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Two reasons. 1. A microwave receiver on earth is much smaller than many solar farms, so we don't use a lot of our finite land for energy generation 2. This coul

Two reasons. 1. A microwave receiver on earth is much smaller than many solar farms, so we don't use a lot of our finite land for energy generation 2. This could actually bring solar to some geographies which aren't well suited to solar due to weather To achieve the energy transition, you need SO MANY solar panels and they will take up a lot of land. That land is most expensive near the very population centers which need the most power. According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts."[1] (this isn't the most precise source I could find, but it's reputable) [1] https://www.nytimes.com/2022/09/20/podcasts/transcript-ezra-klein-interviews-jesse-jenkins.html https://www.nytimes.com/2022/09/20/podcasts/transcript-ezra-...

jacquesm 3y ago
No, it's bs. Because of the double conversion you're never going to outperform solar here on earth and the cost of deployment undoes any advantage you might have had anyway. But: cool project and nice of them to get it bankrolled. > To achieve the energy transition, you need SO MANY solar panels and they will take up a lot of land. This isn't true. > According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts. The estimates are that it's anywhere from 10000 to 30000 square miles which is compared to the 3.8 million square miles of the continental USA an absolute pittance. Doing this by tightbeaming the output of an equivalent solar installation in space is ridiculous.
- joak 3y ago
  Agriculture is 44% of US land, PV on an area <1% is enough to power America. And this 1% can (and should) be deployed in deserts
  - Teever 3y ago
    Not to mention we can make dual use of existing land in the form of roof top solar. The roof of every single new big box store should be covered in solar panels and we should start covering large parking lots in them too. Shaded parking that powers the area would be sweet.
    - beAbU 3y ago
      Come to South Africa where the government is unable to provide reliable electricity to the country. Almost all new developments come with rooftop solar, and quite a few large parking lots in my hometown got covered with solar arrays for shade.
  - natmaka 3y ago
    > Agriculture is 44% of US land, PV on an area <1% is enough to power America There is even a potential convergence: https://en.wikipedia.org/wiki/Agrivoltaics https://en.wikipedia.org/wiki/Agrivoltaics
- grigri907 3y ago
  Sure, on a per square foot basis, but the requirements and limitations are decidedly different. I wouldn't dismiss it out of hand, at least not on a 10-30 year time frame.
  - grigri907 3y ago
    Not to mention you get a 30% boost in solar insolation by snagging the light before it goes through the atmosphere, plus you don't have the resistive losses due to heat building up in a land-based solar cell.
    - jacquesm 3y ago
      And how do you propose to move that power losslessly across your downlink? Those losses will likely be far larger than the 30% gain. Getting rid of waste heat in space is harder than it is on earth!
adrianN 3y ago
> you need SO MANY solar panels and they will take up a lot of land If we used the area that is currently producing ethanol and "biodiesel" for solar panels we would be 80% done with the energy transition.
- jvanderbot 3y ago
  If we used that and the land for cattle feed, maybe we could power space with our microwave transmitters.
ben_w 3y ago
Unfortunately #1 isn't correct. The minimum size of a ground station is fixed by the size of the transmitter, the wavelength, and the distance. One standard design had a 10km contiguous(!) diameter receiver for a 1km diameter transmitter in GEO. You can lower the transmitter to about 6371 km[0] before having to worry about night, but that's only going to give you a factor of about 6 improvement. Different wavelengths can do better antenna gain, but start to need to care about atmospheric absorption. You could in principle also relay the power around the same sort of way we do for data, though that way you get a percentage loss from each relay (might still be worth it, might not, I've not tried to cost it). You're also limited by the maximum safe power density; from what I've heard this is surprisingly close to what you'd get from the yearly average output of PV, which on the one hand means that although it can be an alternative to storage when the question is "what about night and winter?" it is fairly unhelpful when the question is "can we use less land"? > That land is most expensive near the very population centers which need the most power. Yes, but wire-based transmission is better than most give it credit for. And even if it was that bad, if you perfectly solved wireless transmission by any mechanism you like, it's a shorter route from the ground on the opposite side of the planet, to low orbit, around the planet, and back down again (~20e6m), than from geostationary orbit (~36e6m). > According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts."[1] (this isn't the most precise source I could find, but it's reputable) Sounds about right: https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode+Island%2C+Massachusetts+area%29+*+1kW%2Fm%5E2+*+10%25+*+20%25+in+W https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode... (20% efficient cells, 10% capacity factor). Those are 2.9e11, 6.3e10, 5.5e11 W, compared to recent US use of 4.6e11 W. But they're also fairly small states, with 1%, 0.3%, and 2% of the US total population: https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode+Island%2C+Massachusetts+population%29+%2F+usa+population https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode... [0] If I've done the maths right: 1. The radius of the inscribed circle of an equilateral triangle is r = (sqrt(3)/6)*a 2. The altitude of the triangle from any side is h = (sqrt(3)/2)*a = 3*r 3. So the closest you can get the satellites while being sure at you never have a period where the only visible satellites are themselves not in shadow is: triangle_altitude - diameter_earth = 3*r - 2*r = r
- ben_w 3y ago

jacquesm 3y ago

No, it's bs. Because of the double conversion you're never going to outperform solar here on earth and the cost of deployment undoes any advantage you might have had anyway. But: cool project and nice of them to get it bankrolled. > To achieve the energy transition, you need SO MANY solar panels and they will take up a lot of land. This isn't true. > According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts. The estimates are that it's anywhere from 10000 to 30000 square miles which is compared to the 3.8 million square miles of the continental USA an absolute pittance. Doing this by tightbeaming the output of an equivalent solar installation in space is ridiculous.

joak 3y ago
Agriculture is 44% of US land, PV on an area <1% is enough to power America. And this 1% can (and should) be deployed in deserts
- Teever 3y ago
  Not to mention we can make dual use of existing land in the form of roof top solar. The roof of every single new big box store should be covered in solar panels and we should start covering large parking lots in them too. Shaded parking that powers the area would be sweet.
  - beAbU 3y ago
    Come to South Africa where the government is unable to provide reliable electricity to the country. Almost all new developments come with rooftop solar, and quite a few large parking lots in my hometown got covered with solar arrays for shade.
- natmaka 3y ago
  > Agriculture is 44% of US land, PV on an area <1% is enough to power America There is even a potential convergence: https://en.wikipedia.org/wiki/Agrivoltaics https://en.wikipedia.org/wiki/Agrivoltaics
grigri907 3y ago
Sure, on a per square foot basis, but the requirements and limitations are decidedly different. I wouldn't dismiss it out of hand, at least not on a 10-30 year time frame.
- grigri907 3y ago
  Not to mention you get a 30% boost in solar insolation by snagging the light before it goes through the atmosphere, plus you don't have the resistive losses due to heat building up in a land-based solar cell.
  - jacquesm 3y ago
    And how do you propose to move that power losslessly across your downlink? Those losses will likely be far larger than the 30% gain. Getting rid of waste heat in space is harder than it is on earth!

adrianN 3y ago

> you need SO MANY solar panels and they will take up a lot of land If we used the area that is currently producing ethanol and "biodiesel" for solar panels we would be 80% done with the energy transition.

jvanderbot 3y ago
If we used that and the land for cattle feed, maybe we could power space with our microwave transmitters.

ben_w 3y ago

Unfortunately #1 isn't correct. The minimum size of a ground station is fixed by the size of the transmitter, the wavelength, and the distance. One standard design had a 10km contiguous(!) diameter receiver for a 1km diameter transmitter in GEO. You can lower the transmitter to about 6371 km[0] before having to worry about night, but that's only going to give you a factor of about 6 improvement. Different wavelengths can do better antenna gain, but start to need to care about atmospheric absorption. You could in principle also relay the power around the same sort of way we do for data, though that way you get a percentage loss from each relay (might still be worth it, might not, I've not tried to cost it). You're also limited by the maximum safe power density; from what I've heard this is surprisingly close to what you'd get from the yearly average output of PV, which on the one hand means that although it can be an alternative to storage when the question is "what about night and winter?" it is fairly unhelpful when the question is "can we use less land"? > That land is most expensive near the very population centers which need the most power. Yes, but wire-based transmission is better than most give it credit for. And even if it was that bad, if you perfectly solved wireless transmission by any mechanism you like, it's a shorter route from the ground on the opposite side of the planet, to low orbit, around the planet, and back down again (~20e6m), than from geostationary orbit (~36e6m). > According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts."[1] (this isn't the most precise source I could find, but it's reputable) Sounds about right: https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode+Island%2C+Massachusetts+area%29+*+1kW%2Fm%5E2+*+10%25+*+20%25+in+W https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode... (20% efficient cells, 10% capacity factor). Those are 2.9e11, 6.3e10, 5.5e11 W, compared to recent US use of 4.6e11 W. But they're also fairly small states, with 1%, 0.3%, and 2% of the US total population: https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode+Island%2C+Massachusetts+population%29+%2F+usa+population https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode... [0] If I've done the maths right: 1. The radius of the inscribed circle of an equilateral triangle is r = (sqrt(3)/6)*a 2. The altitude of the triangle from any side is h = (sqrt(3)/2)*a = 3*r 3. So the closest you can get the satellites while being sure at you never have a period where the only visible satellites are themselves not in shadow is: triangle_altitude - diameter_earth = 3*r - 2*r = r

ben_w 3y ago