0
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 3, 2019 at 19:05
  • $\begingroup$ Hello @ja72, is my equation correct? $\endgroup$ 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. $\endgroup$ Jan 3, 2019 at 19:13

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Jan 4, 2019 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.