A Comparison of Tolerance Analysis Models for Assembl

ORIGINAL ARTICLE

A comparison of tolerance analysis models for assembly

Chaowang Bo & Zhihong Yang & Linbo Wang &

Hongqian Chen

Received: 18 July 2012 /Accepted: 24 January 2013 /Published online: 13 March 2013# Springer-Verlag London 2013

Abstract Mechanical products are usually made by assem-bling many parts. The dimensional and geometrical varia-tions of each part have to be limited by tolerances so that itcan ensure both a standardized production and a certainlevel of quality defined to satisfy functional requirements.The appropriate allocation of tolerances among the differentparts is the fundamental tool to ensure that assemblies workcorrectly at lower costs. Therefore, to ensure their function-ality, assembly designers have to apply tolerance analysis. Amodel based on either worst case or statistical type analysismay be used. Actually, there are some different models usedor proposed by the literature to make the tolerance analysisof an assembly constituted by rigid parts, but none of themis completely and univocally accepted. None of them hasdone an objective and complete comparison for analyzingthe advantages and the weaknesses and furnishing a criteri-on for the choice and application. This paper briefly intro-duces three of the main models for tolerance analysis, theJacobian, the vector loop, and the torsor. These models arebriefly described and then compared to show their analogiesand differences. Some guidelines are provided as well, withthe purpose of developing a novel approach which is aimedat overcoming some of the limitations of these models.

Keywords Toleranceanalysis . Jacobianmodel .Vector loopmodel . Torsor model

1 Introduction

Increasing competition in industry leads to the adoption ofcost-cutting programs in the design, manufacturing, andassembly of products. Producing high-precision assembliesat lower costs is necessary. Tolerancing decisions can pro-foundly impact the quality and cost of products[1]. The aimof the tolerance analysis is to study the accumulation ofdimensional and/or geometric variations resulting from astack of dimensions and tolerances. The results of the anal-ysis are meaningfully conditioned by the mathematicalmodel adopted.

Some models proposed by the literature carry out thetolerance analysis of an assembly constituted by rigidparts. Requicha introduced the mathematical definitionof the tolerance’s semantic and proposed a solid offsetapproach initially [2, 3]. Since then, numerous modelsare proposed by the literature: the vector loop usesvectors to represent relevant dimensions; the variationalmodel uses homogeneous transformation matrix to rep-resent the variation of an assembly [4]; the matrix usesdisplacement matrix to describe the roto-translationalvariation of a feature [5]; the feature-based approachuses modal interval arithmetic and small degrees offreedom to describe the tolerance specifications[6]; theJacobian uses kinematic chains to formulate the dis-placement matrices; and the torsor uses screw parame-ters to model three-dimensional tolerance zones;Franciosa proposed a method for tolerance analysis tosimulate different assembly sequences[7]. But they stillappear not adequate under many aspects: the schemati-zation of the form deviations, the schematization of thejoints with clearance between the parts, the solution ofcomplex stack-up functions due to the network jointsamong the components, and so on. Moreover, it isdifficult to find literatures in which the differentapproaches are compared systematically with the help

C. Bo : Z. Yang : L. Wang :H. ChenKey Laboratory of High Efficiency and Clean MechanicalManufacture at Shandong University, Ministry of Education,Jinan, China

Z. Yang (*)School of Mechanical Engineering, Shandong University,Jinan 250061, Chinae-mail: [email protected]

Int J Adv Manuf Technol (2013) 68:739–754DOI 10.1007/s00170-013-4795-2

of one or more case studies aimed at highlighting theadvantages and disadvantages. In the literature, somestudies compare the models for tolerance analysis bydealing with their general features [8, 9]. Other studiescompare the main computer-aided tolerancing softwaresthat implement some of the models of the toleranceanalysis, but these studies focus the attention on thegeneral features [10, 11]. A complete comparison ofthe models proposed to solve the tolerance analysis doesnot exist in the literature and, therefore, no guidelinesexist to select the method more appropriate to the spe-cific aims.

The purpose of this paper is to analyze three of the mostsignificant models for the tolerance analysis: Jacobian, vec-tor loop, and torsor. It gives a comprehensive comparison ofthe three models by means of two numerical examples. Itoffers some guidelines for the choice too.

2 Tolerance analysis models

The initial articles and authors about the Jacobian model,the vector loop model, and the torsor model are listed inTable 1.

2.1 Jacobian model

The definitions of different parameters used in the Jacobianmodel are presented in Table 2.Two types of FE pairs in atolerance stack can be distinguished: internal pairs and ki-nematic pairs [12].

Based on small displacements modeling of points usingtransformation matrices of open kinematic chains in robot-ics, a generic dispersion in a pair of functional elements canbe expressed by a set of six virtual joints and coordinateframes (see Fig. 1)[13]. The first three “z” axis of the firstthree frames account for three orthogonal translations, andthe last three “z” axis of the last three frames account forthree orthogonal rotations [14].

The generalization of this model, which enables the explicitand symbolic formulation of the relative position and orientationof any FE in the chain with respect to the base FE, simplyinvolves themultiplication of the transformations on consecutiveFE pairs:

T6n0 ¼ T6

0 � T126 � . . . � T6n

6n�6 ð1Þ

where n represents the total number of FE pairs (both internaland kinematical) involved in the tolerance stack. The globaldeviation of the FR, expressed in datum reference frame(DRF) of the first feature R0, can be expressed as [15]:

Δ s!Δ a!

� �¼ J1J2 . . . J6½ �FEi

. . . J6n�5J6n�4 . . . J6n½ �FEn

h i�

Δ!FEi

��

Δ!FEn

2664

3775

ð2Þwhere Δ s! is the vector of the three small translations of theconsidered point, expressed in the DRF of the first feature R0;Δ~a is the vector of the three small rotations of the considered

Table 1 Related authors and articles of the three models

Models Major authors Articles

Jacobian Lafond and Laperriere [15] Jacobian-based modeling of dispersions affecting pre-defined functional requirements ofmechanical assemblies [15]

Laperriere and Lafond [24] Tolerance analysis and synthesis using virtual joints [24]

Vector loop Chase et al. [25] General 2-D tolerance analysis of mechanical assemblies with small kinematic adjustments [25]

Chase et al. [26] Including geometric feature variations in tolerance analysis of mechanical assemblies [26]

Chase et al. [27] Generalized 3-D tolerance analysis of mechanical assemblies with small kinematic adjustments [27]

Torsor Desrochers [28] Modeling three dimensional tolerance zones using screw parameters [28]

Legoff et al. [29] Geometrical tolerancing in process planning: a tridimensional approach [29]

Table 2 Parameters of the Jacobian model

Parameter Definition

Functionalelement (FE)

Point, curve, or surface that belongs to a part inthe assembly. A FE can be real, for examplethe plane surface of a block, or constructedsuch as the axis of a cylinder or a medianplane

Functionalrequirement(FR)

An important condition to be satisfied betweentwo FEs on different parts, for example afitting condition

Kinematic pair(KP)

Two FEs on different parts make up a kinematicpair if there is a physical or potential contactbetween them

Internal pair (IP) Two FEs each on the same part form an internalpair if they both participate in a contact relationwith some other parts

740 Int J Adv Manuf Technol (2013) 68:739–754

point, expressed in the DRF of the first feature R0;[J1J2…J6]FEi is the 6×6 Jacobian matrix associated with the FE of theith FE pair (internal or kinematical) to which the tolerances are

applied, with i=1 to n; ~ΔFEi is the six-vector of smalldispersions associated with the FE of the ith FE pair (internalor kinematical) to which the tolerances are applied, expressedin the local DRF, with i=1 to n.

For small rotational virtual joints, the ith column of theJacobian matrix Ji is computed as:

Ji ¼ z!i�10 � dn0 � di�1

0

� �z!i�10

" #ð3Þ

while for small translational virtual joints, there is no con-tribution to small rotational displacements of the point ofinterest, and the ith column of the Jacobian matrix Ji iscomputed simply as:

J i ¼ z!i�10

0!

" #ð4Þ

where z!i�10 is the third column of T

!i�1

0 ; and d!i�1

0 is the last

column of T!i�1

0 ; dn0 is the last column of the T!6n

0 .The main steps of the approach are described below[16, 17]:

1. Identify the functional element’s pairs.2. Define the DRF for each functional element and the

virtual joints.3. Create the chain and obtain the overall Jacobian matrix.

4. Once the required stack-up function has been obtained,it may be solved by the usual methods for the worst caseor statistical approaches.

5. Finally, it is necessary to observe that this model isbased on the TTRS criterion and the positional toleranc-ing criterion.

2.2 Vector loop model

Vector-loop-based model uses vectors to represent thedimensions in an assembly [18, 19]. The vectors arearranged in chains or loops representing those dimen-sions that stack together to determine the resultant as-sembly dimensions. There are 6 common joints in 2-Dassemblies and 12 common joints in 3-D assemblies; ateach kinematical joint, a local DRF has to be defined.These joints are used to describe the relative motionsamong mating parts. The tolerances are specified onlyfor the constrained degrees of freedom (dof). Geometrictolerances are considered by adding micro-dof to thejoints just described.

To understand this method better, the basic steps to builda vector loop scheme and to carry out a tolerance analysisare given below [20, 21]:

1. Create assembly graph. The assembly graph is asimplified diagram of the assembly representing theparts, the mating conditions, and the measures toperform.

2. Locate the DRFs for each part. These DRFs are used tolocate features on each part.

3. Locate kinematical joints and create datum paths.4. Create vector loops—using the assembly graph and

the datum paths, vector loops are created. Eachvector loop is created by connecting the datumpaths of the datum traverse by the loops. A vectorloop may be called open or closed if it is related toa measure or not.

5. Derive the equations—the assembly constraints withinvector-loop-based models may be expressed as a con-catenation of homogeneous rigid body transformationmatrices:

R1 � T1 � . . .Ri � Ti � . . .Rn � Tn � Rf ¼ H ð5Þ

where: R1 is the rotational transformation matrix betweenthe x-axis and the first vector; Ri is the rotational transfor-mation matrix between the vectors at node i; Ti is thetranslational matrix of vector i; Rf is the final closure rota-tion with the x-axis; and H is the resultant matrix. If theassembly is described by a closed loop of constraints, H isequal to the identity matrix; otherwise, H is equal to the g

Fig. 1 Virtual joints and coordinate frames to FE pairs [1]

Int J Adv Manuf Technol (2013) 68:739–754 741

vector representing the resultant transformation that willlead to the final gap or clearance and its orientation whenapplied to a DRF.

6. Tolerance analysis—we consider an assembly con-stituted by p-parts. Each part is characterized by thex-vector of the dimensions and by the α-vector ofthe geometrical variables that are known. Whenthese parts are assembled together, the resultantproduct is characterized by the u-vector of the as-sembly variables and by the g vector of the meas-ures required on the assembly. It is possible to getL=J−P+1 closed loops, where J is the number ofthe ties among the parts that looks like:

H x; u; að Þ ¼ 0 ð6Þ

while there is an open loop for each measure to do that lookslike:

g ¼ K x; u; að Þ ð7Þ

The first order Taylor’s series expansion of theclosed loop assembly equations can be written in ma-trix form:

dH ¼ A � dxþ B � duþ F � da ¼ 0 ð8Þ

dg ¼ C � dxþ D � duþ G � da ð9Þwhere Aij=∂Hi/∂xj,Bij=∂Hi/∂uj, Cij=∂Ki/∂xj, Fij=∂Hi/∂αj,Dij=∂Hi/∂uj,Gij=∂Hi/∂αj.

dg ¼ Sx � dxþ Sa � du ð10Þ

where Sx=C−D·B−1·A,Sα=G−D·B−1·F.It is possible to calculate the solution in the worst-case

scenario as:

Δgwc ¼X

k Sxik � txkj j þX

l Sail � ta lj j ð11Þ

while in the statistical scenario (root sum of squares) as:

Δgsc ¼X

k Sxik � txkð Þ2 þX

l Sa il � tað Þ2h i1=2

ð12Þ

The Direct Linearization Method is very simple andrapid, but it is approximated too [22]. When an approximat-ed solution is unacceptable, it is possible to use a numericalsimulation by means of a Monte Carlo technique to improvethe exact solution [18, 23].

2.3 Torsor model

The torsor model uses screw parameters to model three-dimensional tolerance zones [21, 22]. Each actual surface of apart is modeled by a substitute surface. For each of the seventypes of tolerance zone, there are the correspondent screwparameters obtained by annulling the ones that leave the surfaceinvariant in its local frame. Considering a generic feature, if uA,vA, wA are the translation parameters of the point A, and α, β, γare the rotation angles (considered small) as regards to thenominal position, the torsor of point A is given by:

T ¼A

a uAb vAg wA

8<:

9=;

R

ð13Þ

where R is the DRF where the screw parameters are evaluated.Once known the torsor of point A, the torsor of point B may beevaluated.

Therefore, a union of the set of Small DisplacementTorsors that are involved at the joints is used in order toobtain the global behavior of the mechanism. This may bedone by considering that, with the worst-case approach, thecumulative effect of a simple chain of n-elements is simplyexpressed by adding the single components of the torsors:

T0=n ¼ T0=1 þ T1=2 þ . . .þ Tn�1=n ð14ÞThe basic steps of the torsor model are:

1. Identify the elements of the parts and the relationsamong them.

2. Define the parameters of the mechanism—a deviationtorsor has to be associated to each surface of the parts.

3. Compute the cumulative effect of the torsors in-volved in each stack in order to evaluate the func-tional requirements.

3 Model comparison

3.1 First numerical example

To compare the three models previously described, the firstnumerical example shown in Fig. 2 has been used. Theassembly is made of three parts. Part 3 is a cylinder. Thegap g, between the top of the cylinder and part 1, is thecritical assembly feature we wish to control. The case studyhas been solved through both the worst case and the statis-tical approaches. The case study contains all the character-istics and the critical aspects of the problem, but at the sametime it is simple to calculate the exact geometric worst-casevalue of the required range Δg in order to compare theresults of the models. The exact geometric worst-case results


are:5:9974þ1:4629�1:5586 mm for the case considering dimensional

tolerances only and 5:9974þ1:5592�1:7348 mm for the case consid-

ering both dimensional and geometrical tolerances.

3.1.1 Jacobian model

1. Dimensional tolerances onlyThe Jacobian model of the case study is made under the

simplified hypothesis to consider as fixed at 90° the orien-tation of the adjacent sides of part 1 except the inclined one.This simplification is needed to avoid the network in theassembly. The functional requirement g has to be measuredbetween the top of the cylinder and part 1 (see Fig. 2). Onceindicated with x1 to x9 the dimensions of the parts and withu1, u2, u3, u4 the assembly variables (see Fig. 3), thesimplification adopted makes it possible to directly solvethe assembly problem as:

u1 ¼ x1 þ x4 sin x6u2 ¼ x3 � x2 � x1 � x4 sin x6ð Þ tan x6u3 ¼ x1 þ x8u4 ¼ x5 cos x6 þ x3 � x2 � x1 � x4 sin x6ð Þ tan x6

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix7 � x8ð Þ2 � x8 � x4 sin x6 þ x5 sin x6ð Þ2

qð15Þ

There are five functional elements pairs (see Fig. 3): thepoints G and O, O1 and O2, O3 and H are three internalpairs, while O and O1, O2, and O3 are external pairs. Therequired functional requirement g is correspondent to the

functional elements pair associated with the points G andH. It is possible for each FE to locate the virtual joints andthe reference frames and to evaluate the transformation

matrices T10… T30

0 .The matrix of the total transformationis shown in Eq. (16).

Fig. 2 The case study withdimensional and geometricaltolerances [25]

Fig. 3 Functional requirement and the functional elements pairs of thecase study


J ¼ J1 . . . J6½ �FE1. . . J25 . . . J30½ �FE5

h i

¼

0 0 1 0 0 00 �1 0 0 0 01 0 0 0 0 00 x1 þ x8 0 0 0 10 �h� x9 0 1 0 0

hþ x9 0 �x1 � x8 0 1 00 0 1 0 0 00 �1 0 0 0 01 0 0 0 0 00 �x4 sin x6 þ x8 0 0 0 10 �hþ x3 � x9 þ tan x6 x1 � x2 þ x4 sin x6ð Þ 0 1 0 0

h� x3 þ x9� tan x6 x1 � x2 þ x4 sin x6ð Þ 0 x4 sin x6 � x8 0 1 0

0 0 1 0 0 00 �1 0 0 0 01 0 0 0 0 00 �x4 sin x6 þ x8 þ x5 sin x6 0 0 0 1

0�hþ x3 � x9 þ x5 sin x6þ tan x6 x1 � x2 þ x4 sin x6ð Þ 0 1 0 0

h� x3 þ x9 � x5 sin x6� tan x6 x1 � x2 þ x4 sin x6ð Þ 0

x4 sin x6 � x8�x5 sin x6

0 1 0

0 0 1 0 0 00 �1 0 0 0 01 0 0 0 0 00 0 0 0 0 1

0�hþ x3 �m� x9 þ x5 cos x6þ tan x6 x1 � x2 þ x4 sin x6ð Þ 0 1 0 0

h� x3 þmþ x9 � x5 cos x6� tan x6 x1 � x2 þ x4 sin x6ð Þ 0 0 0 1 0

0 0 1 0 0 00 �1 0 0 0 01 0 0 0 0 00 0 0 0 0 10 0 0 1 0 00 0 0 0 1 0

2666666666666666666666666666666666666666666666666666666666666664

3777777777777777777777777777777777777777777777777777777777777775

T

ð16Þ

Once calculated, the Jacobian matrix of the func-tional requirement pair, the stack-up function may beformalized considering that the requirement Δg is eval-uated as the translation of point H along the -Z0 axis(Fig. 3). According to Eqs. (2) and (16), it’s

Δg ¼ �Δz30 ¼ �Δz0 þΔΦ5ðx1 þ x8Þ �Δz6

�ΔΦ11ðx4 sin x6 � x8Þ �Δz12

�ΔΦ17ðx4 sin x6 � x8 � x5 sin x6Þ�Δz18 �Δz24 ð17Þ

where δzi (i=0, 1, 2, 6, 7, 8, 12, 13, 14, 18, 19,20,24,25,26) is the translation along the ith axis andδΦ i ( i= 3, 4, 5, 9,10, 11, 15, 16, 17, 21, 22,23,27,28,29) is the rotation around the ith axis.

However, considering the simplification adopted (fixedangles) due to the need to avoid network, δΦ5=δΦ11=δΦ17=0, then

Δg ¼ �Δz0 �Δz6 �Δz12 �Δz18 �Δz24 ð18Þ

Moreover, with reference to Fig. 3 of the five functionalrequirement pairs, Eq. (15), and considering the nominaldimensions:

Where

m ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix7 � x8ð Þ2 � x8 � x4 sin x6 þ x5 sin x6ð Þ2

qh ¼ x5 cos x6 þ x3 � x2 � x1 � x4 sin x6ð Þ tan x6

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix7 � x8ð Þ2 � x8 � x4 sin x6 þ x5 sin x6ð Þ2

q


Δz0 ¼ �Δx9Δz6 ¼ Δ x3 � x2 � x1 � x4 sin x6ð Þ tan x6ð Þ

¼ 0:3057Δx1 � 0:3057Δx2 þΔx3 þ 0:0894Δx4

� 12:6883Δx6Δz12 ¼ Δ x5 cos x6ð Þ ¼ 0:9563Δx5 � 16:0804Δx6

Δz18 ¼ �Δffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix7� x8ð Þ2 � x8� x4sinx6þ x5sinx6ð Þ2

q¼ �0:3061Δx4 þ 0:3061Δx5 þ 40:0513Δx6�1:4479Δx7 þ 2:4949Δx8

Δz24 ¼ Δx8

ð19Þ

And then is:

Δg ¼ �Δz0 �Δz6 �Δz12 �Δz18 �Δz24¼ �0:3507Δx1 þ 0:3507Δx2 �Δx3 þ 0:2167Δx4�1:2624Δx5 � 11:2826Δx6 þ 1:4479Δx7�3:4949Δx8 �Δx9

ð20Þ

where δxi (i=1…9) is the deviation amplitude of part di-mension xi.

Once the required stack-up function is obtained, it can besolved with the usual methods. For the worst-case approach:

Δg ¼ �P9i¼1

sij j � ti

¼ 0:3507� 0:10þ 0:3507� 0:10þ 0:20þ0:2167� 0:10þ 1:2624� 0:30þ 11:2826� pi=180þ1:4479� 0:20þ 3:4949� 0:10þ 0:30¼ �1:8065 mm

ð21Þ

Fig. 4 Assembly variables and tolerances of vector loop (dimensionaltolerances only)

Fig. 5 Assembly graph of vector loop model

Fig. 6 Datum path of vector loop 1



While for statistical case approach:

Δgsc ¼ �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX9i¼1

si � tið Þ2

vuut ¼ �0:722mm ð22Þ

2. Dimensional and geometrical tolerancesIf the considered case study includes the dimensional

and the geometrical tolerances, nothing changes asregards to the previous case since it has adopted thesimplification of angles in order to avoid the network.The form tolerances (the planeness) do not produce anyeffect because in the Jacobian model the features areconsidered with nominal shape; the position tolerancecan't produce any orientation deviation since the anglesof part 1 are considered fixed.

Therefore, the results are the same as the previouscase where only dimensional tolerances are considered.Moreover, the application of the Envelope Principle orthe Independence Principle does not produce any effectfor the Jacobian model. The application of TTRS doesnot explain well how to handle dependencies of the FEto multiple datums with tolerance specification.

3.1.2 Vector loop model

1. Dimensional tolerances onlyOnce the dimensions of the assembly were indicated

as x1…x9, and the assembly variables as u1, u2, u3 (seeFig. 4), the assembly graph of Fig. 5 has been built. Itshows one joint of “cylinder slider” kind between part 1and part 3 at joint 4, respectively, one joint of parallelcylinder kind between part 2 and part 3 at joint 3, onejoint of planar kind at joint 1 and one joint of edge sliderkind at joint 2 between the part 1 and part 2, and themeasure to perform (g).

A DRF has been assigned to each part; it is centeredin the point O of Fig. 4 for part 1 and O1 for part 2, O3

for part 3. The DRF of part 1 is also considered as theglobal DRF of the assembly. Then, the datum pathshave been created; they are shown in Figs. 6, 7, and 8.

According to Fig. 5, there are L=J−P+1=4−3+1=2closed loops and one open loop. The first closed loopjoints part 1 and part 2 by linking joint 1 and joint 2; thesecond closed loop joints the subassembly part 1–part 2and part 3 by linking joint 1, joint 3, and joint 4; theopen loop joints define the gap.

Then, we can resume the elements of the R and Tmatrices of the loops in Table 3. Once the vector loopsare defined, the relative equations have been generated.

From Eq. (7) and Table 3

g ¼ x9 � u2 � x8 ð23Þ

From the sensitivity analysis:

du ¼ �B�1 � A� dx ¼ Su � dx ð24Þwhere dx=[dx1 dx2 dx3 dx5 dx6 dx7 dx8]

T, du=[du1 du2du3 dΦ1 dΦ2 dΦ3]

T, and

Su ¼

0:3058 � 0:3058 1 0:8669 � 17:0736 � 1:0457 00:3058 � 0:3058 1 1:4725 10:8427 � 1:2262 1:0857�1:0457 1:0457 0 0:3058� 10:0082 0:3058 00 0 0 0 1 0 00 0 0 0:0281 0:8355 � 0:0205 0:08250 0 0 � 0:0281 � 1:8355 0:0205� 0:0825

26666664

37777775

Then, from Eqs. (23) and (24)

dg ¼ dx9 � du2 � dx8

¼ dx9 � ð0:3058dx1 � 0:3058dx2 þ dx3 þ 1:4725dx5

þ 10:8427dx6 � 1:2262dx7 þ 1:0857dx8 Þ � dx8

¼ �0:3058dx1 þ 0:3058dx2 � dx3 � 1:4725dx5

� 10:8427dx6 þ 1:2262dx7 � 2:0857dx8 þ dx9

ð25Þ

where dxi (i=1…9) is the deviation amplitude of partdimension xi.

It is possible to calculate the solution in the worstcase as:

Δg ¼ �P9i¼1

sij j � ti

¼ 0:3058dx1þ 0:3058dx2þ 1:0000dx3þ 1:4725dx5þ10:8427dx6þ 1:2262dx7þ 2:0857dx8þ dx9¼ 0:3058� 0:1þ 0:3058� 0:1þ 1:0000� 0:2þ 1:4725� 0:3þ10:8427� pi=180þ 1:2262� 0:2þ 2:0857� 0:1þ 0:3¼ �1:64595mm

ð26Þ

For statistic case approach:

Δgsc ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX9i¼1

ðsi� tiÞ2vuut ¼ �0:6830mm ð27Þ

2. Dimensional and geometrical tolerancesOnce the dimensions of the assembly are indicated as

x1 ,…, x9, the assembly variables as u1, u2, u3, and thegap between the top side of part 1 and the circle that isthe assembly measure as g, the case study appears asshown in Fig. 2. The DRFs and the datum paths are thesame as the previous case.

The vector loops have no change, but they have totake into consideration the geometrical tolerances. Thegeometrical tolerances have to be translated into the x-vector: the planeness on part 2—α1=0±0.03 mm; thecircular degree on part 2—α2=0±0.05 mm; the circulardegree on part 2—α3=0±0.02 mm; the perpendicularityon part 1—α4=0±0.01 mm.


Then, we can resume the elements of the R and Tmatrices of the loops in Table 4. From Eq. (7) andTable 3, we can get:

g ¼ x9 � u2 � x8 � a3 ð28Þ

As concerning the sensitivity analysis, we have:

du ¼ �B�1Adx� B�1Fda ¼ Sudxþ Sada ð29Þwhere :dx=[dx1 dx2 dx3 dx5 dx6 dx7 dx8]

T, du=[du1 du2du3 dΦ1 dΦ2 dΦ3]

T, dα=[dα1 dα2 dα3 dα4]T, Su is same

to the previous one, and

Sa ¼

1:0457 0 0 00:6433 �1:6847 1:3764 1:3764�0:3058 0 0 00 0 0 0�0:0166 �0:0408 0:0567 0:05670:00166 0:0408 �0:0567 �0:0567

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

9>>>>>>=>>>>>>;

The gap g depends on the x-vector through the sen-sitivity coefficients.

dg ¼ dx9 � du2 � dx8 � da3

¼ �0:3058dx1 þ 0:3058dx2 � dx3 � 1:4725dx5�10:8427dx6 þ 1:2262dx7 � 2:0857dx8 þ dx9

� 0:6433da1 þ 1:6847da2 � 2:3764da3 � 1:3764da4

ð30Þ

It is possible to calculate the solution in the worstcase as:

Δg ¼ �P13i¼1

sij j � ti

¼ 0:3058dx1þ 0:3058dx2þ 1:0000dx3þ 1:4725dx5þ10:8427dx6þ 1:2262dx7þ 2:0857dx8þ dx9þ0:6433 da1 þ 1:6847 da2 þ 2:3764 da3 þ 1:3764 da4

¼ �1:81075mm

ð31Þ


Table 3 Elements of R and T matrices of the loops (dimensionaltolerances only)

Loop 1 Loop 2 Loop 3

Nr. R T Nr. R T Nr. R T

1 0 x2 1 0 x2 1 0 x12 90 x3 2 90 x3 2 90 u23 90+x6 u3 3 90+x6 u3 3 −90 x84 −90 x5 4 −90 x5 4 90 x85 180 x7 5 Φ2 x7 5 0 gap

6 -Φ1 u1 6 180 x8 6 90 x1+x87 −90 x1 7 Φ3 x8 7 90 x98 180 8 90 u2 8 90

9 −90 x110 180

Table 4 Elements of R and T matrices of the loops (dimensional andgeometrical tolerances)

Loop 1 Loop 2 Loop 3

Nr. R T Nr. R T Nr. R T

1 0 x2 1 0 x2 1 0 x12 90 x3 2 90 x3 2 90 u23 90+x6 u3 3 90+x6 u3 3 −90 α4

4 −90 α1 4 −90 α1 4 0 α3

5 0 x5 5 0 x5 5 0 x86 180 x7 6 Φ2 x7 6 90 x87 -Φ1 u1 7 0 α2 7 0 α3

8 −90 x1 8 0 α3 8 0 gap

9 180 9 180 α3 9 90 x1+x810 0 x8 10 90 x911 Φ3 x8 11 90

12 0 α3

13 0 α4

14 90 u215 −90 x116 180


For statistic case approach:

Δgsc ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiX13i¼1

si � tið Þ2vuut ¼ �0:6902mm ð32Þ

3.1.3 Torsor model

1. Dimensional tolerances onlyThe simplification to consider the fixed angles of part

1 (except the inclined side) has been used in order toavoid the network. This simplification may solve theassembly problem.

The first step of the method is to identify the ele-ments of the parts (see Fig. 9) and the relations amongthem; these information are reported in the surfacesgraph of the case study (see Fig. 10). Considering thatthe angles of part 1 are fixed, the network can be solved,and the surfaces graph is simplified as shown in Fig. 11.The cumulative torsor G is expressed as:

GH ¼ �T1:6 þ T1=3 þ T3:1 ¼H

a ub vg Δg

8<:

9=;

R

ð33Þ

where T1,6 is the torsor of feature 1 of part 1; T3,1 is thetorsor of feature 1 of part 3 (the circle); and T1/3 is the

torsor of the link between part 1 and part 3. The func-tional requirement Δg can be expressed by the transla-tion along the global z axis.

The next step is to evaluate the components of thetorsors (indeed it is enough to evaluate the third compo-nents due to translation). For the T1,6 torsor, with refer-ence to Fig. 12 and considering that the case study is a 2Dproblem on x–z plane (i.e., α=0, γ=0, v=0), it is:

T1:6 ¼M

� �b1:6 �� wM1:6

8<:

9=;

R1:6

ð34Þ

where M is the median point of the feature 1.6. consider-ing the point of interest is H.

uHvHwH

8><>:

9>=>; ¼

��

wM1:6

8><>:

9>=>;þ

0 � b1:6� 0 �

�b1:6 � 0

264

375 �

�5

0

�5:9974

8><>:

9>=>;

¼�5:9974b1:6

�wM1:6 þ b1:6

8><>:

9>=>; ð35Þ

T1:6 ¼H

� �5:9974b1:6b1:6 �� wM1:6 þ 5b1:6

8<:

9=;

R1:6

ð36Þ

Fig. 9 Torsor model: elementsand parts of the case study


The local z axis of R1.6 is the reverse of the global Z axis,respectively; therefore, the correspondent translation needsto be inverted. It is:

T1:6 ¼H

� �5:9974b1:6b1:6 �� wM1:6 þ 5b1:6ð Þ

8<:

9=;

R

ð37Þ

Considering the simplification of considering the fixedangles of part 1, β1.6=0 and therefore:

T1:6 ¼H

� 00 �� wM1:6

8<:

9=;

R

ð38Þ

Considering the extreme points remain into the tolerancezone, it results:

T1:6 ¼H

� �0 �� ðt1:6 þ t9Þ=2;þðt1:6 þ t9Þ=2�½ R

�8<: ð39Þ

where t1.6 is the thickness of the tolerance zone of S1.6, andt9 is the dimensional tolerance of the dimension x9. Note thatthis torsor shows the admissible range of variations of thesmall displacements associated to the feature. In the sameway,the torsor T3.1 of feature 1 of part 3 may be computed as:

T3:1 ¼� �t8=2;þt8=2½ �0 �� t8=2;þt8=2½ �

8<:

9=; ð40Þ

where t8 is the dimensional tolerance of the dimension x8.The evaluation of the torsor linking part 1 and part 3 (T1/3)

is very difficult because it needs the solution of the networksamong the components of the assembly. By adopting thesimplification that the angles of part 1 are fixed, it is possibleto solve this problem as the sum of two terms:

T1=3 ¼ T1=2 þ T2=3 ð41Þ

The first term (T1/2) is the torsor of the link between part 1and part 2.

T1=2 ¼

H

� ½�ðt1 þ 0:2924t4 þ 14:3446t6 þ t1:3Þ=2;þðt1 þ 0:2924t4 þ 14:3446t6 þ t1:3Þ=2�

0 �� ½�ð0:3057t1 þ 0:3057t2 þ t3 þ 0:0894t4 þ 12:6883t6Þ=2;

þð0:3057t1 þ 0:3057t2 þ t3 þ 0:0894t4 þ 12:6883t6Þ=2�

8>>><>>>:

9>>>=>>>;

R

ð42Þ

The second term (T2/3) is the torsor of the link between part1–part 2 and part 3.

T2=3 ¼

H

� ½�ð0:2924t4 þ 14:3446t6 þ t8Þ=2;þð0:2924t4 þ 14:3446t6 þ t8Þ=2�

0 �� ½�ð0:3061t4 þ 1:2624t5 þ 23:9708t6 þ 1:4479t7 þ 2:4949t8Þ;

þð0:3061t4 þ 1:2624t5 þ 23:9708t6 þ 1:4479t7 þ 2:4949t8Þ�

8>>><>>>:

9>>>=>>>;

R

ð43Þwhere t1.3 is the thickness of the tolerance zone of S1,3;and t1,…, t8 are the tolerances on the dimensions x1,…,x8. Therefore,the functional requirement is:

Δg ¼ �ð0:3057t1 þ 0:3057t2 þ t3þ 0:3955t4þ1:2624t5 þ 36:6591t6 þ 1:4479t7 þ 2:4949t8 þ t9 þ t1:6Þ=2

ð44Þ

Fig. 10 Torsor model: surface graph of the case study

Fig. 11 Torsor model: simplified surface graph of the case study


Now, it is necessary to relate the thickness of each tolerancezone assigned to each feature to the tolerances required on thecomponents. This is another critical step of the torsor model.However, under the simplified hypothesis adopted (i.e., fixedangles of the part 1) and by considering only dimensionaltolerances, it may have:

t1:6 ¼ 0mm; t1 ¼ 0:20mm; t2 ¼ 0:20mm;t3 ¼ 0:40mm; t4 ¼ 0:20mm; t5 ¼ 0:60mm;t6 ¼ 0:03489; t7 ¼ 0:40mm; t8 ¼ 0:20mm;t9 ¼ 0:60mm

The functional requirement in the worst-case approach is:

$g ¼ �ð0:3057� 0:2þ 0:3057� 0:2þ 0:4þ 0:3955� 0:2þ1:2624� 0:6þ 36:6591� 0:03489þ 1:4479� 0:4þ2:4949� 0:2þ 0:6Þ=2¼ �2:1580mm

ð45Þ

It may be added that the torsor method does not allow toevaluate the results due to a statistical approach since thetorsor’s components are considered the extreme possible inter-vals of the small displacements, and this is not compatiblewith the statistical approach where a probability density func-tion is to each parameter.

2. Dimensional and geometrical tolerancesIf the considered case study both included the dimen-

sional and the geometrical tolerances, Eq. (34) is still valid(under the hypothesis of fixed angles of part 1), and it isalways needed to relate the thickness of the tolerancezones to the tolerances required on the components.Moreover, by using the simplification to consider thefixed angles of part 1, there's no change compared to the

Fig. 12 Torsor model:tolerance zones of the casestudy

Fig. 13 a Part 1. b Part 2. cPart 3. d Definition of theassembly and functionalrequirement for statistical caseapproach


case that only dimensional tolerances are considered.Therefore, the simplification to consider the fixed anglesof part 1 causes the geometrical tolerances has no effecton the results of the case study. Moreover, the applicationof the Envelope Principle or the Independence Principledoes not produce any effect on the torsor model too.

3.2 Second numerical example

The second numerical example shown in Fig. 13 has beenused to compare the three models. This assembly containsthree parts. The functional requirement is the gap shown onthe leftmost side of the assembly (see Fig. 13). Some keytolerances from ta to td are labeled. The relevant parametersare as follows: x1–10mm, x2–5mm, x3 –4 mm, ta–±0.15mm,tb–0.1 mm, tc–±0.1 mm, td–0.1 mm. The exact geometricresults are: 1±0.35 mm for the worst case considering dimen-sional tolerances only and 1±0.26 mm for the case consider-ing both dimensional and geometrical tolerances by 3DCS.

3.2.1 Jacobian model

1. Dimensional tolerances onlyKinematic chains in this case are shown in Fig. 14.

There are six virtual joints and five FE pairs. O0 and O1,O2 and O3, O4, and O5 are internal pairs; O1 and O2, O3,and O4 are external pairs. The [J1J2…J6]FE2, [J1J2…J6]FE4,~ΔFE2and~ΔFE4can be eliminated because O1 andO2, O3,and O4 have the same position and orientation, respective-ly, the contacts between them are assumed perfect. Theform tolerances are not considered here because the realfeatures are considered as coincident with their substituteones in this model. So, the Jacobian matrix is equal to:

J ¼

1 0 0 0 10 0 1 0 0 0 5 0 1 0 0 0 1 00 1 0 �10 0 0 0 1 0 �5 0 0 0 1 0 �1 0 00 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 00 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 00 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 00 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1

26666664

37777775

ð46Þ

The stack-up function may be formalized by consider-ing that the requirement $g is evaluated as the translation ofpoint O0 along the Z0 axis. According to Eqs. (2) and (46),it’s

4g ¼ Δz18 ¼ Δz0 þΔz6 þΔz12

¼ ta� tc� tc ð47Þwhere δzi(i=0, 1, 2, 6, 7, 8, 12, 13, 14) is the translationalong the ith axis.

For the worst-case approach:

4g ¼ �0:35mm ð48Þ


4g ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:152 þ 0:12 þ 0:12

p¼ �0:206mm ð49Þ

2. Dimensional and geometrical tolerancesAccording to Eqs. (2) and (46), the functional re-

quirement is equal to

4g ¼ Δz18 ¼ Δz0 þΔz6 þΔz12¼ taþ tbð Þ � tcþ tdð Þ � tc ¼ taþ tb� 2tc� td

ð50ÞFor the worst-case approach, Δg=±0.45 mm

Δg ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:152 þ 2� 0:052 þ 2� 0:12

p¼ �0:207mm ð51Þ

3.2.2 Vector loop model

Once the dimensions of the assembly were indicated as x1, x2,x3 (see Fig. 13), the assembly graph of Fig. 15 has been built.According to Fig. 15, there are two joints of planar kind andonly one loop that define the gap.

Then, we can resume the elements of the R and T matri-ces of the loops in Table 5. Once the vector loop is defined,the relative equations have been generated.

1. Dimensional tolerances onlyAccording to Eq. (7),

g ¼ x1 � x2 � x3 ð52Þ

Fig. 14 Kinematic chains identification

Fig. 15 Assembly graph of vector loop model



4g ¼ �0:35mm ð53Þ

For the statistical case approach:

4g ¼ �0:206mm ð54Þ2. Dimensional and geometrical tolerances

According to Eq. (7),

g ¼ x1 � x2 � x3 þ a1 � a2 � a3 ð55Þ

where α1, α2, and α3 represent the translations of thejoints due to the geometrical tolerances, parallelism onpart 1—α1=0± tb/2, parallelism on part 3—α2=0± td/2,planeness on part 2— α3=0± td/2..


4g ¼ �0:5mm ð56Þ


4g ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:152 þ 3� 0:05þ 2� 0:012

p¼ �0:28mm ð57Þ

3.2.3 Torsor model

The first step is to identify the elements of the parts and therelations among them; these information are reported in thesurfaces graph (see Fig. 16). The cumulative torsor G isexpressed as:

GH ¼ �T1:1 þ T1 2= þ T2:2 ¼H

a uHb vH

g Δg

8<:

9=;

R

ð58Þ

T1.1 is the torsor of feature 1 of part1; T2.2 is thetorsor of feature 2 of part 2; and T1/2 is the torsor ofthe link between part 1 and part 2. The functionalrequirement Δg can be expressed by the translationalong the global z axis.

The components of the torsors are evaluated:

T1:1 ¼H

a1:1 �� a1:1

� wG1:1

8<:

9=;

R

¼H

0 �� 0� �t1:1=2;þt1:1=2½ �

8<:

9=;

R

T1=2 ¼ T1=3 þ T3=2 ¼H

0 �� t1þt1:2ð Þ=2;þ� t1þ t1:2ð Þ=2½ �

8<:

9=;

R

þH

0 �� t2 þ t3:2ð Þ=2;þ� t2 þ t3:2ð Þ=2½ �

8<:

9=;

R

T2:2 ¼H

0 �� 0� � t3 þ t2:2ð Þ=2;þ t3 þ t2:2ð Þ=2½ �

8<:

9=;

R

ð59Þwhere t1, t2, t3 are the tolerances on the dimensions x1, x2, x3;and t1.1, t1.2, t2.2, t3.2 are the thicknesses of the tolerancezones of S1.1, S1.2, S2.2, S3.2. Therefore, the functional re-quirement is:

Δg ¼ � t1 þ t2 þ t3þt1:1 þ t1:2 þ t2:2 þ t3:2ð Þ=2 ð60ÞIf only dimensional tolerances are considered, it may

have:

t1 ¼ 0:3mm; t2 ¼ 0:2mm; t3 ¼ 0:2mm;t1:1 ¼ t1:2 ¼ t2:2 ¼ t3:2 ¼ 0Δgwc ¼ �0:35mm

ð61Þ

If dimensional and geometrical tolerances are considered,it may have:

t1 ¼ 0:3mm; t2 ¼ 0:2mm; t3 ¼ 0:2mm; t1:1 ¼ 0;t1:2 ¼ 0:1mm; t2:2 ¼ 0; t3:2 ¼ 0:1mmΔgwc ¼ �0:45mm

ð62Þ

Table 5 Elements of R and T matrices of the loops

Dimensional tolerances only Dimensional and geometricaltolerances

Nr R T Nr R T

1 0 x1 1 0 x12 180 x2 2 0 α1

3 0 x3 3 180 x24 0 gap 4 0 α2

5 180 5 0 x36 0 α3

7 0 gap

8 180

Fig. 16 Surface graph


3.3 Comparisons

Table 6 shows the results due to the application of the threeconsidered models to the two cases. The results of vectorloop model are closer to the exact result than the other twomodels, but it is more complex and will cost us more timethan the other two models if there are many dimensional andgeometrical tolerances. The Jacobian model is appropriatefor the 3D assembly and it's more efficient if there are manydimensional tolerances. The torsor model is fit for the ex-treme limits of 3D tolerance zones resulting from a feature'ssmall displacements. Table 7 shows the comparison of thethree models on various aspects.

If both the dimensional tolerances and the geometricaltolerances are applied, the result of Jacobian model is sameor close to the result of Jacobian model only the dimensionaltolerances are applied. To solve the stack-up function, it isneeded to relate the virtual joints’ displacements to thetolerances assigned on the components. However, in theJacobian model, the features are considered with nominalshape, and it can’t handle the form tolerance. Besides, itcan’t explain well how to handle dependencies of FE to

multiple datums with tolerance specification due to theapplication of TTRS.

The three considered models have three common limits.The first deals with the assembly cycle: the three models arenot able to correctly represent the coupling with clearancebetween two parts and consider the functional requirementsin the network. The second deals with the representation ofthe tolerances applied to the assembly’s components: thethree models do not give a complete correspondence amongthe model variables and the part’s tolerances. Moreover, thetranslation of the part’s tolerances into model variables doesnot satisfy the standards (ASME or ISO). The third dealswith the independence principle: the three models do notallow us to apply the independence and/or the envelope ruleto different tolerances of the same parts.

4 Conclusions

This paper makes a brief comparison of three models oftolerance analysis for rigid-parts assembly, the Jacobian,vector loop, and the torsor, in order to point out the

Table 6 Results of the compar-ison among the models appliedto the two cases

Case Approach Analysis methods Exact solution Jacobian Vector loop Torsor

Case 1 Only dim. Worst case +1.4629–1.5586 ±1.8065 ±1.64595 ±2.1580

Statistical – ±0.7220 ±0.6830 –

Dim.+geom. Worst case +1.5992–1.7348 ±1.8065 ±1.81075 ±2.1580

Statistical – ±0.7220 ±0.6902 –

Case 2 Only dim. Worst case ±0.35 ±0.35 ±0.35 ±0.35

Statistical ±0.206 ±0.206 –

Dim.+geom. Worst case ±0.45 ±0.50 ±0.45

Statistical ±0.26 ±0.207 ±0.208 –

Table 7 Comparison of threemodels

aPossiblebNot possiblecEasydDifficult

Aspects of the comparisons Analysis methods Jacobian Vector loop Torsor

Analysis method Worst case −a −a −a

Statistical −a −a –b

Tolerance type Dimensional −a −a −a

Form –b −a –

b

Position −a −a −a

Tolerance parameterization –b

–b

–b

Envelop and independence –b

–b

–b

Stack-up function type Linear −a −a −a

Network –b −a –b

Consider the datum precedence −a −a −a

Tolerance zones interaction −a –b −a

Functional requirement schematization With points −a −a −a

With features −a −a −a

Evaluation of small displacements –d –c –c

Computation of the functional requirements –c –c –d


advantages, weakness, and the scope of application of eachmodel based on the experimental results and the informationavailable from the literature.

Further researches include the definition of a new and moreefficient model able to overcome the limits that have beenhighlighted in this work and consider the functional require-ments in the network. The model can be the unification of theprevious models that can make up for each other, such as theunified Jacobian–torsor model. Also, this model should re-duce the uncertainty of transforming the functional require-ment to the geometrical principle, considering the tolerance ofcomplex features. It should allow simulating different assem-bly sequences and can handle under, fully, and overcon-strained assemblies. A dimensional tolerance assigned to thedistance between two features of a part or of an assembly maybe required with the application of the envelope rule or theindependence principle. To model the form tolerance, it ispossible to introduce a virtual transformation that is assignedto points of the surface to which a form tolerance is assigned.The subjects of further research are to consider dimensionaltolerance with the application of both the envelope principleand the independence principle, to take into account the realfeatures and the interaction of the tolerance zones, to considerjoints with clearance among the assembly components, and toadopt both the worst-case and the statistical approaches tosolve the stack-up functions.

Acknowledgments This work is supported by Independent Innova-tion Foundation of Shandong University (grant 2010TS087) and theNational High Technology Research and Development Program ofChina(No.2012AA040910).

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