6
$\begingroup$

An increase in lift causes an increase in drag. Is it always true? I am confused with two points.

  1. Lift increases with an increase in airspeed but induced drag reduces with an increase in airspeed.
  2. Higher aspect ratio wings generate greater lift but create less induced drag.
$\endgroup$
1
  • 1
    $\begingroup$ This question is not very clear. It should be narrowed down to one specific question. Give some thought as to whether you really mean to be asking about lift, or about lift coefficient. Also about what constraints you intend apply-- constant angle-of-attack? Constant airspeed? Straight-and-level flight? At present it looks like some constraints would apply to some parts of the question and other constraints would apply to other parts of the question. $\endgroup$ Commented Jun 1, 2020 at 15:28

2 Answers 2

10
$\begingroup$

When looking at statements like these, it's important to consider which parameters are varied and which parameters remain fixed.

Let's get this out of the way first by correcting your statements:

  1. Lift increases with increase in airspeed with constant angle of attack but induced drag reduces with increase in airspeed with lift force remaining constant
  2. Higher Aspect ratio wings generates greater lift with constant wing surface area but creates less induced drag with lift force remaining constant.

As you can see, each half of your statement refers to a different scenario. The bottom line is that for lift generation, it's more efficient to give a small change of velocity to a large air mass, than a large change of velocity to a small air mass (because force is generated by momentum change which is linear in velocity, and induced drag is due to the kinetic energy imparted on the air mass which is quadratic in velocity).

When flying fast or with slender (high aspect ratio) wings, you affect a larger air mass every second. You can do two things: either keep your angle of attack constant and get a lot of lift out of the air mass, or in case you don't want to do aerobatic maneuvers, you decrease the angle of attack until lift once more equals weight, and get less induced drag.

$\endgroup$
1
  • $\begingroup$ Thanks for giving more clearity, it helped me a lot. $\endgroup$
    – Rajakr
    Commented Jun 1, 2020 at 15:47
6
$\begingroup$

Lift increases with increase in airspeed but induced drag reduces with increase in airspeed.

If you keep weight ($W$) constant, the total lift ($L$) does not change with change in airspeed ($V$). $L=W$ for quasi-steady flight, right?

What is changed is the lift coefficient, $C_L$. As speed increases, $C_L$ decreases. Induced drag coefficient ($C_{D_i}$) is related to the square of lift coefficient:

$$C_{D_i}=\frac{1}{\pi e A}C_L^2$$

where $e$ is the efficiency factor (related to wing shape), and $A$ is aspect ratio.

The total induced drag ($D_i$) as a function of speed is:

$$D_i=\frac{2W^2}{\rho V^2 S \pi eA}=\frac{2W^2}{\rho V^2 \pi eb^2}$$

where $b$ is the wing span, $\rho$ is air density and $S$ is the wing reference area. As you can see, the induced drag itself decreases with increasing airspeed for constant weight.

Higher Aspect ratio wings generates greater lift but creates less induced drag

Higher aspect ratio wing does not generate greater lift. It just generates more efficient lift, where efficiency here means less induced drag (see first equation, notice how $C_{D_i}$ decreases with increasing $A$). Note: efficiency here ignores structural penalties, skin friction drag and stall characteristics, which are adversely impacted with higher aspect ratio.

$\endgroup$
3
  • $\begingroup$ Can you define the variables more explicitly? What are $W,L,S,e,b$? I guess $V$ is airspeed and $\rho$ is air density? $\endgroup$ Commented Jun 1, 2020 at 18:00
  • $\begingroup$ @WaterMolecule Thanks, it's done. $\endgroup$
    – JZYL
    Commented Jun 1, 2020 at 18:04
  • 2
    $\begingroup$ For people like me who have studied physics, but not much aerodynamics, this clarifies things a lot. +1 $\endgroup$ Commented Jun 1, 2020 at 18:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .