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Old 2nd Aug 2013, 08:32
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Flash131
 
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I previously posted this explanation elsewhere on this forum. Perhaps not the classic explanation of dihedral effect, but it works for me:

Consider the flow over an unswept, untapered wing with no washout. Let’s imagine a streamline (the path followed by an air 'particle') starting at some point on the leading edge of the starboard wing. Let’s further assume that the point is roughly mid span, so we can ignore ‘end-effects’. If the aircraft is slipping neither left nor right, the streamline will have no spanwise component; if we are looking down from on top of the aircraft, the streamline will appear aligned fore and aft.

Now consider what happens if the aircraft slips to the left. Our right wing streamline will now meet the wing trailing edge further outboard than it did with no slip. If the wing has dihedral, the trailing edge ‘seen’ by the streamline will therefore be higher (relative to the leading edge) than when there was no slip. So, in a left slip, the wing surface profile followed by our streamline will appear to be ‘nose-down’ relative to the profile it followed with no slip.The right wing streamline therefore sees a reduction in effective incidence. The equivalent streamline on the left wing leaves the trailing edge further inboard and therefore sees an increase in effective incidence. So the right wing lift decreases and the left wing lift increases, rolling the aircraft such as to oppose the slip.

Note that the above argument is independent of the attitude of the aircraft or the direction of gravity.

Last edited by Flash131; 2nd Aug 2013 at 08:34.
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