4 ms·
There's an equation for that. The effective range of a Gaussian beam is roughly equal to (beam width)^2 / wavelength. I'm not sure what wavelength or beam width
by scythe 5y ago
There's an equation for that. The effective range of a Gaussian beam is roughly equal to (beam width)^2 / wavelength. I'm not sure what wavelength or beam width they used here, though.
- engineer_22 5y ago10Ghz from the article, and photo of the target appears to be 3 meters square
- scythe 5y agoThat means they're right at the Rayleigh length. You'd need a significantly wider beam or higher frequency to send power from orbit. https://en.wikipedia.org/wiki/Rayleigh_length https://en.wikipedia.org/wiki/Rayleigh_length
- DennisP 5y agoIn fact, SPS designs involve receivers measured in square kilometers. But they'd be mostly wire and would only contribute about 0.7 cents/kWh to the cost.