Calculations for Road Maintenance
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1 objectives Calculations for Road Maintenance Provide a review of the mathematics commonly used in roadway maintenance to calculate areas, volumes, rates, and costs. Provide examples of common roadway maintenance math problem and their solutions. Topics of discussion USING UNITS OF MEASUREMENT CALCULATING AREAS CALCULATING VOLUMES CALCULATING RATES AND COSTS MAINTENANCE PROBLEMS CALCULATOR practice CALCULATOR PRACTICE MULTIPLYING ANSWERS 3.14 X SQUARED = X 28 X 42 = Square root of 64 17,
2 Calculator practice CALCULATOR practice ADDING DIVIDING = 14 + ½ = = /34 = 5280/2 = 43560/4 = /15 + ½ + 1/16 = Calculator Practice Calculator practice SUBTRACTING ANSWERS COMBINATIONS ANSWERS = = = (145 * 872) 38= 17 * 1/2000 *2000 = (pi) (2) *6SQUARED = 126, Using units of measurement linear or length 2
3 Linear measurement Length measurements Field exercise Measuring your pace LTAP INSPECTOR S MAINT. TABLES GIVEN: YOUR AVERAGE FOOT STEP WANTED: DETERMINE THE LENGTH OF YOUR AVERAGE FOOT STEP FOR FIELD USE ON THE PROJECT. LAYOUT AND MARK 100 FEET ON A SIDEWALK OR ROAD. SOLUTION: COUNT THE NUMBER OF STEPS IN 100 MEASURED FEET. AVERAGE STEP IS A RELAXED STEP THAT YOU NORMALLY REPEAT. AVERAGE THE NUMBER OF STEPS/4 CALCULATE YOUR AVERAGE STEP LENGTH. MINE IS 2.52 WALK THE 100 FEET FOUR TIMES WITH YOUR AVERAGE STEP Using units of measurement area Subdivision area layout 3
4 Area of a Circle Area of a circle Area of a circle = pi r squared Pi = R = radius Formulas for perimeter, area, & volume are on last page of booklet. Page 22 Page 22 Using units of measurement volume Calculating volumes 4
5 Using units of measurement problem 1 Measurement of areas Given: 3 cubic feet 1 foot square feet 10 miles 7.5 gallons Wanted: What type of measurement is it? Volume, Linear, Area Solution: Volume Linear Linear Area Linear volume Using units of measurement problem 2 Given: Cubic inches in a cubic foot? Square feet in a square yard Cubic feet in a cubic yard. Lineal feet in a mile Square feet in an acre Wanted: How many? Solution: Conversion factors Using units 0f measurement problem 3 Wanted: How many? 30 feet =?yards 27 sq. ft. =? Sq. yds ft. =? Miles 33 feet =? Rods 6 sq. yd. =? sq ft 2 cu yd =? cu ft Solution: 10 yards 3 sq yards 1 mile 2 rods 54 sq ft 54 cu ft FOOTBALL FIELD Measurement conversion 5
6 Problem 2b use your conversion factors GIVEN: 1 FOOTBALL FIELD. WANTED: LENGTH IN FEET LENGTH IN YARDS WIDTH IN YARDS LENGTH IN INCHES NUMBER OF ACRES SOLUTION: 360 FT. 120 YDS YDS INCHES ACRES Weight measurement Measuring weight Sixteen ounces Two thousand pounds Abbreviation for pound Abbreviation for tons Symbol for pound One pound One ton lb. T. # Converting pounds to tons problem 4 Pounds to tons Given: lbs. Wanted: Convert pounds to tons Solution: 56000/2000= 28 Tons 6
7 Converting tons to pounds problem 5 Conversion factors Given: 23 tons Answer: Wanted: Convert to pounds 46,000 lbs. Weight measurement Problem 6 Liquid Measurement Problem 7 Given: 15,000 lbs. 18,500 lbs. 3.5 tons 17 tons Solution: 7.5 tons 9.25 tons 7000 lbs. 34,000 lbs. GIVEN: A TANK HAS A CAPACITY OF 940 CUBIC FEET. WANTED: How many gallons of ashpalt can be placed in the tank? SOLUTION: Gallons Wanted: Convert tons to lbs and lbs to tons Concrete sidewalk problem 7
8 Concrete sidewalk problem Concrete sidewalk problem 8 Calculating quantity of concrete order for sidewalk Problem 8 Asphalt emulsion coverage Problem 9 Given: 4 Sidewalk 5 feet wide, length 100 feet Wanted: Calculate the quantity of ready mix concrete for 100 linear feet of sidewalk in cubic yards you need to order. Solution: 4 /12 /ft X 5 ft x 100 ft./27 cubic feet/ yard cubic yards How much would you order? Asphalt emulsion 1000 gal coverage Highway Maintenance Tables Applications lineal feet covered by 1000 gal tank of emulsion Problem 9 GIVEN:.35 gal/sqyd 22 feet wide SOLUTION: 1169 ft WANTED: Calculating Lineal feet covered by a 1000 gallon Tank Lineal Coverage = 1000gal x (syd/gal) x (1/road width ft.) x (9sft/sqyd) 8
9 Rates of application Rates of application Gallons of Asphalt required per mile for various rates of application Problem 10 Given: Appl rate = 0.40 gal/sqyd Road width = 24 ft Answer 5632 gal Wanted: Gallons of Asphalt per mile = (gal/sqyd) x rd. width x 1 (sqyd/9sqft) x 5280 ft/mile Rates of application Tons of aggregate per mile 9
10 Tons of aggregate per mile Tons of aggregates required per mile for various rates of application problem 11 Given: 300 #/sqyd, 16 ft wide Wanted: How man tons of aggregate per mile? Solution: Tons/mile=(lbs/sqyd)x(rd width)x (1sqyd/9sqft)x(5280ft/mile)x1ton/2000 lbs Answer 1408tons Tons of aggregate per mile cubic Yards per 100 lineal feet cubic Yards per 100 lineal feet Cubic yards of material required per 100 lineal feet Problem 12 Given: Road Width = 24 ft, depth of gravel 6 Wanted:How many cubic yard/100 lineal feet? Solution: Cubic yds/100 ft=depth x (yd/36in) x road width x 100 ft x (1 sqyd/9 sqft) Answer cuyds 10
11 Pounds of aggregate per square yard cubic Yards per 100 lineal feet Pounds of aggregate required per square yard Problem 13 Pounds of aggregate per square yard Given: 8 inches compacted depth Answer and 3200 #/cubic yd. 711 # Wanted: How many pounds of aggregate per square yard? Solution :#/sq yd = depth x (1yd/36 in) x (lbs/cu yd) Compacted weights of various materials Trap rock form of igneous rock that tends to form polygonal vertical fractures 11
12 Weights of different rock products Conversion factors Limestone Pea gravel Shale Sand Conversion factors Questions??????? 12
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