Speed correction

By default, the DEUFRABASE calculation considers personnel cars (PC) for two refernce speeds (90 and 110 km/h) and heavy trucks (HT) for one reference speed (80 km/h).

One can be interested to consider alternative speeds, for example in order to compare DEUFRABASE results with experimental data obtained with an other PC speed (see for example here).

Here, we present two corrections for the LA,eq,1h:

  • changing the PC speed vPC,ref to vPC,new, without changing the HT speed (80 km/h)
  • changing both the PC and HT speeds, vPC,ref to vPC,new and vHT,ref to vHT,new respectively

Changing the PC speed (HT at 80km/h)

Let's consider the LA,eq,1h,PC and LA,eq,1h,HT at the reference speeds, for the PC and the HT respectively. According to the relation that gives the equivalent sound level for a period of T=1 hour (see here), the sound level correction CPC for the PC will be:

C_\text{PC}=10\log\left( \frac{v_\text{PC,new}}{v_\text{PC,ref}}\right)

For a PC reference speed of 90 km/h and a new speed of 100 km/h, the correction will give for example 0.46 dBA.

Summing the contribution of the HT, the new whole equivalent sound level will be given by:

L_\text{A,eq,1h,new}=10\log\left\{ 10^{\left( L_\text{A,eq,1h,PC}+C_\text{PC}\right)/10} +10^{L_\text{A,eq,1h,HT}/10 } \right\}

that can be compared to the 'reference' LA,eq,1h value, without speed correction (i.e. with the reference speed):

L_\text{A,eq,1h,ref}=10\log\left\{ 10^{L_\text{A,eq,1h,PC}/10} +10^{L_\text{A,eq,1h,HT}/10 } \right\}

Manipulating the two last relations in order to remove the HT term, it gives:

L_\text{A,eq,1h,new}=10\log\left\{ 10^{L_\text{A,eq,1h,ref}/10} + \left( 10^{C_\text{PC}/10}-1 \right) \times 10^{L_\text{A,eq,1h,PC}/10}} \right\}

In practice, in order to calculate the 'new' LA,eq,1h (with the PC and HT contributions) with the new PC speed, it requires:

  • to calculate the LA,eq,1h,PC alone, at the reference speed, i.e. without considering the HT contribution,
  • to calculate the whole LA,eq,1h including both PC and HT contribution at the reference speeds.

Changing the both the PC and the HT speed 

Following the same approach, one must consider two speed corrections, one for the PC (CPC) and one for the HT (CHT). The new equivalent sound level will be given by:

L_\text{A,eq,1h,new}=10\log\left\{ 10^{L_\text{A,eq,1h,ref}/10} + \left( 10^{C_\text{PC}/10}-1 \right) \times 10^{L_\text{A,eq,1h,PC}/10}} + \left( 10^{C_\text{HT}/10}-1 \right) \times 10^{L_\text{A,eq,1h,HT}/10}}\right\}

In practice, in order to calculate the 'new' LA,eq,1h (with the PC and HT contributions) with the new PC and HT speeds, it requires:

  • to calculate the LA,eq,1h,PC alone, at the reference speed, i.e. without considering the HT contribution,
  • to calculate the LA,eq,1h,HT alone, at the reference speed, without considering the PC contribution (consider that the whole traffic is made of HT, % of HT = 100%),
  • to calculate the whole LA,eq,1h including both PC and HT contribution at the reference speeds.