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1) Exactly! The FCC requires that unintentional RF emissions must be below a certain point. That's kind of my point. 2) And you're neglecting that cell phones
by bradleyland 14y ago
1) Exactly! The FCC requires that unintentional RF emissions must be below a certain point. That's kind of my point.
2) And you're neglecting that cell phones transmit at 0.3 watts, versus (for example) an FM radio tower that transmits at 1000 watts, or a cell tower transmitting at (around) 500 watts per channel. The point is that the radiation from cell phones is small to begin with. Ground based transmitters are far, far stronger, and planes fly nearest to them at the most critical times.
3) I don't see the significance of the distinction between the PED being 15 ft from the avionics inside the plane versus outside, but I'm open to explanations why it's significant.
- wglb 14y agoI am not sure of the power output of a cell tower--500 watts seems unreasonably large. But let's take your numbers and see where the inverse square law brings us. So multiplying power by the inverse square of the distance in feet for your cell numbers is 0.00133333. 500 watts one mile away gives 1.7935 e-5. The cell signal is 75 times stronger. And it is inside the airplane, whereas the cell tower is outside the metallic skin of the aircraft, whose attenuation effect we have not calculated. I suspect the power output of a cell tower is much less than 500 watts. With gain antennas at that frequency and line-of-sight, it is much likely to be less. Let's take a very popular FM broadcast station: WFMT. Its power is 6000 watts ERP. Its relative strength would be 6.15460 e-7. The cell signal is 2166 times as strong at the cockpit than WFMT. And that is just one cell signal.