With reference to aircraft aerodynamics:

As the aircraft speed increases the aerodynamic load increases in a mathematical relationship proportional to the air density and the square of velocity.

What does the above statement mean in layman's terms? Consider explaining it to a high school student.

Also, what's the simplest explanation for the drag formula too?

Fd=1/2 *Density *Velocity^2 *Coeff. drag *Area

  • $\begingroup$ First, make it relatable. (See my answer to aileron effectiveness question) After the basic concept is clear then you can delve into the underlying equations that help us quantify the phenomenon. $\endgroup$ Aug 1, 2020 at 16:40

5 Answers 5


As the aircraft speed increases the aerodynamic load increases in a mathematical relationship proportional to the air density and the square of velocity.

What does the above statement mean in layman's terms?

  • When a body (here an aircraft) is moving, an airflow is created surrounding the aircraft. This leads to aerodynamics forces which are forces acted by air on the aircraft body. It is common and useful to separate it into different components: Drag (opposite to movement) and Lift (perpendicular to movement).

  • Your statement mean: theses forces become more and more strong when flow speed and gas density increase. To understand this, you can write Newton's second Law (which you have/will probably study in high school) :

$$ \sum\vec{F} = m\vec{a} $$

  • As you see, a force is linked to the mass of the fluid or solid considered. If the density (mass per unit of volume) increase, the mass of air inside a same volume is raising too, as well as aerodynamics loads. To find the square velocity you need to manipulate differential (which you may not study during high school) but you can think about kinetic energy of the flow, which is the energy due to air motion (more that energy is high and more aerodynamics forces will be):

$$E_k = \tfrac{1}{2}mV^2$$

Also, what's the simplest explanation for the drag formula too?

$F_D = \frac{1}{2} \rho V^2 A C_D$

  • This formula express Drag dependency with other fluid quantities. In fact, coefficient $C_D$ comes from dimensional analysis. The goal is to find a metric (without dimension) to evaluate and analysis drag between different shape, body, airfoils, settings, etc..


Even if you have $\rho$ and $V^2$ in this formula, this does not mean that the drag $F_D$ varying in the same way that these quantities. Indeed, coefficient $C_D$ is dependent of other non dimensional parameters such as $R_e$ (Reynolds number), $\alpha_i$ (angle of attack of fluid), etc

 An increase in velocity $V$ will affect $V^2$ but can also lead to a different value of $C_D$..

I tied to popularized some concepts and hope these elements can help you (but I'm aware these can be quite technical for high school level).


Part 1:

As an aircraft travels faster, drag increases more in thicker air (usually lower to the ground) more than it does in thinner air (usually higher in the atmosphere).

Also, doubling speed quadruples drag.

Part 2:

Force of Drag (self-explanatory) =

Density (the thickness of air or how many molecules there are in a cubic meter) * Velocity^2 (the aircraft’s speed multiplied by its own value)

divide this value by 2

then multiply by Coefficient of Drag (basically, a value that shows how easily an object will slide through the air)

finally, multiply by Area (how much surface is actually in contact with the air. Obviously, a bigger object will interact with more air, increasing drag).

Hope this helps!

  • 2
    $\begingroup$ "how many molecules there are in a square meter" – should't that be a cubic meter? $\endgroup$ Aug 2, 2020 at 18:45
  • $\begingroup$ @Peter Kämpf: Slip of the tongue, so to speak. Yes. Cubic meter is correct. $\endgroup$ Aug 2, 2020 at 18:49
  • $\begingroup$ Double velocity = 4 times the drag is the most common sense $\endgroup$ Aug 5, 2020 at 15:49

Trying a less mathematical way to make the equation plausible:

$D = q \times A_{ref}\times c_D $

(with $q = \frac{\rho}{2}v^2$)

$q$ is the so-called "dynamic pressure". That's the increase in pressure you get from stopping the air coming towards your airplane (or car, or whatever) down. As long as you're not getting close to the speed of sound, that's the pressure increase you get at the tip of your aircraft. That air is trying to slow your aircraft down because that air wanted to just go on as it did but you just pushed an aircraft in its face. This pressure doubles as the air density doubles (because there's more mass of air being pushed around), but it quadruples as velocity doubles (because that air is being slowed down a lot harder if it's faster -- the square comes out of the kinetic energy equation).

Now, pressure is force per area. The bigger your aircraft, the more of it you'll get. So if you're multiplying it by your reference area, you get the force that would push you back if that dynamic pressure was acting on your whole reference area. A cube moving straight through air with one of its faces in front is relatively close to generating that much drag.

But of course a decent aircraft is a little better than a cube. It's trying to slip through with as little disturbance as it can, and that's why a good aircraft has a low drag coefficient. So the drag coefficient effectively tells us how much drag the aircraft produces compared to (roughly) a cube moving straight through the air, where one face is as big as $A_{ref}$. For a passenger aircraft these days, $c_D$ is somewhere from 0.02 to 0.03 during cruise.

$D = q \times A_{ref}\times c_D $

$D \approx D_{cube} \times c_D $

The nice thing about this:

$c_D$ is independent of speed, density or size -- it's just a function of the aircaft shape. So if you build a wind tunnel model (which is smaller) and put it in a wind tunnel (which is slower than the real flight), your $c_D$ is mostly the same as for a real aircraft (ignoring Reynolds number effects, they're for another day).

Two things to remember:

1: The reference area is not some fundamental number. For a sphere or a cube, people take the cross-section area, for a car, it's usually the frontal area (i.e. the size of the shadow if you shine a lamp at it from very far ahead, against a wall), and for airplanes people usually use the wing planform area -- but there are slightly different ways of defining that. That's not a problem though, as long as you remember which area was used for the $c_D$ you're working with.

2: For an aircraft in particular, $c_D$ is not constant, because it depends on $c_L$ (the lift coefficient). For a car, that's much easier because it is mostly constant.


The explanation I would like to present is

We are surrrounded by Air Molecules which has a certain pressure exerted on us. Therefore when the aircraft moves forward there is a resistance which is stopping the motion aka air resistance. This is the drag which is the pressure exerted on the body to move forward. When this is overcame the aircraft moves forward.

When the aircraft speed increases the kinetic energy of the air molecules increases thereby increasing the pressure acting on the body which in turn increases the drag or the air resistance. Kinetic energy the velocity is squared thereby it can be understood.

Considering idea gas it can be seen that density is directly proportional to pressure . Therefore this pressure has an effect on force acting on the body with respect to area. Therefore if density is increased pressure increases and force(drag) increases.

PS : - Please correct me if i am wrong.

  • $\begingroup$ This is by far the best answer because it doesn't just "throw a bunch of equations" at the recipient. Unfortunately people often just start sprouting off equations rather than trying to explain the underlying effect like you have done. This shows a lot better understanding rather than just "the equation has a little 2 next to the v"!!! $\endgroup$ Aug 7, 2020 at 17:30

This explanation assumes that the high school students know the concepts of kinetic energy and work

If we look at the kinetic energy of the air in front of the object, we note that:

$$ E = \frac{1}{2} \cdot m \cdot V^2 \tag{1} $$

If we assume that only the air within distance $ds$ times a frontal surface $S$ gets affected, we can write $m$ as:

$$ m = \rho \cdot ds \cdot S \tag{2}$$

If we plug $(2)$ into $(1)$ to get:

$$E = \frac{1}{2} \cdot \rho \cdot ds \cdot S \cdot V^2 \tag{3}$$

We also know that work is equal to (dragging) force $[D]$ times distance:

$$ W = D \cdot ds \tag{4} $$

Due to the energy balance, the energy lost by the air $E$ is equal to the work $W$ done on the object : $$W = E \tag{5}$$ We then subsitute $(4)$ for the left hand side and $(3)$ for the right hand side: $$ D \cdot ds = \frac{1}{2} \cdot \rho \cdot ds \cdot S \cdot V^2 \tag{6} $$ And we can divide both left and right by $ds$ to obtain: $$ D = \frac{1}{2} \cdot \rho \cdot S \cdot V^2 \tag{7} $$

However, this assumes that all the energy contained within our air packet $m$ is completely transferred to the object. And the influence of the object is indeed limited to our air packet $m$ (with size $ds \times S$) . This is usually not the case, and to indicate to what degree this happens we add a correction factor $C_D$ to $(7)$:

$$ D = C_D \cdot \frac{1}{2} \cdot \rho \cdot S \cdot V^2 $$

This also gives you a nice bridge to explain more about the meaning $C_D$ and how it varies for different shapes. In what situation will you have a $C_D$ of 1? What can you do to reduce $C_D$? What does it mean when $C_D$ is larger than 1?

enter image description here

Image from very useful Wikipedia page of Drag Coefficient

  • $\begingroup$ Not really agree with the statement “All the energy contained in the air is transferred to the object.” How energy could be transferred to the solid if it is still motionless and without heat transfert ? (First thermodynamics law) This is more likely that gas kinetic energy is changed in pressure form (energy conservation eq.) leading to aerodynamic forces like drag. $\endgroup$
    – Acsed.
    Aug 3, 2020 at 17:22
  • $\begingroup$ See the image, the case at the top, in the 'idealized' case all air completely stops, and all the kinetic energy is transferred to the plate through the drag force. This allows us to make the energy balance between kinetic energy and work. For the energy balance the intermediate steps don't matter. If you want to make the intermediate step to pressure it's possible, but it only complicates the derivation, and leads to the same result. $\endgroup$
    – ROIMaison
    Aug 3, 2020 at 21:21
  • $\begingroup$ I agree with the formulas you gave, but not really the explanation (I know, I split hairs a little bit here). If the plate do not move, the object does not apply any work nor get some kinetic energy at all. For me, all energy transformation happens inside air flow itself.. $\endgroup$
    – Acsed.
    Aug 3, 2020 at 22:20
  • $\begingroup$ @Acsed. there has to be a transfer to the object. How do you explain the transfer using pressure if there is no movement (in the end pressure is nothing more than force over surface, so the explanation is fundamentally not different)? Please feel free to post your version of the answer, so you have more space to explain. $\endgroup$
    – ROIMaison
    Aug 4, 2020 at 7:16
  • $\begingroup$ « in the end pressure is nothing more than force over surface, so the explanation is fundamentally not different. » Indeed, at the end, we arrived to the same conclusion. I just noticed that applying force is different to transfer energy. By considering the object itself, it is not moving, therefore its kinetic energy is zero and no work could be done from it. On the other side, many things happened in the flow where kinetic energy is changed to pressure form. Anyway, this is a nice debate (even if we need, as you said, more space to explain our views ;) $\endgroup$
    – Acsed.
    Aug 4, 2020 at 16:08

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