Mechatronics 16 (2006) 63–72

Optimal selection of motor and gearhead in mechatronic applications

Fredrik Roos *, Hans Johansson, Jan Wikander

Mechatronics Lab, Department of Machine Design, KTH 100 44, Stockholm, Sweden

Received 24 January 2005; accepted 2 August 2005

Abstract

A method for the selection of motor and gearhead in mechatronic applications is proposed. The method is applicable to any kind ofload and helps to find the optimal motor gearhead combination with respect to output torque, peak power, mass/size and/or cost. Theinput to the method is the load cycle and component data on candidate motors and gearheads. Output is a set of graphs of all motor/gearratio combinations that can drive the given load. From these graphs it is easy to read out the peak power, motor torque and energyefficiency for all feasible motor/gear ratio combinations.� 2005 Elsevier Ltd. All rights reserved.

Keywords: Design methodology; Gears; Motor drives; Servomotors; Servosystems

1. Introduction

The number of electromechanical sub-systems in roadvehicles is increasing fast. Many of the in-vehicle systemsthat traditionally have been powered directly from thecombustion engine via gears, chains or belts are being re-placed with electric systems. This electrification is primarilydriven by the new functionality that can be implementedwith electrically actuated sub-systems, but it is also neces-sary for the transition to electric or hybrid-electric drivetrains. An electrically actuated sub-system can be imple-mented as a mechatronic actuation module, incorporatingall necessary actuator components within one physicalmodule (Fig. 1). This research aims at development of inte-grated design and optimization methods for such mecha-tronic modules. The dimensioning method presented inthis paper focuses on the motor and gearhead within anactuation module, the method is however not delimitedto mechatronic module applications. It is in fact applicableto electric actuation systems in general, but most beneficialwhen applied to computer-controlled, variable speed appli-cations, where a significant part of the motor�s rated power

0957-4158/$ - see front matter � 2005 Elsevier Ltd. All rights reserved.

doi:10.1016/j.mechatronics.2005.08.001

* Corresponding author. Tel.: +46 8 790 96 41; fax: +46 8 20 22 87.E-mail address: fredrikr@md.kth.se (F. Roos).

may be needed to accelerate and decelerate the motor�srotor.

This paper focuses on one of the most important designdecisions in a mechatronic development project: the selec-tion of electric motor and gearhead. This decision is diffi-cult because it affects both the structure of the systemand its dynamic performance. It can also have a significantimpact on the total cost of the system. The method pre-sented here is based on a discrete approach, which aimsat finding the best motor and gearhead combination givena discrete set of components with known data (e.g. frommotor catalogues). However, it is almost always possibleto design application specific components that will be bet-ter [8]. Nevertheless, in low production volume applica-tions, it is usually not an option to design a new motorand gearhead, and therefore necessary to use off-the-shelfcomponents.

Many methods and rules of thumb for the motor/gearratio selection problem are proposed in the literature. Cetin-kunt [3] gives a good overview of the design issues in highperformance servo systems, both from a dynamic andstructural point of view. He especially highlights the con-flict between speed and accuracy.

Many of the classical sizing methods are not applicableto general loads, but require some load characteristics, such

Sensors

EnergySource/Buffer

Controller

Automatic Control

Software Electronics

LoadPower electronicsConverter/Driver

Electromechanicalactuator

Transmission/Linkage

Physical mechatronic module

DiagnosticsCommand from driveror other sub-system

Fig. 1. Block diagram of a mechatronic actuation module.

Nomenclature

gg gear efficiencyh angle [rad]_h angular velocity dh/dt [rad/s]€h angular acceleration d2h/dt2 [rad/s2]s cycle timex angular velocity [rad/s]xm,max max allowed motor velocity [rad/s]xg,max max allowed gear velocity [rad/s]xl,peak load peak angular velocity [rad/s]I current [A]J inertia [kg m2]Kt motor torque constant [N m/A]m mass [kg]n gear ratio

P power [W]R motor winding resistance [X]T torque [N m]Tg,in gearhead torque on motor side [N m]Tg,out gearhead torque on load side (=Tl)Tp motor peak torque rating [N m]Tc motor continuous torque rating [N m]W energy [J]kxk2 root mean square (RMS) normkxk1 infinity norm (maximum)index l loadindex m motorindex g gearhead

64 F. Roos et al. / Mechatronics 16 (2006) 63–72

as purely inertial or constant speed loads (e.g. [4]). In addi-tion, most methods treat the gearhead as ideal. Pash andSeering [5] derive the optimal gear ratio n ¼

ffiffiffiffiffiffiffiffiffiffiffiffiJ l=Jm

pfor

a purely inertial load. This gear ratio maximizes the outputtorque from the actuator, but only in the special case with apurely inertial load and an ideal gearhead. Furthermore,most of the previously published methods are based on amanual iterative approach, making the selection proceduretime consuming and cumbersome, e.g. [12].

Van De Straete et al. [13,14] propose a general methodfor servo drive selection and optimization. Their methodis applicable to all possible loads, and many different typesof electric motors. It is however somewhat complicated andassumes an ideal gearhead, like many other methods.

The method presented in this paper is applicable to allload types and it takes all major limiting phenomena ofthe motor and gearhead into account. The final selectioncriterion is the choice of the user, which is important since

Fig. 2. Electromechanical servo system.

F. Roos et al. / Mechatronics 16 (2006) 63–72 65

different criteria are relevant in different applications. Insome applications the most important aspect may forexample be the weight of the servo drive while in othersit is the cost. This method supports the following selectioncriteria: peak power, output torque, size, weight and energyefficiency. Sometimes other criteria like backlash and accu-racy can be even more important. They are however nottreated here and usually depend more on the componenttype and configuration than the component sizes.

The method assumes that the worst case load cycle isknown (i.e. position and torque as function of time). Inaddition, values on most of the parameters shown inFig. 2 are needed for each of the candidate motors andgearheads. The method then finds all motor/gear ratiocombinations that can drive the specified load.

2. Motor characteristics

Many electric motor types are used in mechatronicapplications. This paper focuses on permanent magnet mo-tors, even though the method also is applicable to othermotor types. Permanent magnet motors can be divided intotwo groups: brushed DC-motors and brushless motors.The phenomena causing the motor limits are the samefor both groups, with the exception of the commutationlimit for brushed motors (Table 1). When checking if a mo-tor can drive a given load, at least three limits have to bechecked: (a) The root-mean square value of the requiredmotor torque has to be lower than the motor�s continuoustorque rating (Tc), (b) the required peak motor torque hasto be lower than the motor�s peak torque rating (Tp) and(c) the required peak motor speed has to be lower thanthe maximum allowed speed of the motor (xm,max).

The torque data in a motor catalogue is only completelyvalid in the neighborhood of the motor�s rated speed. Sincesome of the losses in a motor are speed dependent (e.g.eddy-currents and hysteresis losses in the magnetic circuit)the motor becomes warmer at higher speeds than at lowerwhile producing the same torque. It is therefore possiblefor a motor to produce higher (continuous) torques at

Table 1Permanent magnet motor limits

Continuous torque limit Overheating of winding insulat

Peak torque limit Demagnetization kTmk1 6 Tp

Commutation (DC) kTmxmk1Speed limit Mechanical limit/max supply v

low speeds than at high speeds. This effect is usually not in-cluded in motor data sheets and it is therefore disregardedin this sizing method. It is, however, advisable to check therated speed of the candidate motors. Using a motor with alower rated speed than the application requires may resultin an overheated motor even though the RMS-load is lowerthan the motor�s torque rating.

The motor�s physical speed limit is caused by constraintsin the mechanical structure, but it is often the applicationvoltage that sets the constraint on the maximum motorspeed (motor back EMF = max application voltage). Ifthe motor winding can be adapted to the application it isonly the mechanical speed limit that sets the constrainton the maximum speed. However, information about themechanical speed limit is usually hard to find in motor datasheets. Also data on the commutation limit in DC motorsare rare; therefore will the commutation limit not be in-cluded in the analysis presented in this paper. Hence, thepeak torque limit is assumed to be independent of speed,for all of the above motor types.

3. Gearhead characteristics

The gearhead or gear reducer is often treated as an idealgear ratio when sizing a servo drive. This is obviously asimplified approach, but sometimes it may be goodenough. However, in this method, both gearhead efficiencyand inertia are included in the analysis.

Three gearhead types are frequently used in mechatronicapplications: conventional gear pairs with spur or helicalgears, three-wheel planetary gear trains and harmonicdrives. The choice of gearhead type depends on many fac-tors; perhaps the most important ones are: input speed,backlash, efficiency and cost. The choice of gearhead typeis however not covered by this method and it is assumedthat gearhead data is available.

The phenomena behind the torque limits of a gearheadare far more complex than those for a motor. The limitingfactors of gears are mechanical stresses: tension, compres-sion, bending shear and Hertzian pressure. All major

ion kTmk2 6 Tc where kTmk2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1s

R s0 T

2m dt

qwhere kTmk1 = maxjTmj

oltage kxmk1 6 xm,max where kxmk1 = maxjxmj

66 F. Roos et al. / Mechatronics 16 (2006) 63–72

gearbox components are subjected to cyclic stresses, even ifthe external load is constant. The complexity behind thelimiting factors of a gearhead makes it hard to dimensionand optimize it even for a constant torque and constantspeed load. And it is even harder to optimize the gearheadsize for a dynamic load.

In some gear catalogues a type of root mean cube(RMC) value of the output torque is used as equivalentcontinuous torque. But for infinite life dimensioning(>2 · 106 load cycles/tooth) it seems, according to [1,7],as if the gearhead has to be dimensioned with respect tothe peak load of the load cycle. However this topic is verycomplex and the authors recommend to either follow themanufacturer�s method for equivalent load calculation orto use the peak load as equivalent load. In this paper, thepeak load of the load cycle will be used for gearheaddimensioning, i.e.

T g;eq ¼ kT lk1 ð1Þ

The most important parameter of a gearhead is its torquerating, but given that the candidate gearheads can drivethe load, the most important parameters are: gear ratio n,inertia Jg (reflected to the motor shaft), mass mg and effi-ciency gg. Usually the inertia is small compared to themotor inertia and the weight also tends to be small. How-ever, when the gear ratio is increased, the gearhead tends toget larger while the motor usually decrease in size. Hence,both the weight and inertia of the gearhead can be of sig-nificance. This is obviously dependent of gearhead type.Roos and Spiegelberg [7] compare weight and inertias ofsingle gear pairs with three-wheel planetary gear trains.The required mass and inertia is much lower for three-wheel planetary gears than for the equivalent piniongear-wheel configuration.

The simple gear model used in this method only containsthree parameters: the gear ratio, n, the efficiency gg and thegear inertia Jg. The gear ratio is defined as

n ¼ xm

xl

ð2Þ

and the efficiency

gg ¼T g;out

nT g;in

¼ T l

nT g;in

ð3Þ

A single gear pair can for example have efficiencies up to98%, and a three-wheel planetary gear train up to about97%. To achieve high reduction ratios it is often necessaryto use two or more reduction steps in series. It is for exam-ple hard to design planetary gear trains with a reductionratio above 10, since it requires a very small sun gear orvery large ring gear [11].

The total gearhead efficiency will therefore depend onthe number of gear stages, s:

gtot ¼ gs ð4ÞSince gearhead data sheets normally only contain a singlefigure of the efficiency, this efficiency model assumes cou-

lomb friction only, no viscous friction. In other words,the gear efficiency is assumed to be independent of speed.Moreover, in reality, the efficiency may be a function ofthe torque, as in Harmonic Drives. A low efficiency will re-sult in a warmer gearhead, which may require cooling.According to [6], the thermal factor can normally beignored when selecting a gearhead for intermittent dutyapplications. On the other hand, determining the innertemperature developed by the gears is vital to all continu-ous duty applications.

4. Load and motor torques

The load is here defined as the required output torqueand position, Tl(t) and hl(t) of the gearhead shaft as func-tion of time. Fig. 3 shows Tl(t) and hl(t) of an inertial loadcycle.

The required motor torque to drive the specified load isgiven by

TmðtÞ ¼ ðJm þ J gÞ€hlðtÞnþT lðtÞngg

ð5Þ

which leads to the following equations for the motor tor-que limits:

kTmk2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

s

Z s

0

ðJm þ J gÞ€hlnþT l

ngg

!2

dt

vuut ð6Þ

kTmk1 ¼ max ðJm þ J gÞ€hlnþT l

ngg

�������� ð7Þ

As seen in the expressions above, the necessary motor tor-ques depend on the load as well as the inertia of the motorand gearhead. This means that the torque calculations haveto be made for all candidate motors and gear ratios. Thecalculations can however be simplified in the RMS case(6), by separating the load from the motor parameters asfollows [8]:

kTmk22 ¼ ðJm þ J gÞ2n21

s

Z s

0

€h2

l dt þ1

n2g2g

1

s

Z s

0

T 2l dt

þ 2Jm þ J g

gg

1

s

Z s

0

€hlT l dt

kTmk22 ¼ ðJm þ J gÞ2n2k1 þ1

n2g2g

k2 þ 2Jm þ J g

gg

k3

ð8Þ

where k1, k2 and k3 are three constants representing theload

k1 ¼1

s

Z s

0

€h2

l dt; k2 ¼1

s

Z s

0

T 2l dt; k3 ¼

1

s

Z s

0

€hlT l dt

ð9Þ

For a �constant speed� load with small accelerations, k2alone represents the squared load RMS-torque.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-15

-10

-5

0

5

10

15

time [s]

Load profile

Ang. Velocity [rad/s]

Pos [rad]

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-60

-40

-20

0

20

40

60

time [s]

torq

ue [N

m]

Load Torque

TRMS 37.8528 Nm

TPeak 60 Nm

Fig. 3. Example of an inertial load cycle, torque and angle (position) as function of time.

F. Roos et al. / Mechatronics 16 (2006) 63–72 67

5. Required power and energy

Usually the peak power of the drive system is importantfor the sizing and cost of the inverter (driver electronics).The total motor power consists mainly of two parts: themechanical output power and the power losses in themotor. The losses in the gearhead are here just treated asan increase of the required motor torque by a factor 1/gg.

The heat developed in a motor is generated by themotor�s resistive, magnetic and mechanical losses. Themotor power equals

Pm ¼ P loss þ Pmech ð10Þwhere the motor�s mechanical power Pmech, is given by

Pmech ¼ Tm_hm ¼ ðJm þ J gÞ€hl

_hln2 þT lðtÞgg

_hl ð11Þ

The approach taken in this paper assumes that the mag-netic and friction losses in the motor are small in compa-rison with the resistive losses. Hence, the motor losses,Ploss, are given by

P loss ¼ RmI2 ¼ Rm

T 2m

k2Tð12Þ

The peak input power, for a given motor and gear ratiocombination, is then given by

P peak ¼ kPmk1 ¼ maxðP loss þ PmechÞ ð13ÞFor efficiency evaluation, the total energy consumption bythe different motor/gear ratio combinations over a load

cycle is of interest. Here it is assumed that none of theregenerated energy during braking is stored; all regeneratedelectric energy (negative power) is assumed to be convertedto heat in resistors in the drive circuitry. The total energyrequired to drive the load is given by

W ¼Z s

0

P in dt ð14Þ

where Pin is given by

Pm < 0

Pm P 0

�P in ¼ 0

P in ¼ Pm

ð15Þ

6. The selection procedure

In this section the proposed motor/gearhead selectionand optimization procedure is explained. The method isintended to be implemented in a mathematical numericalsoftware tool; a MatlabTM implementation can be down-loaded from [9]. An overview of the proposed sizing pro-cess is shown in Fig. 4.

6.1. Selection of gearhead type and candidate motors (A)

First a family of gearheads that can drive the given loadhas to be selected. First of all the gearheads need to bemechanically compatible with the candidate motors, otherimportant factors are backlash, available gear ratios, inputspeed, price and so on (see also Section 3). All gearheads in

Select gearhead typeand family that can

drive the load.

Plot results, and selectthe best motor gear

ratio for the application.

Eliminate motor/gearratio combinations thatcannot drive the load.

Tm, rms & Tm, peak .

Derive peak powerand required energy

for the remainingmotor/gear ratios.

Num

eric

al s

oftw

are

C

D

Determine the speedlimit and derive the

maximum gear ratio.

Select CandidateMotors A

B

Fig. 4. Block diagram of the selection procedure.

68 F. Roos et al. / Mechatronics 16 (2006) 63–72

a series usually have approximately the same efficiency.Hence it is possible to use a constant figure of the efficiencygg, for all ratios. The gear inertia is also important tocheck, if it is low in comparison with the candidate motorsit can be neglected. Otherwise, it may be approximated asconstant for all gear ratios.

The choice of candidate motors depends much on theapplication. Generally no larger motors than one that �ex-actly� can drive the load with no gearhead have to be in-cluded. A good idea is to run through the sizing procedureonce with a small number of motors and an ideal gear toget a rough idea of motor sizes and gear ratios of interest.

6.2. Maximum gear ratio (B)

Determine the maximum allowed speed of the motorshaft, xm,max by comparing the speed limits of the motorand gearhead. The motor�s maximum speed limit dependson if the motor�s winding can be changed or not (see Sec-tion 2). If the winding can be adapted (optimized) to theapplication, only the mechanical limit is of interest.

Derive the maximum allowed gear ratio with Eq. (16).Use either one approximation for all of the candidate mo-

tors, or, if the motor is the limiting component and a moreexact analysis is desirable, one for each motor.

nmax ¼minðxm;max;xg;maxÞ

xl;peak

ð16Þ

6.3. Feasible motors and gear ratios (C)

Derive the required RMS-torque as function of gearratio (1 � nmax), for all of the candidate motors (8). Thenfind the range of feasible gear ratios by comparing therequired RMS torque with each motor�s continuous torquerating. Eliminate all data points that not can drive the load,i.e. eliminate all points that not fulfill (17).

T c P kTmk2 ð17Þ

Generally, the peak torque limit also has to be checked.Use Eq. (7) to retrieve the required peak motor torqueand eliminate motor/gear ratio combinations that do notfulfill (18). It is however usually possible to determinebeforehand if the peak torque limit will be a constraintor not, by comparing the RMS and peak torque of the loadalone, e.g. Fig. 3.

T p P kTmk1 ð18Þ

6.4. Visualization of the results (D)

When all motors and gear ratios that can drive the loadhave been derived, it is possible to calculate the peak powerand input energy requirement for those motor/gear ratiocombinations with (13) and (14).

The simplest and perhaps the most comprehensive wayto display the results is to, for each motor, plot the torque,power and energy requirements as functions of gear ratio(Eqs. (8), (13) and (14)).

7. Selection criteria

Depending on application the selection criterion that ismost important may vary. For example, in robotic jointactuators, the weight of the actuator may be most impor-tant, while in other applications the prime considerationis to minimize the required peak power. This method sup-ports the following selection criteria.

7.1. Minimization of motor torque

Minimizing the continuous motor torque is equivalentto maximizing the output torque for a given motor current.Using the motor as torque-efficient as possible allows us toincrease the accelerations in the load cycle and hence in-crease the performance of the system. For a given motorand winding, minimization of the motor torque also mini-mizes the motor current.

3.5

4

4.5

e [N

m]

Motor RMS torque as function of gear ratio, Tl,rms

= 36.4333Nm

PSA 60/4–112PSA 90/6–52PSA 90/6–79PSA 90/6–105

Table 3Gear data, HPG Series [10]

Tg,nom 35–60 N mxg,max 6000 rpmmg 2.3–2.7 kggg >0.85n 5, 11, 21, 33, 45Jg 0.45–0.71 kg cm2

F. Roos et al. / Mechatronics 16 (2006) 63–72 69

7.2. Minimization of weight and size

In many applications, minimization of the actuator�sweight (motor + gearhead) can enhance the overall systemperformance. Optimization of weight and size is the sametype of problem in this discrete sizing method. Usuallylower weight means smaller size and may also lead to lowercosts.

7.3. Peak power/peak current

A large part of the cost of an electric servo system lies inthe driver electronics. Minimizing the required peak powerof the actuator may allow a lower power rating of the in-verter. The supply voltage is usually fixed in an application;therefore minimizing the inverter power rating is equal tominimizing the peak motor (inverter) current. If the win-ding cannot be adapted to the application, this is, for agiven motor, the same as minimizing the motor torque. Itis however not certain that the motor/gearhead combina-tion that requires the lowest torque also requires the lowestcurrent when different motors are compared, since thetorque/current relationship is dependent on the winding.

It gets more complicated if the motor is wounded for theapplication. To minimize the motor current the motorshould be wounded so that the peak motor velocity occursat the supply voltage (this assumes that field weakening notis used). The gear ratio that minimizes the peak power (13)will then also minimize the peak motor (inverter) current,but only under the condition that peak power occurs atthe same instant of the load cycle regardless of gear ratio.If possible, it is nice to avoid modeling the current, sinceit depends on the winding. However, it may exist load cy-cles where the peak power occurs at a different time instantof the load cycle depending on gear ratio. In those cases themotor current must be analyzed instead of just the power.This is however not investigated further in this paper.

7.4. Energy efficiency

A measure of efficiency is to compare the energy con-sumption for the different motors over the load cycle(14). Here it is assumed that none of the regenerated energyis stored, but rather converted to heat in resistors.

Table 2Motor data API-Elmo PSA series ACPM motors [2]

PSA Tc

[N m]Tp

[N m]xm,max

[rpm]mm

[kg]Jm[kg cm2]

R

[X]Kt

[N m/A]

60/4-50 0.76 3.8 8000 3 0.79 15 0.5260/4-75 1.20 6 8000 3.6 0.98 9.0 0.6260/4-112 1.80 9 8000 4.4 1.28 5.9 0.6390/6-52 2.50 12 8000 5.9 2.86 3.9 0.6890/6-79 3.60 18 8000 7.1 3.8 1.45 0.5890/6-105 4.50 22.5 8000 8.2 4.7 1.10 0.61

8. Examples

For simplicity, all motors used in these examples aresynchronous AC motors from Danaher Motion, API-Elmoseries [2]. Essential parameters of the candidate motors areshown in Table 2.

8.1. Inertial load

The goal with this example is to find the best motor/gearhead combination for the inertial load shown inFig. 3. A low backlash harmonic gearhead from HD-systems HPG Series [10] is required (Table 3).

As seen in Tables 2 and 3, the gear inertia in this case isof the same magnitude as the inertia of the smallest candi-date motor. The gear inertia is, for simplicity, set to0.5 kg cm2 for all ratios. Also the efficiency is set to a con-stant, 85%, even though it is in reality a function of outputtorque [10]. The required RMS-torque for each motor isshown in Fig. 5, only motor/gear ratio combinationsthat can drive the load are shown. As seen in the figurethe maximal gear ratio is approximately 45 (max gearinput speed = 6000 rpm). The stars indicate the results ofapplying the classical method of inertial matchingn ¼

ffiffiffiffiffiffiffiffiffiffiffiffiJ l=Jm

pwhile the boxes indicate the minimum motor

torque requirement.

0 5 10 15 20 25 30 35 40 45 501.5

2

2.5

3

Gear ratio

Tor

qu

Fig. 5. Required motor RMS torque. The vertical lines indicate allowedgear ratios (Table 3). The stars indicate the results of the method of inertiabalance, the boxes the minimum of the torque curves.

0 5 10 15 20 25 30 35 40 45 501000

1500

2000

2500

3000

3500

4000

Gear ratio

Pea

k P

ower

[W]

Motor peak power (Pmech + Ploss) as function of gear ratio, Pl,max = 866.25W

PSA 60/4–112PSA 90/6–52PSA 90/6–79PSA 90/6–105

Fig. 6. Required motor peak power, the vertical lines indicate allowedgear ratios.

Table 4Results of the different optimization criteria

Optimization criteria Motor/gear ratio

Min. peak power PSA 90/6-105, n = 11Min. current (fixed winding) PSA 60/4-112, n = 45Max. efficiency PSA 90/6-105, n = 11Weight/size PSA 60/4-112, n = 45Max. output torque PSA 90/6-105, n = 33

70 F. Roos et al. / Mechatronics 16 (2006) 63–72

As seen, the differences between the gear ratios indicatedby the boxes and stars are small, which implies that theclassical method is quite robust to variations in gear inertiaand gear efficiency (non-ideal gears). However, if an idealgear ratio model is used, the required torques will be re-duced by approximately 0.5 N m. This may result in theselection of a motor that cannot drive the load.

The required peak power (13) for the different combina-tions is shown in Fig. 6, and the required energy (14), inFig. 7.

The largest motor (PSA 90/6-105) and a gear ratio of 11minimizes the peak power (compare Figs. 6 and 7). If thewinding is fixed, the peak current rating of the inverter is

0 5 10 15 20 25 30 35 40 45 50100

200

300

400

500

600

700

800

Gear ratio

Ene

rgy

[J]

Consumed energy Wl = 190.575J

PSA 60/4–112PSA 90/6–52PSA 90/6–79PSA 90/6–105

Fig. 7. Required energy to drive the load (14), the horizontal linerepresents the necessary energy to drive the load alone (motor and gearinertia = 0, gear efficiency = 1). The vertical lines indicate allowed gearratios.

minimized when the current is minimized, which meansPSA 60/4-112 and gear ratio 45. But the most energy effi-cient motor/gear ratio combination is still the PSA 90/6-105 with a gear ratio of 11 (Fig. 7).

For performance optimization, i.e. the maximum possi-ble increase of output torque or acceleration, the motorgear ratio combination with the largest difference betweenrequired torque and rated torque should be selected. In thiscase, it is the PSA 90/6-105 at a gear ratio of 33, which hasthe highest additional torque capability (1.4 N m).

To minimize the weight of the servo drive, the smallestmotor shall be selected. The total weight is then4.4 + 2.7 = 7.1 kg. This can be compared with a directdrive configuration (no gear), which requires a motor farabove 20 kg (not included in Table 2). Table 4 shows theresulting motor/gearhead combination for each of the opti-mization criteria.

8.2. Combined load

The load used in this example is a combination of aninertial and friction load (Fig. 8). Since this load cannotbe classified as either an inertial or constant speed load,it is difficult to use any of the classical sizing methods.

The planetary gearhead series used in this example isfrom ZF Machineantriebe GmbH, PGE 25/1 [15]. Dataon these gears can be found in Table 5.

As seen in Table 2 and Fig. 8 none of the candidate mo-tors can drive the load directly all have lower continuoustorque rating than the load RMS torque. Applying the pro-cedure from previous section to this load and motors re-sults in the following plots.

As seen in Fig. 9, the maximum allowed gear ratio is 9.All candidate motors can drive the load even though thesmallest one, PSA60/4-50 needs a gear ratio above 8 todo it (indicated by the arrow in Fig. 9). So if the weightis to be minimized the PSA60/4 should be used. Total mass(motor + gearhead) is then 3 + 1.3 = 4.3 kg. In compari-son, the smallest motor from the same series that can drivethe load directly (gear ratio = 1), has a weight of 9.3 kg(this motor is not included in Table 2 or in the graphs).

The motor peak power is plotted in Fig. 10. As seen inthis figure the minimization of peak power is not so intui-tive. The second largest motor (PSA 60/4-112) has the low-est peak power optimum (1025 W at n = 5). The differencebetween the three smallest motors is however not large,approximately 100 W.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5–100

–50

0

50

100

time [s]

Load profile

Ang. Velocity [rad/s]Pos [rad]

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5–15

–10

–5

0

5

10

15

time [s]

torq

ue [N

m]

Load TorqueT

RMS 5.7079 Nm

TPeak

11.7 Nm

Fig. 8. Combined friction and inertial load.

Table 5Gear data, PGE 25/1 [15]

Tg,nom 12 N mxg,max 6000 rpmmg 1.3 kggg 0.97n 3, 4, 5, 7, 9Jg 0.059 kg cm2 (n = 10)–0.128 kg cm2 (n = 3)

1 2 3 4 5 6 7 8 9

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

Gear ratio

Tor

que

[Nm

]

Motor RMS torque as function of gear ratio, Tl,rms

= 5.7079Nm

PSA 60/4–50PSA 60/4–75PSA 60/4–112PSA 90/6–52

Fig. 9. Motor RMS torque as function of gear ratio, the vertical linesindicate allowed gear ratios.

1 2 3 4 5 6 7 8 91000

1050

1100

1150

1200

1250

1300

1350

1400

1450

1500

Gear ratio

Pea

k P

ower

[W]

Motor peak power (Pmech + Ploss ) as function of gear ratio, Pl,max

= 819W

PSA 60/4–50PSA 60/4–75PSA 60/4–112PSA 90/6–52

Fig. 10. Motor peak power requirements.

F. Roos et al. / Mechatronics 16 (2006) 63–72 71

From an energy efficiency perspective (Fig. 11), the mo-tors are similar; even though the high rotor accelerations athigh gear ratios mean lower efficiencies for the larger mo-tors. The ellipse on the curve representing PSA 90/6-52 inFig. 11 indicates the gear ratio where the deceleration partsof the load cycle changes from requiring positive motorpower to negative power (regenerative drive), hence the dis-continuity in the graph (Eq. (15)). In other words, for gearratios above 5.8, the large kinetic energy stored in the rotor

1 2 3 4 5 6 7 8 9850

900

950

1000

1050

1100

1150

Gear ratio

Ene

rgy

[J]

Consumed energy Wl = 851.7732J

PSA 60/4–50PSA 60/4–75PSA 60/4–112PSA 90/6–52

1 2 3 4 5 6 7 8 9850

900

950

1000

1050

1100

1150

Consumed energy Wl = 851.7732J

Fig. 11. Required energy to drive the load, the vertical lines indicateallowed gear ratios.

72 F. Roos et al. / Mechatronics 16 (2006) 63–72

requires the drive system to produce negative torques in or-der to follow the specified load cycle.

9. Conclusions

The method presented in this paper can find the bestmotor/gear ratio combination for any given load withrespect to weight, size, peak power, torque and efficiency.As input to the method, data of the candidate motorsand gearheads, as well as a load cycle are required.

Including the gearhead inertia and efficiency in the sizingmethod can make a large difference on the motor/gear ratioselection. Applying a classical sizing method with an idealgear model when large gear ratios are required can result ina motor/gear ratio selection that cannot drive the load.

If the winding can be adapted to the application or not,may have a big impact on the choice of motor and gearratio. If it is, the peak current rating of the inverter maytruly be minimized, since the application voltage can beused more efficiently.

Acknowledgements

This research is part of the Green Vehicle/Fuel Celland Hybrid Electric Vehicles (FCHEV) program, jointlyfunded by the Swedish government, Scania CV AB, VolvoAB, Volvo Cars and Saab Automobile.

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