7
$\begingroup$

For an aircraft in straight and level flight, lift-induced drag is the horizontal component of the force perpendicular to the wing chord. Positively cambered aerofoils generate lift starting at small negative angles of attack. So, could the horizontal component of the force generated by the wing point forward?

$\endgroup$
4
  • 2
    $\begingroup$ I'd say induced drag is the horizontal component of the net force, which may not be perpendicular to the wing chord. $\endgroup$
    – Sanchises
    Jan 22 at 19:24
  • 1
    $\begingroup$ If you explore the concept of the "bell-shaped lift distribution" and how it reduces "adverse yaw", I believe you'll find examples of one wingtip creating thrust rather than drag -- not sure exactly how it works-- $\endgroup$ Jan 23 at 14:15
  • 2
    $\begingroup$ After reviewing the answers, I think the flaw in my question is that the total aerodynamic force is not perpendicular to the wing chord. $\endgroup$ Jan 24 at 9:52
  • $\begingroup$ In KSP if the wings are too small and the tail provides up force, yeah, the blue arrow will point forward instead of up and aft, and then you've designed a plane that does the opposite of right itself. $\endgroup$
    – Mazura
    Jan 24 at 17:06

6 Answers 6

13
$\begingroup$

Can induced drag be negative?

Not for the full configuration, but for parts of it.

Induced drag is part of the reaction force when a stream of air is deflected. This reaction force is split into one component, called lift, orthogonal to the initial flow direction and one parallel, called drag.

Regardless of upward or downward lift, this definition will only result in positive drag. The lowest induced drag possible is zero when zero reaction force is created. Any nonzero reaction force creates positive drag.

For a thought experiment, let's split the deflection into tiny segments, each deflecting the stream a bit more. The initial amount of deflection creates almost no drag. The next bit, however, will already start with a small deflection and add its bit to it. Relative to the initial flow direction, here the flow has already an angle and the reaction force, being orthogonal to the local flow angle, will already have positive drag. The further down we now go, each section will add more drag. The drag component of the reaction force will never be negative.

The only situation where local induced drag is positive is when the local flow hits the lift-creating surface such that bending the flow brings it closer to its initial direction of flow. This is possible on the horizontal tail of a longitudinally very stable conventional design which creates a downforce and flies in the downwash of the wing.

$\endgroup$
2
  • $\begingroup$ Thanks. In diagrams showing the pressure vectors over the wing, some of the arrows point forwards but most of them point backwards. The total force always points backwards. I think the flaw in my original argument is that the total aerodynamic force is not perpendicular to the wing chord. $\endgroup$ Jan 24 at 9:51
  • $\begingroup$ @JamesJames If we leave friction out, in separated and fully supersonic flow the total aerodynamic force is indeed perpendicular to the wing chord. Only attached subsonic flow has leading edge thrust which makes the force nearly perpendicular to the direction of movement. $\endgroup$ Jan 26 at 11:16
4
$\begingroup$

By definition, drag is a force slowing the aircraft down. It is the component of all aerodynamic forces that lies parallel to the flight path of the aircraft. So understood, in thist way, NO, it can never be negative. If it was negative, it would be thrust, not drag.

$\endgroup$
9
  • 7
    $\begingroup$ This misses the point of the question. Let's call this hypothetical force "lift-induced thrust". The question remains, could it exist? $\endgroup$ Jan 23 at 13:52
  • $\begingroup$ Well, I guess I was not clear enough, or perhaps the logical connection was too large a leap for some. Because drag is defined as the component of ALL aerodynamic forces that lie parallel to the flight path (or relative wind), it cannot be negative. The sum of all forces created by the relative wind MUST lie in the aft, backwards direction! Simply because the velocity if the air pushing vackwards on the front of all aircraft surfaces is higher than the velocity of the relative wind pushing forwards on the back of the aircraft. I would have thought that was obvious. $\endgroup$ Jan 24 at 17:11
  • $\begingroup$ Unless you want to arbitrarilly define DRAG as some subset of all aerodynamic forces that does not include some of the forces pushing on the front of the airframe. $\endgroup$ Jan 24 at 17:16
  • 1
    $\begingroup$ I understand what you're trying to ask. And the answer is No. ALL aerodynamic forces act normal to the local surface of the airfoil. So ANY Increase in AOA will cause the local force vector pushing at each point on the surface of the airfoil to point further aft (The airfoil is tilting aft to increase AOA). The further aft it points, until it stalls, the larger the component parallel to the relative wind becomes ... $\endgroup$ Jan 25 at 22:40
  • 1
    $\begingroup$ These last two comments are it. That’s all that needs to be said. Make the “rephrase” 2 up the question, and Charles’s comment just above the answer. Delete everything else. Just my opinion, (and not to take away from a few good answers) but there’s a lot of extraneous stuff here that doesn’t really help… $\endgroup$ Jan 26 at 4:33
3
$\begingroup$

For an aircraft in straight and level flight, lift-induced drag is the horizontal component of the force perpendicular to the wing chord.

No answer has yet explicitly pointed out, that this definition of "induced drag" is incorrect. It would be interesting to know where you encountered it. In horizontal flight, lift-induced drag is the horizontal component of the net force generated by the wing.

$\endgroup$
1
  • $\begingroup$ I think you have correctly identified the error in my original argument. The total aerodynamic force is not perpendicular to the wing chord. $\endgroup$ Jan 24 at 9:48
2
$\begingroup$

Not on fixed wing aircraft, but this does happen with rotor wing aircraft and it’s the principal driving mechanism for auto rotation. The section of a rotor blade known as the driven region has an effective force of lift tipped in the direction the rotor spins, driving the rotor blades by means of the air moving through the rotor disc.

$\endgroup$
2
  • 2
    $\begingroup$ Autorotation works pretty much the same way as gliding does in fixed wing aircraft. The induced drag on the blades is aft in direction of relative wind, it's just that the lift itself is tilted forward enough to maintain the autorotation. $\endgroup$
    – Jan Hudec
    Jan 22 at 21:39
  • 1
    $\begingroup$ … you might potentially say that the induced power is negative (depending on the definition; I am not sure about that), but can't say it about induced drag of the individual blades. $\endgroup$
    – Jan Hudec
    Jan 22 at 21:41
0
$\begingroup$

Let's approach this from another direction.

Assume that it's possible to find a wing position that generates thrust rather than drag. Thrust or drag is a matter of airflow over the surface, to generate thrust we must have a net forward airflow. Put an indicated air speed sensor in front of the wing, what does it say? Negative.

Can a plane fly backwards? No, it would stall. Thus our initial premise must be wrong--it's impossible to have a net thrust from the airframe overall. (It is possible to have it for part of the airframe.)

$\endgroup$
9
  • $\begingroup$ This answer misses the distinction between lift-induced drag and parasitic drag. At an angle-of-attack which generates no lift, there will be parasitic drag. But by increasing the angle-of-attack a tiny bit and generating a tiny amount of lift, could there be a small forward-suction force created? Which would reduce total drag, even though net drag remained backwards? That's the question. $\endgroup$ Jan 24 at 9:47
  • $\begingroup$ @JamesJames read about sail boats (they actually use 2 working fluids, air and water, to generate "negative drag"). Some folks around here freak over basic research. To them, if it doesn't look like an airplane, it's not about aviation. $\endgroup$ Jan 24 at 12:08
  • $\begingroup$ @RobertDiGiovanni Sailboats are a different case, you have two fluids with very different properties. Since this is "Aviation" I would assume it's applying to craft entirely in air. $\endgroup$ Jan 24 at 21:33
  • $\begingroup$ @LorenPechtel "Two fluids with very different properties" that both produce a forward force. I think JamesJames is on the right track. I would recommend studying polars on airfoiltools.com. At lower angles of attack (before stall), with sufficient Reynolds number and Camber, forward force from the top suction peak produces enough "- drag" to make net drag much less, yielding some fairly awesome Lift to Drag ratios. $\endgroup$ Jan 25 at 1:21
  • 1
    $\begingroup$ @JamesJames The blackbird works on air vs ground, not pure air. I'm only talking about a pure air environment. $\endgroup$ Jan 26 at 5:22
-1
$\begingroup$

The understanding of the relationship of drag and thrust in aircraft design is crucial to the development of efficient fuel saving aircraft.

From the aircraft reference, thrust is force towards the line of flight, drag is resistance (from the air) to this path. Because they are directly linearly opposed, thrust = - drag is mathematically correct, and can be derived from the steady state formula thrust + drag = 0.

Realizing this (simple) relationship can help explain other phenomena, such as "autorotation", extremely low net drag of airfoils, the benefits of slats, and (putting them all together), the design of soaring birds.

All gliders seek to find an updraft greater than their rate of descent. The broad, highly cambered wing of the eagle is exactly what airliners emulate when preparing to land, allowing them to slow to around 1/3 of their cruising speed.

But what of the slotted wingtips? Could they be using the updraft to provide thrust? This seems to work fairly well for auto rotating helicopters, so...

By changing relative wind away from flight path, drag on a part of the aircraft can produce thrust force (negative drag) to the line of flight. This is accomplished by producing a localized horizontal lift component to the line of flight.

Importantly, to preserve the strict definitions, drag opposes thrust, but localized lift from induced drag can produce a thrust force in the direction of flight.

Now, to get to the point of the question, the Blackbird, and sailing against the wind in general, the key is to extract mechanical energy from a drag force and use it to create a more efficient thrust force in the direction of travel.

Back to thrust = - drag

Let's add in:

thrust = -drag × conversion efficiency

The simplest model would be an anemometer. Mechanical energy is extracted by the difference in drag coefficient between the closed and open end of a cup. You could make a Blackbird using a giant anemometer, and it would roll in any direction when there was sufficient wind.

with the wind, faster than the wind

A bit more challenging. Now we can take the extracted energy and use it to turn a rotating airfoil, a propeller. Airfoils can generate many times more lift force per unit of drag.

So we turn to iceboats, which can easily sail faster than the wind by minimizing drag. Energy input from the wind is amplified by the sail airfoil, creating a thrust force that "runs away" until the total drag (from the entire craft) = thrust (steady state).

The Blackbird propulsion seems to be a form of "autorotation" (with the inner part of the prop blade absorbing drag energy and the outer part lifting).

$\endgroup$
2
  • $\begingroup$ Oh man-- are you going to get me started thinking about downwind-faster-than-the-wind vehicles again? Should I block out the next week of my life on my calendar? Just say no-- $\endgroup$ Jan 25 at 23:02
  • $\begingroup$ Apparently, it's been done. $\endgroup$ Jan 26 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.