Acceleration of moving reference frame

I want to simulate the readings of an accelerometer that is arbitrarily moved through 3D space. In an inertial reference frame $W$, the motion of the accelerometer is described by it's linear acceleration $\vec{a}_{A/W}(t)$ and it's angular velocity $\vec{\omega}_{A/W}(t)$. Let's assume that there's no gravity and at $t=0$ the reference frame of the accelerometer $A$ is equal to $W$.

The question is, how do I compute the acceleration measured by the accelerometer? I know that $\vec{a}_{A/W}(t) = \vec{a}_{A/A}(t) + \vec{\omega}_{A/W}(t) \times \vec{v}_{A/A}(t)$, but the acceleration and velocity of a reference frame with respect to itself should be zero, correct? What am I missing? What does the accelerometer actually measure and in which reference frame?

• An accelerometer measures the force that it feels, in its reference frame - which is non-zero if it is accelerating. – DilithiumMatrix Apr 11 '16 at 15:40
• So it measures changes in linear momentum in an inertial frame. – ja72 Apr 11 '16 at 19:46

To move velocities between two points A and B do the following:

$$\vec{v}_B = \vec{v}_A + \vec{\omega} \times (\vec{r}_B- \vec{r}_A)$$

The derivative of the above moves the material accelerations from A to B

$$\vec{a}_B= \vec{a}_A + \vec{\alpha} \times (\vec{r}_B- \vec{r}_A) + \vec{\omega} \times (\vec{v}_B - \vec{v}_A)$$

There is also the spatial acceleration which is the acceleration of a fixed point in space (the acceleration the material going under this point) which transforms like the velocity

$$\vec{a}'_B = \vec{a}'_A + \vec{\alpha} \times (\vec{r}_B- \vec{r}_A)$$

You convert from spatial to material with the following rule

\begin{aligned} \vec{a}_A & = \vec{a}'_A + \vec{\omega} \times \vec{v}_A \\ \vec{a}_B & = \vec{a}'_B + \vec{\omega} \times \vec{v}_B \end{aligned}

Finally the instantaneous axis of rotation (COR) is found by

$$\vec{r}_{COR} = \vec{r}_A + \frac{\vec{\omega} \times \vec{v}_A}{\| \vec{\omega} \|^2 }$$

I hope this helps.