If you look at this picture, you’ll see the shock on the upper surface is further forward on the wing, compared to the lower shock. Why is that?
(This is at Mach 0.9, as shown at the bottom of the image)
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2$\begingroup$ Because air has accelerated faster on the upper surface than the lower one? $\endgroup$– sophitCommented Mar 13 at 5:50
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As Sophit says -- the air has accelerated more over the upper surface, so the shock happens sooner.
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$\begingroup$ Ah okay, thanks. Just to refresh my memory, don’t the upper surface-normal shocks happen when the pressure recovery starts on the airfoil? $\endgroup$– WyattCommented Mar 13 at 13:57
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$\begingroup$ There is a region of supersonic flow in front of the shock. You can't see the forward boundary of this region. There is some back pressure associated with the pressure recovery. The location of the shock depends on how high the Mach number gets in the supersonic region and also the strength of the back pressure. It is a complex balance that works out differently on the upper vs. lower surface because this airfoil is at angle of attack (and/or is not symmetrical) such that it is producing lift. $\endgroup$ Commented Mar 13 at 16:14
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$\begingroup$ Oh okay, thanks. It makes sense that the bottom surface of an airfoil would have a shock, because the first front 1/4 of the bottom surface makes the same amount of lift as the last 3/4 of the lower surface. This makes the same effect as the upper surface as the high pressure goes to the low pressure (Bernoulli). (Got that from the comments of this answer) $\endgroup$– WyattCommented Mar 13 at 17:10