New Transition Method of a Ducted-Fan Unmanned Aerial Vehicle

Weize Zhang,∗ Quan Quan,† Ruifeng Zhang,‡ and Kai-Yuan Cai§

Beijing University of Aeronautics and Astronautics, 100191 Beijing, People’s Republic of China

DOI: 10.2514/1.C032073

In this research, a winged version of the ISTAR (from “intelligence, surveillance, target acquisition, and

reconnaissance”) unmanned aerial vehicle is modeled first, and then a new transition method is proposed, which

enables the vehicle to fly sideways at low speed for drag reduction.Anonlinear trajectory-following logic is adopted to

correct altitude loss and horizontal drift during the transition. By simulation, the proposed transition method shows

better acceleration performance. The comparison of the induced drag shows that the vehicle should roll to resume

fixed-wing lift as soon as its minimum level-flight speed is achieved to optimize its acceleration performance.

Nomenclature

α, β = angle of attack and angle of side slipacmd = lateral accelerationCd, C

0d = drag coefficient

Cl = lift coefficientClcs, Clα, Cyβ = aerodynamic force curve slopeCy = side force coefficientD, D 0 = dragδi = control surface deflectionF,M = force and moment vectorsG = gravityib = moment of inertia of one bladeJ = inertia matrixL = liftm = vehicle massnb = number of bladesR = direction cosine matrixSwing, Sside = reference areaT = thrustt = timeY = side force

I. Introduction

AMONG various configurations of unmanned aerial vehicles(UAVs), the ducted-fan tail-sitter vertical takeoff and landing

(VTOL) UAV is a simple configuration to realize VTOL function.This advantage has led to the development of several of thesevehicles, such as the GoldenEye, Kestrel, and ISTAR (from“intelligence, surveillance, target acquisition, and reconnaissance”).The ISTAR is themost widely used one. Its general layout consists ofa ducted fan and deflectable vanes at the end of the duct.Different battlefield situationsmay require thevehicle to fly at high

speed (e.g., ground-fire evasion) or at low speed (e.g., battlefieldsurveillance and reconnaissance). In hover and low-speed levelflight, the vehicle’s longitudinal axis is vertical, or with a small angleof inclination [1]. In transition mode, it tilts its body forward toaccelerate. In high-speed level-flight mode, the vehicle’s longitudinalaxis is almost parallel to the horizon [2,3].

ISTARhasmany types of different sizes andweights. Despite suchdifferences, an axisymmetric configuration, as shown in Fig. 1, isshared by all types, which makes it easy to give an aerodynamicmodel so as to simplify control laws [4], but it also has an evidentdisadvantage in high-speed level flight because the lift-to-drag ratio isrelatively low, thus reducing level-flight efficiency.One solution is to add fixed wings. Figure 2 shows a winged

version of an ISTARUAV. Fixed wings enhance its high-speed flightefficiency but generate tremendous drag at a high angle of attack,when the UAV flies at low speed, or when it initializes its transitionflight, with an angle of attack of near 90 deg.An existing solution is touse a variable-incidence wing, for example the GoldenEye, as shownin Fig. 3. However, its pitching mechanism increases the weightand the complexity. To surmount this difficulty, the UAV may flysideways at low speed for drag reduction, because of a smallerwindward side area, and transfer directly from flying sideways tohigh-speed mode after a roll maneuver, as shown in Fig. 4. In fact,before the rollmaneuver is executed, theUAVflies like a helicopter, or awingless versionof ISTAR,because the fixedwing creates no lift at zeroangle of attack. In this case, the control law could be based on thatdeveloped for the wingless version of ISTAR. But after the rollmaneuver, fixed-wing lift is resumed, and a new control law is required.The UAVwill gain two advantages by flying sideways. First of all,

greater acceleration will be achieved because of the drag reduction.Second, the variable-incidence wing is waived off and fixed wing isenough. As described previously, a special control law is required totilt its body in a relatively complicatedway,while eliminating altitudeloss and horizontal drift caused by the roll maneuver. In this research,a nonlinear control law is developed and proved,which allows posingtheUAVat the desired attitude andmaintaining the original trajectory.One possible disadvantage of flying sideways is increased

difficulty of holding a position in a horizontal varying wind. This isequivalent to an ordinary fixed-wing aircraft flying at a bank angle of90 deg. Because this preliminary study is for illustrating a newconcept, such a problem will be discussed in future work.The UAV under consideration has a layout similar to that of the

winged version of ISTAR. It consists of a ducted fan, a rotor, fourvanes at the end of the duct, and two fixed wings, as shown in Fig. 5.The vanes attached under the slipstream of the propeller alwaysproduce control moments at all points of the flight envelope.This paper is organized as follows. Amathematical model is given

in Sec. II, an attitude-control law is developed in Sec. III, a trajectory-following method is adopted to correct horizontal and verticaldeviation in Sec. IV, and the simulation results are given in Sec. V.

II. Mathematical Model

Amodel of the UAVwas created for simulation to verify controllerbehavior. The following sections detail the formulation of this model,with numerical values listed in Appendix A.Let fe1; e2; e3g denote an inertial frame and fex; ey; ezg denote a

body frame rigidly attached to the aircraft whose origin is attached tothe center of gravity, where e1, e2, e3, ex, ey, and ez are the unitvectors of axis x 0, y 0, z 0, x, y, and z, as shown in Fig. 6.

Received 16 August 2012; revision received 8 January 2013; accepted forpublication 6 February 2013; published online 28 June 2013. Copyright ©2013 by the American Institute of Aeronautics and Astronautics, Inc. Allrights reserved. Copies of this paper may be made for personal or internal use,on condition that the copier pay the $10.00 per-copy fee to the CopyrightClearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923; includethe code 1542-3868/13 and $10.00 in correspondence with the CCC.

*Graduate Student, Department of Automatic Control; currently Ph.D.Student, Department of Mechanical Engineering, Mcgill University, Quebec,Canada.

†Assistant Professor, Department of Automatic Control.‡Ph.D., Department of Automatic Control.§Professor, Department of Automatic Control.

1131

JOURNAL OF AIRCRAFT

Vol. 50, No. 4, July–August 2013

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In the following, all vectors are expressed in the body frame, unlessotherwise specified.The vane system has four control surfaces, and each operates

independently. They are numbered 1, 2, 3, and 4, and their deflectioninput is shown in Fig. 7.Equivalent vane deflections (δx, δy, and δz) of roll (z axis), yaw

(y axis), and pitch (x axis) are given as follows:

δz � δ1 � δ2 � δ3 � δ4

δy � δ2 − δ4

δx � δ3 − δ1 (1)

where δ1, δ2, δ3, and δ4 denote the deflection of vane numbers 1, 2, 3,and 4, respectively. The maximum deflection angle of the vanes is setto 28 deg.To overcome singularity points, a quaternion attitude description is

adopted as a global attitude reference [5,6]. This globally nonsingularrepresentation of the orientation is given by the vector �q0 qT �Tsubjected to the constraint q20 � qTq � 1. q is a three-dimensionalvector presented as q � �q1 q2 q3 �T .Note that the quaternion describing the vehicle in hover is� 1 0 0 0 �T . In this case, the axes x, y, and z in the body framecoincide with the axes x 0, y 0, and z 0 in the inertial frame.

Fig. 1 ISTAR UAV.

Fig. 2 Winged version of an ISTAR UAV.

Fig. 3 Variable-incidence wing of the GoldenEye UAV.

Fig. 4 Conventional and new transition flight.

Fig. 5 General layout of the UAV under consideration.

Fig. 6 Inertial frame and body frame.

Fig. 7 Vanes.

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The direction cosine matrix R that transfers vectors in the inertialframe to the body frame can be written as [7]

R�q0q� � I3×3 � 2S�q�2 − 2q0S�q� (2)

where S�·� is the skew-symmetric matrix such that S�u�v � u × v,where × denotes the vector cross product and u, v ∈ R3.It was of primary importance to model the major forces and

moments acting on the vehicle. Simplifying assumptions were madein certain respects.The basic dynamic equations are given as

_p � v

_v � RTF

m

J _Ω � −Ω × JΩ�M

_q � 1

2�q ×Ω� q0Ω�

_q0 � −1

2ΩTq (3)

where p denotes the position vector of the UAV in the inertial frame.The vector v � � v1 v2 v3 �T denotes the velocity components inthe inertial frame along axes x 0, y 0, and z 0, respectively. The vectorΩ � �Ωx Ωy Ωz �T denotes the angular velocity of the vehicleexpressed in the body frame, whereΩx,Ωy, andΩz are, respectively,the pitch, yaw, and roll rates. Here, J denotes the inertia matrix of theairframe, while F and M represent the sum of external forces andmoment vectors acting on the vehicle, expressed in the body frame.The force and moment vectors can be expressed as

F � Faero � Frotor � Fcw � Fcs � Fgrav � Fd (4)

M � Maero �Mrotor �Mcw �Mcs �Mgyro �Md (5)

These component forces and moments are discussed next.

A. Vehicle Aerodynamics

In the following section, all vectors are projected to the body framefor simplification. The projection of the velocity vector to the bodyframe can be written as "

vxvyvz

#� R

"v1v2v3

#(6)

The aerodynamic force consists of the lift, the drag, and the side force.Note that the resultant of duct consists of two contributions: one fromaerodynamics, and one from the crosswind effect of the duct body [4].Here, the aerodynamical contributions of duct and fixed wing areconsidered altogether as a lifting body, and the contribution ofcrosswind effectwill be discussed separately in Sec. II.C. Because thevehicle can maneuver at high angle of attack α and high angle of sideslip β, these two angles are of primary importance. They can beexpressed as follows:

α �

8>>>>><>>>>>:

0 if vy � 0

−tan−1�vyvz

�if vz ≥ 0 and vy ≠ 0

−tan−1�vyvz

�� π if vz < 0 and vy ≠ 0

(7)

β �

8>>>>><>>>>>:

0 if vx � 0

tan−1�vxvz

�if vz ≥ 0 and vx ≠ 0

tan−1�vxvz

�� π if vz < 0 and vx ≠ 0

(8)

Note that Eqs. (7) and (8) are well defined when vz � 0 becausetan−1��∞� � � π

2. Here, Swing is the reference area of the lift and the

drag, and Sside is that of the side force. LetClα andCyβ denote the liftcurve slope and the side-slip force curve slope of the lifting body.Given the fact that the symmetric airfoil is adopted, the coefficients oflift, drag, and side force can be expressed approximately by thesinusoidal expressions written as Eq. (9). The same method isadopted in [4]:

Cl � max

�−Cl;max;min

�Cl;max;

1

2Clα sin�2α�

��Cd � Cd;gain�1 − cos�2α�� � Cd0

Cy � max

�−Cy;max;min

�Cy;max;

1

2Cyβ sin�2β�

��(9)

In [4], this modeling technique is used for an airfoil, but here weassume that it is also acceptable for the entire lifting body. Suchsimplified modeling causes some errors, especially at high angle ofattack and angle of side slip, thus the disturbing moment Md anddisturbing force Fd could not be neglected, which have negativeeffects on attitude and position control. Such disturbance will becompensated by a sliding-mode control and a trajectory-followinglogic, as discussed in following sections.Remark 1: Although Cl is the lift coefficient of the whole lifting

body, including the aerodynamical contributions from the duct andthe fixed wing, it does not take the crosswind effect of the duct intoaccount. This effect will be described separately in Sec. II.C.The minimum level-flight speed could be derived from Cl;max as

follows:

vmin ���������������������������������

mg12Cl;maxρ∞Swing

r(10)

By plugging the numerical values into Eq. (10), one can get aminimum level-flight speed of 23.1 m∕s.The lift L, drag D, and side force Y in the body frame can be

expressed as follows:

L �

24 0

12Clρ∞�v2y � v2z�Swing cos α

12Clρ∞�v2y � v2z�Swing sin α

35 (11)

D �

24 0

12Cdρ∞�v2y � v2z�Swing sin α

− 12Cdρ∞�v2y � v2z�Swing cos α

35 (12)

Y �

24− 1

2Cyρ∞�v2x � v2z�Sside cos β

012Cyρ∞�v2x � v2z�Sside sin β

35 (13)

Figure 8 shows the axes of the body framewith respect toα and β. Thedensity of air ρ∞ is supposed to be constant.Because the vehicle may experience high angle of side slip,

additional drag is generated (the induced drag of side force) and isperpendicular to the side force. By analogy to the discussion of thecomponents of D, the additional drag can be expressed as

C 0d � C 0d;gain�1 − cos�2β�� (14)

D 0 �

24 − 1

2C 0dρ∞�v2x � v2z�Sside sin β

0

− 12C 0dρ∞�v2x � v2z�Sside cos β

35 (15)

In terms of aerodynamic moment, given the symmetric layout of thevehicle, let the vector � 0 0 −z1 �T denote the position vector of the

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aerodynamic center of lift and drag, and let the vector � 0 0 −z2 �Tdenote that of side force and the additional drag. Notice that z1 and z2are not identical, generally. The negative sign means that theseaerodynamic centers are behind the center of gravity, this being theresult of aerodynamically stable design.Therefore, the terms Faero and Maero representing aerodynamic

force and moment can be expressed as

Faero � L�D� Y �D 0

Maero � Maero1 �Maero2 (16)

where

Maero1 � � 0 0 −z1 �T × �L�D�Maero2 � � 0 0 −z2 �T × �Y �D 0�

B. Rotor

The term Frotor representing thrust vector can now be expressed as

T � bω2r Frotor �

24 0

0

T

35 (17)

The term Mrotor representing the rotor torque vector is given asfollows:

Q � κω2r Mrotor �

24 0

0

Q

35 (18)

where b and κ are two positive parameters depending on the densityof air, the size and twist of blades, and other factors [8]. Here, ωrrepresents the angular velocity of the motor.The airflow and the induced velocity through the rotor are negative

because the airflow is accelerated in the negative (−z) direction. Theinduced velocity can be expressed as follows [9]:

vi �1

2

vz −

���������������������v2z �

2T

ρπr2

s !(19)

where r represents the radius of blades.

C. Crosswind Effect of Duct

In the presence of crosswind, the duct is subjected to additionalforce and moment vector, which can be expressed as [4]

Fcw � viρ∞πr2"vxvy0

#(20)

Mcw � Cductρ∞πr

"−vyjvyjvxjvxj0

#(21)

Note that this model is approximate. The vector Mcw measured by[10] and other references such as [1] shows thatMcw varies linearlywith crosswind velocity, not in a quadratic manner as described byEq. (21). This quadratic model is simple and has modeling errors,which contributes to the disturbing moment. But [4] shows that thiserror could be compensated if the control law is properly designed. Arobust sliding mode control law will be shown in Sec. III, and itsrobustness will be proved by assuming that the disturbing moment isbounded.

D. Control Surfaces

The control surfaces (vanes) are exposed in the downwash of therotor. The dynamic pressure qcs can be written as follows:

qcs �1

2ρ∞�vz − vi�2 (22)

LetClcs denote the lift curve slope of the vanes. It is assumed that thedistance between the aerodynamic center of a vane and the z axis, yaxis, and x axis of the body frame are respectively la, lr, and le. Notethat le � lr due to symmetry. By adding a constraint,

δ2 � −δ4 (23)

one can express � δ1 δ2 δ3 δ4 �T with � δx δy δz �T as follows:2664δ1δ2δ3δ4

3775 �

2664−0:5 0 0:50 0:5 0

0:5 0 0:50 −0:5 0

3775" δxδyδz

#(24)

The constraint [Eq. (23)] is necessary because the system [Eq. (1)]has more unknowns than equations; thus, Eq. (1) alone is not enoughto express vane deflections as a function of control inputs.The terms Fcs andMcs representing the force and the moment of

vanes can be written as

Fcs � qcsScsClcs

"−δ2 � δ4δ3 − δ1

0

#(25)

Mcs � qcsScsClcs

24 le 0 0

0 lr 0

0 0 la

3524 δxδyδz

35 (26)

Note that, for vanes, the effect of the drag forces has been neglected.Once the control inputMcs is obtained from the control law, vane

deflections could be derived by inverting Eq. (26) and then pluggingin to Eq. (24).

E. Gravity

Gravitational forces are accounted for as follows, with m as thetotal mass of the UAVand g the acceleration due to gravity:

Fig. 8 Angle of attack and angle of side slip.

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Fgrav � R

24 0

0

−mg

35 � R

24 0

0

−G

35 (27)

F. Gyroscopic Moment

The rotor rotates counterclockwise when seen from the front view.By assuming that the rate of change of the rotor’s angular velocity isnegligible, the expression for this moment vector is given next [4]:

Mgyro � nbibωr

"−ΩyΩx0

#(28)

III. Attitude-Control Design

In this section, a feedback control is developed to stabilize thevehicle at any attitude; the vehicle dynamics of Eq. (3) are used. Toavoid the sign ambiguity described in [3], the Euler angles and theequivalent rotation angle are taken between −π and π, which impliesq0 ∈ � 0; 1 �. Let �qr0 qTr �T denote the quaternion describing theobjective attitude. The vector of control signals isMcs.The error quaternion, which is part of the state vector of the error

system, is defined as

�qe0 qTe � � �qr0 qTr � � �q0 qT � � �qr0 −qTr � � �q0 qT �

where � denotes quaternion multiplication.Because � qe0 qTe � is, in fact, the expression of � q0 qT � in

another inertial frame, the fourth and fifth equation of Eq. (3) are alsoverified for �qe0 qTe �. The error system can be expressed as

_qe �1

2�qe ×Ω� qe0Ω�

_qe0 � −1

2ΩTqe

J _Ω � −Ω × JΩ −Mcom �Mcs �Md (29)

whereMcom � −Maero −Mrotor −Mcw −Mgyro denotes the momentto compensate.Given the presence ofMd, a proportional-derivative sliding mode

controller is proposed to neutralize the disturbance.Theorem 1: Under the following control law:

Mcs � J�−1

2s − Kssign�s� � Kcom

��Mcom (30)

where

s � qe �Ω

sign�s� �

2664sign�s1�sign�s2�sign�s3�

3775

Kcom � J−1Ω × JΩ −1

2�qe × Ω� qe0Ω�

the equilibrium point of system (29)

�qe0 qTe �T � � 1 0 0 0 �T

is globally asymptotically stable.Note that there is an implied gain of 1 with units in the definition of

s, because one cannot add quaternion (with no units) to angularvelocity (with units).The proof of this control law could be found in Appendix B.

Figure 9 shows the simulated result of attitude step change com-mand of �q0 qT �T from �1 0 0 0�T to � 0:6 0:3 0:7 0:245 �T ,with excellent speed of convergence. It takes 1–2 s to convergeto desired attitude. This controller will also be effective in tracking aslowly changing reference, as will be shown in the followingsections.

IV. Trajectory-Following Logic

In this section, our goal is to accomplish the transition withoutaltitude loss or horizontal drift. Any deviation from y 0 � 0 or z 0 � 0should be compensated by a lateral acceleration. In this article, thefollowing trajectory-following logic is adopted [11], as shown inFig. 10.

acmd � 2V2

L1

sin η (31)

The lateral acceleration acmd is perpendicular to velocity vector,pointing toward the desired trajectory. The reference point is on thedesired path at a distance L1 forward of the vehicle. In this research,L1 is set to a value significantly greater than the deviation so that η is asmall angle. Note that Fig. 10 is a simple schema in two dimensions,while the UAV maneuvers in three dimensions. A detailed schemawith coordinate frame as reference will be shown in Figs. 11–13.A significant property of this guidance logic is that the lateral

acceleration changes sign at some points, which guarantees a smoothconvergence to the desired path. Besides, considering the fact that the

Fig. 9 Quaternion attitude step response.

Fig. 10 Trajectory-following logic.

Fig. 11 Schema of the first phase.

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UAV may fly in narrow indoor circumstances, such a control lawoffers quick convergence to a curved path, which allows avoidingindoor obstacles [11].With small-angle approximation, this guidancelogic is equivalent to a proportional-integral-derivative (PID)controller [12], which eliminates static error of trajectory trackingdespite the presence of disturbing force.As mentioned in Sec. I, the transition consists of two phases,

during which the lateral acceleration is generated in respectivelydifferent ways. In the first phase, the vehicle flies sideways. Thegravity G is balanced by the vertical component of side force Y andthrust T (Fig. 11). Because there is no horizontal drift from y 0 � 0,the lateral acceleration defined in Eq. (31) only consists ofcomponents in the x 0-z 0 plane. Such components are generated by theresultant ofG,Y, andT. As it accelerates, it inclines forward, and thusthe angle of inclination increases over time.When the speed reaches athreshold (defined by the minimum level-flight speed with themaximum lift coefficient of the lifting body), the second phasebegins, and the vehicle rolls to transfer to high-speed level flight.Fixed-wing lift is resumed. This roll maneuver causes somehorizontal drift (in the x 0-y 0 plane) and altitude loss (in the x 0-z 0

plane), and thus the lateral acceleration has two components tocorrect the deviations in two planes. In the x 0-y 0 plane, lateralacceleration is generated by the side force Y (Fig. 12), while in thex 0-z 0 plane, it is generated by the resultant of lift L and gravity G(Fig. 13). The skid-to-turn technique is used. The conventionalcoordinated turn is not adopted because the vehicle has a high sideforce curve slope, and thus the skid-to-turn is faster to correct forsmall errors in the x 0-y 0 plane. The high side-force curve slope is dueto the vehicle’s duct: for a standard wingless ISTAR UAV, the ductalone can generate sufficient lift, and because of its axisymmetriclayout, the curve slope of the lift and that of the side force areidentical.In the following, � x 0 y 0 z 0 �T denotes the vector p, and� x 0obj y 0obj z 0obj �T denotes the position of the reference point of thetrajectory-following logic, expressed in the inertial frame. Note that,by small-angle approximation, x 0obj � x 0 � L1.Given that this research focuses on transition flight, it is assumed

that the thrust is a constant. Note that such approximation isreasonable at low speed but may overestimate the thrust at higherspeed because the thrust coefficient generally decreases as speedincreases [9].The desired attitude will be calculated to generate the needed

lateral acceleration. In the first phase, this attitude is characterized bythe angle of inclination γ (Fig. 11), whereas in the second phase, it is

characterized by the yaw angle θ (Fig. 12) and the pitch angle ϕ(Fig. 13).

A. First Phase

The angle of inclination γ is shown on Fig. 11. The lateralacceleration acmd can be expressed as follows:

acmd � 2v21 � v23�����������������������������������������������������

�x 0 − x 0obj�2 � �z 0 − z 0obj�2q sin σtotal (32)

where

σtotal � σ − tan−1�z 0 − z 0objx 0 − x 0obj

�(33)

σ � tan−1�v3v1

�(34)

From force balance:

macmd � mg cos σ − T cos�γ � σ� − Y (35)

Given that β � π2− γ − σ and in view of Eq. (13), Eq. (35) gives

T sin β� Y � mg cos σ −macmd (36)

where

Y � 1

4Cyβρ∞�v21 � v23�Sside sin�2β� (37)

Once acmd is calculated by Eq. (32), β can be obtained from Eqs. (36)and (37) iteratively. Thus, the objective angle of inclination γ is givenby γ � π

2− β − σ. The quaternion representing the desired attitude

can be written as follows:

qr;phase1 ��cos

γ

20 sin

γ

20

�T

(38)

When the first phase ends after the vehicle reaches a threshold speed,the vehicle executes a roll maneuver to level flight.Remark 2: For the conventional transition method, the vehicle

inclines forward, and the angle of inclination could be obtainedwith asimilar method as described previously. The only difference is thatthe vehicle uses lift L instead of side force Y to balance its gravity.

B. Second Phase

The altitude loss and horizontal drift caused by the roll will becompensated by pitch and yaw maneuvers. Note that the pitch angleϕ (Fig. 13) and the yaw angle θ (Fig. 12) are small, and thus these twomaneuvers are commutative and can be considered independently.The lateral acceleration in the x 0-z 0 plane and in the x 0-y 0 plane arenoted acmd1 (Fig. 12) and acmd2 (Fig. 13), respectively. In this case,the direction cosine matrix R is simplified and thus can be expressedas follows:

R �"−θ 1 0

ϕ 0 1

1 θ −ϕ

#(39)

The velocity expressed in the body frame is thus given by

24 vxvyvz

35 � R

24 v1v2v3

35 �

24 −θv1 � v2

ϕv1 � v3v1 � θv2 − ϕv3

35 (40)

Fig. 13 Schema of the second phase in the x 0-y 0 plane.

Fig. 12 Schema of the second phase in the x 0-y 0 plane.

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Similar to the discussion in the first phase, the schema of lateralacceleration and velocity are drawn in the x 0-y 0 plane and the x 0-z 0

plane, respectively.1. x 0-y 0 Plane

acmd1 � 2v21 � v22�����������������������������������������������������

�x 0 − x 0obj�2 � �y 0 − y 0obj�2q sin σtotal1 (41)

where

σ1 � tan−1�v2v1

�σtotal1 � tan−1

�y 0 − y 0objx 0obj − x 0

�� σ1

When expressed in the inertial frame, the vector form of acmd1 is

acmd1vec �"acmd1 sin σ1−acmd1 cos σ1

0

#

2. x 0-z 0 Plane

acmd2 � 2v21 � v23�����������������������������������������������������

�x 0 − x 0obj�2 � �z 0 − z 0obj�2q sin σtotal2 (42)

where

σ2 � tan−1�v3v1

�

σtotal2 � tan−1�z 0 − z 0objx 0obj − x 0

�� σ2

When expressed in the inertial frame, the vector form of acmd2 is

acmd2vec �"acmd2 sin σ2

0

−acmd2 cos σ2

#

To calculate the desired attitude, the objective aerodynamic forces(lift and side force) are essential. Because acmd1 and acmd2 only havelateral components that are perpendicular to velocity, but the gravityG and the thrust T also have such lateral components, these twoforces must be compensated. Given small-angle approximation, andin view of Eqs. (11) and (13), the first and the second components ofaerodynamic force (expressed in body frame) denote respectively thedesired side force and the lift, which is equivalent to the following(other contributions toF are neglected,which are accounted for in thedisturbing force):

L � � 0 1 0 �RFcmd Y � �−1 0 0 �RFcmd (43)

where

Fcmd � m�acmd1vec � acmd2vec� −G − RTT

The desired angle of attack and angle of side slip are expressed as

α � L12ρ∞�v2y � v2z�SwingClα

β � Y12ρ∞�v2x � v2z�SsideCyβ

(44)

Given that vz is significantly greater than vx and vy, Eqs. (7) and (8)can be written as:

α � −vyvz

β � vxvz

(45)

In view of Eqs. (40) and (45), and given that v1 is significantly greaterthan v2 and v3:

α � −ϕv1 � θϕv2 � v3

v1

β � −θv1 � v2v1

(46)

By using the small-angle approximation, the desired pitch angle andyaw angle can be obtained from Eq. (46):

ϕ � −α −v3v1

θ � −β� v2v1

(47)

Here, the three components of velocity are assumed to be derivedfrom onboard sensors. Once the desired angle of attack and angle ofside slip are obtained from Eq. (44), the desired yaw angle and pitchangle (ϕ and θ) are calculated from Eq. (47) accordingly.Considering that the reference attitude quaternion of the roll

maneuver is � 12

12

12

12�T (the z axis and the x axis in the body

frame coincide respectively with the x 0 axis and the y 0 axis in inertialframe), the quaternion representing the desired attitude of the secondphase can be written as follows:

qr;phase2 ���

12

12

12

12

�� qyaw � qpitch

�T

(48)

where

qyaw ��cos

θ

20 sin

θ

20

�qpitch �

�cos

ϕ

2sin

ϕ

20 0

�

Figure 14 shows the control implementation. This diagram is inclusiveof phases 1 and 2, except the roll maneuver, during which only theattitude is under control (closed-loop), but the position is open-loop.

V. Simulation Result

A. Results of New Transition Method

The animation of this simulation has been uploaded online.¶

The transition from hover to forward flight begins at t � 0. Thereis a roll maneuver between phases 1 and 2. During the first step of thissimulation, the roll maneuver is executed at t � 15 s, which is longenough for the vehicle to reach its minimum level-flight speed vmin

derived from Eq. (10); vmin is reached at t � 9 s, and so otherconversion times will be tested in the following sections. The rollmaneuver lasts for 1 s, with the desired attitude quaternion� 12

12

12

12�T . This time interval must be compatible with the

dynamic characteristics of theUAV. For aUAVas agile as this one, 1 sis enough for it to finish the roll maneuver. But for other vehicles,such as a heavier ISTAR model, this time interval could be longer.Figure 15 shows the displacement over time in three directions in

the inertial frame. Figure 16 shows the corresponding velocity, whichis the time derivative of Fig. 15. Thevehicle inclines relatively slowly.The simulation shows that it takes 10 s to achieve a pitch angle ofabout 90 deg. This is the result of the control law because it requiresthe vehicle to maintain altitude, as described in Eq. (36). If it inclinestoo fast, there will not be enough dynamic pressure for the vehicle tobalance its gravity with side force.

¶Data available online at http://www.youtube.com/user/zwzzwz/videos[retrieved 01 July 2012].

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At the beginning, the vehicle does not maintain altitude, butascends slowly until t � 10 s. This is because the attitude control istime-delayed, as shown in Fig. 17; therefore, the angle of inclinationis smaller than the command, thus making the vertical component ofthrust greater than needed. This could be treated as a disturbing forcein the vertical plane. Despite such disturbance, the vehicle descendsback to its original altitude at t � 12 s because the trajectory-following logic is equivalent to a PID controller [12]with small-angleapproximation. In this case, the duct axis of symmetry is nearly

Fig. 14 Control implementation.

Fig. 15 Result from position.

Fig. 16 Result from velocity.

Fig. 17 Results from attitude, with the dotted line representingcommand and the solid line representing simulation.

Fig. 18 Results from vane deflections.

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parallel to the horizontal plane at t � 10 s, and thus the small-angleapproximation is applied.The vane deflections are plotted in Fig. 18. Given the limitation of

the servo rate, an ideal filter is added to eliminate all components over1 Hz in vane deflection command, which means the required servorate is limited to 360 deg ∕s.

B. Comparison Between Conventional and New Transition Method

Because it takes 9 s to reach the minimum level-flight speed, inthe second step, the conversion times are set to 9, 12, 15, and 18 s.The conventional transition method is also simulated for the samevehicle, with the strategy described in Remark 2. A comparison ofdisplacement over time is simulated as shown in Fig. 19. Significantinitial acceleration advantage is demonstrated by the new transitionmethod. This advantage could be ascribed to the low drag due to asmaller windward side by flying sideways, while the conventionalmethod requires the vehicle to fly at an angle of attack of near 90 deg,thus causing tremendous drag.Remark 3: This advantage will disappear at higher speed, as

the conventional method has a higher lift/drag ratio than flying

sideways in level flight. The simulation also shows that the soonerit rolls, the faster it accelerates. This is because, when flyingsideways at high speed, it is equivalent to awingless version of ISTARUAV shown in Fig. 1, and thus it has a lower lift/dragratio in level flight and generates more induced drag. Generally,flying sideways is more efficient at low speed but less efficient at highspeed.The induced drag for different conversion times is plotted in Fig. 20.

Note that the oscillations on the curves are due to the roll maneuver.

VI. Conclusions

The newmethod of transition flight, which tilts the ducted wingedunmanned aerial vehicle (UAV) to fly sideways, enables the UAV toachieve faster acceleration compared to conventional transitionflight, due to its smaller windward side area. The simulation showsthat, by flying sideways, the UAV generates less induced drag at lowspeed but more induced drag at higher speed. Thus, to acceleratefaster, the UAV should roll and convert to high-speed flight modeas soon as its minimum level-flight speed is achieved. The controllaws designed are robust against disturbance. This property is dem-onstrated by the elimination of altitude loss and horizontal driftdespite the disturbance. The simulation shows that the skid-to-turn iseffective in trajectory following.A more accurate model will be developed, including the variation

of thrust, the dynamics of rotor, time delays, and revised aero-dynamical coefficient curves. The negative effects of the horizontalvaryingwindwill also be simulated. Subsequently, flight tests on reala UAV model will be carried out.

Appendix A: Numerical Values

0 5 10 15 200

100

200

300

400

500

600

700

800

Time (S)

Dis

plac

emen

t (M

)

x’

ConventionalRoll at t=9sRoll at t=12 sRoll at t=15 sRoll at t=18 s

Fig. 19 Comparison of displacement (x 0 axis) over time.

0 20 40 60 80 1000

100

200

300

400

500

600

700

v1 (M/S)

Indu

ced

Dra

g (N

)

Conventional

Roll at t=9 s

Roll at t=12 s

Roll at t=15 s

Roll at t=18 s

Fig. 20 Comparison of induced drag for different conversion times.

Table A1

Parameter Value

Mass and inertia

m 100 kgJ diag�25; 30; 28� kg · m2

Duct

Cduct 0.2 m2

r 0.6 mFuselage aerodynamics

Cl;max 1.2Clα 5Swing 2.5 m2

Cy;max 0.5Cyβ 3Sside 0.5 m2

Cd;gain 1.9C 0d;gain 0.5Cd0 0.04z1 0.05 mz2 0.06 m

Control surfaces

lr 0.3 mla 0.3 mle 0.25 mScs 0.06 m

Rotor

b 0:0045 N · s2

κ 0.0015 N · m · s2

ωr 260 rad · s−1

nb 4ib 0.0027 kg · m2

Miscellaneous

ρ∞ 1.225 kg · m−3

g 9.8 N · kg−1

L1 80 mKs 5I3×3Md 10 sin�0:2t��111�T N · m

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Appendix B: Proof of the Controller

First of all, by deriving the following Lyapunov functioncandidate:

V � sTs

in view of Eqs. (3), (5), and (33), one gets

_V � 2sT _s

� 2sT�−1

2s − Kssign�s� � J−1Md

�(B1)

Let

H � �HxHyHz�T � J−1Md (B2)

and

Ks � diag�K1 K2 K3 � �K1 > jHxj; K2 > jHyj; K3 > jHzj�(B3)

Therefore, we get

_V � −sTs� 2sT

2664Hx

Hy

Hz

3775 − 2sT

2664K1sign�s1�K2sign�s2�K3sign�s3�

3775

� −sTs� 2sT

2664Hx

Hy

Hz

3775

− 2�K1js1j � K2js2j � K3js3j�≤ −sTs� 2�Hxjs1j �Hyjs2j �Hzjs3j�− 2�K1js1j � K2js2j � K3js3j�≤ −sTs − 2ΔK�js1j � js2j � js3j�≤ 0 (B4)

where

ΔK � minfK1 − jHxj; K2 − jHyj; K3 − jHzjg

From inequality (B4), one gets that s � 0 is the only invariant setsuch that _V � 0. By using La Salle’s invariance theorem, weconclude that there exists a time ts such that s � 0 for t ≥ ts (i.e.,qe � −Ω), and the equilibrium point is globally asymptoticallystable.When the sliding surface �qe �Ω � 0� is achieved, it then follows

from Eq. (29) that

_qe � −1

2qe0qe

_qe0 �1

2qTe qe

The time derivative of the Lyapunov function candidate W �qTe qe � �1 − qe0�2 gives

_W � −qTe qe < 0

which implies that Ω�t� → 0 and qe�t� → 0 as t→ ∞. The equation

�q0 qT � � �qr0 qTr � � �qe0 qTe �

shows that �q0 qT � → � qr0 qTr � and the equilibrium point isglobally asymptotically stable.

Acknowledgments

This work was supported by the National Natural ScienceFoundation ofChina (61104012), the 973Program (2010CB327904)and the Ph.D. Programs Foundation of Ministry of Education ofChina (20111102120008).

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