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In the book, the example given uses the speed of light for the example, which looks like this.

wavelength = 300,000,000 m/s / 2,400,000,000 Hz
      wavelength = 0.125 m = 12.5 cm


Yet directly preceding that, the book states that RF energy propogates at around 300,000 m/s in the earth's atmosphere. Which would yield the following result.

wavelength = 300,000 m/s / 2,400,000,000 Hz
      wavelength = .000125 = .0125 cm


If the second is the more realistic figure, why is the speed of light used in the example? Are antenna elements commonly .0125 cm in length?

to find the wavelength you need to divide the speed of light by the frequency of the wave. That is a basic physic. The correct speed of light is 300,000,000 m/s. Frequency of 11b/g is 2.4Ghz thus the wavelength is as stated. Antennas are most responsive if its element is multiple of wavelength

The book states you're using the speed of the wave. It then goes on to say that the speed of light is not a realistic figure and that the wave propogates at around 300,000 m/s in the earth's atmosphere. Hence, wouldn't you use 300,000 m/s?

please look at the errat for the book. Btw, what version of the book are you using? I know for fact that in the latest version, there has been a mistake in the number they use for the speed of light. Please refer to the errata

I'm using the latest version. Do you have a link, by chance?

http://www.cwnp.com/products/errata.html

When you get to the CWSP and AP, it is a good idea to first take the printed errata and fix all of the errors. In some cases, I physically cut and paste the errata changes into my book (brings back good memories of second grade :)). It is really nice to know that you have the latest version when reading it with the included errata.


GT Hill