Copyright 2007 - Michael Bush1 Industrial Application of Basic Mathematical Principles Session 12...

Copyright 2007 - Michael Bush

1

Industrial Application of BasicMathematical Principles

Session 12Volume and CubicMeasurement


2

Industrial Application of BasicMathematical Principles Session 12 In-Class

Volume and Cubic Measurement

Volume or Cubic measure refers to measurement of the space occupied by a body.

Each body has three linear dimensions: length, height, and depth.

The principles of volume measure are applied in this unit to three common shapes and the combination of these three shapes: (1) the cube, (2) the rectangular solid, and (3) the cylinder.


3


Developing a Concept of Volume Measure

Volume measure is the product of three linear measurements.

Each measurement must be in the same linear unit.The product is called the volume of the solid or body. Volume is expressed in cubic units.


4



The standard unit of volume or cubic measure is the cubic inch.

The cubic inch is the space occupied by a body. This cube is 1 linear inch long. 1 inch high, and 1 inch deep.

CORNERS ARE ALL

AT RIGHT ANGLES

ONE FOOT(DEPTH)

ONE FOOT(LENGTH)

ONE INCH(LENGTH)

ONE INCH(DEPTH)O

NE

INC

H(H

EIG

HT

)

ON

E F

OO

T(H

EIG

HT

)


5



One cubic foot is the space occupied by a cubical body that is 1’ long by 1’ high by 1’ deep

One cubic yard is the space occupied by a cube that is 1 yd long, 1 yard high, and 1 yard deep.

ONE YARDOR 3 FEET

ONE YARDOR 3 FEET

ONE INCH(LENGTH)

ONE FOOTOR 12 INCHO

NE

FO

OT

OR

12

INC

H

ON

E Y

AR

DO

R 3

FE

ET

ONECUBIC INCH

ONECUBIC FOOT =1,728 CUBIC INCHES

ONE CUBIC YARD = 27 CUBIC FEET


6


Developing a concept of Volume Measure

When expressing a cubic measurement, you can do it several different ways. A cube measuring 6 inches a side can be expressed the following ways

216 inches cubed or 16 cubic inches

216 in³


7


Developing a concept of Volume Measure

Table of Cubic or Volume Measure

Standard unit of measure = 1 cubic inch

1,728 cubic inch = 1 cubic foot

27 cubic foot = 1 cubic yard


8


Expressing Units of Volume Measure

A volume in cubic inches may be expressed in cubic feet by dividing by 1,728 (1,728 cubic inch = 1 cubic foot).

Volumes given in cubic feet may be expressed in cubic yards by dividing by 27 (27 cubic feet = 1 cubic yard)


9


Rule for Expressing Unit of Volume Measure as a larger unit

Divide the given volume by the number of cubic units contained in the required larger units.

Express the quotient in terms of the required larger cubic units.


10


Expressing Unit of Volume Measure as a larger unit

Express 5,184 cubic inch as cubic feet.

31728

5184

Divide the given volume (5,184) by the number of cubic inches contained in one cubic foot (1,728). Express the quotient in terms of the required larger cubic units.

feet cubic 31728

5184


11


Rule for Expressing a larger Unit of Volume Measure as a smaller unit

Multiply the given unit of volume by the number of smaller cubic units contained in one of the required smaller units.

Express the product in terms of the required smaller cubic units.


12


Expressing a larger Unit of Volume Measure as a smaller unit

Express 10 cubic yards in cubic feet.

2702710 x Multiply the given volume (10) by the number of cubic feet contained in one cubic yard (27). Express the product (270) in terms of the required smaller cubic units.

feet cubic x 2702710


13


Rule for Expressing two or more Units of Volume Measure

If the volume expressed in two or more units of measure is to be expressed as a smaller unit.

Multiply those units of measure that are not in terms of the required unit by the number of smaller units equal to the given unit.

Add the remaining units in the original given volume to this product.


14


Rule for Expressing two or more Units of Volume Measure

Express 2 cubic yards, 10 cubic feet in cubic feet.

feet cubic x 54272 Multiply the 2 cubic yards by the number of cubic feet contained in one cubic yard (27). Add to the product (54) the remainder of the given volume (10 cubic feet).

feet cubic 64

feet cubic 10

feet cubic 54


15


Applying Volume Measure to the Cube and Rectangular Solids

In volume measure the three linear dimensions that express length, height, and depth or their equivalents are multiplied to determine the cubical contents of a regular solid.

The product is cubic inches, cubic feet, cubic yards, and etc.

When the area of one surface is extended in a third direction, a solid is formed .


16


Applying Volume Measure to the Cube and Rectangular SolidsIf the original surface is an square and its face is extended to add depth, the resulting figure is a solid.

When all the corners are square and all lengths are equal, it is called a cube or cubical solid.

ORIGINALSURFACESQUARE

DEPTH

HE

IGH

T

LENGTH

DEPTHEQUAL TO

OTHER SIDES

CUBICAL SOLID WHERE ALL SIDES ARE EQUAL AND AT RIGHT ANGLES


17


Rule for computing the Volume of a CubeExpress the dimensions for length, depth, and height in the same linear unit of measure when needed.

Multiply the length x depth x height.

Express the product in terms of units of volume measure.

Express the resulting product, if needed, in lowest terms.


18


Computing the Volume of a Cube

Find the volume of a cube, each side of which is 8 inches long

5128 x 8 x 8 Multiply the given length (8) by the depth (8) by the height (8).

Express the product (512) in terms of volume measure. inches cubic 5128 x 8 x 8


19



Find the volume of a cube that measures 1’ –9” on a side.

926121 x 21 x 21 Multiply the given length (21) x depth (21) x height (21).

Express the product (9261) in terms of volume.

Express 1’ -9” as 21”

inches cubic 926121 x 21 x 21

Express as cu ft and cu in

in cu 621 ft, cu 51728

9261


20



Find the volume of a cube that measures 1’ –9” on a side.

621 1728 x 6423

Multiply the given length (1¾) x depth (1¾) x height (1¾). Express the 23/64 in terms of cubic inch.

Express 1’ -9” as 1¾’

Add the cu in to cu ft in cu 621 ft, cu 5

642354

31 x 431 x 4

31


21


Developing a concept of a Rectangular Solid

A Rectangular Solid resembles a cube except that the faces or sides are rectangular in shape.

The volume of a rectangular solid is equal to the length x depth x height.

DEPTH

LENGTH

HE

IGH

T


22


Rule for finding the Volume of a Rectangular SolidExpress the dimensions for length, depth, and height in the same linear unit of measure when needed.

Multiply the length x depth x height.




23


Computing the Volume of a Rectangular SolidFind the volume the block.

88029 x 10 x 32 Multiply the given length (32) by the depth (10) by the height (9). Express the product (512) in terms of volume measure.

33 in 1152 ,ft 11728

2880

10"2' - 8"

9"

Express all as inch

3in 29 x 10 x 32 880

Express in lowest terms


24


Application of Volume Measure to CylindersThe volume of a cylinder is the number of cubic

units that it contains.

The volume of a cylinder is found by multiplying the area of the base times the length or height.

HEIGHTOR

LENGTH

DIAMETER LENGTH


25


Rule for finding the Volume of a Cylinder

Express the dimensions for length, depth, and height in the same linear unit of measure when needed.

Compute the area of the base.

Multiply the area by the height or length of cylinder.




26


Computing the Volume of a Cylinder

Find the volume of a cylinder 3” in diameter and 10” long, correct to two decimal places.

70.68610 x 7.0686 Multiply the given area (7.0686) by the height (10). Express the product in terms of volume measure. 33 in 70.69in 0.686 7

Compute area of base (.7854xD²)

Express as two places

7.06863 x .7854 2

3in 70.68610 x 7.0686


27


Application of Volume Measure to Irregular FormsIn addition to regular solids, many objects are a

combination of various shapes in modified form.

MODIFIEDFORM

REGULARSOLID

IRREGULAR SOLID

REGULARSOLID

The volume of an irregular solid can be found by dividing it into solids having regular shapes.


28


Application of Volume Measure to Irregular FormsThe volume of each regular or modified solid

form can be computed

MODIFIEDFORM

REGULARSOLID

IRREGULAR SOLID

REGULARSOLID

The sum of the separate volumes equals the volume of the irregular solid. The sum of the parts is equal to the whole.


29


Rule for finding the Volume of an Irregular SolidDivide the solid into regular forms and express the dimensions for length, depth, and height in the same linear unit of measure when needed.

Compute the volume of each regular solid or part of one.

Add the separate volumes.



30


Computing the Volume of an Irregular Solid

Determine the volume of the brass casting, correct to two decimal places.

6"

6"

6"

2"

8"


31



Divide the irregular form into two regular solids.

Compute volume of cube 6” x 6” x 6”

1626 x 6 x 6

6"

6"

6"

2"

8"


32



241.1328 25.1328 216 Add the separate volumes

Express the product in terms of volume measure.

33 in 241.13in 241.1328

Compute volume of cylinder (area of base x height)

Express as two places

25.13288 x 2 x .7854 2

3in 241.1328 25.1328 216


33


Application of Volume Measure to Liquid MeasureConstant reference is made to the

measurements of various liquids used in industry.Liquids are measured by cubical units of measure known as liquid measure.

One common method of determining liquid capacity requires, first, computing the cubical contents of the object. Second, the resulting units of volume measure are then changed to liquid units of measure.


34


Application of Volume Measure to Liquid MeasureThe standard units of liquid measure include the

ounce, pint, quart, gallon, and barrel

Table of Liquid Measure

16 ounces (oz) = 1 pint (pt)

2 pints (pt) = 1 quart (qt)

4 quarts (qt) = 1 gallon (gal)

31½ gallons (gal) = 1 barrel (bbl)


35


Application of Volume Measure to Liquid MeasureThe gallon contains 231 cubic inches of liquid.

With this known value, it is possible to solve problems requiring the use of liquid measure.

231 CUBIC INCHES

ONEGALLON

=


36


Rule for changing units of Volume Measure to units of Liquid Measure

Compute the volume of the object in terms of cubic inches.

Divide the computed volume by 231 (231 cubic inches = 1 gal)

Express the quotient in terms of units of liquid measure.


37


Computing the Liquid Capacity

Express the quotient in terms of liquid measure.

Divide volume in cubic inches (1155) by the number of cubic inches (231) in one gallon

5231

1155

Determine the liquid capacity of a coolant tank whose volume is 1155 cubic inches.

gal 5231

1155


38


Rule for Expressing Larger Unit of Liquid Measure in Smaller Unit

Determine the number of smaller units of liquid measure in one larger unit.

Multiply the given units by this number

Express the product in terms of the required larger cubic units.


39


Expressing Larger Unit of Liquid Measure in Smaller Unit


Multiply the gallons by the quarts per gallon

184 x 214

Express 4½ gallons in quarts.

Determine the number of quarts in one gallon.

4 quarts = 1 gallon

quarts 184 x 214


40



Determine the number of ounces per pint

Multiply the quarts by the pints per quart.

4162 x 2

13

Express 3½ quarts in pints and ounces.

2 pints = 1 quartDetermine the number of pints in one quart.

16 ounces = 1 pint


41



Express the sum in terms of liquid measure.

Multiply the fractional part by the ounces per pint.

416 x 41

ounces 4 pints, 6

Combine the pints and ounces.


42


Rule for Expressing Smaller Units of Liquid Measure in Larger Units

Determine the number of smaller units of liquid measure in one larger unit.

Divide the given units by this number

Express the product in terms of the required larger cubic units.

Where the result is a mixed number, the fractional part can be changed to the next smaller unit.


43


Expressing Smaller Unit of Liquid Measure in Larger Unit


Divide the pints by the pints per gallon

38

24

Express 24 pints in gallons.

Determine the number of pints in one gallon.

8 pints = 1 gallon

gallons 38

24


44



Divide the ounces by the ounces per gallon.

8

32

128

304

Express 304 ounces in gallons, quarts, and pints in gallons.

128 oz = 1 gal

Determine the number of ounces in one gallon.

This is the number of whole gallons. 2 gal


45



This is the whole number of quarts.

2

114 x

8

3Multiply the fractional part by

the number of quarts per gallon.

32 oz = 1 galDetermine the number of quarts in one gallon.

1 quart


46



Combine

This is the whole number of quarts.

12 x 2

1Multiply the fractional part by

the number of pints per quart.

2 pints = 1 quart

Determine the number of pints in one quart.

1 pint

2 gallons, 1 quart, 1 pint


47


Assignment Unit 21 Problem A 1 Page 161Express Each Of The Volumes in Specified Unit

a. 2 cu ft in cu in.

1 cu ft = 1728 cu in

1728 cu in/ cu ft x 2 cu ft = 3456 cu in


48



b. 1½ cu ft in cu in.


1728 cu in/ cu ft x 1½ cu ft = 2592 cu in


49



c. 3⅝ cu ft in cu in.


1728 cu in/ cu ft x 3⅝ cu ft = 6,264 cu in


50



d. 10 cu ft, 19 cu in in cu in.


1728 cu in/ cu ft x 10 cu ft = 17,280 cu in

17,280 cu in + 19 cu in = 17,299 cu in


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e. 3456 cu in in cu ft.


3456 cu in 1728 cu in/ cu ft = 2 cu ft


52



f. 18.144 cu in in cu ft.


18.144 cu in 1728 cu in/ cu ft = 0.0105 cu ft


53


Assignment Unit 21 Problem B 1 A Page 162Determine the Volume of Cube A

A B C

Length 6 8½ 1’-6”

Depth 6 8½ 1’-6”

Height 6 8½ 1’-6”

Volume = Side3

Volume = 63

Volume = 216


54


Assignment Unit 21 Problem B 2 A Page 162Compute the Value of Rectangular Solid A

5"10"

4"

AVolume = L x W x HVolume = 10” x 5” x 4”Volume = 200 in3


55


Assignment Unit 21 Problem C 2 Page 162Determine the Volume

2'-8"

4"

30'-0" 8"

10'-0"

2'-0"

Volume UpperV = l x w x hV = 30 x 2 x 10V = 600 ft³

Volume LowerV = 30 x 2⅔ x ⅔V = 53.367 ft³

Upper + Lower= 653.367 ft³

653.367 ft³ 27 ft³/yd³ =24.20 yd³ =

24 yd³


56


Assignment Unit 21 Problem C 3 A Page 163Compute Volume of Hollow Rectangular Solid

6

12

4

3

1 1/2

1

2

Volume of WholeV = l x w x h

V = 12 x 6 x 4V = 288

Volume of Cut-outV = 2 x 3 x 12V = 72 Volume = Whole – Cut-out

288 – 72 =

216 in³


57


Assignment Unit 21 Problem D 1 A Page 163Determine the Volume of Cylinder A

A B C

Diameter 4 12.5

Radius 1.6

Length 10 24.5 6.4

Volume = .7854D²H

Volume = .7854(4²)10

Volume = 125.66 in³

Volume = .7854x16x10


58


Assignment Unit 21 Problem D 2 A Page 163Determine the Liquid Capacity of Cistern A

Cistern

InsideDiameter

Height

A 4’-0” 6’-0”

B 5’-6” 8’-0”

C 6’-6” 8’-6”

1 cu ft = 7½ gal

Volume = .7854D²HVol. = .7854(4)²6Vol. = 75.398 ft³

1 cu ft = 7½ gal75.398 ft³ x 7½ gal/ft³ =565.5 gal


59


Assignment Unit 21 Problem D 4 A Page 164Determine the Volume

6

2½ 4½

A

WEIGHT OF BRASS = .30 POUND PER CUBIC INCH 10 REQUIRED

Whole VolumeVol. = .7854 D² HVol. = .7854 x 4½² x 6Vol. = 95.426 in³

Volume Cored HoleVol. = .7854 x 2½² x 6Vol. = 29.453 in³

Total VolumeVol. = Whole -

CoreVol. = 95.426 – 29.453Vol. = 65.973 in³

Each WeightWt. = in³ x lb/in³Wt. = 65.973 in³ x .30 lb/in³ =Total WeightTotal Wt. = ea Wt. x Total #Total Wt. = 19.792 lb x 10 =

19.792 lb

197.92 lb


60


Assignment Unit 21 Problem D 4 B Page 164Determine the Volume

Whole VolumeVol. = .7854 D² HVol. = .7854 x 2¾² x 12Vol. = 71.275 in³

Volume Cored HoleVol. = .7854 x 1¼² x 12Vol. = 14.726 in³

Total VolumeVol. = Whole -

CoreVol. = 71.275 – 14.726Vol. = 56.549 in³

Each WeightWt. = in³ x lb/in³Wt. = 56.549 in³ x .32 lb/in³ =Total WeightTotal Wt. = ea Wt. x Total #Total Wt. = 18.096 lb x 24 =

18.096 lb

434.296 lb

12

1-1/

4

2-3/

4

B

WEIGHT OF BRONZE = .32 POUND PER CUBIC INCH 24 REQUIRED


61


Assignment Unit 21 Problem D 6 A Page 164Determine the Volume

2-1/2

6-1/4

6

2-1/2

A

QUANTITY - 204

Volume Large DiameterVol. = .7854 x d² x h Vol. = .7854 x 6¼² x 2½Vol. = 76.699 in³

Volume Small DiameterVol. = .7854 x 4² x

6Vol. = 75.398 in³

Volume of Hole

Vol. = .7854 x 2½² x 8½Vol. = 41.724 in³

Total Volume

Vol. = L + S - HVol. = 110.373 in³Each WeightWt. = Ea x lb/in³Wt. = 9.934 lbTotal WeightEa. Wt. x Total Number9.934 lb x 20 =Total Weight =198.68 lb


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Assignment Unit 21 Problem E 1 Page 164

Width - 12”Height - 6”Length - 1’-4”Quantity - 20Diameter of Cored Holes - 2”Number of Cored Holes - 3Weight of Cast Iron - .28 lb/in³

Note: Cored holes run through entire length

Volume of SolidVol. = l x w x hVol. = 16 x 12 x 6Vol. = 1152 in³

Volume of Cored HoleVol. = .7854D²HVol. = 50.266 in³

Total VolumeSolid – Cored Holes x 3Vol. = 1152 – 3(50.266)Vol. = 1001.203 in³

Wt. = in³ x lb/in³Wt. = 1001.203 in³ x .28 lb/in³Wt. = 280.337 lb

Weight of Each

Total Weight = each x #Total Weight = 280.337 lb x 20Total Weight = 5,606.7 lb


63


Assignment Unit 21 Problem F 1 A Page 165Convert to the indicated units

a. 4 gal qt

b. 6½ gal qt

c. 3¾ gal qt

d. 6½ qt pt

e. 5¼ qt pt

4 qt/gal x 4 gal =4 qt/gal x 6½ gal =4 qt/gal x 3¾ gal =2 pt/qt x 6½ qt =2 pt/qt x 5¼ qt =

16 qt

26 qt

15 qt

13 pt

10½ pt


64


Assignment Unit 21 Problem F 2 A Page 165Determine the liquid capacity

Volume = L x W x HVolume = 23” x 12” x 8½”Volume = 2,346 in3

2,346 in3 ÷ 231 in3/gal =

10.156 gal =

10 gal


65


Volume Measurement Problem 1 Page 141

S

S

S

Side Volume

1. 46 mm

2. 19 in

3. 155.5 mm

4. 27.7 in

5. 125.75 mm

6. 30.26 in

Volume = S³

Volume = 46³

Volume = 97,336 mm³

97,336 mm³


66



S

S

S

Side Volume

1. 46 mm 97,336 mm³

2. 19 in

3. 155.5 mm

4. 27.7 in

5. 125.75 mm

6. 30.26 in

Volume = S³

Volume = 19³

Volume = 6,859 in³

6,859 in³


67



D

Radius Volume

7. 12.0 in

8. 77.3 mm

9. 13.25 in

10. 105.25 mm

11. 62.875 in

12. 18.625 mm

3

πr4 V

3

312xx4 3π

V 373928.21714

V 3in 7,238.246 V

7,238.246 in³


68



D

Radius Volume

7. 12.0 in 7,238.246 in³

8. 77.3 mm

9. 13.25 in

10. 105.25 mm

11. 62.875 in

12. 18.625 mm

3

πr4 V

3

3377xx4 3.π

V 3

2.5804293 V

3mm 41,934,764. V

1,934,764.4 mm³


69



H

R

Radius Height Volume

13. 10.0 in 25.0 in

14. 27 mm 54 mm

15. 36.5 in 67.5 in

16. 14.8 mm 29.6 mm

17. 21.75 in 61.25 in

18. 120.62 mm

398.37 mm V = R²H

V = R²H V = (10.0)²25.0

V = 7,854 in³

7,854 in³


70



H

R

Radius Height Volume

13. 10.0 in 25.0 in 7,854 in³

14. 27 mm 54 mm

15. 36.5 in 67.5 in

16. 14.8 mm 29.6 mm

17. 21.75 in 61.25 in

18. 120.62 mm

398.37 mm V = R²H

V = R²H V = (27)²54 V = 123,672.22 mm³

123,672.22 mm³


71



H

D

Diameter Height Volume

19. 15 in 35.0 in

20. 37 mm 85 mm

21. 52.5 in 105.5 in

22. 39.7 mm 88.2 mm

23. 50.75 in 122.25 in

24. 42.67 mm

184.39 mmV = 0.2618HD²

V = 0.2618HD²V = 0.2618(35.0)15² V = 2,061.675 in³

2,061.675 in³


72



H

D

Diameter Height Volume

19. 15 in 35.0 in 2,061.675 in³

20. 37 mm 85 mm

21. 52.5 in 105.5 in

22. 39.7 mm 88.2 mm

23. 50.75 in 122.25 in

24. 42.67 mm

184.39 mmV = 0.2618HD²

V = 0.2618HD²V = 0.2618(85)37² V = 30,464.357

mm³

30,464.357 mm³

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Copyright 2007 - Michael Bush1 Industrial Application of Basic Mathematical Principles Session 12...

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