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Measurements & Computations in Surveying

Understand the units and computations vital for surveying, from angles to distance measurements in SI and older units. Learn about significant figures and areas calculations.

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Measurements & Computations in Surveying

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  1. Measurements & Computations in Surveying Dr. Dan Trent Mississippi Valley State University February 4, 2013

  2. Objectives of tonight’s lecture • 1 Measurement of distances and angles is the essence of surveying – therefore we need to discuss appropriate units of measure for these and other quantities (area, volume, etc.) • 2 Computation (or data reduction) is also essential to surveying. Surveyors must understand the concept of significant figures in measured and computed quantities

  3. Units of MeasureWhat are they? • SI – stands for Système International. • Most of the world uses this system • Metric system – based on units of 10

  4. Angles • An angle is simply a figure formed by the intersection of two lines. • An angle may be generated by the rotation of a line about a point – from an initial position to a terminal position. • The point of rotation is the vertex of the angle • Angular measurement is concerned with the amount of rotation

  5. Degrees, Minutes, and Seconds • Sexagesimal System. • Most Common • A complete rotation of a line (circle) is divided into 360 degrees of arc • 1 degree = 60 minutes • 1 minute = 60 seconds • Problem Example • 89 59’ 60” • -54  17’ 14”

  6. Angles • Centesimal System • One complete rotation of a line (circle) is divided into 400 grades (or GRADS) • Written as 400g • Each grad is divided into 100 parts called centigrads (1g = 100c) • Each centigrad is divided into 100 parts called centi-centigrads (1c = 100cc) • For an angle represented as 139.4325g the first 2 digits are centigrads and the second 2 are • centi-centigrads

  7. Angles • Important Conversion Formulae • 1g = 0.9 • Circumference of a circle • 2R • There are about 2 radians in a circle • 2 X 3.14 = 6.28 • 6.28 rad = 360 • 1 rad = 57.3 

  8. Distance • Basic unit is the FOOT • Measured in decimals – NOT INCHES • 75.25’ NOT 75’ 3” • In SI the basic unit is the METER • Also measures decimals • Decimeter = .1 meter • Centimeter = .01 meter • Millimeter = .001 meter

  9. Beware • AMERICAN SURVEY FOOT • Obsolete • 1 foot = 0.3048006096 • New standard (1959) • 1 foot = 0.3048 • Difference of about 0.2 m in 100,000m • Or about 2” in 60 miles

  10. Older Units of Measure • In older deeds you may find these units referenced: • 1 foot (ft) = 12 inches (in) • 1 yard (yd) = 3 feet • 1 mile (mi) = 5280 feet = 80 chains (ch) • 1 chain = 66 feet • 1 rod (rd) = 0.25 chain = 16.5 feet • 1 link (lk) = 0.01 chain = 7.92 inches

  11. Metric Distance Measure Review • The meter: • 1 m = 10 dm = 100 cm = 1000 mm • 1 decimeter (dm) = 0.1 meter • 1 centimeter (cm) = 0.01 meter • 1 millimeter (mm) = 0.001 meter • 1 kilometer (km) = 1000 meters

  12. Area • Area is the amount of two dimensional space encompassed within the boundary of a closed figure or shape • Derived from basic unit of length • In the U.S. we use the square foot or sq ft or ft2 • For land area we use the acre (ac) • 1 acre = 43,560 ft2

  13. More Equivalents • 1 Square mile (mi2) = 640 acres (ac) • 1 acre = 10 square chains (sq ch) = 43,560 ft2 • 1 square yard (yd2) = 3 ft X 3 ft = 9 square feet • 1 hectacre (ha) = 100 acres = 10,000 sq meters • 1 square kilometer (km2) = 100 hectacres = 1,000,000 square meters (m2)

  14. Important Conversions • 1 km2 = 0.386 mi2 • 1 hectacre (ha) = 2.47 acres • 1 square meter (m2) = 1.2 yd2 = 10.76 ft2

  15. Volume • U.S. customary unit is cubic feet (ft3) • In surveying we typically are concerned with greater volume so we use cubic yards (yd3) • Cubic meter (m3) is standard in SI • 1 yd3 = 3 ft X 3 ft X 3 ft = 27 ft3 • Use x3 for conversion - Not 9’ • We will calculate volume next week

  16. Conversion to SI metric • 1975 American Congress on Surveying and Mapping (ACSM) • Top 5 • 1976 U.S. Geological Survey (USGS) began using SI units on topographic maps • Problem Example • Convert an area of 125.55 ac to an equivalent area expressed in hectacres

  17. Computations • Significant Figures • A measured distance or angle is never exact • There is no perfect measuring instrument • The “closeness” of the observed value to the true value depends on: • The quality of the measuring instrument • The care taken by the observer taking the measurement

  18. Computations • Significant Figures • In a measured quantity, the number of significant figures is the number of sure or certain numbers plus one estimated digit • Rules • Zeros placed at the end of a decimal number are significant (75.200 has 5 significant figures) • Zeros between other significant digits are significant (17.08 has 4 significant figures)

  19. Computations • Significant Figures • Zeros just to the right of the decimal in numbers smaller than unity (1) are not significant (0.0000123 has 3 significant figures. So does 0.0123 – only 3 significant figures) • Trailing zeros to the right of the digits in a number written without a decimal are generally not significant (35,000 has only 2 significant figures)

  20. Computations • Significant Figures • 25.35 4 significant figures • 0.002535 4 significant figures • 12034 5 significant figures • 120.00 5 significant figures • 12,000 2 significant figures

  21. Computations • Rounding off numbers • Must take into account the number of significant figures in all calculations • Typically we “drop” all digits to the right of our significant figures if the next digit is less than 5 • Typically we “add one” to the last significant figure if the next digit is 5 or more. • Round off 0.1836028 to 2 significant figures • 0.18 • Round off 0.1836028 to 2 significant figures • 0.184

  22. Mistakes and Errors • The difference between a measured distance or angle and its true value may be the result of mistakes or errors • Blunder – A significant mistake caused by human error • Misreading a number on a scale • Measuring the wrong angle

  23. Mistakes and Errors • Systematic and Accidental Errors • Systematic Error – Repetitive errors that are caused by imperfections in the surveying equipment, by the specific method of observation, or by certain environmental factors • Mechanical errors • Cumulative errors

  24. Mistakes and Errors • Systematic and Accidental Errors • Accidental Error – The difference between a true quality and a measurement of that quality that is free from blunders or systematic errors • Accidental errors occur in every measurement • Relatively small and unavoidable errors in observation that are generally beyond the control of the surveyor

  25. Mistakes and Errors • Most Probable Value • If 2 or more measurements of the same quantity made, random errors usually cause different values to be obtained. As long as each measurement is equally reliable, the average value the different measurements is taken to be the TRUE or most probable value. • Sum all measurements and divide by number of measurements • Problem Example • (55.63 + 55.78 + 55.55 + 55.81) ÷ 4 = 55.69

  26. Mistakes and Errors • The 90 Percent Error • The most probable error is that which has an equal chance (50%) of either being exceeded or or not being exceeded in a particular measurement • Expressed as E90 • A distance of 100.00 ft is measured • 90 Percent Error is assumed in one taping operation using a 100-ft tape  0.01 ft • Likelihood is 90% that the tape will fall within the range of 99.99 – 100.01 ft • Still 10% chance of greater than 0.01 ft error • Maximum anticipated error

  27. Mistakes and Errors • The 90 Percent Error    (  ) 2 • E90 = 1.645 X  n(n – 1) •  = sigma, “the sum of” •  = delta, the difference between each individual measurement and the average of n measurements • N = The number of measurements

  28. Mistakes and Errors • The 90 Percent Error • Problem Example • A measurement was taken 5 times • (75.3 + 76.2 + 75.7 + 75.5 + 75.8) ÷ 5 = 75.7m • The value of  (2 ) may be computed by • (75.3 – 75.7)2 = 0.16 • (75.3 – 76.2)2 = 0.25 • (75.3 – 75.7)2 = 0.00 • (75.3 – 75.5)2 = 0.04 • (75.3 – 75.8)2 = 0.01 •  (2 ) = 0.46

  29. Mistakes and Errors • The 90 Percent Error • Problem Example •  (2 ) = 0.46 •  • E90 =1.645 X 0.46/ [ 5 ( 5 – 1 ) ] =  0.25 m • We are 90% sure that true distance is within the range of 75.7  0.25 m

  30. Mistakes and Errors • How accidental errors add up • Problem Example • Lets say we measure a distance of 900 feet with a 100’ tape ( 9 – 100 ft measurements ) • Maximum probable error for measuring 100’ was  0.010 ft • What is the maximum probable error for measuring the total distance of 900 ft with the same tape and the same procedure? • Could we reasonably say 9 X 0.010 = 0.090 ft? • NO! Some errors are likely positive, some negative

  31. Mistakes and Errors • How accidental errors add up • Problem Example • Law of Compensation  • E = E1 X  n • E = the total error of n measurements • E1 = the error for one measurement • n = the number of measurements  • E = 0.010  9 =  0.010 X 3 =  0.030ft

  32. Mistakes and Errors • Overview • The surveyor must constantly be aware of the possibilities for mistakes and errors in survey work • 4 Basic Principles

  33. Accuracy and Precision • NOT THE SAME! • Precision – the degree of perfection used in the survey • Accuracy – the degree of perfection obtained in the results • Surveyor A measures a distance and gets 750.1ft • Surveyor B measures the same distance and gets 749.158 ft • Surveyor B used greater precision • If the true distance was 750.11, Surveyor A was more accurate See board example

  34. Errors of Closure and Relative Accuracy • The difference between a measured quantity and its true, or actual, value is called the error of closure or just closure • Distance from Point A to Point B is determined to be 123.25 m. The same line is measured a 2nd time using the same instruments and procedures and is found to be 123.19 m. • Error of closure is 123.25 – 123.19 = 0.06m • Due to accidental errors

  35. Errors of Closure and Relative Accuracy • Suppose the actual distance was know not be 123.30m. Closure determined as the difference between the average measured value and the known true value. • (123.25 + 123.12) ÷ 2 = 123.22m • Error of closure would be 123.30 – 123.22 = 0.08m

  36. Relative Accuracy • For horizontal distances, the ratio of the error of closure to the actual distance is called the relative accuracy • Degree of accuracy • Order of accuracy • Accuracy ratio • Relative precision • Precision

  37. Relative Accuracy • Relative Accuracy = 1 : D/C • D = distance measured • C = error of closure • A distance of 500 ft is measured with a closure of 0.25 ft • Relative accuracy is 0.25 ft per 500 ft (0.25/500) • Relative accuracy is 1/2000 • Relative accuracy is 1:2000 • For every 2000 ft measured there is an error of 1 ft

  38. Relative Accuracy • Problem Example • A distance of 577.80 ft is measured • A true distance of 577.98 is found • What is relative accuracy of the measurement? • Error of closure is 577.80 – 577.98 = - 0.18 ft • Relative accuracy is 1: D/C • 1: 577.80/0.18 = 1: 3200 • IF survey was 4 times as long, estimated error of closure would be  • 0.18 X  4 = 0.36 ft • Relative accuracy is 1:(4 X 577.80) / 0.36 = 1:6400

  39. Standards of Accuracy • First order accuracy • 1: 1,000,000 • Second order accuracy • Class I accuracy 1: 50,000 • Class II accuracy 1: 20,000 • Third order accuracy • Class I accuracy 1: 10,000 • Class II accuracy 1: 5,000

  40. Choice of Survey Procedure • Surveyor should choose equipment and methods that have a rating or maximum anticipated error closely equal to that for maximum allowable closure

  41. Choice of Survey Procedure • A horizontal control traverse survey is required to close with third order class II accuracy. Total distance of the traverse is about 10,000 ft. What is the required rating or maximum anticipated error per 1000 ft for the survey method to be used? • Relative accuracy for third order class II survey is 1: 5000 • Therefore in 10,000 ft the maximum error of closure is 1/5000 X 10,000 = 2 ft

  42. Choice of Survey Procedure • Relative accuracy for third order class II survey is 1: 5000 • Therefore in 10,000 ft the maximum error of closure is 1/5000 X 10,000 = 2 ft E90 for 1000 ft 2  1000  10,000  2 X  1000 2 E90 for 1000 ft =  1000  1000 0.63 ft

  43. The Assessment • Quiz #1 will be available on the course website at 3:00 p.m. on February 5, 2013. • Bonus period for early submission will end February 7, 2013 at 5:00 p.m. • Drop Dead Deadline is 5:00 p.m. on Friday, February 8, 2013 • Include ANSWERS to Problems – I will take up your work sheets on Monday, February 11, 2013

  44. Submission Guidelines • At the TOP of your paper write: • AT – 403 Your Name Week 2 Quiz • Papers must be word processed using Microsoft WORD • Write out each question, then answer it using complete sentences, correct grammar and spelling below each question • Use 12 point Ariel or 12 point Times New Roman font, double spaced • Email completed paper to Dr. Trent with the subject line heading EXACTLY LIKE THIS: AT 403 – Your Name – Week 2

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