Induced drag is proportional to the square of span loading. The normal method to decrease it would be to increase span. This, however, leads to an exponential weight increase.

So, my question is, if tandem wings are placed with a large longitudinal gap, so as to avoid downwash, would they act as independent wing spans sharing the weight of the aircraft to reduce induced drag? If so, how large would this gap need to be? Would one at the nose and another at the tail be enough?

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    $\begingroup$ The downwash is as high as wide and goes down only a couple of degrees. s $\endgroup$ – Jan Hudec Feb 11 at 16:53

The downwash is about as high as wide, and angled only by a few degrees (more at slow speed). So the longitudinal distance needs to be many times longer than the wing span to make the wings independent.

And it would still be less efficient than increasing the span:

  • If you double the span, you will decrease induced drag four times.
  • If you half the lift, you also decrease the induced drag four times, but then you add the second wing to get your lift back and that also doubles the induced drag, so you have half the induced drag, not quarter.
  • $\begingroup$ That's odd. I thought induced drag was to the square of span loading $\endgroup$ – Abdullah Feb 11 at 17:06
  • $\begingroup$ @Abdullah, yes, inversely proportional to square of span loading. Square of two is four, and inversely proportional, so doubling span will reduce it four times. If you half the lift, you will also reduce it four times, but then you add the other wing to get your lift back, and that also doubles the induced drag, so you get only half induced drag, not quareter like in the first case. Which is logical, because you have only doubled the affected stream tube while doubling the span quadruples the affected stream tube. $\endgroup$ – Jan Hudec Feb 11 at 17:14
  • $\begingroup$ aviation.stackexchange.com/a/36064/45534 $\endgroup$ – Abdullah Feb 11 at 17:21
  • $\begingroup$ @Abdullah, yes, I know that equation. I am basinc my answer on it. $D_i \propto \frac{L^2}{b^2}$. $\frac{L^2}{(2b)^2} = \frac14 \frac{L^2}{b^2}$, but $\frac{(½L)^2}{b^2} + \frac{(½L)^2}{b^2} = \frac{L^2}{4b^2} + \frac{L^2}{4b^2} = \left(\frac14 + \frac14\right) \frac{L^2}{b^2} = \frac24 \frac{L^2}{4b^2} = \frac12 \frac{L^2}{4b^2}$. $\endgroup$ – Jan Hudec Feb 11 at 17:31
  • $\begingroup$ I just realized you're right $\endgroup$ – Abdullah Feb 11 at 17:33

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