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Can someone explain the design of glider wing airfoils and the subsequent pressure distribution over them?

I hypothesize that:

  1. The pressure distribution should form a resultant force in the forward, upward direction, since two components of force are required to maintain equilibrium in the flight of a glider: a) lift - in equal opposition to weight, and b) a forward component of the total reaction generated by the airfoil - in opposition to drag.

  2. Thus I conclude that the angle of attack of a glider airfoil relative to flight direction would be negative, in order to achieve a forward component opposing drag; although it would be positive relative to airflow, such as to generate low pressure over the wings. Either that or the camber is designed such that the two mentioned criterias here, (a) and (b), are upheld.

Should there not be any opposition to drag, I would expect to see the glider decelerate in the forward direction, until the airflow stalls over the wings, and the glider loses lift and falls.

I am also curious to know about the various designs of glider airfoils: a) what are the criterias defining them, and b) how do they differ from those of powered aircraft, if there are any differences at all?

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  • $\begingroup$ The foward component opposing drag is the glider's weight, not forward lift component (which I don't think is technically possible anyway). Take a close look at forces in flight for a gliding/descending aircraft: Lift equals only part of total weight. Drag opposes the remaining one $\endgroup$
    – Radu094
    Commented Sep 24, 2017 at 8:09
  • $\begingroup$ Difference between lift and drag is rather arbitrary, both is caused by more-or-less the same physical phenomena and you can see it as only one (total) aerodynamic force caused by air flowing around the airplane. For practical purposes, this force is usually divided into two components, one, lift, is by definition perpendicular to the direction of motion, second (drag) is parallel. But for the glider, I believe, it is the most easy to think only one total aerodynamic force which is during steady flight always same magnitude and opposite direction to the gravity. $\endgroup$
    – Martin
    Commented Sep 24, 2017 at 12:14
  • $\begingroup$ Yes, & that would be induced drag. There is however profile drag & parasitic drag which will always apply on any body moving through a fluid, namely, air in this case. This drag force will have to be opposed in order to maintain an equilibrium descent along the glide angle; how can this be done is one part of my question above. $\endgroup$
    – Guha.Gubin
    Commented Sep 24, 2017 at 15:38
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    $\begingroup$ It is also worth noting that many glider wings (especially the case for trainers) do not have a uniform shape across the whole wing. Many glider wings have different cross sections to allow for one part of the wing to stall before the rest of the wing. This acts as an alert to the pilot that a stall is imminent but still allowing the pilot to have control using the rest of the wing. $\endgroup$
    – AndyW
    Commented Sep 27, 2017 at 6:07
  • $\begingroup$ related: aviation.stackexchange.com/q/56352/34686 $\endgroup$ Commented Feb 8, 2023 at 16:07

3 Answers 3

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The pressure distribution over a glider's wing is no different from that of a well-designed aircraft. A glider can maintain flight by flying slightly downwards, so the direction of motion is not straight but slightly down. Doing this in rising air will result in sustained flight.

Forces acting on a glider

Look at the sketch above: The flight path angle $\gamma$ is negative and in still air the flight path vector $\text{x}_k$ is parallel to the direction of the airspeed. Due to the inclination of the flight path, the lift vector is tilted forward such that it's horizontal component is exactly equal to the horizontal component of drag, and the vector sum of lift, drag, and weight is zero. This is shown on the right where I moved the vectors into a closed sequence which demonstrates that all forces balance.

From the viewpoint of the glider, the lift is pointing straight up and drag directly backwards, but the weight force is slightly tilted forward. In a way, it seems that the thrust of a glider is its weight force.

Now let the whole air packet in which the glider flies move up. The glider will still sink down within this air but relative to the ground it will gain altitude if the upward air velocity is high enough.

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    $\begingroup$ Great answer as always, but that picture might cause some miunderstandings; Pet peeve, but too many people confuse this: even though flight path angle is negative,the angle ot attack is still positive. $\endgroup$
    – Radu094
    Commented Sep 25, 2017 at 7:12
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    $\begingroup$ @Radu094: Yes, you're right. But to show the AoA I would need to add the airplane coordinate system and complicate things even more, so I decided to leave it off. $\endgroup$ Commented Sep 25, 2017 at 7:18
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    $\begingroup$ @quietflyer I gave the drag vector 5° more tilt - now its angle looks exaggerated, but brings the point across. The colours might make things confusing, so I changed those, too. $\endgroup$ Commented Feb 9, 2023 at 0:45
  • $\begingroup$ As drawn, the lift vector is perpendicular to the line of flight and (enlarged a bit) the weight vector is the hypotenuse of the closed triangle. Though I was once loathe to decompose vectors into vertical and horizontal components, the issue resolves by simply drawing lift, drag and weight vectors as aircraft (not earth) referenced. One can then clearly see, as in paragraph 3, that sin glide $\theta$ × mg is the propulsive force of the glider (opposing drag). $\endgroup$ Commented Feb 9, 2023 at 4:18
  • $\begingroup$ @PeterKämpf -- I would not say the angle of drag vector looks exaggerated now-- it looks close to parallel with the "Xa =Xk" vector, as it should. In the vector triangle, the angle between lift and drag does not look less than 90 degrees-- if anything it still looks very slightly greater than 90 degrees (but could be illusion due to shape of arrowhead). Anyway, the changes helped much, thanks. I'm amazed never found this answer before as it would have been good to link to during when several questions about "what is the thrust/power in a glider", etc came up over the last few years. $\endgroup$ Commented Feb 9, 2023 at 13:19
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I think you are approaching this problem wrong. If you think about how a glider flies, there is no forward force component on the airfoil that opposes drag. By definition, drag is the force that opposes the motion of the craft and lift is perpendicular to the drag force.

Suppose we drop the glider from some height in a typical flight position (ie fuselage parallel to the ground). The only way this glider is going to fly forward (assuming no tailwind) is by dropping the nose, creating a negative angle of attack with respect to the ground. The lift generated by airfoil here will have both horizontal and vertical components relative to the ground. However, you are also losing altitude in this orientation. The point here is, without initial speed (forward momentum) the only way to create forward speed is to convert potential energy into kinetic energy by losing altitude, so a negative angle of attack will create forward speed, but will cost you altitude.

As to your other questions in regard to glider airfoil design vs propelled aircraft airfoiled design, glider airfoil are typically low camber, low lift, and low drag. This is because while higher camber airfoil generate more lift, they also generate significantly higher drag. This would cause the glider to lose forward momentum faster. Since lift is a function of velocity, your glider loses lift during this time and will fall out of the sky. With propulsion, these airfoil can be used to carry heavier loads (since more lift is generated and the propulsion directly counteracts drag forces). Gliders are typically light and therefore require less lift to stay aloft. The low drag profiles allow them to maintain forward momentum and speed without losing much altitude.

In general, you always want the pressure below the wing to be higher than the pressure above the wing. As soon as this condition is not met, the plane/glider will begin to fall from the sky. Also note that just because the pressure below is higher than the pressure above does not guarantee the aircraft will stay aloft. The upward force generated by this pressure (F=P x Area) must exceed the downward force caused by gravity (F=Mass x Gravity).

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This question focuses on a very interesting mechanistic question about what exactly happens when a (model) aircraft is simply dropped from a height.

How does the aircraft start to glide?

If the aircraft is held horizontally, and let go, gravity will cause acceleration. Aerodynamic drag forces now act on the surfaces of the aircraft. Longitudinal stability will start to point the aircraft in the direction of motion because the center of drag and the center of gravity are off-set, causing rotation, but the wings are still stalled. Not yet a glider.

In drawing diagrams, at the start, when released, only gravity and drag act on the plane. The "wind" is striking the underside of the aircraft. If you are a parachute, those are the vectors drawn.

But even parachutes can move sideways by off-setting center of drag and center of gravity. The "sideways" motion, combined with downward motion, is the new flight path.

Mechanisticly, what is drawn is a rotation as well as drag and gravity forces. The "tilt" caused by rotation gives vertical drag a horizontal component, enabling one to change where the parachute will land.

when rotation from drag forces lowers AoA enough to un-stall the aircraft, the wing becomes more efficient at producing upwards and "sideways" forces than vertical drag

The trim of the plane will have rotating forces in balance at an airspeed and AoA sufficient to produce adequate lift.

The glider now flies away in a straight path with the lift, drag, and weight force vectors summing to zero at a constant (steady state) velocity.

Fairly complicated procedure to throw a paper airplane, but illustrating why there is more than just trim to a tail.

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  • $\begingroup$ Last full paragraph (starts with "Note") of this answer is related, also touches on non-steady-state effects. aviation.stackexchange.com/a/56371/34686 Honestly I'd suggest that the present question is best answered by focusing on the steady-state situation, and that the non-steady state situation be addressed by a separate, specific question, if anyone desires to do so. Just my two cents... (We may already have such questions, not sure... ) $\endgroup$ Commented Feb 9, 2023 at 17:35
  • $\begingroup$ Well, 1. mentions "a resultant force in the forward, upward direction". This happens even before the wing is unstalled. The unstalled wing just does it so much better. $\endgroup$ Commented Feb 9, 2023 at 20:07

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