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Why is the drag of a typical plane flying less than the lift?

Lift formula $$L = \frac{1}{2}C_l\rho V ^ 2S$$

$C_l$ lift coefficient
$\rho$ air Density
$V$ air Speed
$S$ area

Drag formula $$D = \frac{1}{2}C_d\rho V ^ 2S$$ $C_d$ drag coefficient
$\rho$ air Density
$V$ air Speed
$S$ area

The two formulas are almost the same, so why is drag less than lift?

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    $\begingroup$ Okay... so I hope you can see that if air density, velocity, and surface area are all the same value, then the reason the values are different for lift and drag is because the lift and drag coefficients are different. And if drag is less than lift, it's because the drag coefficient is smaller than the lift coefficient. So what, exactly, are you unclear about in these equations? $\endgroup$
    – tpg2114
    Mar 7, 2020 at 0:50
  • $\begingroup$ @tpg2114 Why are these two coefficients different? $\endgroup$
    – enbin
    Mar 7, 2020 at 1:17
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    $\begingroup$ @enbinzheng Is there a reason you might think they should be the same? I guess I don't know where the question is coming from or what the underlying physical question is. $\endgroup$
    – tpg2114
    Mar 7, 2020 at 1:19
  • $\begingroup$ @tpg2114 I need to find them for different reasons. You pointed out that the coefficients are different, so why are they different? $\endgroup$
    – enbin
    Mar 7, 2020 at 1:24

6 Answers 6

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The two formulas are almost the same, so why is drag less than lift?

Drag is less than lift because aerospace engineers are generally competent at their jobs and so they design the shape of aircraft such that $C_l>C_d$

Not all shapes have that property. That is why aircraft don’t have random shapes. Their shapes are designed expressly to minimize drag and achieve their other goals.

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I find it helpful to think of another famous problem. You are carrying a large stone with a mass of 100kg. You walk 100m with it. How much work did you do to the stone? The answer, of course, is none at all. The force you applied to hold the stone up is perpendicular to the movement, which means no work is done. This, of course, is counter to our intuition because it sure as heck feels like you did a lot of work, but by the physics definition of work, we did nothing.

In this case you have a solid body holding up the stone. In the case of an aircraft, that column is made of air. However, the solid example points out that, in the ultimate case, there is really no limit to how much upward force you can exert without expending energy. The limiting factor is that we aren't perfect at holding things up with air. There are losses.

As a middle ground, consider a hovercraft. Hovercrafts use a skirt to hold a large volume of air underneath the vehicle. Its easy to see that their lift can be extraordinary. They really only need to replace the energy lost as the air seeps out under the edge of the skirt. These losses are associated with pressure and perimeter length, while the lift is associated with pressure and area. The larger a hovercraft gets, the closer it gets to the ideal.

Aircraft are designed similarity. They lack a skirt, but the skin of the wing is designed to generate as much of a pressure difference between above and below the wing as possible while displacing the minimum amount of air possible.

The full answer, of course, is a fluid mechanics "shut up and calculate" sort of situation. The shape of an airfoil is well understood and we can calculate how it behaves. However, hopefully those comparisons to lifting objects and hovercrafts help to demonstrate why there isn't a particular need for $C_d$ and $C_l$ to be all that similar.

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As another answer said, Drag is less than lift because aerospace engineers are generally competent at their jobs and so they design the shape of aircraft such that $C_l>C_d$

The main feature which achieves this is to have an airfoil which is longer than it is thick. Pressure differences $\rho$ above and below then have a lot more area $A$ to work on than pressure differences fore and aft. Since $F = \rho A$ the vertical force (lift) therefore tends to be stronger.

Streamlining obviously also helps reduce drag, but in general the technical derivation of the various pressure changes really only explains "how" designers make best use of them, rather than "why" they matter.

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The first important thing to note is that drag an lift are orthogonal to each other. The lift pulls the plane up and the drag pulls against the direction of the flight.
So at first glance those forces are independant of each other and as long as the lift is higher than the down force by gravitation, the plane will stay up no matter what the drag does.

That means that a plane could also fly if the drag is greater than the lift. According to https://en.wikipedia.org/wiki/Lift-to-drag_ratio the space shuttle has a lift to drag ratio of 1, which means drag and lift are equal. It is not designed to take off from the ground and it has a very high mass in comparison to regular planes. So if the space shuttle would be very light instead and had enough forward thrust by an engine you could even reduce the lift or increase the drag a bit and still get it off the ground. So you would have effectivly a higher drag then lift.

The following points are important to understand which problems engineers have to manage.
In planes it is obviously the wings that create the lift. If you would minimize the drag on a wing you won't have any lift. To create lift the wing has to have a certain shape that creates a pressure difference between the air above and below the wing. But with the different shape the drag increases as well. So the task of an enineer is, to have sufficient lift while minimizing the drag.
It also depends on the purpose of the plane. The wings of a glider would not be very useful on a military jet. The lift and drag would be too high for the intended speed. The military jet on the other hand depends on the engine to create the forward thrust that is needed for the wing to create sufficient lift.

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Think of the wing as like a wheel. If it had no friction it would have no drag at all.

It has some drag which dissipates energy, but if it is really long and skinny like this <======================> it has arbitrarily low drag. That's because the main cause of energy loss is in the wake vortices that form at the tips of the wing. If the amount of weight per square meter is very low, the speed can also be very low, which minimizes drag.

There are engineering reasons like strength and speed that constrain the "skinnyness" (or "aspect ratio"). But aircraft (like gliders) that do not need much speed, but do need very low drag, are made with a very high aspect ratio. But there's really no lower limit to the amount of drag.

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    $\begingroup$ Tip losses are seldom the most significant. Form drag, skin drag and full-span induced drag are all usually greater contributors. Aspect ratio helps, but it is far from the whole story. $\endgroup$ Mar 8, 2020 at 18:49
  • $\begingroup$ Why can a high aspect ratio reduce drag? $\endgroup$
    – enbin
    Mar 8, 2020 at 19:03
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    $\begingroup$ @enbinzheng: Please take all the time you need and read this e-book. It understands this subject much better than I do. Pay close attention to these subjects: drag, aspect ratio, and power curve. Again, read the e-book. It explains way better than I can. It is a valuable resource. $\endgroup$ Mar 8, 2020 at 21:44
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Note that this answer is just a comparison of the same wing and different angles of attack.

enter image description here

We all know that the wings have an angle of attack. Therefore, the wing is an inclined plane, hereinafter referred to as $A$.

As shown in the figure, the inclination angle of $A$ is $\theta$. The weight of the aircraft is $W$. $W$ has two components, one component $Wt$ is parallel to $A$ downward, and the other component $W_n$ is perpendicular to $A$ downward. The horizontal thrust applied by the aircraft engine is $T$. $T$ also has two components, one is an upward component $T_t$ parallel to $A$, and the other is a downward component $T_n$ perpendicular to $A$. $F$ is the force that $A$ applies to the wing.

It can be known from calculation that when $T = Wtan\theta$, $T_t $ and $W_t$ are two quantities of equal magnitude and opposite directions. They are balanced. On the other hand, the aircraft will accelerate the movement under the action of $T$, and finally make $F$ and $W_n + T_n$ balance.

From the formula $T = Wtan\theta$, as long as $\theta <\frac{\pi}{4}$, then $T <W$. Due to the angle of attack of the wings $\alpha<<\frac{\pi}{4} $, the aircraft can cruise with thrust levels much smaller than the weight of the aircraft.

So we can say that $T$ provides lift. Because in horizontal flight, the lift $L = \frac{T}{tan\theta}= -W$. Without $T$, there is no $L$.

Why is lift greater than drag?

Because $C_d∝tan\theta$, $C_l∝\frac{1}{tan\theta} $, and the angle of attack of the wing $\theta<<\frac{\pi}{4}$, so $C_l> C_d$, lift is greater than drag.

Of course, the actual situation is more complicated, and factors such as stall, airfoil, and low pressure at the top of the wing must be considered.

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    $\begingroup$ This fails to account for the fact that gliders have lift and drag without thrust. It also does not account for the fact that, were I to attach a drogue chute to an aircraft, I could drastically decrease its lift-to-drag ratio without requiring the plane to change angles of attack. I would indeed need more thrust in that situation if I wished to remain in straight and level flight, but L/D ratios work in non-straight-and-level flight as well. They are, in fact, a cornerstone of emergency procedures to deal with engine failures. $\endgroup$
    – Cort Ammon
    Mar 8, 2020 at 8:08
  • $\begingroup$ @CortAmmon Glider also needs thrust: 1. The glider must have an initial speed to fly. This initial speed requires thrust. 2. The glider in the sky also needs the updraft as thrust or the component of gravity as thrust. $\endgroup$
    – enbin
    Mar 8, 2020 at 10:17
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    $\begingroup$ Your non-standard definition of thrust complicates the issue. For anyone coming to read this answer: the definition of "thrust" being used here is any source of energy which put energy into the aircraft at any point in time previous to "now," as opposed to the usual definition which is the force caused by the motor at the current instant. $\endgroup$
    – Cort Ammon
    Mar 8, 2020 at 16:25
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    $\begingroup$ We've discussed this at length on previous questions where you have made the same assertions. Your assertions are similar in nature to "all lift is caused by the big bang." It is true that the system needs to be in an energized state in order to flight in straight and level flight, but that is as far as it can go. We've discussed the case of a meteor, which has lift and drag but your concept of "thrust' will be hard to define for a meteor. $\endgroup$
    – Cort Ammon
    Mar 8, 2020 at 18:23
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    $\begingroup$ And by using "thrust" in such a non-standard way, it can cause confusion for anyone who isn't familiar with your definition. I highly recommend learning the standard meanings for these terms. You will find that airplanes make much more sense when you do. $\endgroup$
    – Cort Ammon
    Mar 8, 2020 at 18:24

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