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

DSTO–TN–0640Next, θ is the angle between x 3 and the projection just mentioned:sin θ = −x 3·z 0 , cos θ =Finally, φ is the angle from y 2 to y 3 in the z 2 direction:√(x 3·x 0 ) 2 + (x 3·y 0 ) 2 . (4.21)sin φ = y 3·z 2 , cos φ = y 3·y 2 . (4.22)Matlab code to calculate the first two of these angles would bepsi = atan2( dot(x3,y0), dot(x3,x0 );theta = atan2( -dot(x3,z0), sqrt((dot(x3,x0))^2 + (dot(x3,y0))^2) );For the last angle φ, we need y 2 and z 2 . These are given byAppropriate Matlab code is theny 2 = y 1 = R z0 (ψ)y 0 , z 2 = R y1 (θ)z 1 = R y1 (θ)z 0 . (4.23)phi = atan2( dot(y3,z2), dot(y3,y2) );The sequence of steps to convert to the six DIS numbers from a lat–long–height andheading–pitch–roll is as follows:1. Use (2.2) to convert lat–long–height (α, ω, h) to (X, Y, Z).2. Find the local NED axes as we did in Section 4.1. That is, start with X, Y , Z in theECEF frame and rotate, first through the longitude and then through the latitude,to give −D, E, N, respectively.3. Rotate the N, E, D vectors, first by the heading around D, then by the pitch aroundthe rotated E, and finally by the roll around the latest rotated N. The resultingvectors are, respectively, the aircraft’s local x, y, z axes (all in the ECEF frame).4. The angles ψ, θ, φ are calculated from (4.20, 4.21, 4.22), where we make the identificationsx 0 = X , y 0 = Y , z 0 = Z ,x 3 = x , y 3 = y , z 3 = z . (4.24)Example 1: An aircraft is flying 10,000m above Adelaide, heading south-east (with nowind component: the heading is taken as the direction in which the nose points), climbingat a 20 ◦ pitch, and holding a 30 ◦ roll. What are the two sets of triplets that the DISstandard requires to specify the aircraft’s locations and orientation?The location is found easily, by converting that aircraft’s lat–long–height at Adelaideto XYZ using (2.2):(X, Y, Z) = (−3.93, 3.48, −3.63) × 10 6 m. (4.25)18