I know when the body rotatates as well as translates IAR or ICR shouldn't be used but I am not able to understand why?


closed as unclear what you're asking by ja72, Jim, Ryan Unger, ACuriousMind, John Rennie Aug 12 '15 at 9:49

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    $\begingroup$ There is always an instantaneous axis of rotation defined. You heard wrong. $\endgroup$ – ja72 Aug 11 '15 at 18:06

Any rigid body in motion can be described as rotating about in instantaneous axis of rotation (IAR) and translating along the same axis at the same time.


A rigid body in moving and at time instant a point A riding on the rigid body has position vector $\vec{r}_A$ and instantaneous linear velocity $\vec{v}_A$ at A. The whole body is rotating with $\vec{\omega}$. We are given 6 motion parameters and one 3 location parameters.

The motion is decomposed as:

  1. The direction of the IAR is given by $$\vec{e} = \frac{\vec{\omega}}{| \vec{\omega} |}$$
  2. The rotational speed (magnitude of rotation) is $$\omega = |\vec{\omega}|$$
  3. The IAR passes through a point $$ \vec{r}_C = \vec{r}_A + \frac{\vec{\omega}\times \vec{v}_A}{|\vec{\omega}|^2}$$
  4. The motion parallel to the IAR is described by the pitch $h$ which is the ratio of the linear speed along the axis to angular speed $$ h = \frac{\vec{\omega} \cdot \vec{v}_A}{|\vec{\omega}|^2}$$
  5. If the above is true then the linear velocity at C is zero and thus $$\vec{v}_A = (\vec{r}_C-\vec{r}_A) \times \vec{\omega} + h \vec{\omega}$$ Lets back substitute and see if we can go full circle: $$ \begin{align} \vec{v}_A & = \frac{\vec{\omega}\times \vec{v}_A}{|\vec{\omega}|^2} \times \vec{\omega} + \frac{\vec{\omega} \cdot \vec{v}_A}{|\vec{\omega}|^2} \vec{\omega} \\ & = \frac{\left(\vec{\omega}\times\vec{v}_{A}\right)\times\vec{\omega}+\left(\vec{\omega}\cdot\vec{v}_{A}\right)\vec{\omega}}{\left|\vec{\omega}\right|^{2}} \\ & = \frac{\vec{v}_{A}\left(\vec{\omega}\cdot\vec{\omega}\right)-\vec{\omega}\left(\vec{v}_{A}\cdot\vec{\omega}\right)+\left(\vec{\omega}\cdot\vec{v}_{A}\right)\vec{\omega}}{\left|\vec{\omega}\right|^{2}} \\ & = \frac{\vec{v}_{A}\left|\vec{\omega}\right|^{2}}{\left|\vec{\omega}\right|^{2}}=\vec{v}_{A} \end{align} $$


  • Given the motion parameters $\vec{\omega}$, $\vec{v}_A$ and of any point A on a moving rigid body, the instant axis of rotation and parallel velocity ratio are uniquely defined by the unit direction vector $\vec{e}$ the closest point on the axis $\vec{r}_C$, the scalar pitch $h$ and the rotation magnitude $\omega$.

  • Given the axis of rotation position $\vec{r}_C$, direction $\vec{e}$, the pitch $h$ and the magnitude $\omega$ the motion of any point A located at $\vec{r}_A$ is fully defined by $\vec{\omega} = \omega \vec{e}$ and $\vec{v}_A =\omega \left((\vec{r}_C-\vec{r}_A) \times \vec{e} + h \vec{e}\right)$


In robotics there is talk about a joint screw axis $\mathbf{s}_A$ such that

$$ \mathbf{v}_A =\mathbf{s}_A \, \omega $$ $$ \begin{bmatrix} \vec{v}_A \\ \vec{\omega} \end{bmatrix} = \begin{vmatrix} (\vec{r}_C-\vec{r}_A) \times \vec{e} + h \vec{e} \\ \vec{e} \end{vmatrix} \,\omega $$ The coordinates of $\mathbf{s}_A$ are the pluecker coordinates of a line (or screw) in space.

NOTE: $\times$ is the vector cross product and $\cdot$ the vector dot product.

  • $\begingroup$ This is one of many of the consequences of Euler's rotation theorem. $\endgroup$ – David Hammen Aug 12 '15 at 0:36
  • $\begingroup$ That, plus the parallel translation which is not included in Euler's theorem. Rotation is the $\vec{r}\times \vec{e}$ part and translation the $h \vec{e}$ part of any linear velocity vector. $\endgroup$ – ja72 Aug 12 '15 at 1:00

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