Investigation on a Power Coupling Steering System for Dual-Motor Drive Tracked Vehicles Based on Speed Control

energies

Article Investigation on a Power Coupling Steering System for Dual-Motor Drive Tracked Vehicles Based on Speed Control

Li Zhai 1,2,* ID , Hong Huang 1,2 and Steven Kavuma 1,2

1 National Engineering Laboratory for Electric Vehicles, Beijing Institute of Technology, Beijing 100081, China; [email protected] (H.H.); [email protected] (S.K.) 2 Co-Innovation Center of Electric Vehicles in Beijing, Beijing Institute of Technology, Beijing 100081, China * Correspondence: [email protected]; Tel.: +86-10-68-91-5202

Received: 24 June 2017; Accepted: 28 July 2017; Published: 1 August 2017

Abstract: Double-motor drive tracked vehicles (2MDTV) are widely used in the tracked vehicle industry due to the development of electric vehicle drive systems. The aim of this paper is to solve the problem of insufﬁcient propulsion motor torque in low-speed, small-radius steering and insufﬁcient power in high-speed large-radius steering. In order to do this a new type of steering system with a coupling device is designed and a closed-loop control strategy based on speed is adopted to improve the lateral stability of the vehicle. The work done entails modeling and simulating the 2MDTV and the proposed control strategy in RecurDyn and Matlab/Simulink. The simulation results show that the 2MDTV with the coupling device outputs more torque and power in both steering cases compared to the 2MDTV without the coupling device, and the steering stability of the vehicle is improved by using the strategy based on speed.

Keywords: track vehicle; dual-motor; dynamic steering; power coupling

1. Introduction The addition of a track to a vehicle improves the vehicle’s traction on soft surfaces, therefore tracks are found on a variety of vehicles such as bulldozers, excavators, tanks, tractors and any vehicle whose application would benefit from the increased traction. Such vehicles are widely used in many areas such as the military, agriculture, construction, mining, and disaster rescue [1]. Although there are many studies on traditional tracked vehicles [2–4], there is still a need for more investigations on electric tracked vehicles. In recent years, with the development of high-power electronic devices, computer control technology and drive motors with high power and high efficiency, electrical drive control technology has made a breakthrough [5]. An electric drive system can be used not only to meet the required dynamic performance of the vehicle, but also to provide energy for electromagnetic weapons [6]. The hybrid electric powertrain with energy management strategies can provide more flexibility to meet driver demand and improve the fuel economy [7]. Moreover, compared to traditional vehicles, electric tracked vehicles’ CO and NOx emissions are reduced [8], making electric drive technology an important research direction in regard to tracked vehicle propulsion. The dual-motor independent drive configuration is the most widely used configuration in electric tracked vehicles [9–11]. This electric propulsion system which is composed of two AC induction motors, final drives and running gears is characterized by complicated multi-variables and nonlinearity [12]. The need for high speed, heavy load and automation has imposed higher and higher requirements on the stability, reliability, controllability and maneuverability of the vehicles [13].

Energies 2017, 10, 1118; doi:10.3390/en10081118 www.mdpi.com/journal/energies Energies 2017, 10, 1118 2 of 17

Skid steering is predominantly used with tracked vehicles, that is to say vehicles are steered by forcing the traction elements on opposite sides of the vehicle to run at different speeds [14]. There have been many other investigations into the steady-state handling characteristics and steering mechanisms [15]. The steady-state handling theory is based on analyzing the sliding motion of the track over the ground however few studies focus on the electric stability control of dynamic steering. The torque of outer motor required for the small radius steering will be quite large during a dynamic steering situation. At higher vehicle speeds the torque output of propulsion motors is limited by their power rating, so they could not necessarily contribute so much to vehicle steering. In order to solve the above problems of the dual-motor independent drive tracked vehicle, it is necessary to design a power coupling steering device. An electro-mechanical drive system with a horizontal axis steering structure is proposed to solve the problem of needing to use large electric motors whose power requirements are far beyond those of the pure drive power [16]. However these needs lots of idle gears and this increases the size of the assembly. A mechanical coupling device is used to decrease the traction power of the unilateral drive motor, which allows the regenerated power to flow from the low-speed side to the high-speed side mechanically [17]. This is quite complex due to the application of three planetary gear systems. A planetary gear coupling device is proposed to improve the efficiency of the inner side motor [18]. Though the efficiency of the motor is higher with the two planetary gear system, it requires higher accuracy for the control system. An electromechanical coupling transmission model with an auxiliary motor was proposed to increase the output torque or power of the motor [19], however the working principle and structural design of the coupling system were not described in detail. There are many studies which focus on the lateral dynamic motion of electric vehicles [20,21], some of which cover electric tracked vehicles [15,22], but few studies take dynamic steering into account. The torque or the power required by the vehicle will be much larger during transients. This paper aimed at solving the following problems by designing a power coupling steering system: (1) shortage of propulsion motor torque at low speeds and small radius steering, and (2) insufficient power at high speed and large radius steering. The lateral stability of the vehicle is improved by employing a closed loop control strategy based on speed. On running the simulation in RecurDyn and Matlab/Simulink, the results show the yaw rate for vehicles with the coupling device increases to the ideal value rapidly and the vehicle achieves better trajectory following.

2. Mathematical Model for Dynamic Steering In order to steer a tracked vehicle, it is necessary to drive one track faster than the other, causing the vehicle to turn to the slower track. One of the ways this is executed is by using a dual-motor independent drive with two motor controllers to control the torque and speed of the two motors which drive the sprocket. This is called an electronic control differential steering system (ECDS) which can improve vehicle’s stability, handling and flexibility. The objective of this paper is to solve the problems of insufficient torque during small radius steering and the shortage of power in high-speed large radius steering. First we selected a tracked vehicle which is the object of our mathematical modeling and analysis. The selected vehicle’s configuration is shown in Table1. Secondly, we analyzed the tracked vehicle’s kinematics and dynamics for stationary steering and dynamic steering in order to study the torque and power required by propulsion motor in both steering cases. Energies 2017, 10, 1118 3 of 17

Table 1. Vehicle parameters.

Parameters Value Vehicle tread, B (m) 1.3 Ground contact length, L (m) 1.7 Rolling resistance coefficient, f 0.04 Transmission efficiency, η 0.9 Drive ratio, ig 6.35 Mass of vehicle, m (kg) 2000 Mass gain coefficient, δ 1.5 Moment of inertia, J (kg/m2) 3000

To do this we took right steering as an example, FL is the tractive force on the outer track, FR is the tractive force on the inner track, RL is the rolling resistance force on the outer track, RR is the rolling resistance on the inner track, Mµ is the steering resistance moment, vL is the track speed on the outer side, vR is the track speed on the inner side and ω is the yaw rate. Steering maneuvers with different radiuses can be shown in Figure1. Mµ is deﬁned as follows:

1 1 µ M = µmgL = max mgL (1) µ 4 4 R 0.925 + 0.15 · B where µ is the coefﬁcient of roll resistance, µmax is the maximum rolling resistance coefﬁcient, R is the steering radius of the vehicle. The rolling resistance of the two tracks which is the same in magnitude but opposite in direction is expressed as follows:

1 R = R = f mg (2) L R 2 According to vehicle dynamics performance indicators: First, maximum speed should be 72 km/h. Second, maximum climbing degree should be 32◦ and climbing speed is 20 km/h. Third, acceleration capability: 0 to 32 km/h less than 8 s. The parameters of the motor are determined as shown in Table2.

Table 2. Motor parameters.

Rated Rated Torque Rated Speed Maximum Maximum Maximum Parameters Power (kW) (N·m) (r/min) Power (kW) Torque (N·m) Speed (r/min) Value 20 64 3000 40 150 9000

In this paper, the dynamic balance corresponding to different steering maneuvers is analyzed to obtain the torque or power required by an unilateral motor. Then we could ﬁnd out whether it can meet the requirements during different steering radius.

2.1. R ≤ 0.5 B Steering The equilibrium relationship between the force and moment of the vehicle when turning from resting, as shown in Figure1a, can be expressed as:

( dv FL − FR + RR − RL = δm dt B B dω (3) (FL + FR) · 2 − (RL + RR) · 2 − Mµ = J dt where FL and FR are tractive force but opposite in direction. It’s worth mentioning that when R = 0, FL = FR, VL = VR, lateral acceleration is zero, dv/dt = 0, and when R = 0.5 B, VR = 0, RR = 0. Energies 2017, 10, 1118 3 of 17

To do this we took right steering as an example, FL is the tractive force on the outer track, FR is the tractive force on the inner track, RL is the rolling resistance force on the outer track, RR is the rolling resistance on the inner track, Mμ is the steering resistance moment, vL is the track speed on the outer side, vR is the track speed on the inner side and ω is the yaw rate. Steering maneuvers with different radiuses can be shown in Figure 1. Mμ is defined as follows: 11μ M ==μmgLmax mgL μ R 440.925+⋅ 0.15 (1) B

where μ is the coefficient of roll resistance, μmax is the maximum rolling resistance coefficient, R is the steering radius of the vehicle. The rolling resistance of the two tracks which is the same in magnitude but opposite in direction is expressed as follows:

==1 RLRRfmg (2) 2 According to vehicle dynamics performance indicators: First, maximum speed should be 72 km/h. Second, maximum climbing degree should be 32° and climbing speed is 20 km/h. Third, acceleration capability: 0 to 32 km/h less than 8 s. The parameters of the motor are determined as shown in Table 2.

Table 2. Motor parameters.

Rated Rated Rated Maximum Maximum Maximum Parameters Power Torque Speed Power (kW) Torque (N·m) Speed (r/min) (kW) (N·m) (r/min) Value 20 64 3000 40 150 9000

In this paper, the dynamic balance corresponding to different steering maneuvers is analyzed to obtain the torque or power required by an unilateral motor. Then we could find out whether it can meet the requirements during different steering radius.

2.1. R ≤ 0.5 B Steering The equilibrium relationship between the force and moment of the vehicle when turning from resting, as shown in Figure 1a, can be expressed as:

 dv FFRR−+−=δ m  LRRL dt  (3) BBdω ()FF+⋅−+⋅−=() RR Mμ J  LR22d LR t

Energieswhere2017, 10 F,L 1118and FR are tractive force but opposite in direction. It's worth mentioning that when 4R of = 170, FL = FR, VL = VR, lateral acceleration is zero, dv/dt = 0, and when R = 0.5 B, VR = 0, RR = 0.

(a) (b)

FigureFigure 1. Force 1. Force analysis: analysis: (a) R (a≤) R0.5 ≤ 0.5B steering; B steering; (b )(bR) >R 0.5> 0.5B Bsteering. steering.

The motor speed on both sides can be expressed as

( 1000i n = g · 3.6 · ω · ( B − R) L 120πrz 2 1000i (4) n = g · 3.6 · ω · ( B + R) R 120πrz 2 where rz is the radius of the sprocket, rz = 150 mm. From Equations (1)–(4), the motor torque and power can be obtained as follows:

( FL,R TL,R = · rz igη (5) TL,R ·nL,R PL,R = 9549 T is the tractive torque but opposite in direction. P is the consumed power used to drive the track. Subscript “L” or “R” means the left or right side of the vehicle.

2.2. R > 0.5 B Steering According to the vehicle dynamic balance, the following formula can be obtained from Figure1b:

( dv FL − FR − RR − RL = δm dt B B dω (6) (FL + FR) · 2 + (RR − RL) · 2 − Mµ = J dt where FL is the tractive force. If we assume the following equation:

µmgL F = − 0.5 f mg = 0 (7) R 4B

We will get the solution R = R0 = 67.54 m. When R > R0, FR is tractive force and is in the same direction with the track. While when R < R0, FR is brake force, generated by motor. The direction between FR and VR is opposite in this situation. The vehicle in this steering case can be divided into static starting steering and driving steering, the motor torque and power required vary with the steering radius and speed. From Equations (1)–(7), Table3 shows the steering power and torque required by dual motors during dynamic steering maneuvers with 0, 0.2 B, 0.5 B steering radii and stationary steering maneuvers with 0, 0.2 B, 0.5 B, 2 B, 5 B, 8 B, 10 B steering radii. From Table3 we can see that when 0 ≤ R ≤ 0.5 B, in both dynamic steering and stationary steering, the motor torque required exceeds the motor’s peak torque (150 N·m). When R = 5 B, the power required by the outer side motor is larger than the motor’s maximum power (40 kW). When R = 8 B or R = 10 B, the power required by both side motors is larger than the motor’s maximum power (40 kW). According to the results we can ﬁnd that during small-radius steering (0 ≤ R ≤ 0.5 B) the torque required by a single motor in dynamic steering is 1.5 times that during stationary steering. During large-radius steering (R = 10 B) the power required by the motor is 1.5 times the motor’s maximum power therefore the motor’s torque and power should increase by 50% at least, which Energies 2017, 10, 1118 5 of 17 results in large size and mass of the motor and other power inverters. As a result, the superiority of ECDS is not highlighted, and although Zhai [1] proposed a single motor and steering motor coupling system, providing larger torque during small-radius steering, and additional power during high speed large-radius steering while the working principle and structural design of the coupled system were not described in detail.

Table 3. Dynamic and stationary steering torque and power required by two motors. Energies 2017, 10, 1118 5 of 17

Situation R (mm) TL (N·m) TR (N·m) nL (r/min) nR (r/min) PL (kW) PR (kW) V (km/h) R = 8 B or R = 10 B, the power required by both side motors is larger than the motor’s maximum 0 272 272 344 344 9.8 9.8 0 powerDynamic (40 kW).0.2 BAccording279 to the results 252 we can find 482 that during 206 small-radius 14.1 steering 5.5 (0 ≤ R ≤ 1.220.5 B) the torque required0.5 B by 258a single motor 181 in dynamic 688 steering 0 is 1.5 times 18.6 that during 0 stationary 3.06 steering. During0 large-radius 192 steering 192 (R = 10 B) 344 the power 344required by6.9 the motor 6.9is 1.5 times 0 the motor's maximum0.2 B power187 therefore 187the motor's torque 482 and power 206 should 9.4 increase 4.0by 50% at 1.22least, which results0.5 inB large size179 and mass 168 of the moto 688r and other 0 power 12.9inverters. As 0 a result, 3.06 the Stationary − − superiority of2 ECDSB is not148 highlighted,127 and although 1720 Zhai [1] 1032 proposed 26.6a single motor13.7 and steering 12.2 5 B 111 −90 3096 3784 43.9 −29.2 30.6 motor coupling8 B system, providing90 larger−69 torque 5847during small-radius 5159 steering, 54.8 and −additional37.2 power 49.0 during high speed10 B large-radius80 steering−59 while 7223the working 6535principle and 60.3 structural−40.5 design of 61.2 the coupled system were not described in detail. 3. Steering System Design 3. Steering System Design 3.1. Steering Coupling Drive System 3.1. Steering Coupling Drive System A steering coupling drive system composed of a new type of center steering motor, two A steering coupling drive system composed of a new type of center steering motor, two electromagneticelectromagnetic clutches, clutches, two two planetaryplanetary gear gear couple couplers,rs, and and two twopropulsion propulsion motors motors is designed is designed and andshown shown in in Figure Figure 2.2 The. The parameters parameters of the of thesteering steering motor motor are determined are determined as listed as in listed Table in 4. TableThe 4. Thesteering steering motor motor can can produce produce large large torque torque at low at low rotation rotation speed. speed. A reverse A reverse mechanism mechanism is designed is designed in in thethe steering steering motormotor to achieve achieve positive positive and and negative negative rotation rotation direction. direction. When When the torque the torque or power or power is is insufﬁcient,insufficient, the the electromagnetic electromagnetic clutch clutch is is engaged engaged and and the the torque torque or orpower power of the of thedriving driving wheel wheel is is satisﬁedsatisfied by theby the coupling coupling of of the the steering steering motor motor andand the propulsion motor. motor.

FigureFigure 2. 2.Steering Steeringcoupling coupling drive system system configuration. conﬁguration.

Table 4. Steering motor parameters.

Rated Rated Rated Maximum Maximum Maximum Parameters Power Torque Speed Power (kW) Torque (N·m) Speed (r/min) (kW) (N·m) (r/min) Value 10 160 600 20 300 1200

Energies 2017, 10, 1118 6 of 17

Table 4. Steering motor parameters.

Rated Rated Torque Rated Speed Maximum Maximum Maximum Parameters Power (kW) (N·m) (r/min) Power (kW) Torque (N·m) Speed (r/min) Value 10 160 600 20 300 1200 Energies 2017, 10, 1118 6 of 17

3.2. Planetary Gear Coupler Design A planetary gear coupler is proposed to couple the torque or power of the propulsion motor and the steering motor. The output torque transmittedtransmitted to the sprocket is increased using the torque coupling modemode during during low-speed low-speed small-radius small-radius steering. steering. The outputThe output power power transmitted transmitted to the sprocket to the issprocket increased is usingincreased the powerusing couplingthe power mode coupling during mode high-speed during large-radius high-speed steering.large-radius Therefore steering. the torqueTherefore or powerthe torque demand or forpower different demand radius for steering differen ist satisﬁedradius steering and the vehicleis satisfied steering and performance the vehicle issteering improved. performance is improved. The planetary geargear couplerscouplers consistsconsists ofof twotwo electromagnetic electromagnetic (EM) (EM) clutches, clutches, two two gear gear pairs pairs and and a planetarya planetary gear gear unit, unit, which which consists consists of aof sun a sun gear, gear, a ring a ring gear, gear, a planet a planet carrier carrier and three and planetthree planet gears, showngears, shown in Figure in 3Figure. Gear 3. pair Gear 1 contains pair 1 contains gear 1 and gear gear 1 and 2 (ring gear gear), 2 (ring whereas gear), gearwhereas pair gear 2 contains pair 2 gearcontains 3 and gear gear 3 4.and It’s gear worth 4. It's noting worth that noting gear 1that and gear gear 1 3 and are mountedgear 3 are on mounted the steering on the motor steering shaft withoutmotor shaft a spline, without transmitting a spline, the transmitting torque occurs the only torque when occurs one of only the EMwhen clutches one of is the engaged. EM clutches The dual is motorengaged. coupling The dual drive motor steering coupling includes drive a single steering side coupling includes mode a single and aside double coupling side coupling mode and mode. a Wedouble take side one coupling side as an mode. example, We astake shown one side in Figure as an 3example,. as shown in Figure 3.

Figure 3. Components of the planetary gear coupler.

When the vehicle is driving straight, the ring gear (R) is fixed by brake 1 while both the EM When the vehicle is driving straight, the ring gear (R) is ﬁxed by brake 1 while both the EM clutch 1 and the EM clutch 2 are disengaged. The power is transmitted to the sun gear (S) of the clutch 1 and the EM clutch 2 are disengaged. The power is transmitted to the sun gear (S) of the planetary gear unit (transmission ratio 6.35) from the propulsion motor, and outputted through the planetary gear unit (transmission ratio 6.35) from the propulsion motor, and outputted through the planet carrier (C) to the sprocket. During steering, if the torque or power does not meet the demand, planet carrier (C) to the sprocket. During steering, if the torque or power does not meet the demand, the torque or the power is coupled by this planetary gear coupling device. The coupling mode the torque or the power is coupled by this planetary gear coupling device. The coupling mode includes includes torque coupling mode and speed coupling mode, as shown in Table 5. torque coupling mode and speed coupling mode, as shown in Table5. Table 5. Coupling mode and the operation of each component. Table 5. Coupling mode and the operation of each component. Mode Braker 1 EM Clutch 1 EM Clutch 2 Steering Motor Propulsion Motor ModeStraight Brakerengaged 1 disconnected EM Clutch 1 disconnected EM Clutch 2 Steeringoff Motor Propulsion working Motor TorqueStraight coupling engaged engaged disconnected disconnected combined disconnected working off working working TorquePower coupling coupling disengaged engaged disconnectedcombined disconnected combined working working working working Power coupling disengaged combined disconnected working working The torque coupling mode is active when the vehicle is turning in a small radius at low speed. In the torque coupling mode the EM clutch 1 is disconnected while the EM clutch 2 is engaged and brake 1 is disengaged. Then the torque of the steering motor and the propulsion motor are coupled to the sun gear (S) through the gear pair 2, transmitted to the sprocket through the planet carrier (C). The power transmission path is shown in Figure 4a. The power coupling mode is active when the vehicle is turning in big radius at high speed. In the power coupling mode the EM clutch 1 is engaged while the EM clutch 2 is disconnected and the brake 1 is engaged. The power of the steering motor is transferred to ring gear (R) through the gear

Energies 2017, 10, 1118 7 of 17

Energies 2017, 10, 1118 7 of 17 The torque coupling mode is active when the vehicle is turning in a small radius at low speed. Inpair the 1 torqueand the coupling power modeof propulsion the EM clutchmotor 1 is is transmitted disconnected to while the sun the gear EM clutch(S). Then 2 is engagedthe power and is brakecoupled 1 is and disengaged. transmitted Then to thethetorque sprocket of thethrough steering the motor planet and carrier the propulsion (C). The power motor transmission are coupled topath the is sun shown gear in (S) Figure through 4b. the gear pair 2, transmitted to the sprocket through the planet carrier (C). The power transmission path is shown in Figure4a.

(a) (b)

FigureFigure 4.4. ((a)) TorqueTorque couplingcoupling transmissiontransmission mode;mode; ((bb)) PowerPower couplingcoupling transmissiontransmission mode. mode.

The torque and power output to the driving sprockets on both sides under different steering The power coupling mode is active when the vehicle is turning in big radius at high speed. In the radius conditions with this coupling device are shown in Table 6. The performance of the vehicle power coupling mode the EM clutch 1 is engaged while the EM clutch 2 is disconnected and the brake 1 without the coupling device is also shown in Table 6. is engaged. The power of the steering motor is transferred to ring gear (R) through the gear pair 1 and theTable power 6. of Motors’ propulsion condition motor comparison is transmitted between to the the vehicle sun gear with (S). and Then without the coupling power is device. coupled and transmitted to the sprocket through the planet carrier (C). The power transmission path is shown in TL TR Ts PL PR Ps nL nR ns FigureRadius4b. Coupling The torque and power(N·m) output (N·m) to the driving(N·m) sprockets(kW) (kW) on both(kW) sides (r/min) under different(r/min) steering(r/min) 0 B without 150 150 0 5.4 5.4 0 0 radius conditions with this coupling device are shown in Table6. The performance344 of344 the vehicle (dynamic) with 150 150 260 5.4 5.4 9.4 344 without the coupling device is also shown in Table6. 0.5 B without 150 150 0 10.8 0 0 0 688 0 (dynamic) with 150 150 108 10.8 0 7.8 688 Table 6. Motors’ condition comparison between the vehicle with and without coupling device. 0 B without 150 150 0 5.4 5.4 0 0 344 344 (stationary) with 150 150 84 5.4 5.4 3 344 Radius Coupling TL (N·m) TR (N·m) Ts (N·m) PL (kW) PR (kW) Ps (kW) nL (r/min) nR (r/min) ns (r/min) 0.50 B B withoutwithout 150 150 150 150 0 0 5.4 10.8 5.4 0 0 0 0 0 344688 344 0 (stationary)(dynamic) withwith 150150 150 150 260 29 5.410.8 5.4 0 9.42.1 344 688 0.52 BB withoutwithout 150 148 150 −127 0 0 10.826.7 0 −13.7 0 0 0 0 6881720 01032 (stationary)(dynamic) withwith 150- 150 - 108 - 10.8- 0 - 7.8- 688 - 0 B without 150 150 0 5.4 5.4 0 0 8 B without 90 −69 0 40.0 −27.1 0 344 4244 344 3744 0 (stationary) with 150 150 84 5.4 5.4 3 344 (stationary) with 90 −69 150 54.8 −37.1 14.8 4244 5129 942 0.5 B without 150 150 0 10.8 0 0 0 688 0 (stationary) with 150 150 29 10.8 0 2.1 688 2 ItB can withoutbe seen from 148 Tables−127 3 and 60 that the26.7 maximum−13.7 torque0 of the propulsion motor 0is 150 (stationary) 1720 1032 N·m, which withcannot meet - the steering - requirement- -- whereas with-- the coupling device a maximum 8 B without 90 −69 0 40.0 −27.1 0 4244 3744 0 torque(stationary) of 280with N·m, can 90 be transmitted−69 from150 the54.8 propulsion−37.1 motor14.8 to sprocket4244 to meet 5129 the dynamic942 or stationary small-radius steering requirements. When the vehicle is turning in a big radius, the peakIt power can be of seen the frommotor Tables is 403 kW and without6 that the coupli maximumng, which torque is not of the enough, propulsion while motorwith coupling, is 150 N ·them, whichvehicle cannot can output meet the a large steering enough requirement power (60 whereas kW at with most) the couplingto meet the device 8 B asteering maximum requirement torque of 280shown N·m, in canTable be 3. transmitted from the propulsion motor to sprocket to meet the dynamic or stationary small-radius steering requirements. When the vehicle is turning in a big radius, the peak power of the 3.3. Planetary Geaer System Design motor is 40 kW without coupling, which is not enough, while with coupling, the vehicle can output a large enough power (60 kW at most) to meet the 8 B steering requirement shown in Table3. 3.3.1. Selection and Preliminary Calculation The planetary gear transmission type 2Z-X (A) (defined by the former Soviet Union), that is NGW type, is selected. The transmission ratio is 6.35. The number of teeth of each gear: sun gear Za

Energies 2017, 10, 1118 8 of 17

3.3. Planetary Geaer System Design

3.3.1. Selection and Preliminary Calculation The planetary gear transmission type 2Z-X (A) (defined by the former Soviet Union), that is NGW type, is selected. The transmission ratio is 6.35. The number of teeth of each gear: sun gear Za = 17, ring gear Zb = 91, planetary gear Zc = 37, so the actual gear ratio i = 6.353 and the gear ratio error is 0.05%. The material of sun gear and planetary gear are 20CrMnTi, level 6 precision. The material of ring gear is 42CrMo, level 7 precision. Calculate the gear module initially according to bending fatigue: s TK K K Y = 3 A F ∑ Fp Fa1 m Km 2 (8) φdzaσFlim where Km is the formula coefficient, KA is the using coefficient, KF ∑ is the comprehensive coefficient, KFp is the bending coefficient, YFa1 is the tooth shape coefficient, φd is the tooth width coefficient. σFlim is the permissible stress. Therefore the gear module is 2.5. The centre distance is 70 mm with a spiral angle of 15.3589◦.

3.3.2. Geometric Size Calculation The geometric sizes of the sun gear (with subscript 1), planet gear (with subscript 2) and ring gear (with subscript 3) are shown in Table7.

Table 7. Geometric sizes of each gear pair (mm).

Reference Circle Basic Circle Tip Circle Root Circle Breadth of Gear Pair Diameter (d) Diameter (db) Diameter (da) Diameter (df) Tooth (b) d = 44.074 d = 41.234 d = 49.074 d = 37.824 b = 43 External gear pair 1 b1 a1 f1 sun d2 = 95.926 db2 = 89.922 da2 = 100.926 df2 = 89.676 bplanet = 48 d = 95.926 d = 89.922 d = 100.926 d = 89.676 b = 48 Internal gear pair 2 b2 a2 f2 planet d3 = 235.926 db3 = 221.159 da3 = 231.320 df3 = 242.176 bring = 43

3.3.3. Checking of the Strength To make sure the coupler can work in extreme condition, the teeth contact stress and the teeth bending strength is checked as shown in Table8. The stress suffered by gear teeth is lower than the allowed value so the coupler can meet the strength requirements.

Table 8. Checking of each gear pair (MPa).

σ σ σ σ Checking Pair σ σ σ σ HP1 HP2 FP1 FP2 Checking H1 H2 F1 F2 (allowed) (allowed) (allowed) (allowed) External gear pair σ ≤ σ 1162.8 1162.8 - - 1566.1 1562.3 - - H1 HP1 (teeth contact stress) σH2 ≤ σHP2 External gear pair σ ≤ σ - - 340.2 295.1 - - 686.5 672.0 F1 FP1 (teeth bending strength) σF2 ≤ σFP2 Internal gear pair σ ≤ σ 434.6 434.6 - - 1818.9 793.6 - - H1 HP1 (teeth contact stress) σH2 ≤ σHP2 Internal gear pair σ ≤ σ - - 233.2 228.9 - - 588.0 383.3 H1 HP1 (teeth bending strength) σH2 ≤ σHP2

3.4. EM Clutch and EM Brake According the maximum torque characteristics of steering motor and planetary gear coupler in Table3, the maximum required torques of EM clutch 1 and EM clutch 2 in Figure3 are 150 and Energies 2017, 10, 1118 9 of 17

130 N·m, respectively. The overall size of the EM clutch 1 and EM clutch 2 is Φ = 90 mm × 100 mm. The maximumEnergies 2017 required, 10, 1118 braking torques and the size of the electromagnetic brake is 1500 N9 ·ofm 17 and Φ = 310 mm × 55 mm, respectively. The important design parameters of the electromagnetic clutches and theclutches electromagnetic and the electromagnetic brakes are proposed brakes are for pr manufacture.oposed for manufacture. Based on theBased above on the calculations, above calculations, the overall size of the coupling device is designed as 468 mm × 364 mm × 378 mm. the overall size of the coupling device is designed as 468 mm × 364 mm × 378 mm. 4. Control Strategy 4. Control Strategy 4.1. Driver Inputs Modeling 4.1. Driver Inputs Modeling The input angle signal of the steering wheel displacement is shown in Figure 5. According to Thedifferent input steering angle signal wheel ofangle the signals, steering the wheel desired displacement turning radius iscan shown be calculated. in Figure 5. According to different steering wheel angle signals, the desired turning radius can be calculated.

- + θ1 θ1 θ2 θ2 θ 0 3 θ θ3 θ4 θ4

Figure 5. Input angle signal of steering wheel. Figure 5. Input angle signal of steering wheel. The displacement of acceleration pedal or brake pedal is assumed to be linear to the output Thetorque displacement of motor. The ofacceleration control coefficient pedal of or acceleration brake pedal (A is) and assumed brake topedal be lineardisplacement to the output signal ( torqueD) of motor.are Thegiven control as follows: coefﬁcient of acceleration (A) and brake pedal displacement signal (D) are given as follows: αα− A A== α−α00 αααmax−α0 max− 0 (9) B = − βββ−β0 (9) B =−βmax−β00 ββ− max 0 where α and β are the displacements of acceleration and brake pedals, respectively. α0 and β0 are the α β free displacementswhere α and β ofare acceleration the displacements and brakeof acceleration pedals, and respectively. brake pedals,αmax respectively.and βmax are0 and the maximum0 are α β displacementsthe free ofdisplacements acceleration of and accelerati brake pedals,on and respectively.brake pedals,A respectively.is in the range max from and 0 tomax 1, are where the the outputmaximum torque of displacements propulsion motor of accelerati is in theon and range brake from pedals, 0 to Trespectively.max. And D Ais is in in thethe rangerange from from 0− to1 1, to 0, wherewhere the output the output torque torque of propulsion of propulsion motor motor is in is thein the range range from fromTmax 0 toto Tmax 0.. And D is in the range from −1 to 0, where the output torque of propulsion motor is in the range from Tmax to 0. 4.2. Torque Distribution Strategy Based on Speed 4.2. Torque Distribution Strategy Based on Speed A torque distribution control strategy based on speed is proposed and shown in Figure6. AccordingA to torqueA, B anddistribution steering control wheel strategy angle signals,based on basedspeed onis proposed the above and calculation, shown in Figure the torque 6. and rotationAccording speed to A desired, B and cansteering be determined. wheel angle Thesignals, coupling based distributionon the above controllercalculation, is the used torque to regulate and rotation speed desired can be determined. The coupling distribution controller is used to regulate the torque of each motor to track the driver inputs based on the real-time rotation speed (nL-real/nR-real) the torque of each motor to track the driver inputs based on the real-time rotation speed (nL-real/nR-real) and the revised torque (∆T1/∆T2) from the slip-ratio controller. The torque generated by each motor is and the revised torque (ΔT1/ΔT2) from the slip-ratio controller. The torque generated by each motor coupledis coupled through through the electromechanical the electromechanical coupling coupling device device and and transmitted transmitted to to the the sprocket. sprocket.

4.2.1. Ideal4.2.1. VehicleIdeal Vehicle Body Body Model Model AccordingAccording to the to dynamic the dynamic analysis, analysis, the the ideal ideal vehicle vehicle model model cancan be su summarizedmmarized as asfollows: follows:

v ( vL,RLR, = ( (Rdes= 0)) 1.81.8B BRdes 0 ωdes = ω = v (10) des  desv (10)  des (Rdes ≠6= 0) 3.6Rdes (0)Rdes 3.6Rdes where ω is the desired yaw rate, VL,R is the left or right side track speed in center steering which can be calculated from the accelerator pedal position, vdes is the desired vehicle speed, Rdes is the ideal steering radius.

Energies 2017, 10, 1118 10 of 17

where ω is the desired yaw rate, VL,R is the left or right side track speed in center steering which can be calculated from the accelerator pedal position, vdes is the desired vehicle speed, Rdes is the ideal Energiessteering2017 radius., 10, 1118 10 of 17

Figure 6. Torque distributiondistribution strategystrategy basedbased onon speed.speed.

TheThe desireddesired vehiclevehicle speedspeed andand idealideal steeringsteering radiusradius cancan bebe determineddetermined byby driverdriver model.model. TheseThese signalsignal areare inputsinputs to to the the ideal ideal vehicle vehicle body body model mo anddel theand desiredthe desired sprocket sprocket rotation rotation speed andspeed torque and aretorque calculated. are calculated.

4.2.2. Coupling Distribution Controller Figure7 7 shows shows thethe workﬂowworkflow ofof the the coupling coupling distributiondistribution controller.controller. TheThe controllercontroller receivesreceives the the signalssignals ((TTdesdes,, ndes)) calculated calculated from from the the ideal vehicle model. Using the fuzzyfuzzy proportionproportion integrationintegration differentiation (PID)(PID) controller controller based based on on the the error erro betweenr between the referencethe reference rotational rotational speed speed and the and actual the rotationalactual rotational speed to speed produce to produce the torque theT11 torque. The fuzzy T11. The rules fuzzy is shown rules in is Figure shown8a–c. in TheFigure torque 8a–c. signal The (torqueTLL) sent signal to the (TLL coupling) sent to judgmentthe coupling module judgment is given module as follows: is given as follows:

TT=+Δ T (11) TLLLL= T1111+ ∆T 1 1 (11)

where ΔT1 is obtained by the slip ratio controller, which will be introduced in next section. Similarly, where ∆T1 is obtained by the slip ratio controller, which will be introduced in next section. Similarly, the torque signal (TRR) sent to coupling judgment module is given as follows: the torque signal (TRR) sent to coupling judgment module is given as follows: =+Δ TTRR 22 T 2 (12) TRR = T22 + ∆T 2 (12) The flowchart of the coupling judgment module is shown in Figure 9. Depending on the differentThe ﬂowchartTLL and T ofRR theobtained, coupling the judgment module module will output is shown different in Figure torque9. Depending commands on to the the different motor Tcontroller.LL and TRR Whenobtained, the required the module torque will and output power different does torquenot exceed commands their peak to the value, motor the controller. steering Whenmotor thedoes required not work, torque Ts = and 0. When power the does required not exceed torque their exceeds peak value,the peak the torque steering of motorthe propulsion does not work,motor,Ts the= 0.steering When themotor required operates torque to generate exceeds theadditional peak torque torque of theto meet propulsion the torque motor, requirements. the steering motorWhen operatesthe power to of generate motor additionalis insufficient, torque the to power meet thecoupling torque mode requirements. is active, Whenthe steering the power motor of motorgenerates is insufﬁcient, the relevant the torque power to meet coupling the coupling mode is active,conditio thens, steeringand output motor enough generates power. the relevant torque to meet the coupling conditions, and output enough power.

Energies 2017, 10, 1118 11 of 17 EnergiesEnergies 2017 2017, 10, 10, 1118, 1118 1111 of of17 17 Energies 2017, 10, 1118 11 of 17

FigureFigure 7. 7. Coupling CouplingCoupling distribution distribution distribution controller. controller. Figure 7. Coupling distribution controller.

(a)( a()a ) (b()b ) (c)( c)

Figure 8. Fuzzy rules: (a) Kp rules (b) Ki rules (c) Kd rules. FigureFigureFigure 8. 8. 8.Fuzzy FuzzyFuzzy rules: rules:rules: (a )(a K)) pK rulespp rules (b ( )b K) i Krulesi rulesrules (c )( ( ccK)) dK rules.dd rules.rules.

Figure 9. Flowchart of the coupling judgment module. FigureFigure 9. 9. Flowchart Flowchart of of the the coupling coupling judgment judgment module. module. 4.2.3. Slip Ratio Controller 4.2.3.4.2.3. Slip SlipUsually, Ratio Ratio Controller trackedController vehicles turns are accompanied by a slip phenomenon: Usually,Usually, tracked tracked vehicles vehicles turns turns are are accompanied accompanied by by a slipa slip phenomenon: phenomenon:

Energies 2017,, 1010,, 11181118 12 of 17 Energies 2017, 10, 1118 12 of 17

− ωω 4.2.3. Slip Ratio Controller  ==−≤ur− ωω r ω sru==−≤ur1 r (ω ) sruuu1 ( ) Usually, tracked vehicles turns are accompanieduu by a slip phenomenon: (13)  ruω − u (13) ( ==−ruω − u ω >  sru==−= u−rω =1− rω ( ()ω≤ >) srus rrωωu 11 u r ω()u  rrrωωω−u u (13) s = rω = 1 − rω (rω > u) where n = rω is the sprocket rotation speed, u is the track longitudinal speed, generally u = V. where n = rω is the sprocket rotation speed, u is the track longitudinal speed, generally u = V. whereInn order = rω isto theimprove sprocket the rotationtracked speed,vehicleu steeringis the track stability longitudinal and use speed,the energy generally efficiently,u = V .the slip In order to improve the tracked vehicle steering stability and use the energy efficiently, the slip ratio Inis ordercontrolled to improve within the a tracked reasonable vehicle range. steering The stability choice andof slip use theratio energy is related efﬁciently, to the the road slip ratio is controlled within a reasonable range. The choice of slip ratio is related to the road ratioparameters, is controlled and normally within a reasonablesdes = 0.15 for range. an asphalt The choice road. of The slip slip ratio ratio is related controller to the is road designed parameters, based parameters, and normally sdes = 0.15 for an asphalt road. The slip ratio controller is designed based andon the normally fuzzy sPID= control 0.15 for anmethod asphalt (similar road. Theto torq slipue ratio controller controller mentioned is designed above) based onto themaintain fuzzy PIDthe on the fuzzy PIDdes control method (similar to torque controller mentioned above) to maintain the controlactual slip method ratio (similar closed toto torque the ideal controller slip ratio, mentioned as shown above) in to Figure maintain 10. the The actual actual slip slip ratio ratio closed is actual slip ratio closed to the ideal slip ratio, as shown in Figure 10. The actual slip ratio is tocalculated the ideal from slip ratio,Equation as shown (13). in Figure 10. The actual slip ratio is calculated from Equation (13). calculated from Equation (13).

sL-des V sL-des V sL-real Slip ratio ∆T1 n Slip ratio sL-real L-real calculator Fuzzy PID ∆T1 nL-real calculator - d Fuzzy PID - d controller 1 Coupling dt controller 1 Coupling dt distribution V distribution V Slip ratio - controller n Slip ratio s - Fuzzy PID controller R-real calculator R-real Fuzzy PID nR-real calculator sR-real d controller 2 d controller 2 dt ∆T2 sR-des dt ∆T2 sR-des

Figure 10. Slip ratio controller. Figure 10. Slip ratio controller. 5. Modeling and Simulation Results 5.5. Modeling Modeling and and Simulation Simulation Results Results The steering co-simulation model shown in Figure 6 is built by multi-body software Recurdyn TheThe steering co-simulationco-simulation modelmodel shown shown in in Figure Figure6 is 6 builtis built by multi-bodyby multi-body software software Recurdyn Recurdyn and and control software Matlab/Simulink. The vehicle model and ground model are developed in andcontrol control software software Matlab/Simulink. Matlab/Simulink. The vehicle The ve modelhicle model and ground and ground model are model developed are developed in Recurdyn in Recurdyn as shown in Figure 11. The driver inputs model, the control model based on speed and Recurdynas shown inas Figureshown 11in. Figure The driver 11. The inputs driver model, inputs the model, control the model control based model on speedbased andon speed the motor and the motor system model are developed in Simulink. The control module in Recurdyn is seamlessly thesystem motor model system are model developed are developed in Simulink. in TheSimulink. control The module control in module Recurdyn in isRecurdyn seamlessly is seamlessly interfaced interfaced with Matlab/Simulink. The parameters of the dynamic simulation parameters are shown interfacedwith Matlab/Simulink. with Matlab/Simulink. The parameters The parameters of the dynamic of the simulationdynamic si parametersmulation parameters are shown are in Tableshown9. in Table 9. The reaction characteristic of motors can be automatically generated by motor controller inThe Table reaction 9. The characteristic reaction characteristic of motors canof motors be automatically can be automatically generated generated by motor by controller motor controller model in model in Simulink and transferred to Recurdyn as input of the motor. modelSimulink in Simulink and transferred and transferred to Recurdyn to Recurdyn as input of as the input motor. of the motor.

(a) (b) (a) (b) Figure 11. (a) Track modeling; (b) Vehicle modeling. FigureFigure 11. 11. ((aa)) Track Track modeling; modeling; ( (bb)) Vehicle Vehicle modeling.

Energies 2017, 10, 1118 13 of 17 Energies 2017, 10, 1118 13 of 17 Table 9. Dynamic simulation parameters.

Table 9. Dynamic simulation parameters. Component Number Inertia (kg·mm²) Stiffness Coefficient Damping Coefficient

2 ComponentSprocket Number2 Inertia 279,379 (kg·mm ) Stiffness 18,000 Coefﬁcient Damping 10 Coefﬁcient SprocketRoad Wheel 10 2 685,898 279,379 12,000 18,000 10 10 RoadCarrier Wheel 106 685,898 5121 15,000 12,000 10 10 CarrierTrack link 198 6 1989 5121 15,0009000 10 10 TrackTrack subsystem link 198 - 1989 - 1,600,000 9000 10,000 10 Track subsystem - - 1,600,000 10,000

5.1. Center Steering 5.1. Center Steering When the steering wheel is turned to the left, the accelerator pedal is pushed to the bottom, the vehicleWhen begins the steering to turn wheelfrom its is static turned position. to the left, The the comparative accelerator trajectory pedal is pushedis shown to in the Figure bottom, 12a. theThe vehicle vehicle begins without to turn coupling from its device static turns position. slowly The due comparative to its insufficient trajectory output is shown torque. in Figure When 12 thea. Thecoupling vehicle device without is used, coupling the yaw device rate turns of the slowly vehicl duee increases to its insufﬁcient rapidly to output the ideal torque. value When (1.3 rad/s) the couplingwhereas device when isthe used, coupling the yaw device rate ofis thenot vehicleused the increases yaw rate rapidly gradually to the increases ideal value to (1.30.8 rad/s rad/s) as whereasshown whenin Figure the coupling 12b. It can device be isseen not from used theFigure yaw 12c rate that gradually the vehicle increases with to the 0.8 rad/scoupling as shown device inoutputs Figure 12a b.larger It can torque be seen in fromthe initial Figure movement 12c that the to vehiclemeet the with needs the of coupling small-radius device dynamic outputs asteering. larger torqueAs the in steering the initial enters movement the steady to meet state, the the needs output of small-radiustorque decrease dynamics and finally steering. tends As theto a steering constant entersvalue. the The steady power state, generated the output by torquesprockets decreases is shown and in ﬁnally Figure tends 12d. toThe a constant power output value. Theis increased power generatedwith coupling, by sprockets but has is a shownsmall value in Figure as a 12whole.d. The power output is increased with coupling, but has a small value as a whole.

-1.25 1.4

-1.3 1.2

1 -1.35

） 0.8 ）

m -1.4 rad/s （ Yaw rate(w/o) Y （ 0.6

w Yaw rate(w) -1.45 0.4

-1.5 0.2 Trajectory (w/o) Trajectory (w) -1.55 0 -0.9 -0.8 -0.7 -0.6 0 1 2 3 4 5 X（m） t（s） (a) (b)

2000 30 Inner side(w/o) Outer side(w/o) 1500 25 Inner side(w) 1000 Outer side(w) 20 500 Inner side(w/o) ） Outer side(w/o) ） 15

0 kW N*m Inner side(w) （

（ 10

Outer side(w) P T -500 5 -1000

-1500 0

-2000 -5 0 1 2 3 4 5 0 1 2 3 4 5 t（s） t（s） (c) (d)

FigureFigure 12. 12.Simulation Simulation results results of centerof center steering: steering: (a) Trajectory; (a) Trajectory; (b) Vehicle (b) Vehicle yaw rate; yaw (c) Sprocketrate; (c) Sprocket torque; (dtorque;) Sprocket (d) power.Sprocket power.

5.2. 0.5 B Steering

Energies 2017, 10, 1118 14 of 17

Energies 2017, 10, 1118 14 of 17 5.2. 0.5 B Steering The steering wheel is manipulated to output R = 0.5 B, the accelerator pedal is pushed to the The steering wheel is manipulated to output R = 0.5 B, the accelerator pedal is pushed to the bottom and the vehicle begins to turn from its static position. The trajectory of the vehicle is shown bottom and the vehicle begins to turn from its static position. The trajectory of the vehicle is shown in Figure 13a. If the coupling device is not added, due to the limited torque, the sprocket cannot in Figure 13a. If the coupling device is not added, due to the limited torque, the sprocket cannot always generate the required torque, so the torque on both sides is always at the upper limit always generate the required torque, so the torque on both sides is always at the upper limit (150 N·m). (150 N·m). In Figure 13b it can be seen that the yaw rate is about 1.2 rad/s when the vehicle is In Figure 13b it can be seen that the yaw rate is about 1.2 rad/s when the vehicle is turning with turning with the coupling device, but only 0.8 rad/s without it. Figure 13c shows that during the the coupling device, but only 0.8 rad/s without it. Figure 13c shows that during the initial dynamic initial dynamic steering phase, the required torque is quite large. The vehicle with the coupling steering phase, the required torque is quite large. The vehicle with the coupling device can satisfy device can satisfy the torque requirement and the torque tends to a smaller constant value in steady the torque requirement and the torque tends to a smaller constant value in steady state. The power state. The power generated by the sprockets is shown in Figure 13d. The power on the outer side generated by the sprockets is shown in Figure 13d. The power on the outer side increases with the increases with the coupling device and the power on the inner side is slightly reduced, but the coupling device and the power on the inner side is slightly reduced, but the overall power is not overall power is not very large. very large.

-0.6 1.4 Trajectory(w/o) -0.8 Trajectory(w) 1.2

Yaw rate(w/o) -1 1 Yaw rate(w)

-1.2 ） 0.8 ） m rad/s （（ Y -1.4 0.6 w -1.6 0.4

-1.8 0.2

-2 0 -1 -0.5 0 0.5 0 1 2 3 4 5 X（m） t（s） (a) (b)

2000 30 Inner side(w/o) 1500 Outer side(w/o) 25 Outer side(w) 1000 Inner side(w) 20 500

） Inner side(w/o) ） 15 Outer side(w/o)

0 kW N*m Inner side(w) （

（ 10 P T -500 Outer side(w) 5 -1000 0 -1500

-2000 -5 0 1 2 3 4 5 0 1 2 3 4 5 t（s） t（s） (c) (d)

FigureFigure 13. 13.Simulation Simulation results results of 0.5of B0.5steering: B steering: (a) Trajectory; (a) Trajectory; (b) Vehicle (b) Vehicle yaw rate; yaw (c) Sprocketrate; (c) Sprocket torque; (dtorque;) Sprocket (d) power.Sprocket power.

5.3.5.3. 2 2 B B Steering Steering TheThe steering steering wheel wheel is is manipulated manipulated to to output outputR R= = 0.5 0.5B ,B the, the accelerator accelerator pedal peda isl is pushed pushed to to the the bottombottom and and the the vehicle vehicle begins begins to to turn turn from from its its static static position. position. The The trajectory trajectory of of the the vehicle vehicle is is shown shown inin Figure Figure 14 14a.a. It It can can be be seen seen that that the the vehicle vehicle understeer understeer is is improved improved with with the the coupling coupling device. device. From From FigureFigure 14 14b,b, we we can can see see that that ﬁnally finally the the yaw yaw rate rate of of the the vehicle vehicle is is 1.2 1.2 rad/s, rad/s, which which is is close close to to the the theoretical value, and the yaw rate of the vehicle without the coupling device is only 0.8 rad/s. As theoretical value, and the yaw rate of the vehicle without the coupling device is only 0.8 rad/s. shown in Figure 14c, since the coupling device is used, the outer side output torque of the vehicle is increased, and the inter side sprocket regenerates braking torque. The power generated by the outer sprockets and regenerated by the inter sprocket are shown in Figure 14d. This time the inner motor

Energies 2017, 10, 1118 15 of 17

As shown in Figure 14c, since the coupling device is used, the outer side output torque of the vehicle isEnergies increased, 2017, 10 and, 1118 the inter side sprocket regenerates braking torque. The power generated by15 the of 17 outer sprockets and regenerated by the inter sprocket are shown in Figure 14d. This time the inner regenerates braking power, while the outer motor output the maximum power. Although the motor regenerates braking power, while the outer motor output the maximum power. Although the theoretical torque and power don’t exceed the maximum value according to Table 3, the torque and theoretical torque and power don’t exceed the maximum value according to Table3, the torque and power of the outer side sprocket in simulation is much larger than theoretically calculated, meaning power of the outer side sprocket in simulation is much larger than theoretically calculated, meaning the steering resistance moment may be larger than we expected. the steering resistance moment may be larger than we expected.

4 1.4 3 1.2 2 1 1

） 0.8 ） 0 m rad/s

（ -1 （ Y 0.6 w -2 Trajectory(w/o) 0.4 -3 Trajectory(w) Yaw rate(w/o) 0.2 -4 Yaw rate(w)

-5 0 -2 0 2 4 6 0 1 2 3 4 5 X（m） t（s） (a) (b)

2000 50 Inner side(w/o) 1500 Outer side(w/o) Inner side(w) 40 1000 Outer side(w) 30 500 ））

0 kW 20 N*m Inner side(w/o) （（ P T -500 Outer side(w/o) 10 Outer side(w) -1000 Inner side(w) 0 -1500

-2000 -10 0 1 2 3 4 5 0 1 2 3 4 5 t（s） t（s） (c) (d)

FigureFigure 14. 14.Simulation Simulation results results of of 2 2B Bsteering: steering: (a ()a Trajectory;) Trajectory; (b ()b Vehicle) Vehicle yaw yaw rate; rate; (c )(c Sprocket) Sprocket torque; torque; (d()d Sprocket) Sprocket power. power.

5.4.5.4. 8 8 B B Steering Steering The steering wheel is manipulated to output R = 8 B, and the accelerator pedal is controlled to The steering wheel is manipulated to output R = 8 B, and the accelerator pedal is controlled to make vehicle accelerate from 0 to 30 km/h to drive steering. The trajectory of the vehicle is shown in make vehicle accelerate from 0 to 30 km/h to drive steering. The trajectory of the vehicle is shown Figure 15a. The understeer characteristic of the vehicle is improved. The yaw rate of the vehicle in Figure 15a. The understeer characteristic of the vehicle is improved. The yaw rate of the vehicle shown in Figure 15b is not as large as that in small-radius steering due to the high speed of the shown in Figure 15b is not as large as that in small-radius steering due to the high speed of the vehicle. vehicle. The power of the outer sprocket is larger thanks to the coupling device, while the power of The power of the outer sprocket is larger thanks to the coupling device, while the power of the vehicle the vehicle without the coupling device is always at maximum value as shown in Figure 15f. without the coupling device is always at maximum value as shown in Figure 15f. Without the coupling Without the coupling device, the vehicle speed decreases rapidly, as shown in Figure 15c. The device, the vehicle speed decreases rapidly, as shown in Figure 15c. The torque and the rotation speed torque and the rotation speed of the sprocket are shown in Figure 15d,e. of the sprocket are shown in Figure 15d,e.

Energies 2017, 10, 1118 16 of 17 Energies 2017, 10, 1118 16 of 17

0.7 12 120 Yaw rate(w/o) Speed(w/o) Yaw rate(w) Speed(w) 0.6 10 100 Trajectory(w/o) Trajectory(w) 0.5 80 8 ））

） 0.4

m 60 6 m/s rad/s （ 0.3 （ Y （ v 40 w 4 0.2 20 0.1 2 0 0 0 -40 -20 0 20 40 60 80 100 0 5 10 15 20 0 5 10 15 20 X（m） t（s） t（s） (a) (b) (c)

0 0 70 Inner side(w/o) Inner side(w/o) Inner side(w/o) Outer side(w/o) -100 Outer side(w/o) 60 Outer side(w/o) -200 Outer side(w) Outer side(w) Outer side(w) -200 Inner side(w) Inner side(w) 50 Inner side(w) -300 ）

） -400 ） 40

-400 kW N*m r/min

（ 30 （

-600 （ P

T -500 n 20 -600 -800 -700 10

-1000 -800 0 0 5 10 15 20 0 5 10 15 20 0 5 10 15 20 t（s） t（s） t（s） (d) (e) (f)

FigureFigure 15. 15.Simulation Simulation resultsresults ofof 88B Bsteering: steering: ((aa)) Trajectory;Trajectory; ((bb)) VehicleVehicle yawyaw rate;rate; (c(c)) VehicleVehicle speed; speed; (d(d)) Sprocket Sprocket torque; torque; (e ()e Sprocket) Sprocket rotation rotation speed speed (f ()f Sprocket) Sprocket power. power.

6. Conclusions 6. Conclusions In this paper, we found that the torque and power required by the propulsion motor are quite In this paper, we found that the torque and power required by the propulsion motor are quite large according to the dynamic analysis of the 2METV. Therefore, a new steering coupling device is large according to the dynamic analysis of the 2METV. Therefore, a new steering coupling device is designed without increasing the peak torque or power of the propulsion motor. The device consists designed without increasing the peak torque or power of the propulsion motor. The device consists of a planetary gear system, two gear pairs, two electromagnetic clutches and a braker. By coupling of a planetary gear system, two gear pairs, two electromagnetic clutches and a braker. By coupling the torque or power of the steering motor and the propulsion motor, the torque or power input to the torque or power of the steering motor and the propulsion motor, the torque or power input to the sprocket is increased to meet the steering performance requirements in different radius. Based the sprocket is increased to meet the steering performance requirements in different radius. Based on on this, a torque-coupled distribution strategy based on speed is proposed. The simulation in this, a torque-coupled distribution strategy based on speed is proposed. The simulation in RecurDyn RecurDyn and Matlab/Simulink shows that the coupling device can improve the output torque or and Matlab/Simulink shows that the coupling device can improve the output torque or power of the power of the sprockets, and using the proposed control strategy can achieve the better trajectory sprockets, and using the proposed control strategy can achieve the better trajectory following. following. Acknowledgments: This work was supported by the National Natural Science Foundation of China for financially supportingAcknowledgments: this project This (51475045). work was supported by the National Natural Science Foundation of China for Authorfinancially Contributions: supporting Lithis Zhai project proposed (51475045). the innovation of the overall system and designed the control methods, HongAuthor Huang Contributions: are responsible Li forZhai the proposed design of coupledthe innovation systems of and th thee overall simulation system of the and system, designed Steven the Kavuma control performed the simulation of control methods. Li Zhai and Hong Huang wrote the manuscript and Steven Kavuma polishedmethods, the Hong manuscript. Huang Allare authorsresponsible read for and the approved design theof coupled manuscript. systems and the simulation of the system, Steven Kavuma performed the simulation of control methods. Li Zhai and Hong Huang wrote the manuscript Conflictsand Steven of Interest:Kavuma Thepolished authors the declare manuscript. no conflict All authors of interest. read and approved the manuscript.

ReferencesConflicts of Interest: The authors declare no conflict of interest.

1.ReferencesZhai, L.; Sun, T.M.; Wang, Q.N.; Wang, J. Lateral stability control of dynamic steering for dual motor drive high speed tracked vehicle. Int. J. Autom. Technol. 2016, 17, 1079–1090. [CrossRef]

2.1. Ketting,Zhai, L.; M.; Sun, Frohner, T.M.; F.;Wang, Wottawah, Q.N.; V.Wang, Drive J. ChainLateral for stability Tracked control Vehicle. of U.S. dynamic Patent steering 6142588 A,for 7 dual November motor 2000.drive 3. LI,high J.; Zhang,speed tracked H.; Meng, vehicle. H.; Tang, Int. J. R. Autom. Research Technol. of tracked 2016, vehicle17, 1079–1090. compliance simulation. Armor. Veh. Engine

2. 2003Ketting,, 92, 4–7.M.; Frohner, F.; Wottawah, V. Drive Chain for Tracked Vehicle. U.S. Patent 6142588 A, 7 4. Kim,November M.S.; Woo, 2000. Y.H. Robust design optimization of the dynamic responses of a tracked vehicle system.

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