Suppose you have a rigid body and you have two local coordinate systems on the body at $P_1$ and $P_2$ and we can write a vector in the $P_2$ coordinate system using $P_1$ via the transformation $v_{p2} = R*v_{p1} + T$.

We know the instantaneous angular velocity vector of the object written with respect to $P_1$, how do we compute the angular velocity vector respect to $P_2$? Additionally how do we compute the orientation of the object with respect to $P_2$ assuming we know its orientation with respect to $P_1$? In other words suppose we know euler angles $(\alpha, \beta, \gamma )$ to rotate the object into the reference frame $P_1$ from a coordinate system located at $P_1$ which is parallel to the global coordinate system and we want to know the euler angles to transform the coordinate system at $P_2$ parallel to the global coordinate system to orient the object the same as it is oriented in $P_1$.

  • $\begingroup$ Angular and tangential accelerations are related like : $$\alpha ={\frac {a_{\perp }}{r}}.$$ So having one, you will deduce the other. $\endgroup$ Commented Apr 11, 2023 at 7:33
  • $\begingroup$ I should have written angular velocity, but basically I have a gyro and I want to compute the orientation on the body using another coordinate frame attached to the body. $\endgroup$ Commented Apr 11, 2023 at 8:40
  • $\begingroup$ Please state your problem better. What kind of body? magnitude and-or direction of angular velocity? what is " the orientation on the body " $\endgroup$
    – trula
    Commented Apr 11, 2023 at 11:59

1 Answer 1


If the two coordinate systems have different orientations, and $R$ is the rotation matrix going from one to the other, then the angular velocity vector is simply transformed as

$$ \omega_{\rm p2} = R\; \omega_{\rm p1}$$

Angular velocity is not location dependent, as it is shared by the entire body and you are just expressing the same quantity in different basis-vectors. The location of the two coordinate systems (and the associated translation $T$) plays no role here.

  • $\begingroup$ I'm not sure I understand this, wouldn't the angular velocity be different since you're adding in a translation also? I mean if you rotate the object based upon the angular velocity around two different points you get two different orientations don't you? $\endgroup$ Commented Apr 12, 2023 at 4:31
  • $\begingroup$ No, you do not. Take a horizontal body and rotate by 30° from one end or the other end. In both situations, the orientation of the body will be the same, 30° from horizontal. A rigid body has a singular angular velocity quantity that can be expressed along different orientations with different vector components, but all cases represent the same rotational velocity. $\endgroup$ Commented Apr 12, 2023 at 4:53
  • $\begingroup$ So if we integrate the same angular velocity from any point on the object the final orientation would be the same? $\endgroup$ Commented Apr 13, 2023 at 5:25
  • 1
    $\begingroup$ ".. angular velocity from any point on the object" - As I said, angular velocity is not location specific. Regardless of where you mount a MEMS gyro, the measurement will be the same. So yes. $\endgroup$ Commented Apr 13, 2023 at 11:45

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