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This question already has an answer here:

While using the Bernoulli's equation to deduce the upthrust on an aircraft , a statement is always made that

The air particles going above th wing have to cover a greater distance in same time than those going below, so they travel at a higher speed which causes the pressure difference.

But why do they have to cover greater distance?
Why do the particles have to meet at the ends of wing again?

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marked as duplicate by RedGrittyBrick, Ben Crowell, John Rennie, Kyle Kanos, knzhou Dec 17 '18 at 13:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ They cover a greater distance because they are traveling along the shape of the wing, which has a longer arc length on top than on bottom. As for the assumption that the time is equal this is perhaps due to continuity. Keep in mind that the analysis is probably assuming laminar flow, no turbulence. That is the ideal scenario that a wing is designed for. $\endgroup$ – ggcg Dec 16 '18 at 19:01
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    $\begingroup$ That is neither her nor there. The point of the analysis is to illustrate how the flow behaves in the ideal situation. You seem to be mixing methods, and states. $\endgroup$ – ggcg Dec 16 '18 at 19:04
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    $\begingroup$ Using all capitals on the internet is considered to be like shouting rudely. $\endgroup$ – StephenG Dec 16 '18 at 19:11
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    $\begingroup$ Possible duplicate of How much effect does the Bernoulli effect have on lift? $\endgroup$ – RedGrittyBrick Dec 16 '18 at 19:23
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    $\begingroup$ @alephzero, then answer the question. I was offering a possible explanation, not an answer. Please, get your points. $\endgroup$ – ggcg Dec 16 '18 at 19:39
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You are right to ask why it is true, because it is not true.

Many people have heard it and repeated it, but it has a name: The "equal time fallacy".

There is an excellent discussion here about why it is not true, and airplanes would not work if it were.

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