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In studying the NASA 247 foot wingspan solar powered Helios aircraft it seemed that evenly spacing the weight would help create a lower bending stress across its span when in flight, as gravity and lift forces would cancel each other out more evenly as compared to a single fuselage.

The Helios did have spaced pods slung underneath, but none at the ends. The aircraft, in turbulence (perhaps an updraft) that caused its wingtips to bend upwards, started a pitch up/pitch down cycle until the wing failed. Could more even spacing of weight on the wing reduce the possibility of this happening again?

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    $\begingroup$ You're still using terminology incorrectly. "Net wing loading" isn't a weight/lift vector sum, it's simple weight per unit area, that has to be offset by lift. Span loading decreases wing flexing moments, it does nothing to change wing loading. $\endgroup$
    – Zeiss Ikon
    Nov 1, 2018 at 16:04
  • $\begingroup$ Edited to reflect more proper "span loading" concept. $\endgroup$
    – Robert DiGiovanni
    Sep 9, 2019 at 9:37
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    $\begingroup$ Does your last edit invalidate the existing answers? It does seem so at first glance, since none of them reference bending stress. Note that this is highly discouraged on this site. $\endgroup$
    – AEhere supports Monica
    Sep 9, 2019 at 9:57
  • $\begingroup$ I actually wanted to delete the question. I have learned much since last year. $\endgroup$
    – Robert DiGiovanni
    Sep 9, 2019 at 10:58

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Net wing loading (weight per unit lifting area) can't be zero in level flight. Lift must equal weight, else the aircraft will accelerate in the direction of the net force.

The loading distribution on the Helios is designed to reduce flexing moment along the span, which is an entirely different thing. As you note, that optimization wasn't perfect, although pitch oscillation wasn't necessarily related to the wing flex that shows in the video.

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weight and aerodynamic lift forces will only cancel out (0 net loading)

Lift and weight cancel out when their vector sum is zero. I.e. $W + L = 0$.

However, wing loading is the ratio of their magnitudes, $\frac{|L|}{|W|}$. Since from the previous equation we have $L = -W$, we can substitute $\frac{|L|}{|W|} = \frac{|-W|}{|W|} = \frac W W$ and trivially see that when weight and lift cancel out, the wing loading is $1$.

This is case whenever the aircraft is flying straight and level, at any speed.

The only situation when the net loading is zero is when lift is zero, which means it is flying along a free-fall trajectory.

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