= Shifted parabolic drag polar
{title2=$C_D=C_{D,\min}+k(C_L-C_{L,*})^2$}
A \b[shifted parabolic drag polar] models <drag coefficient> with its minimum at a possibly nonzero <lift coefficient>:
$$
C_D=m+k(C_L-h)^2,\qquad m>0,\quad k>0.
$$
Here $m=C_{D,\min}$ and $h=C_{L,*}$ in the header notation. The shift separates the coefficient at minimum drag coefficient from the <zero-lift drag coefficient>, which is $C_{D0}=m+kh^2$.
For an original model, take $m=0.018$, $k=0.050$ and $h=0.20$. The coefficients at $C_L=0$, $0.20$ and $0.60$ are 0.020, 0.018 and 0.026. The smallest drag coefficient occurs at $C_L=0.20$, but that point does not maximize the <lift-to-drag ratio>.
Writing $f(C_L)=C_L/C_D$ and differentiating gives
$$
f'(C_L)=\frac{m+kh^2-kC_L^2}{\left[m+k(C_L-h)^2\right]^2}.
$$
For positive lift coefficients the unrestricted optimum is therefore
$$
C_L^{\rm opt}=s=\sqrt{h^2+m/k},\qquad
f_{\max}=\frac{1}{2k(s-h)}=\frac{s+h}{2m}.
$$
The last equality follows by multiplying by $s+h$ and using $s^2-h^2=m/k$. In the example, $s\simeq0.6325$ and the maximum ratio is about 23.12. An allowed lift interval must still contain $s$ before that model optimum is physically attainable.
The model can also be fitted as the ordinary quadratic $C_D=\beta_0+\beta_1C_L+\beta_2C_L^2$. When $\beta_2>0$, completing the square gives $k=\beta_2$, $h=-\beta_1/(2\beta_2)$ and $m=\beta_0-\beta_1^2/(4\beta_2)$. This makes the relation between a fitted quadratic and the shifted parameters explicit.
A <wind tunnel> can sample both sides of the proposed minimum while keeping flow conditions and normalization fixed. A small residual at sampled points does not establish accuracy through <aerodynamic stall>. Nor should $k(C_L-h)^2$ automatically be identified as pure induced drag: the usual ideal finite-wing relation depends on $C_L^2$, and a fitted shifted polar may combine several effects.<Induced Drag Coefficient by NASA Glenn Research Center>[Drag Coefficient and total-drag equations]{ref}
Back to article page
