Acceleration of moving reference frame

Question

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?

Answer 1

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.

Answer 2

If the measurements of the accelerometer are in an inertial frame, you need one which is not accelerating or rotating. If you are using this thing for navigation, your inertial frame should not be rotating with the earth. Since the instrument and the earth are both following the same geodesic around the sun, you probably do not need to be concerned with the centripetal acceleration of the earth.

Answer 3

Function of the Accelerometer

The accelerometer measures the time derivative of its own momentum as it would be measured by an inertial frame with the same velocity. However, it does not display the measured value. It displays the measured value divided by its pre-programmed known rest mass.

$$\frac{dp}{dt} := F = \frac{d(mv)}{dt} = \frac{dm}{dt}v +\frac{dv}{dt}m$$

For low relative velocities between the accelerometer and the measurement frame W, $m = m_0$, $\frac{dm}{dt} = 0$ and we can simplify to $F = m_0\frac{dv}{dt}$. The accelerometer divides by the pre-programmed known rest mass and displays $\frac{dv}{dt}$.

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Predicting the Accelerometer's Display from an Arbitrary Inertial Frame

For an object A rotating around an accelerating center of rotation C at radius r, angular velocity $\omega$, all as measured by W: $$\frac{d\vec v}{dt} = \vec a_A = \vec a_C - r\omega^2 \hat r$$

If we can make the approximation $m = m_0$, we're done: when W looks through her telescope at the accelerometer, that's the value she will read.

If you want to extend the result to relativistic relative velocities (if we run the experiment for a long time, such that v as measured by W is a significant fraction of c):

$$m=\gamma(v) m_0$$ where $$\gamma(v) = \frac{1}{\sqrt{1-v^2/c^2}}$$

then carrying out the time derivative of momentum gets:

$$\frac{d\vec p}{dt} = \frac{\gamma^3 m_0}{c^2}(\vec v \cdot \vec a)\vec v + \gamma m_0 \vec a$$

and W can predict that if W looks through a telescope at the accelerometer display it will read $$\frac{\gamma^3}{c^2}(\vec v \cdot \vec a)\vec v + \gamma \vec a$$

where $\vec a$ is as calculated above. Note that for small relative velocities ($v<<c$), $\gamma \to 1$ so the accelerometer displays $\vec a$ as you would expect.