[Laser] Scintillation and Adaptive Optics

[glennt at charter.net]

Hi Art!

I think the atmosphere & water vapor issues are a bit more complex because the transmissivity varies not only with altitude, but also with wavelength. At some wavelengths the water vapor, CO2 etc. will have almost no effect at all. At other wavelengths the absorption will be so high that any real communication is precluded.

That being said - yeah - a vertical atmosphere is probably fewer miles than a horizontal atmosphere, and at any particular angle (NLS, cloud bounce etc) will be somewhere in between. I think I can see why an atmosphere might be a useful measure of distance if you are an astronomer, but for our commo application it may be imprecise to the point of unusability.

73 de Glenn



FROM {Laser}...

Terry,

Thanks for a very simple, yet elegant description of the problem and it's 'remedy'. It is very helpful. I marked your message in my archive, so I can refer to it again if needed.

I have one question however....

You state that 21 miles is 'about' one atmosphere thick. Isn't a linear 21 mile path more like 3 atmospheres thick? Above 20K feet, the atmosphere gets pretty thin pretty quickly. I'd say a 21 mile line of sight path through the dense lower atmosphere should be considerably more than 1 atmosphere of thickness, shouldn't it?

Since it's also quite a bit cooler at 20K feet an above, the amount of water vapor contained at heights more than 20K feet is very minimal (relative to the lower atmosphere). So, perhaps we might consider a 21 mile line of sight path in the lower atmosphere to be as much as 5 or 6 atmospheres thick when the volume of water vapor is considered also. Or, does water vapor enter into this issue at all? My guess is that water vapor makes the 'air' more dense and since the water vapor isn't evenly distributed, then it is valid to consider the absence of water vapor when discussing the quantification of how many atmospheres the signal passes through.

Regards,

Art