1. Lighting and Acoustics Arch 353 Dr. Abdelrahman Elbakheit References اسس الانارة المعمارية, أستاذ الدكتور محمد عبد الفتاح عبيد.1 2. Architectural Lighting, M. David Egan and Victor Olgyay.

2. Loudness: Loudness have a subjective component and depends on the ear of the person and his mind/age. But also depends on the amplitude and frequency of the sound. Loudness can be described as the impression of sound strength, which changes with the change of sound waves frequency and amplitude. This means we can have two sounds with the same Loudness level but with different frequencies and amplitudes. The unit for loudness is called the Phon, which is the sound at a frequency of 1000Hrz by Decibels above 2x10 -5 N/m 2.(the threshold of hearing). Threshold levels: Here we are trying to establish the base line of sound pressure, sound intensity and sound power. So we can compare any other new sounds with these base lines. The unit of comparison is called the DECIBEL (db), which is a logarithmic scale to measure sound pressure, intensity and power, compared to the base levels of each one of them.

3. Decibels Scale: For Sound Power levels (SPWL): SPWL = 10 log W/W 0 W = Sound Power in Watts W 0 = base level of sound power = 10 -12 Watts. For Sound Intensity Level (SIL): SIL = 10log I/I 0 I = Sound intensity in W/m 2. I 0 = Base level of sound intensity.= 10 -12 Watts/m 2 For Sound Pressure Levels (SPL): SPL =10 log P/P 0 P =Sound Pressure in N/m 2 P 0 = Base level of Sound intensity = 2x10 -5 N/m 2

4. Relationship between Sound Power and Sound intensity: Sound Power PWL = 10 Log W/W 0 = 10 Log W/10 -12 = 10 Log W -10Log10 -12 = 10 Log W +120 Sound Intensity level SIL = 10 Log W/W 0 x4πR 2 =10 Log W- 10 Log10 -12 - 10 Log 4πR 2 =10 Log W -10 Log 4πR 2 +120 =(10 Log W + 120) - 10 Log 4πR 2 =PWL -20 LogR – 10Log4π SIL = PWL -20 Log R – 11 ( for Sound Absorbing floor finish) SIL = PWL -20 Log R – 8 ( for sound reflecting floor finish

5. Adding Sound Level Mathematically: The decibel values of sound is a Logarithmic scale and therefore it is not added arithmetically but in the logarithmic manner. Examples: 1.What is the increase in sound pressure if we added two equal sounds together? Increase in SPL = 10Log (2P 2 /P 0 2 ) – 10Log (P 2 / P 2 0 ) = 10Log 2 + 10Log (P 2 / P 2 0 ) – 10Log (P 2 / P 2 0 ) = 10Log 2 = 10x 0.3 = 3 db increase in SIL = 10Log (2I/I 0 ) – 10Log (I/I 0 ) = 10Log 2 + 10Log (I/I 0 ) – 10Log (I/I 0 ) = 10Log 2 = 3db

6. Adding Sound Level Mathematically Examples: 2.What is the increase in sound intensity if we added three equal sounds together? increase in SIL = 10Log (3I/I 0 ) – 10Log (I/I 0 ) = 10Log 3 + 10Log (I/I 0 ) – 10Log (I/I 0 ) = 10Log 3 = 10 x 0.4771 = 5 db 3. Two sound sources one of 77db and the other 80db they occurred simultaneously what would be the intensity of them together? 77 = 10Log I/I 0 = anilog 7.7 = 5.0119 x10 7 80 = 10Log I/I 0 = anilog 8 = 1x10 8 Therefore SIL = 10 Log ( I 1 + I 2 )/ I 0 = 10 Log ( 5.0119 x10 7 + 1x10 8 ) / 10 -12 ) = 81.76 db

7. Examples: 4. If we have 10 equal sound producing 80db what would be the intensity of each? 80 = 10Log 10 I/I 0 10 I/I 0 = anilog8 I/I 0 = anilog8 / 10 = 1x10 8 / 10 = 10 7 10Log I/I 0 = 10 Log 10 7 = 70 db Sound levels summation calculator: -To use the calculator first determine the difference between the sound levels. -By the resulting difference determine the required increase from the calculator. -Add the increase to the highest sound level. Example: 5. The components of some noise were defined as in the following table, what will be the total sound level for all sound?

8. Example: 5. The components of some noise were defined as in the following table, what will be the total sound level for all sound? Frequency (Hz) 125250500100020004000 SPL(db) 8082.577.5706560 Frequency (Hz) SPL (db) Diff (db) Sum (db) 12580 25082.52.51.984.4 50077.5 1000707.5.7578.256.1585.32 200065 40006051.266.20.0685.38