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I know that $c_{d}=0.03 + 0.095c_{l}^2$.

What is the glide ratio?

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  • $\begingroup$ It looks like a school exercice $\endgroup$
    – Manu H
    Commented Mar 18, 2018 at 12:04

2 Answers 2

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The glide ratio for a given angle of attack is the ratio of lift to drag. Both of these are proportional to the respective coefficients (with a proportionality constant of $\frac12\rho v^2A$), so the glide ratio is simply $c_{\mathrm L}/c_{\mathrm D}$.

Practically, what you're looking for may be the maximal glide ratio. Finding the global maximum of $$\frac{c_{\mathrm L}}{c_{\mathrm D}} = \frac{c_{\mathrm L}}{0.03+0.095 c_{\mathrm L}^2} = \frac1{c_{\mathrm D}}\sqrt{\frac{c_{\mathrm D}-0.03}{0.095}}=\sqrt{\frac1{0.095}\frac1{c_{\mathrm D}}-\frac{0.03}{0.095}\Bigl(\frac1{c_{\mathrm D}}\Bigr)^2}$$ should be a simple matter of algebra -- for the last expression you don't even need calculus, just finding the apex of a parabola.

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    $\begingroup$ Actually, the optimum is when $0.03 = 0.095c_l^2$, so it is at $c_l = 0.562$. $\endgroup$ Commented May 5, 2015 at 7:09
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$C_D = C_{D0} + K \cdot C_L^2$ therefore $C_{D0} = 0.03$ and $K = 0.095$

Maximum lift-to-drag ratio occurs at the condition for minimum drag and at this condition $C_D = 2*C_{D0}$ and $C_L = \sqrt{C_{D0}/K}$.

Therefore $\max(L/D) = C_L/C_D = 0.5 \cdot \sqrt{1/(K \cdot C_{D0})} = 9.4$

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