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Publication Number: FHWAHRT06121 Date: November 2006 
Figure 1. Graph. Plot of measured maximum frost depth to FI
This scatter plot shows the relationship between measured frost depth parenthesis meters end parenthesis on the vertical axis to the annual freezing index parenthesis degreeCelsius days end parenthesis on the horizontal axis. The values on the vertical axis range from zero to 2 point 5 meters, and the values on the horizontal axis range from zero to 2,000 degreeCelsius days. A linear regression line is also provided in the graph with a yintercept of zero and a slope of positive zero point 0 0 1 4.
Figure 2. Graph. Individual distress deduct curves.
This line graph shows the relationship between surface distress amounts parenthesis percentage of occurrence end parenthesis and deduct values for low, moderate, and high severity levels. Distress amounts are on the horizontal axis and range from zero point 1 to 100 on a logarithmic scale. The vertical axis contains deduct values that range from 1 to 100 on a logarithmic scale. The high severity curve is a straight line that is near 1 at zero point 1 percent occurrence and increases to 100 at a percent occurrence of 70. The low severity curve is a straight line that is near 1 at zero point 1 percent occurrence and increases to 30 at a percent occurrence of 70.
Figure 3. Graph. Sample box plot.
This is an example of box plots used in the statistical analysis of the data set. The first quartile, third quartile, and the median natural logarithm I R I values are provided on the vertical axis for each of the BASE categories parenthesis along the horizontal axis end parenthesis.
Figure 4. Scatter Plot. Sample augmented partial residual plot.
This scatter plot is an example the augmented partial residual graphs used in the statistical analysis. It shows augmented partial residuals and the response parenthesis in this example, I R I end parenthesis on the vertical axis with values ranging from negative 2 to positive 2. The horizontal axis contains one of the predictor variables parenthesis in this example, L E S N end parenthesis, which ranges from zero to 10. The cluster of data points has an xcoordinate of less than 4.
Figure 5. Graphs. Assumption validity check for absolute IRI model (before transformation).
This figure contains 4 graphs used to check model assumptions. The graph in the upper left is a histogram of residuals, and it provides the number of occurrences on the vertical axis for various residual bins on the horizontal axis. The vertical axis ranges from zero to 1,750 and the horizontal axis ranges from negative 1 point 375 to positive 3 point 625. The most occurrences were observed between negative point 375 and positive point 625. The graph in the upper right is a scatter plot of residual values parenthesis vertical axis end parenthesis versus predicted values parenthesis horizontal axis end parenthesis. The vertical axis ranges from negative 2 to positive 2. Data points form a fanshaped pattern on the bottom of the cluster. The graph in the lower left is a normal probability plot containing studentresidual values on the vertical axis and normalized rank on the horizontal axis with values ranging from negative 3 to positive 8 and negative 4 to positive 4, respectively. The data points follow the line of equality from negative 4 to a value of positive 2. At normalized rank values greater than 2, the data points move above and away from the line of equality. The graph in the lower right provides a scatter plot of student residual parenthesis vertical axis end parenthesis versus hatdiagonal element parenthesis horizontal axis end parenthesis. Studentresidual values range from negative 3 to positive 7, and the hatdiagonal values range from zero to point zero 7. Most of the data points are clustered around hatdiagonal values of zero to point zero 2 ranging over the entire vertical axis. There is a smaller cluster with hatdiagonal values of point zero 3 to point zero 7 and studentresidual values between negative 2 and positive 2.
Figure 6. Graphs. Assumption validity check for absolute IRI model. (after natural logarithm transformation of the performance measure).
This figure contains 4 graphs used to check model assumptions. The graph in the upper left is a histogram of residuals, and it provides the number of occurrences on the vertical axis for various residual bins on the horizontal axis. The vertical axis ranges from zero to 700 and the horizontal axis ranges from negative 1 point 1 to positive 1 point 3. The most occurrences were observed between negative point 2 and positive point 1. The graph in the upper right is a scatter plot of residual values parenthesis vertical axis end parenthesis versus predicted values parenthesis horizontal axis end parenthesis. The vertical axis ranges from negative 1 to positive 1. Data points are clustered parenthesis with no fan shape parenthesis and centered on zero on the vertical axis. The graph in the lower left is a normal probability plot containing studentresidual values on the vertical axis and normalized rank on the horizontal axis with values ranging from negative 4 to positive 5 and negative 4 to positive 4, respectively. The data points follow the line of equality much closer than those in figure 5. The graph in the lower right provides a scatter plot of student residual parenthesis vertical axis end parenthesis versus hatdiagonal element parenthesis horizontal axis end parenthesis. Studentresidual values range from negative 4 to 5 and the hatdiagonal values range from zero to point zero 7. Most of data points are clustered around hatdiagonal values of zero to point zero 2 ranging over the entire vertical axis. There is a smaller cluster with hatdiagonal values of point zero 3 to point zero 7 and studentresidual values between negative 2 and positive 2.
Figure 7. Scatter plot. Outlierinfluential observation detection plot.
This figure shows a scatter plot of student residual parenthesis vertical axis end parenthesis versus hatdiagonal element parenthesis horizontal axis end parenthesis. Studentresidual values range from negative 4 to positive 5, and the hatdiagonal values range from zero to point zero 7. Most of the data points are clustered around hatdiagonal values of zero to point zero 2 ranging over the entire vertical axis. There is a smaller cluster with hatdiagonal values of point zero 3 to point zero 7, and studentresidual values between negative 2 and positive 2.
Figure 8. Scatter plot. Observed versus predicted values of absolute IRI (shifted) using the robust method.
This is a scatter plot of predicted I R I values parenthesis shifted end parenthesis on the vertical axis versus observed I R I values of the horizontal axis. The vertical axis values range from negative 1 to positive 1 point 5. The horizontal axis values range from negative 1 to positive 1 point 7. Most of the data fall under the line of equality with a tight cluster of data paralleling and just below the line of equality. At observed values of 1 to 2, the predicted values are more spread out and fall away from the line of equality.
Figure 9. Scatter plot. Observed versus predicted values of absolute IRI (shifted) using the GLM method.
This is a scatter plot of predicted I R I values parenthesis shifted end parenthesis on the vertical axis versus observed I R I values of the horizontal axis. The vertical axis values range from negative 1 to positive 1 point 4. The horizontal axis values range from negative 1 to positive 1 point 7. Most of the data form a tight cluster paralleling and centered on the line of equality. At observed values of 1 to 2, the predicted values are more spread out and fall away from the line of equality.
Figure 10. Graph. Example of predicted (without shifting) and observed values for test section 307066.
This scatter plot contains natural logarithm of predicted and observed I R I values on the vertical axis and pavement age on the horizontal axis. Pavement age ranges from zero to 9 years. The observed I R I values increase with age and range from negative point 1 to positive point 11. The predicted I R I values follow the same slope as the observed values, but the values are slightly higher, ranging from negative point zero 5 to positive point 18.
Figure 11. Graph. Example of predicted (shifted) and observed values for test section 307066.
This scatter plot contains natural logarithm of predicted and observed I R I values on the vertical axis and pavement age on the horizontal axis. Pavement age ranges from zero to 9 years. The observed I R I values increase with age and range from negative point 1 to positive point 11. The predicted I R I values follow the same slope as the observed values, ranging from negative point 1 to positive point 12.
Figure 12. Scatter plot. Flexible IRI model without shifting.
This graph shows the natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. The vertical axis values range from negative point 5 to positive 1 point 7, while the horizontal axis values range from negative point 7 to positive 1 point 6. A cluster of data from negative point 3 to positive point 7 on the horizontal axis falls near the line of equality. At values less than negative point 7 the data are above the line and at values greater than point 7 the values are more spread out.
Figure 13. Scatter plot. Flexible IRI model (shifted).
This graph shows natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. Data range from negative point 7 to positive 1 point 5 on the vertical axis and from negative point 7 to positive 1 point 5 on the horizontal axis. Most of the data are centered on the line of equality. Data with horizontal axis values greater than point 7 are more spread out.
Figure 14. Scatter plot. Rigid IRI model without shifting.
This scatter plot shows natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. Data range from zero to 1 point 6 on the vertical axis and negative point 2 to positive 1 point 5 on the horizontal axis. Data between zero and point 9 on the horizontal axis are tightly grouped and centered on the line of equality. Data with values of less than zero on the horizontal axis are tightly spaced and are above the line of equality. Data with values greater than zero point 9 are more spread out.
Figure 15. Scatter plot. Rigid IRI model (shifted).
This scatter plot shows natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. Data range from 0 to 1 point 6 on the vertical axis and negative point 2 to positive 1 point 5 on the horizontal axis. Data less than 1 on the horizontal axis are tightly grouped and centered on the line of equality. Data with values greater than 1 are more spread out.
Figure 16. Scatter plot. Flexible IRI model with linear IRIage relationship.
This scatter plot shows natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. Data range from negative point 8 to positive 1 point 5 on the vertical axis and negative point 8 to positive 1 point 5 on the horizontal axis. Data less than point 9 on the horizontal axis are tightly grouped and centered on the line of equality. Data with values greater than point 9 are more spread out.
Figure 17. Scatter plot. Flexible IRI model with IRIexponential age relationship.
This scatter plot shows natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. Data range from negative point 8 to positive 1 point 4 on the vertical axis and negative point 8 to positive 1 point 6 on the horizontal axis. Data less than point 8 on the horizontal axis are tightly grouped and centered slightly below the line of equality. Data with values greater than point 8 are more spread out.
Figure 18. Scatter plot. Actual and predicted IRI values for test section 011001 using IRIexponential age relationship model.
This scatter plot shows natural logarithm of predicted and observed I R I values on the vertical axis and pavement age on the horizontal axis. Pavement age ranges from 5 to 17 years. The observed I R I values generally increase with age and range from negative point 1 to positive point 3. The predicted I R I values are constant at point 12 for all ages.
Figure 19. Scatter Plot. Rigid IRI model with linear IRIage relationship.
This scatter plot shows natural logarithmic predicted I R I values on the vertical axis versus natural logarithmic observed I R I values on the horizontal axis. Data range from zero to 1 point 6 on the vertical axis and negative point 2 to positive one point 5 on the horizontal axis. Data less than 1 on the horizontal axis are tightly grouped and centered on the line of equality. Data with values greater than 1 are more spread out.
Figure 20. Scatter plot. Rut depth model with linear rut–age relationship.
This scatter plot shows natural logarithmic predicted rut depth values on the vertical axis versus natural logarithmic observed rut depth values on the horizontal axis. Data range from zero to 3 on the vertical axis and from negative point 5 to positive 3 point 1 on the horizontal axis. Most data with horizontal axis values greater than 2 fall below the line of equality. Most data with horizontal axis values less than point 5 fall above the line of equality and are more spread out.
Figure 21. Scatter plot. Rut depth model with rut–natural logarithm age relationship.
This scatter plot shows natural logarithmic predicted rut depth values on the vertical axis versus natural logarithmic observed rut depth values on the horizontal axis. Data range from negative point 5 to positive 2 point 2 on the vertical axis and from negative point 5 to positive 3 point 2 on the horizontal axis. Most data with horizontal axis values greater than 2 fall below the line of equality. Most data with horizontal axis values less than point 5 fall above the line of equality and are more spread out.
Figure 22. Scatter plot. Measured FWPC deduct values.
This scatter plot shows observed F W P C cracking deduct values on the vertical axis versus age, in years, on the horizontal axis. F W P C values range from zero to 100 while pavement age ranges from 1 to 30 years. The data points are scattered across the graph with no discernable pattern. A relatively large number of zero F W P C values range across all pavement ages.
Figure 23. Graph plot. Measured FWPC values (using a subset of test sections).
This plot shows observed F W P C cracking deduct values on the vertical axis versus age, in years, on the horizontal axis. F W P C values range from zero to 100 while pavement age ranges from 1 to 30 years. This graph contains a set of 12 series. Each series represents F W P C data observed on one test section. Each series shows an increase in F W P C with increased age. Eight of the series exhibit approximately equivalent slopes, but these slopes occur at different pavement ages.
Figure 24. Graph plot. Example of logistical analysis to predict distress initiation.
This graph shows the predicted probability of distress occurring versus pavement age. The graph contains a set of five series. Each series represents a climatic scenario and consists of deepfreeze wet region, moderatefreeze wet region, nofreeze wet region, deepfreeze dry region, and moderatefreeze dry region. The probability at year 1 ranges from point 1 to point 25. The probability increases as age increases. The probability is near 1 at year 20.
Figure 25. Graph plot. Observed FWPC deduct values for test section 100102.
This scatter plot shows observed F W P C deduct values on the vertical axis versus pavement age on the horizontal axis. There are five observed F W P C values that are zero with the last one occurring at a pavement age of 2 point 2. This point is identified as the distress initiation age. At pavement ages greater than 2 point 2 the deduct values increase as age increases.
Figure 26. Graph plot. Observed FWPC deduct values for test section 050121,(with regression line).
This scatter plot shows observed F W P C deduct values on the vertical axis versus pavement age on the horizontal axis. None of the observed F W P C deduct values equals zero, and the observed deduct values increase as pavement age increases. The graph also shows a linear regression line. The distress initiation age is identified as the point where the linear regression point intercepts the horizontal axis.
Figure 27. Scatter plot. FWPC model for flexible pavements with linear FWPCage relationship.
This scatter plot shows natural logarithmic predicted F W P C deduct values on the vertical axis versus natural logarithmic observed F W P C deduct values on the horizontal axis. Data are scattered from zero to 4 point 5 on the vertical axis and from zero to 4 point 5 on the horizontal axis. At negative 2 point 3 on the horizontal axis there are relatively numerous data points that range from negative 2 point 3 to positive 4 point 5 on the vertical axis. At negative 2 point 3 on the vertical axis relatively numerous data points range from negative 2 point 3 to positive 4 point 5 on the horizontal axis.
Figure 28. Scatter plot. FWPC model for flexible pavements with FWPCnatural logarithm age relationship.
This scatter plot shows natural logarithmic predicted F W P C deduct values on the vertical axis versus natural logarithmic observed F W P C deduct values on the horizontal axis. Data points are scattered from negative 1 to positive 4 point 5 on the vertical axis and from zero to 4 point 5 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 4 point 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 4 point 5 on the horizontal axis.
Figure 29. Scatter plot. TC model for flexible pavements with linear TCage relationship.
This scatter plot shows natural logarithmic predicted T C deduct values on the vertical axis versus natural logarithmic observed T C deduct values on the xaxis. Data points are scattered from negative point 9 to positive 4 point 5 on the vertical axis and from point 5 to 4 point 5 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 4 point 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 4 point 5 on the horizontal axis.
Figure 30. Scatter plot. TC model for flexible pavements with TCnatural logarithm age relationship.
This scatter plot shows natural logarithmic predicted T C deduct values on the vertical axis versus natural logarithmic observed T C deduct values on the horizontal axis. Data points are scattered from negative 2 to positive 4 point 5 on the vertical axis and from point 5 to 4 point 5 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 4 point 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 4 point 5 on the horizontal axis.
Figure 31. Scatter plot. CB model for rigid pavements with linear CBage relationship.
This scatter plot shows natural logarithmic predicted C B values on the vertical axis versus natural logarithmic observed C B values on the horizontal axis. The data range from negative 2 point 3 to negative point 2 on the vertical axis and from negative 2 point 3 to negative point 5 on the horizontal axis. Most data points are located at negative 2 point 3 on the horizontal axis and range from negative 2 point 3 to negative point 2 on the vertical axis.
Figure 32. Scatter plot. CB model for rigid pavements with CBnatural logarithm age relationship.
This scatter plot shows natural logarithmic predicted C B values on the vertical axis versus natural logarithmic observed C B values on the horizontal axis. The data range from negative 2 point 3 to negative 1 on the vertical axis and from negative 2 point 3 to negative point 5 on the horizontal axis. Most of the data points are located at negative 2 point 3 on the horizontal axis and range from negative 2 point 3 to negative point 2 on the vertical axis.
Figure 33. Scatter plot. LC model for rigid pavements with linear LCage relationship.
This scatter plot shows natural logarithmic predicted L C values on the vertical axis versus natural logarithmic observed L C values on the horizontal axis. Data points are scattered from negative 2 to positive 6 on the vertical axis and from negative 1 point 9 to positive 4 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 3 on the horizontal axis.
Figure 34. Scatter plot. LC model for rigid pavements with LCnatural logarithmage relationship.
This scatter plot shows natural logarithmic predicted L C values on the vertical axis versus natural logarithmic observed L C values on the horizontal axis. Data points are scattered from negative 2 to positive 2 point 6 on the vertical axis and from negative 1 point 9 to positive 4 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 3 on the horizontal axis.
Figure 35. Scatter plot. TC model for rigid pavements with linear TCage relationship.
This scatter plot shows natural logarithmic predicted T C values on the vertical axis versus natural logarithmic observed T C values on the horizontal axis. Data points are scattered from negative 2 to positive 2 point 6 on the vertical axis and from negative 1 point 9 to positive 4 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 3 on the horizontal axis.
Figure 36. Scatter plot. TC model for rigid pavements with TCnatural logarithm age relationship.
This scatter plot shows natural logarithmic predicted T C values on the vertical axis versus natural logarithmic observed T C values on the horizontal axis. Data points are scattered from negative 2 to positive 2 point 1 on the vertical axis and from negative 1 point 9 to positive 4 on the horizontal axis. At negative 2 point 3 on the horizontal axis numerous data points range from negative 2 point 3 to positive 5 on the vertical axis. At negative 2 point 3 on the vertical axis numerous data points range from negative 2 point 3 to positive 3 on the horizontal axis.
Figure 37. Scatter plot. PUMP model for rigid pavements with linear PUMPage relationship.
This scatter plot shows natural logarithmic predicted PUMP values on the vertical axis versus natural logarithmic observed PUMP values on the horizontal axis. The data range from negative 2 point 3 to positive 4 on the vertical axis and from negative 2 point 3 to positive 4 point 5 on the horizontal axis. Numerous data points at negative 2 point 3 on the horizontal axis range from negative 2 point 3 to positive 4 on the vertical axis. Numerous data points at negative 2 point 3 on the vertical axis range from negative 2 point 3 to positive 4 point 5 on the horizontal axis.
Figure 38. Scatter plot. PUMP model for rigid pavements with PUMPnatural logarithm age relations.
This scatter plot shows natural logarithmic predicted PUMP values on the vertical axis versus natural logarithmic observed PUMP values on the horizontal axis. The data range from negative 2 point 3 to positive 4 on the vertical axis and from negative 2 point 3 to positive 4 point 5 on the horizontal axis. Numerous data points at negative 2 point 3 on the horizontal axis range from negative 2 point 3 to positive 4 on the vertical axis. Numerous data points at negative 2 point 3 on the vertical axis range from negative 2 point 3 to positive 4 point 5 on the horizontal axis.
Figure 39. Scatter plot. FLT model for rigid pavements with linear FLTage relationship.
This scatter plot shows natural logarithmic predicted F L T values on the vertical axis versus natural logarithmic observed rut F L T values on the horizontal axis. Data points range from negative point 1 to positive 1 point 5 on the vertical axis and from negative 1 point 6 to positive 2 point 4 on the horizontal axis. Most of the data points with horizontal axis values greater than 1 fall below the line of equality. Most of the data points with horizontal axis values less than zero fall above the line of equality.
Figure 40. Scatter plot. FLT model for rigid pavements with FLTnatural logarithm age relationship.
This scatter plot shows natural logarithmic predicted F L T values on the vertical axis versus natural logarithmic observed rut F L T values on the horizontal axis. Data points range from negative point 1 to positive 1 point 5 on the vertical axis and from negative 1 point 6 to positive 2 point 4 on the horizontal axis. Most of the data points with horizontal axis values greater than 1 fall below the line of equality. Most of the data points with horizontal axis values less than zero fall above the line of equality.
Figure 41. Scatter plot. Regional FI and FTC values.
This scatter plot shows freezethaw cycles on the vertical axis versus freezing index on the horizontal axis, both of which are logarithm scale. Freezethaw cycles range from 5 to 200, while freezing index values range from zero to nearly 2,000. Freezethaw cycles increase with an increase in freezing index for freezing index values less than 400. For freezing index values greater than 400, the number of freezethaw cycles does not increase with increases in freezing index.
Figure 42. Map. Geographic locations of climatic regions.
This map of the United States and Canada shows the location and climatic zone of each test section. The nofreeze test sections are located in the southern third of the United States and along the Pacific Coast. The moderatefreeze sections are located in a band from northern Texas to northern Kansas. Other moderatefreeze sections are found farther north in isolated locations such as southeastern Washington. The deepfreeze region ranges from northern Kansas to Canada.
Figure 43. Scatter plot. Relationship between FI and FTCs.
This scatter plot shows freezethaw cycles on the vertical axis versus freezing index on the horizontal axis on a linear scale. Freezethaw cycles range from 5 to 200 while freezing index values range from zero to nearly 2,000. Freezethaw cycles increase with an increase in freezing index for freezing index values less than 400. For freezing index values greater than 400, the number of freezethaw cycles does not increase with increases in freezing index.
Figure 44. Scatter chart. Mean predicted flexible pavement IRI values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted I R I plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents the established climatic scenarios. All series have linear trends with positive slopes. The predicted values at year 1 are approximately zero point 1 while the predicted values at age 20 range from zero point 59 to zero point 65.
Figure 45. Bar chart. Predicted flexible pavement IRI values at 20 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted I R I plus zero point 1 on the vertical axis parenthesis at 20 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from zero point 59 to zero point 66 with a mean value of zero point 63. The moderatefreeze wet region ranges from zero point 54 to zero point 62 with a mean value of zero point 58. The no freeze wet region ranges from zero point 56 to zero point 64 with a mean value of zero point 60. The deep freeze dry region ranges from zero point 61 to zero point 69 with a mean value zero point 65. The moderatefreeze dry region ranges from zero point 56 to zero point 64 with a mean value of zero point 60.
Figure 46. Scatter graph. Mean predicted rigid pavement IRI values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted I R I plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. All series have linear trends with positive slopes. The predicted values at year 1 range from zero point 1 to zero point 2 while the predicted values at age 20 range from zero point 4 to zero point 6.
Figure 47. Bar chart. Predicted rigid pavement IRI values at 20 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted I R I plus zero point 1 on the vertical axis parenthesis at 20 years end parenthesis and each climatic scenario on the horizontal axis. The bar chart shows the mean predicted value along with the95 percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from zero point 55 to zero point 61 with a mean value of zero point 58. The moderatefreeze wet region ranges from zero point 46 to zero point 54 with a mean value of zero point 50. The nofreeze wet region ranges from zero point 37 to zero point 43 with a mean value of zero point 40. The deepfreeze dry region ranges from zero point 51 to zero point 58 with a mean value of zero point 55. The moderatefreeze dry region ranges from zero point 44 to zero point 50 with a mean value of zero point 47.
Figure 48. Scatter graph. Mean predicted flexible pavement RUT values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted RUT plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. For all series, the predicted values increase with age; however, the rate of increase diminishes as age increases. The predicted values at year 1 range from zero point 8 to 1 point 3 while the predicted values at age 20 range from 1 point 8 to 2 point 2.
Figure 49. Bar chart. Predicted flexible pavement RUT values at 20 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted RUT plus zero point 1 on the vertical axis parenthesis at 20 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from 2 point zero 5 to 2 point 22 with a mean value of 2 point 15. The moderatefreeze wet region ranges from 2 point 15 to 2 point 38 with a mean value of 2 point 26. The nofreeze wet region ranges from 1 point 68 to 1 point 89 with a mean value of 1 point 79. The deepfreeze dry region ranges from 1 point 89 to 2 point zero 9 with a mean value of 1 point 99. The moderatefreeze dry region ranges from 2 point 11 to 2 point 32 with a mean value of 2 point 22.
Figure 50. Chart. Mean predicted flexible pavement FWPC values for each climatic region (BASE =DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted F W P C plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis; however, the rate of increase diminishes as age increases. Distress initiation ranges from 6 to 10 years. The predicted values at age 20 years range from 3 point 5 to 4 point 5.
Figure 51. Bar chart. Predicted flexible pavement FWPC values at 20 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted F W P C plus zero point 1 on the vertical axis parenthesis at 20 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from 4 point 19 to 4 point 61 with a mean value of 4 point 40. The moderatefreeze wet region ranges from 3 point 21 to 3 point 80 with a mean value of 3 point 54. The nofreeze wet region ranges from 4 point 30 to 4 point 80 with a mean value of 4 point 54. The deepfreeze dry region ranges from 4 point zero to 4 point 64 with a mean value of 4 point 36. The moderatefreeze dry region ranges from 3 point 20 to 3 point 75 with a mean value of 3 point 50.
Figure 52. Scatter chart. Meanpredicted flexible pavement TC values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted T C plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of 5 series. Each series represents one of the established climatic scenarios. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis; however, the rate of increase diminishes as age increases. Distress initiation ranges from 6 to 12 years. The predicted values at age 20 years range from 2 point 8 to 4 point 5.
Figure 53. Bar chart. Predicted flexible pavement TC values at 20 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted T C plus zero point 1 on the vertical axis parenthesis at 20 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from 4 point 3 to 4 point 7 with a mean value of 4 point 5. The moderatefreeze wet region ranges from 3 point 5 to 3 point 9 with a mean value of 3 point 6. The nofreeze wet region ranges from 2 point 5 to 3 point zero with a mean value of 2 point 8. The deepfreeze dry region ranges from 4 point 1 to 4 point 6 with a mean value of 4 point 4. The moderatefreeze dry region ranges from 3 point 4 to 3 point 8 with a mean value of 3 point 6.
Figure 54. Scatter graph. Mean predicted rigid pavement LC values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted L C plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis; however, the rate of increase diminishes as age increases. Distress initiation ranges from 6 to 25 years. The predicted values at age 25 years range from negative point 5 to negative point 1.
Figure 55. Bar chart. Predicted rigid pavement LC values at 25 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted LC plus zero point 1 on the vertical axis parenthesis at 25 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from negative point 5 to point 2 with a mean value of negative point 1. The moderatefreeze wet region ranges from negative zero point 9 to point 6 with a mean value of negative point 1. The nofreeze wet region ranges from negative 2 point 3 to negative 1 point 7 with a mean value of negative 2 point 3. The deepfreeze dry region ranges from negative 1 to point 2 with a mean value of negative zero point 4. The moderatefreeze dry region ranges from negative zero point 5 to point 5 with a mean value of negative point 1.
Figure 56. Scatter Graph. Mean predicted rigid pavement TC values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted T C plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis; however, the rate of increase diminishes as age increases. Distress initiation ranges from zero to 11 years. The predicted values at age 25 years range from zero to 1 point 2.
Figure 57. Bar chart. Predicted rigid pavement TC values at 25 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted T C plus zero point 1 on the vertical axis parenthesis at 25 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95 percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from point 1 to 1 point 3 with a mean value of point 8. The moderatefreeze wet region ranges from negative zero point 1 to positive 1 point 2 with a mean value of point 5. The nofreeze wet region ranges from zero point 8 to 1 point 6 with a mean value of negative 1 point 2. The deepfreeze dry region ranges from negative point 5 to positive point 6 with a mean value of zero. The moderatefreeze dry region ranges from zero to 1 with a mean value of point 5.
Figure 58. Scatter chart. Mean predicted rigid pavement FLT values for each climatic region (BASE=DGAB/SG=FINE).
This scatter plot shows natural logarithm predicted F L T plus zero point 1 on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. All series have linear trends with positive slopes. The predicted values at year 1 range from negative zero point 15 to zero point 1, while the predicted values at age 20 range from zero point 3 to zero point 7.
Figure 59. Bar chart. Predicted rigid pavement FLT values at 20 years for each climatic region (BASE=DGAB/SG=FINE).
This bar chart shows natural logarithm predicted F L T plus zero point 1 on the vertical axis parenthesis at 20 years end parenthesis and each climatic scenario on the horizontal axis. The bars show the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from zero point 41 to zero point 61 with a mean value of zero point 51. The moderatefreeze wet region ranges from zero point 3 to zero point 52 with a mean value of zero point 41. The nofreeze wet region ranges from zero point 3 to zero point 54 with a mean value of zero point 41. The deepfreeze dry region ranges from zero point 56 to zero point 80 with a mean value of zero point 68. The moderatefreeze dry region ranges from zero point 20 to zero point 44 with a mean value of zero point 31.
Figure 60. Scatter chart. Flexible pavement IRI for selected sites in each agency.
This scatter plot shows mean predicted I R I values on the vertical axis versus pavement age on the horizontal axis. The plot consists of nine series. Each series represents one of the participating Agencies. For all series, the predicted values increase with age at a rate that increases as age increases. The predicted values at year 1 are 1 and range from 1 point 7 to 1 point 9 at year 20.
Figure 61. Scatter chart. Flexible pavement RUT for selected sites in each agency.
This scatter plot shows predicted RUT on the vertical axis versus pavement age on the horizontal axis. The plot consists of nine series. Each series represents one of the participating Agencies. For all series, the predicted values increase with age; however, the rate of increase diminishes as age increases. The predicted values at year 1 range from 2 to 3 point 2 while the predicted values at age 20 range from 7 to 9 point 5.
Figure 62. Scatter chart. Flexible pavement TC for selected sites in each agency.
This scatter plot shows predicted T C on the vertical axis versus pavement age on the horizontal axis. The plot consists of nine series. Each series represents one of the participating Agencies. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis. Distress initiation ranges from 4 to 11 years. The predicted values at age 20 years range from 25 to 100.
Figure 63. Scatter graph. Flexible pavement FWPC for selected sites in each agency.
This scatter plot shows predicted F W P C on the vertical axis versus pavement age on the horizontal axis. The plot consists of nine series. Each series represents one of the participating Agencies. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis. Distress initiation ranges from 8 to 11 years. The predicted values at age 20 years range from 45 to 70.
Figure 64. Scatter graph. FWPC predictions for sites 1001 and 6027 in Idaho.
This scatter plot shows predicted F W P C on the vertical axis versus pavement age on the horizontal axis. The plot consists of two series. Each series represents one location in Idaho. For site 1 0 0 1, distress initiation occurs at year 7 and the predicted value at year 20 is 70. For site 6 0 2 7, distress initiation occurs at year 9 and the predicted value at year 20 is 25.
Figure 65. Scatter graph. Flexible TC predictions for the environments at sites 0200 and 1004 in Michigan.
This scatter plot shows predicted T C on the vertical axis versus pavement age on the horizontal axis. The plot consists of two series. Each series represents one location in Michigan. For site 0 2 0 0, distress initiation occurs at year 8 and the predicted value at year 20 is 55. For site 1 0 0 4, distress initiation occurs at year 6 and the predicted value at year 20 is 100.
Figure 66. Photo. Road construction in Sweden with deep base section.
This daylight photograph illustrates construction of deepbase road sections in Sweden. The roadway prism is thick and consists mostly of unbound base material.
Figure 67. Photo. Installation of longitudinal drainage to reduce frost heaving.
This daylight photograph shows a workman standing up to his waist in a longitudinal drainage installation. Flexible pipe is being placed in a trench along the edge of pavement with a depth of 1 meter parenthesis 3 feet end parenthesis. Filter fabric lines the trench before the installation of the pipe.
Figure 68. Diagram. Standard pavement section from a Midwestern State.
This line drawing shows a typical pavement cross section. The significance of this diagram is that the cross section is constant over the entire width of the roadway prism. Base and subbase material are placed only under the pavement, and they are not extended to the edge of the prism parenthesis that is, day lighting end parenthesis. It also shows longitudinal edge drains placed below the travel lane and shoulder interface.
Figure 69. Diagram. Primary highway cross section.
This line drawing shows the typical pavement section used to develop life cycle costs for primary highway segments. The section consists of two lanes each 3 point 7 meters parenthesis 12 feet end parenthesis wide with shoulders on both sides that are 2 point 4 meters parenthesis 8 feet end parenthesis wide. The travel lanes have an A C surface course with a thickness of 25 millimeters parenthesis 1 inch end parenthesis, an A C binder course with a thickness of 50 millimeters parenthesis 2 inches end parenthesis, and an A C base course with a thickness of 75 millimeters parenthesis 3 inches end parenthesis. The shoulders have the same A C surface and A C binder course thicknesses, and there is no A C base course. The thickness of the unbound base course is 205 millimeters parenthesis 8 inches end parenthesis under the travel lanes and 280 millimeters parenthesis 11 inches end parenthesis.
Figure 70. Diagram. Interstate highway, left section.
This line drawing shows the left half of a cross section of a typical pavement section used to develop life cycle costs for interstate highway segments. The other half of the drawing appears in figure 71. The section consists of a fourlane divided highway. The lane widths are 3 point 7 meters parenthesis 12 feet end parenthesis. The outside shoulder has a width of 3 meters parenthesis 10 feet end parenthesis, while the inside shoulder has a width of 1 point 2 meters. The A C surface course has a thickness of 25 millimeters parenthesis 1 inch end parenthesis, the A C binder course has a thickness of 75 millimeters parenthesis 3 inches end parenthesis, the A C base course has a thickness of 150 millimeters parenthesis 6 inches end parenthesis, and the unbound base course has a thickness of 305 millimeters parenthesis 12 inches end parenthesis. This layer is consistent under the travel lanes and the shoulders.
Figure 71. Diagram. Interstate highway, right section.
This line drawing shows the right half of a cross section of a typical pavement section used to develop life cycle costs for interstate highway segments. The other half of the drawing appears in figure 70. The section consists of a four lane divided highway. The lane widths are 3 point 7 meters parenthesis 12 feet end parenthesis. The outside shoulder has a width of 3 meters parenthesis 10 feet end parenthesis, while the inside shoulder has a width of 1 point 2 meters. The A C surface course has a thickness of 25 millimeters parenthesis 1 inch end parenthesis, the A C binder course has a thickness of 75 millimeters parenthesis 3 inches end parenthesis, the A C base course has a thickness of 150 millimeters parenthesis 6 inches end parenthesis, and the unbound base course has a thickness of 305 millimeters parenthesis 12 inches end parenthesis. This layer is consistent under the travel lanes and the shoulders.
Figure 72. Distribution chart. Annualized costs for standard primary pavement sections.
This chart shows total present worth on the vertical axis and each climatic scenario on the horizontal axis. The chart shows the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from 450,000 to 550,000 with a mean value of 50,000. The moderatefreeze wet region ranges from 400,000 to 500,000 with a mean value of 450,000. The nofreeze wet region ranges from just over 450,000 to just over 550,000 with a mean value of just over 500,000. The deepfreeze dry region ranges from just over 450,000 to just over 550,000 with a mean value of just over 500,000. The moderatefreeze dry region ranges from 400,000 to 500,000 with a mean value of 450,000.
Figure 73. Distribution chart. Annualized costs for standard Interstate pavement sections.
This chart shows the distribution of the total present worth on the vertical axis and each climatic scenario on the horizontal axis. The chart shows the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deep freeze wet region ranges from 1,400,000 to 1,700,000 with a mean value of 1,550,000. The moderatefreeze wet region ranges from 1,300,000 to 1,600,000 with a mean value of 1,450,000. The nofreeze wet region ranges from just over 1,450,000 to just over 1,750,000 with a mean value of just over 1,600,000. The deepfreeze dry region ranges from 1,450,000 to 1,750,000 with a mean value of just over 1,600,000. The moderatefreeze dry region ranges from 1,300,000 to 1,600,000 with a mean value of 1,450,000.
Figure 74. Distribution chart. Annualized costs for mitigated primary pavement sections.
This chart shows the distribution of the total present worth on the vertical axis and each climatic scenario on the horizontal axis. The chart shows the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from 800,000 to just over 1,000,000 with a mean value of 950,000. The moderatefreeze wet region ranges from 700,000 to 950,000 with a mean value of just over 800,000. The nofreeze wet region ranges from 450,000 to just over 550,000 with a mean value of just over 500,000. The deepfreeze dry region ranges from 800,000 to just over 1,000,000 with a mean value of just over 950,000. The moderatefreeze dry region ranges from 700,000 to 9500,000 with a mean value of just over 800,000.
Figure 75. Distribution chart. Annualized costs for mitigated interstate pavement sections.
This chart shows the distribution of the total present worth on the vertical axis and each climatic scenario on the horizontal axis. The chart shows the mean predicted value along with the 95percent confidence interval for each climate. The 95percent confidence interval for the deepfreeze wet region ranges from 1,800,000 to 2,300,000 with a mean value of 2,050,000. The moderatefreeze wet region ranges from 1,750,000 to 2,200,000 with a mean value of 1,950,000. The nofreeze wet region ranges from 1,450,000 to 1,700,000 with a mean value of just over 1,500,000. The deepfreeze dry region ranges from 1,800,000 to 2,300,000 with a mean value of just over 2,050,000. The moderatefreeze dry region ranges from 1,750,000 to 2,200,000 with a mean value of 1,950,000.
Figure 76. Graph. Comparison of fatigue cracking trends before and after local calibration.
This plot shows fatigue cracking parenthesis as a percentage of wheelpath area end parenthesis on the vertical axis and pavement age parenthesis in years end parenthesis on the horizontal axis. There are three series: trends from the frost model, trends from the N C H R P 137 A model without calibration, and trends from the N CH R P 137 A model with calibration. All three trends are approximately zero at age 1. At year 20, the frost and N C H R P 137 A with calibration trends are approximately 14 while the N C H RP 137 A without calibration is approximately 7.
Figure 77. Graph. Comparison of rutting trends before and after local calibration.
This plot shows rutting cracking parenthesis in inches end parenthesis on the vertical axis and pavement age parenthesis in years end parenthesis on the horizontal axis. There are three series: trends from the frost model, trends from the N C H RP 137 A model without calibration, and trends from the N C H R P 137 A model with calibration. At year 1, the value for the frost model is about zero point 12, the value for the N C H R P 137 A with calibration is zero point 18, and the value for the N C H R P 137 A without calibration is zero point 25. At year 20, the frost model is approximately zero point 28, N C H R P 137 A with calibration is zero point 3, and the N C H R P 137 A without calibration is zero point 61.
Figure 78. Graph. Comparison of ride trends before and after local calibration.
This plot shows the ride index parenthesis inches per mile end parenthesis on the vertical axis and pavement age parenthesis in years end parenthesis on the horizontal axis. There are three series: trends from the frost model, trends from the N C H R P 137 A model without calibration, and trends from the N C H R P 137 A model with calibration. At year 1, the value for the frost model is about 60, the value for the N C H R P 137 A with calibration is 60 and the value for the N C H R P 137 A without calibration is 70. At year 20, the frost model is approximately 110, N C H R P 137 A with calibration is 110, and the N C H R P 137 A without calibration is above 300.
Figure 79. Graph. Individual distress deduct curves.
This graph shows the relationship between surface distress amounts parenthesis percentage of occurrence end parenthesis and deduct values for low, moderate, and high severity levels. Distress amounts are on the horizontal axis and range from zero point 1 to 100 on a logarithmic scale. The vertical axis contains deduct values that range from 1 to 100 on a logarithmic scale. The high severity curve is a straight line that is near 1 at zero point 1 percent occurrence and increases to 100 at a percentage occurrence of 70. The low severity curve is a straight line that is near 1 at zero point 1 percentage occurrence and increases to 30 at a percentage occurrence of 70.
Figure 80. Graph. Example of fatigue cracking trends for different environments.
This scatter plot shows predicted F W P C on the vertical axis versus pavement age on the horizontal axis. The plot consists of five series. Each series represents one of the established climatic scenarios. For all series, the predicted values increase with age parenthesis after distress initiation end parenthesis. Distress initiation ranges from 3 to 7 years. The predicted values at age 30 years range from 50 to 100.
Figure 81. Chart. Fatigue cracking index trend for environmental case wet nofreeze.
This is a graph with the fatigue cracking index on the vertical axis and pavement age on the horizontal axis. From year zero to year 3 the index is 100. At that point it begins dropping with a linear trend as age increases. The index reaches zero at an age of 21.
Figure 82. Chart. Example of shifting trend line to fit index for a given location.
This graph shows the fatigue cracking index on the vertical axis and pavement age on the horizontal axis. From year zero to year 3 the index is 100. At that point it begins dropping with a linear trend with increases in age. The index reaches zero at an age of 21. An additional trend line is provided on the graph with the same slope as the existing index but shifted to the right as appropriate.
Figure 83. Map. Alaska geographic location of analysis test sections.
This map of Alaska shows the location of each test section used in the study. Section 1 0 0 8 is the northernmost section, while 1 0 0 1 and 1 0 0 2 are the southernmost.
Figure 84. Map. Idaho geographic location of analysis test sections.
This map of Idaho shows the location of each test section used in the study. Section 9 0 3 4 is the northernmost section, while 6 0 2 7 is the southernmost.
Figure 85. Map. Illinois geographic location of analysis test sections.
This map of Illinois shows the location of each test section used in the study. Section 1 0 0 2 is the northernmost section, while 1 0 0 3 and 6 0 5 0 are the southernmost.
Figure 86. Map. Indiana geographic location of analysis test sections.
This map of Indiana shows the location of each test section used in the study. Sections 2 0 0 8 and 3 0 0 3 are the northernmost, while 1 0 3 7 and 3 0 3 1 are the southernmost.
Figure 87. Map. Michigan geographic location of analysis test sections.
This map of Michigan shows the location of each test section used in the study. Section 1 0 0 4 is the northernmost, while section 0 2 0 0 is the southernmost.
Figure 88. Map. New York geographic location of analysis test sections.
This map of New York shows the location of each test section used in the study. Section 1 6 4 4 is the northernmost, while section 1 5 9 7 is the southernmost.
Figure 89. Map. North Carolina geographic location of analysis test sections.
This map of North Carolina shows the location of each test section used in the study. Sections 1 0 2 8 and 1 0 3 0 are the northernmost, while section 1 6 4 5 is the southernmost.
Figure 90. Map. Ohio geographic locations of analysis test sections.
This map of Ohio shows the location of each test section used in the study. Sections 0 1 0 0, 0 2 0 0, and 0 8 0 0 are near the center of the State, while 3 0 1 3 is near the southern border.
Figure 91. Map. Pennsylvania geographic locations of analysis test sections.
This map of Pennsylvania shows the location of each test section used in the study. Section 1 5 9 7 is the northernmost, while section 1 6 1 8 is the southernmost.
Figure 92. Diagram. Typical section for rural primary (2 lanes) in Alaska.
This line drawing shows a flexible pavement typical cross section for a rural primary road in Alaska. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis and 1.2meter parenthesis 4foot end parenthesis shoulders. Select material, subbase, aggregate base, and asphalt concrete layers are present in the layer structure. They are consistent across the width of the pavement.
Figure 93. Diagram. Rigid pavement rural interstate typical section for Idaho.
This line drawing shows a rigid pavement typical cross section for rural interstate road in Idaho. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis along with one 1.2meter parenthesis 4 foot end parenthesis and one 3meter parenthesis 10foot end parenthesis shoulder. Geotextile, rock cap, asphalt treated pavement leveling course, and P C C layers are present in the layer structure. They are consistent across the width of pavement.
Figure 94. Diagram. Flexible pavement rural interstate typical section for Idaho.
This line drawing shows a flexible pavement typical cross section for rural interstate road in Idaho. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis along with one 1.2meter parenthesis 4foot end parenthesis and one 3meter parenthesis 10foot end parenthesis shoulder. Geotextile, granular subbase, rock cap, asphalt treated pavement leveling course, and plant mix layers are present in the layer structure. They are consistent across the width of pavement.
Figure 95. Diagram. Rigid pavement rural primary typical section for Idaho.
This line drawing shows a rigid pavement typical cross section for rural primary road in Idaho. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis along with 3meter parenthesis 10foot end parenthesis shoulders. Geotextile, rock cap, asphalt treated pavement leveling course, and P C C layers are present in the layer structure. They are consistent across the width of pavement.
Figure 96. Flexible pavement rural primary typical section for Idaho.
This line drawing shows a flexible pavement typical cross section for rural primary road in Idaho. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis along with 3meter parenthesis 10–foot end parenthesis shoulders. Geotextile, rock cap, asphalt treated pavement leveling course, and plant mix layers are present in the layer structure. They are consistent across the width of pavement.
Figure 97. Diagram. Rigid pavement at LTPP site 163023 in Idaho.
This line drawing shows the cross section for L T P P site 1 6 3 0 2 3 in Idaho. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis along with one 1.2meter parenthesis 4foot end parenthesis and one 3meter parenthesis 10foot end parenthesis shoulder. Select borrow, aggregate base, open graded aggregate base, and P C C layers are present in the layer structure. They are consistent across the width of pavement.
Figure 98. Diagram. Flexible pavement at LTPP site 169032 in Idaho.
This line drawing shows the cross section for L T P P site 1 6 9 0 3 2 in Idaho. The structure consists of 2 lanes with a width of 3.7 meters parenthesis 12 feet end parenthesis along with 3meter parenthesis 10foot end parenthesis shoulders. Rock cap, asphalt treated pavement leveling course, and plant mix layers are present in the layer structure. They are consistent across the width of pavement.
Figure 99. Diagram. Typical portland cement concrete pavement section for New York.
This line drawing shows a rigid pavement typical cross section for New York. The structure consists of 3 lanes with a width of 3.6 meters parenthesis 11 feet end parenthesis along with a 3meter parenthesis 10foot end parenthesis and 1.2meter parenthesis 4foot end parenthesis shoulder. Subbase, permeable base, and P C C layers are present in the layer structure. They are consistent across the width of pavement.
Figure 100. Diagram. Typical hot mix asphalt pavement section for New York.
This line drawing shows a flexible pavement typical cross section for New York. The structure consists of 2 lanes with a width of 3.6 meters parenthesis 11 feet end parenthesis along with a 3meter parenthesis 10foot end parenthesis and 1.2meter parenthesis 4foot end parenthesis shoulder. Select granular subgrade, subbase, permeable base, and A C layers are present in the layer structure. They are consistent across the width of pavement.
Figure 101. Flowchart. Example of NCHRP 137A calibration methodology flowchart.
This flowchart begins with inputting one calibration section and changing transverse cracking calibration factors. Then check if estimated transverse cracking is onethird of actual. If not, change transverse cracking calibration factors and repeat process. Then change faulting calibration factors and check estimated faulting equal to actual faulting. If not, change faulting calibration factors. Then input validation section with different climate and check estimated cracking and estimated faulting to actual. If check fails, change transverse cracking calibration factors and begin process over. Then input validation section with different traffic loading and check estimated cracking and estimated faulting to actual. If check fails, change transverse cracking calibration factors and begin process over. Then input validation section with different base material and check estimated cracking and estimated faulting to actual. If check fails, change transverse cracking calibration factors and begin process over. Then change roughness calibration factors and check estimated I R I to actual. If check fails, change roughness calibration factors. Then input validation section with different climate and check estimated I R I to actual. If check fails, change roughness calibration factors and begin process over. Then input validation section with different traffic loading and check estimated I R I to actual. If check fails, change roughness calibration factors and begin process over. Then input validation section with different base material and check estimated I R I to actual. If check fails, change roughness calibration factors and begin process over. If check passes, output calibration factors.
Equation 1. Equation.
Cooling index equals the sum of mean daily temperature minus 18.33 if the mean daily temperature is greater than 18.33 degrees Celsius.
Equation 2. Equation.
Low severity distress deduct value equals 3.4082 multiplied by the percentage of recorded low severity distress to the power of 0.514.
Equation 3. Equation.
Moderate severity distress deduct value equals 4.457 multiplied by the percentage of recorded moderate severity distress to the power of 0.6107.
Equation 4. Equation.
High severity distress deduct value equals 5.2064 multiplied by the percentage of recorded high severity distress to the power of 0.6956.
Equation 5. Equation.
The total distress deduct value equals the sum of low severity distress deduct, moderate severity distress deduct, and high severity distress deduct values.
Equation 6. Equation.
Residual equals a coefficient, beta subscript 1, multiplied by the explanatory variable added to another coefficient, beta subscript 2, multiplied by the explanatory variable squared.
Equation 7. Equation.
Low severity distress deduct value equals 3.4082 multiplied by the percentage of recorded low severity distress to the power of 0.514.
Equation 8. Equation.
Moderate severity distress deduct value equals 4.457 multiplied by the percentage of recorded moderate severity distress to the power of 0.6107.
Equation 9. Equation.
High severity distress deduct value equals 5.2064 multiplied by the percentage of recorded high severity distress to the power of 0.6956.
Equation 10. Equation.
The total distress deduct value equals the sum of low severity distress deduct, moderate severity distress deduct, and high severity distress deduct values.
Equation 11. Equation.
Plastic strain at depth Z in microstrain equals 0.087 multiplied by number of load repetitions to the power of 0.333 multiplied by vertical stress at depth Z divided by reference stress to the power of 0.333 multiplied by vertical resilient strain at depth Z in microstrain.
Equation 12. Equation.
Natural logarithm of I R I plus 0.1 equals alpha minus delta plus 0.115 multiplied by logarithm of E S A L divided by S N plus 0.0329 multiplied by A G E minus 0.0000433 multiplied by C I plus 0.00000228 multiplied F I plus 0.0000590 multiplied P R E C I P plus 0.000221 multiplied by F T C plus 0.0000859 multiplied by A C T H I C K minus 0.00539 multiplied by logarithm of E S A L divided by S N multiplied by A G E plus 0.00000177 multiplied by A G E multiplied by C I plus 0.00000455 multiplied by A G E multiplied by F I plus 0.643 multiplied by M I R I minus 0.0250 multiplied by M I R I underscore A G E minus 0.00000240 multiplied by A G E multiplied by P R E C I P minus 0.0000109 multiplied by A G E multiplied by F T C.
Equation 13. Equation.
Delta equals alpha plus 0.115 multiplied by logarithm of E S A L divided by S N plus 0.00790 multiplied by M I R I underscore A G E minus 0.0000433 multiplied by C I plus 0.00000228 multiplied F I plus 0.0000590 multiplied P R E C I P plus 0.000221 multiplied by F T C plus 0.0000859 multiplied by A C T H I C K minus 0.00539 multiplied by logarithm of ESAL divided by SN multiplied by M I R I underscore A G E plus 0.00000177 multiplied by M I R I underscore A G E multiplied by C I plus 0.00000455 multiplied by M I R I underscore A G E multiplied by F I plus 0.643 multiplied by M I R I minus 0.00000240 multiplied by M I R I underscore A G E multiplied by P R E C I P minus 0.0000109 multiplied by M I R I underscore A G E multiplied by F T C minus natural logarithm of M I R I plus 0.1.
Equation 14. Equation.
Delta equals negative 0.713 plus 0.115 multiplied by the quotient of the logarithm of 126000 all divided by 5, plus 0.00790 multiplied by 1 minus 0.0000433 multiplied by 205 plus 0.00000228 multiplied by 688, plus 0.0000590 multiplied by 1140, plus 0.000221 multiplied by 80, plus 0.0000859 multiplied by 6.5, minus 0.00539 multiplied by logarithm of 126000 divided by 5 multiplied by 1 plus 0.00000177 multiplied by 1 multiplied by 205, plus 0.00000455 multiplied by 1 multiplied by 688 plus 0.643 multiplied by 1 minus 0.00000240 multiplied by 1 multiplied by 1140, minus 0.0000109 multiplied by 1 multiplied by 80, minus natural logarithm of 1 plus 0.1.
Equation 15. Equation.
Delta equals 0.032.
Equation 16. Equation.
Natural logarithm of I R I plus 0.1 equals negative 0.713 minus 0.032, plus 0.115 multiplied by logarithm of 126000 divided by 5, plus 0.0329 multiplied by 2, minus 0.0000433 multiplied by 205, plus 0.00000228 multiplied 688, plus 0.0000590 multiplied 1140, plus 0.000221 multiplied by 80, plus 0.0000859 multiplied by 6.5, minus 0.00539 multiplied by logarithm of 126000 divided by 5 multiplied by 2, plus 0.00000177 multiplied by 2 multiplied by 205, plus 0.00000455 multiplied by 2 multiplied by 688, plus 0.643 multiplied by 1, minus 0.0250 multiplied by 1, minus 0.00000240 multiplied by 2 multiplied by 1140, minus 0.0000109 multiplied by 2 multiplied by 80.
Equation 17. Equation.
Natural logarithm of I R I plus 0.1 equals 0.123.
Equation 18. Equation.
I R I equals the exponential function E to the power of 0.123, minus 0.1 equals 1.03.
Equation 19. Equation.
Natural logarithm I R I plus 0.1 equals alpha plus 0.501 multiplied by M I R I, plus 0.0985 multiplied by logarithm E S A L divided by D E P T H, plus 0.000321 multiplied by F I, minus 0.0000184 multiplied by C I, plus 0.0000417 multiplied by P R E C I P, plus 0.000776 multiplied by F T C, minus 0.0164 multiplied by M I R I underscore A G E, plus 0.0168 multiplied by A G E, minus 0.00000230 multiplied by F I multiplied by F T C, plus 0.0000000811 multiplied by C I multiplied by F T C, minus 0.000000357 multiplied by F I multiplied by C I, plus 0.00000430 multiplied by F I multiplied by A G E.
Equation 20. Equation.
Natural logarithm I R I plus 0.1 equals negative 0.0510, plus 0.501 multiplied by 1, plus 0.0985 multiplied by logarithm 410000 divided by 9.5, plus 0.000321 multiplied by 688, minus 0.0000184 multiplied by 205, plus 0.0000417 multiplied by 1140, plus 0.000776 multiplied by 80, minus 0.0164 multiplied by 1, plus 0.0168 multiplied by 2, minus 0.00000230 multiplied by 688 multiplied by 80, plus 0.0000000811 multiplied by 205 multiplied by 80, minus 0.000000357 multiplied by 688 multiplied by 205, plus 0.00000430 multiplied by 688 multiplied by 2.
Equation 21. Equation.
Natural logarithm I R I plus 0.1 equals 0.223
Equation 22. Equation.
I R I equals exponential function E to the power of 0..223 minus 0.1 equals 1.15.
Equation 23. Equation.
Natural logarithm of P divided by 1 minus P equals alpha subscript 1 plus 0.917 multiplied by logarithm of E S A L divided by S N, minus 0.0112 multiplied by F T C, plus 0.000106 multiplied by F I, minus 0.000221 multiplied by C I minus 0.00101 multiplied by P R E C I P, plus 0.513 multiplied by A G E, minus 0.0610 multiplied by A G E multiplied by logarithm of E S A L divided by S N, minus 0.000115 multiplied by A G E multiplied by C I, minus 0.000144 multiplied by A G E.
Equation 24. Equation.
Natural logarithm of F W P C plus 0.1 equals alpha subscript 2 minus 0.00365 multiplied by logarithm E S A L divided by S N plus 0.875 multiplied by natural logarithm of A D J underscore A G E plus 0.1, minus 0.000877 multiplied by C I, minus 0.000359 multiplied by F I, minus 0.000171 multiplied by P R E C I P, minus 0.0112 multiplied by F T C, plus 0.000214 multiplied by natural logarithm of A D J underscore A G E, plus 0.1 multiplied by P R E C I P.
Equation 25. Equation.
Natural logarithm of F W P C plus 0.1 equals alpha subscript 2 minus 0.00365 multiplied by logarithm E S A L divided by S N plus 0.875 multiplied by natural logarithm of A D J underscore A G E plus 0.1, minus 0.000877 multiplied by C I, minus 0.000359 multiplied by F I, minus 0.000171 multiplied by P R E C I P, minus 0.0112 multiplied by F T C, plus 0.000214 multiplied by natural logarithm of A D J underscore A G E, plus 0.1 multiplied by P R E C I P.
Equation 26. Equation.
Natural logarithm of lowercase F W P C plus 0.1 equals alpha subscript 2, plus 0.0763 multiplied by logarithm E S A L divided by S N, plus 0.737 multiplied by natural logarithm of A D J underscore A G E, plus 0.1 minus 0.00104 multiplied by C I, minus 0.000412 multiplied by F I, plus 0.0000203 multiplied by P R E C I P, minus 0.0112 multiplied by F T C, plus 0.000207 multiplied by natural logarithm of A D J underscore A G E, plus 0.1 multiplied by P R E C I P.
Equation 27. Equation.
Natural logarithm of 0.7 divided by 1 minus 0.7 equals negative 1.06, plus 0.917 multiplied by logarithm of 126000 divided by 5, minus 0.0112 multiplied by 80, plus 0.000106 multiplied by 688, minus 0.000221 multiplied by 205, minus 0.00101 multiplied by 1140, plus 0.513 multiplied by A G E, minus 0.0610 multiplied by A G E multiplied by logarithm of 126000 divided by 5, minus 0.000115 multiplied by A G E multiplied by 205, minus 0.000144 multiplied by A G E.
Equation 28. Equation.
A G E equals 9.13, which equals crack initiation age.
Equation 29. Equation.
Natural logarithm of F W P C plus 0.1 equals 3.25 minus, 0.00365 multiplied by logarithm 126000 divided by 5, plus 0.875 multiplied by natural logarithm of 2.87 plus 0.1, minus 0.000877 multiplied by 205, minus 0.000359 multiplied 688, minus 0.000171 multiplied by 1140, minus 0.0112 multiplied by 80, plus 0.000214 multiplied by natural logarithm of 2.87 plus 0.1 multiplied by 1140.
Equation 30. Equation.
Natural logarithm of F W P C plus 0.1 equals 2.95.
Equation 31. Equation.
F W P C equals the exponential function E to the power 2.95 minus 0.1 equals 18.9.
Equation 32. Equation.
Natural logarithm of P divided by 1 minus P equals alpha subscript 1 plus 0.723 multiplied by logarithm E S A L divided by S N, minus 0.0134 multiplied by F T C, plus 0.00265 multiplied by F I, plus 0.00130 multiplied by C I, plus 0.000181 multiplied by P R E C I P, plus 0.512 multiplied by A G E, minus 0.0288 multiplied by A G E multiplied by logarithm E S A L divided by S N, minus 0.000186 multiplied by A G E multiplied by C I, minus 0.000192 multiplied by A G E multiplied by F I, minus 0.0000772 multiplied by A G E multiplied by P R E C I P, plus 0.00156 multiplied by A G E multiplied by F T C.
Equation 33. Equation.
Natural logarithm of T C plus 0.1 equals alpha subscript 2 plus, 0.175 multiplied by logarithm of E S A L divided by S N, plus 1.14 multiplied by natural logarithm of A G E plus 0.1, minus 0.000316 multiplied by C I, plus 0.000528 multiplied by F I, minus 0.000110 multiplied by P R E C I P, plus 0.000112 multiplied by natural logarithm of A G E plus 0.1, multiplied by P R E C I P, minus 0.000188 multiplied by C I multiplied by natural logarithm of A G E plus 0.1.
Equation 34. Equation.
Natural logarithm of 0.7 divided by 1 minus 0.7 equals 1.03 plus 0.723 multiplied by logarithm 126000 divided by 5, minus 0.0134 multiplied by 80, plus 0.00265 multiplied by 688, plus 0.00130 multiplied by 205, plus 0.0000181 multiplied by 1140, plus 0.512 multiplied by A G E, minus 0.0288 multiplied by A G E multiplied by logarithm 126000 divided by 5, minus 0.000186 multiplied by A G E multiplied by 205, minus 0.000192 multiplied by A G E multiplied by 688, minus 0.0000772 multiplied by A G E multiplied by 1140, plus 0.00156 multiplied by A G E multiplied by 80.
Equation 35. Equation.
A G E equals 7.54, which equals crack initiation age.
Equation 36. Equation.
Natural logarithm of T C plus 0.1 equals 1.03 plus 0.175 multiplied by logarithm of 126000 divided by 5, plus 1.14 multiplied by natural logarithm of 4.46 plus 0.1, minus 0.000316 multiplied by 205, plus 0.000528 multiplied by 688, minus 0.000110 multiplied by 1140, plus 0.000112 multiplied by natural logarithm of 4.4 6, plus 0.1 multiplied by 1140, minus 0.000188 multiplied by 205 multiplied by natural logarithm of 4.46 plus 0.1.
Equation 37. Equation.
Natural logarithm of T C plus 0.01 equals 3.25.
Equation 38. Equation.
T C equals the exponential function E to the power of 3.25 minus 0.1, which equals 25.6.
Equation 39. Equation.
Natural logarithm of P divided by 1 minus P equals negative 4.01, plus 3.15 multiplied by logarithm E S A L divided by D E P T H, plus 0.0109 multiplied by F T C, plus 0.00139 multiplied by C I, minus 0.000569 multiplied by P R E C I P plus 0.0301 multiplied by A G E, minus 0.0000819 multiplied by A G E multiplied by C I, plus 0.0000345 multiplied by A G E multiplied by F I, plus 0.000103 multiplied by P R E C I P multiplied by A G E.
Equation 40. Equation.
Natural logarithm of L C plus 0.1 equals alpha subscript 1, plus 0.68 multiplied by logarithm E S A L divided by D E P T H, plus 0.00496 multiplied by F T C, plus 0.000644 multiplied by P R E C I P, plus 0.354 multiplied by natural logarithm of A D J underscore A G E, plus 0.01 minus 0.00000766 multiplied by F T C multiplied by P R E C I P.
Equation 41. Equation.
Natural logarithm of 0.55 divided by 1 minus 0.55 equals negative 4.01, plus 3.15 multiplied by logarithm 410000 divided by 9.5, plus 0.0109 multiplied by 80 plus 0.00139 multiplied by 205, minus 0.000569 multiplied by 1140, plus 0.0301 multiplied by A G E, minus 0.0000819 multiplied by A G E multiplied by 205, plus 0.0000345 multiplied by A G E multiplied by 688, plus 0.000103 multiplied by 1140 multiplied by A G E.
Equation 42. Equation.
A G E equals 11.86, which equals crack initiation age.
Equation 43. Equation.
Natural logarithm of L C plus 0.1 equals negative 1.89 plus 0.68 multiplied by logarithm 410000 divided by 9.5, plus 0.00496 multiplied by 80, plus 0.000644 multiplied by 1140, plus 0.354 multiplied by natural logarithm of 2.14 plus 0.01, minus 0.00000766 multiplied by 80 multiplied by 1140.
Equation 44. Equation.
Natural logarithm of L C plus 0.1 equals negative 0.78.
Equation 45. Equation.
L C equals the exponential function E to the power of negative 0.78 minus 0.1, which equals 0.358.
Equation 46. Equation.
Natural logarithm of P divided by 1 minus P equals negative 2.20, plus 3.06 multiplied by logarithm of E S A L divided by D E P T H, plus 0.00658 multiplied by F T C, plus 0.00125 multiplied by C I, minus 0.00207 multiplied by P R E C I P, plus 0.126 multiplied by A G E, minus 0.00016 multiplied by C I multiplied by A G E, minus 0.0000747 multiplied by F I multiplied by A G E, plus 0.000131 multiplied by P R E C I P multiplied by A G E.
Equation 47. Equation.
Natural logarithm of T C plus 0.1 equals alpha subscript 1, plus 0.0102 multiplied by F T C, minus 0.00120 multiplied by F I, plus 0.00223 multiplied by P R E C I P, plus 0.388 multiplied by natural logarithm of A D J underscore A G E plus 0.01 plus, 0.0000153 multiplied by F T C multiplied by F I, minus 0.0000156 multiplied by F T C multiplied by P R E C I P.
Equation 48. Equation.
Natural logarithm of 0.6 divided by 1 minus 0.6 equals negative 2.20, plus 3.06 multiplied by logarithm of 410000 divided by 9.5, plus 0.00658 multiplied by 80, plus 0.00125 multiplied by 205, minus 0.00207 multiplied by 1140, plus 0.126 multiplied by A G E, minus 0.00016 multiplied by 205 multiplied by A G E, minus 0.0000747 multiplied by 688 multiplied by A G E, plus 0.000131 multiplied by 1140 multiplied by A G E.
Equation 49. Equation.
A G E equals 11.86, which equals crack initiation age.
Equation 50. Equation.
Natural logarithm of T C plus 0.1 equals negative 2.28, plus 0.0102 multiplied by 80, minus 0.00120 multiplied by 688, plus 0.00223 multiplied by P R E C I P, plus 0.388 multiplied by natural logarithm of 1.43 plus 0.01, plus 0.0000153 multiplied by 80 multiplied by 688, minus 0.0000156 multiplied by 80 multiplied by 1140.
Equation 51. Equation.
Natural logarithm of T C plus 0.1 equals 0.02.
Equation 52. Equation.
T C equals the exponential function E to the power of 0.02 minus 0.1, which equals 0.92.
Equation 53. Equation.
Natural logarithm of R U T plus 0.1 equals alpha, plus 0.503 multiplied by natural logarithm of A G E, plus 0.1, plus 0.000337 multiplied by C I, plus 0.0000122 multiplied by P R E C I P, plus 0.00348 multiplied by F T C, plus 0.000000298 multiplied by F I multiplied by P R E C I P, minus 0.0844 multiplied by logarithm of E S A L divided by S N multiplied by natural logarithm of A G E plus 0.1, minus 0.000142 multiplied by natural logarithm of A G E, plus 0.1 multiplied by C I, minus 0.0000138 multiplied by natural logarithm of A G E plus 0.1 multiplied by F I, plus beta multiplied by logarithm E S A L divided by S N, plus lambda multiplied by F I.
Equation 54. Equation.
Natural logarithm of R U T plus 0.1 equals 0.367, plus 0.503 multiplied by natural logarithm of 2 plus 0.1, plus 0.000337 multiplied by 205, plus 0.0000122 multiplied by 1140, plus 0.00348 multiplied by 80, plus 0.000000298 multiplied by 688 multiplied by 1140, minus 0.0844 multiplied by logarithm of 126000 divided by 5 multiplied by natural logarithm of 2 plus 0.1, minus 0.000142 multiplied by natural logarithm of 2 plus 0.1, multiplied by 205, minus 0.0000138 multiplied by natural logarithm of 2 plus 0.1, multiplied by 688, plus 0.225 multiplied by logarithm 126000 divided by 5, minus 0.000266 multiplied by 688.
Equation 55. Equation.
Natural logarithm of R U T plus 0.1 equals 1.29.
Equation 56. Equation.
R U T equals the exponential function E to the power of 1.29, minus 0.1, which equals 3.53.
Equation 57. Equation.
Natural logarithm of F A U L T plus 1 equals alpha minus 0.337 multiplied by logarithm E S A L divided by D E P T H, minus 0.0014 multiplied by F T C, minus 0.000551 multiplied by F I, plus 0.00000315 multiplied by F T C multiplied by F I, minus 0.000515 multiplied by C I, minus 0.000792 multiplied by logarithm of E S A L divided by D E P T H, multiplied by C I, plus 0.000000928 multiplied by F I multiplied by C I, minus 0.000842 multiplied by P R E C I P, plus 0.000828 multiplied by logarithm of E S A L divided by D E P T H, multiplied by P R E C I P, plus 0.000000744 multiplied by C I multiplied by P R E C I P, minus 0.00668 multiplied by A G E, plus 0.0509 multiplied by logarithm of E S A L divided by D E P T H, multiplied by A G E, plus 0.00000978 multiplied by F I multiplied by A G E.
Equation 58. Equation.
Natural logarithm of F A U L T plus 1 equals 0.732, minus 0.337 multiplied by logarithm 410000 divided by 9.5, minus 0.0014 multiplied by 80, minus 0.000551 multiplied by 688, plus 0.00000315 multiplied by 80 multiplied by 688, minus 0.000515 multiplied by 205, minus 0.000792 multiplied by logarithm of 410000 divided by 9.5 multiplied by 205, plus 0.000000928 multiplied by 688 multiplied by 205, minus 0.000842 multiplied by 1140, plus 0.000828 multiplied by logarithm of 410000 divided by 9.5 multiplied by 1140, plus 0.000000744 multiplied by 205 multiplied by 1140, minus 0.00668 multiplied by 15, plus 0.0509 multiplied by logarithm of 410000 divided by 9.5 multiplied by 15, plus 0.00000978 multiplied by 688 multiplied by 15.
Equation 59. Equation.
Natural logarithm of F A U L T plus 1 equals 0.367.
Equation 60. Equation.
F A U L T equals the exponential function E to the power of 0.367, minus 1, which equals 0.443.
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