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 multuplying 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 (rhoughly) 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 airplaines people usually use the wing planform area -- but there are slightly different ways of defining that. That's not a huge 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.

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.