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I found the following definition of angular velocity vector of B in A at page 49 of the book "Thomas R. Kane, Peter W. Likins, David A. Levinson - Spacecraft Dynamics - McGraw-Hill (1981)":

enter image description here

The problem is the notation used. If I want the angular velocity vector of B relative to A, I have to write each of the 3 terms in A reference frame ($a_1, a_2, a_3$ unit vectors of A reference frame; $b_1,b_2,b_3$ unit vectors of B reference frame:

$$ \overrightarrow{\omega}_{A\rightarrow B}\Bigr|_{A} = % % \left( \overrightarrow{\omega}_{A\rightarrow B}\Bigr|_{A} \cdot \hat{b}_1\Bigr|_{A} \right) \hat{b}_1\Bigr|_{A} + % % \left( \overrightarrow{\omega}_{A\rightarrow B}\Bigr|_{A} \cdot \hat{b}_2\Bigr|_{A} \right) \hat{b}_2\Bigr|_{A} + % % \left( \overrightarrow{\omega}_{A\rightarrow B}\Bigr|_{A} \cdot \hat{b}_3\Bigr|_{A} \right) \hat{b}_3\Bigr|_{A} = % % % \omega_{b_1,A\rightarrow B}\Bigr|_{A} \hat{b}_1\Bigr|_{A} + % % \omega_{b_2,A\rightarrow B}\Bigr|_{A} \hat{b}_2\Bigr|_{A} + % % \omega_{b_3,A\rightarrow B}\Bigr|_{A} \hat{b}_3\Bigr|_{A} % % $$

where $\omega_{b_i,A\rightarrow B}\Bigr|_{A}$ is the component along $\hat{b}_i$ of the angular velocity vector of B relative to A expressed in A. Instead $\hat{b}_i\Bigr|_{A}$ is the unit vector $\hat{b}_i$ of B expressed in A.

Is this right?

EDIT: in Sec. 1.10 the author said that A and B are 2 rigid bodies which are moving relative to each other:

enter image description here

Thank you in advance.

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    $\begingroup$ The operation is a simple change of coordinates (rotation) and the notation isn't strange but normal for any vector whose components need to be transformed between different frames. $\endgroup$
    – John Alexiou
    Jan 3, 2019 at 19:05
  • $\begingroup$ Hello @ja72, is my equation correct? $\endgroup$
    – Gennaro Arguzzi
    Jan 3, 2019 at 19:08
  • $\begingroup$ Angular rotation does not depend on a point, but only on the orientation so the subscript $A \rightarrow B$ is misleading IMHO. The change of orientation is best done with a 3×3 matrix as a matrix-vector operation because the expression becomes unecessarily long when expressed component by component. $\endgroup$
    – John Alexiou
    Jan 3, 2019 at 19:13

1 Answer 1

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You have two coordinates frames A and B with direction vectors arranged in columns of a 3×3 rotation matrix

$$ \begin{aligned} \boldsymbol{A} & = \left| \matrix{ \boldsymbol{a}_1 & \boldsymbol{a}_2 & \boldsymbol{a}_3 } \right| & \boldsymbol{B} & = \left| \matrix{ \boldsymbol{b}_1 & \boldsymbol{b}_2 & \boldsymbol{b}_3 } \right| \end{aligned}$$

The transformation between these two coordinate frames for any vector $\boldsymbol{\omega}$ is

$$ \begin{aligned} \sideset{^B}{^A} {\boldsymbol{\omega}} & = (\boldsymbol{B}^\top \boldsymbol{A}) \sideset{^A}{^A} {\boldsymbol{\omega}} \\ & = \left| \matrix{ \boldsymbol{b}_1^\top \boldsymbol{a}_1 & \boldsymbol{b}_1^\top \boldsymbol{a}_2 & \boldsymbol{b}_1^\top \boldsymbol{a}_3 \\ \boldsymbol{b}_2^\top \boldsymbol{a}_1 & \boldsymbol{b}_2^\top \boldsymbol{a}_2 & \boldsymbol{b}_2^\top \boldsymbol{a}_3 \\ \boldsymbol{b}_3^\top \boldsymbol{a}_1 & \boldsymbol{b}_3^\top \boldsymbol{a}_2 & \boldsymbol{b}_3^\top \boldsymbol{a}_3 } \right| \sideset{^A}{^A} {\boldsymbol{\omega}} \end{aligned} $$

Where $\boldsymbol{b}^\top \boldsymbol{a} = \boldsymbol{b} \cdot \boldsymbol{a} = b_x a_x + b_y a_y + b_z a_z$ is the inner product of the two vectors.

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  • $\begingroup$ Hello @ja72, if A and B are 2 rigid bodies, what is the meaning of $^A \omega ^A$? If I am sitting on A, I will see each point of A fixed with velocity (linear and angular) equals to zero. See the edited question please. $\endgroup$
    – Gennaro Arguzzi
    Jan 4, 2019 at 9:12
  • $\begingroup$ $^A \omega ^A$ is the angular velocity of A as described by the A coordinate frame. The A coordinate frame has the same orientation as body A in that instant. Personally, I prefer to work in world coordinates only. $\endgroup$
    – John Alexiou
    Jan 4, 2019 at 13:00

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