For a single-lobed pattern the beam solid angle is approximately given by

$$\Omega_A \approx HP_E.HP_H$$

where $HP_E$ and $HP_H$ are the half-power beamwidths in radians of the main beam in the E and H planes. Show that

$$D \approx \frac{41,253}{HP_{E^o}.HP_{H^o}}$$

where $HP_{E^o}$ and $HP_{H^o}$ are the $E$ and $H$ plane half-power beamwidths in degrees.

My "sub-proof"

Give that $D = \frac{4\pi}{\Omega_A}$ and $\Omega_A \approx HP_E.HP_H$ (radians) have:

$D = \frac{4\pi}{HP_E.HP_H} = \frac{4\pi(degree/\pi)}{(degree^2/\pi^2)(HP_{E^o}.HP_{H^o})} = \frac{4\pi^2}{[degree](HP_{E^o}.HP_{H^o})} = \boxed{\frac{39.47}{HP_{E^o}.HP_{H^o}}}$[WRONG]

So what the wrong thinking in my development?



Radian/degree=180/$\pi$, so beam widths in degrees = beam widths in radians times 180/$\pi$, so you get a factor $4\pi$*(180/$\pi$)^2$\approx$41253, not 41.253.

  • $\begingroup$ Yeah. you are right. In my country (Brazil) we used "comma" as a "point" to separate integer of decimal numbers. Therefore this point is in fact a comma in english notation. I already has edited it. Thanks for your answer though. $\endgroup$ – miguel747 Apr 5 '15 at 13:52

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