Toler Ancing State

8/12/2019 Toler Ancing State

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STAT 498 B

Statistical Tolerancing

Fritz Scholz

Spring Quarter 2007


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Objective of Statistical Tolerancing

Concerns itself with mass production, not custom made items.

Dimensions and properties of parts are not exactly what they should be.

Worst case tolerancing can be quite costly.

Manage variation in mechanical assemblies or systems.

Take advantage of statistical independence in variation cancelation.

Also known as statistical error propagation.

Useful when errors and system sensitivities are small.

It is more in the realm of probability than statistics (no inference).

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Exchangeability of 757 Cargo Doors

At issue were the tolerances of gaps and lugs of hinges and their placement on the

hinge lines of aircraft body and door.

10 hinges with 12 lugs/gaps each.

That means that a lot of dimensions have to fit just about right.

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0

0

The Root Sum Square (RSS) paradigm does not work here!

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IBM Collaboration: Disk Drive Tolerances

A

H

C

B

D

S

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Coordination Holes for Aligning Fuselage Panels

ideal

perturbed holes

first hole aligned

third hole rotated

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Main Ingredients: Mean, Variance & Standard Deviation

The dimension or property of interest,X, is treated as a random variable.

X f(x) (density), CDF F(x) = P(Xx) =Z x

f(t) dt.

Mean: =X= E(X) =Z

x

t f(t) dt

Variance: 2 =2X= var(X) = E((X)2) = E(X2)2 =Z x

(t)2 f(t) dt

Standard Deviation: =

var(X)

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Rules forE(X)andvar(X)For constantsa1, . . . , akand random variablesX1, . . . ,Xk

we have forY=a1X1 + . . . + akXk

E(Y) =E(a1X1 + . . . + akXk) =a1E(X1) + . . . + akE(Xk)

For constantsa1, . . . , akandindependentrandom variablesX1, . . . ,Xkwe have

2Y=var(Y) =var(a1X1 + . . . + akXk) =a21var(X1) + . . . + a

2kvar(Xk)

It is this latter property that justifies the existence of the variance concept.

Y=

a21var(X1) + . . . + a

2kvar(Xk)

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Central Limit Theorem (CLT) I

Suppose werandomlyand independently draw random variables X1, . . . ,Xnfromn possibly different populations with respective means andstandard deviations1, . . . ,n and1, . . . ,n

Suppose further that

max21, . . . ,2n

21 + . . . +

2n

0, as n

i.e., none of the variances dominates among all variances

Then Yn= X1+ . . . +Xn has an approximate normal distribution with meanand variance given by

Y=1 + . . . +n and 2Y=

21 + . . . +

2n.

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CLT: Example 1

standard normal population

x1

Density

2 0 2 4

0.0

0.2

0.4

uniform population on (0,1)

x2

Density

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.4

0.8

1.2

a lognormal population

x3

Density

0.0 0.5 1.0 1.5

0

1

2

3

4

5

Weibull population

x4

Density

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

0.0

0.2

0.4

0.6

0.8

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CLT: Example 2

Central Limit Theorem at Work

x1 + x2 + x3 + x4

Density

2 0 2 4 6

0.0

0

0.0

5

0.1

0

0.1

5

0.2

0

0.2

5

0.3

0

9


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CLT: Example 3standard normal population

x1

Density

4 2 0 2 4

0.0

0.2

0.4


x2

Density

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.6

1.2


x3

D

ensity

0.0 0.5 1.0 1.5

0

2

4

Weibull population

x4

D

ensity

0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.4

0.8

Weibull population

x5

De

nsity

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

0.0

0

.4

0.8

10


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CLT: Example 4


x1 + x2 + x3 + x4

Density

2 0 2 4 6

0.0

0

0.1

5

0.30


x2 + x3 + x4 + x5

Density

1 2 3 4 5 6 7

0.0

0.2

0.4

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CLT: Example 5


x1

Density

4 2 0 2 4

0.0

0.2

0.4


x2

Density

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.4

0.8

1.2


x3

Density

0 5 10 15

0.0

0.2

0.4

Weibull population

x4

Density

0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.4

0.8

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CLT: Example 6

Central Limit Theorem at Work (not so good)

x1 + x2 + x3 + x4

Density

0 10 20 30 40

0.0

0

0.0

5

0.1

0

0.1

5

0.2

0

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CLT: Example 7


x1

Density

4 2 0 2 4

0.0

0.2

0

.4


x2

Density

0 5 10 15 20

0.0

0

0.0

2

0.0

4

0.0

6


x3

Density

0.0 0.5 1.0 1.5

0

1

2

3

4

5

Weibull population

x4

Density

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

0.0

0.4

0.8

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CLT: Example 8

Central Limit Theorem at Work (not so good)

x1 + x2 + x3 + x4

Density

20 10 0 10 20 30 40

0.0

0

0.0

2

0.0

4

0.0

6

0.0

8

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What is a Tolerance?

Tolerances recognize thatpart dimensions are not what they should be.

should be= nominal or exact according to engineering design

Exact dimensions allow mass production assembly using interchangeable parts

Variationsaround nominal are controlled by tolerances.

Typical two-sided specification: [Nominal Tolerance, Nominal + Tolerance]

Specifications can beone-sided:[Nominal, Nominal+ Tolerance] or [Nominal Tolerance, Nominal]

Specifications can be asymmetric: [Nominal Tolerance1, Nominal +Tolerance2]

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Simple Examples

Example 1: A disk should have thickness 1/8 with .001 tolerance, i.e.,the disk thickness should be in the range

[.125 .001, .125 + .001] = [.124, .126].

Example 2: A stack often disksshould be1.25high with .01tolerance,i.e., the stack height should be in the range

[1.25 .01, 1.25 + .01] = [1.24, 1.26].

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Disk Stack

1 8 == 0.125 0.001

1.25 0.01

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Worst Case or Arithmetic Tolerancing

The tolerance specification in Example 1,if adhered to,

guaranteesthe tolerance specification in Example 2.

The reasoning is based onworst case or arithmetic tolerancing

The stack ishighestwhen all disks are asthickas possible..126per disk= stack height of10 .126 =1.26.

The stack islowestwhen all disks are asthinas possible..124per disk= stack height of10 .124 =1.24.

This gives the total possible stack height range as[1.24, 1.26].

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disk stack/tolerance stack

.02''

1.25''

.125''

worst case

low stack

worst casehigh stack

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Worst Case or Arithmetic Tolerancing in Reverse

This reasoning can be reversed.

If the stack height has specified end tolerance .01,

and if the disk tolerances are to be the same for all disks (exchangeable),

then we should, by the worst case tolerancing reasoning, assign

.01/10= .001

tolerances to the individual disks (item tolerances).

End tolerances can create very tight and unrealistic item tolerances. Costly!

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Worst Case Analysis or Goal Post Mentality

nominal

nominaltol nominal+tol

tol

Add some structure, aim for the middle

= Statistical Tolerancing

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Statistical Tolerancing Assumption

Statistical tolerancingassumes that disks are chosenat random,not deliberately to make a worst possible stack, one way or the other.

The disk thickness variation within tolerances is described by a distribution.

Thehistogram, summarizing these thicknesses, is often assumed to be

normalor Gaussianwith centerD at the middle of the tolerance rangeand with standard deviation such that

3standard deviations = tolerance.or

D= 13

TOLD so that [D 3D, D + 3D] = tolerance interval

The normality assumption is a simplification, butis not essential.

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Normal Histogram/Distribution of Disk Thicknesses

Histogram of 10,000 Thicknesses

disk thickness

Density

0.1235 0.1240 0.1245 0.1250 0.1255 0.1260 0.1265

0

200

400

600

800

1000

1200

18 16

99.73% .135%.135%

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Why Does Statistical Tolerancing Work

Under the normal population model= we will seeabout13.5out of10, 000disks with thickness .126.

The chance of randomly selecting such a fat or fatter disk is .00135 = 13.5/10, 000

The chance of having such bad (thick) luckten times in a row is

.00135 . . . .00135= (.00135)10 =2.01 1029 (!!!)

Choosing thicknesses at random from this normal population we (justifiably)hope thatthick and thin will averageout to some extent.

Makeindependentvariationwork for you, not against you!If life gives youlemons, makelemonade! Turn a negative into a positive!

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The Insurance Principle of Averaging

We look forward to the day when everyone will receivemore than the average wage.

Australian Minister of Labour, 1973

The etymology of average derives from the Arabic: awaryahmeaning shipwreck, damaged goods, and linking it to the custom

of averaging the losses of damaged cargo across all merchants

You get the good with the bad

Havarie in German means: shipwreck

Awerij in Dutch/Afrikaans means: average, damage to ship or cargo

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Distribution of Stack Heights

Choosing many stacks S=D1 + . . . +D10 of ten disks eachwe get anormalpopulation of stack heights,

with meanE(S) =E(D1) + . . . + E(D10) =10 .125 =1.25,

and standard deviationS=

2D1+ . . . +

2D10

=

10D=

10 .001/3=.00105

thusSranges over

1.25 3 10 .001/3=1.25 10 .001 =1.25 .00316

.00316 =

10 .001 10 .001 =.01

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Normal Histogram/Distribution of Stacks

Histogram of 10,000 Stack Heights

stack height of 10 disks

Density

1.240 1.245 1.250 1.255 1.260

0

100

200

300

400

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Root Sum Square (RSS) Method

ForS=D1 + . . . +D10, with independentdisk thicknessesDi, we have

S=

var(D1 + . . . +D10) =2D1+ . . . +2D10

Interpreting TOLi=TOLDi= 3Di andTOLS=3Swe have

TOLS=3S = 32

D1+ . . . +2D10= (3D1)

2 + . . . + (3D10)2

=

TOL21 + . . . + TOL210=

10 TOLD

S 3Scontains99.73%of theSvalues, becauseSN(S,2S).

This is referred to as theRoot Sum Square (RSS) Methodof tolerance stacking.

Contrast with arithmetic or worst case tolerance stacking

TOLS=TOL1 + . . . + TOL

10=10 TOLD

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Some Comments on Notation

NumericallyTOLi=TOLi are the same, they are just different in what they

represent: statistical variation range versus worst case variation range.

Again,TOLSandTOLSrepresent statistical and worst case variation ranges,

but they are not the same since

TOL21 + . . . + TOL

210=

(TOL1)

2 + . . . + (TOL10)2 TOL1 + . . . + TOL10

We get= only in the trivial cases when n=1 or

when n>1 and TOL1=. . .=TOLn=0.

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S i i l T l i B fi


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Statistical Tolerancing Benefits

= stack height variation is much tighter than specified

could try to relax the tolerances on the disks,

relaxed tolerances= lower cost of part manufacture

take advantage of tighter assembly tolerances= easier assembly

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RSS f G l


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RSS for Generaln

When we stackn disks, replace10 by n above:

TOLS= n TOLD or TOLD= 1n TOLS

As opposed to the worst case tolerancing relationships

TOLS=n

TOLD or TOL

D=

1

n TOLS

More generally when theTOLDi are not all the same

TOLS=

TOL21 + . . . + TOL

2n or TOL

S=TOL

D1

+ . . . + TOLDn

Reverse engineering TOLS TOLDi or TOLS TOLDi not so obvious.

Reduce the largestTOLDi to get greatest impact on TOLS. TOL

Di???

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B d i i


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Benderizing

As much as RSS gives advantages over worst case or arithmetic tolerancing

it was found that the RSS tolerance buildup was often optimistic in practice.

A simple remedy was proposed by Bender (1962) and it was called Benderizing.

It consists in multiplying the RSS expression by 1.5, i.e., use

TOLS=1.5 TOL21 + . . . + TOL2nThis still only grows on the order of

n, but provides a safety cushion.

The motivation? When shop mechanics were asked about the dimension accuracy

they could maintain, they would respond based on experience memory.

It was reasoned that a mechanics experience covers mainly a 2range.To adjustTOLi=2ito TOLi=3ithe factor3/2=1.5was applied.

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U if P t V i ti


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Uniform Part Variation

Suppose that the normal variation does not adequately represent

the variation of the manufactured disks.

Assume that disk thicknesses varyuniformlyover

[nominalTOLD, nominal + TOLD] = [TOLD, + TOLD] due to tool wear.

E(D) = and2D =

Z +TOLDTOLD

1

2TOLD(t)2 dt substituting(t)/TOLD=x

= TOL2D

Z 1

1

1

2x2 dx with dt/TOLD=dx

= TOL2D

x3

6

1

1=TOL2D

13

6 (1)

3

6

=

TOL2D3

= D=TOLD/

3 or 3D=

3 TOLD=cTOLD, c=

3=1.732.

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U if P t V i ti I t TOL


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Uniform Part Variation Impact onTOLS

Forn 3the distribution ofS is approximately normal, i.e.,SN(S,2S)see next slide.

Thus most ( 99.73%) of theSvariation is withinS 3S

TOLS=3S= (3D1)2 + . . . + (3Dn)

2

S=

n D = TOLS=3S=

n3 D=

n

3 TOLD=

ncTOLD,

i.e., we have a uniform distributionpenalty factorc=

3=1.732.

Recall that under normal part variation we had: TOLS=

nTOLD.

Here the inflation factor is motivated differently from Benderizing.

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CLT f S f U if R d V i bl


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CLT for Sums of Uniform Random Variables

U1 ++U2 ++U3

Density

0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.0

0.2

0.4

0.6

0.8

U1 ++++U4

Density

0.5 1.0 1.5 2.0 2.5 3.0 3.5

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

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Uniform Part Variation Comparison ith Worst Case


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Uniform Part Variation: Comparison with Worst Case

Compare this to the worst case tolerancing

TOLD=TOLS

nor TOLS=n TOLD,

TOLS=

3

nTOLD < TOLS=nTOL

D when3


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Motivating the 3 cT LinkBothT and capture the variability/scale of a distribution.

Increasing that scale by a factor should increase andTby that same factor.

Tcaptures (almost) all of the variation range.

is a mathematically convenient scale measure, because of RSS rule.

For a normal distribution 3captures almost all of the variation range.There it makes sense to equateT=3.

For other distributions we need a factor c to make that correspondence T=3/c,i.e., 3/ccaptures (almost) all of the variation in the distribution.

= 3 =cT. The penalty or inflation factorc is found via calculus.39

Distribution Inflation Factors 1


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normal density

c = 1

uniform density

c = 1.732

triangular density

c = 1.225

trapezoidal density: k = .5

c = 1.369

elliptical density

c = 1.5

half cosine wave density

c = 1.306

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Student t density: df = 4

c = 1

Student t density: df = 10

c = 1

beta density

= = 3

c = 1.134

beta density

= = .6

c = 2.023

beta density (parabolic)

= = 2c = 1.342

DIN - histogram density

c = 1.512

p = .7 , g = .4

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Details on Distribution Inflation Factors 1


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The factorsc are chosen such that for finite range densities we have3 D=c TOLD

cnormal=1

Finite range densities can always be scaled to a range[1, 1],except for beta where[0, 1]is the conventional standard interval.

cuniform=

3, ctriangular=

1.5, celliptical=1.5, ccos=3 1 8/

2

ctrapezoidal=

3(1 + k2)/2 where2kis the range of the middle flat part.

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The beta density takes the following form:

f(z) = (a + b)(a)(b)

za1(1 z)b1 for 0 z 1, andg(z) =0 else

Fora=b the beta density is symmetric around.5 cbeta=3/

2a + 1.

The histogram or DIN density takes the following form

f(z) =

p2g for |z| g,

1p2(1g) forg< |z| 1

0 else

cDIN=

3[(1 p)(1 + g) + g2]

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RSS with Mixed Distribution Inflation Factors


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RSS with Mixed Distribution Inflation Factors

Assume that disk thicknessesDihave different tolerance specifications

i TOLi, i=1, . . . , n and with possibly different distribution factorsc1, . . . , cn

Again the stack dimensionS=D1 + . . . +Dn is approximately normally distributed

with mean and standard deviation given by

S=1 + . . . +n and S=

21 + . . . +

2n

By way of3i=ciTOLiwe get for Sthe tolerance rangeS TOLS, where

TOLS=3S=

(31)

2 + . . . + (3n)2 =

(c1TOL1)

2 + . . . + (cnTOLn)2

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Statistical Tolerancing by Simulation


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Statistical Tolerancing by Simulation

Randomly generate part dimensions according to appropriate distributionsover respective tolerance ranges

Calculate the resulting critical assembly dimension, i.e., draw ten thicknessesfrom a distribution of thicknesses and compute the stack height (sum).

Repeat the above many times, Nsim=1000(or Nsim 1000) times.

Form the histogram of the 1000(or more) critical dimensions.

Compare histogram with specified limits on the critical assembly dimension(stack height).

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Statistical Tolerancing by Simulation & Iteration


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Statistical Tolerancing by Simulation & Iteration

If histogram has lots of room within assembly specification or tolerance limits

relax tolerances on the aggregating parts.

If histogram violates assembly specification or tolerance limits significantly,tighten tolerances on the aggregating parts.

Repeat process until satisfied. Opportunity for Experimental Design.

Vectorize part dimension generation= critical dimension generation.

All this can be done on a computer (e.g.,using R) in a matter of seconds

and can save a lot of waste and rework.

There are commercial tools, e.g., VSA

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Is Linear Tolerance Stack Special?


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Is Linear Tolerance Stack Special?

height=thickness1 + . . . + thicknessn or Y=X1 + . . . +Xn

From here it is a little step to Y=a0 + a1 X1 + . . . + an Xn,where a0, a1, . . . , an are known multipliers or coefficients.

They are constant as opposed to the random quantities Xi.

For example, Y=16 + 3 X1 + 2 X2 + 7 X3 + (2)X4

Call X1, . . . ,Xn inputs or input dimensions and Y output dimension.

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Crankcase Tolerance Chain


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Crankcase Tolerance Chain

G

L2

L1

L3

L4

L5

L6

G=L1 L2 L3 L4 L5 L6=L1 (L2 + . . . +L6)

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Input/Output Black Box


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Of more general interest and applicability would be I/O relationsof the following type Y= f(X1, . . . ,Xn)


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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

X1 ...

Xn

Y= f(X1, . . . ,Xn)

fdescribes what you have to do with the inputs Xito arrive at an output Y.

The propagation of variation in theXicauses what variation in the output Y?

49

Smooth Functions f


51/117

Smooth Functions f

When the outputYvaries smoothly with small changes in theXi, then

Y a0 + a1 X1 + . . . + an Xnfor all small perturbations in X1, . . . ,Xnaround1, . . . ,n.

The above approximation forY= f(X1, . . . ,Xn)comes fromthe one-term Taylor expansion of f around1, . . . ,n.

Y= f(X1, . . . ,Xn) f(1, . . . ,n) +n

i=1

f(1, . . . ,n)

i(Xi i)

using

ai=f(1, . . . ,n)

iand a0= f(1, . . . ,n)

n

i=1

f(1, . . . ,n)

ii

50

Good Linearization Example


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Good Linearization Example

x

y

1.0 1.2 1.4 1.6 1.8 2.0

2

3

4

5

6

7

51

Medium Linearization Example


53/117

Medium Linearization Example

x

y

1.0 1.5 2.0 2.5

0

2

4

6

52

Poor Linearization Example


54/117

Poor Linearization Example

x

y

1.0 1.5 2.0 2.5 3.0 3.5 4.0

4

2

0

2

4

6

53

The Sensitivity Coefficients or Derivatives


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The Sensitivity Coefficients or Derivatives

The sensitivity coefficientaican then be determined by calculus or

numerically by experimenting with the black box, making small changes in Xineariwhile holding the otherXs fixed at their s and assessing

the rate of change inYin each case, i.e., for each i=1, . . . , n.

The previous analysis can proceed, once we realize that

2aiXi= a2i2Xi= (aii)

2 and 2a0= 0.

2Y = 2a0+a1X1+...+anXn

= 2a0+2a1X1+ . . . +

2anXn=

a1X1

2+ . . . +

anXn

254

The General Tolerance Stack Formula


56/117

The General Tolerance Stack Formula

and by 3 Xi= ciTXi

(3Y)2 =

3a1X1

2+ . . . +

3anXn

2=

a1c1TX1

2+ . . . +

ancnTXn

2CLT= YN(Y,2Y), i.e., most variation of Yis withinY 3Y

TOLY=3Y=

a1c1TOLX1

2+ . . . +

ancnTOLXn

2

I have seen engineers applying

TOLY=3Y=

TOL21 + . . . + TOL

2n

regardless of theaiandci. RSS was a magic bullet they did not understand.

55

Simulation for General f


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Simulation for General f

Simulation ofY= f(X1, . . . ,Xn)is an option as well.

A normal distribution for the inputs Xiis not essential.

The CLT still gives us normal outputs, most of the time.

The latter depends on the sensitivities/derivatives of fand the relative variations of the inputs.

56

Sensitivities and CLT


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Sensitivities and CLT

Recall the crucial condition forY=X1 + . . . +Xn N(Y,2Y)

max21, . . . ,

2n

21 + . . . +

2n

0, as n

ForY=a0

+ a1X

1+ . . . + a

nX

n N(

Y,2

Y)this translates to

max

a2121, . . . , a

2n

2n

a21

21 + . . . + a

2n

2n

0, as n

A largeaican mess things up, i.e., makea2i

2i dominant.

A smallaican dampen the effect of a large or otherwise dominant 2i.

57

Mean Shifts


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Mean Shifts

So far we have assumed that the distributions of part dimensions

were centered on the middle of the tolerance interval.

Why should there be that much precision in centering when the

actual inputs or part dimensions can be quite variable?

It makes sense to allow for some kind of mean shift or targeting error

while still insisting on having all or most part dimensions within

specified tolerance ranges.

58

Two Strategies of Dealing with Mean Shifts


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Two Strategies of Dealing with Mean Shifts

Two strategies of dealing with mean shifts:

1. stack these shifts in worst case fashion arithmetically

2. stack these shifts statistically via RSS

In either case combine this in worst case fashion or arithmetically with the

RSS part variation stack.

The reason for the last worst case stacking step is that the mean shifts

represent persistent effects that do not get played out independently and

repeatedly for each produced part dimension.

59

Mean Shifts Stacked Arithmetically


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ea S ts Stac ed t et ca y

probability

density

0.6 0.4 0.2 0.0 0.2 0.4 0.6

part dimensionX1

part dimensionX2

part dimension

X3

assembly stack

Y == X1 ++ X2 ++ X3

mean shifts add

in worst case fashion

60

Mean Shifts Stacked via RSS


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probability

density

0.6 0.4 0.2 0.0 0.2 0.4 0.6

part dimensionX1

part dimensionX2

part dimension

X3

assembly stack

Y == X1 ++X2 ++X3

mean shifts add

in RSS fashion

61

Mean Shifts within Tolerance Interval


63/117

For the part variation to stay within tolerance there has to be a tradeoff

between variability and mean shift.

62

Mean Shifts, Variability & Cpk


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, y pk

The capability indexCpkmeasures the distance of the mean to the closest tolerance limit in relation to 3.

If the tolerance interval is given by[L, U]then

Cpk=minU

3 ,

L

3

Cpk= 1 means that we have somewhere between .135% to .27% of partdimensions falling out of tolerance.

However, this does not control the mean shift. We could have U andCpk=1. Then all part dimensions would be nearU= worst case stacking.

63

Bounded Mean Shifts


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Bound the mean shifti, typically as a fraction of the tolerance Ti:

i= iTi 0

i

1.

But maintainCpk 1

iTi + 3i Ti = 3i (1 i)Ti

64

Arithmetically Stacking Mean Shifts


66/117

y g

= Hybrid tolerance stacking formulaarithmeticallycombiningarithmetically combined mean shiftsand

statistical tolerancing

TOLY= 1|a1|TOLX1+ . . . +n|an|TOLXn

+

(1 1)2

a

2

1c

2

1TOL

2

X1+ . . . + (1 n)2

a

2

nc

2

nTOL

2

Xn

This grows on the order ofn and not n, but with a reduction factor.

1=. . .= n=0 = RSS stacking.

1=. . .= n=1 = Worst case arithmetical stacking.

65

RSS Stacking of Mean Shifts


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g

= Hybrid tolerance stacking formula

arithmeticallycombiningRSS combined mean shiftsand

statistical tolerancing

TOLY=21c

21a

21TOL

2X1

+ . . . +2nc2na

2nTOLXn

2

+

(1 1)2

a21c

21TOL

2X1+ . . . + (1 n)

2

a2nc

2nTOL

2Xn

The ciare the penalty factors for the distributions governing the mean shifts.Theciare the penalty factors for the distributions governing part variation.

What is the interpretation of1=. . .= n=1?Consistent part dimensions with system outputY=E(Y) TOLY.

66

Distributions with Mean Shift I


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shifted normal density

c = 1

shifted uniform density

c = 1.732

shifted triangular density

c = 1.225

shifted trapezoidal density: a = .5

c = 1.369

shifted elliptical density

c = 1.5

shifted half cosine wave density

c = 1.306

67

Distributions with Mean Shift II


69/117

shifted Student t density: df = 4

c = 1

shifted Student t density: df = 10

c = 1

shifted beta density

= = 3

c = 1.134

shifted beta density

= = .6

c = 2.023

shifted beta density (parabolic)

= = 2

c = 1.342

DIN - histogram density

p = .7 , g = .4

c = 1.512

68

Other Variants


70/117

So far we have accommodated mean shifts at the price of reduced part

dimension variability in order to maintainCpk 1.

Rather than dividing upTOLinto mean shift and a3range (by squeezing

down3 to maintain Cpk 1) we can increaseTOLto the sum of the originalTOL =3plus the mean shift represented as a fraction of the increasedTOL, i.e.,

TOLi=3i +iTOLi or TOLi= 3i1

i

= TOLi1

i

.

For details on how the stacking formulas change see the provided reports.

69

Actuator


71/117

70

Actuator Case Study


72/117

The following geometric problem arose in an actuator design situation.

In the abstract: we have a triangle with legsA,R and B.

The angle betweenA and R is denoted by.

We have the following tolerance specificationsA

A0

TA andR

R0

TR.

The legB, representing the actuator, can be adjusted such that the angle agrees

exactly with a specified value0.

Once = 0

is achieved the actuator is in its neutral position.

From thereB can extend or contract by an amount thus changing the angleto a maximum and minimum value maxandmin, respectively.

71

The Question of Interest


73/117

A=A0andR=R0= nominal values formaxandmin, denoted bymax,0andmin,0, respectively.

The question of interest is:

How much variation ofmaxandminaroundmax,0andmin,0can we expect

due to the variations inA and R over their respective tolerance ranges

A0 TAandR0 TR?

72

Geometric Considerations


74/117

GivenA,R and 0the length of the (neutral position) actuator length is

B=B(A,R) =

A2 +R2 2AR cos(0).Extending/contracting the actuator byx= from the neutral position

=

x=2 arctan

(sx A)(sx R)sx(

sx

Bx)

,whereBx=B(A,R) +xand sx= (A +R +Bx)/2.

Note that corresponds tomaxandcorresponds tomin.

xis affected byA and R in quite a variety of ways

= max= max(A,R) and min= min(A,R).

73

Statistical Tolerancing via Simulation


75/117

The simplest way of dealing with the variation behavior of= maxand

= mindue to variation inA and R is through simulation =

R.

GetN-vectors ofA and R values from N(A, (TA/3)2)and N(R, (TR/3)

2).

Calculate the correspondingly adjustedB=B(A,R)vector and from that

theN-vectors ofmaxandmin, respectively.

HereA=A0,R=R0andA=TA/3,R=TR/3normal distribution.

The results usingN= 1, 000, 000simulations is shown on the next slide.

It usedtheta.simNNand took just a few seconds to run.

Vertical bars on either side of the histograms = estimated 3 = T limits.

It is easy to change the distributions describing the variation in A and R.

74

(A,R) (N,N)Simulation Output, Nsim=106


76/117

( , ) (N ,N ) p , sim

max 0

Den

sity

15.0 15.2 15.4 15.6 15.8

0.0

1.0

2.0

3.0 T1 ==0.359

omax,, 0 0 ==15.325

o

0 ==55o

min 0

Density

16.6 16.4 16.2 16.0 15.8 15.6 15.4

0.0

0.5

1.0

1.5

2.0

2.5

T2 ==0.467o

min,, 0 0 ==15.999o

0 ==55o

75

Statistical Tolerancing via RSS


77/117

T1=a2max,A T2A+ a

2max,R

T2R and T2=a2min,A

T2A+ a2min,R

T2R,

where

amax,A=maxA

, amax,R=maxR

, amin,A=minA

, and amin,R=minR

All derivatives are evaluated at the nominal values(A0,R0)of (A,R).

These RSS formulae come from the linearization ofx(A,R)near(A0,R0), i.e.,

x(A,R) = x(A0,R0) + (AA0)x

AA=A0,R=R0 + (RR0)

x

RA=A0,R=R0 ,

which is then taken as an approximation forx(A,R)near(A,R) = (A0,R0).

76

Approximation Quality


78/117

The approximation quality depends on the smoothness of the functionx

with respect toA andR at (A0,R0).

The approximation quality also depends on the tolerances TAandTR.

TAandTRdetermine over what rangexis approximated.

WhenTAorTRget too large, quadratic terms may come into play normality???

All this assumes of course that x is differentiable near(A,R) = (A0,R0).

There are tolerance situation where differentiability is an issue and in that case the

RSS paradigm does not work.

77

The Derivatives


79/117

x

A

= 1

1 +(sx

A)(sx

R)

sx(sxBx)

A

(sx A)(sx R)

sx(sx Bx)and

x

R=

1

1 +(sxA)(sxR)

sx(sxBx)

R

(sx A)(sx R)

sx(sx Bx) .

Next we have

A

(sx A)(sx R)

sx(sx Bx) =

2

(sx A)(sx R)

sx(sx Bx)

1

A

(sx A)(sx R)sx(sx Bx)

and

R

(sx A)(sx R)

sx(sx Bx) =2

(sx A)(sx R)

sx(sx Bx)

1

R

(sx A)(sx R)sx(sx Bx) .

78

More Derivatives


80/117

We also have the following list of derivative expressions

BxA

= A R cos(0)A2 +R2 2AR cos(0)

and BxR

= R A cos(0)A2 +R2 2AR cos(0)

(sx A)A

=1

2

A R cos(0)

B 1

and

(sx R)A

=1

2

A R cos(0)

B+ 1

(sx A)R

=1

2

R A cos(0)

B+ 1

and

(sx R)R

=1

2

R A cos(0)

B 1

sxA

=1

2

A R cos(0)

B+ 1

and

sxR

=1

2

R A cos(0)

B+ 1

(sx Bx)A

=1

2

1 A R cos(0)

B

and

(sx Bx)R

=1

2

1 R A cos(0)

B

.

79

And More Derivatives


81/117

A

(sx A)(sx R)sx(sx

Bx)

= 1

s2x(sx Bx)2

(sx R) A

(sx A) + (sx A) A

(sx R)

sx(sx Bx)

(sx A)(sx R)

(sx Bx) A

sx + sx

A(sx Bx)

R

(sx A)(sx R)sx(sx Bx)

= 1

s2x(sx Bx)2 (sx R)

R

(sx

A) + (sx

A)

R

(sx

R)sx(sx Bx)(sx A)(sx R)

(sx Bx)

Rsx + sx

R(sx Bx)

.

80

And More Derivatives


82/117

Rather than just using these expressions as they are it is advisable to simplify them

somewhat to avoid significance loss in the calculations.

Thus we obtained the following reduced expressions:

(sx

R)

A(sx

A) + (

sx

A)

A(sx

R) =

R

2 [1

cos(

0)] +

x

2B[A

R cos

(

0)]

(sx Bx) A

sx + sx

A(sx Bx) = R

2[1 + cos(0)]

x

2B[A R cos(0)]

(sx

R)

R

(sx

A) + (sx

A)

R

(sx

R) =

A

2

[1

cos(0)] +

x

2B

[R

A cos(0)]

(sx Bx) R

sx + sx

R(sx Bx) = A

2[1 + cos(0)]

x

2B[R A cos(0)].

81

RSS Calculations


83/117

The R function deriv.theta produced the following derivatives for A0= 12.8,

R0=6,0=55, and =1.6

maxA

= .00006636499 and minA

= .004038650and

maxR

= 0.04473785 and minR

=0.05810921.

The RSS calculation using normal variation for A and R then gives the following

values forT1andT2based onTA=.12and TR=.14

T1=0.3588609 and T2=0.4669441,

which agree remarkably well with the simulated quantities.

The derivatives ofmaxandminwith respect toA are smaller than

the derivatives with respect to R by at least an order of magnitude.

Important when considering other distributions governing the variation of A and R.

82

Numerical Differentiation


84/117

The derivation of the derivatives was quite laborious, but R code is compact.

Useful in understanding the variation propagation in the tolerance analysis.

An obvious alternative approach is numerical differentiation.

It requires the evaluation of the function x, used in the simulation anyway.

The respective derivatives are approximated numerically at(A,R) = (A0,R0)by

difference quotients for very small values of

xA

A=A0,R=R0

x(A0 +,R0)x(A0,R0)

xR

A=A0,R=R0

x(A0,R0 +)x(A0,R0)

.

83

Numerical Differentiation Example


85/117

For =.00001the R functionderiv.numericgives

maxA

A=A0,R=R0

.00006636269 and minA

A=A0,R=R0

.004038651

and

maxR

A=A0,R=R0

0.04473777 and minR

A=A0,R=R0

0.05810908 .

These agree very well with the derivatives obtained previously via calculus.

84

Revisit RSS for Linear Combinations


86/117

A linear combinationYof independent, normal variation terms Xi

Y=a0 + a1X1 + . . . + anXn with known constantsa0, a1, . . . , an,

is normally distributed.

Most of the Yvariation falls within 3Yof its meanY= a0 + a1X1+ . . . + anXn.

2Y= 2a1X1

+ . . . +2anXn= a21

2X1

+ . . . + a2nXn2 .

ForXi N equate3Xi= Ti, i.e., most of theXivariation falls withini 3Xi

= general RSS tolerance stacking formula

TY=3Y=

a21(3X1)2 + . . . + a2n(3Xn)

2 =

a21T2

1 + . . . + a2nT

2n

applicable for linear approximations to smooth functions of normal inputs.

85

CLT and Adjustment Factors


87/117

Y=a0 + a1X1 + . . . + anXn with known constantsa0, a1, . . . , an,

is approximately normally distributed provided

max

a21

2X1

a212

X1+ . . . + a2n

2Xn

, . . . ,a2n

2Xn

a212

X1+ . . . + a2n

2Xn

is small,

i.e., none of the a2i 2i terms dominates the others.

Making use of adjustment factors, chosen such that 3i=ciTi, get

TY=3Y=

a21(3X1)

2 + . . . + a2n(3Xn)2 =

c21a

21T

21 + . . . + c

2na

2nT

2n .

applicable for linear approximations to smooth functions of any random inputs,subject to above CLT condition.

TheTishould be small for linearization to be reasonable.

86

Simulations with Other Distributions forAandR


88/117

The next few slides show simulations with0=55and =1.6and

(A,R) (U(12.8 .12, 12.8 + .12),N(6, (.14/3)2)usingsim.thetaUN

(A,R) (N(12.8, (.12/3)2),U(6 .14, 6 + .14))usingsim.thetaNU

(A,R) (U(12.8 .12, 12.8 + .12),U(6 .14, 6 + .14))usingsim.thetaUU

(A,R) (U(12.8.012, 12.8+.012),U(6.014, 6+.014)) using sim.thetaUU

87

(A,R) (U,N)Simulation Output, Nsim=1064


89/117

max 0

Density

15.0 15.2 15.4 15.6 15.8

0

1

2

3

4

T1 ==0.359o

max,, 0 0 ==15.325o

0 ==55o

min 0

Density

16.6 16.4 16.2 16.0 15.8 15.6 15.4

0.0

1.0

2.0

3.0

T2 ==0.469o

min,, 0 0 ==15.999o

0 ==55o

88

(A,R) (N,U)Simulation Output, Nsim=106


90/117

max 0

Density

14.5 15.0 15.5 16.0

0.0

0.5

1.0

1.5

2.0

T1 == 0.622o

max,, 0 0 == 15.325o

0 == 55o

min 0

Density

17.0 16.5 16.0 15.5 15.0

0.0

0.5

1.0

1.5

2.0

T2 == 0.809o

min,, 0 0 == 15.999o

0 == 55o

89

(A,R) (U,U)Simulation Output, Nsim=106


91/117

max 0

Density

14.5 15.0 15.5 16.0

0.0

0.5

1.0

1.5

2.0

T1 == 0.622o

max,, 0 0 == 15.325o

0 == 55o

min 0

Density

17.0 16.5 16.0 15.5 15.0

0.0

0.5

1.0

1.5

2.0

T2 == 0.81o

min,, 0 0 == 15.999o

0 == 55o

90

(A,R) (U,U)Simulation Output, Nsim=106


92/117

max 0

Density

15.25 15.30 15.35 15.40

0

5

10

15

20

T1 == 0.0622o

max,, 0 0 == 15.325o

0 == 55o

min 0

Density

16.10 16.05 16.00 15.95 15.90

0

5

10

15

20

T2 == 0.0809o

min,, 0 0 == 15.999o

0 == 55o

91

RSS Calculation with Inflation Factors


93/117

Applying the RSS formula assuming a uniform distribution for both A and R we get

T1=

(.00006636269)2

3 .122

+ (.04473777)2

3 .142

360

2 = 0.6215642and

T2=

(.004038651)2 3 .122 + (.05810908)2 3 .142 360

2 =0.8087691

using the inflation factor c=

3and the numerical derivatives in both cases.

Reasonable agreement with the values.622and.81from simulation.

Not surprising when linearization is good. We are simply using the variance rules.

However, T1andT2do not capture the variation range of x, since the CLT fails.

Tightening the tolerances in last case= echoes the uniform distribution ofR.Linearity was not good with wider tolerances= tilted uniform.

92

theta.simUUUU


94/117

Here we let 4 inputs vary with result shown on next slide.

A U(12.8 .22, 12.8 + .22)

R

U(6

.15, 6 + .15)

U(1.6 .05, 1.6 + .05)

0

U(55

.5, 55 + .5)

Try other tolerances in these uniform distributions.

93

VaryingA,R,0andUniformly


95/117

max 0

Density

14 15 16 17

0.0

0.2

0.4

0.6

0.8

1.0

T1 == 1.38o

max,, 0 0 == 15.325o

0 == 55o

min 0

Densit

y

18 17 16 15 14

0.0

0.2

0.4

0.6

0.8

1.0

T2 == 1.59o

min,, 0 0 == 15.999o

0 == 55

o

94

Final Comments


96/117

This actuator example has been very instructive. It showed

the importance of dominant variability by a single input

the effect of the CLT when sufficiently many contributing inputs are involved

the importance of simulation

the importance of derivatives

the effect of the variability ranges on the linearization approximation quality.

95

Voltage Amplifier


97/117

96

Output Voltage V0


98/117

The amplified output voltage is a function of 6 variables,

2 input voltagesE1, andE2and 4 resistances R1, . . . ,R4

V0= f(E1,E2,R1,R2,R3,R4) =

E1

1 +R2R1

1 +R3R4

E2

R2R1

Nominal values:

E1=1V,E2= 1V,R1=10,R2=100,R3=10, andR4=100.

= V0=20V.97

The Derivatives


99/117

V0

E1 =

1 +R2R1

1 +R3R4 ,

V0

E2 = R2

R1

V0

R1=

E1

1 +R3R4 R2

R21

+E2 R2

R21

, V0

R2=

E1R1

1 +R3R4 E2

R1

V0

R3

=

E1

1 +R2R11 +R3R4

2 1

R4

, V0

R4

=E1

1 +R2R1

1 +R3R42

R3

R24

98

V.amp.simN2U4(del=.1)


100/117

> V.amp.simN2U4(del=.1)

$V0

[1] 20

$delta

[1] 0.1

$derivatives

[1] 10.000000000 -10.000000000 -1.909090909 0.190909091

+ -0.090909091 0.009090909

$sigmas

[1] 0.33314890 0.33326565 1.10195837 1.10243208

+ 0.05248074 0.05246410

$nominals

[1] 1 -1 10 100 10 100

99



101/117

V0

Density

14 16 18 20 22 24 26

0.0

0

0.0

5

0.1

0

0.1

5

0.2

0

0.2

5

Ei~N((i,, ((i))2)), Ri~U((i i , i ++ i))

== 0.1

100



102/117

> V.amp.simN2U4(del=.05)

$V0

[1] 20

$delta

[1] 0.05

$derivatives

[1] 10.000000000 -10.000000000 -1.909090909 0.190909091+ -0.090909091 0.009090909

$sigmas

[1] 0.16657056 0.16676230 0.55079156 0.55108759

+ 0.02627634 0.02624854

$nominals

[1] 1 -1 10 100 10 100

101



103/117

V0

Density

17 18 19 20 21 22 23

0.0

0.1

0.2

0.3

0.4

0.5

Ei~N((i,, ((i))2)), Ri~U((i i , i ++ i))

== 0.05

102

V.amp.simU6(del=.1)


104/117

> V.amp.simU6(del=.1)

$V0

[1] 20

$delta

[1] 0.1

$derivatives

[1] 10.000000000 -10.000000000 -1.909090909 0.190909091+ -0.090909091 0.009090909

$sigmas

[1] 0.57739282 0.57698360 1.10221137 1.10199967

+ 0.05251682 0.05253420

$nominals

[1] 1 -1 10 100 10 100

103

V.amp.simU6(del=.1)


105/117

V0

Densit

y

14 16 18 20 22 24 26

0.0

0

0.0

5

0.1

0

0.1

5

0.2

0

Ei~U((i i , i ++ i)) , Ri~U((i i , i ++ i))

== 0.1

104

V.amp.simN6(del=.1)


106/117

> V.amp.simN6(del=.1)

$V0

[1] 20

$delta

[1] 0.1

$derivatives

[1] 10.000000000 -10.000000000 -1.909090909 0.190909091

+ -0.090909091 0.009090909

$sigmas

[1] 0.33348276 0.33332256 0.63653780 0.63714909

+ 0.03031808 0.03029352

$nominals

[1] 1 -1 10 100 10 100

105

V.amp.simN6(del=.1)


107/117

V0

Density

16 18 20 22

0.0

0.1

0.2

0.3

0.4

Ei~N((i,, ((i))2)), Ri~N((i,, ((i))

2))

== 0.1

106

V.amp.simN6(del=.05)


108/117

> V.amp.simN6(del=.05)

$V0

[1] 20

$delta

[1] 0.05

$derivatives

[1] 10.000000000 -10.000000000 -1.909090909 0.190909091

+ -0.090909091 0.009090909

$sigmas

[1] 0.16656830 0.16669687 0.31840687 0.31774622

+ 0.01514106 0.01513453

$nominals

[1] 1 -1 10 100 10 100

107

V.amp.simN6(del=.05)


109/117

V0

Density

18 19 20 21

0.0

0.2

0.4

0.6

0.8

Ei~N((i,, ((i))2)), Ri~N((i,, ((i))

2))

== 0.05

108

Some Final Comments


110/117

R3andR4appear to have negligible effect.

Normal variations on all 6 inputs produce approximately normal V0distributions.

The linearizations appears to be a mild issue here.

Ei N andRi Ushow much stronger deviations from normality,but not too bad as far as the TV0= 3V0 range is concerned.Distributions appear nearly triangular, because of dominance of R1andR2.

ForEi UandRi U the distribution seems similar to previous case.The main termsR1andR2are not as dominant compared to E1andE2.

109

References on Statistical Tolerancing


111/117

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