posted on Mar, 14 2011 @ 01:00 PM
As an engineer, I don't understand an awful lot of stuff written about HAARP.
We know it's a set of HF radio transmitters operating on two frequencies in the HF band, with an output of ~3.6MW which is fed to a double set (one
for each frequency) phased arrray antenna which is flat on the ground, making it's boresight 90 degrees with the surface.
Look up any textbook on RF antenna design. You’ll find that the output from a phased array antenna can be steered. The extent to which it can be
steered from the normal (boresight) is about +/- 45 degrees from the normal, at which point the ERP is about 6 dB down. i.e., half the power. Don’t
take my word for it, look it up in any decent textbook.
Yes, HAARP has been used for years. Essentially is is chucking 3.6MW electromagnetic radiation at HF (not microwave) upwards, at 90 degrees from the
ground, +/- 45 degrees, max. (We just discussed the basic funcamental engineering and you are free to check basic theory yourself)
3.6MW is about half, yes half, the amount of energy that you’d get from , say, a hot air balloon burner. Check out any convenient balloon
That’s all. Period.
OK, you have 3.6MW available at the transmitter. Assuming it's some kind of a weapon, you have to deliver that energy to a target.
Let’s be really generous and suggest that the HF beam can bounce between the ground and the ionosphere all the way round form the USA to Japan.
Let’s be generous and assume it only diverges by 10 degrees.
Lets be generous and say it looses only 5% of its energy at each reflection.
The ionosphere is 50-90km up. Call it 90km
Distance between USA and Japan is about 9000km
If we steer the beam at 45 degrees, we can get to Japan in about 100 reflections
Ignore the fact that, steered at 45 degrees, the beam power is 6dB down. For fun, say the beam power is 100% at transmission. it will be 95% after
the first reflection, 90.25% after the second, 85.7375 after the third, until after the 100th, we have 0.59% of the power. About 21 kW. The same power
as two decent electric shower units.
Now, the path length along all those reflections is about is about 13000km. If the beam diverges 10 degrees, it will be spread over a circle 2300km in
diameter. That’s over 4 million square km. Or about 0.5milliwatts per square km.
Now you have to deliver that power through a few hundred meters of water, then a few km of rock, to where the fault is which you want to make slip.
How much HF radio wave is losst through a few cm of water, let alone a few hundred meters?
If you really think you can do any kind of damage with milliwatts of RF power, you'd be better advised to stop using your cellphone: you might
trigger an earthquake where you are standing.
Take each point I have made and see if you can refute any of them (with supportable references, please).