Tolerance Stack-Up Analysis Complete Paper

TOLERANCE STACK-UP ANALYSIS Gary Whitmire

Page 1 December, 2004

Many engineering design offices observe the principle of 100% interchangeability when considering the effect of component tolerances upon the assembly and the functioning of the finished product. In such cases decisions on the component tolerance are based upon the assumption that the extreme assembly conditions can and will be met in practice. Where the tolerances for the assembled product are small, this policy leads to an impractical tolerance for the individual parts, despite the fact that in many cases the probability of encountering parts at their extreme conditions is very improbable. Where safety is of paramount importance, it is understandable that designers should insist upon infallible interchangeability, interchangeability, no matter the cost. However, there are many cases where insistence upon 100% interchangeability is not warranted and an occasional failure to assemble or to function would not be serious, particularly when judged in the light of the overal1 economic gain in production. It is often comparatively simple to estimate the maximum cumulative effect upon the assembly and thus ascertain if the functioning of the assembly is maintained. One way to ensure that all assemblies fit the designated size range is to determine the assembly tolerance by simply adding the tolerances of the individual parts. This method is safe and is usually practical for assemblies consisting of only two or three parts. However, for assemblies made up of several parts, the totals are usually too large to be acceptable. One method for overcoming this problem is to provide the widest practical component tolerances based upon the statistical fact that it is unlikely that all maximum-tolerance parts or all minimumtolerance parts would ever be brought together in the same assembly. This paper presents a methodology for calculating a tolerance stack-up analysis to predict worst case assembly variation for 100% and 99.73% interchangeability based on preferred component tolerances, or to develop component tolerances that will support assembly design goals.

You Will Learn How To: •

Master basic tolerance concepts including the difference between tolerance and variation, how bilateral, unilateral, and limit dimension tolerances differ.

•

Do basic stack-up analysis techniques including the loop method.

•

Determine if there is 100% interchangeability. interchangeability.

•

Determine if there is interchangeability at ± three sigma or 99.73% interchangeability.

•

Get interchangeability when the tolerances are too tight for manufacturing.

•

Determine if the assembly will function at its worst condition.

•

Determine if larger tolerance zones will meet the design requirements.

•

Reconstruct the interchangeability conditions, even years later.

•

Determine if statistical interchangeability will give a lower cost product.

•

Do a formal tolerance analysis for documenting the design conditions.

Assumptions and Understandings for Doing a Tolerance Analysis •

It is understood that changes in temperature affect the size of parts. Engineering standards all over the world state that all dimensions apply at 68 °F (20 °C).

•

When doing a tolerance analysis, it is assumed that all manufactured parts will meet the dimensional requirements stated on the drawing.

Genium Publishing Corporation



•

When doing a tolerance analysis, it is assumed that all parts are rigid in the free state and that they retain that shape when assembled. However, it is understood that a part may not be rigid in its free state due to length, size, shape or type of material.

•

It is assumed that all parts are machined to the middle of their dimensions and controlled by a quality control program.

Why Do a Tolerance Analysis •

A tolerance analysis is typically done to determine if the parts will assemble 100% of the time, or will assemble 99.73% (± 3 sigma) of the time statistically.

•

A tolerance analysis is typically done to determine if the parts will function properly at worst condition.

•

A tolerance analysis will determine if the drawing tolerance can be larger.

•

A tolerance analysis will determine if statistical tolerancing will give a lower cost product.

•

A fully documented tolerance analysis provides a record of the design dimensional requirements that can be reviewed at a later date in case of a product failure.

•

It impresses the boss, because not everyone knows how to fully document a dimensional design condition.

The Three Types of Tolerance Analysis (See Figure 1) 1. Radial Stack – Normally involves diameters or radial directions. 2. Linear Stack – Involves dimensions that are in the X, Y, or Z direction. 3. Assembly Stack – Involves radial or linear directions of more than one part.


TOLERANCE STACK-UP ANALYSIS


Gary Whitmire

0

0 3

4X .875

A

4X R

A

0 1 2

RADIAL ASSEMBLY

LINEAR .75 .62 .375 .12

0

SECTION A-A Figure 1

The Tolerance Analysis Form is shown in Figures 2 and 3. This form provides a way to standardize and increase the accuracy of the tolerance analysis.




Gary Whitmire

Assembly #

Rev.

+

Direction Direction Dimensions Dimensions

A

B

Today's Date

/

/

Rev.

Page

RMS

+ Tolerance

Square the Tolerance in "D"

Drawing Number & Revision

D

E

F

C

1 2 3 4 5 6 7 8 9

+

+

TOTAL

TOTAL

+ Algebraic Sum

+

+

+

+ + (For 3

100% interchangeability

Figure 2


sigma)

of



Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

1

Rev.

Page

2

Form Instruction, See next page 3

4 +


A

B

A through F

RMS

+ Tolerance



D

E

F

C

1 2 3 4 5 6 7 8

5

9

6

+

9

8 +

TOTAL

TOTAL

1 0

+

1 1

+

Algebraic + Sum 7 100% interchangeability

+

1 2

+ + (For 3

Figure 3


sigma)

of



Instructions for Completing the Form 1. Insert the next assembly number and revision letter. 2. Insert the starting date of the tolerance analysis. If the tolerance analysis is changed at a later date, make it revision "A." The page number starts with 1 of 1; for more pages, 1 of 2, etc. 3. This is the area where a sketch is made or a cut-and-paste is added. This entire system can be added to a CAD format. 4. Column A numbers will correspond to the dimension numbers on the sketch - not mandatory.

Column B lists all of the dimensions going in the positive direction. Column C lists all of the dimensions going in the negative direction. Column D lists all of the plus/minus tolerances for both columns B and C. Column E lists all of the results obtained by squaring the plus/minus tolerance in column D. Column F lists all of the drawing numbers and revisions used in the analysis. 5. This is the total of column B. Copy the total into the box below (follow the arrow). 6. This is the total of column C. Copy the total into the box below (follow the arrow). 7. Insert the algebraic sum of the totals of columns B and C. This must be a positive number. If it is not, the positive and negative directions need to be reversed, or there is interference in the assembly. 8. This is the total of column D. Copy the total into the box below (follow the arrow). 9. This is the total of column E. Copy the total into the area under the square root symbol. 10. Take the square root of the total of column E and insert it above the square root symbol. 11. Insert the algebraic sum to the right (follow the arrow). 12. Insert the square root to the right (follow the arrow).

Step-by-Step Analysis #1 (See Figure 4 for the completed analysis) The Figure 4 drawing analysis determines the minimum/maximum snap ring groove width. 1. Draw or attach a drawing to the tolerance analysis form. 2. Establish the START and END points for the tolerance analysis. 3. Convert all dimensions involved in the analysis to an equal plus/minus tolerance as shown below:

1.901 +.000, -.005 = 1.8985 ± .0025 .43 - .41 = .42 ± .01 .220 - .215 = .2175 ± .0025 4. Number the dimensions - not mandatory. 5. Determine the POSITIVE and NEGATIVE direction of the analysis as shown below: a. Consider the tolerance analysis as a loop.




b. Starting at the START of the loop, whichever (left or right / up or down) direction will result in a dimension going through the START/END gap must be the positive direction for all dimensions going in the same direction. In Figure 4, the 2.62 ± .01 dimension is the positive dimension. The other three dimensions are negative dimensions. c.

Mark these dimensions with positive and negative signs.

6. Tabulate the POSITIVE, NEGATIVE dimensions and the PLUS/MINUS tolerance in columns B, C, and D. The order is not important. 7. Total the B, C, and D columns. 8. Establish the ALGEBRAIC SUM of columns B and C. 9. Bring down the total from column D.

The ALGEBRAIC SUM and the PLUS/MINUS tolerance is the 100% condition of the snap ring gap.




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

START

END

.084 + .025 = .109 Max. Gap .084 - .025 = .059 Min. Gap

.084 ± .025 #3

+ 2.62 ± .01 #1 _

#2 _ .43 .41

+.000 1.901 - .005

#4 .220 .215


RMS

Tolerance

Square the Tolerance in “D”


C

D

E

F

1

1.8985

.0025

2

.42

.01

A

3

B

.01

2.62 .2175

4

.0025

5 6 7 8 9

2.62

2.536

.025

TOTAL

TOTAL 2.62 2.536

Algebraic Sum

.084

.025


(For 3 sigma)

Figure 4

Figure 5 is a workshop problem for the reader and Figure 6 is the solution.


of



Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

Determine the unknown minimum / maximum length

+.008 1.000 .000

unknown length

.190 .186


A

B

C

RMS

Tolerance



D

E

F

1 2 3 4 5 6 7 8 9

TOTAL TOTAL

Algebraic Sum


(For 3 sigma)

Figure 5


of



Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

Determine the unknown minimum / maximum length

END

+.008 1.000 .000

.816 + .006 = .822 .816 - .006 = .810

START

.190 .186


A

B

1 2

RMS

Tolerance



C

D

E

F

.188

.002 .004

1.004

3 4 5 6 7 8 9

1.004

.188

TOTAL

.006

TOTAL

1.004

Algebraic Sum

.188 .816

.006


(For 3 sigma)

Figure 6


of



This next subject explains the theory of statistical dimensioning, and outlines the procedure for using this system. Most designs use arithmetic dimensioning which means dimensioning such that the tolerances of component parts of an assembly are combined to give the high and low limits. This procedure assumes that all component parts will be produced within the drawing specifications, and that the resulting assembly fit is equally likely to be anywhere between the cumulative tolerance limits. This results in the following: (a) The component parts have very close tolerances in order to provide the desired fit. (b) Secondary machining operations are needed. (c) Adjustments must be designed into the assembly in order to assure proper fit and function.

An alternative to arithmetic dimensioning is statistical dimensioning which is the assigning of dimensions and tolerances to assembly components on the basis that the assembly tolerance is equated to the square root of the sum of the squares of the individual tolerances. This is not as technical as it sounds. This approach assumes that the resultant assembly fit is concentrated in the middle of the cumulative tolerance limits rather than equally likely anywhere between the cumulative tolerance limits.

Statistical Process Control (SPC) Process control and quality control must be maintained on any statistical dimension. The ASME Y14.5-1994 standard on dimensioning and tolerancing has developed a symbol and a general note (shown below) to be placed next to any dimension that must be controlled by the process and monitored by quality control. The ASME Y14.5-1994 standard and symbol does not state the rules to follow for SPC. Drawing note: FEATURES IDENTIFIED AS STATISTICALLY TOLERANCED ST SHALL BE PRODUCED WITH STATISTICAL PROCESS CONTROLS. Optional note: FEATURES IDENTIFIED AS STATISTICALLY TOLERANCED ST SHALL BE PRODUCED WITH STATISTICAL PROCESS CONTROLS, OR TO THE MORE RESTRICTIVE ARITHMETIC LIMITS.

Benefits of Statistical Tolerancing Statistical tolerancing reduces costs and /or improves the quality and performance of the units by producing components closer to average fits. Statistical dimensioning should not replace present design methods unless benefits can be realized. This technique is applicable where: (a) Limited space requires a close assembly tolerance. (b) Selective assembly would otherwise be required. (c) Adjustments or complex designs would be required to reduce tolerance accumulations. (d) Fits with a narrow range of clearance would be required. (e) Close tolerances on the component parts would be required if arithmetic dimensioning was used. (f) Secondary operations or slow machining operations would be necessary to hold tolerances for processes such as machining heat-treated parts. (g) 100% interchangeability is not possible. (h) The result is a cheaper manufacturing process.




The successful application of statistical dimensioning depends on: (a) A design analysis based on normal distribution. This means manufacturing to the middle of the dimension. (b) The use of process controls to produce parts to a near normal distribution within the drawing specifications.

Responsibility It is the responsibility of everyone connected with statistical dimensioning to recognize the following: (a) Engineers must understand not only the basic application techniques, but also the necessity of producing parts to the mean dimension with a normal distribution. (b) Manufacturing Engineering must design tools to produce parts around the mean dimension as required by the drawing specifications. (c) Quality Control must consult with Manufacturing Engineering to provide the necessary inspection equipment, assist manufacturing in setting controls, audit the lot as a final inspection on the controlled dimensions, and, where necessary, determine process capability information using the controlled dimensions and forward the results to Manufacturing Engineering.

Normal Distribution A normal distribution is formed when manufacturing processes produce dimensions in a random manner about the mean, with a majority of the dimensions close to the mean and a decreasing number occurring away from the mean. If the dimensions, from a reasonably stable process, are measured and recorded according to size, a plot of the resulting frequency distribution will approximate the shape shown in Figure 7.




mean

X 68.27%

1σ

1σ

94.45%

2σ

2σ 99.73%

3σ

3σ Figure 7

Statistical Step-by-Step Analysis #1 (See Figure 8 for the completed analysis) This step-by-step instruction produces a snap ring groove of acceptable size 99.73% of the time. 1. Square each plus/minus tolerance shown in column D, and put the answers in column E. 2. Calculate the total amounts in column E. Put this result within the square root symbol (follow the arrow). This is called "THE SUM OF THE SQUARE." 3. Calculate the square root of the total amount in column E. This result goes above the square root symbol. This is called "ROOT MEAN SQUARE (RMS). Move this answer to the right. 4. Bring over to the right the algebraic sum (follow the arrow).

At this time, do not try to understand the answer of .084 ± .0146.




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

START

END

99.73%

.084 .0146 #3

+ 2.6200 #1 _

#4


.2175

.0100 #2 _

1.8985 .0025

.4200 .0100

.0025

RMS

Tolerance



C

D

E

F

1

1.8985

.0025

.00000625

2

.42

.01

.0001

.01

.0001

.0025

.00000625

A

3

B

2.62 .2175

4 5 6 7 8 9

2.62

2.536

.025

TOTAL

.0002125

TOTAL 2.62

.0146

2.536

Algebraic Sum

.084

.0002125

.025


.084

.0146 (For 3 sigma)

Figure 8


of



Loosening Up the Original Tolerance As previously developed, the 100% condition for the snap ring gap is .084 ± .025. If the drawing is changed so that all dimensions have the tolerance increased by ± .0125, and you recalculate the tolerance allowance to produce an acceptable snap ring gap 99.73% of the time, the result is the same .084 ± .025. The ± .0125 tolerance was established using the formula: One over the square root of N, where N is the number of dimensions. See Figure 9.




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

This page illustrates the four tolerance zones can be increase and still meet the original design requirements shown on page 14 of .084 ± .025. This is done by using the 99.73% probability. The formula for the percent of the part tolerance is 1 = 1 = .5 ; .5 X .025 = .0125 N

END

4

START

.084 .025 #3

+ 2.6200 #1 _

#4

A

B

.0125 #2 _

1.8985 .0125

.2175


99.73%

.0125

RMS

Tolerance



F

C

D

1

1.8985

.0125

E .00015625

2

.42

.0125 .0125

.00015625 .00015625

.2175

.0125

.00015625

3

2.62

4

.4200 .0125

5 6 7 8 9

2.62

2.536

.05

TOTAL

.000625

TOTAL

2.62 2.536 Algebraic Sum

.084

.025 .000625

.05


.084

.025 (For 3 sigma)

Figure 9


of



Gary Whitmire

Figure 10 is a Workshop Problem for the Reader and Figure 11 is the solution.

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

Establish the min./max. gap for 100% and 99.73%

.043±.004

.043 ±.004 .113 ±.004

.375 ±.005 .062 ± .002 .080 ±.003


A

B

C

RMS

Tolerance



D

E

F

1 2 3 4 5 6 7 8 9

TOTAL TOTAL

Algebraic Sum


(For 3 sigma)

Figure 10


of



Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

Establish the min./max. gap for 100% and 99.73% 100% .134±.022 And

99.73% .134±.009 .043±.004

START END

.043±.004 .113±.004

+ .375±.005

.062 ± .002 .080±.003


A

B

1 2

RMS

Tolerance


Drawing Number & R evision

F

C

D

.043

.004

E .000016

.005

.000025

.375

3

.080

.003

.000009

4

.062

.002

.000004

5

.113

.000016

6

.043

.004 .004

.241

.022

.000016

7 8 9

.375

TOTAL

.000086

TOTAL

.375 .241

Algebraic Sum

.134

.0093 .000086

.022


.134

.009 (For 3 sigma)

Figure 11


of



Gary Whitmire

Workshop Problem – Using Figure 11, repeat the process shown in Figure 9. The solution is given in Figure 12.

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

This page illustrates the six tolerance zones that can be increase and still meet the original design requirements shown on page 17 of .134 ± .022. This is done by using the 99.73% probability. The formula for the percent of the part tolerance is 1 = 1 = .408 ; .408 X .022 = .009. N

6

Original 100% ± tolerance

.043±.009

.134±.022 .043±.009 .113±.009

.375±.009 .062 ± .009 .080±.009


A

B

1 2

RMS

Tolerance



F

C

D

.043

.009

E .00008

.009

.00008

.375

3

.080

.009

.00008

4

.062

.009

.00008

5

.113 .043

.009

.00008

.009

.00008

.241

.054

6 7 8 9

.375

TOTAL

.00048

TOTAL

.375 .241 Algebraic Sum

.134

.022 .00048

.054


.134

.022 (For 3 sigma)

Figure 12


of



I have never found the information on the next few pages in a technical book. I learned these principles from IBM 35 years ago when I was working for them. In the next example, shown in Figure 13, the top surface of one part must be .0005 to .018 below the top surface of another part.

.0005 - .018

.125

Unknown Length .250

Figure 13

Analysis (a) The average clearance is (.0005 + .018) / 2 = .009 (.00925). Therefore, the required mean length (labeled “Unknown Length” in Figure 13) is .009 + .125 + .250 = .384. The total tolerance to meet the design requirements of .018 - .0005 is: .018 - .0005 = .0175 or ± .00875. The three dimensions (.125, .250, and .384) will have an individual tolerance of ± .00875/3 = .0029. The better number, shown in Figure 14, is .0028. (b) When considering statistical tolerancing, the assembly tolerance as stated above is ± . 00875. Figure 15 illustrates the ratio of statistical addition to arithmetic addition using the formula one divided by the square root of 3 = 58%. Thus 58% of the assembly tolerance that can be assigned to the three dimensions, is an approximation. Therefor .58 X .0175 = .010 or ± .005 for each dimension. Recheck this using the square root of the sum of the squares to determine that the design requirements have been met. 2

2

2

Square root of .005 + .005 + .005 =.00866 X 2 = .0173 (an approximation of .0175), so, all is good because .0173 is less than .0175. (c) All tolerance values have been adjusted using the most economical distribution based on available process information. Note that the values used to determine the statistical sums are squared. Thus changes in the largest tolerances will have the greatest effect on the final distribution of tolerances.




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

TO MEET 100% DESIGN REQUIREMENTS On the previous page, the assembly tolerance ±.00875. .00875/3 = ± .0029 for all three dimensions shown below. This is an approximation.

The algebraic sum shown below is +.009 ± .0087 +.009 + .0087 = + .0179 (this is less than .018) (GOOD) +.009 - .0087 = +.0003 ( this is less than .0005) (BAD) The .0003 is .0002 less than .0005, so the three tolerances need to be adjusted to .0028 = +.0174 to +.0006

START

END

.0028 .1250±.0029

.2500

.0029 .0028


A

B

C

1

.1250

2

.2500

3

.3840

.3840

.0028 .0029

RMS


Tolerance

D .0029 .0029 .0029

E


F

.0028 .0028 .0028

4 5 6 7 8 9

.0084 .384

.375

.0087

TOTAL

TOTAL


.009

.0084 .0087


(For 3 sigma)

Figure 14


of


STATISTICAL SUM = ARITHMETIC SUM 100


N(tol.) 2 = N(tol.)

N N N 1 X = = N N N N N To illustrate the ratio of statistical addition to arithmetic addition of the formula above:

80

INVERSE OF ROOT X

T60 N E C R E P40

20

0

1

2

3

4 5 6 7 8 9 10 11 12 13 NUMBER OF DIMENSIONS

14

Figure 15

Figure 16 changes the tolerance from ± .0028 to ± .0049. Using statistical tolerancing, assume that meeting the design requirements 99.73% of the time is acceptable.




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page

TO MEET 99.73% OF THE DESIGN REQUIREMENTS For statistical tolerancing take 58% X .0175 = .010 or ± .005; actually ± .0049 is better. +.009 + .0085 = + .0175 (this is less than .018) (GOOD) +.009 - .0085 = +.0005 ( perfect) (GOOD)

START

END

.1250 ± .0049 .3842 .250


A

B

1 2 3

.0049

RMS

Tolerance



F

C

D

.1250 .2500

.0049

E .000024

.0049 .0049

.000024 .000024

.3840

.0049

4 5 6 7 8 9

.384

.375

.0147

TOTAL

.000072

TOTAL


.009

.0085 .000027

.0147


.009

.0085 (For 3 sigma)

Figure 16


of



Figure 17 shows a functional clearance gap of .005 to .019. The average gap is (.005 + .019) / 2 = .012, the mean dimension for the internal opening is .125 + .500 + .375 + .250 + .250, and the average gap of .012 = 1.512. The gap tolerance is .019 - .005 = .014 or ± .007.

.005 - .019 GAP .125 ± .001 1.512 ± .0025 this was determined below

.500 ± .005 .375 ± .003 .250 ± .002 .250 ± .002

Figure 17

In the previous example, all parts had the same tolerance. In this example, the parts have different tolerances, so the formula is different. The tolerance for the internal opening 1.512 dimension will be X as established by the following calculation: 2

2

2

2

2

2

.007 = .001 + .005 + .003 + .002 + .002 + X 2

.007 = .000043 + X

2

.000049 = .000043 + X 2

X = .000049 - .000043 2

X = .000006 X = .000006 X = ± .002449 or ± .0024 The plus/minus is an approximation. A better tolerance is ± .0025. Therefore, the dimension for the opening is 1.512 ± .0025. The problem is that the five inside parts add up to be 1.500 ± .020. There is no way the required gap of .005 to .019 can be met. By changing all of the tolerances to ± .0012 as shown in Figure 18, the design requirements can be met, but this is more expensive.




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page 1 of 1

The gap tolerance is .019 - .005 = .014 or ± .007. This ± .007 / 6 = ± .0012. The other tolerance combinations can be used as long as the total ± .007 is maintained. For 100% interchangeability: The maximum gap is .012 + .007 = .019 The minimum gap is .012 - .007 = .005 END

.005 - .019 GAP

START

.125 ± .0012 .500 ± .0012

+1.512 ± .0012 .375 ± .0012 .250 ± .0012 .250 ± .0012

+

Direction Dimensions

Tolerance

B

C .125

D .0012

2

.500

..0012

3

.375

4

.250

.0012 .0012

5

.250

.0012

A 1

6

RMS

+




E

F

.0012

1.512

7 8 9

+ 1.512

1.500

+ .007

TOTAL

TOTAL

+ 1.512 1.500 Algebraic Sum

+

.012

+

+ .007 100% interchangeability

+ for

+ 3 sigma

Figure 18

This is arrived at by going back to the original tolerance and converting to statistical tolerancing (see Figure 19).




Gary Whitmire

Assembly #

Rev.

Today's Date

/

/

Rev.

Page 1 of 1

By using the original tolerance of each part: For 99.73% interchangeability The maximum gap is .012 + .007 =.019 The minimum gap is .012 - .007 =.005

END

.005 - .019 GAP

START

.125 ± .001 .500 ± .005

+1.512 ± .0025 .375 ± .003 .250 ± .002 .250 ± .002

+

RMS

+



Tolerance


B

C .125

D .001

E .000001

2

.500

.005

.000025

3

.375

4

.250

.003 .002

.000009 .000004

5

.250

.002

.000004

.0025

.000006

A 1

6

1.512


F

7 8 9

+ 1.512

1.500

+ .0155

TOTAL

.000049

TOTAL

+ 1.512 Algebraic Sum

.007

1.500

+

.012

.000049

+ .0155 100% interchangeability

Figure 19

This completes the paper on “How to Do a Tolerance Analysis.” I hope it will be useful.


+

+ .012 +

.007 for 3 sigma

Tolerance Stack-Up Analysis Complete Paper

Recommend Documents