copyright 2007 - michael bush1 industrial application of basic mathematical principles session 12...
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Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles
Session 12Volume and CubicMeasurement
Copyright 2007 - Michael Bush
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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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Developing a Concept of Volume Measure
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
)
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Developing a Concept of Volume Measure
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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)
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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 .
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Computing the Volume of a Cube
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Computing the Volume of a Cube
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Express the product in terms of units of volume measure.
Express the resulting product, if needed, in lowest terms.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Express the product in terms of units of volume measure.
Express the resulting product, if needed, in lowest terms.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Express the product in terms of units of volume measure.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Computing the Volume of an Irregular Solid
Determine the volume of the brass casting, correct to two decimal places.
6"
6"
6"
2"
8"
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Computing the Volume of an Irregular Solid
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"
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Computing the Volume of an Irregular Solid
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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)
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
=
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Larger Unit of Liquid Measure in Smaller Unit
Express the quotient in terms of liquid measure.
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Larger Unit of Liquid Measure in Smaller Unit
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Larger Unit of Liquid Measure in Smaller Unit
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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.
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Smaller Unit of Liquid Measure in Larger Unit
Express the quotient in terms of liquid measure.
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Smaller Unit of Liquid Measure in Larger Unit
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Smaller Unit of Liquid Measure in Larger Unit
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Expressing Smaller Unit of Liquid Measure in Larger Unit
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Assignment Unit 21 Problem A 1 Page 161Express Each Of The Volumes in Specified Unit
b. 1½ cu ft in cu in.
1 cu ft = 1728 cu in
1728 cu in/ cu ft x 1½ cu ft = 2592 cu in
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Assignment Unit 21 Problem A 1 Page 161Express Each Of The Volumes in Specified Unit
c. 3⅝ cu ft in cu in.
1 cu ft = 1728 cu in
1728 cu in/ cu ft x 3⅝ cu ft = 6,264 cu in
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Assignment Unit 21 Problem A 1 Page 161Express Each Of The Volumes in Specified Unit
d. 10 cu ft, 19 cu in in cu in.
1 cu ft = 1728 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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Assignment Unit 21 Problem A 1 Page 161Express Each Of The Volumes in Specified Unit
e. 3456 cu in in cu ft.
1 cu ft = 1728 cu in
3456 cu in 1728 cu in/ cu ft = 2 cu ft
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Assignment Unit 21 Problem A 1 Page 161Express Each Of The Volumes in Specified Unit
f. 18.144 cu in in cu ft.
1 cu ft = 1728 cu in
18.144 cu in 1728 cu in/ cu ft = 0.0105 cu ft
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 2 Page 141
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 7 Page 142
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
Copyright 2007 - Michael Bush
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 8 Page 142
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 13 Page 143
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
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Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 14 Page 143
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
Copyright 2007 - Michael Bush
71
Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 19 Page 144
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
Copyright 2007 - Michael Bush
72
Industrial Application of BasicMathematical Principles Session 12 In-Class
Volume Measurement Problem 20 Page 144
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