Ever wondered how lift is generated on the wing of an aircraft or how a propeller generates thrust? There’s a legend that goes like this: The particles of the air want to stay together, but as the particle that goes along the upper part of the wing needs to go a longer way, it needs to be faster, and according to Bernoulli’s Principle, faster air means lower pressure, leading to lift. Nice explanation, isn’t it? So very romantic, the particles that just won’t let themselves be separated.
It’s just that…it’s completely wrong! Air particles aren’t married to each other, and it may even be that they are completely mixed up — in case of turbulent flow — with the wing still generating lift! In addition, Bernoulli’s Principle does not account for friction in the air, and thus by itself cannot explain drag.
For our further analysis of our quadrocopter model, we need to model the behaviour of the propeller and the engine that drives it. Modelling the behaviour of the propeller involves the use of the formulae for thrust [latex]F_T[/latex] and power [latex]P[/latex]
\begin{align} F_T &= c_T \rho D^4 n^2 \\ P &= c_P \rho D^5 n^3 \end{align}and working with these. Specifically, we need to understand the nature of the thrust and power coefficients [latex]c_T[/latex] and [latex]c_P[/latex].
So, let’s have a look how lift and drag on an airfoil come to be, how we can use dimensional analysis to get the typical formulae for describing them, and where to find what’s missing.
Reaction Forces: When a Ball meets a Wall
Let’s first recap a bit of physics basics. Remember Newton’s Third Law of Motion?
For every action, there is an equal and opposite reaction.
Let’s imagine we are sitting in a car at rest, with the motor running. If we now press the accelerator, the engine will provide power that is transferred to the wheels, which in turn will exert a force on the street. Now, the wheel is obviously accelerated to turn so that its bottom moves towards the back of our car — the opposite direction of where we want to go!
So it cannot be the wheel that provides our forward acceleration, because the wheel enacts a force in the wrong direction. Instead, it’s the street surface which makes us move: The force accelerating the wheel makes the wheel enact that force on the street surface, and the street surface will enact an opposite force of equal size on the wheel, accelerating us forward.
In this example, the action is the force acting from the wheel on the street surface, and the reaction is the force acting from the street surface on the wheel and — in consequence — on the whole car. And essentially, that also explains lift and drag. So, thanks for reading, have a good night! …
Of course, it’s a bit more involved. For a better understanding, let’s imagine a ball being thrown at a wall, as indicated in the figure to the right. When the ball hits the wall, it cannot proceed further in the same direction, so its velocity vector and with it its momentum must change. The change of velocity is indicated in the figure by [latex]\Delta v[/latex].
As we know from Newton’s Second Law of Motion, a change of velocity requires a force acting on the ball. It is reasonable to assume that this force is enacted at the contact area of the ball to the wall.
Again, applying Newton’s Third Law of Motion, there must be an opposite force with the same magnitude in the system. But where is it?
Well, there is only one other object touching the contact area, and that is the wall. Thus, the reverse force must be acting on the wall. If we hadn’t fixed the wall in our reference system, we would see it being propelled backwards.
But, how big is this force? Well, in our example, we cannot tell exactly. However, we know the change of the force over time in a qualitative manner:
- When the ball first touches the wall, the force is practically non-existent.
- Then, when the ball is compressed more and more, it acts against being compressed with increasing force, until it is maximally compressed and at rest.
- From then on, it is accelerated again in its new direction, and as it decompresses in the process, the force decreases again.
The total change in momentum until the time [latex]t[/latex]:
\begin{equation} \Delta \vec{p}\left(t\right) = \int_0^t \vec{F}\left(\tau\right) d\tau \end{equation}If you want to check the validity of this equation, just consider that [latex]\Delta \vec{p} = m \Delta \vec{v}[/latex] and [latex]F = m \vec{a} = m \frac{d}{dt} \vec{v}[/latex]. If you look closely, you can see that Conservation of Momentum follows from Newton’s Third Law of Motion, according to which for the total forces we have [latex]\vec{F} = 0[/latex], and thus for the total momentum in the system, [latex]\frac{d}{dt} \vec{p}=0[/latex] must hold.
We call [latex]\vec{F}\left(t\right)[/latex] the reaction force, as it is a reaction to the change of momentum of part of the system (the ball).
However, although we cannot exactly determine the force, we can derive something about the change of momentum. Looking at the velocity triangle in the figure above, we see that [latex]\Delta v[/latex] is proportional to the magnitudes of [latex]v_1[/latex] and [latex]v_2[/latex] (just try scaling the triangle in your head). We know that the latter are equal, so the change in velocity is proportional to the magnitude [latex]v_1[/latex]. It is also proportional to the mass of the ball, so we have
\begin{equation} \Delta \vec{p} \sim m v_1 \end{equation}
Now, if we were to constantly throw balls onto the wall, all having the same mass and initial velocity, we could determine the mean force acting on the wall. Let’s assume that over a time interval of length [latex]\Delta t[/latex] we would throw a mass of [latex]\Delta m[/latex] onto the wall. Then the change in momentum over that time would be
\begin{equation} \Delta \vec{p}_{tot} \sim \Delta m v_1 \end{equation}
Let’s further assume that we throw the balls at a constant frequency, so that we get a mean mass-flow rate of [latex]\dot{m} \approx \frac{\Delta m}{\Delta t}[/latex]. The mean force over the same time would then be
\begin{equation} \vec{F} = \frac{\Delta \vec{p}_{tot}}{\Delta t} \sim \frac{\Delta m}{\Delta t} v_1 = \dot{m} v_1 \end{equation}
If we now think of the balls as being a gas of density [latex]\rho[/latex] moving at velocity [latex]v_1[/latex] through a tube with cross-section [latex]A[/latex], the mass flow rate is given by
\begin{equation} \dot{m} = \rho A v_1 \end{equation}
If we plug all of this together, we get the force to be
\begin{equation} \vec{F} \sim \rho A \left(v_1\right)^2 \end{equation}
which already looks eerily like the usual formula for lift and drag:
\begin{equation} \label{eq:aerodynamic-force} F = c \rho A v^2 \end{equation}
Reaction Forces on an Wing
Let’s have a look at the cross-section of a wing and how the air flows around it:
The drawing above is only a schematic, but there are some important aspects of the airflow visualised there. What we see is that the air flows in a laminar fashion around the airfoil (that’s what we call the cross-section of a wing). An airflow is laminar if all the sheets of air are nicely separated and there is no turbulence.
We also see that the air on the left side is quite parallel to the horizontal plane, while on the right side it briefly flows downward, before assimilating to the free-flow again. As we have now learned, this indicates a change in momentum, which means that there is a force in play. As there should be: We would expect the wing to generate lift, an upwards force, so the resulting change in momentum should be downwards.
But where does this change in momentum come from? Think about what would happen if there was no change in momentum: The air would simply flow through the wing. The air cannot do that, so it must change direction to avoid the wing. It will again change direction when the wing “moves away” from the flow again, as the pressure of the air flowing beside it will push it that way. That change of direction implies a change of momentum, and that change of momentum must come with a force acting on the air.
So finally, we know where lift and drag come from: They represent the sum of all these forces along the wing, or the net force, and these forces result from the air having to follow the slope of the airfoil. There is a general convention to define lift and drag according to the direction of the free-stream:
- Lift is the part of the force perpendicular to the direction of the free-stream, with positive lift pointing upwards, and
- drag is the part of the force parallel to the direction of the free-stream, with positive drag pointing in the direction of the free-stream flow.
Again, we ask ourselves: How big are these forces? And again, we cannot give that number exactly just from this basic model — specifically, if we consider friction –, but we can characterise it quite well using dimensional analysis.
Dimensional Analysis of Lift and Drag
Now it’s time to use what we learned in a previous article about dimensional analysis — but this time we apply it to our wing. From our previous considerations, we already have identified a few parameters that may influence lift and drag:
- the density of the air in the free-stream [latex]\rho_\infty[/latex],
- the size and form of the wing, represented by its projected surface area [latex]A[/latex], and
- the free-stream velocity of the air [latex]v_\infty[/latex].
Geometrically, the so-called angle of attack [latex]\alpha[/latex] will also play a role, as shown in the figure above. This is the angle between the direction of the free stream and the chord line of the airfoil — the theoretical line from the leading to the trailing edge.
If we are working at velocities close to the speed of sound, we need to consider compressibility. Thus, another parameter is the speed of sound in the free stream [latex]a_\infty[/latex]. The speed of sound may differ between the free stream and around the wing, as it depends on the density of the air, and this may well be influenced by the flow around the wing. Therefore, we only consider the free-stream speed of sound — the speeds of sound elsewhere in the stream would be expressible based on this.
Further, we also need to consider internal friction in the air — otherwise there will be no drag at all, according to a finding that is known as D’Alembert’s Paradox. The friction will be represented by the free-stream viscosity [latex]\mu_\infty[/latex]. The friction between two sheets of fluid is proportional to the size of the contact surface and the gradient of velocity between both. The viscosity is the proportionality constant.
So, we finally have the following parameters influencing lift and drag, with their dimensions ([latex]M[/latex] for mass, [latex]L[/latex] for length, and [latex]T[/latex] for time):
Quantity | Symbol | Dimension |
---|---|---|
Lift/Drag | [latex]F[/latex] | [latex]\frac{ML}{T^2}[/latex] |
Angle of Attack | [latex]\alpha[/latex] | [latex]1[/latex] |
Area of the Wing | [latex]A[/latex] | [latex]L^2[/latex] |
Free-Stream Velocity | [latex]v_\infty[/latex] | [latex]\frac{L}{T}[/latex] |
Density of the Air | [latex]\rho_\infty[/latex] | [latex]\frac{M}{L^3}[/latex] |
Viscosity | [latex]\mu[/latex] | [latex]\frac{M}{LT}[/latex] |
Speed of Sound | [latex]a_\infty[/latex] | [latex]\frac{L}{T}[/latex] |
Obviously, the lift and drag forces are functions of the other parameters:
\begin{equation} F = f\left(\alpha,A,v_\infty,\rho_\infty,\mu,a_\infty\right) \end{equation}
The dimensions used in these parameters are mass [latex]M[/latex], length [latex]L[/latex] and time [latex]M[/latex]. We’ll use the density [latex]\rho_\infty[/latex], the free-stream velocity [latex]v_\infty[/latex] and the wing area [latex]A[/latex] to represent them:
Dimension | Symbol | Representation |
---|---|---|
Mass | [latex]M[/latex] | [latex]\rho_\infty \sqrt{A^3}[/latex] |
Length | [latex]L[/latex] | [latex]\sqrt{A}[/latex] |
Time | [latex]T[/latex] | [latex]\frac{\sqrt{A}}{v_\infty}[/latex] |
Using these, we find the following expressions with equivalent dimensions:
Quantity | Dimension | Equivalent Expression |
---|---|---|
Lift/Drag | [latex]\frac{ML}{T^2}[/latex] | [latex]\rho_\infty A {v_\infty}^2[/latex] |
Angle of Attack | [latex]1[/latex] | [latex]1[/latex] |
Viscosity | [latex]\frac{M}{LT}[/latex] | [latex]\rho_\infty \sqrt{A} v_\infty[/latex] |
Speed of Sound | [latex]\frac{L}{T}[/latex] | [latex]v_\infty[/latex] |
We could now use [latex]\frac{\mu}{\rho \sqrt{A} v_\infty}[/latex] as the dimensionless quantity to represent friction. If you are interested in aerodynamics, that expression may look familiar to you. The Reynolds Number is often used to represent the influence of friction in fluid flows, but it looks slightly different:
\begin{equation} {Re}_\infty := \frac{\rho_\infty l v_\infty}{\mu_\infty} \end{equation}So, first of all, nominator and denominator are reversed — larger numbers indicate less friction –, and instead of [latex]\sqrt{A}[/latex] we have the so-called characteristic length [latex]l[/latex] in there. In aerodynamics, the chord length [latex]c[/latex] is usually used for [latex]l[/latex], so we might not want to deviate from that. Thus, we will use the Reynolds-Number as a dimensionless representation of viscosity.
For the speed of sound, we’d get [latex]\frac{a_\infty}{v_\infty}[/latex], which is the inverse of the Mach Number. So, we’d rather use the Mach Number [latex]M_\infty:=\frac{v_\infty}{a_\infty}[/latex] directly. Thus, our relationship has the following form:
\begin{equation} \frac{F}{\rho {v_\infty}^2 A} = C\left(\alpha,{Re}_\infty,M_\infty\right) \end{equation}
with
\begin{align} {Re}_\infty &:=& \frac{\rho c v_\infty}{\mu_\infty} \\ M_\infty &:=& \frac{v_\infty}{a_\infty} \end{align}This is the formula in the form that you’ll find in most aerodynamics literature. The value of [latex]C\left(\alpha,{Re}_\infty,M_\infty\right)[/latex] is called the coefficient of lift when calculating lift and the coefficient of drag when calculating drag.
So, where we originally had six parameters to vary in an experiment ([latex]\alpha[/latex], [latex]c[/latex] resp. [latex]A[/latex], [latex]v_\infty[/latex], [latex]a_\infty[/latex], [latex]\rho_\infty[/latex] and [latex]\mu_\infty[/latex]), we are now left with three parameters to vary ([latex]\alpha[/latex], [latex]{Re}_\infty[/latex] and [latex]M_\infty[/latex]), and we have a formula that will help us deal with the rest of the variation without additional experiments.
Propeller Aerodynamics
On our multicopter, we don’t have wings. We have propellers, which could be described as rotating wings. Lift turns to thrust, drag turns to torque, torque turns to required power.
The performance of a propeller is determined by a relationship between these quantities:
- Thrust force [latex]F[/latex],
- propeller torque [latex]Q[/latex],
- propeller power (due to torque) [latex]P[/latex],
- rotational frequency of the propeller [latex]n[/latex],
- propeller diameter [latex]D[/latex],
- velocity of the air in free-stream [latex]v_\infty[/latex],
- density of the air in free-stream [latex]\rho_\infty[/latex],
- viscosity of the air in free-stream [latex]\mu_\infty[/latex],
- speed of sound in free-stream [latex]a_\infty[/latex].
Now, here is an exercise for you: See if you can find the usual descriptions of [latex]F[/latex] , [latex]Q[/latex] and [latex]P[/latex] based on dimension analysis:
\begin{eqnarray}
F &:= c_t\left(J,Re,M\right) \rho D^4 n^2 \
Q &:= c_q\left(J,Re,M\right) \rho D^5 n^2 \
P &:= c_p\left(J,Re,M\right) \rho D^5 n^3
\end{eqnarray}
You will again come across the Reynolds Number [latex]Re[/latex] and the Mach Number [latex]M[/latex]. I’ll give you a few hints for these:
- The Reynolds Number for propellers is usually based on the chord and tangential velocity at 75% of propeller radius for the characteristic length and velocity. The symbol used for the former is typically [latex]c_\frac{3}{4}[/latex]. Yes, you’ll have to add that to the list of dimensional parameters.
- The Mach Number is typically based on the velocity of the propeller tip.
In addition, you’ll find the advance ratio relevant:
\begin{equation} J := \frac{v_\infty}{n D} \end{equation}
With some basic geometry knowledge you should be able to figure these out. If you get stuck, I recommend looking at this paper by Robert Deters, Gavin Ananda and Michael Selig of the University of Illinois – Urbana-Champaign (UIUC) ( Citation: Deters, Krishnan & al., 2014 Deters, R., Krishnan, G. & Selig, M. (2014). Reynolds number effects on the performance of small-scale propellers. American Institute of Aeronautics and Astronautics. https://doi.org/10.2514/6.2014-2151 )
Where to find out more
Now that we found out in general what the aerodynamic coefficients can do for us, we’d of course like to know how to get them. Well, there’s always measurement. If you don’t want to do it yourself, the University of Illinois – Urbana-Champaign (UIUC) provides a pretty extensive source of data for airfoils and propellers:
Or perhaps you also want to go back to the origins of flight, to the systematic experiments with large numbers of airfoils at NACA, the National Advisory Committee for Aeronautics, a predecessor of today’s NASA ( Citation: Jacobs, Ward & al., 1933 Jacobs, E., Ward, K. & Pinkerton, R. (1933). The characteristics of 78 related airfoil sections from tests in the variable-density wind tunnel (460). National Advisory Committee for Aeronautics : Washington, DC. Retrieved from https://ntrs.nasa.gov/search.jsp?R=19930091108 ) ?
However, we can also try to model and simulate the behaviour of airflow around objects. If we assume friction to be negligible (non-viscous or inviscid flow), there’s a whole theory of fluid dynamics modelled around that assumption: potential flow. (Actually, that also requires that the flow is free of rotation.) This is quite a powerful approach as it allows to model fluid flow as a potential field (similar to gravity), and many complicated flows can be described by superposition of basic flows such as uniform flow, flow sources/sinks or circular vortices.
Also, thin airfoil theory makes heavy use of potential flow. It gives us some pretty good first-order estimates of the lift coefficient and the coefficient of the so-called pitch moment for thin airfoils. An airfoil is considered thin if its thickness is small compared to its chord length.
One of the most well-known results from that is the estimated gradient of the lift coefficient of a thin 2D airfoil for small angles of attack (with the angle of attack [latex]\alpha[/latex] being given in radians):
\begin{equation} \frac{\partial}{\partial \alpha} c_l = 2 \pi \end{equation}
However, modelling flow around airfoils using potential flow has one important drawback: there is no drag at all! As drag due to friction is not modelled, it cannot be considered. But even if we sum up the forces due to change in momentum over the airfoil, there is no component along the direction of the free-stream flow.
That means: In potential flow, airfoils generate lift, but no drag. This realisation is known as D’Alembert’s Paradox, after the 18th-century French mathematician Jean le Rond d’Alembert.
Still, potential flow gets us pretty far towards a solution for viscous flow: We can approximate a first solution, and then add corrections, e.g., using boundary layer models. The boundary layer is the area around an object within which the velocity gradient is large enough for friction to matter. By first determining a potential flow solution, the thickness of the boundary layer can be estimated. Then, another potential flow solution is found, increasing the thickness of the airfoil by that of the boundary layer. This is an iterative process which goes on until a good, stable solution is found. Using that solution, it is then possible to derive the amount of drag produced inside the boundary layer.
Then, there’s induced drag, which comes from the fact that our wings are not infinitely wide — although we can approximate infinitely wide wings by making them long and slender. This is what is done with sail planes, who have wings with pretty high aspect ratios. To actually estimate the lift and drag of a real wing, we can use lifting line theory, which again uses a simple model based on potential flow to transform lift and drag data on 2D-airfoils into lift and drag data for a finite wing.
And for the really complicated cases, we can try to explicitly solve the Navier-Stokes equations numerically using computational fluid dynamics (CFD). With these, we can create a virtual wind tunnel, and approximate the air flow quite well. However, there is a lot of computational power involved, and the setup requires much more intricate models of our objects than the other approaches.
Of course there are also tools available for the simpler methods. Modelling the performance of 2D-airfoils in inviscid and viscous, subsonic flow is supported by the well-known XFOIL software. The tool XFLR5 extends this to wing design using lifting-line and vortex sheet methods.
Conclusion
So, now we know where lift, drag, propeller thrust, torque and power come from, and how we can characterise them. These formulae allow us to do a good bit of work, and at least get some pretty good estimates for the performance of a wing — if we know the coefficients. They do not allow us to directly derive these coefficients, but of course there are different methods to approximate them for individual forms of wings: measurements, thin-airfoil theory, potential flow, boundary-layer methods, lifting-line theory or CFD.
Now we are well-equipped for modelling the aerodynamics part of our multicopter engines. Next time, we’ll look into modelling our engine with a DC motor.
Bibliography
- Deters, Krishnan & Selig (2014)
- Deters, R., Krishnan, G. & Selig, M. (2014). Reynolds number effects on the performance of small-scale propellers. American Institute of Aeronautics and Astronautics. https://doi.org/10.2514/6.2014-2151
- Jacobs, Ward & Pinkerton (1933)
- Jacobs, E., Ward, K. & Pinkerton, R. (1933). The characteristics of 78 related airfoil sections from tests in the variable-density wind tunnel (460). National Advisory Committee for Aeronautics : Washington, DC. Retrieved from https://ntrs.nasa.gov/search.jsp?R=19930091108