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Can I use the z accelerations to calculate the load factor in that axis? Please explain how or why not.

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It sure can; in fact, they are equivalent (with an offset of 1G).

In the body frame, the vertical equation of motion is:

$$Z+mg\cos\theta\cos\phi=m(\dot{w}+pv-qu)$$

Accelerometers can't measure inertial or gravitational forces, so vertical acceleration as measured is $N_z=Z/m$.

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  • $\begingroup$ What do you mean by "accelerometers can't measure inertial or gravitational forces"? That's exactly what they measure. $\endgroup$
    – Zeus
    Commented Nov 11, 2019 at 23:19
  • $\begingroup$ @Zeus I mean exactly what I wrote. An accelerometer under free fall will measure no force. $\endgroup$
    – JZYL
    Commented Nov 11, 2019 at 23:29
  • $\begingroup$ Yes, and there will be no load. This is precisely because it measures both the gravitational force and the inertial force of the acceleration, which add up to zero. A static accelerometer will measure the gravitational force (acting on its sensor) at its location and will not show zero, despite zero acceleration. $\endgroup$
    – Zeus
    Commented Nov 11, 2019 at 23:38
  • $\begingroup$ @Zeus A static accelerometer on a table top shows 1G because it's measuring the normal force (Z in the expression). If it's measuring the gravity component as well, then it should read zero. What I wrote and what you wrote above are equivalent. $\endgroup$
    – JZYL
    Commented Nov 11, 2019 at 23:44
  • $\begingroup$ Well, you can say it's a philosophical question, because you can always claim that the accelerometer 'actually' measures the reaction of its internal spring (if we assume a classic weight accelerometer). But this is an 'implementation detail'; it's meant to measure the gravitational (and inertial) force. $\endgroup$
    – Zeus
    Commented Nov 11, 2019 at 23:54

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