lectric vector (i.e., no polarization loss) is given by equa- tion 8.6, where V is the voltage at the center of the dipole, E is the field strength in volts per meter, and a is the wavelength in meters. Chapter 8 Radio Wave Propagation (8.6) If the dipole is connected to a feeder of matching impedance, E is halved. If polarization and feeder losses are taken into account, the input to a receiver can be calculated. A useful formula for calculating free-space attenuation, derived from the power density equation, 8.1 and from equation 8.6, and published in ITU-R recommendations, is given by equation 8.7, where Lfr is the free-space loss in decibels, f is the frequency in megahertz, and d is distance in kilometers. Lfr=32.44 + 20log f + 20logd (8.7) It will be noted that equation 8.7 introduces a frequency component into the calculation. As an example, the power density from a 500-watt transmitter working into a lossless isotropic antenna at a distance of 1 kilometer is 0.039 W-3/m 2 and at 16 kilometers it is 0.155 W-6/m². This is a power ratio of 24 dB, which is to be expected because there is a 6-dB power loss every time that the distance is doubled. The ITU-R formula (equation 8.7) gives a free-space loss of 72.4 dB at 1 km and 96.5 dB at 16 km. Again, that is a dif- ference of 24 dB. Figure 8.2 shows the free-space loss for frequency and distance based on the equation's formula. 3 X 106 3 X 105 3x 104 100.0 Frequency (GHz) Fig. 8.2 Free-space loss versus distance for frequencies from 1 MHz to 100 GHz (units on the diagonal lines are in dB). The Propagation Medium The Propagation Medium Almost all propagation involves transmission through some of the Earth's atmosphere, and the structure of the atmosphere and features within it are shown in Figure 8.3. Most meteo- rological activity and cloud formation occur in the first 10 km, and jet aircraft cruise at between 10 and 15 km. Air pressure falls with height, and at 30 km radiation from the Sun is sufficient to generate some free electrons, but the first distinct i