Tài liệu Handbook of Mechanical Engineering Calculations P26 doc

26.1
SECTION 26
MET ALWORKING AND
NONMET ALLIC MATERIALS
PROCESSING
ECONOMICS OF MACHINING 26.2
Estimating Cutting Time with
Different Tool Materials 26.2
Comparing Finish Machining Time and
Costs with Different Tool Materials
26.6
Finding Minimum Cost and Maximum
Production Tool Life for Disposable
Tools 26.10
Computing Minimum Cost and
Maximum Production Tool Life for
Regrindable Tools 26.11
MACHINING PROCESS CALCULATIONS
26.12
Total Element Time and Total
Operation Time 26.12
Cutting Speeds for Various Materials
26.13
Depth of Cut and Cutting Time for a
Keyway 26.14
Milling-Machine Table Feed and
Cutter Approach 26.15
Dimensions of Tapers and Dovetails
26.15
Angle and Length of Cut from Given
Dimensions 26.16

Tool Feed Rate and Cutting Time
26.17
True Unit Time, Minimum Lot Size,
and Tool-Change Time 26.18
Time Required for Turning Operations
26.18
Time and Power to Drill, Bore,
Countersink, and Ream 26.20
Time Required for Facing Operations
26.20
Threading and Tapping Time 26.22
Turret-Lathe Power Input 26.23
Time to Cut a Thread on an Engine
Lathe 26.24
Time to Tap with a Drilling Machine
26.25
Milling Cutting Speed, Time, Feed,
Teeth Number, and Horsepower
26.26
Gang-, Multiple-, and For-Milling
Cutting Time 26.28
Shaper and Planer Cutting Speed,
Strokes, Cycle Time, Power 26.29
Grinding Feed and Work Time 26.30
Broaching Time and Production Rate
26.31
Hobbing, Splining, and Serrating Time
26.31
Time to Saw Metal with Power and
Band Saws 26.32

Oxyacetylene Cutting Time and Gas
Consumption 26.33
Comparison of Oxyacetylene and
Electric-Arc Welding 26.35
Presswork Force for Shearing and
Bending 26.36
Mechanical-Press Midstroke Capacity
26.36
Stripping Springs for Pressworking
Metals 26.37
Blanking, Drawing, and Necking
Metals 26.37
Metal Plating Time and Weight 26.38
Shrink- and Expansion-Fit Analyses
26.39
Press-Fit Force, Stress, and Slippage
Torque 26.40
Learning-Curve Analysis and
Construction 26.43
Learning-Curve Evaluation of
Manufacturing Time 26.44
Determining Brinell Hardness 26.47
Economical Cutting Speeds and
Production Rates 26.47
Optimum Lot Size in Manufacturing
26.49
Precision Dimensions at Various
Temperatures 26.50
Horsepower Required for
Metalworking 26.51

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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
26.2 DESIGN ENGINEERING
Cutting Speed for Lowest-Cost
Machining
26.53
Reorder Quantity for Out-of-Stock
Parts
26.54
Savings with More Machinable
Materials 26.55
Time Required for Thread Milling
26.55
Drill Penetration Rate and Centerless
Grinder Feed Rate 26.56
Bending, Dimpling, and Drawing
Metal Parts 26.56
Blank Diameters for Round Shells
26.60
Breakeven Considerations in
Manufacturing Operations 26.60
Calculating Geometric Dimensions of
Drawn Parts 26.62
Analyzing Stainless-Steel Molding
Methods 26.67
Reducing Machining Costs by
Designing with Shims
26.69

Analyzing Taper Fits for
Manufacturing and Design
26.73
Designing Parts for Expected Life
26.77
Wear Life of Rolling Surfaces 26.79
Factor of Safety and Allowable Stress
in Design 26.81
Rupture Factor and Allowable Stress
in Design 26.84
Force and Shrink Fit Stress,
Interference, and Torque 26.85
Selecting Bolt Diameter for Bolted
Pressurized Joint 26.87
Determining Required Tightening
Torque for a Bolted Joint 26.91
Selecting Safe Stress and Materials
for Plastic Gears 26.92
Economics of Machining
ESTIMATING CUTTING TIME AND COST WITH
DIFFERENT TOOL MATERIALS
A 9-in (22.86-cm) diameter steel shaft is to be ‘‘heavy roughed’’ with either of two
cutting tools—high-speed steel (HSS), or cemented carbide. The work material is
AISI 1050 having a hardness of 200 BHN. Feed rate is 0.125 in/r (3.17 mm/r);
depth of cut
ϭ 1.0 in (25.4 mm); tool life is based on 0.030-in (0.726-mm) flank
wear. Choose the most effective tool to use if the tool signature is:
Ϫ6, 10, 6, 6,
15, 15,
1

⁄
16
R; the tool-changing time ϭ 4 min for both tools; the cost of a sharp
tool
ϭ $0.50 for HSS and $2.00 for cemented carbide; and M ϭ machine labor
plus overhead rate, $ / min
ϭ 15 cents for each type of tool.
Calculation Procedure:
1. Determine the minimum-cost tool life for each type of tool material
Analyses of the economics metal of cutting with different types of cutting-tool
materials are often plotted on two bases—Figs. 1 and 2. Figure 1 shows the ma-
chining cost, tool cost, and nonproductive cost added to show the total cost per
piece. In Fig. 2, the machine time, tool-changing time, and nonproductive time are
added and plotted as the total time per piece.
Studies show that the cutting speed and production rate resulting from minimum-
cost tool life of approximately the same value is much higher for carbide tools than
for high-speed steel tools—150 ft/min (45.7 m/min) cutting speed for carbide tools
vs. 30 ft / min (9.14 m/min) for high-speed steel tools. These two values of cutting
speed will be used in this procedure.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.3
SI Values
200 fpm 60.9m/min
400 121.9
600 182.9
800 243.8
1000 304.8

1200 365.8
1400 426.7
FIGURE 1 Total cost per piece is found by adding the plots of ma-
chining costs, tool costs, and nonproductive costs. (T. E. Hayes and
American Machinist.)
The minimum-cost tool life, T
c
, is a function of the slope, n, of the tool-life
curve, Fig. 3. It can be said that n is one of the controlling influences on Hi-E
cutting conditions.* Thus, for high-speed steel, the expression for T
c
is:
1 t
T ϭϪ1 ϩ TCT
ͩͪͩ ͪ
c
nM
where T
c
ϭ minimum-cost tool life, min; n ϭ slope of tool-life curve; M ϭ machine
labor plus overhead rate, $ / min; TCT
ϭ tool-changing time, min. Substituting,
*The Hi-E term was originally coined by Thomas E. Hayes, Service Engineer, Metallurgical Products
Department, General Electric Company, and first published in his article, ‘‘How to Cut Costs with Carbides
by ‘Hi-E’ Machining.’’
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.4 DESIGN ENGINEERING

SI Values
200 fpm 60.9m/min
400 121.9
600 182.9
800 243.8
1000 304.8
1200 365.8
1400 426.7
FIGURE 2 Total time per piece is found by adding the plots of ma-
chine times, tool-changing time, and nonproductive time. (T. E. Hayes
and American Machinist.)
1 0.50
T ϭϪ1 ϩ 4
ͩͪͩͪ
c
125 0.15
ϭ 51.3 min
For cemented carbide, we have
1 t
T ϭϭ Ϫ1 ϩ TCT
ͩͪͩ ͪ
c
nM
12
ϭϪ1 ϩ 4
ͩͪͩͪ
0.25 0.15
ϭ 52 min
Thus, the T
c

, values for both tools are approximately the same.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.5
FIGURE 3 A combination of the total cost per piece and total
time per piece plots on a single graph forms the Hi-E range
between their respective minimum points. (Brierley and Siek-
mann.)
2. Compute the tool life for maximum productive rate
The tool life for maximum productive rate T
p
, min, is given by
1
T ϭϪ1 TCT
ͩͪ
p
n
where symbols are as before.
Substituting for high-speed steel we have
1
T ϭϪ1 ϭ 28 min
p
0.125
Entering Fig. 3 at 28 min and projecting to the HSS plot, we find that the cutting
speed should be 33 ft/min (10.1 m / min).
Using the same relation for cemented carbide, we find, entering Fig. 3 at 12
minute and projecting up to the cemented-carbide plot, the cutting speed to be 220
ft/min (67.1 m/min).

3. Tabulate the results of the calculations
List the cutting conditions for each type of tool material, as in Table 1. Studying
the results in Table 1 shows that only about 20 percent as much time is required
per piece with cemented-carbide tools as with HSS tools, and the total cost per
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.6 DESIGN ENGINEERING
TABLE 1 Operation of the Job Illustrated in Figure 1 at
Minimum Cost-Cutting Conditions Results in the Following
Economic Comparison. Machining Costs are Halved and
Production is Tripled*
Cutting conditions HSS
Cemented
carbide
Machine time per piece 45 min 9.1 min
Nonproductive time per piece 10 min 10 min
Labor plus overhead rate $0.15 $0.15
Machine cost per piece $6.75 $1.36
Nonproductive cost per piece $1.50 $1.50
Tool cost per piece $0.50 $2.00
Total cost per piece $8.75 $4.86
Total time per piece 55 min 19.1 min
Pieces per hour 1.1 3.1
*Brierley and Siekmann.
piece is only about 55 percent of that of HSS. Thus, the higher tool cost results in
greater productivity (3.1 pieces per hour vs. 1.1 pieces per hour).
Related Calculations. This procedure is the work of Robert G. Brierley, Tool
Applications Specialist, Metallurgical Products Department, General Electric Com-

pany and H. J. Siekmann, Vice President, Marketing, Martin Metals Company,
Division of Martin Marietta Corporation. If reflects the Hi-E approach used at
General Electric Company, plus the basics of metalworking physics.
The Hi-E range is shown in Fig. 4, which depicts a combination of the tool cost
per piece and total time per piece plotted on a single graph. The Hi-E range is
between the respective minimum points.
Since tool-life plots are important in the Hi-E analyses of machining economics,
the value of n is of much interest. Although n varies slightly as machining condi-
tions are changed, Brierley and Siekmann cite the following values for practical
everyday use to satisfy the calculations for the Hi-E range: For high-speed steel,
n
ϭ 0.125 and ([1/n] Ϫ 1) ϭ 7; for carbide, n ϭ 0.25 to 0.30 and ([1/n] Ϫ 1) ϭ
3 for the 0.25 value; for cemented oxide or ceramic tools, n ϭ 0.50 to 0.70 and
([1/n]
Ϫ 1) ϭ 1 for the 0.50 value. More exact values can be obtained from
tabulations available from ASTME.
The procedure given here was presented by the above two authors in their book
Machining Principles and Cost Control, McGraw-Hill.
COMPARING FINISH MACHINING TIME AND
COSTS FOR DIFFERENT TOOL MATERIALS
Compare machining costs and times for cemented-carbide and cemented-oxide tools
for a high-speed finishing operation using the data given in Fig. 5 and the equations
in the previous procedure. Tabulate the results for comparison.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.7
0.125 ipr 3.175 mm
1.000 in. 25.4 mm

0.030 in. 0.762 mm
FIGURE 4 Heavy roughing of a steel shaft with carbide widens the Hi-E range compared with
using high-speed steel. (Brierley and Siekmann.)
Calculation Procedure:
1. Find the minimum-cost tool life for each tool material
Use the T
c
equation of step 1 of the previous procedure with the same symbols.
Then, for cemented carbide,
1 t
T ϭϪ1 ϩ TCT
ͩͪͩ ͪ
c
nM
1 0.25
ϭϪ1 ϩ 1
ͩͪͩ ͪ
0.3 0.15
ϭ 6.22 min
Likewise, using the same equation for cemented oxide,
1 t
T ϭϪ1 ϩ TCT
ͩͪͩ ͪ
c
nM
1 0.375
ϭϪ1 ϩ 1
ͩͪͩ ͪ
0.7 0.15
ϭ 1.57 min

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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.8 DESIGN ENGINEERING
SI Values
0.010 ipr 0.254 mm
1.000 in. 25.4 mm
0.030 in. 0.762 mm
FIGURE 5 A high-speed finishing operation switched to cemented oxide. (Brierley and Siek-
mann.)
2. Determine the tool life for the maximum productive rate
As in step 1, above, use the equation and symbols from step 2 in the previous
procedure. Thus, for cemented-carbide tools,
1
T ϭϪ1 TCT ϭ 2.33 min
ͩͪ
p
n
Projecting from 2.33 min on the horizontal scale in Fig. 5, we find the cutting speed
to be 1150 ft /min (350.5 m / min).
For cemented-oxide tools,
1
T ϭϪ1 TCT
ͩͪ
p
n
ϭ 0.45 ϭϾ20,000 ft/min
Plotting from 0.45 min, we find that the cutting speed would exceed 20,000 ft/min
(6096 m / min)

3. Summarize the calculations in tabular form
Table 2 summarizes the calculations for these two tooling materials. As you can
see, there is a significant difference in the machine time per piece: 1 7.2 min vs.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.9
TABLE 2 Minimum Cost-Cutting Conditions Using Cemented
Oxide Rather Than Carbide Halve the Machining Costs of This
Finishing Operation While Production Is Doubled*
Cutting conditions
Cemented
carbide
Cemented
oxide
Machine time per piece 17.2 min 1.63 min
Nonproductive time per piece 10 min 10 min
Labor plus overhead rate $0.15 $0.15
Machine cost per piece $2.50 $0.245
Nonproductive cost per piece $1.50 $1.50
Tool cost per piece $0.25 $0.375
Total cost per piece $4.25 $2.120
Total time per piece 27.2 min 11.63 min
Pieces per hour 2.2 5.4
*Brierley and Siekmann.
1.63 min. Likewise, the cost is at a 10-times ratio: $0.245 vs. $2.50, and the piece
output is more than double: 5.4 pieces per hour vs. 2.2 pieces per hour. As in the
previous procedure, the more expensive tool significantly increases the output while
reducing production costs.

Related Calculations. This procedure, like the previous one, is the work of
Brierley and Siekmann. Full citation information is given in the previous procedure.
In building their approach to the economics of machining, Brierley and Siek-
mann give a number of key equations that lead up to the steps presented in this
and the previous procedure. These equations are: (1) Machining cost
ϭ (machining
time per piece)(labor
ϩ overhead rate); (2) Machining time ϭ [(length of piece
cut)(cut)]/(feed)(rpm of cutter); (3) Tool cost
ϭ (tool-changing cost ϩ tool-grinding
cost per edge
ϩ tool depreciation per edge ϩ tool inventory cost)/(production per
edge); (4) Cost to change the tool
ϭ (tool-changing time)(the machine operator’s
rate
ϩ overhead); (5) Tool-grinding cost per edge ϭ [(grinding time)(grinder’s rate
ϩ overhead)]/(edges per grind); (6) Brazed-tool depreciation cost per edge ϭ (cost
of tool) / (number of regrinds
ϩ 1); (7) For disposable-insert toolholder or milling-
cutter head, Tool depreciation cost per edge
ϭ [(cost of disposable insert/number
of cutting edges per insert)
ϩ (cost of holder or head)]/[(number of inserts in life
of holder) (number of edges per insert)]; (8) For on-end insert toolholder or re-
gindable inserted-blade milling-cutter head, Tool depreciation cost per edge
ϭ (cost
of insert) / [(number of regrinds per insert)(number of edges per grind)]
ϩ (cost of
holder or head)/[(number of in life of holder or head)(number of regrinds per
insert)(number of edges per grind)]; (9) Tool inventory cost

ϭ (number of tools at
machine
ϩ number of tools in grinding room)(cost per tool)(inventory cost rate);
(10) Nonproductive cost
ϭ load and unload time ϩ (other noncutting time)(operator
labor
ϩ overhead rate); (11) Total machining time ϭ machine time from Eq. (1) ϩ
tool changing time ϩ nonproductive time.
Using the above eleven equations and the relations given in Figs. 3, 4, and 5,
the economics of machining can be planned in a preliminary way for a given
machine. Then the Hi-E approach and advances in it should be considered for in-
depth analysis of the economics of a given machining application.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.10 DESIGN ENGINEERING
FINDING MINIMUM COST AND MAXIMUM
PRODUCTION TOOL LIFE FOR
DISPOSABLE TOOLS
Find the minimum cost and maximum production tool life for a disposable tool
having the following characteristics: price of insert plus toolholder depreciation,
P
ϭ $1.80; total cutting edges in the life of the insert, E ϭ 6; machine operator’s
rate, MR
ϭ $4.00/h; machine overhead rate, MO ϭ $8.00/h; tool-changing time,
TCT
ϭ 1 min; the constant for the slope of the tool-life line for carbide tools, n ϭ
3.5.
Calculation Procedure:

1. Find the minimum-cost tool life
The expression for the minimum-cost tool life, T
c
,isgivenby
1 t
T ϭϪ1 ϩ TCT
ͩͪͩ ͪ
c
nM
where
price of insert
ϩ toolholder depreciation 1.80
t ϭϭϭ0.30
total cutting edges in life of insert 6
labor per hour
ϩ overhead per hour 4.00 ϩ 8.00
M ϭϭϭ0.20
60 60
TCT
ϭ tool-changing time (min) ϭ 1
1
Ϫ 1 ϭ a constant (3.5) based on the slope
ͩͪ
n
of the tool-life line
Substituting,
0.30
T (min) ϭ 3.5 ϩ 1
ͩͪ
c

0.20
T (min)
ϭ 3.5 ϫ 2.5
c
T (min) ϭ 8.75
c
2. Calculate the maximum production tool life for this tool
To solve for T
p
, we need only the constant, n, and the tool-changing time. Or,
1
T ϭϪ1 TCT
ͩͪ
p
n
T
ϭ 3.5 ϫ 1
p
T ϭ 3.5
p
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.11
3. Show an alternative approach to the calculation of T
c
and T
p
Convert the known quantities to desired values. Thus, (a) Total tool cost per cutting

edge, P/E
ϭ t, $1.80 /6 ϭ $0.30; (b) cost of labor plus overhead per hour ϭ MR
ϩ RO ϭ MRO ϭ $4.00 ϩ $8.00 ϭ $12.00; cost of labor plus overhead per minute
ϭ MRO/60 ϭ M ϭ $12.00/60 ϭ $0.20; (c) combined tool cost plus tool-changing
cost per edge
ϭ [t ϩ (M ϫ TCT )] ϭ [0.30 ϩ (0.20 ϫ 1)] ϭ $0.50.
Now we are ready to calculate the tool life for minimum part cost. Since the
standard formula for carbide tools is 3.5(C / M )
ϭ TC ϭ 3.5(0.50 / 0.20) ϭ 8.75
min.
The next step is to convert the tool-life values for T
c
and T
p
to cutting speeds.
Further, it may be desirable to convert cutting speeds from linear dimensions to
revolutions per minute for the operator’s convenience.
Related Calculations. The computations shown here can be done on a pre-
printed form or by using a computer program specially prepared for this purpose.
Brierley and Siekmann, in the previously cited reference, present preprinted forms
for this purpose. The procedure given here is from that reference. Anyone seeking
to use the Hi-E method, or its advancements, will find much help in the preprinted
forms available to them. The approach given here is valuable for anyone seeking
the most economical machining tools.
COMPUTING MINIMUM COST AND MAXIMUM
PRODUCTION TOOL LIFE FOR REGRINDABLE
TOOLS
Find the minimum cost and maximum production tool life for regrindable brazed-
type tools, on-end slugs, or others which are normally resharpened by grinding.
The variables for the tool are: Price of tool, P

ϭ $3.50; tool cutting edges in life
of tool, E
ϭ 7; tool grinder’s rate, GR ϭ $4.00/h; toolroom overhead rate, GO ϭ
$8.00/h; grinding time per edge, GT ϭ 5 min; machine operator’s rate, MR ϭ
$4.00/h; machine overhead rate, MO ϭ $8.00 / h; tool-changing time, TCT ϭ min.
Calculation Procedures:
1. Find the total tool cost per cutting edge
(a) The tool cost per cutting edge, P / E
ϭ $3.50/7 ϭ $0.50. (b) Grinding cost per
minute
ϭ (GR ϩ GO)/60 ϭ ($4.00 ϩ $8.00) / 60 ϭ $0.20. (c) Grinding cost per
cutting edge
ϭ [(GC /m)GT]GC/E ϭ $0.20 ϫ 5 ϭ $1.00. (d ) Total cost per edge
(tool cost
ϩ grinding cost) ϭ [(P/E) ϩ (GC /E) ϭ t] ϭ 0.50 ϩ 1.00 ϭ 1.50.
2. Find the total cost of labor plus overhead
Cost per hour
ϭ MR ϩ MO ϭ MRO ϭ $4.00 ϩ $8.00 ϭ $12.00. (b) Cost per
minute
ϭ M ϭ MRO /60 ϭ $12.00/60 ϭ $0.20.
3. Find the combined tool cost plus tool-changing cost per edge
Combined cost
ϭ C ϭ [t ϩ M(TCT)] ϭ $1.50 ϩ ($0.20 ϫ 3) ϭ $2.10.
4. Calculate tool life for minimum part cost
Use the standard formula for carbide tools
ϭ T
p
ϭ (3.5C /M) ϭ 3.5(2.10/0.20) ϭ
36.75 min.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.12 DESIGN ENGINEERING
5. Calculate tool life for maximum production rate
The standard formula for carbide tools is 3.5(TCT )
ϭ T
p
ϭ 3.5 ϫ 3 ϭ 10.5 min.
6. Convert tool life to cutting speed
Desired tool life
Minimum part cost, T
c
36.75 min
Maximum production rate, T
p
10.5 min
Related Calculations. The greater cost of getting a new cutting edge on the
job as reflected in the higher first cost of the tool, the added cost of grinding, and
the longer time to change cutting edges results in a longer tool life for minimum
part cost. This illustrates the need for tools that are low in cost per edge and easily
and quickly changed if low unit costs are achieved. As in the previous procedure,
the next step is to convert tool-life values to cutting speeds and then to revolutions
per minute to suit the diameter or diameters being machined.
This procedure is the work of Brierley and Siekmann, as reported in the refer-
ence cited earlier.
Machining Process Calculations
TOTAL ELEMENT TIME AND TOTAL OPERATION
TIME
The observed times for a turret-lathe operation are as follows: (1) material to bar

stop, 0.0012 h; (2) index turret, 0.0010 h; (3) point material, 0.0005 h; (4) index
turret, 0.0012 h; (5) turn 0.300-in (0.8-cm) diameter part, 0.0075 h; (6) clear hex-
agonal turret, 0.0009 h; (7) advance cross-slide tool, 0.0008 h; (8) cutoff part,
0.0030 h; (9) aside with part, 0.0005 h. What is the total element time? What is
the total operation time if 450 parts are processed? Pointing of the material was
later found unnecessary. What effect does this have on the element and operation
total time?
Calculation Procedure:
1. Compute the total element time
Compute the total element time by finding the sum of each of the observed times
in the operation, or sum steps 1 through 9: 0.0012
ϩ 0.0010 ϩ 0.0005 ϩ 0.0012
ϩ 0.0075 ϩ 0.0009 ϩ 0.0008 ϩ 0.0030 ϩ 0.0005 ϭ 0.0166 h ϭ 0.0166 (60 min /
h)
ϭ 0.996 minute per element.
2. Compute the total operation time
The total operation time
ϭ (element time, h)(number of parts processed). Or,
(0.0166)(450)
ϭ 7.47 h.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.13
3. Compute the time savings on deletion of one step
When one step is deleted, two or more times are usually saved. These times are
the machine preparation and machine working times. In this process, they are steps
2 and 3. Subtract the sum of these times from the total element time, or 0.0166
Ϫ

(0.0010 ϩ 0.0005) ϭ 0.0151 h. Thus the total element time decreases by 0.0015
h. The total operation time will now be (0.0151)(450)
ϭ 6.795 h, or a reduction of
(0.0015)(450)
ϭ 0.6750 h. Checking shows 7.470 Ϫ 6.795 ϭ 0.675 h.
Related Calculations. Use this procedure for any multiple-step metalworking
operation in which one or more parts are processed. These processes may be turn-
ing, boring, facing, threading, tapping, drilling, milling, profiling, shaping, grinding,
broaching, hobbing, cutting, etc. The time elements used may be from observed or
historical data.
Recent introduction of international quality-control specifications by the Inter-
national Organization for Standardization (ISO) will require greater accuracy in all
manufacturing calculations. The best-known set of specifications at this time is ISO
9000 covering quality standards and management procedures. All engineers and
designers everywhere should familiarize themselves with ISO 9000 and related
requirements so that their products have the highest quality standards. Only then
will their designs survive in the competitive world of international commerce and
trading.
CUTTING SPEEDS FOR VARIOUS MATERIALS
What spindle rpm is needed to produce a cutting speed of 150 ft/min (0.8 m / s)
on a 2-in (5.1-cm) diameter bar? What is the cutting speed of a tool passing through
2.5-in (6.4-cm) diameter material at 200 r/min? Compare the required rpm of a
turret-lathe cutter with the available spindle speeds.
Calculation Procedure:
1. Compute the required spindle rpm
In a rotating tool, the spindle rpm R
ϭ 12C /
␲
d, where C ϭ cutting speed, ft /min;
d

ϭ work diameter, in. For this machine, R ϭ 12(150)/
␲
(2) ϭ 286 r/min.
2. Compute the tool cutting speed
For a rotating tool, C
ϭ R
␲
d/12. Thus, for this tool, C ϭ (200)(
␲
)(2.5)/12 ϭ 131
ft/min (0.7 m/s).
The cutting-speed equation is sometimes simplified to C
ϭ Rd/4. Using this
equation for the above machine, we see C
ϭ 200(2.5) / 4 ϭ 125 ft / min (0.6 m / s).
In general, it is wiser to use the exact equation.
3. Compare the required rpm with the available rpm
Consult the machine nameplate, American Machinist’s Handbook, or a manufac-
turer’s catalog to determine the available spindle rpm for a given machine. Thus,
one Warner and Swasey turret lathe has spindle speed of 282 compared with the
286 r/min required in step 1. The part could be cut at this lower spindle speed, but
the time required would be slightly greater because the available spindle speed is
286
Ϫ 282 ϭ 4 r/min less than the computed spindle speed.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.14 DESIGN ENGINEERING
FIGURE 6 Keyway dimensions.

When preparing job-time estimates, be certain to use the available spindle speed,
because this is frequently less than the computed spindle speed. As a result, the
actual cutting time will be longer when the available spindle speed is lower.
Related Calculations. Use this procedure for a cutting tool having a rotating
cutter, such as a lathe, boring mill, automatic screw machine, etc. Tables of cutting
speeds for various materials (metals, plastics, etc.) are available in the American
Machinist’s Handbook, as are tables of spindle rpm and cutting speed.
DEPTH OF CUT AND CUTTING TIME FOR A
KEYWAY
What depth of cut is needed for a
3
⁄
4
-in (1.9-cm) wide keyway in a 3-in (7.6-cm)
diameter shaft? The keyway length is 2 in (5.1 cm). How long will it take to mill
this keyway with a 24-tooth cutter turning at 130 r/min if the feed is 0.005 per
tooth?
Calculation Procedure:
1. Sketch the shaft and keyway
Figure 6 shows the shaft and keyway. Note that the depth of cut D in
ϭ W/2 ϩ
A, where W ϭ keyway width, in; A ϭ distance from the key horizontal centerline
to the top of the shaft, in.
2. Compute the distance from the centerline to the shaft top
For a machined keyway, A
ϭ [d Ϫ (d
2
Ϫ W
2
)

0.5
]/2, where d ϭ shaft diameter, in.
With the given dimensions, A
ϭ [3 Ϫ (3
2
Ϫ 0.75
2
)
0.5
]/2 ϭ 0.045 in (1.1 mm).
3. Compute the depth of cut for the keyway
The depth of cut D ϭ W /2 ϩ A ϭ 0.75/2 ϩ 0.045 ϭ 0.420 in (1.1 cm).
4. Compute the keyway cutting time
For a single milling cutter, cutting time, min ϭ length of cut, in /[(feed per tooth)
ϫ (number of teeth on cutter)(cutter rpm)]. Thus, for this keyway, cutting time ϭ
2.0/[(0.005)(24)(130)] ϭ 0.128 min.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.15
Related Calculations. Use this procedure for square or rectangular keyways.
For Woodruff key-seat milling, use the same cutting-time equation as in step 4. A
Woodruff key seat is almost a semicircle, being one-half the width of the key less
than a semicircle. Thus, a
9
⁄
16
-in (1.4-cm) deep Woodruff key seat containing a
3

⁄
8
-
in (1.0-cm) wide key will be (
3
⁄
8
)/2 ϭ
3
⁄
16
in (0.5 cm) less than a semicircle. The
key seat would be cut with a cutter having a radius of
9
⁄
16
ϩ
3
⁄
16
ϭ
12
⁄
16
,or
3
⁄
4
in
(1.9 cm).

MILLING MACHINE TABLE FEED AND CUTTER
APPROACH
A 12-tooth milling cutter turns at 40 r /min and has a feed of 0.006 per tooth per
revolution. What table feed is needed? If this cutter is 8 in (20.3 cm) in diameter
and is facing a 2-in (5.1-cm) wide part, determine the cutter approach.
Calculation Procedure:
1. Compute the required table feed
For a milling machine, the table feed F
T
in/min ϭ ƒ
t
nR, where ƒ
t
ϭ feed per tooth
per revolution; n
ϭ number of teeth in cutter; R ϭ cutter rpm. For this cutter, F
T
ϭ (0.006) ϫ (12)(400) ϭ 28.8 in/min (1.2 cm/s).
2. Compute the cutter approach
The approach of a milling cutter A
c
in ϭ 0.5D
c
ϪϪw
2
)
0.5
, where D
c
ϭ

2
0.5(D
c
cutter diameter, in; w ϭ width of face of cut, in. For this cutter, A
c
ϭ 0.5(8) Ϫ
0.5(8
2
Ϫ 2
2
)
0.5
ϭ 0.53 in (1.3 cm).
Related Calculations. Use this procedure for any milling cutter whose dimen-
sions and speed are known. These cutters can be used for metals, plastics, and other
nonmetallic materials.
DIMENSIONS OF TAPER AND DOVETAILS
What are the taper per foot (TPF) and taper per inch (TPI) of an 18-in (45.7-cm)
long part having a large diameter d
l
of 3 in (7.6 cm) and a small diameter of d
s
of
1.5 in (3.8 cm)? What is the length of a part with the same large and small diameters
as the above part if the TPF is 3 in/ft (25 cm/m)? Determine the dimensions of
the dovetail in Fig. 7 if B
ϭ 2.15 in (5.15 cm), C ϭ 0.60 in (1.5 cm), and a ϭ
30Њ.A
3
⁄

8
-in (1.0-cm) diameter plug is used to measure the dovetail.
Calculation Procedure:
1. Compute the taper of the part
For a round part TPF in/ft
ϭ 12(d
l
Ϫ d
s
)/L, where L ϭ length of part, in; other
symbols as defined above. Thus for this part, TPF
ϭ 12(3.0 Ϫ 1.5)/18 ϭ 1 in/ft
(8.3 cm / m). And TPI in/in
ϭ (d
l
Ϫ d
s
)/L
2
, or (3.0 Ϫ 1.5)/18 ϭ 0.0833 in/in
(0.0833 cm / cm).
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.16 DESIGN ENGINEERING
FIGURE 7 Dovetail dimensions.
The taper of round parts may also be expressed as the angle measured from the
shaft centerline, that is, one-half the included angle between the tapered surfaces
of the shaft.

2. Compute the length of the tapered part
Converting the first equation of step 1 gives L
ϭ 12(d
l
Ϫ d
s
)/TPF. Or, L ϭ 12(3.0
Ϫ 1.5) /3.0 ϭ 6 in (15.2 cm).
3. Compute the dimensions of the dovetail
For external and internal dovetails, Fig. 7, with all dimensions except the angles in
inches, A
ϭ B ϩ CF ϭ I ϩ HF; B ϭ A Ϫ CF ϭ G Ϫ HF; E ϭ P cot (90 ϩ a/
2)
ϩ P; D ϭ P cot (90 Ϫ a /2) ϩ P; F ϭ 2 tan a; Z ϭ A Ϫ D. Note that P ϭ
diameter of plug used to measure dovetail, in.
With the given dimensions, A
ϭ B ϩ CF,orA ϭ 2.15 ϩ (0.60)(2 ϫ 0.577) ϭ
2.84 in (7.2 cm). Since the plug P is
3
⁄
8
in (1.0 cm) in diameter, D ϭ P cot (90 Ϫ
a/2) ϩ P ϭ 0.375 cot (90 Ϫ
30
⁄
2
) ϩ 0.375 ϭ 1.025 in (2.6 cm). Then Z ϭ A Ϫ D
ϭ 2.840 Ϫ 1.025 ϭ 1.815 in (4.6 cm). Also E ϭ P cot (90 ϩ a/2) ϩ P ϭ 0.375
cot (90
ϩ

30
⁄
2
) ϩ 0.375 ϭ 0.591 in (1.5 cm).
With flat-cornered dovetails, as at I and G, and H
ϭ
1
⁄
8
in (0.3 cm), A ϭ I ϩ
HF. Solving for I,wegetI ϭ A Ϫ HF ϭ 2.84 Ϫ (0.125)(2 ϫ 0.577) ϭ 2.696 in
(6.8 cm). Then G
ϭ B ϩ HF ϭ 2.15 ϩ (0.125)(2 ϫ 0.577) ϭ 2.294 in (5.8 cm).
Related Calculations. Use this procedure for tapers and dovetails in any me-
tallic and nonmetallic material. When a large number of tapers and dovetails must
be computed, use the appropriate tables in the American Machinist’s Handbook.
ANGLE AND LENGTH OF CUT FROM
GIVEN DIMENSIONS
At what angle must a cutting tool be set to cut the part in Fig. 8? How long is the
cut in this part?
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.17
FIGURE 8 Length of cut of a part.
Calculation Procedure:
1. Compute the angle of the cut
Use trigonometry to compute the angle of the cut. Thus, tan a
ϭ opposite side/

adjacent side
ϭ (8 Ϫ 5)/6 ϭ 0.5. From a table of trigonometric functions, a ϭ
cutting angle ϭ 26Њ34Ј , closely.
2. Compute the length of the cut
Use trigonometry to compute the length of cut. Thus, sin a
ϭ opposite side / hy-
potenuse, or 0.4472
ϭ (8 Ϫ 5)/hypotenuse; length of cut ϭ length of hypotenuse
ϭ 3 /0.4472 ϭ 6.7 in (17.0 cm).
Related Calculations. Use this general procedure to compute the angle and
length of cut for any metallic or nonmetallic part.
TOOL FEED RATE AND CUTTING TIME
A part 3.0 in (7.6 cm) long is turned at 100 r/min. What is the feed rate if the
cutting time is 1.5 min? How long will it take to cut a 7.0-in (17.8-cm) long part
turning at 350 r/min if the feed is 0.020 in / r (0.51 mm / r)? How long will it take
to drill a 5-in (12.7-cm) deep hole with a drill speed of 1000 r/min and a feed of
0.0025 in / r (0.06 mm/r)?
Calculation Procedure:
1. Compute the tool feed rate
For a tool cutting a rotating part, ƒ
ϭ L/(Rt), where t ϭ cutting time, min. For this
part, ƒ
ϭ 3.0 /[(100)(1.5)] ϭ 0.02 in/r (0.51 mm / r).
2. Compute the cutting time for the part
Transpose the equation in step 1 to yield t
ϭ L /(Rƒ), or t ϭ 7.0/[(350)(0.020)] ϭ
1.0 min.
3. Compute the drilling time for the part
Drilling time is computed using the equation of step 2, or t
ϭ 5.0 /[(1000)(0.0025)]

ϭ 2.0 min.
Related Calculations. Use this procedure to compute the tool feed, cutting
time, and drilling time in any metallic or nonmetallic material. Where many com-
putations must be made, use the feed-rate and cutting-time tables in the American
Machinist’s Handbook.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.18 DESIGN ENGINEERING
TRUE UNIT TIME, MINIMUM LOT SIZE, AND
TOOL-CHANGE TIME
What is the machine unit time to work 25 parts if the setup time is 75 min and the
unit standard time is 5.0 min? If one machine tool has a setup standard time of 9
min and a unit standard time of 5.0 min, how many pieces must be handled if a
machine with a setup standard of 60 min and a unit standard time of 2.0 min is to
be more economical? Determine the minimum lot size for an operation requiring
3 h to set up if the unit standard time is 2.0 min and the maximum increase in the
unit standard may not exceed 15 percent. Find the unit time to change a lathe
cutting tool if the operator takes 5 min to change the tool and the tool cuts 1.0
min/cycle and has a life of 3 h.
Calculation Procedure:
1. Compute the true unit time
The true unit time for a machine T
u
ϭ S
u
/N ϩ U
s
, where S

u
ϭ setup time, min; N
ϭ number of pieces in lot; U
s
ϭ unit standard time, min. For this machine, T
u
ϭ
75/75 ϩ 5.0 ϭ 6.0 m in.
2. Determine the most economical machine
Call one machine X, the other Y. Then (unit standard time of X, min)(number of
pieces)
ϩ (setup time of X, min) ϭ (unit standard time of Y, min)(number of pieces)
ϩ (setup time of Y, min). For these two machines, since the number of pieces Z is
unknown, 5.0Z
ϩ 9 ϭ 2.0Z ϩ 60. So Z ϭ 17 pieces. Thus, machine Y will be more
economical when 17 more pieces are made.
3. Compute the minimum lot size
The minimum lot size M
ϭ S
u
/(U
s
K), where K ϭ allowable increase in unit-
standard time, percent. For this run, M
ϭ (3 ϫ 60)/[(2.0)(0.15)] ϭ 600 pieces.
4. Compute the unit tool-changing time
The unit tool-changing time U
t
to change from dull to sharp tools is U
t

ϭ T
c
C
t
/l,
where T
c
ϭ total time to change tool, min; C
t
ϭ time tool is in use during cutting
cycle, min; l
ϭ life of tool, min. For this lathe, U
t
ϭ (5)(1) /[(3)(60)] ϭ 0.0278
min.
Related Calculations. Use these general procedures to find true unit time, the
most economical machine, minimum lot size, and unit tool-changing time for any
type of machine tool—drill, lathe, milling machine, hobs, shapers, thread chasers,
etc.
TIME REQUIRED FOR TURNING OPERATIONS
Determine the time to turn a 3-in (7.6-cm) diameter brass bar down to a 2
1
⁄
2
-in
(6.4-cm) diameter with a spindle speed of 200 r/min and a feed of 0.20 in (0.51
mm) per revolution if the length of cut is 4 in (10.2 cm). Show how the turning-
time relation can be used for relief turning, pointing of bars, internal and external
chamfering, hollow mill work, knurling, and forming operations.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.19
FIGURE 9 Turning operations.
Calculation Procedure:
1. Compute the turning time
For a turning operation, the time to turn T
t
min ϭ L /(ƒR), where L ϭ length of
cut, in; ƒ
ϭ feed, in/r; R ϭ work rpm. For this part, T
t
ϭ 4 / [(0.02)(200)] ϭ 1.00
min.
2. Develop the turning relation for other operations
For relief turning use the same relation as in step 1. Length of cut is the length of
the relief, Fig. 9. A small amount of time is also required to handfeed the tool to
the minor diameter of the relief. This time is best obtained by observation of the
operations.
The time required to point a bar, called pointing, is computed by using the
relation in step 1. The length of cut is the distance from the end of the bar to the
end of the tapered point, measured parallel to the axis of the bar, Fig. 9.
Use the relation in step 1 to compute the time to cut an internal or external
chamfer. The length of cut of a chamfer is the horizontal distance L, Fig. 9.
A hollow mill reduces the external diameter of a part. The cutting time is com-
puted by using the relation in step 1. The length of cut is shown in Fig. 9.
Compute the time to knurl, using the relation in step 1. The length of cut is
shown in Fig. 9.
Compute the time for forming, using the relation in step 1. Length of cut is

shown in Fig. 9.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.20 DESIGN ENGINEERING
TIME AND POWER TO DRILL, BORE,
COUNTERSINK, AND REAM
Determine the time and power required to drill a 3-in (7.6-cm) deep hole in an
aluminum casting if a
3
⁄
4
-in (1.9-cm) diameter drill turning at 1000 r/min is used
and the feed is 0.030 in (0.8 mm) per revolution. Show how the drilling relations
can be used for boring, countersinking, and reaming. How long will it take to drill
a hole through a 6-in (15.2-cm) thick piece of steel if the cone height of the drill
is 0.5 in (1.3 cm), the feed is 0.002 in/r (0.05 mm/r), and the drill speed is 100
r/min?
Calculation Procedure:
1. Compute the time required for drilling
The time required to drill T
d
min ϭ L/ƒR, where L ϭ depth of hole ϭ length of
cut, in. In most drilling calculations, the height of the drill cone (point) is ignored.
(Where the cone height is used, follow the procedure in step 4.) For this hole, T
d
ϭ 3 /[(0.030) ϫ (1000)] ϭ 0.10 min.
2. Compute the power required to drill the hole
The power required to drill, in hp, is hp

ϭ 1.3LƒCK, where C ϭ cutting speed,
ft/min, sometimes termed surface feet per minute sfpm
ϭ
␲
DR/12; K ϭ power
constant from Table 3. For an aluminum casting, K
ϭ 3. Then hp ϭ
(1.3)(3)(0.030)(
␲
ϫ 0.75 ϫ 1000/12)(3) ϭ 66.0 hp (49.2 kW). The factor 1.3 is
used to account for dull tools and for overcoming friction in the machine.
3. Adapt the drill relations to other operations
The time and power required for boring are found from the two relations given
above. The length of the cut
ϭ length of the bore. Also use these relations for
undercutting, sometimes called internal relieving and for counterboring. These same
relations are also valid for countersinking, center drilling, start or spot drilling, and
reaming. In reaming, the length of cut is the total depth of the hole reamed.
4. Compute the time for drilling a deep hole
With parts having a depth of 6 in (15.2 cm) or more, compute the drilling time
from T
d
ϭ (L ϩ h)/(ƒR), where h ϭ cone height, in. For this hole, T
d
ϭ (6 ϩ
0.5)/[(0.002)(100)] ϭ 32.25 min. This compares with T
d
ϭ L /ƒR ϭ 6/
[(0.002)(100)]
ϭ 30 min when the height of the drill cone is ignored.

TIME REQUIRED FOR FACING OPERATIONS
How long will it take to face a part on a lathe if the length of cut is 4 in (10.2
cm), the feed is 0.020 in/r (0.51 mm/r) and the spindle speed is 50 r/min? Deter-
mine the facing time if the same part is faced by an eight-tooth milling cutter
turning at 1000 r/min and having a feed of 0.005 in (0.13 mm) per tooth per
revolution. What table feed is required if the cutter is turning at 50 r/min? What
is the feed per tooth with a table feed of 4.0 in/min (1.7 mm/s)? What added table
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.21
TABLE 3 Power Constants for Machining
travel is needed when a 4-in (10.2-cm) diameter cutter is cutting a 4-in (10.2-cm)
wide piece of work?
Calculation Procedure:
1. Compute the lathe facing time
For lathe facing, the time to face T
f
min ϭ L /(ƒR), where the symbols are the same
as given for previous calculation procedures in this section. For this part, T
f
ϭ 4/
[(0.02)(50)]
ϭ 4.0 min.
2. Compute the facing time using a milling cutter
With a milling cutter, T
f
ϭ L /(ƒ
t

nR), where ƒ
t
ϭ feed per tooth, in / r; n ϭ number
of teeth on cutter; other symbols as before. For this part, T
f
ϭ 4/[(0.005)(8) ϫ
(1000)] ϭ 0.10 min.
3. Compute the required table feed
In a milling machine, the table feed F
t
in/min ϭ ƒ
t
nR. For this machine, F
t
ϭ
(0.005) ϫ (8)(50) ϭ 2.0 in/min (0.85 mm/s).
4. Compute the feed per tooth
For a milling machine, the feed per tooth, in/r, ƒ
t
ϭ F
t
/Rn. In this machine, ƒ
t
ϭ
4.0/[(50)(8)] ϭ 0.01 in/r (0.25 mm/r).
5. Compute the added table travel
In face milling, the added table travel A
t
in ϭ 0.5[D
c

Ϫ ( D Ϫ W
2
)
0.5
], where the
2
c
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.22 DESIGN ENGINEERING
symbols are the same as given earlier. For this cutter and work, A
t
ϭ 0.5[ 4 Ϫ (4
2
Ϫ 4
2
)
0.5
] ϭ 2.0 in (5.1 cm).
THREADING AND TAPPING TIME
How long will it take to cut a 4-in (10.2-cm) long thread at 100 r/min if the rod
will have 12 threads per inch and a button die is used? The die is backed off at
200 r/min. What would the threading time be if a self-opening die were used
instead of a button die? What will the threading time be for a single-pointed thread-
ing tool if the part being threaded is aluminum and the back-off speed is twice the
threading speed? The rod is 1 in (2.5 cm) in diameter. How long will it take to tap
a 2-in (5.1-cm) deep hole with a 1-14 solid tap turning at 100 r /min? How long
will it take to mill-thread a 1-in (2.5-cm) diameter bolt having 15 threads per inch

3 in (7.6 cm) long if a 4-in (10.2-cm) diameter 20-flute thread-milling hob turning
at 80 r/min with a 0.003 in (0.08-mm) feed is used?
Calculation Procedure:
1. Compute the button-die threading time
For a multiple-pointed tool, the time to thread T
t
ϭ Ln
t
/R, where L ϭ length of
cut
ϭ length of thread measured parallel to thread longitudinal axis, in; n
t
ϭ number
of threads per inch. For this button die, T
t
ϭ (4)(12) / 100 ϭ 0.48 min. This is the
time required to cut the thread.
Compute the back-off time B min from B
ϭ Ln
t
/R
B
, where R
B
ϭ back-off rpm,
or B
ϭ (4)(12)/200 ϭ 0.24 min. Hence, the total time to cut and back-off ϭ T
t
ϩ
B ϭ 0.48 ϩ 0.24 ϭ 0.72 min.

2. Compute the self-opening die threading time
With a self-opening die, the die opens automatically when it reaches the end of the
cut thread and is withdrawn instantly. Therefore, the back-off time is negligible.
Hence, the time to thread
ϭ T
t
ϭ Ln
t
/R ϭ (4)(12)/100 ϭ 0.48 min. One cut is
usually sufficient to make a suitable thread.
3. Compute the single-pointed tool cutting time
With a single-pointed tool, more than one cut is usually necessary. Table 4 lists the
number of cuts needed with a single-pointed tool working on various materials.
The maximum cutting speed for threading and tapping is also listed.
Table 4 shows that four cuts are needed for an aluminum rod when a single-
pointed tool is used. Before computing the cutting time, compute the cutting speed
to determine whether it is within the recommended range given in Table 4. From
a previous calculation procedure, C
ϭ R
␲
d/12, or C ϭ (100)(
␲
)(1.0)/12 ϭ 26.2
ft/min (13.3 cm/s). Since this is less than the maximum recommended speed of
30 r / min, Table 4, the work speed is acceptable.
Compute the time to thread from T
t
ϭ Ln
t
c/R, where c ϭ number of cuts to

thread, from Table 4. For this part, T
t
ϭ (4)(12)(4) /100 ϭ 1.92 min.
If the tool is backed off at twice the threading speed, and the back-off time B
ϭ Ln
t
c/R
B
, B ϭ (4)(12)(4) /200 ϭ 0.96 min. Hence, the total time to thread and
back off
ϭ T
t
ϩ B ϭ 1.92 ϩ 0.96 ϭ 2.88 min. In some shapes, a single-pointed
tool may not be backed off; the tool may instead be repositioned. The time required
for this approximates the back-off time.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.23
TABLE 4 Number of Cuts and Cutting Speed for Dies
and Taps
4. Compute the tapping time
The time to tap T
t
min ϭ Ln
t
/R. With a solid tap, the tool is backed out at twice
the tapping speed. With a collapsing tap, the tap is withdrawn almost instantly
without reversing the machine or tap.

For this hole, T
t
ϭ (2)(12)/100 ϭ 0.28 min. The back-off time B ϭ Ln
t
/R
B
ϭ
(2)(14)/200 ϭ 0.14 min. Hence, the total time to tap and back off ϭ T
t
ϩ B ϭ
0.28 ϩ 0.14 ϭ 0.42 min.
The maximum spindle speed for tapping should not exceed 250 r/min. Use the
cutting-speed values given in Table 4 in computing the desirable speed for various
materials.
5. Compute the thread-milling time
The time for thread milling is T
t
ϭ L /(ƒnR), where L ϭ length of cut, in ϭ cir-
cumference of work, in; ƒ
ϭ feed per flute, in; n ϭ number of flutes on hob; R ϭ
hob rpm. For this bolt, T
t
ϭ 3.1416 /[(0.003)(20)(80)] ϭ 0.655 min.
Note that neither the length of the threaded portion nor the number of threads
per inch enters into the calculation. The thread hob covers the entire length of the
threaded portion and completes the threading in one revolution of the work head.
TURRET-LATHE POWER INPUT
How much power is required to drive a turret lathe making a
1
⁄

2
-in (1.3-cm) deep
cut in cast iron if the feed is 0.015 in / r (0.38 mm/r), the part is 2.0 in (5.1 cm) in
diameter, and its speed is 382 r/min? How many 1.5-in (3.8-cm) long parts can be
cut from a 10-ft (3.0-m) long bar if a
1
⁄
4
-in (6.4-mm) cutoff tool is used? Allow
for end squaring.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
26.24 DESIGN ENGINEERING
TABLE 5 Turret-Lathe Power Constant
Calculation Procedure:
1. Compute the surface speed of the part
The cutting, or surface, speed, as given in a previous calculation procedure, is
C
ϭ R
␲
d/12, or C ϭ (382)(
␲
)(2.0)/12 ϭ 200 ft/min (1.0 m/s).
2. Compute the power input required
For a turret lathe, the hp input hp ϭ 1.33DƒCK, where D ϭ cut depth, in; ƒ ϭ
feed, in / r; K ϭ material constant from Table 5. For cast iron, K ϭ 3.0. Then hp ϭ
(1.33)(0.5)(0.015)(200)(3.0) ϭ 5.98, say 6.0 hp (4.5 kW).
3. Compute the number of parts that can be cut

Allow 2 in (5.1 cm) on the bar end for checking and
1
⁄
2
in (1.3 cm) on the opposite
end for squaring. With an original length of 10 ft
ϭ 120 in (304.8 cm), this leaves
120
Ϫ 2.5 ϭ 117.5 in (298.5 cm) for cutting.
Each part cut will be 1.5 in (3.8 cm) long
ϩ 0.25 in (6.4 mm) for the cutoff,
or 1.75 in (4.4 cm) of stock. Hence, the number of pieces which can be cut
ϭ
117.5/1.75 ϭ 67.1, or 67 pieces.
Related Calculations. Use this procedure to find the turret-lathe power input
for any of the materials, and similar materials, listed in Table 5. The parts cutoff
computation can be used for any material—metallic or nonmetallic. Be sure to
allow for the width of the cutoff tool.
TIME TO CUT A THREAD ON AN ENGINE LATHE
How long will it take an engine lathe to cut an acme thread having a length of 5
in (12.7 cm), a major diameter of 2 in (5.1 cm), four threads per inch (1.575 threads
per centimeter), a depth of 0.1350 in (3.4 mm), a cutting speed of 70 ft/min (0.4
m/s), and a depth of cut of 0.005 in (0.1 mm) per pass if the material cut is medium
steel? How many passes of the tool are required?
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING
METALWORKING AND NONMETALLIC MATERIALS PROCESSING 26.25
TABLE 6 Thread Cutting Speeds

Calculation Procedure:
1. Compute the cutting time
For an acme, square, or worm thread cut on an engine lathe, the total cutting time
T
t
min, excluding the tool positioning time, is found from T
t
ϭ Ld
t
Dn
t
/(4Cd
c
),
where L
ϭ thread length, in, measured parallel to the thread longitudinal axis; d
t
ϭ thread major diameter, in; D ϭ depth of thread, in; n
t
ϭ number of threads per
inch; C
ϭ cutting speed, ft/min; d
c
ϭ depth of cut per pass, in.
For this acme thread, T
t
ϭ (5)(2)(0.1350)(4)/[4(70)(0.005)] ϭ 3.85 min. To this
must be added the time required to position the tool for each pass. This equation
is also valid for SI units.
2. Compute the number of tool passes required

The depth of cut per pass is 0.005 in (0.1 mm). A total depth of 0.1350 in (3.4
mm) must be cut. Therefore, the number of passes required
ϭ total depth cut, in/
depth of cut per pass, in
ϭ 0.1350 /0.005 ϭ 27 passes.
Related Calculations. Use this procedure for threads cut in ferrous and non-
ferrous metals. Table 6 shows typical cutting speeds.
TIME TO TAP WITH A DRILLING MACHINE
How long will it take to tap a 4-in (10.2-cm) deep hole with a 1
1
⁄
2
-in (3.8-cm)
diameter tap having six threads per inch (2.36 thread per centimeter) if the tap turns
at 75 r/min?
Calculation Procedure:
1. Compute the tap surface speed
By the method of previous calculation procedure, C
ϭ R
␲
d/12 ϭ (75)(
␲
)(1.5)/12
ϭ 29.5 ft/min (0.15 m / s).
2. Compute the time to tap the hole and withdraw the tool
For tapping with a drilling machine, T
t
ϭ Dn
t
D

c
␲
/(8C ), where D ϭ depth of cut
ϭ depth of hole tapped, in; n
t
ϭ number of threads per inch; D
c
ϭ cutter diameter,
in
ϭ tap diameter, in. For this hole, T
t
ϭ (4)(6)(1.5)
␲
/[8(29.5)] ϭ 0.48 min, which
is the time required to tap and withdraw the tool.
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METALWORKING AND NONMETALLIC MATERIALS PROCESSING