2 ms·
Note that this math is misleading for calculating how much things will heat up, because effective antenna aperture of the thing that you'd heat up increases wit
by mlyle 5y ago
Note that this math is misleading for calculating how much things will heat up, because effective antenna aperture of the thing that you'd heat up increases with the square of frequency. So the square terms of frequency cancel each other out.
It's still small, but the best simple approximation is just to take the 100W and divide by the proportion of the antenna pattern your target occupies.
A 100W source on 2GHz or 60GHz at 10 meters distance is going to dump about the same amount of energy into you. If the transmit antenna has 3dB of gain, that's going to be about 50mW.
- jdiez17 5y agoThat's interesting. I'm not quite sure how that works but I'll look into it. If you have some resources, I'd appreciate it!
- mlyle 5y agoSee the derivation section in your link. It's assuming an isotropic antenna at the receiver, whose effective area gets smaller as the frequency increases. But people do not get smaller as the frequency increases: they occupy the same fraction of the 10m radius sphere around the antenna.