，，，，

1.China Helicopter Research and Development Institute，Jingdezhen 333001，P.R.China；2.Key Laboratory of Fundamental Science for National Defense?Advanced Design Technology of Flight Vehicle，Nanjing University of Aeronautics and Astronautics，Nanjing 210016，P.R.China

Abstract: In the passive turning state，the helicopter turns through the tail rotor force and the friction of the ground to the tire.In practice，it is found that the helicopter will turn difficultly under low aircraft ground speed or static state.This paper takes a certain type of helicopter as the research object，and establishes the dynamic model of helicopter ground turning motion based on the basic theory of dynamics.This model takes into account the six-degree-of-freedom motion model of the helicopter body，the motion model of the landing gear buffer，the tire mechanics model and the friction characteristics of the strut friction disc.The dynamic simulation of the helicopter right angle turn and static turn is carried out，and the influence of parameters such as tail rotor pull，taxi speed，tail wheel stability distance on the dynamic response of the turn is studied.The results show that under the same ground taxing speed，the tail wheel angle increases with the increase of tail rotor force.When the tail rotor force is the same，the tail wheel angle increases with the increase of ground taxing speed.When the helicopter is completely static，it is the most difficult to turn，which requires much bigger force of the tail rotor to turn.In addition，the change of the stability distance of the tail wheel has an obvious influence on the turning.When the stability distance is doubled，the tail rotor force will be reduced by 30% to the same angle of the tail wheel.

Key words：helicopter ground motion；dynamics analysis；passive turning；landing gear

0 Introduction

When the rear three-point layout of helicopter is turning on the ground，it is mainly turned by dif?ferential braking or the tail rotor force［1-4］.This pa?per mainly studies the situation that the helicopter turns by the tail rotor force at low speed.The force of the tail rotor drives the helicopter and tires to pro?duce lateral motion，which deflects the tail wheel and the axis of rotation，thus realizing the helicopter turning，which belongs to the passive turning mode［5］.In practice，its rocker arm has a large turn friction torque for the shimmy need of the tail wheel［6］.Therefore，the turning process is divided into two steps：（1）When the tail rotor force is small，the tail rotor provides lateral load to drive the tires produce side-slip motion.At this time，the torque generated by the lateral force of the tail wheel cannot overcome the friction torque，and the tail wheel state does not deflect.（2）When the tail rotor force increases to a higher level，the deflection of helicopter will intensify.The torque generated by the lateral force of tail wheel can overcome the fric?tion torque and drive the rocker arm to rotate，the wheel will turn a angle，and the helicopter enters the normal turning state［7-10］.

In order to analyze the motion of helicopter un?der different speeds and tail rotor force，the dynamic equation of helicopter tail wheel passive turning is established in this paper.In the model，many factors are considered to solve the force and motion charac?teristics of the helicopter in the whole process of pas?sive turning，such as the overall layout of the heli?copter，the structural layout of the tail landing gear，the dynamic characteristics of the tire，the speed of the helicopter，and the tail rotor force.

1 Mathematical Modeling of Heli?copter Ground Turning Motion

1.1 Basic hypothesis of model

In order to make the model accurate and practi?cal，the model should be simplified and assumed ap?propriately［11］to reduce the uncertainty factors caused by some parameters that are difficult to deter?mine and the accumulation of errors caused by too many degrees of freedom and constraints.And the following assumptions are made for the model［12-13］：

（1）The helicopter is divided into two parts：elastic support mass and inelastic support mass.The elastic support mass includes the helicopter body supported by the buffer，the outer cylinder of the buffer.While the inelastic support mass includes the buffer piston rod，the wheel，the brake device and other auxiliary mechanisms.

（2）The motion of the elastic support mass is six degrees of freedom，and the mass is concentrat?ed at the center of mass of the fuselage.The motion of the inelastic support mass only considers the translational degrees of freedom in three directions，ignoring the lateral deformation and torsional defor?mation of the buffer.

（3）Each wheel has an independent degree of rotational freedom.

（4）Ignoring the aerodynamic load of the in?elastic support mass，the main rotor force and tail rotor force are treated as concentrated forces，which act on the center of gravity of the helicopter and the center of the tail rotor respectively.

According to the geometric characteristics of the helicopter，the ground turning diagram is estab?lished，as shown in Fig.1.

Fig.1 Helicopter ground turning diagram

1.2 Definition and transformation of coordi?nate system

When the helicopter is moving on the ground，it is subject to gravity，the rotor pull，the tail rotor lateral force，and the force of ground on the tires.Different forces need to be expressed in different co?ordinate systems.The definition of the coordinate system is as follows.

As shown in Fig.2，in the ground coordinate systemOi?Xi YiZi，the center of mass of helicopter in the initial state is taken as the coordinate origin，the axisXiis parallel to the center line of runway and points to the forward direction of the helicopter，the axisYiis perpendicular to the axisXiin the hori?zontal plane，and the axisZiis vertically downward.In Fig.2，θis the pitch angle，ψis the roll angle and?is the yaw angle.

Fig.2 Relationship between ground inertial coordinate sys?tem and airframe coordinate system

In the airframe coordinateOb?XbYbZb，the cen?ter of mass of helicopter is taken as the coordinate origin，fixed with the helicopter，the axisXbpoints forward along the axis of the helicopter body，the axisYbis perpendicular to the axisXbin the horizon?tal plane，and the axisZbfollows the right-hand rule and goes down vertically.

In the wheel coordinate systemOj?XjYjZj，the rotation center of the wheel is taken as the coordi?nate origin，the axisXjpoints to the forward direc?tion of the wheel，the axisZjis perpendicular to the axisXjand vertically downwards，and the axisYjpoints to the right according to the right-hand rule.

1.3 Helicopter dynamics model

When the helicopter is moving on the ground，it is subject to gravity，the rotor force，the tail rotor lateral force，and the force of ground on the tires.The forces that the ground on the tires include longi?tudinal friction，lateral friction，vertical support re?action force，and righting torque.

The translational equation of the helicopter body center of mass is established in the ground iner?tial coordinate system，which can be written as

whereTibis the transformation matrix from the body coordinate system to the inertial coordinate system；Tbjthe transformation matrix from the wheel coordi?nate system to the body coordinate system，in whichjis thejlanding gear；Gthe gravity matrix in the body coordinate system；Tthe tail rotor force matrix in the body coordinate system；mthe body mass；the acceleration vector of the body mass center；Pthe force matrix of rotor ten?sion acting on the body；and [Fxj Fyj Fzj]Tthe force of each landing gear on the body.

The equation of helicopter body rotation around the center of mass can be written as

whereIis the matrix of the moment of inertia of the helicopter body；ωthe matrix of the angular veloci?ty of the helicopter body rotating around the center of mass；Rbjthe matrix of the distance between the connection point of each landing gear and the air?frame to the center of mass；［Mxj Myj Mzj］the ac?tion torque of each landing gear on the body；MPthe moment matrix of rotor tension acting on the body；MTthe tail rotor moment matrix in the body coordi?nate system.

Angular velocityω=(ωx，ωy，ωz)T，the rela?tionship between the angular velocity component and the derivative of the attitude angle is given by

The force analysis of the inelastic support mass including the piston rod of the landing gear buffer and the wheel is carried out.According to Newton’s second law，the dynamic equation of the inelastic support mass is listed as

whereMgjis the inelastic support mass of a single landing gear；the acceleration vector of the inelastic support mass；Tjithe transformation matrix from the ground coordinate system to the wheel coordinate system；[Fxjk，F(xiàn)yjk，F(xiàn)zjk]Tthe tire force of each wheel；Wgithe gravity of the inelastic support mass；andthe reaction force exerted by the helicopter body on the landing gear.

1.4 Mathematical model of related forces

1.4.1 Mechanical model of landing gear buffer

The paper chooses single chamber oil pneumat?ic shock absorber as the landing gear buffer.The buffer strut forcefsis composed of air spring forcefa，oil damping forcefdand structural restriction force.Then the total buffer strut force can be writ?ten by［14-15］

whereKSis the limited stiffness of the force and compression structures of the buffer related to the structural restriction force；Sthe buffer stroke；S0the stroke when the buffer is fully extended；andSmaxthe stroke when the buffer reaches its maximum compression.

The air spring forcefais related to the inflation volume，inflation pressure，and buffer compres?sion，and has a non-linear positive correlation with the buffer compression.It can be expressed as

whereAais the air pressure area of the piston；P0the initial inflation pressure of the buffer；V0the ini?tial inflation volume of the buffer；Patmthe local at?mospheric pressure；andnthe polytropic index of air，generally ranging from 1.0 to 1.3.

The oil damping forcefdis related to the com?pression speed of the buffer，the area of the oil hole and the effective oil pressure area of the buffer.The direction of the oil damping force is opposite in the positive stroke and the reverse stroke.The oil damp?ing forcefdis expressed by

whereρois the filling oil density of the buffer；AozandAocare the effective oil pressure areas of the main oil cavity and the side oil cavity of the buffer，respectively；CdzandCdcthe oil hole shrinkage coef?ficients of the main oil cavity and the side oil cavity，respectively；AhzandAhcthe areas of the main oil hole and the side oil hole，respectively .

1.4.2 Gravity and rotor force

The gravity of the helicopter acts on the center of mass of the helicopter and does not have torque effect on the helicopter，which is expressed in the ground inertial coordinate system and can be given by

wheregis the acceleration of gravity，g=9.81 m/s2.

The main rotor force generally acts in the longi?tudinal symmetry plane of the helicopter，and its line of action generally does not pass through the center of mass，and there is a certain distanceepfrom the center of mass.The rotor force is not paral?lel to the direction of gravity，but has a forward in?clination angleη1，which causes the helicopter to move forward in the ground taxiing process.The ro?tor force is expressed in the airframe coordinate sys?tem as

The rotor force torque is expressed in the air?frame coordinate system as

1.4.3 Mechanical model of tail rotor force

The tail rotor force acts on the tail beam of the helicopter，which is perpendicular to the symmetri?cal plane of the helicopter body，and has a certain distanceeTfrom the center of mass，which gener?ates a lateral turning torque for the helicopter.The tail rotor force is not in the longitudinal plane of symmetry of the helicopter.Viewed from the tail of the helicopter，there is an upward angle of 20° with the longitudinal plane of symmetry，takingη2.The tail rotor force and torque are expressed in the air?frame coordinate system as

Due to the high-speed rotation of the main ro?tor，the helicopter produces a reaction torque.In or?der to balance this torque，in addition to the forceTrequired for turning，the tail rotor also has a forceT1to balance the reaction torque.The tail rotor force and torque are expressed in the airframe coordinate system as

1.4.4 Tire mechanics model

The NASA tr64 tire model is adopted for this tire study.The characteriastics of the tire are consid?ered in this model concluding longitudinal slip，side slip，vertical compression，the vertical support reac?tion force longitudinal friction and lateral friction［3］.Fig.3 shows the NASA tr64 semi empirical tire model.

Fig.3 Wheel coordinate system and force diagram

(1)Vertical support reaction force

whereKTis the tire vertical vibration stiffness coeffi?cient，CTthe tire vertical vibration damping coeffi?cient，δthe tire compression，andthe tire com?pression speed.

(2)Longitudinal friction

Longitudinal friction is related to the combina?tion coefficient of the tire and the vertical load on the tire，which can be expressed as

whereμxis the longitudinal sliding friction coeffi?cient.Although the relationship between the longitu?dinal sliding friction coefficient and the longitudinal slip rate is complicated，the coefficient can be given by empirical formula in the calculation of this chap?ter（Fig.4）.

In Fig.4，μtis the heading friction coefficient of the tire，and the slip rate is expressed as

Fig.4 Relationship between slip ratio and longitudinal fric?tion coefficient of tire

whereVxis the longitudinal velocity of the wheel，ωthe rotational angular velocity of the wheel，Rethe rotational radius of the wheel and can be ex?pressed as

(3)Lateral friction

The lateral friction of the tire is related to the lateral and longitudinal velocity of the tire，the coef?ficient of lateral friction and the vertical load of the tire，which can be expressed as

whereμais the lateral friction coefficient of the tire，andΦcan be expressed as

whereVxiandVyiare the longitudinal sub-velocities and lateral sub-velocities of each tire.According to the above equation，the direction of tire lateral force changes with the direction of the tire lateral sub-ve?locities.

(4)Rolling resistance torque

During the rolling process of the wheel，the tire will also be affected by the rolling resistance torque，which can be expressed as

whereμris the coefficient of rolling resistance torque.

The longitudinal resistance torque of the wheel can be written as

When the helicopter brakes，the braking torqueMbis superimposed with the longitudinal resistance torque of the wheel.Under the action of the two torques，the rotation speed of the helicopter wheel slows down.

1.4.5 Mechanics model of friction shimmy damper

In this paper，the helicopter tail landing gear is a strut landing gear，and the friction shimmy damp?er is installed between the outer cylinder and the in?ner cylinder［3］，as shown in Fig.5.

Fig.5 Schematic diagram of tail landing gear

The helicopter tail landing gear adopts dry fric?tion shimmy damper，as shown in Fig.6，and the damping torque of friction can be expressed as

Fig.6 Schematic diagram of friction shimmy damper

whereDis the outer diameter of the friction shimmy damper；dthe inner diameter of the friction shimmy damper；μthe friction coefficient of the friction shimmy damper；andFznthe positive pressure of the friction shimmy damper.

The dynamic friction coefficient and the static friction coefficient are different.In general，the dy?namic friction coefficient is smaller than the static friction coefficient.When using the dynamic friction coefficientμ1，rotation resistance torqueMz1is ob?tained，and when the static friction coefficientμ2is taken，rotation resistance torqueMz2is obtained.The disc motion relationship is shown in Fig.7.As?suming thatM1andM2are the rotational torques of the tire lateral force acting on the friction disc，they increase gradually from static state.When the rota?tional torqueM1of the tire lateral force at the fric?tion disc is greater than the resistance torqueMz2，the friction disc starts to rotate and stops when the rotational torqueM1is less than the dynamic friction torqueMz1at the friction disc.M2cannot rotate the friction disc.The red part in Fig.7 rotates，and the blue part does not rotate.

Fig.7 Rotation state of friction disc

2 Simulation Model

It is convenient to change the parameters of the model and simulate under different working condi?tions by using MATLAB to establish the helicopter model，which is convenient for researchers to achieve rapidly design.

According to the helicopter ground turning dy?namic model in Section 1，the mechanical model of the buffer，the rotor tension model，the tire force model and the friction damper model are established in Matlab.After that，these subsystems are built in?to a whole helicopter model.Table 1 gives the pa?rameters of helicopter model.

Table 1 Helicopter parameters

Fig.8 shows the simulation model built accord?ing to the mathmatical model of buffer in Section 1.

Fig.8 Mechanical model of landing gear buffer

Fig.9 is the simulation model of the tail landing gear friction disc of the pendulum reducer according to Section 1.4.5.

Fig.9 Mechanics model of friction shimmy damper

Fig.10 shows the tire force model established according to Section 1.4.4.

Fig.10 Tire mechanics model

The whole helicopter model is composed of above part models，as shown in Fig.11.

Fig.11 Helicopter model

The thrust of helicopter ground taxiing is pro?vided by the main rotor，the main rotor force for?ward inclination angle isθ，and the thrust can be changed by adjusting the inclination angle.Mean?while，the thrust will also affect the helicopter land?ing gear’s pressure on the ground，but the main ro?tor force remains unchanged in this process.Table 2 gives the deflection angle to main rotor correspond?ing to taxiing speed.

Table 2 Deflection angle to main rotor corresponding to taxiing speed

3 Analysis of Simulation Results

3.1 Comparison of different tail rotor force

The taxiing speedvof helicopter is 2 m/s，and the critical tail rotor force is 3 055 N.Therefore，the tail rotor forceF=3 300，3 500，and 3 700 N are selected for right angle turning simulation.

(1)The tail rotor force 3 300 N

At the 2 s of the simulation，the tail rotor force is linearly increased to 3 300 N to make the helicop?ter turn.Due to the large friction torque at the tail friction shimmy damper，the helicopter cannot re?turn completely by the lateral force of the tail wheel.Therefore，when the fuselage turns 90° and the tail rotor force slowly stops，the reverse force of the tail rotor is applied to align the tail wheel until it stops.

As shown in Fig.12，the tail wheel begins to deflect at 3 s after the simulation starts，and stops deflecting at 6.3 s，maintaining at 13.8°，and the he?licopter starts to make a steady turn.At 17 s，the fu?selage has turned 90°，the tail rotor force gradually stops and increases in reverse，and the tail wheel starts to turn back at 17.2 s.At 17.9 s，the tail wheel deflection is 0°，the aligning is completed，and the helicopter begins to taxi straight.

Fig.12 Tail wheel deflection angle(F=3 300 N,v=2 m/s)

As shown in Fig.13，the helicopter has a rela?tively small curvature during the initial turn，and the turning radius is about 19 m.

Fig.13 Projection curve center of gravity on the ground(F=3 300 N,v=2 m/s)

As shown in Fig.14，at the beginning of turn?ing，the tail wheel lateral force is linearly related to the tail rotor force.When the tail rotor force reaches its maximum，the single tail wheel lateral force also reaches the peak value of 4 259 N.Then the helicop?ter starts to turn，and when the fuselage reaches 90°，the tail rotor force decreases first and then in?creases in reverse，making the tail wheel align.

Fig.14 Tail wheel lateral force(F=3 300 N,v=2 m/s)

It can be seen from Fig.15 that the lateral force of the right main wheel rapidly increases to 583 N at 2 s during the right-angle turning.This is because when the reverse torque is applied，the reverse torque is mainly balanced by the friction between the tire and the ground in the condition of without the force of the tail rotor.With the increase of the tail rotor force，the lateral force of the right main wheel decreases and increases inversely，reaching the peak value of 559 N.After the tail wheel de?flects，the lateral force increases slowly.When the fuselage reaches 90°，the tail rotor exerts a reverse force，and the lateral force of the main wheel in?creases rapidly，and decreases rapidly when the tail wheel returns to normal.Due to the existence of re?verse torque，it cannot be reduced to zero and is sta?bilized at 809 N.Table 3 summarizes the above re?sults.

Fig.15 Main wheel lateral force(F=3 300 N,v=2 m/s)

Table 3 Simulation data corresponding to tail rotor force 3 300 N

(2)The tail rotor force 3 500 N

At 2 s of the simulation，the tail rotor force is linearly increased to 3 500 N to make the helicopter turn.The simulation data are shown in Table 4.

Table 4 Simulation data corresponding to tail rotor force 3 500 N

Table 5 Simulation data corresponding to tail rotor force 3 700 N

(3)The tail rotor force 3 700 N

At 2 s of the simulation，the tail rotor force is linearly increased to 3 700 N，which makes the heli?copter turn.The simulation data are shown in Ta?ble 5.

According to the above three working condi?tions of Tables 3，4 and 5，it can be concluded that the greater the tail rotor force，the greater the tail wheel angle，the greater the lateral force at the main wheel and the tail wheel as the tail rotor force in?crease.The increase of tail rotor force has a signifi?cant impact on the rotation angle of the tail wheel.

3.2 Comparison of different turning speeds

In this set of simulations，right-angle turning of the helicopter is simulated with three groups of taxi?ing speeds of 1，2，4 m/s.The tail rotor force is 3 300 N，and the tail wheel stability distance is 72 mm.

(1)Taxiing speed 1 m/s

At 2 s of the simulation，the tail rotor force is linearly increased to 3 300 N to make the helicopter turn.

As shown in Fig.16，the tail wheel begins to deflect at 2.8 s after the beginning of the simulation，and stops deflecting at 6.4 s，maintaining at 13.57°，and the helicopter begins to make a steady turn.At 23 s，the fuselage has turned 90°，the tail rotor force gradually stops and increases in reverse，and the tail wheel begins to align at 23.2 s.At 24 s，the tail wheel deflection angel is 0°，the aligning is complet?ed，and the helicopter starts to taxi in a straight line.

Fig.16 Tail wheel deflection angle(F=3 300 N,v=1 m/s)

As shown in Fig.17，the trajectory of the cen?ter of gravity is close to a quarter circle on the ground，and the turning radius is 20 m.

Fig.17 Projection curve of center of gravity on the ground(F=3 300 N,v=1 m/s)

As shown in Fig.18，at the beginning of the turn，the tail wheel lateral force is linearly related to the tail rotor force.When the tail rotor force reaches the maximum，the lateral force of the single tail wheel also reaches the peak value of 4 271 N.When the fuselage reaches 90°，the lateral force of the sin?gle tail wheel first decreases and then increases in re?verse，making the tail wheel align.

Fig.18 Tail wheel lateral force(F=3 300 N,v=1 m/s)

In the process of right-angle turning，the lateral force of the right main wheel rapidly increases to 636 N within 2 s，as shown in Fig.19.As the tail ro?tor force increases，it first decreases and then in?creases in reverse，reaching the peak value of 700 N.After that，due to the sudden change of the tail rotor force，the lateral force of the right main wheel is stabilized at 849 N when the process of tail wheel aligning is finished.The simulation data correspond?ing to taxiing speed of 1 m/s are given in Table 6.

Fig.19 Tail wheel lateral force(F=3 300 N,v=1 m/s)

Table 6 Simulation data corresponding to taxiing speed of 1 m/s

(2)Taxiing speed 4 m/s

In this simulation，the taxiing speed is set to 4 m/s，and the tail rotor force was linearly increased to 3 300 N at 2 s of the simulation to make the heli?copter turn.The simulation data corresponding to taxxing speed of 4 m/s are given in Table 7.

Table 7 Simulation data corresponding to taxiing speed of 4 m/s

The simulation data of taxiing speed of 2 m/s is shown in Table 3.It can be concluded from Ta?bles 3，6，and 7 that when the tail rotor force is the same，the greater the taxing speed，the easier it is to turn.The influence of taxing speed on the lateral force of tail wheel is not obvious.

3.3 Stationary turn

For the helicopter in a completely static state，it is impossible to calculate the lateral force and fric?tional resistance of the tire by the relationship of the wheel deflection angle，therefore，an empirical for?mula for calculating the friction force of the station?ary tire is referred to.The turning of the landing gear is subject to the uniformly distributed friction force of the ground against the tires.The uniformly distributed friction force produces two frictional re?sistance torques：The uniformly distributed friction torque and the torque generated by the equivalent concentrated friction force to the piston rod，which can be expressed as

whereMjis the uniformly distributed friction torque，PVNthe vertical load of the landing gear strut，athe half length of the tire touching the ground，bthe stability distance，andd′the wheel di?ameter.

In the static state，the helicopter completely re?lies on the tail rotor force to make the tail wheel ro?tate around the axis of the strut.Initially，the tail ro?tor force is linearly increased to 4 655 N，mean?while，the tail wheel begins to rotate under the ac?tion of driving force.Through the simulation，it is found that this kind of turning is unstable.After the helicopter moves，it is switched to the dynamic sim?ulation model，the criterion for switching is that when the tail wheel starts to deflect a small angle，the tail wheel slips slightly，which means a smooth transition to the dynamic simulation model.And the tail rotor force is set to 4 357 N after switching.

It can be concluded from Fig.20 that the trajec?tory of each point can be obviously divided into two sections.In the first section，the static turning theo?ry is applied，and the tail wheel is completely taxi?ing on the ground.The second section is switched to the dynamic turning state，and the helicopter turns on the ground with a turning radius of 7 m.

Fig.20 Projection of each point of fuselage on the ground

As shown in Fig.21，the tail wheel angle in?creases linearly when the tail rotor force is applied within 1 s to 2 s，and the rotation angle increases to 18°，and remains unchanged after removing the tail rotor force from 2 s to 2.5 s.At 2.8 s，the torque transferred from the tail wheel to the tail friction plate is greater than the static friction torque of the tail wheel，and the tail wheel begins to deflect.At 4.2 s，the tail wheel stops deflecting and the tail wheel deflection angle remains at 28°.When the simulation runs to 11 s，the tail wheel starts to align，and the process is completed at 11.8 s，then it enters the straight-line taxiing state.

Fig.21 Tail wheel deflection angle

As shown in Fig.22，when switching to the dy?namic turning model，there is a sudden change in the lateral force at 2.5 s.Later，with the increase of the tail rotor force，the lateral force gradually in?creases and finally reaches the maximum value of 5 115 N，and the lateral force reaches 3 815 N after the steady-state turning.

Fig.22 Lateral force of tail wheel

As shown in Fig.23，when switching to the dy?namic model，there is a sudden change in the lateral force of the main wheel，too.In the process of rightangle turning，the lateral force of the right main wheel first increases with the increase of the tail ro?tor force，finally reaching the peak value of 2 962 N.Then the helicopter begins to turn.When the tail ro?tor exerts the force in reverse to make the tail wheel align，the lateral force of the right main wheel rapid?ly reduced to zero.The data of ground static rightangle turning are given in Table 8.

Fig.23 Lateral force of main wheel

Table 8 Data of ground static right?angle turning

3.4 Comparison of different tail wheel stability distance

In this set of simulations，the tail wheel stabili?ty distance is doubled to 144 mm，and after obtain?ing the critical tail rotor force，the three groups of tail rotor force are set to 2 300，2 500，and 2 700 N，respectively.

(1)Tail rotor force 2 300 N

At 2 s of the simulation，the tail rotor force is linearly increased to 2 300 N to make the helicopter turn.The simulation data corresponding to tail rotor force 2 300 N are given in Table 9.

Table 9 Simulation data corresponding to tail rotor force 2 300 N

(2)Tail rotor force 2 500 N

At 2 s of the simulation，the tail rotor force is linearly increased to 2 500 N to make the helicopter turn.The simulation data corresponding to tail rotor force 2 500 N are given in Table 10.

Table 10 Simulation data corresponding to tail rotor force 2 500 N

(3)Tail rotor force 2 700 N

At the 2 s of the simulation，the tail rotor force is linearly increased to 2700 N to make the helicop?ter turn.The simulation data corresponding to tail rotor force 2 700 N are given in Table 11.

Table 11 Simulation data corresponding to tail rotor force 2 700 N

Comparing Tables 3，4，5 with Tables 9，10，11，it can be concluded that the stability distance has a decisive effect on the tail wheel deflection an?gle.When the tail wheel stability distance is dou?bled，the tail rotor force required for the same angle of the tail wheel is reduced by one-third，and the lat?eral force of the tail wheel is reduced by half.

4 Conclusions

（1）The six-degree-of-freedom dynamic model of helicopter taxiing and turning is established，in?cluding the airframe dynamics model，the mathe?matical model of related forces and the tire mechan?ics model.

（2）The influence of speed and tail rotor force on helicopter turning radius is analyzed.When the taxing speed is the same，the angle of the tail wheel increases with the increase of the force of the tail ro?tor.When the tail rotor force is the same，the tail wheel angle increases with the increase of taxing speed.

（3）It is analyzed that in the static state，the he?licopter only uses the tail rotor force to make a turn on the ground，and it needs to provide bigger tail ro?tor force to turn the tail wheel than other states by nearly 50%.It is concluded that it is difficult to turn in the static state.

（4）The influence of different tail wheel stabili?ty distance on helicopter turning is discussed，and it is concluded that the stability distance has a signifi?cant effect on tail wheel deflection.When the stable distance of the tail wheel is doubled，the tail rotor tension required for the tail wheel to rotate at the same angle is reduced by about 30%，and the later?al force of the tail wheel is reduced by nearly 50%.