Novice From United Kingdom, joined Aug 2012, 89 posts, RR: 0 Posted (7 months 2 weeks ago) and read 3204 times:
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At VR the aircraft should be smoothly rotated to a body angle of 15 degrees in accordance with the recommended Boeing rate of 3 degrees per second.
I understand that the 15 degrees must be measured from the horizon though what does the 3 degrees per second be measured from and what does it mean?
tdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80 Reply 2, posted (7 months 1 week 6 days 20 hours ago) and read 3103 times:
Quoting Novice (Thread starter): I understand that the 15 degrees must be measured from the horizon
It's from *horizontal*, not the horizon. If you're anywhere with terrain features, 15 degrees above horizontal will be less than 15 degrees above the horizon.
Yukon880 From United States of America, joined Sep 2011, 113 posts, RR: 2 Reply 3, posted (7 months 1 week 6 days 20 hours ago) and read 3094 times:
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Quoting tdscanuck (Reply 2): It's from *horizontal*, not the horizon. If you're anywhere with terrain features, 15 degrees above horizontal will be less than 15 degrees above the horizon
Not if you're rotating off the last half of Runway 33 at ANC.
Sorry Tom, I couldn't resist!
Jetlagged From United Kingdom, joined Jan 2005, 2452 posts, RR: 17 Reply 4, posted (7 months 1 week 6 days 9 hours ago) and read 2944 times:
Quoting Novice (Thread starter): I understand that the 15 degrees must be measured from the horizon though what does the 3 degrees per second be measured from and what does it mean?
Pitch angle and rate are as indicated by the pitch scale on the PFD. There is no pitch rate indication, but to achieve 15 deg at 3 deg/sec will take 5 seconds. The pitch target is displayed on the PFD and the pilot will aim to get there smoothly in about 5 seconds. The exact pitch rate doesn't matter, the point is to achieve a steady and consistent pitch rate. Too fast or erratic and you risk a tail strike. Too slow and you may compromise terrain clearance and will be accelerating rapidly through V2.
The pitch target isn't always 15 deg though. It depends on the aircraft type and the takeoff conditions.
The glass isn't half empty, or half full, it's twice as big as it needs to be.
XFSUgimpLB41X From United States of America, joined Aug 2000, 3952 posts, RR: 36 Reply 5, posted (7 months 1 week 6 days 3 hours ago) and read 2864 times:
Quoting Jetlagged (Reply 4): The pitch target isn't always 15 deg though. It depends on the aircraft type and the takeoff conditions.
Bingo. It was routine on the 757 to end up at 20 degrees nose up when somewhat light. On a very heavy 737-800 you'll be going to about 13.
tb727 From United States of America, joined Jun 2005, 1375 posts, RR: 4 Reply 6, posted (7 months 1 week 6 days 2 hours ago) and read 2844 times:
Quoting Jetlagged (Reply 4): The pitch target isn't always 15 deg though. It depends on the aircraft type and the takeoff conditions.
And on the 727, we have a target pitch for 3 engines and 2 engines figured for every takeoff. Depending on weight and takeoff flap config it can vary from 7 to 22 degrees for 2 and 3 engine. Our rotation is a little slower at 2-3 degrees per second then after liftoff you just pitch for V2+10 to 1000', target is a just kind of a guide to get you close to it.
Quoting tdscanuck (Reply 2): It's from *horizontal*, not the horizon. If you're anywhere with terrain features, 15 degrees above horizontal will be less than 15 degrees above the horizon.
Just to make this clear then, the horizon is where the earth and sky met therefore where the top of the terrain and the sky met. So for instance going by that definition if there where a mountain in front of the runway, the horizon viewed by the pilot on the runway would be a much higher angle/degrees than the 15 degrees measured from the horizontal?
Also is horizontal measured from the longitudinal axises of the plane?
tdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80 Reply 9, posted (7 months 1 week 18 hours ago) and read 2185 times:
Quoting Novice (Reply 7): Just to make this clear then, the horizon is where the earth and sky met therefore where the top of the terrain and the sky met.
Yes.
Quoting Novice (Reply 7): So for instance going by that definition if there where a mountain in front of the runway, the horizon viewed by the pilot on the runway would be a much higher angle/degrees than the 15 degrees measured from the horizontal?
Whether it's higher than 15 degrees depends on the terrain. But yes, the horizon (where the earth meets the sky) would be at a higher angle than horizontal. As a result, 15 degrees above horizontal would be less than 15 degrees above the horizon.
Quoting Novice (Reply 7): Also is horizontal measured from the longitudinal axises of the plane?
Horizontal is defined relative to the geoid (approximately mean sea level). The pitch angle is measured between horizontal and, usually, the plane's longitudinal axis. There are a couple of instances in flight dynamics when you measure relative to the flight path and/or the stability axes (which generally aren't the same as the longitudinal axis) but those aren't presented to the pilot.
3 degrees per second isn't quick...it would take 2 minutes to go through a full revolution at that rate. Put another way, it's half the speed that a sweep second hand moves.
Quoting thegeek (Reply 8): Gives that uncomfortable moment when you are pushing into your seat just as you take off.
That's not caused directly by the rotation rate...the normal acceleration just due to pitch rate is pretty small. The vast majority of the normal acceleration bump at rotation is caused by the AoA increase of the wing increasing lift and accelerating the aircraft upwards.
The distinction between forces. What Tom is saying, I think, is that the normal force you feel is less the centripetal force from the pitch rate and more due to the wings producing progressively more lift as alpha increases.
Why do Aerospace Engineering students have to turn things in on time?
tdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80 Reply 12, posted (7 months 1 week 7 hours ago) and read 2053 times:
Quoting thegeek (Reply 10): Quoting tdscanuck (Reply 9):
The vast majority of the normal acceleration bump at rotation is caused by the AoA increase
Am I missing something? AoA increase = rotation rate given a level trajectory on the runway.
Yes. The point is that the vertical force you feel isn't the centripedal force due to rotation rate, it's force due to vertical acceleration from lift. Changing rotation rate won't make much change to the vertical acceleration you feel (it will change how fast that force comes on, i.e. the third derivative or jitter).
Quoting BMI727 (Reply 11): The distinction between forces. What Tom is saying, I think, is that the normal force you feel is less the centripetal force from the pitch rate and more due to the wings producing progressively more lift as alpha increases.
Exactly. Centripedal force due to pitch rate is very small compared to vertical acceleration due to lift.
tdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80 Reply 15, posted (7 months 6 days 19 hours ago) and read 1845 times:
Quoting thegeek (Reply 14):
My point was that isn't the peak AoA higher than it would be with a slower rotation rate, perhaps starting earlier?
Maybe, but given that there's not much lag between AoA, lift increase, and vertical acceleration change and that the pitch rate is pretty slow, I can't see it being very significant. The angle you rotate to is going to make a lot more difference than the rate you rotate at.
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