air to ground system's antenna's offboresight angle in horizontal and vertical planes

I want to compute offboresight angle in horizontal and vertical planes from an airborne transmitter (TX) pointing towards the ground to a notional receiver (EX) location.

An example is shown below (3D figure and side view of the 3D figure). Here D is the airborne TX at latitude (degrees), longitude (degrees), and altitude (meter) latD, lonD, and hD. The notional RX point A has an latitude (degrees), longitude (degrees), and altitude (meter) latA, lonA, and hA. In the 3D figure, HB and HA are Earth radii where B is the vertical projection of airborne at point D to ground (so HB abd BD are aligned and DB is perpendicular to BC which is tangent to Earth).

The green arrow DN shows the boresight of the airborne TX and it is along DB (pointing to ground). I want to find the offboresight angle in horizontal and vertical planes.

Vertical offboresight angle is an angle formed between the lines DA and DB in the side view cross section picture of the 3D figure. The code below gives "el" as the angle between DE and DB; since DB and BC are 90 deg and BC and DE are parallel, thus DB and DE are also at 90 deg. Hence, the vertical angle <ADB = 90 - <EDA . Hence I subtract "el" from 90 to get the vertical angle.

[az, el,slantRange]=geodetic2aer(latA, lonA, hA, lanD, lonD, hD)
verticalAngle = 90 - abs(el);

I don't know how to get the horizontal angle. If I assume airborne flies along DE, then horizontal angle would be <MDB (angle between DM and DB) where DM is the projection of DA to the horizontal antenna pattern whose boresight is DN. I don't know how to compute this anlge.

Answer 1

[Altaïr]

Hey @MatG,

The image might be a bit misleading, so introducing a point J along BH, at a distance equal to the projection of BA along DH, could help clarify things. Here's a revised image:

The assumption that ∠ADJ equals 90° - ∠EDA might not be entirely accurate, as points A, J, D, and E do not necessarily lie on the same plane. DE and JC being parallel suggests they could be coplanar only if A and C are coincident. In this 3D scenario, ∠ADJ is the off-boresight angle, while ∠CDJ and ∠MDJ are the vertical and horizontal boresights, respectively. Here's a procedure I suggest to estimate these angles, assuming the coordinates of A and D are known with respect to the earth's center H, and the direction of travel (along DE but not equal to DE in magnitude) is known:

1. Determine the off-boresight angle ∠ADJ, which is the angle between vectors DA and DH (since DH and DJ vectors share the same direction).
2. Use ∠ADJ to find the components of DA along and perpendicular to DJ (or DH) in the plane containing A, D, and J. Let DF be the component perpendicular to DJ.
3. Project DF along and perpendicular to the travel direction (along DE) to obtain vectors DE and DG, respectively.
4. Calculate DM as the resultant of DG and DJ, and calculate DC as the resultant of DE and DJ.
5. The vertical and horizontal boresights, ∠CDJ and ∠MDJ, are the angles made by vectors DC and DM with DJ, respectively.

These MATLAB Answers might be helpful:

• Projecting a vector onto another - https://www.mathworks.com/matlabcentral/answers/2216
• Angle between to vectors - https://www.mathworks.com/matlabcentral/answers/101590