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Today I encountered a problem. There is a book saying, "Why does the lift coefficient of a swept-wing airplane decrease with the increase in the weight of the airplane?" I cannot think of the reason. Isn't the lift only coefficient related to the state of the wing?

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    $\begingroup$ Welcome to Av.se! You might consider using a more descriptive title for your question; as is, reading the title doesn’t really indicate what specifically your question is about. $\endgroup$ – Ralph J May 4 '18 at 4:24
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A swept wing has less torsional strength than a straight wing, ie there's less structure at 90d to the root supporting the tip of a swept wing.
Thus a swept wing twists when it's loaded, which unloads the tips and thus generates less lift at higher weights.

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  • $\begingroup$ Unless it's swept forward! <grin> bing.com/images/… $\endgroup$ – Charles Bretana May 4 '18 at 14:03
  • $\begingroup$ No @Charles Bretana forward swept wings have very different torsional and stress-strain characteristics and because different sweep angles are optimized for different Mach numbers, the lift generated is not really proportional or comparable to back swept wings. $\endgroup$ – Jihyun May 4 '18 at 16:47
  • $\begingroup$ Who said it was? All I am saying is that it does not twist in a manner which "unloads the wing tips and thus generates less lift at higher weights". In fact, unless constructed to eliminate twist due to the airloads, it does exactly the opposite. But it's a moot point because forward swept wings are the exception, not the general rule. $\endgroup$ – Charles Bretana May 4 '18 at 23:21

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