# 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? Commented Jan 21, 2020 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? Commented Jan 21, 2020 at 6:52

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