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In the answer to this question, Peter Kampf stated that:

"In the majority of cases it is aerodynamically advantageous to vary sweep with relative thickness, but it is not structurally and economically efficient."

Why is this the case?

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Here's a high speed reason:

TL;DR: They counter each other's effect on the wave drag

Thickness versus wave drag

Basically, a thicker airfoil has a bigger curvature, and will accelerate the airflow around it more. This leads to more lift for the same speed. However, at cruise speed, closer to M=1, sonic effects will occur. This will lead to increased drag.

This can seen from the figure below, with increasing $t/c$, the value of $c_d$ is higher, and the rapid increase associated to wave drag starts earlier. The value of $M$ at which this occurs is called the Drag Divergence Mach number, $M_{DD}$.

CDW vs t/c Source: NASA supercritical airfoils: A matrix of family-related airfoils

Sweep versus wave drag

When a sweep angle is present, we can decompose the velocity in two components, of which only the one perpendicular to the chordline ($V_\bot$) is effecting lift (and thus wave drag).

By increasing the sweep, we can reduce the value of $V_\bot$, allowing the airfoil to experience a lower value of $M$, thus delaying the drag rise.

CDW vs sweep

Source: Wing Shape Multidisciplinary Design Optimization

Combined effects

As you can see, for wave drag, the two counteract eachother. If you have a certain value of $t/c$ and a certain $\Lambda$, this will lead to a certain $M_{DD}$. If for some reason, you want to increase the value of $t/c$, the wave drag increase will come earlier, leading to a lower $M_{DD}$. To maintain the same $M_{DD}$, you could counteract the effects of a larger $t/c$ by increasing $\Lambda$.

This is clearly shown by the emperic Korn-Lock-Mason method, which states:

$$ M_{DD} = \frac{K_A}{cos \Lambda} - \frac{t/c}{cos^2 \Lambda} -\frac{c_l}{cos ^3\Lambda} $$

This also shows the effects of lift on the value of $M_{DD}$.

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