Assuming there are two aircraft, each of the same density and each the same shape, am I correct in understanding that the smaller aircraft would have a lower take-off speed? I have explained how I have come to this conclusion below;

Given both aircraft are the same density, the weight of each aircraft would be proportional to the length cubed. Therefore;

$W = A l^3$

Likewise, given both aircraft have the same shape, the surface area would be proportional to length squared. Therefore;

$S = Bl^2$

The lift equation is of course;

$L = \frac{1}2C\rho v^2 S$, where;

  • $C$ = Co-efficient of lift
  • $\rho$ = Air density
  • $v$ = Velocity of aircraft
  • $S$ = Surface area of wings

Given the shape of both aircraft are the same, $C$ should be the same for both aircraft. The same will be true for $\rho$, as they are travelling through the same air density, so these (and $\frac{1}2$) can be replaced by a constant, $D$. Therefore;

$L = Dv^2S$

As $S=Bl^2$, $L = DBv^2l^2$, where $D$ and $B$ can be combined to form a new constant, $E$. Therefore;

$L = Ev^2l^2$

At take-off speed, lift will equal weight. Therefore;

$Al^3=Ev^2l^2$, so $v^2=\frac{Al^3}{El^2}$. With $\frac{A}{E}$ forming a new constant, $F$, this simplifies to $v^2=Fl$. Therefore;

$v \propto \sqrt{l}$.

Does this therefore mean that a 1:100 scale model of an aircraft would have a takeoff speed 10 times lower than that of the original aircraft, or is there something that I have not taken into account?


It is true in general that smaller aircraft have lower takeoff speeds than larger ones, but the relationship is complicated by a variety of factors, as follows.

A small aircraft (say, an Ercoupe) has a relatively small speed range over which it operates: takeoff speed ~70 MPH, maximum speed ~110 MPH and its wing profile is a compromise between low and high speed performance. That compromise allows it to get away without the flaps, slats and/or leading-edge droop that large planes use to configure their wings for landings and takeoff (150-175 MPH) and then reconfigure them for high speed cruise (550-575 MPH). This means that the large plane's wing area and coefficient of lift are not constant: they are radically different at takeoff than they are for cruise in a big plane.

  • $\begingroup$ As so often, the best answer has the lowest score. +1 $\endgroup$ – Peter Kämpf Mar 31 at 19:07
  • $\begingroup$ thank you peter!!! $\endgroup$ – niels nielsen Mar 31 at 22:41

Twice as big airplane has 8 times bigger mass, but just 4 times bigger wing area, giving at the same speed just 4 times bigger air lift. So it needs $\sqrt{2}$ higher speed, as the lift is proportional to $v^2$. Edit: Similarly, 1:100 scale model would need $\sqrt{(10^6/10^4)}=10$ times lower speed.

Note that all is considered as a rough estimation, based on pure geometrical similarity. If material strength and total flight mass are considered, the exponent $a$ in $m=L^a$ will be less then 3,but more then 2.

Another factor is that bigger airplanes can afford variable wing geometry due flaps, slots and similar.


The formulas are correct, the conclusion is also definitely correct. A paper airplane flies at a lower minimum speed than a commercial airplane, even though its density is higher! This would seem to be an advantage in favor of small flying objects but actually the opposite is true:

  1. First of all you DO WANT to fly faster - you will get to the destination earlier ;-)
  2. Small, light objects flying much faster than their stall speed are very unstable because some unbalanced or asymmetric force from the air flow ($L$ is your example) can make them tumble or flip, since it is much bigger than their gravity and inertia (which is $W$ in your notation), so keep in mind that $W<<L$ is highly unstable.

There is a just a small inadequacy in your argument. There are limits because of the fixed strength of materials, so in practice the assumption of fixed density may be wrong. I am not saying it is wrong, but that requires further examination at engineering level.

  • $\begingroup$ Thanks for your answer. Yes, of course that makes sense that the density is likely to be different for a smaller aircraft than a larger one, I just kept this constant in this example to keep things simple by keeping all external variables the same. $\endgroup$ – PhysicsGuy123 Mar 30 at 19:59
  • $\begingroup$ Why would your point 2 be true? Proof, please! $\endgroup$ – Peter Kämpf Mar 31 at 19:05
  • $\begingroup$ @PeterKämpf, this is physics SE, not aviation, here an equation like $W << L$ are considered as solid enough. But just as a hint for your curiosity, it is the reason why the airplanes have a tail, and rockets that are meant to fly through the atmosphere have fins, and also why bullets are made to spin in the rifles barrel. $\endgroup$ – Kostas Mar 31 at 20:20
  • $\begingroup$ No need to be miffed! However, you never explain why the tail on a light aircraft would be less effective. To be honest: That line "light objects flying much faster than their stall speed are very unstable" is complete bunk. The forces on the wing grow in proportion to the forces on the tail as airspeed rises. $\endgroup$ – Peter Kämpf Mar 31 at 22:08
  • $\begingroup$ Point 2 does make sense to me. After all, when you take into account the moment of inertia of the aircraft - $I \propto mr^2$, so $I = A l^3 l^2 = A l^5$. Torque is $T = F d$, and as $d \propto l$, and given the force from the wings will be proportional to the surface area, $T = B l^2 l = B l^3$. Angular acceleration is of course $\frac{T}{I}$, so $\alpha = \frac {B l^3}{A l^5}$, so $\alpha \propto l^{-2}$. This therefore suggests to that a smaller aircraft is likely to be more unstable, given they will be more affected by external factors that cause a force on the wings/horizontal stabiliser. $\endgroup$ – PhysicsGuy123 Apr 1 at 16:50

Your conclusion is generally correct, but if you have ever been inside a big airliner, you will have noticed that it is not a solid piece of metal, but hollow on the inside. So the assumption that mass scales with the cube of length is not entirely correct.

If you look at existing designs, mass scales with an exponent of length that is between 2.2 and 2.4 given that the remaining parameters stay comparable (please do not compare jets with piston aircraft, for example!). So there is indeed an increase in wing loading as scale goes up, but a very moderate one. The cube law does help the larger airplane, however, to achieve a much higher range and limits its cargo capacity by mass rather than volume.

As @NielsNielsen points out correctly, larger aircraft change their wing shape a lot in order to combine high wing loading with low landing speeds. They can afford to do this because the thrust of jets (and even turboprops) does not drop with speed as much as that of a piston aircraft does. That allows them to cover a higher speed range, and for high speed a high wing loading is a big advantage.

  • $\begingroup$ Thanks for your answer. As per my response to @Kostas answer, I appreciate that mass is not going to be proportional to length cubed, but I just chose this as a rough approximation, to keep things simpler and keep external variables the same, because as you say, the true exponent would vary between aircraft, and would depend on things like whether it is a cargo or passenger aircraft and so on... $\endgroup$ – PhysicsGuy123 Apr 1 at 16:22

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