motion reliability analysis of the delta parallel robot

Research ArticleMotion Reliability Analysis of the Delta Parallel Robotconsidering Mechanism Errors

Yu Li 1 Deyong Shang 1 Xun Fan1 and Yue Liu2

1College of Mechanical and Information Engineering China University of Mining and Technology (Beijing) Beijing 100083 China2Department of Mechanical Engineering Tsinghua University Beijing 100084 China

Correspondence should be addressed to Deyong Shang shangdeyongcumtbeducn

Received 8 October 2018 Revised 13 March 2019 Accepted 23 April 2019 Published 2 May 2019

Academic Editor Xuping Zhang

Copyright copy 2019 Yu Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Delta parallel robots are widely used in assembly detection packaging sorting precision positioning and other fields With thewidespread use of robots people have increasing requirements for motion accuracy and reliability This paper considers theinfluence of various mechanism errors on the motion accuracy and analyzes the motion reliability of the mechanism Firstly weestablish a kinematic model of the robot and obtain the relationship between the position of the end effector and the structuralparameters based on the improved DndashH transform rule Secondly an error model considering the dimension error the errorof revolute joint clearance driving error and the error of spherical joint clearance is established Finally taking an actual robotas an example the comprehensive influence of mechanism errors on motion accuracy and reliability in different directions isquantitatively analyzed It is shown that the driving error is a key factor determining the motion accuracy and reliability Theinfluence of mechanism errors on motion reliability is different in different directions The influence of mechanism errors onreliability is small in the vertical direction while it is great in the horizontal direction Therefore we should strictly control themechanism errors especially the driving angle to ensure the motion accuracy and reliability This research has significance forerror compensation motion reliability analysis and reliability prediction in robots and the conclusions can be extended to similarmechanisms

1 Introduction

Compared with series robots parallel robots have the advan-tages of large carrying capacity good motion performancefast moving speed etc [1ndash4] Delta parallel robots [5] arewidely used in assembly detection packaging aerospacemedical devices precision positioning and other fieldsPeoplersquos demand for accuracy in parallel robots is alsoincreasing The motion accuracy of parallel robots is a keyproblem in their design and manufacturing [6 7] North-eastern University professor Sun Zhili [8] and his teammade a thorough research on error modeling of parallelrobot and analyzed the motion accuracy Popponen andArai [9] proposed an accuracy analysis method for linearparallel mechanisms based on position and clearance errorsof hinges In 2009 Chebbi et al [10] studied the 3-UPUparallel robot and constructed a model of the mechanismMohan [11] presented the end effector pose error modeling

and motion accuracy analysis of a planar 2PRPndashPPRparallel manipulator with an unsymmetrical (U-shaped)fixed base The error model was established based on thescrew theory with considerations of configuration (geo-metrical) errors Based on a novel six-degree-of-freedom(DOF) parallel manipulator mechanism Fu [12] studiedthe forward and inverse kinematics The paper addressed thekinematic accuracy problem of the proposed new parallelrobot with three legs due to the location of the U joint errorsclearance and driving errors Chouaibirsquos paper showed thatthe orientation error variation depends essentially on theparallelogram configuration of the passive legs out of its plane[13]

Aiming at the error analysis and motion reliabilityanalysis of the Delta parallel robot currently most studiesmainly consider the influence of dimension error of thecomponents Some studies consider the clearance errorsbut the main consideration is the error of revolute joint

HindawiMathematical Problems in EngineeringVolume 2019 Article ID 3501921 10 pageshttpsdoiorg10115520193501921

2 Mathematical Problems in Engineering

Table 1 Homogeneous transformation matrix between local coordinate systems

Local coordinate system Variable 120579119894 Homogeneous transformation matrix(119909119910119911)1 1205791198941 OT1 = R(119910 120572119894)M(119909 1199031)R(119911 minus1205791198941)(119909119910119911)2 1205791198942 1T2 = R(119909 119886)M(119911 1205791198941)R(119911 minus1205791198942)(119909119910119911)3 1205791198943 2T3 = R(119910 1205791198943)(119909119910119911)4 119873119900119899119890 3T4 = M(119909 119888)R(119910 minus1205791198943)R(119911 1205791198942)(119909119910119911)5 1205794 4T5 = M(119909 minus1199032)R(119910 minus120572119894)R(119910 1205794)(119909119910119911)6 1205795 5T6 = M(119910 minus119889)R(119909 1205795)

clearance in rotating structures Because parallelmechanismsare used in high-speed situations the error of sphericaljoint clearance will reduce the kinematic accuracy andaggravate wear and affect the motion stability of mecha-nisms In addition the existing research mainly considersthe individual influence of each error on the kinematicaccuracy of the mechanism The influence on the motionerror and reliability of the mechanism in different directionsis not considered under the comprehensive effect of variouserrors

According to the above situation the kinematic modelof the robot is established by the DndashH matrix transfor-mation method An error model considering dimensionerror the error of revolute joint clearance driving errorand the error of spherical joint clearance is establishedThe influence of each error source on the motion erroris analyzed In addition based on the above analysis thecomprehensive influence of mechanism errors on motionaccuracy and reliability in different directions is analyzedA reliability analysis provides an important theoretical basisfor improving the kinematic performance of the mechanismand is of great significance to robot mechanism optimization[14]

2 The Coordinate System of the DeltaParallel Mechanism

The structural model of the Delta parallel robot is shown inFigure 1 and the kinematic chain of the Delta parallel robotis shown in Figure 2

The global coordinate system O minus XYZ is located at thecenter point of the fixed platform at point OThe 119909 axis pointsto theOA1 the119910 axis is perpendicular to the platform surfaceand the negative direction of the 119910 axis is the gravity directionof the moving platform

The position of the rotating pair on the fixed platform andthe moving platform is arranged in a triangle of which thecircumradius is respectively 1199031 and 1199032

The length of the driving arm AiBi is 119886 and the lengthof the parallelogram mechanism (the driven arm BiCi) is 119888The length of PN of the mechanism is 119889 (PN is the distancebetween the center point P of the moving platform and theorigin N of the local coordinate system) 120572119894 is the anglebetween the OAi and the global coordinate 119909 axis which isshown in the diagram When the corresponding kinematic

Figure 1 The structural model of the robot

chain 119894 is 1 2 3 120572119894 is 0o 120o and 240o respectivelyThe inputangle (driving angle) is 1205791198941 1205794 1205795 and the rotation angle ofdriven arm is 1205791198942 The swing angle of driven arm is 1205791198943 PointN is the origin of (119909119910119911)6 in the local coordinate system PointT is the working position

3 Modeling of Mechanism Position Error

According to the structural characteristics of the delta par-allel robot it can be seen that the sources of the positionerror of the mechanism are mainly in the following cate-gories

(1) Dimension error(2) The error of revolute joint clearance(3) Driving error(4) The error of spherical joint clearance

31 Considering Dimension Error According to the homo-geneous transformation rule we can get the homogeneoustransformation matrix between local coordinate systems asshown in Table 1

Therefore the homogeneous transformation matrix OT6of the coordinate system (119909119910119911)6 relative to the global coor-dinate system (XYZ)O can be expressed in two ways asfollows

Mathematical Problems in Engineering 3

O

Y

ZX

P

N

a

c

d

T

Ｒ1

Ｒ2

Ｒ3

Ｒ4

Ｒ5

Ｒ6

Ｓ1

Ｓ2

Ｓ4Ｓ5

Ｓ6

Ｔ3

Ｔ5

Ｔ6

i

r1

r2

Ｃ

4

5

i3

Ｃ

i2

i1Ｃ

Ｃ

Ｃ

Figure 2 Motion diagram of kinematic chain

OT6 = 6prod119896=1

119896minus1T119896

= [[[[[[

cos 1205794 sin 1205795 sin 1205794 cos 1205795 sin 1205794 119886 cos120572119894 cos 1205791198941 + 119888 (cos120572119894 cos 1205791198942 cos 1205791198943 minus sin120572119894 sin 1205791198943) + (1199031 minus 1199032) cos1205721198940 cos 1205795 minus sin 1205795 minus119889 minus 119888 sin 1205791198942 cos 1205791198943 minus 119886 sin 1205791198941minus sin 1205794 sin 1205795 cos 1205794 cos 1205795 cos 1205794 minus119886 sin120572119894 cos 1205791198941 minus 119888 (cos120572119894 sin 1205791198943 + sin120572119894 cos 1205791198942 cos 1205791198943) + (1199032 minus 1199031) sin1205721198940 0 0 1]]]]]]

(1)

OT6 = [[[[[[

cos 120579119910 sin 120579119910 cos 120579119909 sin 120579119910 cos 120579119909 1199090 cos 120579119909 minus sin 120579119909 119910minus sin 120579119910 sin 120579119909 cos 120579119910 cos 120579119909 cos 120579119910 1199110 0 0 1]]]]]]

(2)

where 120579119909and 120579119910 are the angles of the rotating mechanismaround the 119909 axis and the 119910 axis of the global coordinatesystem respectively It is easy to get 120579119909 = 1205795 120579119910 = 1205794

Let (1) and (2) be transformed as followsR(119910 120572119894)minus1OT6119877(119909 1205795)minus1M(119910 minus119889)minus1R(119910 1205794)minus1R(119910minus120572119894)minus1 we can get

T = [[[[[[

1 0 0 119909 cos120572119894 minus 119911 sin1205721198940 1 0 119889 + 1199100 0 1 119909 sin120572119894 + 119911 cos1205721198940 0 0 1]]]]]]

(3)

T = [[[[[[

1 0 0 1199031 minus 1199032 + 119888 cos 1205791198942 cos 1205791198943 + 119886 cos 12057911989410 1 0 minus119888 sin 1205791198942 cos 1205791198943 minus 119886 sin 12057911989410 0 1 minus119888 sin 12057911989430 0 0 1]]]]]]

(4)

From (3) and (4) we can obtain

119909 = minus119888 sin 1205791198943 sin120572119894 + 1199031 cos120572119894 minus 1199032 cos120572119894+ 119886 cos 1205791198941 cos120572119894 + 119888 cos 1205791198942 cos 1205791198943 cos120572119894119910 = minus119886 sin 1205791198941 minus 119888 sin 1205791198942 cos 1205791198943 minus 119889


The sleeve hole

The shaft

The error circle

２Ｃ

２1

２2

Figure 3 Model of revolute joint

119911 = minus119888 sin 1205791198943 cos120572119894 minus 1199031 sin120572119894 + 1199032 sin120572119894minus 119886 cos 1205791198941 sin120572119894 minus 119888 cos 1205791198942cos 1205791198943 sin120572119894(5)

Equation (5) is the relationship between the end outputposition and the structural parameters in the ideal case Infact the parameters of each structure have certain deviationsdue to the influence of various factors

The Taylor series expansion is carried out on the upperform The expression is carried out as follows

Δ119909o = 3sum119894=1

(cos120572119894Δ1199031198941 minus cos120572119894Δ1199031198942 + cos 1205791198941cos120572119894Δ119886119894+ (cos 1205791198942 cos 1205791198943 cos120572119894 minus sin 1205791198943 sin120572119894) Δ119888119894+ (minus1199031198941 sin120572119894 + 1199031198942 sin120572119894 minus 119886119894 cos 1205791198941 sin120572119894minus 119888119894 cos 1205791198942 cos 1205791198943 sin120572119894 minus 119888119894 sin 1205791198943 cos120572119894) Δ120572119894)

Δ119910o = minus 3sum119894=1

(sin 1205791198941Δ119886119894 + sin 1205791198942cos 1205791198943Δ119888119894 + Δ119889)Δ119911o = 3sum

119894=1

(minussin120572119894Δ1199031198941 + sin120572119894Δ1199031198942 minus cos 1205791198941sin120572119894Δ119886119894minus (cos 1205791198942 cos 1205791198943 sin120572119894 + sin 1205791198943cos120572119894) Δ119888119894+ (minus1199031198941 cos120572119894 + 1199031198942 cos120572119894 minus 119886119894 cos 1205791198941 cos120572119894minus 119888119894 cos 1205791198942 cos 1205791198943 cos120572119894 + 119888119894 sin 1205791198943 sin120572119894) Δ120572119894)

(6)

The relation between the input error and the output errorof the mechanism is obtained from (6)

According to the above equation the error sources areΔ1199031198941 Δ1199031198942 Δ119886119894 Δ119888119894 Δ120572119894 etc Moreover there are also errors inrevolute joints driving errors and errors in spherical joints

32 Considering the Error of Revolute Joint Clearance Theexistence of clearances in these joints is inevitable due tomachining tolerances wear and material deformation The

revolute joint of the Delta robot is composed of a sleevehole and a shaft The error of the revolute joint is thedifference between the radii of the two components that isthe clearance error is the radial error of the revolute jointTherefore the planar model is used to describe the clearanceof the revolute joint as shown in Figure 3

The error circle is the circle with a radius of Rei From themodel of revolute joint we can get

Rei = R1 minus R2 (7)where R1and R2 are radii of the sleeve hole and the shaft

This paper only considers the clearance error between therevolute joint of the platform and the driving arm Accordingto the effective length theory [15] the clearance error Rei ofthe revolute joint is equivalent to the change in length ofthe driving arm Δ119886 Namely Rei = Δ119886 Therefore the errorcaused by the revolute joint is as follows

Δ119909R = 3sum119894=1

cos 1205791198941 cos120572119894Rei

Δ119910R = minus 3sum119894=1

sin 1205791198941Rei

Δ119911R = minus 3sum119894=1

cos 1205791198941 sin120572119894Rei(8)

33 Considering Driving Error The input angle 1205791198941 is thedriving angle of the mechanism at any moment The first-order Taylor series expansion of (5) is at 1205791198941 Therefore theerror caused by driving error is as follows

Δ119909D = 3sum119894=1

minus 119886 sin 1205791198941 cos120572119894Δ1205791198941Δ119910D = 3sum

119894=1

minus 119886 cos 1205791198941Δ1205791198941Δ119911D = 3sum

119894=1

119886 sin 1205791198941 sin120572119894Δ1205791198941(9)


O

The spherical shell

The sphere

O

x

z

y

Figure 4 Model of spherical joint

34 Considering the Error of Spherical Joint Clearance Thestructure of spherical joint is composed of spherical shelland sphere The structure of the spherical joint is shownin Figure 4 When the robot works the sphere moves inthe spherical shell and the motion model at this time is aspatial model Ideally the spherical center of the sphere islocated at point O Because the clearance is inevitable thespherical center is actually located at point O1015840 In two casesthe distance between the center of the sphere and the originis the clearance of the spherical joint

The error of the spherical joint affects the output errorof the Delta parallel robot The spherical joint constrains thetranslational freedom in three directions while it does notlimit the rotational degree of freedom Therefore the errorcaused by the clearance between the spherical joint is onlythree translational errors [16] Namely the errors in the 119909 119910and 119911 directions are Δ119909 Δ119910 and Δ119911 respectively

When the clearance error of the spherical joint is 120588 theangle between the projection in the 119909119900119910 plane of 120588 and the 119909axis is 120572 and the angle between the clearance direction andthe 119911 axis is 120573The errors of the spherical joint in the 119909 119910 and119911 directions are as followsΔ119909Q = 120588 sin120573 cos120572

Δ119910Q = 120588 sin120573 sin120572Δ119911Q = 120588 cos120573

(10)

According to the structural diagram of the Delta parallelmechanism it is known that the clearance 120588 of spherical jointis equivalent to the change in length of the parallelogrammechanism (driven arm) rod Δ119888 Namely 120588 = Δ119888 Thereforethe position error of the end effector caused by spherical jointis equivalent to the output error considering Δ119888 So it can beobtained

Δ119909Q = 3sum119894=1

(cos 1205791198942 cos 1205791198943 cos120572119894 minus sin 1205791198943 sin120572119894) 120588Δ119910Q = minus 3sum

119894=1

sin 1205791198942 cos 1205791198943120588

Δ119911Q = minus 3sum119894=1

(cos 1205791198942 cos 1205791198943 sin120572119894 + sin 1205791198943 cos120572119894) 120588(11)

From the above analysis the total position error of theend effector in the 119909 119910 and 119911 directions is the superpositionof the four kinds of errors above at any time So it can beobtained

Δ119902 = Δ119902O + Δ119902R + Δ119902D + Δ119902Q = 119899sum119894=1

119870119902119904119894Δ119904119894119902 = 119909 119910 119911 (12)

In (12) 119870119902119904119894 is the error transfer coefficient of Δ119904119894The value of119870119902119904119894 is usually determined by the posture and

structural parameters of the mechanism together The valueof 119870119902119904119894 can quantitatively reflect the degree of the influenceof error Δ119904119894 on the motion accuracy of the robot

4 Motion Reliability Analysis ofthe Mechanism

Due to the existence of random factors such as tolerance anddriving the value of the error in the error model is not a fixedsingle value but is distributed randomly Motion reliabilityanalysis is to calculate the probability that the output errorof the end effector falls within the allowable range Namelythat is to obtain the motion reliability of the mechanism

The size distribution in the mechanical manufacturingindustry generally conforms to the normal distribution [17]By referring to the relevant literature [18] it is normalized andverified that the joint clearance obeys the normal distribu-tion

The error of the rods the error of revolute joint clearanceand the spherical joint clearance the angle error etc belongto the dimension errors Therefore each error source of themechanism can be seen as a normal distribution It is easy toknow that the position error of the end effector also obeys thenormal distribution in the 119909 119910 and 119911 directions at any time


The expressions of mean value and standard deviation are asfollows

120583119904 = 119899sum119894=1

119870119902119904119894120583119904119894 (13)

120590119904 = radic119863119904 = radic 119899sum119894=1

11987011990221199041198941205902119904119894 (14)

where 120583119904119894 and 120590119904119894 are the mean value and standard deviationof each original input error

By statistical knowledge and by analogy with the theoryof stress intensity distribution interference and reliabilityengineering the allowable error distribution is similar tothe intensity distribution and the actual error distribution issimilar to the stress distributionThe actual error distributionobeys the normal distribution 119873 sim (120583119904 1205901199042)

In most cases the random errors affecting the motionaccuracy of the mechanism are independent or weaklycorrelated and they generally have a normal distribution

When calculating the reliability of a mechanismrsquos motionaccuracy there aremany randomerror components that needto be considered Their mean and standard deviations are1205831 1205832 120583119899 and1205901 1205902 120590119899 respectively According to theoperation rules of normal distribution and variance themeanand standard deviation of the synthesized output error are

120583119904 = 1205831 + 1205832 + sdot sdot sdot + 120583119899 (15)

120590119904 = radic12059012 + 12059022 + sdot sdot sdot + 1205901198992 (16)

When the allowable error is 120576 120576 isin [1205761 1205762] then thereliability is

119877 = 119875 (1205761 lt Δ119902 lt 1205762) (17)

119877 = 119875 (1205761 lt Δ119902 lt 1205762)= 119875 (Δ119902 lt 1205762) minus 119875 (Δ119902 lt 1205762) = Φ(1205762 minus 120583120590 )

minus Φ(1205761 minus 120583120590 )(18)

Because the output error of the mechanism is a series ofrandom values that conform to normal distribution based onthe stress-strength interference theory and reliability theorythe allowable range about the mechanism does not needto be determined The allowable range is not a fixed valuebut variable For the sake of uniformity and convenience ofcalculation the allowable error can be regarded as havingnormal distribution

The mean value of the allowable error is 120583120576 and thestandard deviation is 120590120576 then the allowable error 120576 is subjectto the normal distribution 119873 sim (120583120576 1205901205762)

When the allowable error 120576 obeys the normal distributionlet the function be 119866(119885)119866(119885) = 120576 minus Δ119902 gt 0 where 120576 is the maximum value ofallowable error

The above equation indicates that the output error issmaller than the allowable error Since both the output errorΔ119902 and the allowable error 120576 obey the normal distribution itis easy to know that their probability density functions are asfollows

1198911 (1199091) = Δ119902 = 1radic2120587120590119904 exp[minus12 (119909 minus 120583119904120590119904 )2] (19)

1198912 (1199092) = 120576 = 1radic2120587120590120576 exp[minus12 (119910 minus 120583120576120590120576 )2] (20)

In the above equation120583119904120590119904120583120576 and120590120576 are themean valueand standard deviation of Δ119902 and 120576 respectively Accordingto the stress-strength interference model the reliability canbe expressed as follows

119877 = 119875 (120576 gt Δ119902)= intinfinminusinfin

1198912 (1199092) [int1199092minusinfin

1198911 (1199091) d1199091] d1199092= intinfin0

119891 (119911) d119911= intinfin0

1radic2120587120590119911 exp[minus12 (119911 minus 120583119911120590119911 )2] dz(21)

Turning it into a standard normal distribution let 120583 =(119885 minus 120583119911)120590119911Then the following equation can be obtained

119877 = 119875 (120576 gt Δ119902) = intinfin0

119891 (119911) d119911 == intinfinminus120573

1radic2120587exp [minus121199062] d119906 = Φ (120573) (22)

where 119891(119911) = (1radic2120587120590119911)exp[minus(12)((119911 minus 120583119911)120590119911)2] 120573 =120583119911120590119911 = (120583120576 minus 120583119904)radic1205901205762 + 1205901199042 119885 = 120590 minus Δ119902The reliability coefficient can be obtained from the defini-

tion of reliability and the coupling equation of the reliabilitycalculation

119911119877 = 120583120576 minus 120583119904radic1205901205762 + 1205901199042 (23)

Then themotion reliability of parallelmechanism in the119909119910 and 119911 directions is obtained

119877119902 = Φ (119911119877) 119902 = 119909 119910 119911 (24)

where Φ(∙)is the standard normal distribution function

5 Numerical Example

This paper takes the robot of ChenXing (Tianjin) automationequipment company in our laboratory as an example Themodel of this robot is D3PMndash1000 It is mainly composed ofa fixed platform moving platform driving arm and drivenarm The robotrsquos workspace covers a diameter of 500ndash1400


Table 2 Structural parameters of the robot

Structural parameters Numerical valueRadius of the fixed platform (1199031mm) 125Radius of the moving platform (1199032mm) 50The length of the driving arm (119886mm) 400The length of the driven arm (119888mm) 950

Figure 5 The diagram of D3PMndash1000 robot

mm and its grasping speed is 75ndash150 times per minute Itsmaximum acceleration and maximum velocity are 100 ms2and 8 ms respectively It has the characteristics of movingin the 119909 119910 and 119911 directions and is widely used in the fieldof high-speed sorting and packaging The robot is shown inFigure 5

Through the product instruction manual and actualmeasurement the structural parameters of this robot areshown in Table 2

The distribution angles of the driving arm are 1205721 = 0o1205722 = 120o and 1205723 = 240o respectively The trajectory ofthe center point of the mechanism end actuator is as follows119909 = 30 sin(120o119905) (mm) 119910 = minus500 minus 20119905 (mm) and 119911 =minus30 cos (120o) + 30 (mm) and the exercise time is 3 s(1) The expression of each error source shows thatthe error transfer coefficient of the seven error sources(Δ1199031198941 Δ1199031198942 Δ119886119894 Δ119888119894 Δ120572119894 119877119890119894 120588) is small and the errors arecompensatory on the three kinematic chains they are not themain factors affecting themotion accuracy of themechanismThe error transfer coefficient of the driving angle Δ1205791198941 isobviously greater than the other error transfer coefficientswhich has a greater impact on the position error of themechanism so it should be strictly controlled

The error transfer coefficient of the driving angle Δ1205791198941 isshown in Figures 6ndash8

Similarly the error transfer coefficient graphs of othererror sources can be obtained but are omitted here(2) When 119905=1s the motion reliability analysis of the endeffector of the mechanism is made (the permissible precisionof the output error is 120576 120576 sim 119873(1 012))

Driving angle 1Driving angle 2Driving angle 3

minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)

Figure 6 The error transfer coefficient of the driving angle Δ1205791198941 inthe 119909 direction


minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)


The original errors are independent and consistent withnormal distribution From the real discrete distributiondata of each original error the distribution law of eachoriginal error can be determined according to Monte Carlo


Table 3 The mean value and standard deviation of the input errors

Original input errors (119894=123) Mean value Standard deviationDimension error of the fixed platformΔ1199031198941mm 0 001

Dimension error of the moving platformΔ1199031198942mm 0 001

Length error of the driving arm Δ119886119894mm 0 001Length error of the driven arm Δ119888119894mm 0 001Installation error Δ120572119894rad 0 001Driving error Δ1205791198941rad 001 001The error of revolute joint clearance119877119890119894mm 001 001

The error of spherical joint clearance120588mm 001 001


minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)


theory [19] and then themean and standard deviation can becalculated Finally the reliability can be obtained accordingto the distribution parameters and their mean value andstandard deviation are shown in Table 3

According to the mean value and standard deviation ofeach original error source in Table 3 and the error transfercoefficient of each error the distribution law of the errorcaused by each error can be determined When the meanvalue and standard deviation of the total error are deter-mined the motion reliability of the robot can be calculated

Taking 119905=1s the error transfer coefficient of the errorsources on each chain is shown in Table 4

The error transfer coefficients of the error sources inTable 4 show that the error transfer coefficients of Δ1199031198941 Δ1199031198942Δ119886119894 Δ119888119894 Δ120572119894 119877119890119894 and 120588 are relatively small and have certaincompensations in the 119909 119910 and 119911 directions while the error

source of Δ1205791198941 has a relatively large influence on the outputerror

The error transfer coefficients in Table 4 provide the datasource for determining the mean value and the variance ofthe total error by the superposition principle

According to 120583119904 = sum119899119894=1119870119902119904119894120583119904119894 and 120590119904 = radic119863119904 =radicsum119899119894=111987011990221199041198941205902119904119894 the mean value and variance of output errorcan be deduced from the original error The mean value andthe variance of output error in the 119909 119910 and 119911 directions areshown in Table 5(3) From the above data the calculation of motionreliability is as follows

From Table 5 the following is obtainedThe mean of output error in the 119909 direction 120583119909= 010The mean of output error in the 119910 direction 120583119910= ndash396The mean of output error in the 119911 direction 120583119911= 016The variance of output error in the 119909 direction 1205901199092= 107

then 120590119909= 103The variance of output error in the 119910 direction 1205901199102= 513


then 120590119911= 082When 119905 = 1 s the reliability of the mechanism in the 119909

direction is

119877119909 = Φ( 120583120576 minus 120583119909radic1205901205762 + 1205901199092) = Φ( 1 minus 010radic012 + 1032)= Φ (087) = 08079

(25)

The reliability of the mechanism in the 119910 direction is

119877119910 = Φ( 120583120576 minus 120583119910radic1205901205762 + 1205901199102) = Φ( 1 minus (minus396)radic012 + 2262)= Φ (219) = 09857

(26)


Table 4 Error transfer coefficient of each error source on each kinematic chain119870119902119904119894 Δ1199031198941 Δ1199031198942 Δ119886119894 Δ119888119894119909 direction

119894 = 1 1 ndash1 06889 ndash02080119894 = 2 ndash05 05 ndash02883 01760119894 = 3 ndash05 05 ndash03444 02434

119910 direction119894 = 1 0 0 ndash07249 ndash09716119894 = 2 0 0 ndash08171 ndash09360119894 = 3 0 0 ndash07249 ndash09716

119911 direction119894 = 1 0 0 0 01125119894 = 2 ndash0866 0866 ndash04993 03048119894 = 3 0866 ndash0866 05966 ndash01239119870119902119904119894 Δ120572119894 Δ1205791198941 119877119890119894 120588

119909 direction119894 = 1 450000 ndash1449759 06889 ndash02080119894 = 2 220829 817097 ndash02883 01760119894 = 3 697518 724879 ndash03444 02434

119910 direction119894 = 1 0 ndash1377752 ndash07249 ndash09716119894 = 2 0 1153001 ndash08171 ndash09360119894 = 3 0 ndash1377752 ndash07249 ndash09716

119911 direction119894 = 1 ndash545617 0 0 01125119894 = 2 ndash127496 1415253 ndash04993 03048119894 = 3 ndash116903 ndash1255528 05966 ndash01239

Table 5 The mean value and the variance of output error

The direction The mean value The variance119909 direction 010 107119910 direction ndash396 513119911 direction 016 068



(27)

Therefore the reliability of the mechanism is

119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)

When the mechanism errors are not considered we canobtain that the mean value and the standard deviation ofoutput error are 1205831015840= 0 1205901015840= 0



(29)

Setting Δ119877119878 as the difference between the calculatedvalues of reliability in two cases we can obtain Δ119877119878 = 1198771015840119878 minus119877119878 = 09999 minus 08791 = 01208 = 1208

That is to say when there are mechanism errors themotion reliability of the mechanism will decrease by about12 The error transfer coefficient of the driving arm angleΔ1205791198941 is obviously greater than the other error transfercoefficients which has a greater impact on the positionerror of the mechanism Therefore the mechanism errorsespecially the errors of driving arm angle have a signif-icant effect on the motion accuracy and reliability of therobot

When considering the mechanism errors the reliabilitycalculation result obtained in the 119910 direction is basicallythe same as the result without considering the mechanismerror and both are around 099 Setting Δ119877119910 as the differencebetween the calculated values of reliability we can obtainΔ119877119910 = 1198771015840119910minus119877119910 = 1198771015840119878minus119877119910 = 09999minus09857 = 00142 = 142So the motion accuracy of the mechanism in the 119910 directionis relatively easy to control

The results of this paper can be applied to other high-speed and high-precision mechanisms such as high-speedparallel machine tools In order to improve the manufactur-ing precision the influence of the mechanism errors shouldbe considered In addition the reliability analysis of themechanism can provide a reference for setting reliabilitystandards for the parallel mechanism


6 Conclusions

The paper establishes a kinematic model and the errormodel for the Delta robot and analyzes the motion reliabilityof the robot considering mechanism errors According tothe simulation results and computational analysis drivingerror is the main factor affecting motion accuracy of themechanism The influence of driving error on the motionreliability of themechanism is decisive so it should be strictlycontrolledThe effect of the mechanism errors on the motionreliability varies with the direction The mechanism errorshave little influence on the reliability of the mechanism inthe 119910 direction while the mechanism errors have greaterinfluence on the reliability in the 119909 and 119911 directions

Additionally the influence of each error source on themotion error and reliability of the mechanism is consideredcomprehensively Under the comprehensive influence ofmechanism errors the reliability of themechanism is reducedby about 12 Therefore the mechanism errors should beminimized the driving angle especially should be controlledto improve the motion accuracy of the robot The numericalexample in this paper shows that the calculationmethods andresults of motion reliability provide a reference for reliabilityanalysis of other parallel mechanisms

Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research was supported by the National Natural Sci-ence Foundation of China (Youth Science Foundation) (No11802035) and the Fundamental Research Funds for theCentral Universities (No 2011YJ01)

References

[1] K Der-Ming ldquoDirect displacement analysis of a Stewart plat-formmechanismrdquoMechanism and MachineTheory vol 34 no3 pp 453ndash465 1999

[2] M Mazare M Taghizadeh and M Rasool Najafi ldquoKinematicanalysis and design of a 3-DOF translational parallel robotrdquoInternational Journal of Automation and Computing vol 14 no4 pp 432ndash441 2017

[3] K H Hunt ldquoStructural kinematics of in- parallel-actuatedrobot - armsrdquo SME Journal of Mechanism Transmission andAutomation in Design vol 105 no 4 pp 705ndash712 1983

[4] M H Korayem and H Tourajizadeh ldquoMaximum DLCC ofspatial cable robot for a predefined trajectory within theworkspace using closed loop optimal control approachrdquo Journalof Intelligent amp Robotic Systems vol 63 no 1 pp 75ndash99 2011

[5] Z Dai X Sheng J Hu et al Design and Implementation ofBezier Curve Trajectory Planning in DELTA Parallel Robots

Springer International Publishing Berlin Germany 2015[6] H Qiu Y Li and Y Li ldquoA new method and device for motion

accuracy measurement of NC machine tools Part 1 Principleand equipmentrdquoThe International Journal of Machine Tools andManufacture vol 41 no 4 pp 521ndash534 2001

[7] L W Tsai C G Walsh and R E Stamper ldquoKinematics ofnovel three DOF translational platformsrdquo in Proceedings ofIEEE International Conference on Robotics and Automationpp 3446ndash3451 IEEE (Robotics amp Automation Society) StaffMinneapolis Minnesota 1996

[8] X Dongtao and Z Sun ldquoKinematic reliability analysis ofthe modified delta parallel mechanism based on monte carlomethodrdquoMachinery Design ampManufacture vol 10 pp 167ndash1692016

[9] T Ropponen and T Arai ldquoAccuracy analysis of a modifiedstewart platform manipulatorrdquo in Proceedings of the IEEEInternational Confererce on Robotics and Automation vol 1 pp521ndash525 1995

[10] A-H Chebbi Z Affi and L Romdhane ldquoPrediction of the poseerrors produced by joints clearance for a 3-UPU parallel robotrdquoMechanism and Machine Theory vol 44 no 9 pp 1768ndash17832009

[11] S Mohan ldquoErroranalysisand control scheme for the errorcorrection in trajectory-trackingofa planar 2PRP-PPRrdquoMecha-tronics vol 46 pp 70ndash83 2017

[12] J Fu F Gao W Chen Y Pan and R Lin ldquoKinematic accuracyresearch of a novel six-degree-of-freedom parallel robot withthree legsrdquo Mechanism and Machine Theory vol 102 pp 86ndash102 2016

[13] Y Chouaibi AHChebbi Z Affi andL Romdhane ldquoAnalyticalmodeling and analysis of the clearance induced orientationerror of the RAF translational parallel manipulatorrdquo Roboticavol 34 no 8 pp 1898ndash1921 2016

[14] S-MWang andK F Ehmann ldquoErrormodel and accuracy anal-ysis of a six-DOF Stewart Platformrdquo Journal of ManufacturingScience and Engineering vol 124 no 2 pp 286ndash295 2002

[15] S J Lee ldquoDetermination of probabilistic properties of velocitiesand accelerations in kinematic chains with uncertaintyrdquo Journalof Mechanical Design vol 113 no 1 pp 84ndash90 1991

[16] J Wang J Bai M Gao H Zheng and T Li ldquoAccuracy analysisof joint-clearances in a Stewart platformrdquo Journal of TsinghuaUniversity vol 42 no 6 pp 758ndash761 2002

[17] Y Musheng Research amp Application on Statistical Tolerance inManufacturing Quality Control Nanjing University of Scienceand Technology 2009

[18] X Li X Ding and G S Chirikjian ldquoAnalysis of angular-error uncertainty in planar multiple-loop structures with jointclearancesrdquoMechanism and MachineTheory vol 91 pp 69ndash852015

[19] Huchuan Fault Tree-Monte CarloMethod Simulation and Anal-ysis amp Sampling Scheme Research of Reliability Test for ElectricMultiple Unit (EMU) China Academy of Railway Science 2013

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Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

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Dierential EquationsInternational Journal of

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Table 1 Homogeneous transformation matrix between local coordinate systems

Local coordinate system Variable 120579119894 Homogeneous transformation matrix(119909119910119911)1 1205791198941 OT1 = R(119910 120572119894)M(119909 1199031)R(119911 minus1205791198941)(119909119910119911)2 1205791198942 1T2 = R(119909 119886)M(119911 1205791198941)R(119911 minus1205791198942)(119909119910119911)3 1205791198943 2T3 = R(119910 1205791198943)(119909119910119911)4 119873119900119899119890 3T4 = M(119909 119888)R(119910 minus1205791198943)R(119911 1205791198942)(119909119910119911)5 1205794 4T5 = M(119909 minus1199032)R(119910 minus120572119894)R(119910 1205794)(119909119910119911)6 1205795 5T6 = M(119910 minus119889)R(119909 1205795)

clearance in rotating structures Because parallelmechanismsare used in high-speed situations the error of sphericaljoint clearance will reduce the kinematic accuracy andaggravate wear and affect the motion stability of mecha-nisms In addition the existing research mainly considersthe individual influence of each error on the kinematicaccuracy of the mechanism The influence on the motionerror and reliability of the mechanism in different directionsis not considered under the comprehensive effect of variouserrors

According to the above situation the kinematic modelof the robot is established by the DndashH matrix transfor-mation method An error model considering dimensionerror the error of revolute joint clearance driving errorand the error of spherical joint clearance is establishedThe influence of each error source on the motion erroris analyzed In addition based on the above analysis thecomprehensive influence of mechanism errors on motionaccuracy and reliability in different directions is analyzedA reliability analysis provides an important theoretical basisfor improving the kinematic performance of the mechanismand is of great significance to robot mechanism optimization[14]

2 The Coordinate System of the DeltaParallel Mechanism

The structural model of the Delta parallel robot is shown inFigure 1 and the kinematic chain of the Delta parallel robotis shown in Figure 2

The global coordinate system O minus XYZ is located at thecenter point of the fixed platform at point OThe 119909 axis pointsto theOA1 the119910 axis is perpendicular to the platform surfaceand the negative direction of the 119910 axis is the gravity directionof the moving platform

The position of the rotating pair on the fixed platform andthe moving platform is arranged in a triangle of which thecircumradius is respectively 1199031 and 1199032

The length of the driving arm AiBi is 119886 and the lengthof the parallelogram mechanism (the driven arm BiCi) is 119888The length of PN of the mechanism is 119889 (PN is the distancebetween the center point P of the moving platform and theorigin N of the local coordinate system) 120572119894 is the anglebetween the OAi and the global coordinate 119909 axis which isshown in the diagram When the corresponding kinematic

Figure 1 The structural model of the robot

chain 119894 is 1 2 3 120572119894 is 0o 120o and 240o respectivelyThe inputangle (driving angle) is 1205791198941 1205794 1205795 and the rotation angle ofdriven arm is 1205791198942 The swing angle of driven arm is 1205791198943 PointN is the origin of (119909119910119911)6 in the local coordinate system PointT is the working position

3 Modeling of Mechanism Position Error

According to the structural characteristics of the delta par-allel robot it can be seen that the sources of the positionerror of the mechanism are mainly in the following cate-gories

(1) Dimension error(2) The error of revolute joint clearance(3) Driving error(4) The error of spherical joint clearance

31 Considering Dimension Error According to the homo-geneous transformation rule we can get the homogeneoustransformation matrix between local coordinate systems asshown in Table 1

Therefore the homogeneous transformation matrix OT6of the coordinate system (119909119910119911)6 relative to the global coor-dinate system (XYZ)O can be expressed in two ways asfollows


O

Y

ZX

P

N

a

c

d

T

Ｒ1

Ｒ2

Ｒ3

Ｒ4

Ｒ5

Ｒ6

Ｓ1

Ｓ2

Ｓ4Ｓ5

Ｓ6

Ｔ3

Ｔ5

Ｔ6

i

r1

r2

Ｃ

4

5

i3

Ｃ

i2

i1Ｃ

Ｃ

Ｃ


OT6 = 6prod119896=1

119896minus1T119896

= [[[[[[


(1)

OT6 = [[[[[[


(2)



T = [[[[[[


(3)

T = [[[[[[


(4)




The sleeve hole

The shaft

The error circle

２Ｃ

２1

２2





Δ119909o = 3sum119894=1


Δ119910o = minus 3sum119894=1


119894=1


(6)








Δ119909R = 3sum119894=1

cos 1205791198941 cos120572119894Rei

Δ119910R = minus 3sum119894=1

sin 1205791198941Rei

Δ119911R = minus 3sum119894=1

cos 1205791198941 sin120572119894Rei(8)


Δ119909D = 3sum119894=1


119894=1

minus 119886 cos 1205791198941Δ1205791198941Δ119911D = 3sum

119894=1

119886 sin 1205791198941 sin120572119894Δ1205791198941(9)


O

The spherical shell

The sphere

O

x

z

y





Δ119910Q = 120588 sin120573 sin120572Δ119911Q = 120588 cos120573

(10)


Δ119909Q = 3sum119894=1


119894=1

sin 1205791198942 cos 1205791198943120588

Δ119911Q = minus 3sum119894=1



Δ119902 = Δ119902O + Δ119902R + Δ119902D + Δ119902Q = 119899sum119894=1

119870119902119904119894Δ119904119894119902 = 119909 119910 119911 (12)









120583119904 = 119899sum119894=1

119870119902119904119894120583119904119894 (13)

120590119904 = radic119863119904 = radic 119899sum119894=1

11987011990221199041198941205902119904119894 (14)





120583119904 = 1205831 + 1205832 + sdot sdot sdot + 120583119899 (15)



119877 = 119875 (1205761 lt Δ119902 lt 1205762) (17)


minus Φ(1205761 minus 120583120590 )(18)









1198912 (1199092) [int1199092minusinfin

1198911 (1199091) d1199091] d1199092= intinfin0

119891 (119911) d119911= intinfin0



119877 = 119875 (120576 gt Δ119902) = intinfin0

119891 (119911) d119911 == intinfinminus120573




119911119877 = 120583120576 minus 120583119904radic1205901205762 + 1205901199042 (23)


119877119902 = Φ (119911119877) 119902 = 119909 119910 119911 (24)


5 Numerical Example












minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)



minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)










minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018









O

Y

ZX

P

N

a

c

d

T

Ｒ1

Ｒ2

Ｒ3

Ｒ4

Ｒ5

Ｒ6

Ｓ1

Ｓ2

Ｓ4Ｓ5

Ｓ6

Ｔ3

Ｔ5

Ｔ6

i

r1

r2

Ｃ

4

5

i3

Ｃ

i2

i1Ｃ

Ｃ

Ｃ


OT6 = 6prod119896=1

119896minus1T119896

= [[[[[[


(1)

OT6 = [[[[[[


(2)



T = [[[[[[


(3)

T = [[[[[[


(4)




The sleeve hole

The shaft

The error circle

２Ｃ

２1

２2





Δ119909o = 3sum119894=1


Δ119910o = minus 3sum119894=1


119894=1


(6)








Δ119909R = 3sum119894=1

cos 1205791198941 cos120572119894Rei

Δ119910R = minus 3sum119894=1

sin 1205791198941Rei

Δ119911R = minus 3sum119894=1

cos 1205791198941 sin120572119894Rei(8)


Δ119909D = 3sum119894=1


119894=1

minus 119886 cos 1205791198941Δ1205791198941Δ119911D = 3sum

119894=1

119886 sin 1205791198941 sin120572119894Δ1205791198941(9)


O

The spherical shell

The sphere

O

x

z

y





Δ119910Q = 120588 sin120573 sin120572Δ119911Q = 120588 cos120573

(10)


Δ119909Q = 3sum119894=1


119894=1

sin 1205791198942 cos 1205791198943120588

Δ119911Q = minus 3sum119894=1



Δ119902 = Δ119902O + Δ119902R + Δ119902D + Δ119902Q = 119899sum119894=1

119870119902119904119894Δ119904119894119902 = 119909 119910 119911 (12)









120583119904 = 119899sum119894=1

119870119902119904119894120583119904119894 (13)

120590119904 = radic119863119904 = radic 119899sum119894=1

11987011990221199041198941205902119904119894 (14)





120583119904 = 1205831 + 1205832 + sdot sdot sdot + 120583119899 (15)



119877 = 119875 (1205761 lt Δ119902 lt 1205762) (17)


minus Φ(1205761 minus 120583120590 )(18)









1198912 (1199092) [int1199092minusinfin

1198911 (1199091) d1199091] d1199092= intinfin0

119891 (119911) d119911= intinfin0



119877 = 119875 (120576 gt Δ119902) = intinfin0

119891 (119911) d119911 == intinfinminus120573




119911119877 = 120583120576 minus 120583119904radic1205901205762 + 1205901199042 (23)


119877119902 = Φ (119911119877) 119902 = 119909 119910 119911 (24)


5 Numerical Example












minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)



minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)










minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018









The sleeve hole

The shaft

The error circle

２Ｃ

２1

２2





Δ119909o = 3sum119894=1


Δ119910o = minus 3sum119894=1


119894=1


(6)








Δ119909R = 3sum119894=1

cos 1205791198941 cos120572119894Rei

Δ119910R = minus 3sum119894=1

sin 1205791198941Rei

Δ119911R = minus 3sum119894=1

cos 1205791198941 sin120572119894Rei(8)


Δ119909D = 3sum119894=1


119894=1

minus 119886 cos 1205791198941Δ1205791198941Δ119911D = 3sum

119894=1

119886 sin 1205791198941 sin120572119894Δ1205791198941(9)


O

The spherical shell

The sphere

O

x

z

y





Δ119910Q = 120588 sin120573 sin120572Δ119911Q = 120588 cos120573

(10)


Δ119909Q = 3sum119894=1


119894=1

sin 1205791198942 cos 1205791198943120588

Δ119911Q = minus 3sum119894=1



Δ119902 = Δ119902O + Δ119902R + Δ119902D + Δ119902Q = 119899sum119894=1

119870119902119904119894Δ119904119894119902 = 119909 119910 119911 (12)









120583119904 = 119899sum119894=1

119870119902119904119894120583119904119894 (13)

120590119904 = radic119863119904 = radic 119899sum119894=1

11987011990221199041198941205902119904119894 (14)





120583119904 = 1205831 + 1205832 + sdot sdot sdot + 120583119899 (15)



119877 = 119875 (1205761 lt Δ119902 lt 1205762) (17)


minus Φ(1205761 minus 120583120590 )(18)









1198912 (1199092) [int1199092minusinfin

1198911 (1199091) d1199091] d1199092= intinfin0

119891 (119911) d119911= intinfin0



119877 = 119875 (120576 gt Δ119902) = intinfin0

119891 (119911) d119911 == intinfinminus120573




119911119877 = 120583120576 minus 120583119904radic1205901205762 + 1205901199042 (23)


119877119902 = Φ (119911119877) 119902 = 119909 119910 119911 (24)


5 Numerical Example












minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)



minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)










minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018









O

The spherical shell

The sphere

O

x

z

y





Δ119910Q = 120588 sin120573 sin120572Δ119911Q = 120588 cos120573

(10)


Δ119909Q = 3sum119894=1


119894=1

sin 1205791198942 cos 1205791198943120588

Δ119911Q = minus 3sum119894=1



Δ119902 = Δ119902O + Δ119902R + Δ119902D + Δ119902Q = 119899sum119894=1

119870119902119904119894Δ119904119894119902 = 119909 119910 119911 (12)









120583119904 = 119899sum119894=1

119870119902119904119894120583119904119894 (13)

120590119904 = radic119863119904 = radic 119899sum119894=1

11987011990221199041198941205902119904119894 (14)





120583119904 = 1205831 + 1205832 + sdot sdot sdot + 120583119899 (15)



119877 = 119875 (1205761 lt Δ119902 lt 1205762) (17)


minus Φ(1205761 minus 120583120590 )(18)









1198912 (1199092) [int1199092minusinfin

1198911 (1199091) d1199091] d1199092= intinfin0

119891 (119911) d119911= intinfin0



119877 = 119875 (120576 gt Δ119902) = intinfin0

119891 (119911) d119911 == intinfinminus120573




119911119877 = 120583120576 minus 120583119904radic1205901205762 + 1205901199042 (23)


119877119902 = Φ (119911119877) 119902 = 119909 119910 119911 (24)


5 Numerical Example












minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)



minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)










minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018










120583119904 = 119899sum119894=1

119870119902119904119894120583119904119894 (13)

120590119904 = radic119863119904 = radic 119899sum119894=1

11987011990221199041198941205902119904119894 (14)





120583119904 = 1205831 + 1205832 + sdot sdot sdot + 120583119899 (15)



119877 = 119875 (1205761 lt Δ119902 lt 1205762) (17)


minus Φ(1205761 minus 120583120590 )(18)









1198912 (1199092) [int1199092minusinfin

1198911 (1199091) d1199091] d1199092= intinfin0

119891 (119911) d119911= intinfin0



119877 = 119875 (120576 gt Δ119902) = intinfin0

119891 (119911) d119911 == intinfinminus120573




119911119877 = 120583120576 minus 120583119904radic1205901205762 + 1205901199042 (23)


119877119902 = Φ (119911119877) 119902 = 119909 119910 119911 (24)


5 Numerical Example












minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)



minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)










minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018


















minus200

minus150

minus100

minus50

0

50

100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)



minus160

minus150

minus140

minus130

minus120

minus110

minus100

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)










minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018















minus150

minus100

minus50

0

50

100

150

The e

rror

tran

sfer c

oeffi

cien

t (m

mr

ad)

05 1 15 2 25 30Time (s)













direction is


(25)



(26)













(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018




















(27)


119877119878 = 1119899 119899sum119894=1

119877119894 = 13 (119877119909 + 119877119910 + 119877119911) = 08791 (28)




(29)






6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018









6 Conclusions



Data Availability




Acknowledgments


References




























Journal of












Journal of







Volume 2018






Volume 2018















Journal of












Journal of







Volume 2018






Volume 2018








motion reliability analysis of the delta parallel robot

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