1
$\begingroup$

Using WiFi or bluetooth frequencies 5.8GHz and 2.4-2.485GHz respectively, I want to place a hermetically sealed circuit at the bottom of a gold fish bowl, and want to know if the water will significantly attenuate WiFi or Bluetooth.

Gold fish bowl would be probably 6" to 1', or more.

So, I want to know rate of signal absorption per distance through intervening medium of water.

$\endgroup$
2
  • $\begingroup$ Why don't you look up the permittivity of H2O at those wavelengths? In addition, do you have the radiative pattern over 4pi steradians for your device? How do you know how much energy is being transmitted straight down thru the glass, ignoring the water entirely? $\endgroup$ Jul 22 '16 at 12:44
  • $\begingroup$ Isn't the reason why microwave ovens work and the 2.4GHz band is "open" because water (and water vapor) seriously absorb (attenuate) this frequency? $\endgroup$
    – hazzey
    Jul 22 '16 at 19:07
4
$\begingroup$

There is information out there, but probably the best way to find out is to actually try it. Grab a small Bluetooth or Wifi device and seal it in a waterproof enclosure or bag, and submerge it to the bottom of the fish bowl under water. See what happens. This should provide more valuable insights than the simple theoretical RF attenuation of water per meter, since in the fish bowl you have an irregular geometry and also the device will have a short path for RF out the bottom of the bowl.

Some references that might be useful as well:

A forum thread reported that Bluetooth under water doesn't work.

A white paper from Laird Technologies says, "The 5 GHz wave form is attenuated by common building materials to a greater degree than the 2.4 GHz wave form. On the other hand, the 2.4 GHz wave form is optimally absorbed by water." (Optimizing Operation at 5 GHz, p. 4)

$\endgroup$
1
  • $\begingroup$ Thanks. Yeah, I'll just try it. From the physics forum, I learned that pure (like from distilled) is best. So, I will use a very small sphere of water. $\endgroup$
    – Doug Null
    Jul 22 '16 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.