A modeling approach for suitability evaluation of traffic noise prediction under mixed traffic situation in mid-sized Indian cities

Abstract

An efficient traffic noise prediction model is a decision-making tool for creating a peaceful environment. Despite the escalating concerns regarding traffic noise in India, limited research focuses on traffic noise prediction to address the unique road traffic scenarios prevalent in the country. Therefore, in the present study, suitability evaluation of the most extensively used prediction models, Federal Highway Administration (FHWA) and Calculation of Road Traffic Noise (CRTN), is carried out. Subsequently, a robust regression model tailored for accurate traffic noise prediction is formulated. However, to modify and enhance the applicability of the model, a novel reference energy mean emission level equation is developed for each vehicle category predominantly plying on Indian roadways. The FHWA approach yielded mean absolute error (MAE) and root mean square error (RMSE) values of 2.5 and 2.6, respectively. In contrast, the CRTN approach exhibited higher errors with MAE and RMSE values of 3.4 and 3.6, respectively. In addition, assessing all other performance metrics demonstrates the better applicability and efficiency of the modified FHWA model with optimum accuracy. The number of heavy vehicles, average traffic speed, and traffic flow were observed to be the vital factors influencing traffic noise generation. Furthermore, using the proposed model results, a regression model for different time durations that can be extensively used in mid-sized Indian cities is established. This research contributes significantly to the field, providing valuable insights for town planners, highway engineers, and authorities to formulate effective noise mitigation strategies in urban planning.

Appendix

Sample calculation of hourly equivalent noise level (L_eq) with hypothetical sampling location and traffic data.

Sampling Station: say XYZ

Time duration: say 6.00 PM–7.00 PM.

The linear regression model developed for the evening time period is as follows:

$${{\text{L}}}_{{\text{predicted}}}=12.177 \times {\mathrm{Log}}_{10}\sum {10}^{\frac{{\text{Leqi}}}{10}}-11.375$$

Sampling distance (D): 7.5 m

Reference distance (D₀): 10 m

Volume and speed correction = A_VS = 10 × Log₁₀(D₀ × V/S) – 25

Distance correction = A_D = 10 × Log₁₀ (D₀/D)^1+α

Ground cover correction = As = 0 (for hard surface)

Equivalent noise level of ith vehicle L_eqi = L₀ + A_VS + A_D + A_S

Firstly, calculating L_eqi for all the vehicle categories on the basis of assumed Traffic Volume (V) and Average Speed (S).

Car or Jeep:

Traffic Volume (V) of Car or Jeep = say 334

Average Speed of Car or Jeep (S) = say 38 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 29.586 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 21.689 \\ & = \, 29.586 \, \times {\text{ Log}}_{10}\left( {38} \right) \, + \, 21.689 \\ & = \, 68.42 \\ \;\;\;\;{{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right)-- \, 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, x \, 334/38} \right) \, -- \, 25 \\ & = \, - 5.56 \\ \;\;\;\;\;{{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ \;\;\;\;\;\;\;\;\;\;\; & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ \;\;\;\;\;\;{\text{As }} & = \, 0 \\ \end{aligned}$$

Equivalent noise level for Car or Jeep:

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = {{\text{L}}_0} + {{\text{A}}_{{\text{VS}}}} + {{\text{A}}_{\text{D}}} + {{\text{A}}_{\text{S}}} \\ & = \, 68.42 \, + \, \left( { - 5.56} \right) \, + \, 1.24 + \, 0 \\ & = \, 64.1 \\ \end{aligned}$$

Scooter or Bike (Petrol Driven):

Traffic Volume (V) of Scooter or Bike (Petrol Driven) = say 1021

Average Speed of Scooter or Bike (Petrol Driven) (S) = say 40 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 31.298 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 16.819 \\ & = \, 31.298 \times {\text{ Log}}_{10}\left( {40} \right) \, + \, 16.819 \\ & = \, 66.96 \\ {{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) - 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, x \, 1021/40} \right) - 25 \\ & = \, - 0.93 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As }} & = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for Scooter or Bike (Petrol Driven):

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = \, {{\text{L}}_0} + \, {{\text{A}}_{{\text{VS}}}} + \, {{\text{A}}_{\text{D}}} + \, {{\text{A}}_{\text{S}}} \\ & = \, 66.96 + \, \left( { - 0.93} \right) \, + \, 1.24 + \, 0 \\ & = \, 67.2 \\ \end{aligned}$$

LCV (Light Commercial Vehicle):

Traffic Volume (V) of LCV (Light Commercial Vehicle) = say 57

Average Speed of LCV (Light Commercial Vehicle) (S) = say 25 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 36.354 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 20.490 \\ & = \, 36.354 \, \times {\text{ Log}}_{10}\left( {25} \right) \, + \, 20.490 \\ & = \, 71.31 \\ {{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) - 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 57/25} \right) - 25 \\ & = \, - 11.42 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As}} & \, = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for LCV (Light Commercial Vehicle):

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = {{\text{L}}_0} + {{\text{A}}_{{\text{VS}}}} + \, {{\text{A}}_{\text{D}}} + \, {{\text{A}}_{\text{S}}} \\ & = \, 71.31 \, + \, \left( { - 11.42} \right) \, + \, 1.24 \, + \, 0 \\ & = \, 61.1 \\ \end{aligned}$$

Bus:

Traffic Volume (V) of Bus = say 70

Average Speed of Bus (S) = say 22 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 37.623 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 19.479 \\ & = \, 37.623 \, \times {\text{ Log}}_{10}\left( {22} \right) \, + \, 19.479 \\ & = \, 69.98 \\ {{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) - 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 70/22} \right) - 25 \\ & = \, - 9.97 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As}} & \, = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for Bus:

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = \, {{\text{L}}_0} + \, {{\text{A}}_{{\text{VS}}}} + \, {{\text{A}}_{\text{D}}} + \, {{\text{A}}_{\text{S}}} \\ & = \, 69.98 \, + \, \left( { - 9.97} \right) \, + \, 1.24 + \, 0 \\ & = \, 61.2 \\ \end{aligned}$$

Truck:

Traffic Volume (V) of Truck = say 121

Average Speed of Truck (S) = say 22 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 39.591 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 18.132 \\ & = \, 39.591 \, \times {\text{ Log}}_{10}\left( {22} \right) \, + \, 18.132 \\ & = \, 71.27 \\ {{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) \, -- \, 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 121/22} \right) \, -- \, 25 \\ & = \, - 7.59 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As}} & \, = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for Truck:

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = \, {{\text{L}}_0} + \, {{\text{A}}_{{\text{VS}}}} + \, {{\text{A}}_{\text{D}}} + \, {{\text{A}}_{\text{S}}} \\ & = \, 71.27 \, + \, \left( { - 7.59} \right) \, + \, 1.24 \, + \, 0 \\ & = \, 64.9 \\ \end{aligned}$$

3Wheeler or Auto Rickshaw (Diesel Driven):

Traffic Volume (V) of 3Wheeler or Auto Rickshaw (Diesel Driven) = say 389

Average Speed of 3Wheeler or Auto Rickshaw (Diesel Driven) (S) = say 28 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 3.234 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 61.510 \\ & = \, 3.234 \, \times {\text{ Log}}_{10}\left( {28} \right) \, + \, 61.510 \\ & = \, 66.19 \\ {{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) \, -- \, 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 389/28} \right) \, -- \, 25 \\ & = \, - 3.57 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As}} & \, = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for 3Wheeler or Auto Rickshaw (Diesel Driven):

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = {{\text{L}}_0} + \, {{\text{A}}_{{\text{VS}}}} + \, {{\text{A}}_{\text{D}}} + \, {{\text{A}}_{\text{S}}} \\ & = \, 66.19 \, + \, \left( { - 3.57} \right) \, + \, 1.24 \, + \, 0 \\ & \; = \, 63.8 \\ \end{aligned}$$

Tractor or Trailor:

Traffic Volume (V) of Tractor or Trailor = say 26

Average Speed of Tractor or Trailor (S) = say 20 km/h

$$\begin{aligned} {{\text{L}}_0} & = 12.285 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 63.152 \\ & = 12.285 \, \times {\text{ Log}}_{10}\left( {20} \right) \, + \, 63.152 \\ & = \, 79.13 \\ \;{{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) - 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 26/20} \right) - 25 \\ & = \, - 13.86 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As}} & \, = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for Tractor or Trailor:

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = {{\text{L}}_0} + {{\text{A}}_{{\text{VS}}}} + {{\text{A}}_{\text{D}}} + {{\text{A}}_{\text{S}}} \\ & = \, 79.13 \, + \, \left( { - 13.86} \right) \, + \, 1.24 \, + \, 0 \\ & = \, 66.5 \\ \end{aligned}$$

E-Rickshaw:

Traffic Volume (V) of E-Rickshaw = say 90

Average Speed of E-Rickshaw (S) = say 15 km/h

$$\begin{aligned} {{\text{L}}_0} & = \, 19.7311 \, \times {\text{ Log}}_{10}\left( {\text{S}} \right) \, + \, 42.918 \\ & = \, 19.7311 \, \times {\text{ Log}}_{10}\left( {15} \right) \, + \, 42.918 \\ & = \, 66.12 \\ {{\text{A}}_{{\text{VS}}}} & = \, 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) - 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 90/15} \right) - 25 \\ & = \, - 7.21 \\ {{\text{A}}_{\text{D}}} & = \, 10 \, \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = \, 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ & = \, 1.24 \\ {\text{As}} & \, = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for E-Rickshaw:

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = \, {{\text{L}}_0} + \, {{\text{A}}_{{\text{VS}}}} + \, {{\text{A}}_{\text{D}}} + \, {{\text{A}}_{\text{S}}} \\ & = \, 66.12 \, + \, \left( { - 7.21} \right) \, + \, 1.24 \, + \, 0 \\ & = \, 60.1 \\ \end{aligned}$$

E-Bike:

Traffic Volume (V) of E-Bike = say 21

Average Speed of E-Bike (S) = say 25 km/h

$$\begin{aligned} {{\text{L}}_0} & = 21.1210 \times {\text{Log}}_{10}\left( {\text{S}} \right) + 32.598 \\ & = 21.1210 \times {\text{Lo}}{{\text{g}}_{10}}\left( {25} \right) + 32.598 \\ & = 62.12 \\ {{\text{A}}_{{\text{VS}}}} & = 10 \, \times {\text{ Log}}_{10}\left( {{{\text{D}}_0} \times {\text{ V/S}}} \right) - 25 \\ & = 10 \, \times {\text{ Log}}_{10}\left( {10 \, \times \, 21/25} \right) - 25 \\ & = - 15.75 \\ {{\text{A}}_{\text{D}}} & = 10 \times {\text{ Log}}_{10}{\left( {{{\text{D}}_0}/{\text{D}}} \right)^{1 + \alpha }} \\ & = 10 \, \times {\text{ Log}}_{10}{\left( {10/7.5} \right)^{1 + 0}} \\ \; & = \, 1.24 \\ {\text{As }} & = \, 0 \\ \end{aligned}$$

Hence, equivalent noise level for E-Bike:

$$\begin{aligned} {{\text{L}}_{{\text{eqi}}}} & = {{\text{L}}_0} + {{\text{A}}_{{\text{VS}}}} + {{\text{A}}_{\text{D}}} + {{\text{A}}_{\text{S}}} \\ & = \, 62.12 \, + \, \left( { - 15.75} \right) \, + \, 1.24 \, + \, 0 \\ & = \, 47.6 \\ \end{aligned}$$

Hence, the predicted noise level for evening time will be calculated using following equation.

$${{\text{L}}}_{{\text{predicted}}}=12.177 \times {\mathrm{Log}}_{10}\sum {10}^{\frac{{\text{Leqi}}}{10}}-11.375$$

$$=12.177 \times {\mathrm{Log}}_{10}\sum {10}^{\frac{64.1}{10}}{+10}^{\frac{67.2}{10}}{+10}^{\frac{61.1}{10}}{+10}^{\frac{61.2}{10}}{+10}^{\frac{64.9}{10}}{+10}^{\frac{63.8}{10}}{+10}^{\frac{66.5}{10}}{+10}^{\frac{60.1}{10}}{+10}^{\frac{47.6}{10}}-11.375$$

 = 77.9 dBA.