Does wing with bell lift distribution has negative lift at tips during cruising flight?

Looking at this diagram, tips has very small lift but it is not negative.

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https://ntrs.nasa.gov/citations/20180004462 open pdf

  • $\begingroup$ +1 for illustrations $\endgroup$
    – Kolom
    Aug 10, 2020 at 19:31

2 Answers 2


1. Summary

As your first plot shows, the optimal Bell-Shaped Lift Distribution (BSLD) is positive throughout its span for an untwisted planform. What's different is that the downwash near the wingtip is negative for the BSLD, thereby giving rise to an apparent thrust and decreasing the overall induced drag compared to the Elliptical-Shaped Lift Distribution (ESLD).

However, for a non-optimal BSLD, you may get negative lift near the wingtips. And when there is geometric/aerodynamic twist, then you will almost certainly get negative lift over some portion of the span at AOAs lower than your design point.

2. Reasoning

Inspired by an Aerocrafty blog (be cautious that the derivation in that article contains some minor errors), we can derive an approximate BSLD without resorting to complex analysis. From Lifting-Line Theory, section lift ($L^{'}$) and upwash ($w$) are given as follows:

$$L^{'}(y_0) = \rho_\infty V_\infty \Gamma(y_0)$$

$$w(y_0) = -\frac{1}{4\pi} \int_{-b/2}^{b/2}\frac{(d\Gamma/dy)}{y_0-y}dy$$

where $y_0$ is the lateral span coordinate, $b/2$ is half span, $\Gamma(y_0)$ is the circulation distribution of the bound vortex, $\rho_\infty$ is the free-stream density, and $V_\infty$ is the free-stream airspeed.

Through a transformation of coordinates (from Cartesian to polar):


We can write any general circulation distribution as a Fourier sine series:

$$\Gamma(y) = \Gamma(\theta) = \sum_{n=1}^{\infty} {A_n\sin{n\theta}}$$

  • For an ESLD, the circulation distribution is only the first term, with $A_1=\Gamma_{e_0}$, where $\Gamma_{e_0}$ is the circulation at wing root:

$$\Gamma_e(\theta) = \Gamma_{e_0}\sin{\theta}$$

  • For a BSLD, we will only keep the first two terms and truncate the rest (noting that due to symmetry, only odd terms remain):

$$\Gamma_b(\theta) \approx A_1\sin{\theta} + A_3\sin{3\theta}$$

Now we can solve for the unknown coefficients $A_1$ and $A_3$ in the BSLD distribution by requiring that a comparative ESLD must have the same total lift ($L$) and root bending moment ($B$):

$$L = \int_{-b/2}^{b/2} {L^{'}dy} = L_e = L_b \tag{a}$$

$$B = \int_{-b/2}^{b/2} {yL^{'}(y)dy} = B_e = B_b \tag{b}$$

That's two equations and two unknowns. The solutions are:

$$A_1 = \frac{b_e}{b_b}\Gamma_{e_0}$$

$$A_3 = \frac{5\Gamma_{e_0}}{3} \frac{b_e}{b_b} \left( \frac{b_e}{b_b}-1 \right)$$

where $b_e$ is the wing span in the ESLD case satisfying (a) and (b), while $b_b$ is the wing span in the BSLD case. That is, we are expressing BSLD as a ratio of span ratio compared to an Elliptical case with the same total lift and root bending moment. Notice that when $b_e/b_b = 1$, we retrieve the ESLD exactly.

2.1 Validation

In the beginning, I mentioned that this derivation is an approximation. But it's not a bad one. Calculating the total induced drag with this formulation, we get:

$$D_i = \frac{2L^2}{\rho_\infty \pi b^2 V_\infty^2} \left[\frac{25}{3} \left( \frac{b_e}{b_b} \right)^4 - \frac{50}{3} \left( \frac{b_e}{b_b} \right)^3 + \frac{28}{3} \left( \frac{b_e}{b_b} \right)^2 \right]$$

Compare this with the exact result from NACA TN-2249:

$$D_i = \frac{2L^2}{\rho_\infty \pi b^2 V_\infty^2} \left[8\left( \frac{b_e}{b_b} \right)^4 - 16\left( \frac{b_e}{b_b} \right)^3 + 9\left( \frac{b_e}{b_b} \right)^2 \right]$$

2.2 Results

Time to get back to the question we began with, can the lift (or circulation) distribution be negative on any part of the span? The plot below shows the circulation distribution of our approximate solution for various ratios:

Bell-Shaped Distribution

For our approximate formulation, the most optimal BSLD occurs at $b_e/b_b = 0.8$ ($b_b/b_e = 1.25$ in the plot). As you can see, it's all positive up to the wing tip.

However, as we keep increasing the span (while maintaining total lift and root bending moment), at some point the distribution goes negative near the wingtip. Eventually the bending moment at certain points along the span will exceed that at the wing root (Ref NACA TN-2249).

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Farhan
    Aug 11, 2020 at 22:31

It depends on the angle of attack. The design case of the bell shaped lift distribution is maximum lift. At cruise, angle of attack is lower, lowering the lift coefficient for the whole wing and also at the tips. In cruise lift at the tips is negative.

In Ludwig Prandtl's 1933 paper "Über Tragflügel kleinsten induzierten Widerstandes" he excluded negative lift at the tips because his algorithm would had equated negative lift with negative spar mass. This lift distribution has been used by Al Bowers in his paper "On Wings of the Minimum Induced Drag" (NASA/TP—2016–219072) which is the source of your illustrations.

Why would negative lift be sensible? The Horten brothers used aggressive wing twist in their flying wings and included negative lift at the tips which would only become slightly positive at very high angle of attack, right before stall. Negative wing tip lift means that a downward aileron deflection reduces induced drag while the upward moving aileron increases induced drag on their respective tips, which creates proverse yaw instead of the adverse yaw which is common with ailerons on lift-producing wing tips. By using negative lift on the outermost wing section, the Horten brothers could achieve good flying characteristics even without any fin or vertical surface.

Since the bell distribution of Ludwig Prandtl shown by Al Bowers is for maximum lift and maximum bending moment, it also will show negative lift at practical angles of attack, so it too makes use of negative tip lift to produce proverse yaw. The answer, therefore, depends on the angle of attack at which the lift distribution is plotted, and in cruising flight the answer is yes, it does have negative lift at the tips.

  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Farhan
    Aug 11, 2020 at 22:31

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