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I was reading an article about drones and happened to glance at an included 3-view of the General Atomics Predator series UAV. The inverted-V tail appears to leave, at most, around 6" of clearance to the ground.

enter image description here

On takeoff or landing, how is there enough clearance to allow the operator to rotate? Is the wing designed to have a high angle of incidence on the ground?

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The tail arrangement was initially chosen to protect the propeller from touching the ground first. The configuration of what later became the Predator was initially designed to be launched from a submarine, so it needed to fit into a torpedo tube when folded. When this configuration was scaled up, the general arrangement was kept. Now the tail surfaces are much larger relative to the propeller, so the initial reason for their location is not easily discernible.

The high aspect ratio $AR$ of the Predator's unswept wing gives it a much higher lift curve slope than that of the swept wing of an airliner, so less rotation is needed to increase lift for take-off. Also, the highly cambered airfoil has a very low, negative zero-lift angle of attack, so at level attitude the wing already produces substantial lift. This was chosen to fly at optimum endurance speed with a horizontal fuselage attitude. The lift coefficient $c_L$ for optimum endurance is $$c_L = \sqrt{3\cdot\pi\cdot AR\cdot\epsilon\cdot c_{D0}}$$ At that lift coefficient, induced drag $c_{Di} = \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$ is three times bigger than zero-lift drag $c_{D0}$, which means endurance is longest when the aircraft flies at a high lift coefficient and low speed. Basically, the Predator is a point design, flying almost always at the same attitude and lift coefficient. Ground attitude equals flight attitude, and a rotation beyond a few degrees will make it stall.

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  • $\begingroup$ just a small correction: Predator and other UAV's fly at near-constant CL for optimum endurance = true. But the speed that they fly of course changes with aircraft weight (as fuel is burnt), and endurance UAVs have a high fuel-fraction. $\endgroup$ – Gürkan Çetin Jun 2 '15 at 18:09
  • $\begingroup$ @GürkanÇetin: Right, speed at the end of the mission is maybe 75% of what it was right after take-off. Changed the last paragraph. $\endgroup$ – Peter Kämpf Jun 2 '15 at 18:31
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On takeoff or landing, how is there enough clearance to allow the operator to rotate?

The Predator just doesn't rotate enough to hit the tail: See this takeoff video.

Also, it lands very flat.

The drawing in your question shows that there is 5.3° of rotation angle before tailstrike occurs.

Also, from the drawing, the angle of incidence doesn't look remarkable.

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  • $\begingroup$ Makes sense, 5.3° just seems very, very slight, with no margin for error. $\endgroup$ – egid Jun 23 '14 at 2:47
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    $\begingroup$ There may be another factor at work here. For example, the aircraft I'm familiar with, classic 747s, rotate to, typically, 12 or 13 degrees nose up. Since you don't rotate until you have flying speed, the aircraft actually comes off the ground before you reach the nose up target. $\endgroup$ – Terry Jun 24 '14 at 19:58
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    $\begingroup$ 5.3 seems slight with no margin for error - but remember that a Predator is fairly small, slow, light and stable compared to a big widebody jet: it doesn't need a lot of runway, so it has plenty of time to assume the "flat" trajectory, and the lack of high speed and weight mean it doesn't have to flare much on landing... it can quite happily saunter up to the runway and just slowly drop onto it. Given a long enough runway and nice flat areas on the glideslope/climbout, a 747 could land and take off virtually flat, too. $\endgroup$ – Jon Story Nov 27 '14 at 13:03
  • $\begingroup$ @JonStory: Only up to a point; the 747 has a high enough wing loading that you'd eventually run into the maximum-tyre-speed limit if you tried to land or take off too flat. $\endgroup$ – Sean Jun 12 '19 at 22:28

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