 2) Science Fiction. Both of those are expensive as hell, by the way. Cooling via radiation follows Stefan–Boltzmann: P = εσAT⁴. Let's assume a good surface (emissivity ~0.9) at 300 K (27 °C) at 400 W per square meter per side. A flat panel radiating from both faces into deep space gets 800 W/m, not including the losses from, say, the Sun, or from IR coming off the Earth. Now, input power. Sunlight in orbit ~1,360 W/m², assume ~22% cell efficiency, we got 300 W/m². So each 1 MW compute, 3,300 m² of solar panel and minimum 1,200–1,500 m² of radiator. In case ya didn't know - 1 MW is tiny from a present-day-datacenter perspective. It's like 8 racks. So we're talking orbital megastructures here, many many many square kilometers, and this is with all the best case assumptions, and magic radiator panels that never see the sun, or the earth, or the moon. This is just the basic numbers here, by the way. There's a garbage truck full of other unsolvable problems if you poke your head in there. Aside from the "Avoid Regulations" aspect, and the "Everything That Burns Deorbiting is Depreciation" aka "The Starlink Trick", I'm not sure what the hell the draw is. |
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| ▲ | m4rtink 9 hours ago | parent | next [-] |
| Cooling in space is hard but not impossible - while current (IMHO stupid without advanced in space infra) space data center projects work with a couple MW, many advanced space propulsion concepts might have to reject hundreds of MW if not a couple GW. For that you might need more advanced stuff like liquid droplet radiators (https://en.wikipedia.org/wiki/Liquid_droplet_radiator), heat sinks & pulsed operation. Still, it should be eventually doable. As for space data centers - I think the main issue is the complete lack of in space infrastructure for resource mining, processing and manufacturing & maintenance. It is kinda like building your first practical steam locomotive & the deciding to build directly an airliner. No suitable materils, experience, work force, material sources, etc. We eventually went from locomotives to airliner, in an incremental manner & expanding the supporting infrastructure to support the ever more ambitious projects. |
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| ▲ | DanHulton 6 hours ago | parent | next [-] | | It might be eventually doable, as an experiment or as a flex, sure. But it's never going to come close to being cost-reasonable versus the equivalent infrastructure here on earth. | | |
| ▲ | toephu2 4 hours ago | parent | next [-] | | > But it's never going to come close never? I doubt that. Technology will improve over time. Eventually I bet it will become cheaper. Have you tried building in the U.S.? Why do you think it's so expensive to build in the U.S.? It's due to regulation and red tape. | | |
| ▲ | unrented7977 3 hours ago | parent [-] | | Technology improves, but the laws of physics are constant and absolute (on human timescales). Thermodynamics says no today, no tomorrow, and no 100 years from now. That's not ever going to change. | | |
| ▲ | bagels 3 hours ago | parent [-] | | What was it the 27th law that says radiating heat in to space does not work? I'm skeptical of the whole thing too, but it's not an impossible engineering challenge, just an expensive one. |
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| ▲ | asdff 6 hours ago | parent | prev [-] | | The security advantages are enormous since access to space is so tightly restricted and controlled, compared to the access potential of a land based data center. Only risk in space is maybe we start WWIII with china and the US directly trading blows. On the ground, any insurgent group can disable your infrastructure. Nothing is truly safe on the surface of the earth. Anyone can strap a bomb on a drone now. See examples from the currently active wars. | | |
| ▲ | 7e 5 hours ago | parent | next [-] | | It costs at least 50x more to put a GPU in space than it does on Earth. For that price you can have dozens more capacity in bunkers, under the sea, or on remote islands. Do you think your insurgents are going to get all two dozen? They could travel to the far corners of the earth, destroying 22 of them, and you'd still be ahead. Further, I wouldn't be surprised if a satellite with such a monstrous solar and radiator footprint wouldn't be susceptible to a laser based attack from the ground; either frying it or pushing it into an unstable orbit by vaporizing a few bits. | | |
| ▲ | jryle70 2 hours ago | parent [-] | | It costs much more than 50 times because there are no GPUs in space yet. Google is only planning to have some sort of space data centers mid 2030s, if everything works out. A big if, but if they don't start now then we'll never know. |
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| ▲ | ForHackernews 6 hours ago | parent | prev [-] | | What? No, exactly the opposite. It's very easy to jam radio signals and much harder to cut wires. There's a reason the drones on the front lines in Ukraine are dragging fibre optic lines these days. | | |
| ▲ | asdff 5 hours ago | parent | next [-] | | There's probably so many ways to get around that with space based technology. I can quickly imagine several methods. It depends on what the system is for which might be a good method to use. Method 1: same as how u2 planes dumped their data: air drop physical media containing data and catch it in the air. Method 2: laser based emission to specific detectors. Method 3: baseball style communication: station is under observation and manipulates in some way to serve as a signalling language. Method 4: numbers station Method 5: bill yourself as an isp and have some coded syntax that can be supplied in plain sight with the rest of isp traffic. | | |
| ▲ | scheme271 2 hours ago | parent | next [-] | | Method 1 runs into the problem of how to replace that media. The DC is in space so it's not like with U2 planes where they landed. Also, I think you mean the keyhole satellites and not U2 planes, since a plane lands at a secure site and can offload media then. | |
| ▲ | sunbum 4 hours ago | parent | prev [-] | | Method 1: Sure datacenters with latency measured in several hours sure are useful, and can also be intercepted
Method 2: Can be jammed by drone with laserpointer.
Method 3: Let me just transfer gigabytes of data via physical signalling
Method 4: Can still be jammed
Method 5: Can in fact still be jammed? | | |
| ▲ | scottyah 3 hours ago | parent [-] | | > Sure datacenters with latency measured in several hours sure are useful Training models takes weeks, are you really worried about a couple hours? > Can be jammed by drone with laserpointer Ok, you're just joking |
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| ▲ | alexnewman 6 hours ago | parent | prev [-] | | But more and more Ukraine is using long range drones guided by SpaceX |
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| ▲ | ruszki 7 hours ago | parent | prev [-] | | According to Wikipedia this reduces weight and not the required area. Also AI said the same thing, but I can't trust in it this blindly. So, how smaller would be the required surface area? |
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| ▲ | SyzygyRhythm 2 hours ago | parent | prev | next [-] |
| No point in running them at room temperature. GPUs, etc. run fine at 95 C. If you run your cooling loop at 70 C instead, you get 70% more cooling compared to 27 C. At any rate, 1 MW for a single satellite is fine. Just launch several thousand of those and you get to real numbers. Also, there's no need to talk about "magic" radiators. You orient them so they're at a knife edge to both the Sun and Earth. This is not difficult (the Moon is irrelevant). |
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| ▲ | octoberfranklin an hour ago | parent [-] | | No point in running them at room temperature. GPUs, etc. run fine at 95 C. They run fine for a short while, but not nearly as long. Heat accelerates all aging processes. It's how they artificially age chips in order to calculate MTBF. |
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| ▲ | xur17 6 hours ago | parent | prev | next [-] |
| > Now, input power. Sunlight in orbit ~1,360 W/m², assume ~22% cell efficiency, we got 300 W/m². So each 1 MW compute, 3,300 m² of solar panel and minimum 1,200–1,500 m² of radiator. We need > 2x more solar panels than we need radiators. Doesn't this imply radiation isn't really the limiting factor here? |
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| ▲ | snovv_crash 6 hours ago | parent | next [-] | | Getting the energy back from the solar panel is easy via copper cables. Getting the heat back out there to the radiators is a bit harder, you needed fluids and pumps and heat exchangers which have lots of moving parts and need maintenance. | | |
| ▲ | alexnewman 6 hours ago | parent [-] | | This has all been covered over and over again. It’s actually not that big of a deal | | |
| ▲ | datadrivenangel 5 hours ago | parent [-] | | Doing it in a cost and weight effective way is still a big deal, because if it's not within ~10x the cost of ground based data centers, not enough people will use it to justify building it. | | |
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| ▲ | rayiner 6 hours ago | parent | prev [-] | | [flagged] |
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| ▲ | mike_ivanov 9 hours ago | parent | prev | next [-] |
| Radiating 1MW at 500K (227C) with a 0.4MW heat pump takes about 200 m^2 flat sheet surface. Inputs - solar+nuclear for double fun. So - quite feasible. |
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| ▲ | lopsotronic 6 hours ago | parent | next [-] | | Moves 1 MW of heat with 0.4 MW of work? I.e. 2.5 COP {coefficient of performance). That's insane, and I mean that in a good way. Could you dig me up a cite for that? That's thumping the Carnot limit: [[T_cold / (T_hot − T_cold)]]. 2.5, while rejecting at 500 K, cold side's at least 357 K (eeehhhhhhh 84 °C) . . . and that's an absolutely perfect Carnot machine. At 50% Carnot -- a pretty good heat pump, real world performance is 40-60 -- cold side's at 417 K (144 °C). 417k, feeding your GPU coolant loops. | | |
| ▲ | hex4def6 2 hours ago | parent [-] | | Think you have an error -- it's t_hot / (t_hot - t_cold) With those numbers, ideal carnot would be 500/(500-357) = 3.5. Multi-stage could potentially get you to a COP of 2 or so. So 0.5MW. | | |
| ▲ | lopsotronic 19 minutes ago | parent [-] | | I believe that's Carnot COP for a heat pump used for heat+. I used the refrigeration version, T_cold / (T_hot - T_cold), which I'm 80 percent sure is the right one here. Depends on which heat you want Heat adding to hot side: COP_heat = Q_hot / W = T_hot / (T_hot − T_cold). Heat leaving the cold side: COP_cool = Q_cold / W = T_cold / (T_hot − T_cold). Another one (more common in the day to day, for me at least): heat-engine efficiency, η = 1 - T_cold / T_hot. Cycle forward to make work from heat. |
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| ▲ | ChickeNES 8 hours ago | parent | prev | next [-] | | And if you look at SpaceX's Starmind sats, they will have a 160 m^2 liquid radiator for 175kw/250kw peak compute. | |
| ▲ | cyberax 6 hours ago | parent | prev [-] | | Do we even _have_ semiconductors that can work at 220C? And if you're thinking about using some kind of refrigeration cycle, its efficiency is going to be bad. | | |
| ▲ | tristanj 6 hours ago | parent [-] | | 1) The chips don't reach 220C. The 220C is the temperature at the hot end of the heat pump. The chips are on the cold end of the heat pump. 2) The International Space Station has used a dual-loop ammonia/water-based heat pump to cool the station temperatures. It's been in place for several decades. Heat pumps are a proven technology. Other satellites have also used heat pumps, such as SES-17 in geostationary orbit https://www.esa.int/Applications/Connectivity_and_Secure_Com... | | |
| ▲ | andruby 5 hours ago | parent | next [-] | | > The 220C is the temperature at the hot end of the heat pump. The chips are on the cold end of the heat pump. If we want the heat pump's cold end at about 40–65°C, then for each 1MW of GPU heat, we need another 1MW of heat pump power. Now you need 2MW of solar power. Good news is that the radiator at 227C (500K) can emit about 5× more heat per square meter than at 57C (330K) | |
| ▲ | lopsotronic 6 hours ago | parent | prev | next [-] | | ISS's two external cooling loops hold about 540 kg of ammonia combined, together they dump 70 kW. | |
| ▲ | cyberax 4 hours ago | parent | prev [-] | | Of course. But their hot ends are nowhere near 220C. I don't think such heat pumps even exist right now except in labs. Looks like some experimental pumps within this region have CoP around 30%: https://www.sciencedirect.com/science/article/abs/pii/S03605... So you'll need a lot of additional energy to run the pumps. Which will require additional radiator area. |
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| ▲ | spullara 6 hours ago | parent | prev | next [-] |
| Glad you are on the case before these companies foolishly waste their money sending datacenters to space. |
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| ▲ | foota 6 hours ago | parent | prev | next [-] |
| Getting this all up into orbit it obviously the hard part, but if you're already building so much solar capacity the cooling actually doesn't seem unreasonable? |
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| ▲ | JoeAltmaier 5 hours ago | parent | prev | next [-] |
| Surface area is a materials problem? Folded microstructure, atomic-scale textured surface or some other science-fiction solution could have square kilometers of surface area in a shoebox. |
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| ▲ | AlotOfReading 4 hours ago | parent | next [-] | | Imagine you have two blackbody radiators with the same bulk properties, except one has surface area shenanigans like aerogels. In the far field as a whole, it seems like both should radiate essentially the same regardless of the internal details. You can shape emissive direction, or improve efficiency of non-ideal materials, but even ideal materials don't fix the issues pointed out by the parent. | | |
| ▲ | peri-cl an hour ago | parent [-] | | Right; it's only area exposed to the exterior that counts. A physical object can't thermally radiate more power than a perfect blackbody spanning its convex hull. (This follows because a physical object can't absorb more light than a perfect blackbody spanning its convex hull. A perfect blackbody by definition absorbs 100% of incident light, which is a hard upper bound. Any line incident on an object is also incident on its convex hull). (Consider an isothermal object that emits more power than a blackbody in the shape of its hull at the same temperature. If you were to place that object in a closed system at thermal equilibrium, the interior of an insulated emissive sphere—combining assumptions, it would emit more power than it absorbs, in violation of the 2nd law. Starting from an isothermal system, the object would grow colder, and the enclosing container hotter). |
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| ▲ | sgsjchs 4 hours ago | parent | prev [-] | | It needs to be facing open space instead of other parts of itself, otherwise the radiation is just reabsorbed. |
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| ▲ | Forrest7778 7 hours ago | parent | prev | next [-] |
| Great read, thanks for sharing. I am interested in reading some more about the other unsolvable problems that exist in this space, do you have any recommendations that you wouldn't mind pointing me at? It would be greatly appreciated, and thank you :) |
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| ▲ | tintor 6 hours ago | parent | prev | next [-] |
| Is it possible for one side of panel to be used for solar power and other side for radiating heat? |
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| ▲ | jupp0r 6 hours ago | parent | prev | next [-] |
| Wouldn't this be solved like similar problems on earth by making small structures with large surface areas? |
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| ▲ | foota 6 hours ago | parent [-] | | I don't think so. Large surface area helps with convective cooling I think by increasing the surface area that participates in heat exchange with the air (or other thermally conducting material), radiative cooling wouldn't benefit from this because you can't concentrate light beyond the source that it's emitted from (etendue). Though I do wonder if it would be possible to have some kind of internal heat pump driven by electrical power to juice up the temperature of the radiators to increase the power being radiated away? E.g., run a heat pump to increase the temperature of a working fluid and then run high temperature radiators? I think it would work and I don't immediately see that it would violate the laws of thermodynamics? (this is ignoring all practically, I'm sure the engineering would be devilishly hard, although if you're already shooting for the moon you might as well throw in some artificial gravity to boot, it's not like the robots get motion sickness) | | |
| ▲ | echoangle 4 hours ago | parent [-] | | You can use heatpumps to increase radiator temperature but then you need a heatpump and need to power it. But the principle is sound. |
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| ▲ | 6 hours ago | parent | prev | next [-] |
| [deleted] |
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| ▲ | ww520 6 hours ago | parent | prev | next [-] |
| Don’t you get 4 faces to radiate away, assuming a long rectangular tube. |
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| ▲ | JackSlateur 8 hours ago | parent | prev | next [-] |
| It would be a shame if a rock came out of nowhere and hit that many square kilometers stucture Luckily, there are almost no rock in space. |
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| ▲ | tessierashpool 9 hours ago | parent | prev [-] |
| there are two arguments for it. one is marketing. the other is that you could make tiny datacenters and flood the sky with them. in effect, not datacenters at all, but some kind of dataswarm coordinating at literal lightspeed via lasers. they'd still be wildly expensive to deploy, and probably litter the orbit zone with fast-moving debris. |
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| ▲ | nolok 9 hours ago | parent [-] | | Your "other" makes no sense. It doesn't matter if you make a few big or a lot smaller, in space you will still need the same space for the same amount of megawatt. Or did you miss the scale of parent's post ? Because in that dream scenario of "let's ignore all the issues except that" and "the earth and the sun don't have any impact", it's still 3 THOUSANDS square meters for a MW of 8 racks. You want to go smaller and go one rack only sure, it's still hundreds of square meters. Check the size of current orbital structure for a point of reference, you can't dwarf those and call it a "dataswarm of tiny datacenters flooding the sky". |
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