Tolerance Design and Analysis Part II

Dr Tafesse GebresenbetAAiT AAUAAiT, AAU

References

C li C M T l D i h db k f d l i 1. Creveling, C.M., Tolerance Design; a handbook for developing optimal specifications,1996, Addison Wesley

2. Fortini, Dimensioning for interchangeable Manufacture, 2. Fortini, Dimensioning for interchangeable Manufacture, 1967,Industrial press

3. Ken Chase, Basic tools for tolerance analysis of Mechanical A bli Assemblies, 2001

4. Bjorke, Computer aided tolerancing, 2nd edition, 1992, Library of Congressof Congress

5. Kai Yang & Basem Al Haik, Design for Six Sigma: A road map for product development.2003, McGraw‐Hill

6. Paul J. Drake, Jr., Dimensioning and Tolerancing Handbook, McGraw‐Hill, 1999

Dr. Tafesse Gebresenbet 2

Lecture outline

T l A l i (P II)Tolerance Analysis (Part II)Comparison of Tolerance stack‐up modelsStatistics to predict rejects Statistics to predict rejects Percent contribution Numerical examplespTolerance analysis of assemblies Tolerance accumulationMontecarlo simulation Models for Tolerance accumulations


Tolerance analysis

One of the effective tools of Dimensional variation management is tolerance analysis Tolerance analysis is a quantities tool for predicting the

l ti f i ti i bl b f i t kaccumulation of variation in an assembly by performing a stack‐up analysis

1. Identifying dimensions which chain together to control a critical assembly dimension or feature critical assembly dimension or feature

2. The mean, or average, assembly dimension is determined by summing the mean of dimensions in the chain

3. The variation in the assembly dimension is estimated by 3. The variation in the assembly dimension is estimated by summing the corresponding component variations. “stack‐up”

4. The predicted assembly variation is compared to the engineering limits to estimate the number of rejects, or

f i bli nonconforming assemblies 5. Design or production changes may be made after evaluating

the results of the analysis.

4Dr. Tafesse Gebresenbet

Tolerance analysis

The four most popular models of tolerance stack‐ up are listed below with their advantages and their limitations.


Comparison of tolerance stack‐up models

The two most common stack‐up models are: Worst Case (WC).

Computes the extreme limits by summing absolute values of tolerances to obtain the worst combination of over and undersize partspIf the worst case is within assembly tolerance limits, there will be no rejected assemblies.For given assembly limits, WC will require the tightest part tolerance. Thus it is the most costly



Statistical (RSS)Adds variations by root‐sum‐square (RSS), since it

id b biliti f th ibl bi ti th considers probabilities of the possible combinations, the predicted limits are more reasonableRSS predicts the statistical distribution of the assembly p yfeature, from which percent rejects can be estimated. It can also account for static mean shifts.


Comparison of tolerance stackup modelsE l Example.

An assembly has nine component of equal precision, such that the same tolerance Ti may be assumed for each. Predict assembly variationsvariations.

WC: TASM = Σ | Ti | = 9 x 0.01 = ± 0.09

RSS: TASM = ( ) 03.001.09 22 ±=×=∑ iT

( ± denotes a symmetric range of variations) Th f WC di t h i ti th RSS Th Therefore, WC predicts much more variation than RSS. The difference is even greater as the number of components dimension in the chain increases)


Comparison of tolerance stackup modelsE l Example.

An assembly has nine component of equal precision, such that the same tolerance Ti may be assumed for each. Predict assembly variationsvariations.

WC: TASM = Σ | Ti | = 9 x 0.01 = ± 0.09

RSS: TASM = ( ) 03.001.09 22 ±=×=∑ iT

( ± denotes a symmetric range of variations) Th f WC di t h i ti th RSS Th Therefore, WC predicts much more variation than RSS. The difference is even greater as the number of components dimension in the chain increases)



Example Suppose TASM =± 0.09 is specified as a design requirement. The stack‐up analysis is reversedThe stack up analysis is reversed.

The required component tolerance are determined from h bl lthe assembly tolerance,

WC: 01009.0±== ASM

iT

TWC:

RSS:

01.099

±iT

03.0309.0

9±=== ASM

iT

T

Here we find that the WC requires tighter tolerances than RSS to meet an assembly requirement RSS to meet an assembly requirement


Statistics to Predict Rejects

All manufacturing processes produce random variations in each dimension. If h t d k t k f h If one measures each part and keeps track of how many are produced at each size, you could make a frequency plot, as shown below

Normal or Gaussian distributionμ‐mean σ‐ standard deviation σ standard deviation UL & LL – upper and lower limits of size ±3σ ‐ process capability

Frequency plot of size distribution for a process with random error


Statistics to Predict Rejects 1 Perform a tolerance stack‐up analysis to calculate the mean and standard 1. Perform a tolerance stack up analysis to calculate the mean and standard

deviation of the assembly dimension X, which has design requirements XULand XLL

2. Calculate the number of standard deviations from the mean to each limit:

X

LLLL

X

XULXUL

XXZZ σσ

−== −

3. Using standard normal tables, look up the fraction of assemblies lying between ZLL and ZUL ( the area under the curve). This is the predicted yield, or faction of assemblies which will meet requirements. The fraction lying

t id th li it i ( i ld) Th th di t d j t ll outside the limits is (1.0 – yield). These are the predicted rejects, usually expressed in parts per million 9ppm). (note that the standard tables list only positive Z, since the normal distribution is symmetric)


Percent contribution

Another valuable, yet simple evaluation tool is the percent contribution.By calculating the percent contribution that each variation By calculating the percent contribution that each variation contributes to the resultant assembly variations, designers and manufacturing personnel can decide where to concentrate their quality improvement effortsconcentrate their quality improvement efforts.The contribution is just the ratio of a component standard deviation to the total assembly standard deviation

WC:ASMTTCont 100% =

RSS2

2100%

ASM

iContσ

σ=


ExampleE l Example:

A clearance must be maintained between the rotating shaft and bushing shown in the figure. The minimum clearance must not be less than 0.001 inches. The max is not specified. Nominal dimension and tolerance for each part are The max is not specified. Nominal dimension and tolerance for each part are given in the table below .

The first step involves converting the given dimensions and tolerances to centered dimensions and symmetric tolerances.This is a requirement for statistical tolerance analysisThe resulting centered dimensions and symmetric tolerances are listed in the last two columns.


ExampleThe next step is to calculate the mean clearance and variation The next step is to calculate the mean clearance and variation about the mean. The variation about the mean is calculated both by WC and RSS stack‐up, for comparison up, for comparison Mean Clearance:

WC variation:

inSBC 0032..07478.07510.0 =−=−=

inTTT SBc 0016.00006.00010.0 =+=+=WC variation:

RSS variation:

SBc ...

inTTT SBc 00117.00006.00010.0 2222 =+=+=

(Note that the component tolerances are summed )The predicted range of tolerance is C= 0.0032±0.00117 in (RSS) or, Cmax=0 0044, Cmin=0 00203 in Cmax=0.0044, Cmin=0.00203 in Note that Cmax and Cmin are not absolute limits. They represent the ±3σ limits of variations. It is the overall process capability of this assembly process calculated from process capabilities of each of y p p pthe component dimensions is the chain


Example

The next question to answer is how many assemblies will have a clearance less than 0.002 in ?To answer this question, we first must calculate ZLL in terms of q , LLdimensionless σ units The corresponding yield is obtained by table lookup in a math table or by using a spreadsheet, such as Microsoft EXCEL:y g p ,

σC = inTc 00039.03

=

ZLL = σσ

087.300039.0

0032.0002.0−=

−=

−

C

CLL

The results from EXCEL are Yield = NORMSDIST(ZLL) = 0.001011 Or , 99.8989 percent good assemblies 1011 ppm (parts per million

rejects) (Here only ZLL was only needed, since there was no upper limit specified)


How to account for mean shifts?

I i i i i i l l l i It is common practice in statistical tolerance analysis to assume that the mean of the distribution is stationary, located at the midpoint b/n the LL and UL. y pThis is generally not true.

All processes shift with time due to numerous causes such as; tool wear, thermal expansion, operators error, fixture errors, setup errors, setup differences from batch to batch, material properties differences etc.batch to batch, material properties differences etc.

A shift in the nominal dimension of any part in the chain can throw the whole assembly off center by a can throw the whole assembly off center by a corresponding amount.


How to account for mean shifts?Wh th f th di t ib ti hift ff t it When the mean of the distribution shifts off‐center, it can cause serious problems. More of the tail of the distribution is shoved beyond the limit, increasing the number of rejects.

Te slope of the curve steepens as you move the mean toward the limit, so the rejects can increase dramatically

Mean shift can become the dominant sources of rejects, and no j ,company can afford to ignore them.

Normal distribution with a mean shift causes an increase in rejects


How to account for mean shifts?

Th ki d f hif There are two kinds of mean shifts 1. Static mean shifts, which occurs once, and affect every part

produced thereafter with a fixed error, causing fixed errorp , g

2. Dynamic mean shifts which occur gradually overtime. They may drift in one direction, or back and forth.

l l d l l lOvertime, large‐scale production requires multiple setups, multi‐cavity molds, multiple suppliers etc.

To model the effective or static mean shifts, one simply alters the p ymean value of one or more of the component dimensions.

If you have data of actual mean shifts, that is even better. When you calculate the distance from the mean to LL and UL in σ units, you calculate the distance from the mean to LL and UL in σ units, you can calculate the rejects at each limit.


How to account for mean shifts?M d li d i hift i lt i th t l Modeling dynamic mean shifts requires altering the tolerance stack‐up model.

Instead of estimating the standard deviation σi of the dimensional tolerances form Ti = 3 σi, as in conventional RSS tolerance analysis, a modified form is used to account for higher quality level processes:

where Cp is the process capability indexiPi iCT σ3=

LLULC p−

=

If the UL and LL correspond to ±3σ of the process, then the difference UL‐LL= 6σ, and Cp will be 1.0. Thus, a Cp of 1.0

σ6p

, p , p fcorresponds to a moderate quality level of ±3σ .

If the tolerance corresponds to ± 6σ, UL‐LL = 12σ, and Cp =2.0, corresponding to an “extremely high quality level” of ± 6σcorresponding to an extremely high quality level of ± 6σ


How to account for mean shifts?Th i Si d l f t l t k t f b th hi h The six Sigma model for tolerance stack‐up accounts for both high quality and dynamic mean shift by altering the stack‐up equation to include the Cp and a drift factor k for each dimension in the chain.

∑⎥⎥⎦

⎤

⎢⎢⎣

⎡

−=

2

)1(3 ip

iASM kC

T

i

σ

As Cp increase, the contribution of that dimension decreases, causing σASM to decrease. The drift factor kmeasures how much the mean of a distribution has been observed to drift during production .

The figure shows that k corresponds to the shift in the mean as a percent of the tolerance. If there Is no data, it is usually set to k=0.25. The six sigma model uses a drift factor k and ± 6σ Limits to simulate high quality levels.

Dr. Tafesse Gebresenbet

Also note That k is a fraction b/n 0 & 1.0

21

Tolerance allocation methods T l All ti M th dTolerance Allocation Methods

The rational allocation of component tolerances requires the establishment of some rule upon which to base the allocation. The following are examples of useful rules The following are examples of useful rules:

Allocation By Proportional ScalingThe designer begins by assigning reasonable component tolerances based on process or design guidelines.He then sums the component tolerances to see if they meet He then sums the component tolerances to see if they meet the specified assembly tolerance. If not, he scales the component tolerances by a constant proportionality factor proportionality factor. In this way the relative magnitudes of the component tolerances are preserved.


Examples

Th f ll i l i b d h h f d h i The following example is based on the shaft and housing assembly shown in figure below. Initial tolerances for parts B, D, E, and F are selected from tolerance guidelines p gfor the turning process, such as figure 2.


ExamplesE h di i i bj t t i tiEach dimension is subject to variation.Variation accumulate through the chain, causing the clearance to vary as the resultant of sum of variations. h i l d l li i f h The nominal and process tolerance limits for each one are

listed in the Table corresponding to the figure.The design requirement for the clearance U is give below. Th d l li i f l U d i d b The upper rand lower limits of clearance U are determined by the designer, from performance requirements.Such assembly requirements are called Key Characteristics. Th i i l bl f hi h ff They represent critical assembly features, which affect performance

Thus Design requirement (U) = 0.020 ± 0.015 in

Initial design tolerance for dimensions B, D, E and F are selected t a des g to e a ce o d e s o s , , a d a e se ectedfrom a tolerance chart (Figure 2 below)


Examples

Fig. 2. Tolerance range of machining processesTolerance range of machining processes


ExamplesTh t l f B D E d F h f th iddl f The tolerances for B, D, E and F were chosen from the middle of the range of the turning process, corresponding to the nominal size analysis .Th di i f th b i d t i i i d The dimensions of the bearings and retaining ring are vendor supplied, thus tolerances for A, C and G are fixed


ExamplesTh t t i t l l t th l d i ti b t th The next step is to calculate the mean clearance and variation about the mean. The variation has been calculated both by WC and RSS stack up for comparisons. Thus, the mean (average) clearance is:

U= 020 +/‐ 015 = ‐A + B ‐ C + D ‐ E + F ‐ G U= .020 +/ .015 = A + B C + D E + F G Average Clearance = ‐.0505 + 8.000 ‐ .509 + .400 ‐ 7.705 + .400 ‐ .509 = .0265

WC variation WC variation

0 0015+0 008 +0 0025 +0 002 +0 006 +0 002 +0 0025

=+++++= GFDCBAU TTTTTTT

= 0.0015+0.008 +0.0025 +0.002 +0.006 +0..002 +0.0025= 0.0245

RSS variation

0025000200060002000250008000150 2222222

2222222

++++++=

++++++= GFEDCBAU TTTTTTTT

Dr. Tafesse Gebresenbet

01108.00025.0002.0006.0002.00025.0008.00015.0

=++++++

27

ExamplesPart‐per million rejects

rejectsppmUU

ZU

ULUL 679.103.2

00369.00265.0035.0

⇒=−

=−

= σσ

Th ib i h b l l d f ll

rejectsppmUUZU

LLLL 0030.082.5

00369.00265.0005.0

⇒−=−

=−

= σσ

The percent contribution has been calculated for all seven dimension for both WC and RSS. A plot of the results is shown below. RSS is greater because it is the square of the ratio of the variation

ASMTTCont 100% =

2

2100%

ASM

iContσ

σ=


ExamplesIn this example it was observed that there is a mean shift of 0 0065 In this example it was observed that there is a mean shift of 0.0065 from the target value i.e, = 0.0265. Thus 0.020 in , midway between LL and UL. The analysis illustrates the effect of the mean shift – a large increase The analysis illustrates the effect of the mean shift a large increase in rejects at the upper limit and reduced reject at the lower limit To correct the problem, we must modify one or more nominal values of the dimensions B, D, E or F, since A, C and G are fixed.Correcting the problem is more challenging. Simply changing a callout on a drawing to center the mean will nom make it happen.The mean value of a single dimension is the average of many produced

parts. Machinist cannot tell what the mean is until they have made many parts. They must account for tool wear, temperature changes setup errors etc The cause of the problem must be changes, setup errors, etc. The cause of the problem must be identified and corrected. It may require tooling careful monitoring of the target value and the parameters affecting errors.


ExamplesI i f l l i f ll h di i i h It requires careful evaluation of all the dimension in the chain to see which is cost effective to modify. In this case, we have chosen to increase dimension E by 0.0065 in, to a value of 7.715 in. The results are:

Mean σASM ZLL Rejects ZUL RejectsσASM LL j UL j

0.020 in 0.01108 in ‐4.06 24 ppm 4.06 24 ppm

This would be a good solution, if we could successfully hold the mean value by better fixturing more frequent tool sharpening, statistical process control and the like.sharpening, statistical process control and the like.


ExamplesS h f h b ll d Suppose the mean of the process cannot be controlled sufficiently. In that case we may choose to adjust the tolerance of one or more dimensions. The largest contributions are dimensions B on the shaft and E on the housing. Thus we reduce them both to 0.004 in with the results 4

Mean σASM ZLL Rejects ZUL Rejects

6 i i

This corresponds to an effective quality level of ± 3.63σ, that is,

0.0265 in 0.00247 in ‐8.72 0 ppm 3.45 284 ppm

This corresponds to an effective quality level of ± 3.63σ, that is, for a two‐tailed, centered distribution having the same number of total rejects (142 at each limit)


Examples –Mixing normal and uniform distributions

S h h f ( B) i j bb d h i h Suppose the shaft (part B) is jobbed out to a new shop, with which we have no previous experience. We are uncertain how much variation to expect. How shall we account for the uncertainty?We could do worst case analysis, but that would penalize the entire assembly for just one part of unknown quality. y j p q yWe could instead resort to a uniform distribution, applied to dimension B, leaving the others normalThe uniform distribution is sometimes called the ‘equal The uniform distribution is sometimes called the equal likelihood’ distribution. It is rectangular in shape. There are no trials, as with the normalEvery size b/n the upper and lower tolerance limits has an equal probability of occurring. The uniform distribution is conservative. And it predicts greater variation than the normal, but not as great as worst case.



F if di t ib ti th t l li it t ± For a uniform distribution, the tolerance limits are not ± 3σ, as they are for the normal. They are equal to . Thus the stack‐up equation becomes

σ3±

22

33 ∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛= jiASM

TTσ

Where the first summation is the squares of the σi for the normal distributions and the second sum is for the uniform

33 ∑∑ ⎟⎠

⎜⎝⎠⎝

f fdistributions. For the example problem, dimension B has upper and lower limits of 8.008 and 7.992 in, respectively, corresponding to the σ3±99 p y p ghe limits. We assume the assembly distribution has been centered and the only change is that B is uniform rather than gnormal.



S b tit ti th t l f B i th d ti d Substituting the tolerance for B in the second summation and tolerance for each of the other dimension in the first summation, the results are:

Mean σASM ZLL Rejects ZUL Rejects

0.020 in 0.00528 in ‐2.84 2243 ppm 2.84 2243 ppm

The predicted rejects assume that the resulting distribution of assembly clearance U is normal. This is generally true if f y g y fthere are five or more dimensions in the stack. Even if all the component dimensions were uniform, the resultant would still approximate a normal distribution. ppHowever, if one non‐normal dimension has a much larger variation than the sum of all the others in the stack, the assembly distribution would be non‐normal


Examples –Six sigma analysis

Si Si l i f l d if i h Six Sigma analysis accounts for long‐term drift in the mean, or dynamic mean shift, in manufactured parts. It uses the process capability index Cp and drift factor k to simulate the long term spreading of the distributions as discussed earlier, In the following, Six Sigma is applied to two models of the above In the following, Six Sigma is applied to two models of the above example.

1. The first uses Cp =1.0 for comparisons directly with RSS, corresponding to ±3σ quality level with and without drift corresponding to ±3σ quality level, with and without drift correction.

2. The second case uses Cp=2.0 for comparison of ±6σ quality levels with the ±3σ.

The results are presented in the Table below , alongside WC and RSS results for comparisons. All centered cases used the modified pnominal's to center the distribution mean.


Examples –Six sigma analysis Comparison of Tolerance analysis models for the above exampleComparison of Tolerance analysis models for the above example

Non centered cases used a mean shift of 0.0065 in. The RSS –uniform results were not presented before, as this case applied p , ppuniform distribution to all seven dimension for comparison to WC.


Concluding remarksLimitations of Common Assembly ModelsThe common models for assembly tolerance accumulation have distinct limitations when applied to tolerance allocation: 1. The Worst Limit model results in component tolerances which

are tight and costly to produce.2. Statistical models allow looser tolerances, but often predict

higher assembly yields than actually occur in production.3. For assemblies with small numbers of components, or which

have one component tolerance which is much greater than the remaining components, the common Statistical model can give component tolerances which are tighter than if computed by a Worst Limit model.

4. Statistical models assume manufacturing variations follow a Normal or classic bell‐shaped distribution, symmetrically positioned at the midpoint of the tolerance limits.


Concluding remarks

) Th d k i ibl k 5) They do not take into account possible skewness or bias which are common in manufactured parts. Bias results in a shift in the nominal dimension. It is particularly harmful, since it can accumulate in an assembly and cause unexpectedly high rejection rates. Virtually all manufacturing processes exhibit bias Virtually all manufacturing processes exhibit bias. Some processes are more prone to bias than others. Bias can occur from tooling or fixture errors, setup errors or tool wear.

It may be deliberately introduced during setup to compensate for tool wear or to allow for re‐work.

It may naturally occur in a process as in thermal shrinkage of It may naturally occur in a process as in thermal shrinkage of molded parts. Bias is as critical a factor in an assembly model as is the process capability or variance.


Tolerance analysis of AssembliesM t C l Si l tiMonte Carlo Simulation

Monte Carlo Simulation is a powerful tool for tolerance analysis of mechanical assemblies.

It can be used for both nonlinear assembly functions and non‐normal distributions.

It is based on the use of a random number generator to simulate the effects of manufacturing variations on assemblies (see Figure below). ( g )

The biggest disadvantage of the Monte Carlo method is that in order to get accurate estimates it is necessary to generate very large number of samples and it is computationally very large number of samples, and it is computationally intensive.

Also if the distributions of the independent variables pchange, the whole analysis must be redone.


Tolerance analysis of AssembliesM C l i l i Monte Carlo simulation


Tolerance AccumulationModels for Tolerance Accumulation

To obtain the tolerance representation models we must calculate the effect of individual tolerances at specific point(s) in mechanical assemblies.

Homogeneous Matrix Transformation Based MethodsThis is a tolerance representation model for assemblies that is similar to tolerance analysis based methods used in robotic applications to tolerance analysis based methods used in robotic applications.

A methodology to represent standard Y14.5M‐1982 tolerances using homogeneous matrix transformations is described.

Matrix transformations represent both nominal relations between parts and the variations caused by geometric deviations allowed by the tolerances.

Th l i l l t th t ti ti l ti t f th l ti f th thThe analysis calculates the statistical estimate of the location of the nth

part in an assembly starting from the first part or a fixture.

Most types of tolerance specifications can be represented and analysis can be done using this representation This approach is well suited for be done using this representation . This approach is well suited for CAD integration


Tolerance AccumulationDi Li i i M h d (DLM)Direct Linearization Method (DLM)

This method uses the matrix methods and the root mean sum squares analysis to establish vector loop based models sum squares analysis to establish vector loop based models that estimate the tolerance stack‐up in the assemblies.

It linearizes the geometric variations in form and gorientation in terms of angular and positional variation.

The variations in the mechanical assemblies are classified as dimensional variations (lengths and angles), geometric form and feature variations (position, roundness, angularity, etc) and kinematic variations (small adjustments between mating parts).

It is assumed that there is one at least contact point in the bl assembly.


Tolerance analysis of AssembliesVi l J i M h d U i J bi T f iVirtual Joint Methods Using Jacobian Transformations

Tolerances are variations of the functional elements, and b id d can be considered

as small linear and angular displacements of a Functional Element (FE) with respect to their nominal Functional Element (FE) with respect to their nominal position, orientation and form.

The functional elements are assembly parts on which the The functional elements are assembly parts on which the tolerances are given.


Tolerance analysis of AssembliesC i d D f F d M d lConstrained Degree of Freedom Model

This model specifically supports analysis of tolerances i th l in the loops.

We can assume that in spatial analysis of mechanisms, h j i t ki ti i t i t ll b t each joint, or kinematic pair, restricts all but one or

two degrees of relative freedom (six in all) between adjacent bodies.

In other words, depending upon the type of coupling, a joint completely restricts k relative freedoms (say), and completely allows 6‐k relative freedoms..


Tolerance AccumulationV i l T l R iVectorial Tolerance Representation

This method is one of the easiest as it can convert bi t i l t l i t t i ambiguous geometrical tolerances into parametric

tolerances.

It b il i t t d i t CAD b d t [W b It can be easily integrated into CAD based systems [Weber, C., Britten, W., Thome,1998].

This type of representation overcomes the main This type of representation overcomes the main drawback in geometrical tolerancing schemes, which is the absence of a sound mathematical representation..


Tolerance AccumulationG S M h dGap‐Space MethodAssembly analysis is performed by integrating the density function over a tolerance region or assembly density function over a tolerance region or assembly region described in Gap‐space. This method was first proposed in [Zou, Z., Morse, E, 2001].

Most of the other tolerancing models have the restriction that the parts are in contact but the gap‐restriction that the parts are in contact but the gapspace model does not have this restriction


Tolerance AccumulationA t d S T h iAugmented Space TechniqueThis method evaluates the assemblability of a product based on tolerances and adjustable displacements.j p

The propagations of tolerances are represented in terms of ellipsoids and are calculated using jacobiansterms of ellipsoids and are calculated using jacobiansand sweep operations (Minkowski addition).

This means that in a multi loop assembly each and every loop is taken individually.

The tolerance sweeps, and the clearance sweeps are plotted and the assemblability is calculated based on the intersection of both these sweepsthe intersection of both these sweeps.


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Tolerance Design and Analysis Part II