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After the well known crashes I looked at the question of stabilizer trimming. In case of trim runaway the pilot is supposed to "grasp and hold" the wheel manually steering the stabilizer. Hence my question: how many turns of the wheel to achieve a 2.5 degrees deflection of the stabilizer?

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    $\begingroup$ Good question. Neither FCOM, FCTM nor QRH answer how much one wheel rotation changes the trim, as far as I could find. The only reference to an amount of rotation is this: "Use force to cause the disconnect clutch to disengage. Approximately 1/2 turn of the stabilizer trim wheel may be needed." from the QRH, but this does not answer the question. $\endgroup$ – Bianfable Sep 29 at 14:28
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About 45 Rotations

In this YouTube video you can see Mentour pilot demonstrating manual trimming of a Boeing 737 NG. They start at about 11:30 with:

we have 4 units nose down now

and at 12:30 they say:

now we are at about 3 degrees

I counted 18 full rotations of the trim wheel in this time. This would result in

$$ \frac{18}{4^\circ - 3^\circ} \times 2.5^\circ = 45 $$

rotations for 2.5 degrees of trim change.

Note that the trim indicator is not very precise, so the number is only an estimate:

737 Trim Wheel (frame from the video)

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  • $\begingroup$ Thanks! I think that "units" are not exactly degrees, but the estimate is good. So when MCAS is active the wheel will turn 45 turns in 9 seconds!Terrific $\endgroup$ – Ahmed Sep 29 at 18:23
  • $\begingroup$ @Ahmed I always wondered if those units are actually degrees. I asked that question now. $\endgroup$ – Bianfable Sep 30 at 8:50
  • $\begingroup$ OK, it was unclear in the papers I read. Thanks $\endgroup$ – Ahmed Sep 30 at 15:13

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