Antenna gain is often expressed in the following form,

$G = \frac{4\pi A_{e}}{\lambda^{2}}$,

where $A_{e}$ is the effective area of the antenna and $\lambda$ is the operating wavelength. However, using the antenna equation, the effective area can be expressed in terms of the main beam width (3dB width) $\Omega$,

$A_{e} = \frac{\lambda^{2}}{\Omega}$.

Assuming both antennas operate at the same wavelength, the following is true,

$G_{B} = G_{A}\frac{\Omega_{B}}{\Omega_{A}}$.

I hope this is helpful.

P.s. this essentially follows from the definition of gain.

The radiation pattern of any dipole antenna looks similar to what you are showing in the 2D plots - but in your interpretation of the 3D pattern you have the axes wrong.

A dipole antenna with the main axis vertical will transmit power in the horizontal plane, with less and less power as you go further away (inverse square law). If you measure the power as a function of angle $\theta$ to the horizontal, you will get a $\cos^2\theta$ relationship. The 3D pattern can be found on many sites - it is shown on http://www.antenna-theory.com/antennas/dipole.php as

Your datasheet is a bit confusing: when it shows the gain "in the vertical plane", 0 degrees corresponds to the horizontal direction. It would have been so much better if they had rotated the graph 90 degrees...

Why does it have this shape? Well, it follows directly from the fact that electrons in the antenna are accelerating in the vertical direction, and that the power depends on the direction of the acceleration. A derivation for this is given in the wikipedia article on the Larmor Equation. Intuitively, you need to "see" the electron moving in order to see its radiation - when it is moving straight towards you, you don't see the lateral motion. And since EM waves require a lateral (transverse) E vector, there can be no EM radiation in the direction that the antenna is pointing.

As for the gain of the antenna: a dipole can have a gain of 3 dB in the horizontal plane. This follows from the fact that it is "radiating only to half of space" - with twice as much power being available at the peak (this is a hand waving argument but you were asking about "intuition".) Antennas that are more directional (restrict the power to a narrower angle) can have greater gains. The datasheet says this antenna has a gain of 2 dB: a bit less than 3 because of losses.

## Best Answer

You can't prove this, it's not true. It is only a very good approximation: it fits with the boundary conditions. For example, in a half wave dipole, by symmetry, the current needs to be maximum in the middle at the feeder inputs and nought at the ends of the antenna (no current flows out into space!). It's also found that calculations grounded on this assumption fit well with experimental results and that numerical models of the antenna show something very lika a sinusoidal distribution. Similar boundary conditions apply for antennas other than the half wave dipole, but the principle is the same: the sinusoid is an approximation.

Currents at all different points in the antenna act, through the retarded potential, on the currents at all other parts of the antenna. Charge also is shuttled between different points in the antenna. The system is highly coupled, and, as such, one can really only find the true current density either numerically or experimentally. A simple numerical model assumes the antenna comprises short current lengths (wherein the current is approximately constant) spaced by time varying charges. Models like this one tell us that the current distribution is, to a good approximation, sinusoidal.