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What is the relation between sweep angle and lift-to-drag ratio?

Ans does depend on wing planform?

For subsonic speeds.

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  • $\begingroup$ I thought it didn't change L/D, but instead shifted the same behavior to a higher speed speed, and of course shockwaves. $\endgroup$ – DKNguyen Nov 16 '20 at 15:56
  • $\begingroup$ are you sure about that? $\endgroup$ – user52248 Nov 16 '20 at 17:05
  • $\begingroup$ No, if I was I would have put it as an answer. $\endgroup$ – DKNguyen Nov 17 '20 at 1:00
  • $\begingroup$ no one answer ? $\endgroup$ – user52248 Nov 19 '20 at 6:26
  • $\begingroup$ Yeah, seems oddly quiet. $\endgroup$ – DKNguyen Nov 19 '20 at 14:08
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This is my take on this.

We assume that an infinite-span swept wing is under compressible flow, and that the flow acts on it as in Figure 1 below:

Swept wing flow visualization

With that, we can now utilize the equation of motion to find the relation between the sweep angle and the coefficient of lift, which in turn, will make the relation with the ratio L/D

From the equation of motion, we have:

$$(1-M^2_{\infty}\cos^2\Lambda)\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2}$$

Now, only the perpendicular component acts we have that:

$$(u'_c)_n = \frac{\partial \phi}{\partial x} = \frac{1}{\sqrt{1-M^2_{\infty}\cos^2\Lambda}} \frac{\partial \Phi}{\partial x} = \frac{(u'_i)_n}{\sqrt{1-M^2_{\infty}\cos^2\Lambda}}$$ $$C_{p_c} = -2\frac{\cos\Lambda(u'_c)_n}{U_\infty} = -2\frac{\cos\Lambda(u'_i)_n}{U_\infty\sqrt{1-M^2_{\infty}\cos^2\Lambda}}$$

Now, since:

$$C_L = \frac{1}{c}\oint C_p dx$$

We have that:

$$C_L = \frac{C_{Li} }{\sqrt{1-M^2_{\infty}\cos^2\Lambda}}$$

The above equation shows that the lift coefficient is sweep angle dependent, therefore, the ratio L/D is also sweep angle dependent.

This is a very complex side of aerodynamics and understanding it completely requires some study. I recommend studying Chapter 11 from "Fundamentals of Aerodynamics" by J. Anderson (a classic) and Chapter 8 from "Aerodynamics for Engineering Students" by E.L. Hughton.

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  • $\begingroup$ Does sweep reduce L/D or not? $\endgroup$ – user52248 Nov 20 '20 at 16:22
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    $\begingroup$ @Сократ Second last paragraph, but assumes things about the Cd relative to C_L $\endgroup$ – DKNguyen Nov 20 '20 at 16:49
  • $\begingroup$ @DKNuyen all gliders use zero sweep,maybe that show that sweep reduce L/D ratio?gliders most import things is to has L/D as much as posssible.. $\endgroup$ – user52248 Nov 22 '20 at 6:15
  • $\begingroup$ please consider typing the equations out with mathjax, rather than posting images. it is helpful both for people that use screen readers and for the search results. $\endgroup$ – Federico Nov 22 '20 at 7:38

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