Difference between revisions of "Talk:2812: Solar Panel Placement"
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Can someone smarter than me do the math on running power lines to a panel on the sun? How long until it would pay for itself?[[Special:Contributions/172.70.214.125|172.70.214.125]] 05:08, 8 August 2023 (UTC) | Can someone smarter than me do the math on running power lines to a panel on the sun? How long until it would pay for itself?[[Special:Contributions/172.70.214.125|172.70.214.125]] 05:08, 8 August 2023 (UTC) | ||
| + | :$22 million / $0.20 per kWH = 110 million kWH, divide by hours in a year and you get about 12549 kW. Google says the sun is 150 million KM away. IDK the exact details, but a calculator I found online suggests a copper cable with a cross-sectional area of 10m^2 can handle that amount of power transported 150 million KM. The density of copper's about 9 cm/g^3. 150 million km * 10 m^2 * 9 g/cm^3 = 1.35 * 10^16 kg. The cost to get one kilogram to Low-Earth Orbit according to google is >$10,000, let's just use that. Total cost to get to LEO: $1.35 * 10^20. Divide by profit per year => 6.1 trillion years. [[User:Tiln|Tiln]] ([[User talk:Tiln|talk]]) 07:10, 8 August 2023 (UTC) | ||
Revision as of 07:10, 8 August 2023
Can someone smarter than me do the math on running power lines to a panel on the sun? How long until it would pay for itself?172.70.214.125 05:08, 8 August 2023 (UTC)
- $22 million / $0.20 per kWH = 110 million kWH, divide by hours in a year and you get about 12549 kW. Google says the sun is 150 million KM away. IDK the exact details, but a calculator I found online suggests a copper cable with a cross-sectional area of 10m^2 can handle that amount of power transported 150 million KM. The density of copper's about 9 cm/g^3. 150 million km * 10 m^2 * 9 g/cm^3 = 1.35 * 10^16 kg. The cost to get one kilogram to Low-Earth Orbit according to google is >$10,000, let's just use that. Total cost to get to LEO: $1.35 * 10^20. Divide by profit per year => 6.1 trillion years. Tiln (talk) 07:10, 8 August 2023 (UTC)
