The formula for drag force is $F_D=\frac{1}{2}\rho v^2 C_d A$. Why is the $\frac{1}{2}$ significant here? I think that the drag coefficient $C_d$ already serves the purpose.
Is it for a historical reason?
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3There is a good discussion of this factor in https://physics.stackexchange.com/q/487220/ – G.Lang Jan 11 '21 at 10:35
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2Does this answer your question? Why is the drag force proportional to $v^2$ and defined with a factor of $1/2$? – Agnius Vasiliauskas Jan 11 '21 at 10:55
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I believe it was initially shown that the drag force is proportional (under certain conditions) to the dynamic pressure $q$ and the frontal (projected) area $A$. The proportionality constant of this relationship is what we call a drag coefficient $C_D$.
$$F_D\propto qA\quad\Leftrightarrow\quad F_D=C_DqA$$
Nice and neat. It turns out that this paramter $C_D$ depends largely on frontal shape in particular but also on several other factors involved with the flow over the surface.
Now since the dynamic pressure already is defined as $q=\frac12\rho v^2$, there comes the half.
$$F_D=C_DqA=C_D\,\underbrace{\frac12 \rho v^2}_qA$$
The drag coefficient could have been defined to include the half. But that would remove the visually clear involvement of the dynamic pressure in the equation. Also, since the drag coefficient now is involved in many other relationships throughout aerodynamical science, changing its definition in order to simplify this equation might complexify other relationships.
Steeven
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