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I give some physics lessons to a friend. She asked me a question that I am unable to answer. Could you help me ?

A plane has a weight of $2\times10^6$kg. The surface of the wing is $1200 \text{m}^2$. We assume an air density equal to $1\text{kg}/\text{m}^3$. The speed of air under the wing is $100\text{m}/\text{s}$. What should be the speed of air over the wing so that the plane remains in the air ?

I searched for expressions of lift (as for instance on Wikipedia http://en.wikipedia.org/wiki/Lift_(force)) and tried to link it with the speed of fluid over and under the wing but didn't find anything relevant with the only pieces of information I have in the exercise. In my opinion, the speed of air will depend on the shape of the wing... And if we don't know anything about the shape, we are unable to answer the question. Am I wrong ?

Any idea ?

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Considering the tag "homework" I know the solution that was expected. Bernoulli law:

$$\frac{\rho v_{under}^2}{2} + p_{under} = \frac{\rho v_{over}^2}{2} + p_{over}$$

$v_{over}$ and $v_{under}$ are the air flow speeds over and under the wing respectively, $p$ is pressure, $\rho$ is the air's density. The desired lift force should be equal to the plane's weight:

$$Mg = F = S (p_{under} - p_{over})$$

$S$ is the wing's area, $M$ is the plane's mass.

You know everything (except $v_{over}$), so you can solve the system to find it.

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  • $\begingroup$ I indeed suspect this is the answer the question was looking for. But this would give a very incomplete picture to physics students on how aircraft fly! $\endgroup$
    – Rody Oldenhuis
    Aug 10, 2012 at 13:56
  • $\begingroup$ @RodyOldenhuis At first I though that too, but now I think it isn't terribly wrong. What's unphysical in this model? Probably that suits better for chat. $\endgroup$
    – Yrogirg
    Aug 10, 2012 at 14:05
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    $\begingroup$ It isn't wrong per se, merely incomplete. For example, how would you factor in downwash -- air being driven downwards by the wing if the wing has a downward curvature or is at an angle. Downward flow means the incoming airflow's momentum is changed, which means there is an upward force - Bernoulli doesn't account for that (Chat is OK, but this needed to be seen near the question :) $\endgroup$
    – Rody Oldenhuis
    Aug 10, 2012 at 14:11
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    $\begingroup$ Indeed, but how is that reflected by Benoulli? My point is: the question is based on too simplistic an understanding of an aircraft. But clearly, it is the intended answer. If you're to be a physicist this would give you an itch so bad you would step up to that professor and don't leave until he admits you're right! :p $\endgroup$
    – Rody Oldenhuis
    Aug 10, 2012 at 15:22
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    $\begingroup$ @Rody: The relationship between the Bernoulli effect and downwash is very simple: you can't have one without the other. Regarding inverted flight, it's simply a matter of the angle of attack (which directs the downwash), and that affects the point at which the air approaches the rounded leading edge, which affects the difference in distance traveled by the air over one side of the wing versus the other. This has been thoroughly discussed. $\endgroup$
    – Mike Dunlavey
    Aug 18, 2012 at 14:18
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You are correct, you do not have sufficient information to answer this question with anything but a rough estimate. The simplest equation one can use to calculate a wing's lift is

$L = \frac12 C_L \rho S V^2$

where $\rho$ is the air density, $S$ the wing surface area, $V$ the airspeed and $C_L$ the so-called lift coefficient. The lift coefficient is a dimensionless scale factor (usually determined empirically, but it can be computed from first principles), which depends on the shape and size of the wing, its surface properties, planform, etc. etc. etc.

Also, by the way the question has been posed, I think the one who asked the question (your friend? his/her prof?) has an incomplete notion of how lift is generated. A lot more than just speed differences are involved in (and very relevant for) how much lift actually results (see various explanations on your own link), or this better one.

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  • $\begingroup$ Thanks for your comment. Actually, this was a question she had to her exam. She failed and she has to pass this exam again so I help her. We are doing again all the exercises and we were blocked on this one. This is strange because the exam is supposed to be about basis of physics, as mechanics, waves and so on... I translated the question from French so maybe it is not well written but I can assure that the only mistakes I may have made are grammar ones. ;-) $\endgroup$
    – AnSy
    Aug 10, 2012 at 13:08
  • $\begingroup$ @Bagheera Half of what you hear in a classroom is crap. Education is figuring out which half is which :) $\endgroup$
    – Rody Oldenhuis
    Aug 10, 2012 at 13:57
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I did a little back-of-the-envelope calculating.

A Cessna 172 has a wing loading of 13 lb/ft^2, which works out to 63.6 kg/m^2. So that is how much lift it generates at its clean stall speed of 50 kts. (A knot is 1 nautical mile per hour, which is 1 minute of latitude per hour.)

Your situation has wing loading of 2*10^6 / 1200 m^2 or 1667 kg/m^2.

That is a wing loading 26 times higher than the Cessna. Since lift is proportional to speed squared, the wing would have to be traveling 5.1 times as fast, or 256 kt. That would be its stall speed. Compare that to a landing speed of roughly 190kt for the space shuttle, and 140 for a commercial jet.

That plane probably needs a much bigger wing, if it's that heavy.

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