3
$\begingroup$

The explanation for lift that I find easiest to visualize is that air is deflected downward by the wing. This downward action causes an equal and opposite reaction of an upward force against the plane, as in accordance with Newton's third law.

As you can see from the image below the red lines both end lower than they started.

enter image description here

Air at sea level weighs 1.293 kg per cubic meter. The acceleration due to gravity is 9.8m/s².

That comes up to 9.8N of force being needed to keep a 1kg aircraft in the sky. With perfect efficiency this equals 9.8 watts.

If trying to hover VTOL aircraft, can you multiply how much air a propeller can move, by how fast it moves it to predict lift force? Since force is mass x acceleration?

Or do fluid dynamics work differently as the air exiting the propeller will spread out instead of all going straight down? If that is the case can using a duct with a properly fitted bell nozzle solve this problem? Or am I thinking into realms that I shouldn't be?

$\endgroup$
7
$\begingroup$

First off, your calculation for 9.8 W does not seem to be based on any actual physics. Check the units! I can't tell if you did $P=F\cdot m$ or $P=\frac{F}{m}$, but the result is either the nonsensical quantity of joule-kilogram per meter, or acceleration respectively.

A correct calculation based on keeping something aloft by propelling air downwards would be as follows. To keep something aloft, you need to expel mass downwards at a certain speed. The force to keep the object aloft is $$F=m_{object}g$$ The force generated by downwards momentum transfer is $$F=\dot{m}v$$ with $\dot{m}$ indicating mass flow (kilogram per second) of the air (not the mass of the object). The energy flow (power) required to impart this momentum on the airflow is $$ P=\frac{1}{2}\dot{m}v^2$$ Here we can draw an important conclusion. The power requirement is arbitrarily small, by increasing the mass flow and decreasing the downwards velocity. Immediately, you can see why helicopters have big rotors and airplane wings are so large: they want to affect as much air mass as possible for increased efficiency. Even flying faster increases lift efficiency (reduced induced drag) by affecting more air per unit of time. Of course, at some point other sources of drag will dominate.


The above calculations assume that force is purely generated by creating a downwards mass flow. This is not exact reality. For example, put your aircraft on the ground, and it does not sink through it even though no mass flow is created. When flying close to the ground, some of the airflow downwards creates a pressure due to hitting the ground below, which helps you keep aloft as well (this is called the ground effect). Other than that, viscosity plays a small role by generating a force opposite the downwards air movement (although viscosity is required for lift generation with an airfoil). However, to answer the main question: equating lift by how much air is moved is a very good approximation.


Let's finally address your idea of using a Bell nozzle. This is a nozzle that is used on supersonic propellants (the 'throat' of the nozzle marks the transition from subsonic to supersonic), and is used to increase the speed as much as possible. This is very inefficient in terms of energy, but since rockets need to carry all the propellant mass (whereas an airplane gets 'free' mass flow by travelling through air), and need extra propellant to carry that propellant up, speed wins instead of mass, and rocket engines are optimised for nozzle velocity.

$\endgroup$
  • $\begingroup$ +1 Excellent answer! $\endgroup$ – Peter Kämpf Jul 22 '18 at 12:20
  • $\begingroup$ @PeterKämpf Thank you. Do you perhaps know how much viscosity practically influences lift? (outside maintaining the Kutta condition) $\endgroup$ – Sanchises Jul 22 '18 at 13:07
  • $\begingroup$ @Sanchises The math I was trying to do when I came up with 9.8w was force in newtons equals mass(kg) * acceleration(m/s). Since one newton equals 1 watt I thought the conversion would work out. Your answer though was very good and I understand this a lot better thank you! $\endgroup$ – YAHsaves Jul 23 '18 at 23:48
  • 1
    $\begingroup$ @YAHsaves Whoever told you one newton equals one watt was wrong. Power is force times velocity. One watt equals one newton-meter per second. Always check your units! $\endgroup$ – Sanchises Jul 24 '18 at 5:36
  • $\begingroup$ To answer your question from July 22 (sorry to have missed it!): Viscosity reduces the lift curve slope and limits maximum lift. Think of the boundary layer as an additional part of the wing which changes the shape of the airfoil. On the suction side this part becomes thicker as air slows down past the suction peak, so the angle of attack of the airfoil + boundary layer is lower than the geometric angle of attack of the naked airfoil. This becomes really dramatic when the flow separates from the airfoil. $\endgroup$ – Peter Kämpf Dec 4 '18 at 6:32
4
$\begingroup$

The point is not that the air ends up lower when leaving the wing, but that an impulse has been imparted on it such that it continues to move downwards behind the wing.

Proof: Read this answer on wake formation, and watch the animated GIF in this answer.

The next question is: How much air is involved?

That is much harder to pin down precisely because the boundaries of the air that is influenced by the wing are theoretically at infinity. But there is a practical approximation that is very similar to Froude's theory for propellers. Just take what flows through a circle with the diameter of the wingspan. This answer uses this idea to calculate induced drag, and it reaches the correct result.

So there you have it: Yes, lift can be equated to how much air is accelerated.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.