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Session 8

Session 8. PRECISION, ACCURACY, AND TOLERANCE. Session 8 - Objectives. After studying this unit you should be able to determine the degree of precision of any measurement number. round sums and differences of measurement numbers to proper degrees of precision.

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Session 8

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  1. Session 8 PRECISION, ACCURACY, AND TOLERANCE

  2. Session 8 - Objectives • After studying this unit you should be able to • determine the degree of precision of any measurement number. • round sums and differences of measurement numbers to proper • degrees of precision. • determine the number of significant digits of measurement numbers. • round products and quotients of measurement numbers to proper • degrees of accuracy. • compute absolute and relative error between true and measured • values. • compute maximum and minimum clearances and interferences of • bilateral and unilateral tolerance-dimensioned parts. • compute total tolerance, maximum limits, and minimum limits of • lengths. • solve practical applied problems involving tolerances and limits.

  3. Session 8 7–1 Exact and Approximate (Measurement) Numbers 220 7–2 Degree of Precision of Measuring Instruments 221 7–3 Common Linear Measuring Instruments 222 7–4 Degree of Precision of a Measurement Number 223 7–5 Degrees of Precision in Adding and Subtracting Measurement Numbers 224 7–6 Significant Digits 224 7–7 Accuracy 225 7–8 Accuracy in Multiplying and Dividing Measurement Numbers 226 7–9 Absolute and Relative Error 227 7–10 Tolerance (Linear) 228 7–11 Unilateral and Bilateral Tolerance with Clearance and Interference Fits 229 UNIT EXERCISE AND PROBLEM REVIEW 232

  4. Measurement – Page 220 Measurement is used to communicate size, quantity, position, and time The ability to measure with tools and instruments and to compute with measurements is required in most occupations. Measurement is the comparison of a quantity with a standard unit

  5. Measurement – Page 220 Linear measurement is a means of expressing the distance between two points; it is the measurement of lengths. A linear measurement has two parts: a unit of length and a multiplier. 2.5 inches 15¼ miles Multiplier unit of length multiplier unit of length

  6. Measurement – Page 221 Measurements such as 2.5 inches or 15.25 miles are examples of denominate numbers since they refer to a specific unit of measure Compound denominate numbers have more that one unit of measure For example 7 feet 2 inches or 6 yards 1 foot 9 inches

  7. Measurement – Page 221 • All measurements are approximations • Degree of precision of a measurement number depends on number of decimal places used • Number becomes more precise as number of decimal places increases • Measurement 4.923 inches is precise to nearest tenth (0.001) inch • Measurement 4.92 inches is precise to nearest hundredth (0.01) inch

  8. Common Linear Measuring Instruments Page 222 • Types of instruments include: • Tape measure • Folding rule • Steel tape • Steel rules • Vernier, dial, and electronic calipers • Standard, digital, and electronic micrometers

  9. Degree of Precision – Page 223 • The degree of precision of a measurement number depends on the number of decimal places used. • The number becomes more precise as the number of decimal places increases. For example, 4.923 inches is more precise than 4.92 inches. • Range of a measurement includes all values represented by the number

  10. Example – Page 223 What is the degree of precision and the range for 2 inches? • The degree of precision of 2 inches is to the nearest inch • The range of values includes all numbers equal to or greater than 1.5 inches or less than 2.5 inches

  11. Example – Page 223 What is the degree of precision and the range for 2.0 inches? • The degree of precision of 2.0 inches is to the nearest 0.1 inch. • The range of values includes all numbers equal to or greater than 1.95 inches and less than 2.05 inches.

  12. Example – Page 223 What is the degree of precision and the range for 2.00 inches? • The degree of precision of 2.00 inches is to the nearest 0.01 inch • The range of values includes all numbers equal to or greater than 1.995 inches or less than 2.005 inches

  13. Example – Page 223 What is the degree of precision and the range for 2.000 inches? • The degree of precision of 2.000 inches is to the nearest 0.001 inch • The range of values includes all numbers equal to or greater than 1.9995 inches or less than 2.0005 inches

  14. Exercise 7–4 – Page 224 For each measurement find: a. the degree of precision b. the range 2. 1.62 in a. The degree of precision is to the nearest 0.01 inch. b. The range of values includes all numbers equal to or greater than 1.615 inches and less than 1.625 inches.

  15. Exercise 7–4 – Page 224 For each measurement find: a. the degree of precision b. the range 4. 7.08 mm a. The degree of precision is to the nearest 0.01 mm. b. The range of values includes all numbers equal to or greater than 7.075 mm and less than 7.085 mm.

  16. Exercise 7–4 – Page 224 For each measurement find: a. the degree of precision b. the range 6. 9.1837 in a. The degree of precision is to the nearest 0.0001 in. b. The range of values includes all numbers equal to or greater than 9.18365 in and less than 9.18375 in.

  17. Exercise 7–4 – Page 224 For each measurement find: a. the degree of precision b. the range 8. 36.0 mm a. The degree of precision is to the nearest 0. 1 mm. b. The range of values includes all numbers equal to or greater than 35.95 mm and less than 36.05 mm.

  18. Exercise 7–4 – Page 224 For each measurement find: a. the degree of precision b. the range 10. 23.00 in a. The degree of precision is to the nearest 0. 01 in. b. The range of values includes all numbers equal to or greater than 23.095 in and less than 23.005 in.

  19. Exercise 7–4 – Page 224 For each measurement find: a. the degree of precision b. the range 12. 14.01070 in a. The degree of precision is to the nearest 0. 00001 in. b. The range of values includes all numbers equal to or greater than 14.010695 in and less than 14.010705 in.

  20. Adding and Subtracting – Page 224 • When adding or subtracting measurements, all measurements must be expressed in the same kind of units. • Often measurement numbers of different degrees of precision are added or subtracted.

  21. Adding and Subtracting – Page 224 • When adding and subtracting numbers, there is a tendency to make answers more precise than they are. • An answer that has more decimal places than it should gives a false degree of precision. • The sum or difference cannot be more precise than least precise measurement number used in computations. • Round the answer to the least precise measurement number used in the computations.

  22. Example – Page 224 Add 15.63 in + 2.7 in + 0.348 in and round answer to degree of precision of least precise number • Since 2.7 in is the least precise measurement, round the answer to 1 decimal place

  23. Example – Page 224 2. 3.0928 cm - 0.532 cm = 2.5608 cm Since the least precise number is 0.532 cm, round the answer to 3 decimal places. 3.0928 cm - 0.532 cm = 2.561 cm

  24. Example – Page 224 3. 73 ft + 34.21 ft = 107.21 ft Since the least precise number is 73 ft, round the answer to the nearest whole number. 73 ft + 34.21 ft = 107 ft

  25. Example – Page 224 4. 73.0 ft + 34.21 ft = 107.21 ft Notice that this example is identical to last example, except the first measurement is 73.0 ft instead of 73 ft. Since the least precise number is 73.0 ft, round the answer to one decimal place. 73.0 ft + 34.21 ft = 107.2 ft

  26. Exercise 7–5 – Page 224 Add or subtract the following measurement numbers. Round answers to the degree of precision of the least precise number 2. 14.863 mm - 5.0943 mm = 9.769 mm 4. 0.0009 in + 0.001 in = 0.002 in 6. 31.708 cm2 - 27.69 cm2 = 4.02 cm2

  27. Exercise 7–5 – Page 224 Add or subtract the following measurement numbers. Round answers to the degree of precision of the least precise number 8. 0.0187 cm3 + 0.70 cm3 = 0.72 cm3 10. 6.01 lb + 15.93 lb + 18.0 lb = 39.9 lb 12. 84.987 mm + 39.01 mm – 77 mm = 47mm

  28. Significant Digits – Page 224 It is important to understand what is meant by significant digits and to apply significant digits in measurement calculations. A measurement number has all of its digits significant if all digits, except the last, are exact and the last digit has an error of no more than half the unit of measurement in the last digit. For example, 6.28 inches has 3 significant digits when measured to the nearest hundredth of an inch.

  29. Significant Digits – Page 225 Rules for determining the number of significant digits in a given measurement: • All nonzero digits are significant • Zeros between nonzero digits are significant • Final zeros in a decimal or mixed decimal are significant • Zeros used as place holders are not significant unless they are identified as significant (usually with a bar above it)

  30. Examples – Page 225 812 has 3 significant digits • (all nonzero digits are significant) 7.139 has 4 significant digits • (all nonzero digits are significant) 14.3005 has 6 significant digits • (zeros between nonzero digits are significant) 9.300 has 4 significant digits • (final zeros of a decimal are significant

  31. Examples – Page 225 0.008 has 1 significant digits • (zeros used as place holders are not significant) 23,000 has 2 significant digits • (zeros used as place holders are not significant) 23.000 has 3 significant digits • (final zeros of a decimal are significant

  32. Examples – Page 225 Identify the number of significant digits in the following 2. 3.950 (4) 8. 0.00187 (3) 4. 147.005 (6) 10. 8.020 (4) 6. 7.004 (4) 12. 8,603.0 (5)

  33. Exercise 7–6 – Page 225 Determine the number of significant digits for the following measurements 2. 0.0378 (3) 8. 137.000 (6) 4. 0.020 (2) 10. 8.005 (4) 6. 9,700 (2) 12. 385.007 (6)

  34. EXERCISE 7–6 Determine the number of significant digits for the following measurements 14. 1.040 (4) 18. 87,200.00 (7) 16. 87,195 (5) 20. 4,000,100 (5)

  35. Accuracy – Page 225 • Determined by the number of significant digits in a measurement • The greater the number of significant digits, the more accurate the number • Product or quotient cannot be more accurate than the least accurate measurement used in the computations

  36. Examples – Page 226 2.09 is accurate to 3 significant digits 0.1250 is accurate to 4 significant digits 0.0087 is accurate to 2 significant digits 50,000 is accurate to 1 significant digit 68.9520 is accurate to 6 significant digits

  37. Examples – Page 226 When measurement numbers have the same number of significant digits, the number that begins with the largest digit is the most accurate Consider the following measurement numbers: 3,700, 4,100, and 2,900 all have 2 significant digits. 4,100 would be considered the most accurate

  38. Exercise 7–7 – Page 226 For each of the following groups of measurement numbers, identify the number that is most accurate. 2. 18.6; 1.860; 0.186 (1.860) 4. 0.0009; 0.004; 0.44 (0.44) 6. 0.27; 50,720; 52.6 (50,720)

  39. Exercise 7–7 – Page 226 For each of the following groups of measurement numbers, identify the number that is most accurate. 8. 39.03; 436; 0.0235 (39.03) 10. 0.930; 0.0086; 5.31 (0.930; 5.31) 12. 86,000; 9,300; 435 (435)

  40. Exercise 7–7 – Page 226 For each of the following groups of measurement numbers, identify the number that is most accurate. 14. 5.0003; 5.030; 5.00 (5.0003) 16. 0.1229; 7.063; 20.125 (20.125)

  41. Multiplying and Dividing – Page 226 • When multiplying and dividing numbers, there is a tendency to make answers more accurate than they actually are. • An answer that has more significant digits than it should gives a false impression of accuracy. • A product or quotient cannot be more accurate than the least accurate measurement number used in the computations. • Round the answer to the least accurate measurement number used in the computations.

  42. Examples – Page 226 Product of 3.896 in × 63.6 in = 247.7856 Since least accurate number is 63.6, answer must be rounded to 3 significant digits, or 248 in Product of 7,500 mi x 2.250 = 16,875 mi Since least accurate number is 7,500, round the answer to 2 significant digits, or 17,000.

  43. Examples – Page 226 Quotient of 0.009 mm  0.4876 mm = 0.018457752 mm Since least accurate number is 0.009, answer must be rounded to 1 significant digits, or 0.02 mm. 802,000 lb ÷ 430.78 x 1.494 = 2,781.438321 Since the least accurate number is 802,000, round the answer to 3 significant digits, or 2,780.

  44. Exercise 7–8 – Page 227 Multiply or divide the following measurement numbers. Round answers to the same number of significant digits as the least accurate number. 2. 1.85 in x 3.7 6.8 in 4. 9.085 cm ÷ 1.07 8.49 cm 6. 51.9 ÷ 0.97623 53.2

  45. Exercise 7–8 – Page 227 Multiply or divide the following measurement numbers. Round answers to the same number of significant digits as the least accurate number. 8. 830.367 x 9.455 7851 10. 71,200 x 19.470 x 0.168 233,000 12. 30,000 ÷ 154.9 ÷ 80.03 2

  46. Exercise 7–8 – Page 227 Multiply or divide the following measurement numbers. Round answers to the same number of significant digits as the least accurate number. 14. 15.635 x 0.415 x 10.07 65.3 16. 52.3 x 6.890 x 0.0073 2.6

  47. Absolute and Relative Error – Page 227 • Absolute error = True Value – Measured Value if true value is larger: • Absolute error = Measured Value – True Value, if measured value is larger • Percent Error = Relative error × 100% (Percent error is relative error expressed as a percent)

  48. Example – Page 227 • The actual or true value of the diameter of a shaft is 1.7056 inches. The shaft is measured as 1.7040 inches. Compute the absolute error, relative error, and percent error. • Absolute error = True value – Measured value = 1.7056 – 1.7040 = 0.0016 inch Ans • Percent error = 0.00094 × 100% = 0.094% Ans

  49. Exercise 7–9 – Page 228 • 2. The actual or true value of an item is 0. 53 mm. The item is measured as 0.52 mm. • Absolute error = True value – Measured value = 0.53 – 0.52 = 0.01 mm • Percent error = 0.02 × 100% = 2%

  50. Exercise 7–9 – Page 228 • 4. The actual or true value of a distance is 485 mi. The distance is measured as 482 mi. • Absolute error = True value – Measured value = 485 – 482 = 3.00 mi • Percent error = 0.00619 × 100% = 0.619%

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