Steve, it's more or less a geometry problem, or perhaps a trigonometry problem. Flying slow is not an option in a 727 or an MD-11, and I would venture to say for any large turbine aricraft. I believe the object of this "maneuver" is to minimize landing distance, so I think I could safely assume that we would be flying at the lowest safe speed to begin with.SteveC said:O.K., then going back to the premise of the original question (descent below GS inside MM) would require either:
1. Higher descent rate (airspeed remains constant, steeper approach path), or
2. Lower approach airspeed (descent rate remains constant, steeper approach path).
Sound reasonable?
Third proposed technique:
1. Reduce power,
2. increase pitch slightly to maintain descent rate,
3. accept lower airspeed, resulting in steeper approach path.
Now we are out of parameters by being too slow???
How am I doing?
(Thanks for walking me through this. Takes me a while to catch up with you guys some times.)
Draw a triangle, with the first point being the point where the glideslope intersects the Middle Marker, about 200 feet above the ground. The second point would be on the runway, 1,000' beyond the threshold, the point where the 3° glidepath hits the runway. The third point is a point directly under the first, on the ground. The hypotenuse of this triangle, the long side, the side that goes from the airplane to the touchdown point, can be viewed as the velocity vector. If the airplane is travelling at 120 knots, that vector is 120 knots, and it's in the direction of that 3° glidepath. The hypotenuse can also be considered to be the sum of the other two vectors that form that triangle. One vector goes straight down, or vertical. The other goes from that point towards the aimpoint on the runway. Using trigonometry, and knowing the angles and dimensions, we could calculate the vertical vector.
Now, if we change the triangle, making the second point the threshold of the runway rather that 1,000 feet down the runway, we have a different shaped triangle. Using trigonometry this time, we find that the vertical component of the 120 kt vector, the hypotenuse, is larger, because the corresponding angle is larger.
I have a feeling that even I wouldn't understand what I just said were I not visualizing the triangles in my mind. It would be far simpler if I had a white board!
I might try to snap a digital photo of a whiteboard and figure out a way to include it in a post, but not tonight.
Until then, the best I can tell you is A) decreasing airspeed is not an option and B) lowering the nose will result in an increased descent rate.
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