Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

For context, a hectare (10000 m²) directly facing the sun receives 10 megawatts of sunlight at 1kW/m². The city of Kiel is at 54°19' north, so at the equinox a hectare near it receives about cos(54°19') ≈ 58% of that: 5.8 megawatts. If you were to cover the whole field with horizontal solar panels with a mainstream 22% efficiency, the peak electrical power generated that day would be 1.28 megawatts. But you'd need 2.20 peak megawatts of solar modules to do it, because the modules are rated based on directly facing the sun and getting 1kW/m² of sunlight, which they could never do in this situation. According to the prices on https://www.solarserver.de/photovoltaik-preis-pv-modul-preis... those modules are currently 0.100€/Wp, so that's 280 kiloeuros of solar panels. This is about an order of magnitude more expensive than the land under them.

The average produced would only be on the order of 100–200kW, because Germany's capacity factor is for shit. Kiel is in an especially terrible place, with only about 2.7kWh/m² per day on average, according to https://solargis.com/resources/free-maps-and-gis-data?locali....

In real life people tilt the panels in a solar farm toward the sun and leave space between rows of panels so they don't shade each other even in the winter. That is because, even at today's historically record low prices, the panels cost enormously more than the land they're sitting on, so it makes sense to economize a bit on panels even at the cost of needing more land.





Consider applying for YC's Winter 2026 batch! Applications are open till Nov 10

Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: