# Can a gravity-compensated accelerometer still measure or detect free-fall?

Take the simplest example of a plane aligned perfectly to the NED inertial reference frame at all time. The plane is restricted to only thrust or elevate upward or down with zero pitch, yaw, and roll (so we can forget about rotational matrices between the frames). The {nose, wing, belly} of the plane points {North, East, Down} respectively at all time. The $$z$$-accelerometer is attached to the plane CG, also pointing positively downward.

If I allow the plane to hover stationarily in midair like a helicopter, the $$z$$-accelerometer would register an opposing acceleration from the normal force (-1g). I can deduct that value to zero out $$a^b_z$$

$$a^b_z \equiv \textrm{data measured by the } z \textrm{-accelerometer} = -1g$$ $$a^i_z \equiv z\text{-acceleration observed in the inertial frame}$$ $$a^i_z = a^b_z - a_{\text{normal}} = -1g - (-1g) = 0 \;\;\;\longleftarrow \text{OK}$$

When the plane thrusts or elevates upwards by a magnitude of $$0.5g$$ in the $$-\hat{\mathbf{k}}$$ direction, the plane has to counter gravity to provide this total upward acceleration. Hence, accelerometer registers $$a^b_z = -0.5g - 1g = -1.5g$$. In the perspective of an inertial observer:

$$a^i_z = a^b_z - a_{\text{normal}} = -1.5g -(-1g) = -0.5g \;\;\;\longleftarrow \text{ OK}$$

Vice versa, if we drop or lower the plane by a magnitude of $$0.5g$$ in the $$+\hat{\mathbf{k}}$$ direction, the plane actually has to pull itself up slightly with the accelerometer measuring $$a^b_z = -0.5g$$

$$a^i_z = a^b_z - a_{\text{normal}} = -0.5g -(-1g) = +0.5g \;\;\;\longleftarrow \text{ OK}$$

If we shut down the engine and somehow still maintain the plane's heading as the plane enters free-fall like a book, the accelerometer would simply measure the gravitation acceleration $$a^b_z = +1g$$ in the positive $$\hat{\mathbf{k}}$$ direction. If we plug this value in the the same equation:

$$a^i_z = a^b_z - a_{\text{normal}} = +1g -(-1g) = +2g \;\;\;\longleftarrow \text{WHAT?!}$$

The compensating factor of $$-1g$$ could fool a blind inertial observer to interpret the motion as a plane lowering itself by $$2g$$. How can we fix this error or can we differentiate free-fall acceleration from object's intended acceleration?

• the accelerometer detects all acceleration explainthatstuff.com/accelerometers.html . It generally does not know if the acceleration is caused by gravity or thrust. The software generally need to make guess/inference how much of that acceleration is due to gravity and how much is due to thrust. If you have the initial state at rest with only gravity and no acceleration, it might be possible to incrementally "delete" the gravity part and end up with only the net thrust acceleration. Commented Dec 13, 2022 at 3:51

There is a simple mistake here. If you drop the plane and let it free fall (without air resistance), the accelerometer measures $$0g$$, so your equation looks like:
$$a^i_z = a^b_z - a_{\text{normal}} = 0g -(-1g) = 1g$$
In other words, because you have defined "zero" acceleration $$a^i_z$$ as the accelerometer fixed to the floor of the plane under $$-1g$$ of gravity, removing the influence of gravity by allowing the plane to free fall will register as $$+1g$$ upward.
In an absolute sense, your proper acceleration $$a^b_z$$ was $$-1g$$ when hovering stationary in a gravitational field, and during free fall is $$0g$$.
The common thread is, if I am under the influence of no forces other than gravity, I am on a geodesic path, and I register zero proper acceleration. In other words, I am following the path that the spacetime curvature wants me to take. In fact, standing on Earth's surface as we are now is not a geodesic, because spacetime wants me to keep falling toward the center of the Earth. The solid ground prevents this, so an accelerometer in my hand will register the non-zero value of $$-1g$$.