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## The code for solving Wahba's problem

version 1.4 (18.7 KB) by

Code for testing Davenport's (eigen-valued) and Markley's (SVD) solutions to Wahba's problem - 1965

Updated

Wahba's problem was published in 1965, SIAM Review, Vol 7, No 3.
Wabha's problem in short is determining ones (the body's) attitude using a number or co-registered vectors in a reference frame and observation vectors in body coordinates.

Basically the problem is minimizing the following cost function to get R, the rotation matrix (or attitude quaternion):
L = 0.5 SUM a_i (b_i - R r_i)^2
where
a_i - is the weights (a in the code)
b_i - observations in body coordinates (rb in the code)
r_i - known database of co-registered datapoints in a reference coordinates (rr in the code)

the above is equivalent to solving in quaternion from:
L = lambda_0 - trace(RB) = lambda_0 - q' K q
where
q - is the attitude quaternion; and
K - is calculated as below

Please follow the code. You will see the Equations were simply implemented as in the article.

There is one change however, the author
preferred using Zipfel's order for the quaternion representation, thus:
q = [ q0 q1 q2 q3 ] = [ cos (the/2) e(1)sin(the/2) e(1)sin(the/2) e(1)sin(the/2) ]

Note that Body orientations can similarly be found with Gupta's 1998 method or
by simply co-registering a known star-database distance matrix based on observed
angular-distances. Note that Gupta's work seems to be based on Hyslop 1987 work.

The main source document used was "Humble Problems" by F. Landis Markley - 2006. It was freely downloadable at the time of writing from: http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20060012294_2006013132.pdf

distance matrix traversing method).

Note that to turn points the Tensor R is transpose(dcm) if you are using an allowable coordinate system. See the 1st or 2nd Edition of "Modeling and Simulation of Aerospace Vehicle Dynamics" - Zipfel.

AlexG

### AlexG (view profile)

Why are the diagonal elements of the K matrix switched from the version in "Humble Problems"?
Also, I found something I cannot understand. While using your files I defined my measurements in the Reference frame rr and then a rotation matrix R_zyx=R_x(-phi)*R_y(-theta)*R_z(-psi), which is meant to be the rotation from Reference to Body frames. According to what I understand, it's supposed to take rr and apply R_zyx to derive rb=R_zyx*rr, but that yields wrong results. In fact, Using R_xyz=R_zyx', the rotation matrix from Body to Reference frame provides the right result (how?!). If I rotate using the quaternion method like you do, it works fine. But I'd like to understand what I'm missing about the rotation matrix in this case. Thank you.

Petri

### Petri (view profile)

Dankie, goeie oplossing.