2. Basics of helicopter aerodynamics

The basic flight regimes of helicopter include hover, climb, descent, and forward flight, and the analysis and study of these flight regimes can be approached by the actuator disk theory, where an infinite number of zero thickness blades support the thrust force generated by the rotation of the blades [1]. The air is assumed to be incompressible and the flow remains in the same direction (one-dimensional), which for most flight conditions is appropriate. The helicopter main rotor generates a vertical force in opposition to the helicopter's weight and a horizontal propulsive force for forward flight. Also, the main and tail rotors generate the forces and moments to control the attitude and position of the helicopter in three-dimensional space.

#### 2.1. Hovering flight

The cross sections in Figure 5 denote: the plane far upstream of the rotor, where in the hovering case the air velocity is null (section 0–0); the planes just above and below the rotor

Figure 5. The helicopter in hovering flight.

If the two rotors are mounted either side of the fuselage, on pylons or wing tips, the configu-

Another aircraft type that should be mentioned is the autogiro (invented by Huan de la Cievra), which is a hybrid between a helicopter and a fixed wing airplane. It uses a propeller for the forward propulsion and has freely spinning nonpowered main rotor that provides lift.

The basic flight regimes of helicopter include hover, climb, descent, and forward flight, and the analysis and study of these flight regimes can be approached by the actuator disk theory, where an infinite number of zero thickness blades support the thrust force generated by the rotation of the blades [1]. The air is assumed to be incompressible and the flow remains in the same direction (one-dimensional), which for most flight conditions is appropriate. The helicopter main rotor generates a vertical force in opposition to the helicopter's weight and a horizontal propulsive force for forward flight. Also, the main and tail rotors generate the forces and moments to control the attitude and position of the helicopter in three-dimensional space.

The cross sections in Figure 5 denote: the plane far upstream of the rotor, where in the hovering case the air velocity is null (section 0–0); the planes just above and below the rotor

ration is referred to as side by side (Figure 4).

Figure 4. The side by side rotors.

Figure 3. The coaxial rotors (a) and the intermeshing blades (b).

22 Flight Physics - Models, Techniques and Technologies

2. Basics of helicopter aerodynamics

2.1. Hovering flight

disk (sections 1–1, and 2–2); the far wake section, denoted by ∞. At the plane of rotor, the velocity through the rotor disk is vi (named the induced velocity) and in the far wake the air velocity is w. For a control volume surrounding the rotor and its wake, as shown in Figure 5 and dS ! ¼n ! �dS the unit normal area vector (the unit normal vector <sup>n</sup> ! is oriented outward the control volume), according to the Reynolds Transport Theorem, for any extensive parameter B, where B = b � m, the following equation is valid

$$
\left(\frac{dB}{dt}\right)\_{system} = \frac{\partial}{\partial t} \iiint\limits\_{control} \rho b dV + \iint\limits\_{control} (\rho b) \overrightarrow{V} \cdot d\overrightarrow{S} \tag{1}
$$

where V ! is the local velocity, m is the mass of fluid, and ρ is the fluid density. For a steady flow, the above equation becomes

$$
\left(\frac{dB}{dt}\right)\_{system} = \iint\_{control} (\rho b) \vec{V} \cdot d\vec{S} \tag{2}
$$

The conservation of mass (this case corresponds to B = m and b = 1)

$$
\left(\frac{dm}{dt}\right)\_{system} = \iint\limits\_{source} (\rho) \vec{V} \cdot d\vec{S} \tag{3}
$$

This equation requires the condition that the total amount of mass entering a control volume equals the total amount of mass leaving it. For steady-flow processes, we are not interested in the amount m of mass that flows in or out the control volume, but we are interested in amount of mass flowing per unit time, that is the mass flow rate, m\_ , well the conservation of fluid mass applied to this finite control volume can be rewritten as

$$-\iint\limits\_{surface2} \rho v\_l dS + \iint\limits\_{surface\text{v}} \rho u d\mathcal{S} = 0\tag{4}$$

Therefore,

$$
\rho v\_i A = \rho w A\_{\circ} \tag{5}
$$

The conservation of fluid momentum (this case corresponds to B ¼ mV ! and b ¼V ! )

$$
\left(\frac{dm\ \overrightarrow{V}}{dt}\right)\_{system} = \iint\limits\_{control} \left(\rho \overrightarrow{V}\right) \overrightarrow{V} \cdot d\overrightarrow{S} \tag{6}
$$

The principle of conservation of fluid momentum gives the relationship between the rotor thrust and the time rate of change of fluid momentum out of the control volume. The left part of Eq. (6) represent the sum of all forces that operate upon the control volume, namely the helicopter rotor thrust force, T. In projection on rotational axis, Eq. (6) becomes

$$T = w \iint\_{surfac\bullet} (\rho w) d\mathcal{S} = w\dot{m} \tag{7}$$

where m\_ is the mass flow rate in the control volume.

The conservation of energy (in this case <sup>B</sup> <sup>¼</sup> <sup>E</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> mV<sup>2</sup> and <sup>b</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> <sup>V</sup><sup>2</sup> )

$$
\left(\frac{dE}{dt}\right)\_{\text{sistem}} = \iint\_{\text{control}} \left(\rho \frac{1}{2} V^2\right) \vec{V} \cdot d\vec{S} \tag{8}
$$

The work done on the helicopter rotor is equal to the gain in energy of the fluid per unit time, and dE/dt represents the power consumed by the rotor, being equal to T � vi, therefore,

$$T \cdot \upsilon\_i = \iint\limits\_{control} \left(\rho \frac{1}{2} V^2\right) \overrightarrow{V} \cdot d\overrightarrow{S} = \frac{1}{2} w^2 \dot{m} \tag{9}$$

Taking into account that <sup>T</sup> <sup>¼</sup> mw\_ , we have mwv \_ <sup>i</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> <sup>w</sup><sup>2</sup>m\_ or vi <sup>¼</sup> <sup>1</sup> <sup>2</sup> w.

The conservation of mass (this case corresponds to B = m and b = 1)

24 Flight Physics - Models, Techniques and Technologies

applied to this finite control volume can be rewritten as

Therefore,

� ðð

surface2

The conservation of fluid momentum (this case corresponds to B ¼ mV

dm V! dt !

system ¼ ðð

helicopter rotor thrust force, T. In projection on rotational axis, Eq. (6) becomes

ðð

surface∞

control surface

ρ 1 2 V2 � �

T ¼ w

dE dt � �

sistem ¼ ðð

where m\_ is the mass flow rate in the control volume.

The conservation of energy (in this case <sup>B</sup> <sup>¼</sup> <sup>E</sup> <sup>¼</sup> <sup>1</sup>

dm dt � �

system ¼ ðð

ρvidS þ

control surface

ðð

ρwdS ¼ 0 (4)

ρviA ¼ ρwA<sup>∞</sup> (5)

!

<sup>ρ</sup><sup>w</sup> � �dS <sup>¼</sup> wm\_ (7)

<sup>2</sup> <sup>V</sup><sup>2</sup> ) and b ¼V ! )

surface∞

control surface

The principle of conservation of fluid momentum gives the relationship between the rotor thrust and the time rate of change of fluid momentum out of the control volume. The left part of Eq. (6) represent the sum of all forces that operate upon the control volume, namely the

ρV � � ! V ! �dS !

<sup>2</sup> mV<sup>2</sup> and <sup>b</sup> <sup>¼</sup> <sup>1</sup>

V ! �dS !

This equation requires the condition that the total amount of mass entering a control volume equals the total amount of mass leaving it. For steady-flow processes, we are not interested in the amount m of mass that flows in or out the control volume, but we are interested in amount of mass flowing per unit time, that is the mass flow rate, m\_ , well the conservation of fluid mass

ρ � �V ! �dS !

(3)

(6)

(8)

From the equation of continuity <sup>ρ</sup>viA <sup>=</sup> <sup>ρ</sup>wA∞, it follows that <sup>A</sup><sup>∞</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> <sup>A</sup> and obviously, <sup>r</sup><sup>∞</sup> <sup>¼</sup> <sup>R</sup>ffiffi 2 p therefore, the ratio of the rotor to the radius of the wake is <sup>R</sup>=r<sup>∞</sup> <sup>¼</sup> ffiffiffi 2 <sup>p</sup> .

Replacing the velocity w in the vena contracta (section ∞) in the expression of thrust force T, it follows that

$$T = \dot{m}w = \dot{m}(2v\_i) = \rho A v\_i (2v\_i) = 2\rho A v\_i^2 \tag{10}$$

The induced velocity at the plane of the rotor disk is vhover,

$$
\upsilon\_h = \upsilon\_i = \sqrt{\frac{T}{A} \frac{1}{2\rho}}\tag{11}
$$

This expression shows that induced velocity is dependent explicitly on the disk loading T/A, which is an important parameter in the helicopter design.

The power required to hover is the product between thrust T and induced velocity vi,

$$P = T \cdot \upsilon\_i = T \sqrt{\frac{T}{A} \frac{1}{2\rho}} = \frac{T^{\frac{3}{2}}}{\sqrt{2\rho A}} \tag{12}$$

This power, called the ideal power, forms the majority of the power consumed in hover, which is itself a high power-consuming helicopter flight regime.

In assessing rotor performance and compare calculations for different rotors, nondimensional quantities are useful. The induced velocity is normalized using the rotor tip speed, RΩ, where R is the rotor radius and Ω is the angular velocity,

$$
\lambda\_h = \frac{v\_i}{R\Omega} \tag{13}
$$

The parameter λ<sup>h</sup> is called the induced inflow ratio in hover.

The thrust force is also normalized like the lift for the fixed-wing, that is, the product of a pressure and an area, where the pressure is the dynamic pressure, considered at the rotor blade tips and the area is the total disk area, A = πR<sup>2</sup> , so, the thrust coefficient is defined by

$$C\_T = \frac{T}{\frac{1}{2}\rho (R\Omega)^2 \cdot A} \tag{14}$$

The inclusion on the half in the denominator is consistent with the lift coefficient definition for a fixed-wing aircraft. The rotor power, CP, and rotor torque, CQ, are defined as

$$\mathbf{C} = \frac{P}{\frac{1}{2}\rho (R\Omega)^3 \cdot A}; \quad \mathbf{C}\_Q = \frac{P}{\frac{1}{2}\rho (R\Omega)^2 \cdot R \cdot A} \tag{15}$$

Taking into account that power is related to torque by P = Ω � Q, then numerically CP = CQ.

Starting from the definition of the induced inflow ratio in hover, λh, it follows that

 $\lambda\_h = \frac{v\_i}{R\Omega} = \frac{1}{R\Omega}$  $\sqrt{\frac{T}{2\rho A}} = \sqrt{\frac{T}{4\frac{1}{2}\rho A (R\Omega)^2}} = \frac{1}{2}\sqrt{C\_T}$  $\text{ therefore } C\_T = 4\lambda\_h^2.$ 

The rotor power coefficient can be represented as

$$\mathbf{C}\_{P} = \frac{\mathbf{T} \cdot \boldsymbol{\upsilon}\_{i}}{\frac{1}{2}\rho(\mathbf{R}\boldsymbol{\Omega})^{3} \cdot \mathbf{A}} = \frac{\mathbf{T}}{\frac{1}{2}\rho(\mathbf{R}\boldsymbol{\Omega})^{2} \cdot \mathbf{A}} \frac{\boldsymbol{\upsilon}\_{i}}{(\mathbf{R}\boldsymbol{\Omega})} = \mathbf{C}\_{T} \cdot \boldsymbol{\lambda}\_{i\prime} \text{ or } \mathbf{C}\_{P} = \frac{1}{2}\mathbf{C}\_{T}^{\frac{3}{2}}\tag{16}$$

#### 2.2. Vertical climb

Considering the helicopter in climb, one can see that the flow enters the stream tube far upstream of the rotor and then passes through the rotor itself, finally passing away from the rotor forming the wake (Figure 6). When the helicopter leaves the hovering condition and moves in a vertical direction, the flow remains symmetrical about the thrust force line, which is normal to the rotor disk. The flow becomes very complex in a medium descent rate condition, but in climb, the mathematical approach is close to that used in the hover conditions.

The air enters the stream tube with velocity Vc and then acquires an additional velocity vi as it passes through the helicopter rotor disk, and finally, it forms the wake with a velocity Vc + vi . Applying the principles of conservation for mass, momentum, and energy like in the hover we get:

$$
\dot{m} = \rho A (V\_{\mathbb{C}} + \upsilon\_i); \quad T = \dot{m}w; \quad \dot{w} = 2\upsilon\_i \tag{17}
$$

Therefore, T ¼ mw\_ ¼ ρA Vð Þ� <sup>C</sup> þ vi 2vi and dividing by 2ρA it follows that

$$\frac{T}{2\rho A} = (V\_{\mathcal{C}} + \upsilon\_i)\upsilon\_i = V\_{\mathcal{C}} \cdot \upsilon\_i + \upsilon\_i^2 \tag{18}$$

The left part of the above equation represents the square of induced velocity in hover, v<sup>2</sup> <sup>h</sup>, and replacing it, we get

$$\left(\upsilon\_{\text{h}}^{2} = V\_{\text{C}} \cdot \upsilon\_{i} + \upsilon\_{i}^{2}\right) \quad \text{or} \quad \left(\frac{\upsilon\_{i}}{\upsilon\_{\text{h}}}\right)^{2} + \frac{V\_{\text{C}}}{\upsilon\_{\text{h}}} \cdot \left(\frac{\upsilon\_{i}}{\upsilon\_{\text{h}}}\right) - 1 = 0\tag{19}$$

The ratio vi/vh must always be positive in the climb, so the valid solution is

Figure 6. The axial climbing flight.

CT <sup>¼</sup> <sup>T</sup> 1

Taking into account that power is related to torque by P = Ω � Q, then numerically CP = CQ.

<sup>p</sup> , therefore CT <sup>¼</sup> <sup>4</sup>λ<sup>2</sup>

vi

h.

ð Þ <sup>R</sup><sup>Ω</sup> <sup>¼</sup> CT � <sup>λ</sup>i, or CP <sup>¼</sup> <sup>1</sup>

m\_ ¼ ρA Vð Þ <sup>C</sup> þ vi ; T ¼ mw; w \_ ¼ 2vi (17)

Starting from the definition of the induced inflow ratio in hover, λh, it follows that

<sup>2</sup> <sup>ρ</sup>ð Þ <sup>R</sup><sup>Ω</sup> <sup>2</sup> � <sup>A</sup>

but in climb, the mathematical approach is close to that used in the hover conditions.

Therefore, T ¼ mw\_ ¼ ρA Vð Þ� <sup>C</sup> þ vi 2vi and dividing by 2ρA it follows that

T

<sup>h</sup> <sup>¼</sup> VC � vi <sup>þ</sup> <sup>v</sup><sup>2</sup>

v2

Considering the helicopter in climb, one can see that the flow enters the stream tube far upstream of the rotor and then passes through the rotor itself, finally passing away from the rotor forming the wake (Figure 6). When the helicopter leaves the hovering condition and moves in a vertical direction, the flow remains symmetrical about the thrust force line, which is normal to the rotor disk. The flow becomes very complex in a medium descent rate condition,

The air enters the stream tube with velocity Vc and then acquires an additional velocity vi as it passes through the helicopter rotor disk, and finally, it forms the wake with a velocity Vc + vi

Applying the principles of conservation for mass, momentum, and energy like in the hover we get:

<sup>2</sup>ρ<sup>A</sup> <sup>¼</sup> ð Þ VC <sup>þ</sup> vi vi <sup>¼</sup> VC � vi <sup>þ</sup> <sup>v</sup><sup>2</sup>

vi vh � �<sup>2</sup> þ VC vh � vi vh � �

The left part of the above equation represents the square of induced velocity in hover, v<sup>2</sup>

<sup>i</sup> or

The ratio vi/vh must always be positive in the climb, so the valid solution is

a fixed-wing aircraft. The rotor power, CP, and rotor torque, CQ, are defined as

<sup>2</sup> <sup>ρ</sup>ð Þ <sup>R</sup><sup>Ω</sup> <sup>3</sup> � <sup>A</sup>

2 ffiffiffiffiffiffi CT

<sup>2</sup> <sup>ρ</sup>ð Þ <sup>R</sup><sup>Ω</sup> <sup>3</sup> � <sup>A</sup> <sup>¼</sup> <sup>T</sup> 1

CP <sup>¼</sup> <sup>P</sup> 1

<sup>λ</sup><sup>h</sup> <sup>¼</sup> vi

<sup>R</sup><sup>Ω</sup> <sup>¼</sup> <sup>1</sup> RΩ

2.2. Vertical climb

replacing it, we get

ffiffiffiffiffiffi T 2ρA q

26 Flight Physics - Models, Techniques and Technologies

<sup>¼</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>T</sup> 4� 1 <sup>2</sup>ρA Rð Þ <sup>Ω</sup> <sup>2</sup> <sup>q</sup> <sup>¼</sup> <sup>1</sup>

The rotor power coefficient can be represented as

CP <sup>¼</sup> <sup>T</sup> � vi 1

The inclusion on the half in the denominator is consistent with the lift coefficient definition for

; CQ <sup>¼</sup> <sup>P</sup> 1

<sup>2</sup> <sup>ρ</sup>ð Þ <sup>R</sup><sup>Ω</sup> <sup>2</sup> � <sup>A</sup> (14)

<sup>2</sup> <sup>ρ</sup>ð Þ <sup>R</sup><sup>Ω</sup> <sup>2</sup> � <sup>R</sup> � <sup>A</sup> (15)

<sup>i</sup> (18)

� 1 ¼ 0 (19)

<sup>T</sup> (16)

.

<sup>h</sup>, and

$$\frac{\upsilon\_i}{\upsilon\_h} = -\frac{1}{2}\frac{V\_\mathbb{C}}{\upsilon\_h} + \sqrt{\frac{1}{4}\left(\frac{V\_\mathbb{C}}{\upsilon\_h}\right)^2 + 1} \tag{20}$$

The power consumed is given by the product of the thrust and the total velocity through the rotor disk, that is

$$P = T(V\_c + \upsilon\_i) = T \cdot V\_{\mathbb{C}} + T \cdot \upsilon\_i = P\_{c \text{ lim } b} + P\_i \tag{21}$$

#### 2.3. Vertical descent

In the vertical descent, the air enters the stream tube from below the rotor with velocity VD and passes through the rotor disk with the velocity VD � vi, the wake being formed with velocity VD � w, as it is shown in Figure 7. The mass flow rate in vertical descent is m\_ ¼ ρA Vð Þ <sup>D</sup> þ vi , where VD is negative, and the conservation of momentum gives the thrust force

$$T = \iint\limits\_{\text{control}\atop \text{surface}} \left( \rho \overrightarrow{V} \right) \overrightarrow{V} \cdot d\overrightarrow{S} = -\dot{m} \cdot w \tag{22}$$

Figure 7. The stream tube in descent.

Even if the sign of thrust is negative, that does not mean that the thrust is negative, because the assumed sign convention consists of positive velocity w, in down direction. According to the conservation energy principle, it follows that

$$T \cdot (V\_D - v\_i) = -\frac{1}{2}\dot{m}w(2V\_D - w) \tag{23}$$

Replacing the expression of thrust T, namely T ¼ �mw\_ , in the above equation, we have

$$-\dot{m}w(V\_D - \upsilon\_i) = -\frac{1}{2}\dot{m}w(2V\_D - \overline{w})\tag{24}$$

therefore, vi <sup>¼</sup> <sup>w</sup> 2.

Similarly, to climb case, having the expression of the mass flow rate m\_ ¼ ρA Vð Þ <sup>D</sup> þ vi , where velocity VD is negative and vi is positive, we can write

$$T = -\dot{m}w = -\rho A (V\_D + \upsilon\_i) \cdot 2\upsilon\_i = -2\rho A (V\_D + \upsilon\_i)\upsilon\_i \tag{25}$$

so

$$\frac{T}{2\rho A} = -V\_D \cdot \upsilon\_i - \upsilon\_i^2 \tag{26}$$

Dividing by v<sup>2</sup> <sup>h</sup> <sup>¼</sup> <sup>T</sup> <sup>2</sup>ρ<sup>A</sup> the above equation becomes

$$
\left(\frac{\upsilon\_i}{\upsilon\_l}\right)^2 + \frac{V\_D}{\upsilon\_l} \left(\frac{\upsilon\_i}{\upsilon\_l}\right) + 1 = 0\tag{27}
$$

with the solutions

Even if the sign of thrust is negative, that does not mean that the thrust is negative, because the assumed sign convention consists of positive velocity w, in down direction. According to the

> 1 2

> > 1 2

Similarly, to climb case, having the expression of the mass flow rate m\_ ¼ ρA Vð Þ <sup>D</sup> þ vi , where

<sup>2</sup>ρ<sup>A</sup> ¼ �VD � vi � <sup>v</sup><sup>2</sup>

T ¼ �mw\_ ¼ �ρA Vð Þ� <sup>D</sup> þ vi 2vi ¼ �2ρA Vð Þ <sup>D</sup> þ vi vi (25)

mw\_ ð Þ 2VD � w (23)

mw\_ ð Þ 2VD � w (24)

<sup>i</sup> (26)

T � ð Þ¼� VD � vi

�mw V \_ ð Þ¼� <sup>D</sup> � vi

T

Replacing the expression of thrust T, namely T ¼ �mw\_ , in the above equation, we have

conservation energy principle, it follows that

velocity VD is negative and vi is positive, we can write

therefore, vi <sup>¼</sup> <sup>w</sup>

so

2.

Figure 7. The stream tube in descent.

28 Flight Physics - Models, Techniques and Technologies

$$\frac{\upsilon\_i}{\upsilon\_h} = -\frac{1}{2}\frac{V\_D}{\upsilon\_h} \pm \sqrt{\frac{1}{4}\left(\frac{V\_D}{\upsilon\_h}\right)^2 - 1} \tag{28}$$

In order to have real solutions, the following condition must be accomplished

$$\frac{1}{4} \left( \frac{V\_D}{v\_h} \right)^2 - 1 \ge 0 \tag{29}$$

That means, |VD|>2vh. The valid solution is

$$\frac{\upsilon\_{\text{i}}}{\upsilon\_{\text{h}}} = -\frac{1}{2}\frac{V\_{D}}{\upsilon\_{\text{h}}} - \sqrt{\frac{1}{4}\left(\frac{V\_{D}}{\upsilon\_{\text{h}}}\right)^{2} - 1} \tag{30}$$

In the region of flight that corresponds to �2 ≤ VD/vh ≤ 0, the control volume cannot be defined and the velocity curve can be defined experimentally. An approximation of the velocity in this region, called vortex ring state, could be [1]

$$\frac{\upsilon\_{l}}{\upsilon\_{l}} = k + k\_{1} \left(\frac{V\_{D}}{\upsilon\_{l}}\right) + k\_{2} \left(\frac{V\_{D}}{\upsilon\_{l}}\right)^{2} + k\_{3} \left(\frac{V\_{D}}{\upsilon\_{l}}\right)^{3} + k\_{4} \left(\frac{V\_{D}}{\upsilon\_{l}}\right)^{4} \tag{31}$$

with k = 0.974, k<sup>1</sup> = � 1.125, k<sup>2</sup> = � 1.372, k<sup>3</sup> = � 1.718, and k<sup>4</sup> = � 0.655.

Figure 8 shows the graphical results from this analysis, made in the Maple soft program.

In the normal working state of the rotor, if the climb velocity increases, the induced velocity decreases and also, in the windmill brake state if the descent velocity increases the induced velocity decreases and asymptotes to zero at high descent rates. In the vortex ring region, the induced velocity is approximated, because momentum theory cannot be applied. The flow in this region is unsteady and turbulent having upward and downward velocities. During normal powered flight, the rotor generates an induced airflow going downward and there is a recirculation of air at the blade tips, having the form of vortices, which exist because higher pressure air from below the rotor blade escapes into the lower pressure area above the blade. The rate of descent that is required to get into the vortex ring state varies with the speed of the induced airflow. Although vortices are always present around the edge of the rotor disk, under certain airflow conditions, they will intensify and, coupled with a stall spreading outward from the blade root, result in a sudden loss of rotor thrust. Vortex ring can only occur when the following conditions are present: power on, giving an induced flow down through rotor disk; a rate of descent, producing an external airflow directly opposing the induced flow; low forward speed.

Figure 8. Induced velocity variation.

#### 2.4. Power required in axial climbing and descending flight

In a climb or descent, the power ratio is

$$\frac{P}{P\_h} = \frac{V\_{\mathbb{C},D} + \upsilon\_i}{\upsilon\_h} = \frac{V\_{\mathbb{C},D}}{\upsilon\_h} + \frac{\upsilon\_i}{\upsilon\_h} \tag{32}$$

Using Eqs. (20) and (30), and substituting in the above equation, it follows that

• For a climb: <sup>P</sup> Ph <sup>¼</sup> <sup>1</sup> 2 VC vh þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4 VC vh � �<sup>2</sup> þ 1 r , which is valid for VC vh ≥ 0;

$$\bullet \quad \text{For a descent:} \frac{p}{P\_h} = \frac{1}{2} \frac{V\_D}{v\_h} - \sqrt{\frac{1}{4} \left(\frac{V\_D}{v\_h}\right)^2 - 1}, \text{ which is valid for } \frac{V\_C}{v\_h} \le -2;$$

For the vortex ring state, we can use the approximation (31) for the induced velocity ration, therefore in this case, the power ratio is

$$\frac{P}{P\_h} = \frac{V\_D}{\upsilon\_h} + \frac{\upsilon\_i}{\upsilon\_h} = \frac{V\_D}{\upsilon\_h} + k + k\_1 \left(\frac{V\_D}{\upsilon\_h}\right) + k\_2 \left(\frac{V\_D}{\upsilon\_h}\right)^2 + k\_3 \left(\frac{V\_D}{\upsilon\_h}\right)^3 + k\_4 \left(\frac{V\_D}{\upsilon\_h}\right)^4 \tag{33}$$

Using the same Maple soft program like for induced velocity, we obtain the following picture for the power ratio, P/Ph.

Figure 9. Power required as a function of climb and descent velocity.

According to the power to power in hover ratio values, shown in Figure 9, the power required to climb is always greater than the power required to hover, namely this ratio is greater than unity. In descent flight, the rotor extracts power from the air and uses less power than to hover.

#### 2.5. Induced velocity in forward flight

2.4. Power required in axial climbing and descending flight

1 4 VC vh � �<sup>2</sup>

r

1 4 VD vh � �<sup>2</sup>

r

<sup>¼</sup> VD vh

P Ph

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

þ k þ k<sup>1</sup>

<sup>¼</sup> VC,D <sup>þ</sup> vi vh

Using Eqs. (20) and (30), and substituting in the above equation, it follows that

þ 1

� 1

VD vh � �

For the vortex ring state, we can use the approximation (31) for the induced velocity ration,

þ k<sup>2</sup>

Using the same Maple soft program like for induced velocity, we obtain the following picture

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>¼</sup> VC,D vh þ vi vh

, which is valid for VC

, which is valid for VC

VD vh � �<sup>2</sup>

þ k<sup>3</sup>

vh ≥ 0;

vh ≤ � 2;

VD vh � �<sup>3</sup>

þ k<sup>4</sup>

VD vh � �<sup>4</sup> (32)

(33)

In a climb or descent, the power ratio is

30 Flight Physics - Models, Techniques and Technologies

Figure 8. Induced velocity variation.

Ph <sup>¼</sup> <sup>1</sup> 2 VC vh þ

therefore in this case, the power ratio is

<sup>¼</sup> VD vh þ vi vh

Ph <sup>¼</sup> <sup>1</sup> 2 VD vh �

• For a climb: <sup>P</sup>

• For a descent: <sup>P</sup>

P Ph

for the power ratio, P/Ph.

In forward flight, the rotor must be tilted (Figure 10) in order to have a propulsive force to propel the helicopter forward, with a velocity V∞. This velocity has two components: one component normal to the rotor disk plane, V<sup>∞</sup> sin α, and another component, parallel to the rotor disk plane, V<sup>∞</sup> cos α.

The rotor thrust, T, is given by T ¼ 2ρAvi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð Þ V<sup>∞</sup> cos α <sup>2</sup> <sup>þ</sup> ð Þ <sup>V</sup><sup>∞</sup> sin <sup>α</sup> <sup>þ</sup> vi 2 q and the induced velocity in forward flight can be written as

$$w\_i = \frac{\frac{T}{2\rho A}}{\sqrt{\left(V\_\infty \cos a\right)^2 + \left(V\_\infty \sin a + v\_i\right)^2}} = \frac{v\_h^2}{\sqrt{\left(V\_\infty \cos a\right)^2 + \left(V\_\infty \sin a + v\_i\right)^2}}\tag{34}$$

In order to get an analytical solution for the induced velocity, it is necessary to define two coefficients: the advance ratio, μ, and the inflow ratio, λ, as it follows

<sup>μ</sup> <sup>¼</sup> <sup>V</sup><sup>∞</sup> cos <sup>α</sup> <sup>R</sup><sup>Ω</sup> (35)

$$
\lambda = \frac{V\_{\text{ov}}\sin\alpha + v\_i}{R\Omega} = \frac{V\_{\text{ov}}\sin\alpha}{R\Omega} + \frac{v\_i}{R\Omega} = \mu\tan\alpha + \lambda\_i \tag{36}
$$

Dividing Eq. (34) to RΩ, we get

$$\frac{\upsilon\_i}{R\Omega} = \lambda\_i = \frac{\frac{v\_h^{\epsilon}}{\left(R\Omega\right)^2}}{\sqrt{\left(\frac{V\_\bullet - \cos\alpha}{R\Omega}\right)^2 + \left(\frac{V\_\bullet - \sin\alpha + v\_i}{R\Omega}\right)^2}} = \frac{\lambda\_h^2}{\sqrt{\mu^2 + \lambda^2}}\tag{37}$$

This expression leads to the following equation for the inflow ratio, λ,

$$\lambda = \mu \tan \alpha + \frac{\lambda\_h^2}{\sqrt{\mu^2 + \lambda^2}}, \text{ or } \frac{\lambda}{\lambda\_h} = \frac{\mu}{\lambda\_h} \tan \alpha + \frac{1}{\sqrt{\left(\frac{\mu}{\lambda\_h}\right)^2 + \left(\frac{\lambda}{\lambda\_h}\right)^2}}\tag{38}$$

The above equation can be very easy to be solved in Maple soft. In Figure 11, three curves are shown, in coordinates <sup>μ</sup> <sup>λ</sup><sup>h</sup> and <sup>λ</sup> λh , representing three values of angle α, namely α = 0 deg, α = 4 deg, and α = 6 deg. The command plot used in Maple was "implicitplot" for the equation <sup>y</sup> <sup>¼</sup> <sup>x</sup> � tan <sup>α</sup> <sup>þ</sup> <sup>1</sup>ffiffiffiffiffiffiffiffiffi <sup>x</sup>2þy<sup>2</sup> <sup>p</sup> .

The inflow ratio, λ/λ<sup>h</sup> increases with the increase of the rotor disk angles of attack.

Figure 10. Rotor in forward flight.

Figure 11. Inflow ratio λ/λ<sup>h</sup> as a function of forward speed ratio.

## 3. Helicopter systems

#### 3.1. Main rotor systems

<sup>μ</sup> <sup>¼</sup> <sup>V</sup><sup>∞</sup> cos <sup>α</sup>

v2 h ð Þ <sup>R</sup><sup>Ω</sup> <sup>2</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

> λ λh ¼ μ λh

The above equation can be very easy to be solved in Maple soft. In Figure 11, three curves are

deg, and α = 6 deg. The command plot used in Maple was "implicitplot" for the equation

The inflow ratio, λ/λ<sup>h</sup> increases with the increase of the rotor disk angles of attack.

RΩ þ

<sup>þ</sup> <sup>V</sup><sup>∞</sup> sin <sup>α</sup>þvi RΩ � �<sup>2</sup> <sup>q</sup> <sup>¼</sup> <sup>λ</sup><sup>2</sup>

tan α þ

, representing three values of angle α, namely α = 0 deg, α = 4

vi

<sup>R</sup><sup>Ω</sup> <sup>¼</sup> <sup>V</sup><sup>∞</sup> sin <sup>α</sup>

V<sup>∞</sup> cos α RΩ � �<sup>2</sup>

This expression leads to the following equation for the inflow ratio, λ,

λ2 h ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>μ</sup><sup>2</sup> <sup>þ</sup> <sup>λ</sup><sup>2</sup> <sup>q</sup> ; or

<sup>λ</sup> <sup>¼</sup> <sup>V</sup><sup>∞</sup> sin <sup>α</sup> <sup>þ</sup> vi

Dividing Eq. (34) to RΩ, we get

32 Flight Physics - Models, Techniques and Technologies

shown, in coordinates <sup>μ</sup>

Figure 10. Rotor in forward flight.

<sup>x</sup>2þy<sup>2</sup> <sup>p</sup> .

<sup>y</sup> <sup>¼</sup> <sup>x</sup> � tan <sup>α</sup> <sup>þ</sup> <sup>1</sup>ffiffiffiffiffiffiffiffiffi

vi <sup>R</sup><sup>Ω</sup> <sup>¼</sup> <sup>λ</sup><sup>i</sup> <sup>¼</sup>

λ ¼ μ tan α þ

<sup>λ</sup><sup>h</sup> and <sup>λ</sup> λh <sup>R</sup><sup>Ω</sup> (35)

h ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>μ</sup><sup>2</sup> <sup>þ</sup> <sup>λ</sup><sup>2</sup>

1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi μ λh � �<sup>2</sup>

<sup>þ</sup> <sup>λ</sup> λh

<sup>R</sup><sup>Ω</sup> <sup>¼</sup> <sup>μ</sup> tan <sup>α</sup> <sup>þ</sup> <sup>λ</sup><sup>i</sup> (36)

<sup>q</sup> (37)

� �<sup>2</sup> <sup>r</sup> (38)

The primary way to distinguish between different main rotor systems is represented by the movement of the blade relative to the main rotor hub. The main categories are fully articulated, semi rigid, and rigid. In hovering flight, the blades flap up and lag back with respect to the hub and reach equilibrium position under the action of aerodynamic and centrifugal forces. In forward flight, the asymmetry of the dynamic pressure over the disk produces aerodynamic forces that are the functions of the blade azimuth position. The hinges allow each blade to independently flap and lead or lag with respect to the hub plane. The lead-lag hinge allows inplane motion of the blade due to the Coriolis and radius of gyration changing in flapping movement. Transition from hover to forward flight introduces additional aerodynamic forces and effects that are not found when the helicopter is in stationary hover. Due to the difference in relative airspeed between the advancing and retreating blades, the lift is constantly changing through each revolution of the rotor.

Figure 12 shows the flapping, lead-lag, and feathering motion of a rotor blade.

In a fully articulated rotor, each main rotor blade is free to move up and down (flapping), to move forth and back (dragging), and to twist about the spanwise axis (feathering). Semi rigid rotor has, normally, two blades attached rigidly to the main rotor hub and is free to tilt and rock independently of the main rotor mast, one blade flaps up and other flaps down.

The rigid rotor system cannot flap or drag, but it can be feathered. The natural frequency of the rigid rotor is high, so the stability is difficult to be achieved.

#### 3.2. Anti-torque system

The single rotor helicopters require a separate rotor to overcome the effect of torque reaction, namely the tendency for the helicopter to turn in the opposite direction to that of the main rotor. The anti-torque pedals are operated by the pilot's feet and vary the force produced by the tail rotor to oppose torque reaction.

#### 3.3. Swash plate assembly

It has the purpose to transmit cyclic and collective control movements to the main rotor blades and consists of a stationary plate and a rotating plate. The stationary plate is attached to the

Figure 12. Blade movement axis.

main rotor mast and the rotating plate is attached to the stationary plate by a bearing surface and rotates at the same speed as the main rotor blades.

#### 3.4. Trim

Figure 12 shows the flapping, lead-lag, and feathering motion of a rotor blade.

rigid rotor is high, so the stability is difficult to be achieved.

3.2. Anti-torque system

3.3. Swash plate assembly

Figure 12. Blade movement axis.

the tail rotor to oppose torque reaction.

34 Flight Physics - Models, Techniques and Technologies

rock independently of the main rotor mast, one blade flaps up and other flaps down.

In a fully articulated rotor, each main rotor blade is free to move up and down (flapping), to move forth and back (dragging), and to twist about the spanwise axis (feathering). Semi rigid rotor has, normally, two blades attached rigidly to the main rotor hub and is free to tilt and

The rigid rotor system cannot flap or drag, but it can be feathered. The natural frequency of the

The single rotor helicopters require a separate rotor to overcome the effect of torque reaction, namely the tendency for the helicopter to turn in the opposite direction to that of the main rotor. The anti-torque pedals are operated by the pilot's feet and vary the force produced by

It has the purpose to transmit cyclic and collective control movements to the main rotor blades and consists of a stationary plate and a rotating plate. The stationary plate is attached to the The neutral position of the cyclic stick changes as the helicopter moves off from to hover in forward flight. Trim control can adjust the mechanical feel in flight by changing the neutral position of the stick.

#### 3.5. Collective and cyclic pitch control

Collective pitch lever controls the lift produced by the rotor, while the cyclic pitch controls the pitch angle of the rotor blades in their cyclic rotation. This tilts the main rotor tip-path plane to allow forward, backward, or lateral movement of the helicopter.
