Value chain · CLM-1465
In plain terms
If the antennas in a row are spaced too far apart, extra beams as strong as the main one go out in other directions. How close they must be depends on how far the beam is swung. Unswung, up to a whole wavelength apart is safe. Swung 30 degrees, the antennas must be closer than two-thirds of a wavelength.
The claim as recorded
A straight, even row swung to angle theta0 also radiates full-strength beams in each direction theta with sin(theta) = sin(theta0) + m / d for a non-zero whole number m, with d the spacing in wavelengths. None of these lies in a real direction while d < 1 / (1 + |sin theta0|). The limit is 1 wavelength with no swing, 0.667 wavelengths at a swing of 30 degrees, 0.586 at 45 degrees and 0.5 at 90 degrees, so a row spaced half a wavelength apart has no such beam at any swing short of 90 degrees, and a closer row has none at any swing.
Evidence behind this: T0: derived from the same sum as CLM-1463, and checked in CALC-142 against the page model, whose sum of the waves finds each extra beam where the formula puts it. A held 1994 student design report states the same limit, d / lambda = 1 / (1 + sin theta_s) (SRC-418, T5, PDF 72 [printed 66], eqn 4-4). No textbook is held.
Scope and limits
The onset is where the peak of the extra beam enters view. In a short row the extra beam is wide, and part of it shows earlier: for 8 elements swung 30 degrees the power towards the far edge is already 0.32 of the main beam at 0.62 wavelengths and 0.71 at 0.64. Real elements radiate less to the side, which weakens an extra beam near the edge but does not remove it. A two-dimensional layout, such as the UK design's helix (CLM-1177), has its own condition. At exactly 90 degrees a half-wavelength row has a second beam at the opposite edge.
What it assumes
Far field; a straight, evenly spaced row; elements that radiate equally in every direction in the plane drawn, so an extra beam is as strong as the main one.
Where it comes from
This is the project's own working, shown in full:
CALC-142 step 3. The waves also all arrive together where d sin(theta) - c dt is a whole number of wavelengths, which gives sin(theta) = sin(theta0) + m / d. The first such beam enters from the far side at sin(theta) = -1 when d = 1 / (1 + sin theta0). Swung 30 degrees, the page model's sum of the waves finds the extra beam at the formula's angle at 0.67, 0.70, 0.85 and 1.00 wavelengths, and none at 0.50 and 0.60. A bisection on d over the extra-beam directions returns 1.0000, 0.8520, 0.6667, 0.5858, 0.5359 and 0.5000 at 0, 10, 30, 45, 60 and 90 degrees, the same as 1 / (1 + sin theta0) to four places; it checks the step from the directions to the limit, not the directions.
Basis: spacing in wavelengths; swing angle from the row's normal
Related records
- Answers question
- QST-013 not published in this release
Cite this
energyaperture.com/c/clm-1465
CLM-1465, Energy Aperture, as of 2026-10-17.