Construction & maintenance noise
NZS 6803 provides calculation methods for noise from construction activity making reference to BS 5228. This page describes some of the key elements of the calculations.
Reference data for construction equipment and processes are given in BS 5228 as sound pressure levels at 10 m. Data quoted in other places are sometimes given in an alternative format as a sound power level, often when associated with computer models. Sound power is a physical characteristic of a noise source, which results in different sound pressure levels at different locations. The relationship between the two quantities is shown below:
| Sound power level | Propagation | Sound pressure level at 10 m | |
|---|---|---|---|
| (Quantity used in computer models) | (Reference distance 10 m) | (Quantity given in NZS 6803/BS 5228) | |
| Equation | Lw = | 20×log10(distance) + 8 | + Lp |
| Example | Lw = | + 20×log10(10m) + 8 | + 81 |
| Result | 109 = | + 20 + 8 |
+ 81 |
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As can be seen above the sound power level and the sound pressure level at 10 m are related by a simple equation (28 dB difference). It doesn’t matter which quantity is used providing that subsequent calculations use the corresponding version of the equations.
All calculations should be A-weighted data. In older documents this is shown as 'dBA'.
NZS 6803 requires assessment of sound levels over a representative period between fifteen minutes and one hour to account for the variable nature of construction sound. If any specific item of equipment is not operating for the full period then the level predicted for that item of equipment is reduced. This is shown in the example below where one hour is selected as a representative period and a bulldozer is operating for only 45 minutes in the hour.
| Equipment operating | Percentage operating time | Correction | Corrected level | |
|---|---|---|---|---|
| Equation | Lp | P | - 10×log10(P/100) | |
| Example | 70 dB | 75% | - 10×log10(75/100) | |
| Result | 70 dB | 75% | - 1.2 dB | = 68.8 dB |
The calculation models each item of equipment as a ‘point’ source and the sound levels decrease 6 dB each time the distance from the source doubles (20×log10(distance)).
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For source data in terms of the sound pressure level at 10 m, the façade level at the nearest house is calculated as follows:
| Sound pressure level @ 10 m | Propagation | Façade reflection | Façade level | |
|---|---|---|---|---|
| Equation | + Lp @ 10 m | - 20×log10(distance/10) | + 3 | = Lp @ Façade |
| Example | + 81 dB | - 20×log10(50/10) | + 3 | = Lp @ Façade |
| Result | + 81 dB | - 20×0.7 | + 3 | = 70 dB |
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Noise surveys are peformed at a location 1m from the facade of a buidling. Consequently, sound is reflected off the facade, and results in an increased level being measured. A 3dB facade correction is required to be included in calculations to allow for this effect.
Where there are multiple items of the same equipment the following equation is used to calculate the cumulative sound level:
| Lp@façade,total = | Lp@façade,source,1 + 10×log10(number of sources) |
Sound from multiple items of different equipment is combined using the following equation:
| Lp@façade,total = | 10×log10(100.1×Lp@façade,source,1 + 100.1×Lp@façade,source,2 +… |
| + 100.1×Lp@façade,source,N) |
Where there is a moving source the BS 5228 data relates to the equipment at the nearest (loudest) point rather than being an average value. For slow moving sources in a constrained area, BS 5228 makes allowance for the times when the equipment is further away from the houses and therefore quieter. This is done by effectively reducing the operating time.
The effective reduction in the operating time is related to the ratio of the distance to the house and the distance the equipment traverses. BS 5228 gives correction factors for various ratios, and the following equation approximately fits that data. This should be capped at a value of 1 so that the effective operating time cannot exceed 100%.
|
Distance traversed by |
Minimum distance to house |
Correction factor | |
|---|---|---|---|
| Equation | ltr | dmin | +0.07+0.6(ltr/dmin) |
| Example | 100m | 50m | +0.07+0.6(100/50) = 0.43 |
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The correction factor is multiplied by the actual percentage operating time to obtain the effective percentage operating time. The predicted level is then adjusted in the same manner as described above for the percentage operating time.




