# Solid angle for different E/H plane beamwidths

The solid angle formula calculates the surface area on a unit sphere, from projecting a rectangular patch onto the surface of a sphere.

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This is calculated using the azimuth angle $$\phi$$ and the elevation angle $$\theta$$:

$$\Omega = \int_{0}^{\phi} \int_{0}^{\theta} \sin{\theta'} d\theta' d\phi$$

However an antenna beam pattern will project a curved patch onto a sphere.

Case 1: Circle projection

First consider a linearly polarized, symmetrical parabolic dish antenna. This ideal geometry ensures the E/H planes have the same beamwidth. A circle is projected onto the unit sphere.

Note this is equivalent to the surface area of a hemisphere between the circle and the sphere.

$$\Omega = \int_{0}^{2\pi} \int_{0}^{\theta} \sin{\theta'} d\theta' d\phi = 2\pi(1-\cos{\theta})$$

Case 2: Ellipse projection

Now consider any linearly polarized antenna, such that the E/H planes have different beamwidths. I presume this would be projecting an ellipse onto the unit sphere.

I presume the solid angle can be calculated from the surface area of a semi-ellipsoid. A solid angle is a fraction of the surface area of the unit sphere, the ellipsoid is not coincident with a sphere (other than the trivial case), so this can't be correct.

Some questions:

• How can the solid angle formula be used to derive the (elliptical) solid angle of an antenna with a beamwidth of $$\phi$$ degrees in the E-plane and $$\theta$$ degrees in the H-plane?
• Are there any approximations to this?
• I think things are made worse by the addition of the sphere analogy. If you must project the beam onto a unit sphere, you'll find the area is the same as the solid angle. But there's no need to invoke a sphere to find solid angles. In your first case - there is no hemisphere, just a portion of the unit sphere. In the second case, there's definitely no ellipsoid; an ellipsoid isn't coimcident with a sphere at all (except in the trivial case). How about "just" integrating to find the solid angle of the elliptical beam itself, with your first formula? – tomnexus Jan 21 at 6:33
• You're right - an ellipsoid is not coincident with the sphere. I will remove that part. Instead the solid angle on the elliptical beam should be some fraction of the surface area of a hemisphere. How would I go about integrating this? – pymekrolimus Jan 21 at 6:52

$${4 \pi \over 100} = {0.04 \pi}$$