252 G. James et al. FCG = FF FR = FF FRχ χN = I I I I χA χN χB χN χC χN χD χN ⎡ ⎢ ⎢ ⎣ FA FB FC FD ⎤ ⎥ ⎥ ⎦ =T ∗Fbj; (24.17) Note that the order of the entries in the Fbj vector are as follows: Fbj = ⎡ ⎢ ⎣ FbjAX (t0) · · · FbjAX (tn) . . . . . . . . . FbjDZ (t0) . . . FbjDZ (tn) ⎤ ⎥ ⎦; (24.18) where A, B, C, and D denote the corner; and X, Y, and Z denote direction of force. And the order of entries in the FCG matrix is as follows: FCG = FF FR = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ FFX (t0) · · · FFX (tn) FFY (t0) . . . FFY (tn) FFZ (t0) . . . FFZ (tn) FRX (t0) · · · FRX (tn) FRY (t0) . . . FRY (tn) FRZ (t0) . . . FRZ (tn) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ; (24.19) where FF denotes translational force at the CG, and FR denotes rotational moment at the CG. At this point, the Below-JEL forces are a rank deficient partition that drives the CG accelerations via rigid body modes. Hence in order to solve Eq. (24.17) for Fbj, an SVD pseudo-inversion is used. The resulting answer will be of rank 6. However, a full rank rigid body plus flex body solution will result from later steps. Another solution approach is to assume that Fbj is defined to be a linear combination of the basis vectors (12 x 6 left singular vectors of TT denoted as U), which can be defined with a Singular Value Decomposition (SVD): U∗S ∗VT =TT. (24.20) Fbj =U∗α; (24.21) where α is the 6 x n matrix of scale factors for the basis vectors of T to create Fbj. Note that system-level mass properties [MOI, POI, and CG location,] as well as sensor geometry are known and represent the only parameters that are extracted from an FEM. Step #6 – Define Below-JEL Null Space Transformation Vectors for Below-JEL Flex Forces The Below-JEL forces (Fbj) developed in the previous step are necessarily a rank deficient partition that drives the CG accelerations via rigid body modes. Hence the approach implemented in this step estimates the non-rigid body partition of the Below-JEL forces (Ff6) to complete the rank and drive the flex-body part of the measured accelerations: FT =Fbj +Ff6. (24.22)
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