AARO GoFast Case Resolution and Methodology

UNCLASSIFIED
12
UNCLASSIFIED
𝑅(𝛽)𝑦 ∙ 𝐿𝑂𝑆 = [
cos (−35°) 0 sin (−35°)
0 1 0
− sin(−35°) 0 cos(−35°)
] · [
6297
0
0
] (9𝑎)
= [
5158
0
3612
] (9𝑏)
The rotation about the z-axis by -57° to account for the sensor azimuth is given (8a-8b).
𝑅𝑧(𝛾) ∙ 𝐿𝑂𝑆 = [
cos (−57°) −sin (−57°) 0
sin (−57°) cos(−57°) 0
0 0 1
] ∙ [
5158
0
3612
] (10𝑎)
= [
2809
−4326
3612
] (10𝑏)
And finally, (9a-9b) show the rotation about the z-axis -9.6° for the aircraft yaw relative to
position 1.
𝑅𝑧(𝛾) ∙ 𝐿𝑂𝑆 = [
cos (−9.6°) −sin (−9.6°) 0
sin (−9.6°) cos(−9.6°) 0
0 0 1
] · [
2809
−4326
3612
] (11𝑎)
= [
2049
−4734
3612
] (11𝑏)
This means the UAP was 2,049 m ahead of, 4,734 m to the left of, and 3,612 m below the F/A-
18’s position at t2. We can now apply the coordinates for the F/A-18 from (7) and (8) to find the
UAP location at t2. Adding the UAP’s relative coordinates from (9b) to the aircraft’s Δx and Δy
displacement from t1 to t2 gives the UAP position.
[2,049, −4,734, 3,612] + [2,461, −207, 0] = [4,510, −4,941, 3,612] (12)
The UAP was 3,612 m below the F/A-18, or at an altitude of 4,008 m (13,150 ft), very close to
the altitude at t1 indicating the UAP moved in a mostly level path.
Results
With the location of the UAP known at t1 and t2, the distance between the locations was
calculated using the cartesian coordinate distance formula as shown in (13a-13c).
𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2 + (𝑧2 − 𝑧1)2 (13𝑎)