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Part 1

Observing how induced drag is derived, induced drag is loss of kinetic energy due to total vorticity that is produced by wing.

I want to find induced drag of finite wing with L=const over span.

Finite wing with constant lift distribution is just math concept,not existing in reality, cant be tested in any experiment because even wing has lots of twist,the wingtip area cannot satisfy Γ=const, too much pressure leaking.

This is formula for local induced angle of attack, if L=const over span, circulation Γ=const, dΓ/dy=0, implies alpha= 0 thus Dind.=0

-I am intersted will in Treffz plane get same result, has equation "dΓ/dy" inside?- enter image description here

Part 2

By definition airfoil has zero induced drag, I want to check this mathematically.

Airfoil is infinite wing with L=const over span, I think this integral (picture below) for alpha induced is valid. Same like in part 1 if dΓ/dy=0, Dind=0.

But when I use this formula: Di= L* 2 / (0.5 x ρ x π x V* 2 x b* 2) and put infinity instead b and L, (infinity wingspan will give infinity lift), I get Di=∞/∞

If I use L'Hospital's rule will I get Dind=0?

For specific case where L=W, indeed this formula get Dind=0, but I want to analyze in general case. enter image description here

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Even if lift distribution were constant spanwise, lift would go anyway to zero at the end of the wing because... well, there's no wing anymore and there you would have a huge variation of lift i.e. huge generation of induced drag. Actually this sudden variation of lift would be so big to produce a much bigger induced drag in respect to a smoother (ideally elliptical) spanwise lift distribution.

Mathematically, the integral for the induced drag (or induced speed or induced angle) is extended from $-\infty$ to $+\infty$, but since the integrand is zero outside the wing, then the integration is normally written between $±b$. Anyway the integration (which is not a normal integration but a Cauchy principal value integration) is done approaching the wing's tips from $\infty$ and therefore what happen just outside the tip does matter. In particular, just outside the wing it must be $\Gamma=0$ because there there's no wing and lift is null.

So in order for $\Gamma$ to be:

  1. zero at the tip;
  2. and constant everywhere, i.e. with $\frac{d\Gamma}{dy}=0$

then it must be $\Gamma=0$ everywhere. This is the only lift distribution which is mathematically compatible with 1. and 2.

In the equation for the induced drag, $b$ should be on the denominator side:

$D_i = \frac{L²}{q \pi e b²}$

Therefore if $b\rightarrow \infty$ then $D_i \rightarrow 0$.

The value of $L$ has nothing to do with this result because for an infinite wing (aka airfoil) the whole maths for the induced drag just doesn't exist in the first place: the integration of the pressure in the drag's direction is just zero for an airfoil and therefore there's no need to elaborate any theory for it. So, when we deal with the induced drag theory, the wing cannot be infinite by mathematical definition and therefore neither the lift.

Wings with (almost) constant spanwise lift distribution can actually be built playing with the chord distribution and/or the twist. NASA played with the twist in the past and got a quite constant spanwise lift distribution:

lift distribution on a highly twisted wing

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    $\begingroup$ At the end of the wing, lift is constant not zero. This just mathematical concept, 100% abstract, just like airfoil. Integration shows whenever you have dΓ/dy=0, induced drag become zero. Yes b is on the denominator side, but infinity wingspan will produce infinity lift, if you don't introduce a specific condition L=W. $\endgroup$
    – user74815926
    Dec 1, 2022 at 10:42
  • $\begingroup$ "Wings with (almost) constant spanwise lift distribution can actually be built playing with the chord distribution" Key word is almost, this wing has no dΓ/dy=0, especially at the wingtip area. Formula dont know what is one mm out of wingtip, dont take this in consideration. $\endgroup$
    – user74815926
    Dec 1, 2022 at 10:53
  • $\begingroup$ I updated my answer, let me know if it's clear. $\endgroup$
    – sophit
    Dec 1, 2022 at 12:36
  • $\begingroup$ "these are the only two mathematically valid solution of the problem." You mean integral is not undefined or indeterminate expressions? $\endgroup$
    – user74815926
    Dec 1, 2022 at 21:39
  • $\begingroup$ I mean that the only possibility for 1) $\Gamma$ @ tip to be zero and 2) $\frac{d\Gamma}{dy}$ to be zero, is that $\Gamma$ must be zero everywhere. Just basic mathematics, nothing to do with induced drag or whatever. If a variable is zero somewhere and its derivative is zero too, then this variable can only be zero. Think about the definition of derivative: the variation of a quantity. If this variation is zero then the quantity doesn't change and if this quantity is zero somewhere then it must be zero everywhere, by definition of null derivative. $\endgroup$
    – sophit
    Dec 1, 2022 at 22:02

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