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For a given speed, a swept wing is a fatter alternative to a straight, thin wing. But why? Does not the sweep lengthen the wing and make it heavier? Is the swept wing chosen for the extra fuel tank volume?

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Could you quantify the justification? Like, let's say we have this beam of depth x............

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  • $\begingroup$ You add an edit to your question, I add an edit to my answer. $\endgroup$ – Peter Kämpf Feb 15 at 8:40
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Yes, sweep makes a wing heavier and less efficient. But making the airfoil of a straight wing thinner runs into diminishing returns and also will drive up mass.

Since a thin wing needs to create the same lift as a thicker one, only the local speed increase from thickness can be tackled by making the airfoil thinner. The speed increase necessary for lift creation needs to be maintained, so the thickness contribution becomes a minor factor for the wings. However, a thinner wing requires a heavier spar to carry the same lift loads, so more lift needs to be created by the thinner wing to carry that extra weight. Now further thinning will drive up the speed increase from lift which will eat up the gains from thinning the wing.

If you look at parametric weight equations for airliner wings, the relative thickness at the wing root $\delta$ is part of them and generally occurs in a factor like $$m_{Wing}\propto\frac{1}{\delta^{\,0.36}}$$ Now it is important to stress that this is valid for the range of thicknesses from which that relation had been derived, something between $\delta$ = 14% and 16%. If you drop below 14%, the exponent will most likely go up since the mass of the spar alone should be proportional to $\frac{1}{\delta}$ and becomes more dominant the thinner the wing becomes. But if root thickness is decreased from 15% to 14%, that means we stay within the limits of the statistical model, the wing mass increase should be 2.5%.

The same model gives the influence of sweep on wing mass as $$m_{Wing}\propto\frac{1}{cos\varphi^{\,0.57}}$$ Based on 40° wing sweep, every decrease in spar thickness of 1% requires a sweep decrease of about 2.9° in order to keep wing weight constant, based on that relation. If that 1% thickness decrease allows you to desweep the wing by 3° or more, you come out ahead.

As always, a compromise is the best solution. Airliners of the propeller era had thicker wings than jet airliners. Some sweep and some thickness reduction will allow to push the critical Mach number to the highest value. Tank volume is indeed an important factor and is responsible for the comparatively low aspect ratios of early jet airliners.

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  • $\begingroup$ Does the exponent of the second equation also change? $\endgroup$ – Abdullah Feb 15 at 9:53
  • $\begingroup$ @Abdullah All these exponents depend heavily on the composition of the equation. When - for example - dive speed is part of the mix, it helps to "hide" the influence of sweep since both are closely correlated. So the exponent will already change between equations (Roskam - Douglas - Boeing - Datcom - LTH, and I'm sure the Russians have their own version, too) and equally when a different mix of planes is used. $\endgroup$ – Peter Kämpf Feb 15 at 12:26
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Also, you have to think that the air does not hit the wing at a 90* angle to the leading edge on a swept wing. The shape of the wing ribs is not necessarily the shape that the air is being in contact with. It is probably going to be a longer trip than the shape of the rib in certain areas of the wing, like close to the tip, where both leading and trailing edges have some sweep to them.

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