# Drag equation

In fluid dynamics, the drag equation is a formula used to calculate the force of drag experienced by an object due to movement through a fully enclosing fluid. The equation is:

${\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,u^{2}\,C_{D}\,A}$
${\displaystyle F_{D}}$ is the drag force, which is by definition the force component in the direction of the flow velocity,
${\displaystyle \rho }$ is the mass density of the fluid,[1]
${\displaystyle u}$ is the flow velocity relative to the object,
${\displaystyle A}$ is the reference area, and
${\displaystyle C_{D}}$ is the drag coefficient – a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. In general, ${\displaystyle C_{D}}$ depends on the Reynolds number.

The equation is attributed to Lord Rayleigh, who originally used L2 in place of A (with L being some linear dimension).[2]

The reference area A is typically defined as the area of the orthographic projection of the object on a plane perpendicular to the direction of motion. For non-hollow objects with simple shape, such as a sphere, this is exactly the same as a cross sectional area. For other objects (for instance, a rolling tube or the body of a cyclist), A may be significantly larger than the area of any cross section along any plane perpendicular to the direction of motion. Airfoils use the square of the chord length as the reference area; since airfoil chords are usually defined with a length of 1, the reference area is also 1. Aircraft use the wing area (or rotor-blade area) as the reference area, which makes for an easy comparison to lift. Airships and bodies of revolution use the volumetric coefficient of drag, in which the reference area is the square of the cube root of the airship's volume. Sometimes different reference areas are given for the same object in which case a drag coefficient corresponding to each of these different areas must be given.

For sharp-cornered bluff bodies, like square cylinders and plates held transverse to the flow direction, this equation is applicable with the drag coefficient as a constant value when the Reynolds number is greater than 1000.[3] For smooth bodies, like a circular cylinder, the drag coefficient may vary significantly until Reynolds numbers up to 107 (ten million).[4]

## Discussion

The equation is easier understood for the idealized situation where all of the fluid impinges on the reference area and comes to a complete stop, building up stagnation pressure over the whole area. No real object exactly corresponds to this behavior. CD is the ratio of drag for any real object to that of the ideal object. In practice a rough un-streamlined body (a bluff body) will have a CD around 1, more or less. Smoother objects can have much lower values of CD. The equation is precise – it simply provides the definition of CD (drag coefficient), which varies with the Reynolds number and is found by experiment.

Of particular importance is the ${\displaystyle u^{2}}$  dependence on flow velocity, meaning that fluid drag increases with the square of flow velocity. When flow velocity is doubled, for example, not only does the fluid strike with twice the flow velocity, but twice the mass of fluid strikes per second. Therefore, the change of momentum per second is multiplied by four. Force is equivalent to the change of momentum divided by time. This is in contrast with solid-on-solid friction, which generally has very little flow velocity dependence.

## Relation with dynamic pressure

The drag force can also be specified as,

${\displaystyle F_{D}\propto P_{d}A}$

where, PD is pressure exerted by fluid on area A. Here the pressure PD is referred to as dynamic pressure due to kinetic energy of fluid experiencing relative flow velocity u. This is defined in similar form as kinetic energy equation:

${\displaystyle P_{d}={\frac {1}{2}}\rho u^{2}}$

## Derivation

The drag equation may be derived to within a multiplicative constant by the method of dimensional analysis. If a moving fluid meets an object, it exerts a force on the object. Suppose that the variables involved – under some conditions – are the:

• speed u,
• fluid density ρ,
• viscosity ν of the fluid,
• size of the body, expressed in terms of its frontal area A, and
• drag force FD.

Using the algorithm of the Buckingham π theorem, these five variables can be reduced to two dimensionless parameters:

Alternatively, the dimensionless parameters via direct manipulation of the variables.

That this is so becomes apparent when the drag force FD is expressed as part of a function of the other variables in the problem:

${\displaystyle f_{a}(F_{D},\,u,\,A,\,\rho ,\,\nu )\,=\,0.\,}$

This rather odd form of expression is used because it does not assume a one-to-one relationship. Here, fa is some (as-yet-unknown) function that takes five arguments. Now the right-hand side is zero in any system of units; so it should be possible to express the relationship described by fa in terms of only dimensionless groups.

There are many ways of combining the five arguments of fa to form dimensionless groups, but the Buckingham π theorem states that there will be two such groups. The most appropriate are the Reynolds number, given by

${\displaystyle \mathrm {Re} \,=\,{\frac {u\,{\sqrt {A}}}{\nu }}}$

and the drag coefficient, given by

${\displaystyle C_{D}\,=\,{\frac {F_{D}}{{\frac {1}{2}}\,\rho \,A\,u^{2}}}.}$

Thus the function of five variables may be replaced by another function of only two variables:

${\displaystyle f_{b}\left({\frac {F_{D}}{{\frac {1}{2}}\,\rho \,A\,u^{2}}},\,{\frac {u\,{\sqrt {A}}}{\nu }}\right)\,=\,0.}$

where fb is some function of two arguments. The original law is then reduced to a law involving only these two numbers.

Because the only unknown in the above equation is the drag force FD, it is possible to express it as

${\displaystyle {\frac {F_{D}}{{\frac {1}{2}}\,\rho \,A\,u^{2}}}\,=\,f_{c}\left({\frac {u\,{\sqrt {A}}}{\nu }}\right)}$

or

${\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,A\,u^{2}\,f_{c}(R_{e}),\,}$      and with     ${\displaystyle C_{D}\,=\,f_{c}(R_{e}).}$

Thus the force is simply ½ ρ A u2 times some (as-yet-unknown) function fc of the Reynolds number Re – a considerably simpler system than the original five-argument function given above.

Dimensional analysis thus makes a very complex problem (trying to determine the behavior of a function of five variables) a much simpler one: the determination of the drag as a function of only one variable, the Reynolds number.

The analysis also gives other information for free, so to speak. The analysis shows that, other things being equal, the drag force will be proportional to the density of the fluid. This kind of information often proves to be extremely valuable, especially in the early stages of a research project.

## Experimental methods

To empirically determine the Reynolds number dependence, instead of experimenting on huge bodies with fast-flowing fluids (such as real-size airplanes in wind-tunnels), one may just as well experiment on small models with more viscous and higher flow velocity, because these two systems are similar.