My dad was an airline pilot and, during his training, his instructor told him that, in absolutely ideal conditions, (in practice obviously impossible) that the actual material composing the wing surface was not important, as long as the air pressure over the wing was kept perfectly, absolutely even, along it's length and that the air velocity was kept absolutely constant, .

So please think of this as a thought experiment, or performed in a wind tunnel instead of real conditions.

Ignore all the functions a modern commercial aircraft wing needs a large main spar for, carrying fuel, taking the weight of the engine, acting as a support for control surfaces , acceleration during the take off run, and so on.

If it did not need inherent strength to do all the above, and the airflow was perfect, absolutely constant, would a paper wing surface generate the lift required?

I do appreciate that the slightest degree of pressure difference not connected with actual lift generation across the wing would cause the wing to collapse, and that drag vortices at the wing tips would have to be taken as negligible, but, as a total thought experiment alone, would it work?

EDIT I think the point the instructor was trying to make was that given the above conditions, that any skin material would generate exactly the same amount of lift and I have a feeling he was right, although in real life you obviously need the strength of modern wings. END EDIT

  • $\begingroup$ How thick of a paper wing are you talking about? The wings still have to hold the weight of the aircraft while in the air so a single sheet of paper would never work. But paper stacked 4 feet thick like a regular wing? Maybe. The wing spar does a lot more than hold the weight of the things attached to the wing. $\endgroup$
    – tpg2114
    Jul 24 '15 at 19:31
  • $\begingroup$ And also -- are you referring to a paper skin, ie. the wing surface itself, or the entire wing structure (skin, ribs, spars, etc)? $\endgroup$
    – tpg2114
    Jul 24 '15 at 19:33
  • $\begingroup$ @tpg2114 for all I know, it could have been a wind up for first day rookie pilots, so I was never sure the instructor was serious. But in general, my idea was that the material has to be much lighter than current aluminium, the wing does what all wings have to do, just generate enough lift (downward momentum) to take the weight of the plane. Maybe a flying wing aircraft rather than a jumbo would have been a better example. $\endgroup$
    – user81619
    Jul 24 '15 at 19:39
  • $\begingroup$ @tpg2114 no just the skin, please ignore the other bits, I think the instructor was right in principle, but in practice, obviously no chance. $\endgroup$
    – user81619
    Jul 24 '15 at 19:41
  • 1
    $\begingroup$ You know that planes started out with paper (or cloth) covered wings right? They had internal (and external) structure to provide load bearing strength, but the paper / cloth itself was providing (over a limited distance) the surface over which the force was generated. Does that relate to your question? $\endgroup$
    – Floris
    Jul 24 '15 at 19:57

According to the wiki page the 747 has a max takeoff "weight" of about 350,000 kg (depending on the model) and a wing surface of about 500 m2. That means that a force of 3.5 MN must be carried by 5 million square cm, or 0.7 N per square cm.

If you can somehow split this evenly between the top and bottom surface, then you need to come up with a paper surface that can carry the equivalent of a 35 g mass (about 1 oz) on every square cm of its surface.

Clearly, unless you have a relatively dense-meshed (honey comb?) structure right below the skin, your paper will not be able to withstand that kind of force over a very large distance. To support this kind of pressure the paper will need to have curvature - as you can see clearly in this photo of a "thin skinned" (model) plane source

enter image description here

The curvature, the tensile strength, and the pressure difference, are all related. For simplicity, in a spherical object (balloon) of radius $r$ you can see that the tension relates to the pressure difference by looking at the circumference ($2\pi \; R$) that has to support the force (pressure times area, $\pi R^2 \Delta P$). It follows that the tension in the surface is

$$2\pi R T = \pi R^2 P\\ T = \frac{Pr}{2}$$

Here, $R$ is the radius of curvature. The larger the radius of curvature (the flatter the surface - i.e. the more "tightly stretched") the greater the tension needed to support the weight.

According to slide 4 of these lecture notes :

enter image description here

the paper used in their test had a tensile strength of about 7 kg/cm, or 7000 N/m. According to the above formula, this means it needs a radius of curvature less than

$$R_{max} = \frac{2T}{P} = \frac{2\cdot 7000}{0.35\cdot 10^4} = 4 \mathrm{\;m}$$

So the radius of curvature must be less than 4 m if we can divide the pressure over both surfaces; if we magically wanted to create a single curved wing out of paper that could carry this force, it would have to have a radius of curvature of less than 2 m. If you have curvature in only one direction (more likely - in fact, you will be bending some of the paper "the wrong way" which makes things much worse) you need to halve the radius of curvature to 2 m even for double sided support. The width of the 747 wing is considerably greater than that: so you can only do this if you have multiple struts along the wing to support the paper. In fact, if you assume you will allow a 10 cm deviation from "flat", then the spacing of the struts (for 2 m radius of curvature) will be

$$ d = 2\sqrt{2 R h} = 2\sqrt{0.4} \approx 1.2 \mathrm{m}$$

This is not a completely implausible number. I would not want to build a plane like that... but it is not completely implausible.

That is quite surprising.

  • $\begingroup$ Thanks very much taking the time for a very comprehensive answer Floris, I worded the question badly, I should have stressed that the airfoil section and wing area would be identical to the real aircraft. I have flown planes myself and I felt strongly it should work, in theory only. Thanks again for your time. $\endgroup$
    – user81619
    Jul 24 '15 at 21:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy