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The equation has some notation that is difficult to find the meaning for. It is equation (3) in the paper 'Quaternion Averaging' by F. Landis Markley, et al. on page 3 under 'The Average Quaternion'. Here is the equation (forgive the poor Latex):

Using the definition of the Frobenius norm, the orthogonality of $A(\mathbf{q})$ and $A(\mathbf{q}_i)$, and some properties of the matrix trace (denoted by $\text{Tr}$) gives

$$ \begin{eqnarray} \lVert A(\mathbf{q}) - A(\mathbf{q}_i) \rVert ^2_F && = \text{Tr } \{[A(\mathbf{q}) - A(\mathbf{q}_i)]^T[A(\mathbf{q}) - A(\mathbf{q}_i)]\} \\ && = 6-2 \text{ Tr }[A(\mathbf{q})A^T(\mathbf{q}_i)] \end{eqnarray} $$

Specifically, with reference to $A(\mathbf{q})$, the brackets suggest this is a function, but $A$ would usually be the symbol for a matrix. $\mathbf{q}$ is likely a quaternion, so maybe a quaternion matrix?

Also, with reference to $A(\mathbf{q}_i)$, the $_i$ would be used in summation, but there is a lack of summation in the equation itself that would use it. Elsewhere in the paper it suggests $\mathbf{q}_i$ is a set of n attitude estimates, so a set of quaternions?

Curly braces would usually denote a set, but is it hard to tell whether ${[A(\mathbf{q}) - A(\mathbf{q}_i)]^T[A(\mathbf{q}) - A(\mathbf{q}_i)]}$ is a set.

Likewise, square brackets may denote any number of things, including equivalence class, or the floor, and so on.

coldnumber
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AshRubigo
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  • I just wanted to point out that this equation in the Markley paper is not needed for the final calculation, it is only used to show how the final calculation is arrived at. For the final calculation, you only need Eqs. (12) and (13). These are shown implemented here: https://math.stackexchange.com/a/3435296/365886 – Luke Hutchison Aug 25 '20 at 11:05

1 Answers1

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Each $\mathbf q_i$ is an attitude estimate ("in quaternion form," so I'd say it is a quaternion).

$A(\mathbf{q})$ is a matrix. More specifically, the paper says on page 3, it is the attitude matrix of a quaternion $\mathbf{q}$.

The fact that the equation doesn't have a summation and the use of an arbitrary index $i$ also tell you that the expression is valid for any attitude estimate. The equation is plugged into a summation in equation (2).

The expression $[A(\mathbf{q}) - A(\mathbf{q}_i)]^T[A(\mathbf{q}) - A(\mathbf{q}_i)]$ is the product of two matrices; one is the transpose of the matrix $A(\mathbf{q}) - A(\mathbf{q}_i)$ and the other is the matrix $A(\mathbf{q}) - A(\mathbf{q}_i)$. Square brackets are used, I think, to avoid nested parentheses.

The trace is a function of a matrix; a reason for the use of the curly braces could be that there are parenthesis and square brackets in the argument.

I found another paper that explains the notation (page 2, especially): http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20030093641.pdf

coldnumber
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  • So it is equation (2) that is the average and not equation (3) (being only part of it)? And the paper you linked is invaluably useful, thank you. Also, you're saying because $A(\textbf{q})$ is a function and includes parentheses, square and curly brackets are used? Are you sure? This seems unconventional. – AshRubigo Aug 04 '15 at 08:15
  • Yes, equation (3) gives you the average and uses equation (2), which is the Frobenius norm of the difference of the matrices. The parenthesis, brackets, braces thing is just notation, I think. You could very well just write it as $\text{Tr } ((A(\mathbf{q}) - A(\mathbf{q}_i))^T(A(\mathbf{q}) - A(\mathbf{q}_i)))$, but as you can probably see, this makes it harder to read. The matrix $A(\mathbf{q})$ is a function of $\mathbf{q}$, and the trace Tr is a function of the matrix inside the curly brackets. Does that make sense? – coldnumber Aug 04 '15 at 08:22
  • It does, mixing brackets inappropriately dilutes the meaning of the very specific symbology, which causes confusion, although it does seem to be a lose-lose. One more thing, could you confirm that attitude matrices are in fact versors? – AshRubigo Aug 04 '15 at 08:36
  • I had to look up the term "versor"; wikipedia tells me that a versor is a quaternion of norm 1, in which case I'd say no. Equation (6) of your paper shows that the attitude matrix is a $3\times 3$ matrix (because it's the sum of three $3\times 3$ matrices). – coldnumber Aug 04 '15 at 08:41
  • My confusion comes from here, but is beyond the scope of this question. Thanks again. – AshRubigo Aug 04 '15 at 08:50
  • You're welcome. – coldnumber Aug 04 '15 at 08:52