I've often thought of the similarity between sound and RF propagation with respect to diffraction around terrain, since longer wavelengths should more easily bend around terrain while shorter wavelengths are more line-of-sight. Hiking the nearby Appalachian Trail in PA, you can have one side of a hill exposed to traffic noise and cell towers from a nearby Interstate while the other side isn't. Hiking over the top of the ridge I always wonder whether the traffic noise and cellphone signal fade at around the same time, since their wavelengths are similar.
850 MHz cellular -> 3e8 / 850e6 = 0.35 m 1000 kHz sound -> 343 / 1000 = 0.34 m Nat From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bill Hooper Sent: Tuesday, 2008 January 15 15:08 To: U.S. Metric Association Subject: [USMA:40067] Re: Stuart & Sons Pianos On 2008 Jan 15 , at 10:58 AM, Bill Hooper wrote: The (one metre) wavelength for electromagnetic waves (light, radio, etc.) would be: f = (3 x 10^9 m/s)/(1 m) = 3 x 10^9 Hz = 3 GHz This is my week for making stupid errors and correcting myself. In my note on frequency and wavelength of light waves and the like, I wrote the above. The figure I used for the speed of light is incorrect. It should be 3 x 10^8 m/s. That makes the calculation wrong, too. The whole thing should read: f = (3 x 10^8 m/s)/(1 m) = 3 x 10^8 Hz = 0.3 GHz (or 300 MHz) This error does not affect my argument that a 1 m wavelength wave has a very different frequency depending on whether it is a sound wave or a radio wave. I guess I can't always rely on my memory for things like common constants. Bill Hooper 1810 mm tall Fernandina Beach, Florida, USA ========================== SImplification Begins With SI. ==========================
