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

Ryan Hirani
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  • Highly related to this other recent question/answer: https://aviation.stackexchange.com/questions/71511/how-does-the-load-factor-vary-when-the-aircraft-pitches-up-do – quiet flyer Nov 12 '19 at 01:27

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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 $N_z=Z/m$.

JZYL
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  • What do you mean by "accelerometers can't measure inertial or gravitational forces"? That's exactly what they measure. – Zeus Nov 11 '19 at 23:19
  • @Zeus I mean exactly what I wrote. An accelerometer under free fall will measure no force. – JZYL Nov 11 '19 at 23:29
  • 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. – Zeus Nov 11 '19 at 23:38
  • @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. – JZYL Nov 11 '19 at 23:44
  • 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. – Zeus Nov 11 '19 at 23:54
  • @Zeus This is not merely an implementation detail. Inertial forces (e.g. coriolis, centripetal) aren't actually forces; that's why they are called fictitious forces. Gravitational forces are special too, hence the elevator equivalence principle. – JZYL Nov 12 '19 at 00:13
  • They are not 'actually' forces, but an accelerometers measures exactly them, rather than the acceleration per se. In an orbit/turn, it will indicate a force away from the centre (centrifugal), instead of the centripetal acceleration that causes it. It is a matter of our interpretation to move from one to the other. – Zeus Nov 12 '19 at 04:34
  • @Zeus I don't want to dwell on this since it's offering no value to the question. Obviously, the inertial forces cannot exist without the external forces. But the point is, you "measure" one or the other. I choose to express measurement as body forces minus gravity here. Moved to chat: https://chat.stackexchange.com/rooms/100915/discussion-between-jimmy-and-zeus – JZYL Nov 12 '19 at 05:17