Using Rotations to Build Aerospace Coordinate Systems - Defence ...

DSTO–TN–0640Equation (2.2) together with the appropriate heights then gives approximately:⎡ ⎤−4.67⎡ ⎤−3.94r them←C= ⎣ 2.57⎦ × 10 6 metres, r us←C= ⎣ 3.49⎦ × 10 6 metres. (4.13)−3.55−3.65Hence⎡ ⎤−7.25r them←us= r them←C− r us←C= ⎣−9.21⎦ × 10 5 metres. (4.14)0.92Now we need to find x, y, z at our location. First we find the local north, east, anddown vectors N, E, D by (4.2) or (4.3), where the heights play no part in the calculation.These were given in (4.8). These vectors must be rotated to produce our own aircraftaxes. First, place our aircraft in the local NED frame flying north, straight and level. Itsaircraft axes then match the local NED axes:x 0 = N , y 0 = E , z 0 = D . (4.15)Next rotate the aircraft axes around z 0 by 45 ◦ , and then rotate 20 ◦ around the new y-axis y 1 , taking care to get the angle signs right by way of the right hand rule for rotation.The sequence is:x 1 = R z0 (45 ◦ )x 0y 1 = R z0 (45 ◦ )y 0z 1 = R z0 (45 ◦ )z 0 = z 0x = R y1 (20 ◦ )x 1y = R y1 (20 ◦ )y 1 = y 1z = R y1 (20 ◦ )z 1 . (4.16)The final vectors are approximately⎡ ⎤−0.94⎡ ⎤−0.17⎡ ⎤0.31x = ⎣−0.06⎦ , y = ⎣−0.80⎦ , z = ⎣−0.60⎦ . (4.17)0.35−0.58 0.74Dotting these with r them←usin (4.11) gives the set of components we require, so that theother aircraft turns out to have coordinates (765, 802, 393) km in the frame of our aircraft.(Note that these numbers result from keeping more significant figures than are shown inthe numbers above.) As a simple check, the length of this vector should be the distanceof the aircraft:distance of aircraft = √ 765 2 + 802 2 + 393 2 km ≃ 1176 km, (4.18)which is reasonable. To actually sight the aircraft, we first need to turn our head to ourright by approximately tan−1 802765 , or 46◦ . Again this makes sense, since the aircraft isapproximately due east and we are already flying north-east. After having turned ourhead through 46 ◦ , we need to turn it downwards by tan−1 √ 393765 2 +802 2, or about 20◦ .16