Pyramidal horn flare angle and length

A horn is a flared waveguide. Waves from the throat reach the edge of the aperture by a longer path than at the centre, so the aperture field is not in phase. The allowed path difference δ fixes the relation between flare angle θ, axial length L and aperture A.

                 . P (aperture edge)
              .  |
  L + δ    .     | A/2
        .  θ/2   |
  O  .___________| C (aperture centre)
  (apex)    L

Derivation (applies to each plane of the pyramidal horn): From the right triangle OCP, with OC = L, CP = A/2, OP = L + δ:

cos(θ/2) = L / (L + δ)                     ...(1)
tan(θ/2) = (A/2) / L = A / (2L)            ...(2)

(L + δ)² = L² + (A/2)²
L² + 2Lδ + δ² = L² + A²/4
Since δ << L, drop δ²:
2Lδ = A²/4
L = A² / (8δ)                              ...(3)

Flare angle:
θ = 2 tan⁻¹(A / 2L) = 2 cos⁻¹[L / (L + δ)]  ...(4)

Putting (3) into (2):

tan(θ/2) = A / (2 · A²/8δ) = 4δ / A
θ = 2 tan⁻¹(4δ / A)

Optimum horn: for the most directivity with a given length, the path difference is limited to:

• E-plane: δ ≤ 0.25λ (E-field must be nearly in phase)
• H-plane: δ ≤ 0.4λ (field tapers to zero at the edges, so more phase error is allowed)

So for a pyramidal horn with E-plane aperture B and H-plane aperture A:

L_E = B² / (8 × 0.25λ) = B² / (2λ),   θ_E = 2 tan⁻¹(B / 2L)
L_H = A² / (8 × 0.4λ)  = A² / (3.2λ), θ_H = 2 tan⁻¹(A / 2L)

The horn length is the larger of the two; the other flare is then fixed by geometry.

Example: E-plane aperture B = 10λ, δ = 0.25λ: L = 100λ²/(8 × 0.25λ) = 50λ; θ/2 = tan⁻¹(10/100) = 5.71°, so θ ≈ 11.4°.