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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1Good 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. – Bianfable Sep 29 '19 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:
(frame from the video)
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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 – Ahmed Sep 29 '19 at 18:23
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@Ahmed I always wondered if those units are actually degrees. I asked that question now. – Bianfable Sep 30 '19 at 08:50
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