html5-img
1 / 63

Session 8

Session 8. PRECISION, ACCURACY, AND TOLERANCE. 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

nayda-griffith
Télécharger la présentation

Session 8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Session 8 PRECISION, ACCURACY, AND TOLERANCE

  2. 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

  3. MEASUREMENT • Measurement is used to communicate size, quantity, position, and time • Measurement is the comparison of a quantity with a standard unit • Measurements such as 2.5 inches or 15.25 miles are examples of denominate numbers since they refer to a special unit of measure • Compound denominate numbers have more that one unit of measure

  4. MEASUREMENT • 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

  5. LINEAR MEASURING INSTRUMENTS • Types of instruments include: • Tape measure • Folding rule • Steel tape • Steel rules • Vernier, dial, and electronic calipers • Standard, digital, and electronic micrometers

  6. RANGE OF A MEASUREMENT • 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

  7. EXAMPLE 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

  8. EXAMPLE 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.

  9. EXAMPLE 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

  10. EXERCISE 7–4 For each measurement find: a. the degree of precision b. the range 2. 1.62 in a. The degree of precision is 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.

  11. EXERCISE 7–4 For each measurement find: a. the degree of precision b. the range 4. 7.08 mm a. The degree of precision is 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.

  12. EXERCISE 7–4 For each measurement find: a. the degree of precision b. the range 6. 9.1837 in a. The degree of precision is 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.

  13. EXERCISE 7–4 For each measurement find: a. the degree of precision b. the range 8. 36.0 mm a. The degree of precision is 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.

  14. EXERCISE 7–4 For each measurement find: a. the degree of precision b. the range 10. 23.00 in a. The degree of precision is 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.

  15. ADDING AND SUBTRACTING • 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.

  16. ADDING AND SUBTRACTING • 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.

  17. ADDING AND SUBTRACTING • Example: 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

  18. EXAMPLE 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

  19. EXAMPLE 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

  20. EXAMPLE 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

  21. EXERCISE 7–5 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

  22. SIGNIFICANT DIGITS 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.

  23. SIGNIFICANT DIGITS 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)

  24. SIGNIFICANT DIGITS • Examples: 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

  25. SIGNIFICANT DIGITS • Examples: 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 -

  26. EXAMPLE 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)

  27. EXERCISE 7–6 Determine the number of significant digits for the following measurements 2. 0.0378 (3) 8. 137.000 (6) 4. 0.020 (2) 10. 8.020 (4) 6. 9,700 (2) 12. 8.005 (4)

  28. EXERCISE 7–6 Determine the number of significant digits for the following measurements 12. 385.007 (6) 16. 87,195 (5) 14. 1.040 (4) 18. 87,200.00 (7) 20. 4,000,100 (7)

  29. ACCURACY • 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

  30. ACCURACY • Examples: • Number 2.09 is accurate to 3 significant digits • Number 0.1250 is accurate to 4 significant digits • Number 0.0087 is accurate to 2 significant digits • Number 50,000 is accurate to 1 significant digit • Number 68.9520 is accurate to 6 significant digits • Note: When measurement numbers have the same number of significant digits, the number that begins with the largest digit is the most accurate

  31. EXERCISE 7–7 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)

  32. EXERCISE 7–7 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)

  33. EXERCISE 7–7 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)

  34. MULTIPLYING AND DIVIDING • 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.

  35. ACCURACY • Examples: • Product of 3.896 in × 63.6 in = 247.7856, but 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, but since least accurate number is 7,500, round the answer to 2 significant digits, or 17,000.

  36. ACCURACY • Examples: • Quotient of 0.009 mm  0.4876 mm = 0.018457752 mm, but 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, but since the least accurate number is 802,000, round the answer to 3 significant digits, or 2,780.

  37. EXERCISE 7–8 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) 4. 9.085 cm ÷ 1.07 (8.49) 6. 51.9 ÷ 0.97623 (0.976)

  38. EXERCISE 7–8 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 (2.0)

  39. EXERCISE 7–8 Multiply or divide the following measurement numbers. Round answers to the same number of significant digits as the least accurate number. 14. 15.635 0.415 10.07 (0.415) 16. 52.3 6.890 0.0073 (52.3)

  40. ABSOLUTE AND RELATIVE ERROR • 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)

  41. ABSOLUTE AND RELATIVE ERROR • Example: 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

  42. EXERCISE 7–9 • 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 Ans • Percent error = 0.02 × 100% = 2% Ans

  43. EXERCISE 7–9 • 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 Ans • Percent error = 0.00619 × 100% = 0.619% Ans

  44. EXERCISE 7–9 • 6. The actual or true value of a weight is 6056 kg. The weight is measured as 6100 kg. • Absolute error = True value – Measured value = 6056 – 6100 = 44 kg Ans • Percent error = 0.0068 × 100% = 0.68% Ans

  45. EXERCISE 7–9 • 8. The actual or true value of a bore is 0.9347 in. The bore is measured as 0.9341 in. • Absolute error = True value – Measured value = 0.9347 – 0.9341 = 0.0006 in Ans • Percent error = 0.0006 × 100% = 0.06% Ans

  46. EXERCISE 7–9 • 10. The actual or true value of a distance is 27.2 ft. The distance is measured as 26.9 ft. • Absolute error = True value – Measured value = 27.2 – 26.9 = 0.3 ft Ans • Percent error = 0.01 × 100% = 1% Ans

  47. TOLERANCE (LINEAR) • Tolerance (linear) is the amount of variation permitted for a given length • Tolerance is equal to the difference between maximum and minimum limits of a given length

  48. TOLERANCE (LINEAR) • Example: The maximum permitted length (limit) of a tapered shaft is 134.2 millimeters. The minimum permitted length (limit) is 133.4 millimeters. Find the tolerance • Tolerance = maximum limit – minimum limit = 134.2 mm – 133.4 mm = 0.8 mm Ans

  49. TOLERANCE (LINEAR) Example: The maximum permitted depth (limit) of the dado joint shown is 21/32 inch. The tolerance is 1/16 inch. Find the minimum permitted depth (limit). The minimum limit equals the maximum limit minus the tolerance. 21/32 – 1/16 = 19/32

  50. EXERCISE 7–10 Refer to the tables in Figures 7–9 and determine the tolerance, maximum limit, or minimum limit as required for each problem. • 2. Tolerance = max limit – min limit • = 7’ 9 1/16” – 7’ 8 15/16” • = 1/8” Ans • 4. Max limit = min limit + tolerance • = 0.904” + 0.007” • = 0.911” Ans

More Related