Copyright Michael Bush1 Industrial Application of Basic Mathematical Principles Session 12 Volume and Cubic Measurement
Copyright Michael Bush2 Industrial Application of Basic Mathematical 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 Michael Bush3 Industrial Application of Basic Mathematical 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 Michael Bush4 Industrial Application of Basic Mathematical 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.
Copyright Michael Bush5 Industrial Application of Basic Mathematical 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.
Copyright Michael Bush6 Industrial Application of Basic Mathematical 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
Copyright Michael Bush7 Industrial Application of Basic Mathematical 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
Copyright Michael Bush8 Industrial Application of Basic Mathematical 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 Michael Bush9 Industrial Application of Basic Mathematical 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 Michael Bush10 Industrial Application of Basic Mathematical Principles Session 12 In-Class Expressing Unit of Volume Measure as a larger unit Express 5,184 cubic inch as cubic feet. 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.
Copyright Michael Bush11 Industrial Application of Basic Mathematical 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 Michael Bush12 Industrial Application of Basic Mathematical Principles Session 12 In-Class Expressing a larger Unit of Volume Measure as a smaller unit Express 10 cubic yards in cubic feet. 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.
Copyright Michael Bush13 Industrial Application of Basic Mathematical 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 Michael Bush14 Industrial Application of Basic Mathematical 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. 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).
Copyright Michael Bush15 Industrial Application of Basic Mathematical 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 Michael Bush16 Industrial Application of Basic Mathematical Principles Session 12 In-Class Applying Volume Measure to the Cube and Rectangular Solids If 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.
Copyright Michael Bush17 Industrial Application of Basic Mathematical Principles Session 12 In-Class Rule for computing the Volume of a Cube Express 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 Michael Bush18 Industrial Application of Basic Mathematical 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 Multiply the given length (8) by the depth (8) by the height (8). Express the product (512) in terms of volume measure.
Copyright Michael Bush19 Industrial Application of Basic Mathematical Principles Session 12 In-Class Computing the Volume of a Cube Find the volume of a cube that measures 1’ –9” on a side. Multiply the given length (21) x depth (21) x height (21). Express the product (9261) in terms of volume. Express 1’ -9” as 21” Express as cu ft and cu in
Copyright Michael Bush20 Industrial Application of Basic Mathematical Principles Session 12 In-Class Computing the Volume of a Cube Find the volume of a cube that measures 1’ –9” on a side. 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
Copyright Michael Bush21 Industrial Application of Basic Mathematical 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.
Copyright Michael Bush22 Industrial Application of Basic Mathematical Principles Session 12 In-Class Rule for finding the Volume of a Rectangular Solid Express 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 Michael Bush23 Industrial Application of Basic Mathematical Principles Session 12 In-Class Computing the Volume of a Rectangular Solid Find the volume the block. Multiply the given length (32) by the depth (10) by the height (9). Express the product (512) in terms of volume measure. Express all as inch Express in lowest terms
Copyright Michael Bush24 Industrial Application of Basic Mathematical Principles Session 12 In-Class Application of Volume Measure to Cylinders The 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.
Copyright Michael Bush25 Industrial Application of Basic Mathematical 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 Michael Bush26 Industrial Application of Basic Mathematical 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. Multiply the given area (7.0686) by the height (10). Express the product in terms of volume measure. Compute area of base (.7854xD²) Express as two places
Copyright Michael Bush27 Industrial Application of Basic Mathematical Principles Session 12 In-Class Application of Volume Measure to Irregular Forms In addition to regular solids, many objects are a combination of various shapes in modified form. The volume of an irregular solid can be found by dividing it into solids having regular shapes.
Copyright Michael Bush28 Industrial Application of Basic Mathematical Principles Session 12 In-Class Application of Volume Measure to Irregular Forms The volume of each regular or modified solid form can be computed The sum of the separate volumes equals the volume of the irregular solid. The sum of the parts is equal to the whole.
Copyright Michael Bush29 Industrial Application of Basic Mathematical Principles Session 12 In-Class Rule for finding the Volume of an Irregular Solid Divide 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 Michael Bush30 Industrial Application of Basic Mathematical Principles Session 12 In-Class Computing the Volume of an Irregular Solid Determine the volume of the brass casting, correct to two decimal places.
Copyright Michael Bush31 Industrial Application of Basic Mathematical 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”
Copyright Michael Bush32 Industrial Application of Basic Mathematical Principles Session 12 In-Class Computing the Volume of an Irregular Solid Add the separate volumes Express the product in terms of volume measure. Compute volume of cylinder (area of base x height) Express as two places
Copyright Michael Bush33 Industrial Application of Basic Mathematical Principles Session 12 In-Class Application of Volume Measure to Liquid Measure Constant 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 Michael Bush34 Industrial Application of Basic Mathematical Principles Session 12 In-Class Application of Volume Measure to Liquid Measure The 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)
Copyright Michael Bush35 Industrial Application of Basic Mathematical Principles Session 12 In-Class Application of Volume Measure to Liquid Measure The gallon contains 231 cubic inches of liquid. With this known value, it is possible to solve problems requiring the use of liquid measure.
Copyright Michael Bush36 Industrial Application of Basic Mathematical 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 Michael Bush37 Industrial Application of Basic Mathematical 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 Determine the liquid capacity of a coolant tank whose volume is 1155 cubic inches.
Copyright Michael Bush38 Industrial Application of Basic Mathematical 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 Michael Bush39 Industrial Application of Basic Mathematical 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 Express 4½ gallons in quarts. Determine the number of quarts in one gallon. 4 quarts = 1 gallon
Copyright Michael Bush40 Industrial Application of Basic Mathematical 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. Express 3½ quarts in pints and ounces. 2 pints = 1 quartDetermine the number of pints in one quart. 16 ounces = 1 pint
Copyright Michael Bush41 Industrial Application of Basic Mathematical 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. Combine the pints and ounces.
Copyright Michael Bush42 Industrial Application of Basic Mathematical 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 Michael Bush43 Industrial Application of Basic Mathematical 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 Express 24 pints in gallons. Determine the number of pints in one gallon. 8 pints = 1 gallon
Copyright Michael Bush44 Industrial Application of Basic Mathematical Principles Session 12 In-Class Expressing Smaller Unit of Liquid Measure in Larger Unit Divide the ounces by the ounces per gallon. 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 Michael Bush45 Industrial Application of Basic Mathematical Principles Session 12 In-Class Expressing Smaller Unit of Liquid Measure in Larger Unit This is the whole number of quarts. Multiply the fractional part by the number of quarts per gallon. 32 oz = 1 gal Determine the number of quarts in one gallon. 1 quart
Copyright Michael Bush46 Industrial Application of Basic Mathematical Principles Session 12 In-Class Expressing Smaller Unit of Liquid Measure in Larger Unit Combine This is the whole number of quarts. Multiply 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 Michael Bush47 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem A 1 Page 161 Express 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
Copyright Michael Bush48 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem A 1 Page 161 Express 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 Michael Bush49 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem A 1 Page 161 Express 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 Michael Bush50 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem A 1 Page 161 Express 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 Michael Bush51 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem A 1 Page 161 Express Each Of The Volumes in Specified Unit e cu in in cu ft. 1 cu ft = 1728 cu in 3456 cu in 1728 cu in/ cu ft = 2 cu ft
Copyright Michael Bush52 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem A 1 Page 161 Express Each Of The Volumes in Specified Unit f cu in in cu ft. 1 cu ft = 1728 cu in cu in 1728 cu in/ cu ft = cu ft
Copyright Michael Bush53 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem B 1 A Page 162 Determine the Volume of Cube A ABC Length68½8½1’-6” Depth68½8½1’-6” Height68½8½1’-6” Volume = Side 3 Volume = 6 3 Volume = 216
Copyright Michael Bush54 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem B 2 A Page 162 Compute the Value of Rectangular Solid A Volume = L x W x H Volume = 10” x 5” x 4” Volume = 200 in 3
Copyright Michael Bush55 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem C 2 Page 162 Determine the Volume Volume Upper V = l x w x h V = 30 x 2 x 10 V = 600 ft³ Volume Lower V = 30 x 2 ⅔ x ⅔ V = ft³ Upper + Lower= ft³ ft³ 27 ft³/yd³ = yd³ =24 yd³
Copyright Michael Bush56 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem C 3 A Page 163 Compute Volume of Hollow Rectangular Solid Volume of Whole V = l x w x h V = 12 x 6 x 4 V = 288 Volume of Cut-out V = 2 x 3 x 12 V = 72 Volume = Whole – Cut-out 288 – 72 =216 in³
Copyright Michael Bush57 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem D 1 A Page 163 Determine the Volume of Cylinder A ABC Diameter412.5 Radius1.6 Length Volume =.7854D²H Volume =.7854(4²)10 Volume = in³ Volume =.7854x16x10
Copyright Michael Bush58 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem D 2 A Page 163 Determine the Liquid Capacity of Cistern A Cistern Inside Diameter Height A4’-0”6’-0” B5’-6”8’-0” C6’-6”8’-6” 1 cu ft = 7½ gal Volume =.7854D²H Vol. =.7854(4)²6 Vol. = ft³ 1 cu ft = 7½ gal ft³ x 7½ gal/ft³ = gal
Copyright Michael Bush59 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem D 4 A Page 164 Determine the Volume Whole Volume Vol. =.7854 D² H Vol. =.7854 x 4½² x 6 Vol. = in³ Volume Cored Hole Vol. =.7854 x 2½² x 6 Vol. = in³ Total Volume Vol. = Whole - Core Vol. = – Vol. = in³ Each Weight Wt. = in³ x lb/in³ Wt. = in³ x.30 lb/in³ = Total Weight Total Wt. = ea Wt. x Total # Total Wt. = lb x 10 = lb lb
Copyright Michael Bush60 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem D 4 B Page 164 Determine the Volume Whole Volume Vol. =.7854 D² H Vol. =.7854 x 2¾² x 12 Vol. = in³ Volume Cored Hole Vol. =.7854 x 1¼² x 12 Vol. = in³ Total Volume Vol. = Whole - Core Vol. = – Vol. = in³ Each Weight Wt. = in³ x lb/in³ Wt. = in³ x.32 lb/in³ = Total Weight Total Wt. = ea Wt. x Total # Total Wt. = lb x 24 = lb lb
Copyright Michael Bush61 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem D 6 A Page 164 Determine the Volume Volume Large Diameter Vol. =.7854 x d² x h Vol. =.7854 x 6¼² x 2½ Vol. = in³ Volume Small Diameter Vol. =.7854 x 4² x 6 Vol. = in³ Volume of Hole Vol. =.7854 x 2½² x 8½ Vol. = in³ Total Volume Vol. = L + S - H Vol. = in³ Each Weight Wt. = Ea x lb/in³ Wt. = lb Total Weight Ea. Wt. x Total Number lb x 20 = Total Weight = lb
Copyright Michael Bush62 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem E 1 Page 164 Width - 12” Height - 6” Length - 1’-4” Quantity - 20 Diameter of Cored Holes - 2” Number of Cored Holes - 3 Weight of Cast Iron -.28 lb/in³ Note: Cored holes run through entire length Volume of Solid Vol. = l x w x h Vol. = 16 x 12 x 6 Vol. = 1152 in³ Volume of Cored Hole Vol. =.7854D²H Vol. = in³ Total Volume Solid – Cored Holes x 3 Vol. = 1152 – 3(50.266) Vol. = in³ Wt. = in³ x lb/in³ Wt. = in³ x.28 lb/in³ Wt. = lb Weight of Each Total Weight = each x # Total Weight = lb x 20 Total Weight = 5,606.7 lb
Copyright Michael Bush63 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem F 1 A Page 165 Convert to the indicated units a.4 galqt b.6½ galqt c.3¾ galqt d.6½ qtpt e.5¼ qtpt 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 Michael Bush64 Industrial Application of Basic Mathematical Principles Session 12 In-Class Assignment Unit 21 Problem F 2 A Page 165 Determine the liquid capacity Volume = L x W x H Volume = 23” x 12” x 8½” Volume = 2,346 in 3 2,346 in 3 ÷ 231 in 3 /gal = gal =10 gal
Copyright Michael Bush65 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 1 Page 141 SideVolume mm in mm in mm in Volume = S³ Volume = 46³ Volume = 97,336 mm³ 97,336 mm³
Copyright Michael Bush66 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 2 Page 141 SideVolume mm97,336 mm³ in mm in mm in Volume = S³ Volume = 19³ Volume = 6,859 in³ 6,859 in³
Copyright Michael Bush67 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 7 Page 142 RadiusVolume in mm in mm in mm 7, in³
Copyright Michael Bush68 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 8 Page 142 RadiusVolume in7, in³ mm in mm in mm 1,934,764.4 mm³
Copyright Michael Bush69 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 13 Page 143 RadiusHeightVolume in25.0 in mm54 mm in67.5 in mm29.6 mm in61.25 in mm mm V = R²H V = (10.0)²25.0V = 7,854 in³ 7,854 in³
Copyright Michael Bush70 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 14 Page 143 RadiusHeightVolume in25.0 in7,854 in³ mm54 mm in67.5 in mm29.6 mm in61.25 in mm mm V = R²H V = (27)²54V = 123, mm³ 123, mm³
Copyright Michael Bush71 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 19 Page 144 DiameterHeightVolume in35.0 in mm85 mm in105.5 in mm88.2 mm in in mm mm V = HD² V = (35.0)15²V = 2, in³ 2, in³
Copyright Michael Bush72 Industrial Application of Basic Mathematical Principles Session 12 In-Class Volume Measurement Problem 20 Page 144 DiameterHeightVolume in35.0 in2, in³ mm85 mm in105.5 in mm88.2 mm in in mm mm V = HD² V = (85)37²V = 30, mm³ 30, mm³