Coordinate Dimensioning Table of Contents Return to the Previous Slide Slide 1QuitMaster Table of ContentsGlossary 86 57.6 56.6 100 99 9.6 9.4 25.5 25.4

Coordinate Dimensioning Table of ContentsReturn to the Previous Slide Slide 1 QuitMaster Table of ContentsGlossary

86

57.656.6

100 99

9.69.4

25.525.4

3634

41.341.1

20.00 - 20.13

0.14 A C

B

M42 X 1.5 - 6g

0.1 M B M

0.06 A

6.66.1

9.69.4

44.6044.45

0.08 M A

C

A

0.02

31.831.6

0.1 B

8X 7.9 - 8.1

8X 45°

0.14 M A C M

44Engineering Graphics PrinciplesEngineering Graphics Principles

CoordinateCoordinateDimensioning andDimensioning and

TolerancingTolerancing

UnitUnit


86

57.656.6

100 99

9.69.4

25.525.4

3634

41.341.1

20.00 - 20.13

0.14 A C

B

M42 X 1.5 - 6g

0.1 M B M

0.06 A

6.66.1

9.69.4

44.6044.45

0.08 M A

C

A

0.02

31.831.6

0.1 B

8X 7.9 - 8.1

8X 45°

0.14 M A C M

COORDINATE DIMENSIONING and

TOLERANCINGBasic Hole, Basic Shaft

and Using Tables

COORDINATE DIMENSIONING and

TOLERANCINGBasic Hole, Basic Shaft

and Using Tables

ENGINEERINGENGINEERING GRAPHICSGRAPHICS


Calculating Mating Part FitsCalculating Mating Part Fits

•While there may be several ways to calculate the dimensions and tolerances for While there may be several ways to calculate the dimensions and tolerances for mating part fits, they all eventually narrow down to essentially one concept—functionmating part fits, they all eventually narrow down to essentially one concept—function . . The major questions deal with the purpose to be served by the assembly of parts that fit The major questions deal with the purpose to be served by the assembly of parts that fit together. What it is supposed to do, and how the elements in the assembly are supposed together. What it is supposed to do, and how the elements in the assembly are supposed to perform in relation to one another. Answering these questions regarding mating part to perform in relation to one another. Answering these questions regarding mating part fits, greatly diminishes the challenge of determining the appropriate values for fits, greatly diminishes the challenge of determining the appropriate values for dimensions and tolerances. dimensions and tolerances.

We will explore the applications of the principles of the Basic Hole and Basic Shaft We will explore the applications of the principles of the Basic Hole and Basic Shaft systems. They can be extremely helpful in instances where the basic parameters are systems. They can be extremely helpful in instances where the basic parameters are known and the designer must establish the values on the drawing. It should be fun for known and the designer must establish the values on the drawing. It should be fun for you to try on your own.you to try on your own.


BASIC HOLE SYSTEMBASIC HOLE SYSTEM


Tolerances—Basic Hole and Shaft SystemsTolerances—Basic Hole and Shaft Systems

•To calculate a fit between mating parts, and place the dimension values on the drawing, To calculate a fit between mating parts, and place the dimension values on the drawing, the first thing you have to know is the purpose for fitting the parts together. There are the first thing you have to know is the purpose for fitting the parts together. There are standard fit tables that identify many functions that are served by mating part standard fit tables that identify many functions that are served by mating part assemblies. assemblies.

•Knowing the function, you can identify the nominal size and the limit tolerance Knowing the function, you can identify the nominal size and the limit tolerance range of the mating parts.range of the mating parts. Let’s presume a half-inch shaft fitting into a half-inch hole. Let’s presume a half-inch shaft fitting into a half-inch hole. Next, you must determine the intentional difference between the mating parts—positive Next, you must determine the intentional difference between the mating parts—positive or negative allowance—along with the tolerance values for both the shaft and hole or negative allowance—along with the tolerance values for both the shaft and hole diameters. diameters. All of these values are available in the standard fit tablesAll of these values are available in the standard fit tables . When these . When these parameters are known, you can employ what is called the parameters are known, you can employ what is called the Basic Hole SystemBasic Hole System or or Basic Basic Shaft SystemShaft System to compute the dimension values and the appropriate tolerances to be to compute the dimension values and the appropriate tolerances to be assigned to each of the individual part drawings.assigned to each of the individual part drawings.

•The Basic Hole System is a procedure for calculating and placing mating part The Basic Hole System is a procedure for calculating and placing mating part dimensions and tolerance information on the drawing. If followed, you can know ahead dimensions and tolerance information on the drawing. If followed, you can know ahead of time exactly what the tolerance accumulation, or “build-up” will be under any of time exactly what the tolerance accumulation, or “build-up” will be under any increment of size variation in either of the mating parts. You can also be assured that if increment of size variation in either of the mating parts. You can also be assured that if the parts are manufactured according to the specifications on the drawing, they will fit the parts are manufactured according to the specifications on the drawing, they will fit together and function as intended.together and function as intended.


Calculating Standard and Non-Standard Fits Calculating Standard and Non-Standard Fits (32a)(32a)

The Basic Hole and Basic Shaft systems are excellent tools to have in your tool box. Most authors The Basic Hole and Basic Shaft systems are excellent tools to have in your tool box. Most authors will wax eloquent (and endlessly) about these procedural tools. I have tried to reduce them down to will wax eloquent (and endlessly) about these procedural tools. I have tried to reduce them down to three simple steps. Four basic parameters must be known to use these systems:three simple steps. Four basic parameters must be known to use these systems: Basic Size, Basic Size, Allowance, Hole Tolerance, and Shaft ToleranceAllowance, Hole Tolerance, and Shaft Tolerance . .

– Basic size is the hole or shaft at MMC. All tolerance values are generated from that Basic size is the hole or shaft at MMC. All tolerance values are generated from that value. value.

– Allowance is the minimum clearance or maximum interference. It can be positive Allowance is the minimum clearance or maximum interference. It can be positive (clearance) or negative (interference). (clearance) or negative (interference).

BASIC HOLE SYSTEM – BASIC HOLE SYSTEM – Allowance based on the hole at MMCAllowance based on the hole at MMC

1. The basic size is always equal to the 1. The basic size is always equal to the minimumminimum holehole size (Hole MMC) size (Hole MMC)

2. 2. Subtract the allowanceSubtract the allowance from the basic size to get the from the basic size to get the maximum shaftmaximum shaft size (Shaft MMC) size (Shaft MMC)

3. Apply the appropriate tolerances to the hole and to the shaft (Hole & Shaft LMC)3. Apply the appropriate tolerances to the hole and to the shaft (Hole & Shaft LMC)

BASIC SHAFT SYSTEM – BASIC SHAFT SYSTEM – Allowance base on the shaft at MMCAllowance base on the shaft at MMC

1. Basic size is always equal to the 1. Basic size is always equal to the maximum shaftmaximum shaft size size

2. 2. Add the allowanceAdd the allowance to the basic size to get the to the basic size to get the minimum holeminimum hole size size

3. Apply the appropriate tolerances to the shaft and the hole 3. Apply the appropriate tolerances to the shaft and the hole


Basic Hole SystemBasic Hole System

The basic size is determined by converting the nominal size into its decimal The basic size is determined by converting the nominal size into its decimal equivalent. Remember, the basic hole system mandates that the basic size must be equivalent. Remember, the basic hole system mandates that the basic size must be equal to the minimum hole size. When you have converted the value to a decimal equal to the minimum hole size. When you have converted the value to a decimal equivalent, place the value as the low limit for the hole size.equivalent, place the value as the low limit for the hole size.

Basic Size =Basic Size = . . 9/16” (.5625)9/16” (.5625) Allowance =Allowance = .0006 .0006Shaft Tolerance =Shaft Tolerance = -.0009-.0009Hole Tolerance = Hole Tolerance = +.0010+.0010



Basic Size =Basic Size = .5625 .5625Allowance =Allowance = .0006 .0006Shaft Tolerance =Shaft Tolerance = -.0009-.0009Hole Tolerance = Hole Tolerance = +.0010+.0010

The basic hole system always establishes the basic size as the minimum hole size. The basic hole system always establishes the basic size as the minimum hole size. (Remember, the smallest size limit (Remember, the smallest size limit alwaysalways goes on the bottom in a stacked limits form.) goes on the bottom in a stacked limits form.)

.5625




Subtract the Subtract the allowanceallowance (the intentional difference between mating parts) from the (the intentional difference between mating parts) from the basic size to obtain the maximum shaft size (Maximum Material Condition). basic size to obtain the maximum shaft size (Maximum Material Condition). Place the value as the upper limit of the shaft dimension.Place the value as the upper limit of the shaft dimension.

.5625

Basic SizeAllowance

Max Shaft .5625 -.0006




Subtract the Subtract the allowanceallowance (the intentional difference between mating parts) from the (the intentional difference between mating parts) from the basic size to obtain the maximum shaft size (Maximum Material Condition). basic size to obtain the maximum shaft size (Maximum Material Condition). Place the value as the upper limit of the shaft dimension.Place the value as the upper limit of the shaft dimension.

.5625.5619

.5625–.0006 .5619

Basic SizeAllowanceShaft MMC

Shaft MMC




Apply the Apply the tolerancestolerances to both the hole and the shaft to complete the dimensions; to both the hole and the shaft to complete the dimensions; addadd the hole tolerance to the hole lower limit, and the hole tolerance to the hole lower limit, and subtractsubtract the shaft tolerance the shaft tolerance from the shaft upper limit. Do the arithmetic and apply the values.from the shaft upper limit. Do the arithmetic and apply the values.

.5625.5619

Shaft Hole .5621 .5625– .0009 +.0010

.5625–.0006 .5619

Shaft MMC




Apply the Apply the tolerancestolerances to both the hole and the shaft to complete the dimensions; to both the hole and the shaft to complete the dimensions; addadd the hole tolerance to the hole lower limit, and the hole tolerance to the hole lower limit, and subtractsubtract the shaft tolerance the shaft tolerance from the shaft upper limit. Do the arithmetic and apply the values.from the shaft upper limit. Do the arithmetic and apply the values.

.5635

.5625.5619.5612

Shaft Tol Hole Tol .5621 .5625– .0009 +.0010 .5612 .5635

-.0006 -.0013

+.0010 -.0000

.5625 -.0006 .5619

Shaft MMC

.5625 .5625Values Referenced to Basic Size:



Basic Size =Basic Size = 1-1/8” (1.125) 1-1/8” (1.125)Allowance =Allowance = .0025.0025Shaft Tolerance =Shaft Tolerance = .0015.0015Hole Tolerance = Hole Tolerance = .0020.0020

STOP!STOP! Do NOT advance to the next slide. Let’s have you solve a problem on your own. Retrieve Do NOT advance to the next slide. Let’s have you solve a problem on your own. Retrieve problem problem #38#38 in your book, and, using the basic hole system, apply the dimensions on the drawing. Use stacked limits in your book, and, using the basic hole system, apply the dimensions on the drawing. Use stacked limits when placing the information on the sheet. Then, when you have finished, check your work against the next when placing the information on the sheet. Then, when you have finished, check your work against the next slide.slide.

Shaft Hole



Nominal Size = Nominal Size = 1-1/8”1-1/8”Basic Size (BS) =Basic Size (BS) = 1.12501.1250Allowance (A) =Allowance (A) = .0025.0025Shaft Tolerance (ST) =Shaft Tolerance (ST) = .0015.0015Hole Tolerance (HT) = Hole Tolerance (HT) = .0020.0020

How did you do? Are your dimension values the same as these? Are you comfortable with How did you do? Are your dimension values the same as these? Are you comfortable with the process? If not, go back to the beginning of this topic and review the major concepts. the process? If not, go back to the beginning of this topic and review the major concepts. Understanding this procedure will assist you as we move into the next concept.Understanding this procedure will assist you as we move into the next concept.

1.12701.1250

1.12251.1210

Shaft Hole 1.1225 1.1250 Basic Size – .0015 -.0025 Allowance 1.1210 1.1225 Max Shaft


BASIC SHAFT SYSTEMBASIC SHAFT SYSTEM



Basic Shaft SystemBasic Shaft SystemSTOP!STOP! Do NOT proceed to the next slide. Using the Do NOT proceed to the next slide. Using the Basic Shaft SystemBasic Shaft System and the following data, complete the and the following data, complete the drawing by placing the appropriate dimensions on both the shaft and the hole. Use the drawing by placing the appropriate dimensions on both the shaft and the hole. Use the stacked limitsstacked limits expression expression for the dimensions. (Relying on your memory of the rules for application of the Basic Shaft System, complete for the dimensions. (Relying on your memory of the rules for application of the Basic Shaft System, complete the problem before advancing through the next series of slides. Usethe problem before advancing through the next series of slides. Use Problem #39 Problem #39 in your book.)in your book.)

Shaft stays at nominal size and hole adjusts for MMC allowanceShaft stays at nominal size and hole adjusts for MMC allowance


Basic Shaft SystemBasic Shaft System


The basic shaft system establishes the basic size as the maximum shaft size. Place The basic shaft system establishes the basic size as the maximum shaft size. Place the value as the upper shaft limit in the stacked orientation for limits dimensioning.the value as the upper shaft limit in the stacked orientation for limits dimensioning.(Remember, the upper size limit (Remember, the upper size limit alwaysalways goes on top in a stacked expression.) goes on top in a stacked expression.)



Nominal Size = Nominal Size = 9/16” 9/16”Basic Size =Basic Size = .5625 .5625Allowance =Allowance = .0004 .0004Shaft Tolerance =Shaft Tolerance = -.0008-.0008Hole Tolerance = Hole Tolerance = +.0010+.0010

The basic shaft system establishes the basic size as the maximum shaft size. Place the The basic shaft system establishes the basic size as the maximum shaft size. Place the value as the upper shaft limit (MMC) in the stacked orientation for limits dimensioning. value as the upper shaft limit (MMC) in the stacked orientation for limits dimensioning. (Remember, the upper size limit (Remember, the upper size limit alwaysalways goes on top in a stacked tolerance expression.) goes on top in a stacked tolerance expression.)

.5625




.5625

The lower limit of the hole size (MMC) is obtained by The lower limit of the hole size (MMC) is obtained by addingadding the allowance to the the allowance to the basic size. Do the addition and place the dimension as the lower limit of the hole size.basic size. Do the addition and place the dimension as the lower limit of the hole size.

.5625 +.0004




.5625.5629

.5625 +.0004 .5629

The lower limit of the hole size (MMC) is obtained by The lower limit of the hole size (MMC) is obtained by addingadding the allowance to the the allowance to the basic size. Do the addition and place the dimension as the lower limit of the hole size. basic size. Do the addition and place the dimension as the lower limit of the hole size. The process is completed by applying the tolerances to the respective components by The process is completed by applying the tolerances to the respective components by subtracting the shaft tolerance from the shaft upper limit, and adding the hole tolerance subtracting the shaft tolerance from the shaft upper limit, and adding the hole tolerance to the hole dimension lower limit. Apply your results to the drawing.to the hole dimension lower limit. Apply your results to the drawing.



Nominal Size = Nominal Size = 9/16”9/16”Basic Size =Basic Size = .5625 .5625Allowance =Allowance = .0004 .0004Shaft Tolerance =Shaft Tolerance = -.0008-.0008Hole Tolerance = Hole Tolerance = +.0010+.0010

The dimensioning process of the mating parts is now complete. Do you understand The dimensioning process of the mating parts is now complete. Do you understand the procedures that we followed? If you don’t, review them again from the the procedures that we followed? If you don’t, review them again from the beginning of this series of slides.beginning of this series of slides.

.5625

.5617.5639.5629

Shaft Hole .5625 .5629–.0008 +.0010 .5617 Shaft LMC .5639 Hole LMC




The dimensioning of the mating parts is completed by applying the tolerances to the The dimensioning of the mating parts is completed by applying the tolerances to the respective components. respective components. SubtractSubtract the shaft tolerance from the upper limit, and the shaft tolerance from the upper limit, and addadd the hole tolerance to the lower limit. Apply them to the drawing.the hole tolerance to the lower limit. Apply them to the drawing.

.5625

.5617.5639.5629


.5635

.5625.5619.5612


.5625

.5617.5639.5629


Basic Hole Compared to Basic Shaft SystemBasic Hole Compared to Basic Shaft System

The same basic size (.5625)The same basic size (.5625)was used for both calculations.was used for both calculations.Compare the two results forCompare the two results foractual sizes of the shaft and actual sizes of the shaft and the hole.the hole.


STANDARD FITSSTANDARD FITS


American National Standard Fit TablesAmerican National Standard Fit Tables

• Tremendous advantages have been gained by standardizing fits, and publishing them Tremendous advantages have been gained by standardizing fits, and publishing them in tables. They are the means for establishing tolerances for any given size, type, or in tables. They are the means for establishing tolerances for any given size, type, or class of fit. Each of the possible fit options within the tables are prescribed class of fit. Each of the possible fit options within the tables are prescribed according to their intended function. according to their intended function.

• If the designer or engineer knows the functional requirements for the fit, a match can If the designer or engineer knows the functional requirements for the fit, a match can easily be found that conforms to the specific need. The standard provides choices for easily be found that conforms to the specific need. The standard provides choices for a series of types and classes of fits on a unilateral hole basis such that the fit a series of types and classes of fits on a unilateral hole basis such that the fit produced by mating parts in any one class will produce approximately similar produced by mating parts in any one class will produce approximately similar performance throughout the entire range of sizes. performance throughout the entire range of sizes.

• The standard fit tables are consistent with the The standard fit tables are consistent with the basic hole systembasic hole system. It is important that . It is important that you understand how the tables themselves can be used to satisfy functional you understand how the tables themselves can be used to satisfy functional specifications and provide dimensional data to be used on engineering drawings. specifications and provide dimensional data to be used on engineering drawings.

• We will examine a portion of only one page of the tables, and use the information to We will examine a portion of only one page of the tables, and use the information to solve a simple problem dealing with a specific class of running and sliding fit. You solve a simple problem dealing with a specific class of running and sliding fit. You may want to review the example application several times to assure yourself that you may want to review the example application several times to assure yourself that you understand the procedure.understand the procedure.


American National Standard Fit TablesAmerican National Standard Fit Tables

•The tables reflecting standard fits between mating parts may be found in virtually any The tables reflecting standard fits between mating parts may be found in virtually any design handbook, Machinery’s Handbook, and in any reputable drafting manual.design handbook, Machinery’s Handbook, and in any reputable drafting manual.

• Standard fits fall into 3 major Standard fits fall into 3 major categoriescategories and 5 basic and 5 basic typestypes of fits. Within each of these of fits. Within each of these divisions there are several divisions there are several classificationsclassifications of fits. The first category, of fits. The first category, Running Running andand sliding sliding fits, for instance, has nine different classifications of this specific type of fit. The second fits, for instance, has nine different classifications of this specific type of fit. The second category, category, Location Location fits, is a broad category of standardized fits, within which there are 3 fits, is a broad category of standardized fits, within which there are 3 major sub-categories: major sub-categories: clearance,clearance, transition,transition, and and interferenceinterference fits. Finally, there is a fits. Finally, there is a category for category for InterferenceInterference, or , or ForceForce fits consisting of 5 different classifications. fits consisting of 5 different classifications.

The five divisions of standard fits are designated by letters:The five divisions of standard fits are designated by letters:

CategoryCategory TypeType ClassificationClassification

Sliding Sliding RC = Running/Sliding (Rotational or Reciprocal Motion) Clearance Fits RC = Running/Sliding (Rotational or Reciprocal Motion) Clearance Fits (RC 1 - RC 9)(RC 1 - RC 9)

LC = Location Clearance Fits LC = Location Clearance Fits (LC 1 - LC 11) (LC 1 - LC 11)

LocationLocation LT = Location/Transition Fits (Clearance through Interference) LT = Location/Transition Fits (Clearance through Interference) (LT 1 - LT 6) (LT 1 - LT 6)

LN = Location Interference Fits LN = Location Interference Fits (LN 1 - LN 3) (LN 1 - LN 3)

ForceForce FN = Force or Shrink Fits FN = Force or Shrink Fits (FN 1 - FN 5)(FN 1 - FN 5)

The specification is complete when the category and the classification are combined, such The specification is complete when the category and the classification are combined, such as RC 3 or FN 5, and coupled with a specific diameter size.as RC 3 or FN 5, and coupled with a specific diameter size.


Running and Sliding Fits* -- American National StandardRunning and Sliding Fits* -- American National Standard

RC 1 Close sliding fits are intended for the accurate location of parts which must assemble without perceptible play.RC 2 Sliding fits are intended for accurate location, but with greater maximum clearance than class RC 1. Parts made to this fit move and turn easily but are not intended to run freely, and in the larger sizes may seize with small temperature changes.RC 3 Precision running fits are about the closest fits which can be expected to run freely, and are intended for precision work at slow speeds and light journal pressures, but are not suitable where appreciable temperature differences are likely to be encountered.RC 4 Close running fits are intended chiefly for running fits on accurate machinery with moderate surface speeds and journal pressures, where accurate location and minimum play are desired.RC 5 RC 6

Basic hole system. Limits are in thousandths of an inch.Class RC 1

Lim

its o

f C

lear

ance Standard

Limits

HoleH5

Shaftg4

Class RC 2

Lim

its o

f C

lear

ance Standard

Limits

HoleH6

Shaftg5

Class RC 3

Lim

its o

f C

lear

ance Standard

Limits

HoleH7

Shaftf6

Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

Class RC 5

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shafte7

Class RC 6

Lim

its o

f C

lear

ance Standard

Limits

HoleH9

Shafte8

Medium running fits are intended for higher running speeds, or heavy journal pressures, or both.

*From ANSI B4.1--1967 (R 1987). Larger diameters and RC 7 through RC 9 not included in this presentation.*From ANSI B4.1--1967 (R 1987). Larger diameters and RC 7 through RC 9 not included in this presentation.

NominalSize Range

inInches

0 - 0.12 0.1 + 0.2 - 0.1 0.1 + 0.25 - 0.1 0.3 + 0.4 - 0.3 0.3 + 0.6 - 0.3 0.6 + 0.6 - 0.6 0.6 + 1.0 - 0.6 0.45 - 0 - 0.25 0.55 - 0 - 0.3 0.95 - 0 - 0.55 1.3 - 0 - 0.7 1.6 - 0 - 1.0 2.2 - 0 - 1.20.12 - 0.24 0.15 + 0.2 - 0.15 0.15 + 0.3 - 0.15 0.4 + 0.5 - 0.4 0.4 + 0.7 - 0.4 0.8 + 0.7 - 0.8 0.8 + 1.2 - 0.8 0.5 - 0 - 0.3 0.65 - 0 - 0.35 1.12 - 0 - 0.7 1.6 - 0 - 0.9 2.0 - 0 - 1.3 2.7 - 0 - 1.50.24 - 0.40 0.2 + 0.25 - 0.2 0.2 + 0.4 - 0.2 0.5 + 0.6 - 0.5 0.5 + 0.9 - 0.5 1.0 + 0.9 - 1.0 1.0 + 1.4 - 1.0 0.6 - 0 - 0.35 0.85 - 0 - 0.45 1.5 - 0 - 0.9 2.0 - 0 - 1.1 2.5 - 0 - 1.6 3.3 - 0 - 1.90.40 - 0.71 0.25 + 0.3 - 0.25 0.25 + 0.4 - 0.25 0.6 + 0.7 - 0.6 0.6 + 1.0 - 0.6 1.2 + 1.0 - 1.2 1.2 + 1.6 - 1.2 0.75 - 0 - 0.45 0.95 - 0 - 0.55 1.7 - 0 - 1.0 2.3 - 0 - 1.3 2.9 - 0 - 1.9 3.8 - 0 - 2.20.71 - 1.19 0.3 + 0.4 - 0.3 0.3 + 0.5 - 0.3 0.8 + 0.8 - 0.8 0.8 + 1.2 - 0.8 1.6 + 1.2 - 1.6 1.6 + 2.0 - 1.6 0.95 - 0 - 0.55 1.2 - 0 - 0.7 2.1 - 0 - 1.3 2.8 - 0 - 1.6 3.6 - 0 - 2.4 4.8 - 0 - 2.81.19 - 1.97

1.97 - 3.15


Using the Standard Fit TablesUsing the Standard Fit Tablesto Assign Dimensions to Mating Partsto Assign Dimensions to Mating Parts

Using the standard fit tables, determine the limits for a nominal 9/16 (.5625) inch diameter Using the standard fit tables, determine the limits for a nominal 9/16 (.5625) inch diameter RC 4 fit between a shaft and a hole. Place the dimensions on the drawing using stacked RC 4 fit between a shaft and a hole. Place the dimensions on the drawing using stacked limits form.limits form.



First, identify the size range in the left-most column in the fit table for RC fits. I have First, identify the size range in the left-most column in the fit table for RC fits. I have highlighted the row that identifies the size range that will include the designated highlighted the row that identifies the size range that will include the designated diameter in the problem. (Note that this column is specified in inches. All other diameter in the problem. (Note that this column is specified in inches. All other columns in the table are specified in columns in the table are specified in thousandthsthousandths of an inch.) of an inch.)

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15



First, identify the size range in the left-most column in the fit table for RC fits. I have highlighted First, identify the size range in the left-most column in the fit table for RC fits. I have highlighted the row that identifies the size range that will include the designated diameter in the problem. (Note the row that identifies the size range that will include the designated diameter in the problem. (Note that this column is specified in inches. All other columns in the table are specified in that this column is specified in inches. All other columns in the table are specified in thousandthsthousandths of an inch.)of an inch.) It is important that you remember that this is the It is important that you remember that this is the onlyonly column in the table that has values column in the table that has values expressed in expressed in inchesinches..The remaining columns are expressed in thousandths of an inch.The remaining columns are expressed in thousandths of an inch.

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15



Next, read across the table on the row that is highlighted to identify the parameters Next, read across the table on the row that is highlighted to identify the parameters for an RC 4 fit. Because of the lack of space, I have excluded the columns for fits for an RC 4 fit. Because of the lack of space, I have excluded the columns for fits other than the RC 4. You may wish to return to a previous slide where you can other than the RC 4. You may wish to return to a previous slide where you can review a more complete representative page from the standards.review a more complete representative page from the standards.

Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15



The first column for a class RC 4 fit establishes the The first column for a class RC 4 fit establishes the limits of clearancelimits of clearance. The number on . The number on top is the top is the allowanceallowance at MMC. Converted to inches (by moving the decimal 3 places to the at MMC. Converted to inches (by moving the decimal 3 places to the left), this number would be .0006, six tenths of one-thousandth, or six ten-thousandths. left), this number would be .0006, six tenths of one-thousandth, or six ten-thousandths. The number below (2.3) is the maximum clearance at LMC, or twenty-three ten-The number below (2.3) is the maximum clearance at LMC, or twenty-three ten-thousandths (.0023).thousandths (.0023).

Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15



Notice that the Notice that the HoleHole size, at its size, at its lowerlower limit, reads “- 0”. Therefore, use the basic limit, reads “- 0”. Therefore, use the basic size (nominal 9/16 inch, converted to its decimal equivalent, .5625) as the lower size (nominal 9/16 inch, converted to its decimal equivalent, .5625) as the lower limit for the dimension of the hole. limit for the dimension of the hole.

Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625



Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625

For the upper size limit for the hole you will notice that the value on the chart is +1.0. This For the upper size limit for the hole you will notice that the value on the chart is +1.0. This value when converted into inches equals one thousandth of an inch—or ten ten-value when converted into inches equals one thousandth of an inch—or ten ten-thousandths. Add the required .0010 to the basic size for the upper limit of the hole (.5625 thousandths. Add the required .0010 to the basic size for the upper limit of the hole (.5625 + .0010 = .5635). This is the upper size limit of the hole, which can now be added to the + .0010 = .5635). This is the upper size limit of the hole, which can now be added to the drawing. drawing. It is added to the basic size value. It is added to the basic size value.

.5635



Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625

To determine the upper size limit of the shaft, we must subtract (note the sign in the table To determine the upper size limit of the shaft, we must subtract (note the sign in the table for the upper size limit of the shaft) 0.6 thousandths (or six ten-thousandths) from the basic for the upper size limit of the shaft) 0.6 thousandths (or six ten-thousandths) from the basic size. The result will be .5619 (.5625 - .0006 = .5619). Add this value to the drawing as the size. The result will be .5619 (.5625 - .0006 = .5619). Add this value to the drawing as the upper size limit on the shaft.upper size limit on the shaft.

.5635.5619



Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625

.5635.5619

Finally, determine the lower limit of the shaft diameter by subtracting 1.3 thousandths Finally, determine the lower limit of the shaft diameter by subtracting 1.3 thousandths (or thirteen ten-thousandths) from the (or thirteen ten-thousandths) from the basic sizebasic size, and include the value in the drawing., and include the value in the drawing.

.5612



Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625

.5635.5619.5612

Now, look back to the column labeled “Limits of Clearance”. These values reveal the Now, look back to the column labeled “Limits of Clearance”. These values reveal the total amount of variation that can be allowed on a 9/16” RC 4 fit--when the parts are at total amount of variation that can be allowed on a 9/16” RC 4 fit--when the parts are at maximum material condition and at least material condition. maximum material condition and at least material condition.

On the On the drawingdrawing, compare the dimensions diagonally., compare the dimensions diagonally.

Min ClearanceMin Clearance Max ClearanceMax Clearance .0006.0006 .0023.0023



Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625

.5635.5619.5612

At maximum material condition (MMC—largest shaft; smallest hole), the difference At maximum material condition (MMC—largest shaft; smallest hole), the difference is .0006, or 0.6 thousandths. is .0006, or 0.6 thousandths.



Class RC 4

Lim

its o

f C

lear

ance Standard

Limits

HoleH8

Shaftf7

0.3 + 0.6 - 0.31.3 - 0 - 0.70.4 + 0.7 - 0.41.6 - 0 - 0.90.5 + 0.9 - 0.52.0 - 0 - 1.10.6 + 1.0 - 0.62.3 - 0 - 1.30.8 + 1.2 - 0.82.8 - 0 - 1.6

NominalSize Range

inInches

0 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 0.71 - 1.19 1.19 - 1.97

1.97 - 3.15

.5625

.5635.5619.5612

At least material condition (LMC—smallest shaft; largest hole), there is a clearance At least material condition (LMC—smallest shaft; largest hole), there is a clearance of .0023, or 2.3 thousandths.of .0023, or 2.3 thousandths.

-.0006 -.0013

+.0010 -.0000

.5625 .5625Values Referenced to Basic Size:


End of Unit Number FourEnd of Unit Number Four

Documents

Coordinate Dimensioning Table of Contents Return to the Previous Slide Slide 1QuitMaster Table of ContentsGlossary 86 57.6 56.6 100 99 9.6 9.4 25.5 25.4