# What is the formula for coupling between non-parallel wires?

I want to solve the equations for wire antennas where the wires are not all parallel. The classic method for parallel wires is to compute the magnetic potential at points on one wire due to current segments of the other wires (and its own) and solve a matrix equation. Been there done that successfully for parallel wires. The magnetic vector potential has the same direction as the current segment that gave rise to it. That math says that two wires or segments at right angles don't couple as the magnetic vector potential due to one wire is at right angles to the other wire. OK. Now consider two short dipoles a few wavelengths apart, one transmitting,the other receiving. Now rotate the transmitting one 45 degrees. That puts the other about 3dB down on its polar pattern. Now rotate the receiving dipole 45 degrees the other way.Another 3dB loss. But now they are wires at right angles to eachother (but not to the line drawn between them, the direction of propagation) So why is the received signal only 6dB down and not zero? There is something missing in the math. Wires at right angles DO couple, as two +/_ 45 deg skewed dipoles demonstrate. The angle of the wire to the line drawn between them has to have something to do with it. Cannot find any formula in any book that elucidates this. Help?

• NEC-2 simulation shows that a short dipole in free space induces no current in an orthogonal identical short dipole a wavelength away. It seems the relationship should follow the cosine of the angle between the dipoles. Please, cite your reference, re: "...the received signal only 6dB down and not zero". – Brian K1LI May 23 '20 at 9:42
• Rotating one dipole 45 degrees and the other 45 results in an asymmetry not present when rotating just one Of them 90 degrees.The near tips will be closer than the far tips, allowing capacitive coupling between the High voltage nodes. – hotpaw2 May 23 '20 at 12:44
• Depending on the two axis of rotation, planar or offset axis. And where the rotation axis is located, centered or offset, on one or both dipoles. – hotpaw2 May 23 '20 at 12:51
• Rotate 45 degrees, along what axis? And if one 45 degree rotation is a 3 dB loss, why would another be 3 dB again? It's simple to show this reasoning is flawed: adding 6 more 45 degree rotations brings the dipoles back to their original orientation, but quite obviously does not add another 24 dB of loss. – Phil Frost - W8II May 23 '20 at 15:50

If each short (~0.1$$\lambda$$) dipole is rotated by 45$$^o$$ about its own center, then neither dipole lies in the null of the other. The current excited in the "receiving" dipole is $$(cos(45^o))^2$$, or 1/2 of the current observed when the two dipoles are broadside to each other. This is confirmed by NEC-2 modeling.