NOISE REDUCTION IN SPACESNOISE REDUCTION IN SPACESBYBY

THE ABSORPTION OF SOUND ENERGYTHE ABSORPTION OF SOUND ENERGY

ABSORPTION OF SOUNDABSORPTION OF SOUND

A WHEN PHYSICAL SOUND ENERGY STRIKES A WHEN PHYSICAL SOUND ENERGY STRIKES A SURFACEA SURFACE

1 The sound bounces to another direction if the 1 The sound bounces to another direction if the surface is reflective.surface is reflective.

2 The energy will cease if the material is absorptive.2 The energy will cease if the material is absorptive.

3 A combination of 1 & 2, depending upon the 3 A combination of 1 & 2, depending upon the reflective / absorptive nature of the surface.reflective / absorptive nature of the surface.

4 A portion of the energy will travel through the 4 A portion of the energy will travel through the surface, depending upon the mass and elasticity of surface, depending upon the mass and elasticity of the material. the material.

5 The energy will scatter if the surface is refractive. 5 The energy will scatter if the surface is refractive. B REFLECTION OF SOUND ENERGYB REFLECTION OF SOUND ENERGY

1 Energy that reflects from a smooth surface 1 Energy that reflects from a smooth surface bounces away at the same angle with which it struck bounces away at the same angle with which it struck the surface, opposite the direction it came.the surface, opposite the direction it came.

2 Angles of incidence and reflection:2 Angles of incidence and reflection:

incidence reflectionincidence reflection

3 If sound energy reflects from one surface, moves 3 If sound energy reflects from one surface, moves to another and reflects, then moves to another and to another and reflects, then moves to another and reflects, repeating until it decays, a garble of reflects, repeating until it decays, a garble of audibility is created within a space because of time audibility is created within a space because of time lag, called REVERBERATION. lag, called REVERBERATION.

4 Since sound travels at a constant speed through a 4 Since sound travels at a constant speed through a given medium, like air, a time variance occurs given medium, like air, a time variance occurs between the original sound-to-recipient, and the between the original sound-to-recipient, and the collective reflected sounds-to-recipient. The collective reflected sounds-to-recipient. The recipient hears the sound numerous times at different recipient hears the sound numerous times at different periods in the form of an echo, or garble, and the periods in the form of an echo, or garble, and the sound may not be intelligible. sound may not be intelligible.

5 If a recipient hears the same sound several times 5 If a recipient hears the same sound several times because of reverberation, and the travel distance because of reverberation, and the travel distance from direct path is different to reflected path, an echo from direct path is different to reflected path, an echo will occur if the path difference is in excess of 68 feet. will occur if the path difference is in excess of 68 feet. Or if the path difference is as much as 50 feet, the Or if the path difference is as much as 50 feet, the sounds may not be intelligiblesounds may not be intelligible..

6 6 Reverberation time is the number of seconds Reverberation time is the number of seconds required for a sound to decay. required for a sound to decay.

7 In a situation involving a human voice, since 7 In a situation involving a human voice, since intelligibility of words is constructed upon changes intelligibility of words is constructed upon changes of pitch and loudness of syllables that form words, of pitch and loudness of syllables that form words, if if the reverberation time within a room is more than 1 ½ the reverberation time within a room is more than 1 ½ secondsseconds, the echoed syllables will mix, and the , the echoed syllables will mix, and the speech will not be intelligible, particularly if the speech will not be intelligible, particularly if the speech is rapid or the overall pitch is low. speech is rapid or the overall pitch is low.

SOUND RAY DIAGRAMMING is a graphic process SOUND RAY DIAGRAMMING is a graphic process of indicating the paths of sound from a point of indicating the paths of sound from a point source by constructing straight lines that source by constructing straight lines that indicate the travel path until it contacts a surface indicate the travel path until it contacts a surface and reflects and reflects

Convex irregular shapes condense the effective Convex irregular shapes condense the effective ceiling area to help reflect sound energy. The ceiling area to help reflect sound energy. The specific shape is designed to cover a specific specific shape is designed to cover a specific area.area.

A uniform convex ceiling area will reinforce A uniform convex ceiling area will reinforce sound projection with a uniform pattern of sound projection with a uniform pattern of reflected energy.reflected energy.

REFLECTION – continuedREFLECTION – continued8 The reflection of sound is useful to reinforce 8 The reflection of sound is useful to reinforce sounds within a space, utilizing elements of the sounds within a space, utilizing elements of the surrounding envelope when effective without surrounding envelope when effective without creating echoes. creating echoes.

9 The surfaces under consideration where reflection 9 The surfaces under consideration where reflection is useful should be made of reflective materials, and is useful should be made of reflective materials, and arranged in such way as to direct the reflected sound arranged in such way as to direct the reflected sound toward the audience. toward the audience.

10 Reflected sounds that bounces from one surface 10 Reflected sounds that bounces from one surface to another, then to another, and another before to another, then to another, and another before reaching the audience become a nuisance in causing reaching the audience become a nuisance in causing reverberation.reverberation.

11 Sound energy that goes directly to the audience, 11 Sound energy that goes directly to the audience, or is reflected off a surface to the audience will be or is reflected off a surface to the audience will be absorbed, and will not bounce. absorbed, and will not bounce.

C ABSORPTIONC ABSORPTION1 The reverberation of sound is important to some 1 The reverberation of sound is important to some types of facilities, such as spaces where an added types of facilities, such as spaces where an added ‘depth’ of sound or ring is desirable for dramatic ‘depth’ of sound or ring is desirable for dramatic affect. affect.

2 Where it is desirable to limit reverberation for the 2 Where it is desirable to limit reverberation for the sake of audible clarity, such as the speaking voice, sake of audible clarity, such as the speaking voice, surfaces that do not contribute to reinforcement for surfaces that do not contribute to reinforcement for the benefit of the audience should be absorptive or the benefit of the audience should be absorptive or refractive.refractive.

3 Sound energy that strikes an absorptive surface 3 Sound energy that strikes an absorptive surface will die there and not bounce to cause reverberation. will die there and not bounce to cause reverberation.

4 Spaces that have a variance in the number of 4 Spaces that have a variance in the number of people in an audience, present a potential problem people in an audience, present a potential problem with reverberation, since people, because of the with reverberation, since people, because of the softness of the body, clothing, etc. are absorptive.softness of the body, clothing, etc. are absorptive.

5 If all the surfaces in a room are reflective and the 5 If all the surfaces in a room are reflective and the room is filled with people, the reflective surfaces are room is filled with people, the reflective surfaces are limited to walls & ceiling, because audiences are limited to walls & ceiling, because audiences are absorptive, and audiences cover a floor.absorptive, and audiences cover a floor.

6 If in the same room, only half the people are 6 If in the same room, only half the people are present, an added floor area is available to cause present, an added floor area is available to cause reverberation. But in rooms with fixed seats, such reverberation. But in rooms with fixed seats, such as auditoriums and churches, it is important for the as auditoriums and churches, it is important for the seatbacks and cushions to be of absorptive material seatbacks and cushions to be of absorptive material so that a partially filled room will have the same so that a partially filled room will have the same absorptive area as a room filled with people. absorptive area as a room filled with people.

7 Floors beneath seating areas that are reflective do 7 Floors beneath seating areas that are reflective do not contribute an appreciative amount of sound not contribute an appreciative amount of sound reflection within the room if the empty seats are reflection within the room if the empty seats are absorptive, but will contribute to a slight absorptive, but will contribute to a slight reverberation and give depth to sound. reverberation and give depth to sound.

D NOISE REDUCTION DUE TO ABSORPTION OF D NOISE REDUCTION DUE TO ABSORPTION OF SOUNDSOUND

1 Noise in the form of unwanted sound caused by 1 Noise in the form of unwanted sound caused by reverberation can be reduced by installing surface reverberation can be reduced by installing surface materials that stop the reflection of energy.materials that stop the reflection of energy.

2 All construction materials used on finish surfaces 2 All construction materials used on finish surfaces are laboratory tested for their ability to absorb sound are laboratory tested for their ability to absorb sound energy. An absorption coefficient is a fractional energy. An absorption coefficient is a fractional number that represents the percentage of sound number that represents the percentage of sound energy that will be absorbed by a material.energy that will be absorbed by a material.

3 Absorption coefficients are determined for a 3 Absorption coefficients are determined for a material at various sound frequencies, ranging from material at various sound frequencies, ranging from low sounds at 125 hertz to high sounds at 4000 hertz. low sounds at 125 hertz to high sounds at 4000 hertz.

4 Consider a constant sound source such as a horn. 4 Consider a constant sound source such as a horn. Say the direct sound loudness of the horn is 30 Say the direct sound loudness of the horn is 30 decibels at a point 10 feet from the horn. Then place decibels at a point 10 feet from the horn. Then place the horn within a room, say 30 feet square, that has the horn within a room, say 30 feet square, that has hard, sound reflecting surfaces. Sound the horn and hard, sound reflecting surfaces. Sound the horn and measure the loudness at the same 10’ distance.measure the loudness at the same 10’ distance.

5 The level of sound will be higher than the initial 30 5 The level of sound will be higher than the initial 30 decibels because the measuring device records the decibels because the measuring device records the direct sound PLUS the reflected sound from walls, direct sound PLUS the reflected sound from walls, floor, ceiling. floor, ceiling.

6 Any room has a characteristic called 6 Any room has a characteristic called TOTAL ROOM TOTAL ROOM ABSORPTIONABSORPTION which is a function of the summation which is a function of the summation of the absorption values of its surfaces. Surface of the absorption values of its surfaces. Surface absorption is measured in units called SABINS, and absorption is measured in units called SABINS, and equals area x the absorption coefficient. TOTAL equals area x the absorption coefficient. TOTAL ROOM ABSORPTION equals the summation of the ROOM ABSORPTION equals the summation of the surface absorptions: surface absorptions:

SURFACE ABSORPTIONSURFACE ABSORPTIONAbsorption = surface area x coefficient of Absorption = surface area x coefficient of

absorption, orabsorption, or a = S x alphaa = S x alpha, , wherewhere

a = sound absorption in units of Sabinsa = sound absorption in units of SabinsS = area of surface in square feetS = area of surface in square feetalphaalpha = coefficient of absorption = coefficient of absorption

TOTAL ROOM ABSORPTIONTOTAL ROOM ABSORPTION

A = A = ∑ S x ∑ S x alphaalpha, where, where A = total room absorption in SabinsA = total room absorption in Sabins ∑ ∑ S x S x alpha alpha = the sum of all the surface = the sum of all the surface

absorptions absorptions that make up the enclosure of the that make up the enclosure of the room.room.

ROOM NOISE REDUCTIONROOM NOISE REDUCTION, , NR = 10 log [ aNR = 10 log [ a22 / a / a11 ] ]

wherewhere aa11 = absorption in Sabins BEFORE treatment is done = absorption in Sabins BEFORE treatment is done aa22 = absorption in Sabins AFTER treatment is done = absorption in Sabins AFTER treatment is done

NOISE REDUCTION by the absorption of sound NOISE REDUCTION by the absorption of sound within the room is merely reducing the amount of within the room is merely reducing the amount of sound that bounces from its source to adjacent floor, sound that bounces from its source to adjacent floor, wall, ceilings, and any object within the room that wall, ceilings, and any object within the room that would reflect sound.would reflect sound.

It is simply a process to make the room quieter.It is simply a process to make the room quieter.Recollect the times you have been to a Recollect the times you have been to a

restaurant, and the noise inside was practically restaurant, and the noise inside was practically unbearable. Some examples of this in Lubbock are unbearable. Some examples of this in Lubbock are Abuelos, Garcias, The Crab Shack, and perhaps Abuelos, Garcias, The Crab Shack, and perhaps numerous others I have not visited where the floors, numerous others I have not visited where the floors, walls, and ceilings are hard and sound reflective.walls, and ceilings are hard and sound reflective.

A situation called the “cocktail hour affect” A situation called the “cocktail hour affect” happens. One group of people visiting cannot hear happens. One group of people visiting cannot hear their conversation, so they speak louder; another their conversation, so they speak louder; another group does the same thing; and another group; and group does the same thing; and another group; and so on, until a din of noise occurs within the space.so on, until a din of noise occurs within the space.

It is sound that reflects from the interior It is sound that reflects from the interior surfaces. In order to reduce the noise, treat the surfaces. In order to reduce the noise, treat the exposed surfaces so they will ABSORB sound rather exposed surfaces so they will ABSORB sound rather than reflect.than reflect.

But realize that no more than about 11 to 12 But realize that no more than about 11 to 12 decibel sound reduction can be realized, since no decibel sound reduction can be realized, since no amount of absorption will reduce airborne sound. amount of absorption will reduce airborne sound.

Consider a room 30’ x 30’ x 10’ in size. One wall is Consider a room 30’ x 30’ x 10’ in size. One wall is glass to the exterior, floor to ceiling. Assume the glass to the exterior, floor to ceiling. Assume the doors into the room have the same absorptive doors into the room have the same absorptive coefficient as the walls.coefficient as the walls.

The absorptive qualities of the room are thus:The absorptive qualities of the room are thus: The absorption coefficients, The absorption coefficients, alphaalpha,, are as follows: (at are as follows: (at

500 Hertz)500 Hertz) unfinished concrete floorunfinished concrete floor = .02= .02 unfinished gyp. bd. wallsunfinished gyp. bd. walls = .05= .05 unfinished gyp. bd. ceiling unfinished gyp. bd. ceiling = .05= .05 glass wallglass wall = .18= .18

Surface absorptions are as follows:Surface absorptions are as follows: floor = 30 x 30 x .02 floor = 30 x 30 x .02 = 18.00 sabins= 18.00 sabins ceiling = 30 x 30 x .05 ceiling = 30 x 30 x .05 = 45.00 sabins= 45.00 sabins walls = 3 (30 x 10) x .05 walls = 3 (30 x 10) x .05 = 45.00 sabins= 45.00 sabins glass = 30 x 10 x .18 glass = 30 x 10 x .18 = 54.00 sabins= 54.00 sabins

TOTAL ROOM ABSORPTION TOTAL ROOM ABSORPTION = 162.00 sabins= 162.00 sabins

Consider that in the preceding room, reverberation Consider that in the preceding room, reverberation would probably be a problem, since all the surfaces would probably be a problem, since all the surfaces are reflective. However, since the room is are reflective. However, since the room is relatively small, the distance involved for echoes relatively small, the distance involved for echoes may not be severe. may not be severe.

REVERBERATION TIMEREVERBERATION TIME, the number of seconds it , the number of seconds it takes for a sound to decay, is a function of the takes for a sound to decay, is a function of the volume of the room and the area of the room:volume of the room and the area of the room:

T = .05 x [ V / A ] where .05 is a constantT = .05 x [ V / A ] where .05 is a constant

V = volume of room in cu.ft.V = volume of room in cu.ft.

A = room absorption in sabinsA = room absorption in sabins

Reverberation Time in the example room Reverberation Time in the example room T = .05 x [ (30x30x10) / 162 ] = .05 x [ 9000 / T = .05 x [ (30x30x10) / 162 ] = .05 x [ 9000 / 162 ]162 ]

T = .05 x 55.55 = T = .05 x 55.55 = 2.78 seconds2.78 seconds the the room would have a definite ring.room would have a definite ring.

Consider the room as a classroom, and Consider the room as a classroom, and carpet is carpet is added to the flooradded to the floor – which may have the affect of the – which may have the affect of the absorption rate of a classroom full of people . . .absorption rate of a classroom full of people . . .

Absorption coefficient, Absorption coefficient, alphaalpha,, of carpet = .14 of carpet = .14Absorption = 30 x 30 x .14 = 126Absorption = 30 x 30 x .14 = 126 If the walls, ceiling, & glass remain the same, thenIf the walls, ceiling, & glass remain the same, then

New ROOM ABSORPTIONNew ROOM ABSORPTION = 126 + 45 + 45 + 54 = = 126 + 45 + 45 + 54 = 270 sabins270 sabins

NOISE REDUCTIONNOISE REDUCTION = 10 log [ 270 / 162 ] = 10 log = 10 log [ 270 / 162 ] = 10 log 1.671.67

= 10 x .22 = = 10 x .22 = 2.2 decibels2.2 decibels

And And REVERBERATION TIMEREVERBERATION TIME = .05 x (9000 / 270) = .05 = .05 x (9000 / 270) = .05 x 33.3x 33.3

= = 1.67 seconds1.67 seconds. .

Noise is reduced slightly, & reverberation time still Noise is reduced slightly, & reverberation time still highhigh

NEXT,NEXT, add add acoustic tile to the ceiling, acoustic tile to the ceiling, which in most which in most cases where speech is important, is a no-no, as the cases where speech is important, is a no-no, as the ceiling is a handy place to reinforce soundceiling is a handy place to reinforce sound for an for an audience - but this example is to demonstrate audience - but this example is to demonstrate NOISE NOISE REDUCTION.REDUCTION.

Absorption coefficient, Absorption coefficient, alphaalpha,, for ceiling tile = .80 for ceiling tile = .80 Ceiling surface absorption = 30 x 30 x .8 = 720 Ceiling surface absorption = 30 x 30 x .8 = 720 sabinssabins

New ROOM ABSORPTIONNew ROOM ABSORPTION = 126 + 720 + 45 + 54 = = 126 + 720 + 45 + 54 = 945 945 sabinssabins..

NOISE REDUCTIONNOISE REDUCTION = 10 log [ 945 / 162 ] = 10 log 5.83 = 10 log [ 945 / 162 ] = 10 log 5.83

= 10 x .77 = = 10 x .77 = 7.7 decibels7.7 decibels

And And REVERBERATION TIMEREVERBERATION TIME = .05 X (9000 / 945) = .05 x = .05 X (9000 / 945) = .05 x 9.529.52 = = .47 seconds.47 seconds

Noise Reduction and Reverberation time is reduced Noise Reduction and Reverberation time is reduced significantlysignificantly

NEXT, NEXT, add plywood paneling to the wallsadd plywood paneling to the walls. The side . The side walls are normally good for reinforcing sound to an walls are normally good for reinforcing sound to an audience. The wall opposite the speaker should be audience. The wall opposite the speaker should be absorbent or refractive (broken surface to scatter absorbent or refractive (broken surface to scatter reflected sound).reflected sound).

Absorption coefficient, Absorption coefficient, aa, for paneling = .17, for paneling = .17Surface absorption for walls = 30x10 x 3 x .17 = 153 sabinsSurface absorption for walls = 30x10 x 3 x .17 = 153 sabins

New ROOM ABSORPTIONNew ROOM ABSORPTION = 153+126+720+54 = = 153+126+720+54 = 1,053 1,053 sabinssabins

NOISE REDUCTIONNOISE REDUCTION = 10 log [ 1053 / 162 ] = 10 log 6.5 = 10 log [ 1053 / 162 ] = 10 log 6.5

= 10 x .81 = = 10 x .81 = 8.1 decibels 8.1 decibels ( 7.7)( 7.7)

REVERBERATION TIMEREVERBERATION TIME = .05 X ( 9000 / 1053 ) = .05 X = .05 X ( 9000 / 1053 ) = .05 X 8.558.55 = = .43 second .43 second ( .47 ) ( .47 )

Wall paneling does not make a significant change.Wall paneling does not make a significant change.

FINALLY, add drapes to the windowsFINALLY, add drapes to the windows. Slatted . Slatted blinds do not add significantly to absorption of blinds do not add significantly to absorption of sound, but refract energy to reduce reflection. sound, but refract energy to reduce reflection.

Absorption coefficient, Absorption coefficient, aa, for drapes = .49 , for drapes = .49 Surface absorption for drapes = 30x10 x .49 = 147 sabinsSurface absorption for drapes = 30x10 x .49 = 147 sabins

New New ROOM ABSORPTIONROOM ABSORPTION = = 153+126+720+147 = 153+126+720+147 = 1146 1146 sabins.sabins.

NOISE REDUCTIONNOISE REDUCTION = 10 log [ 1146 / 162 ] = 10 log 7.07 = 10 log [ 1146 / 162 ] = 10 log 7.07

= 10 x .85 = = 10 x .85 = 8.5 decibels8.5 decibels

REVERBERATION TIME REVERBERATION TIME = .05 x (9000 / 1146) = .05 x 7.85= .05 x (9000 / 1146) = .05 x 7.85

= = .39 second.39 second

In both conditions from beginning left to right, it In both conditions from beginning left to right, it is significant that the progressive curves flatten is significant that the progressive curves flatten quickly. Even if all the surfaces were 100 % quickly. Even if all the surfaces were 100 % sound absorptive, nothing can be done to reduce sound absorptive, nothing can be done to reduce the direct sound through air. Noise reduction by the direct sound through air. Noise reduction by absorption would probably never be above 11 to absorption would probably never be above 11 to 12 decibels.12 decibels.

900 + 900 + 1200 = 3000 = 900 + 900 + 1200 = 3000 = surface area of room and surface area of room and would be the room absorption if all surfaces were would be the room absorption if all surfaces were 100 % absorptive100 % absorptive – which is purely hypothetical. – which is purely hypothetical.

Noise Reduction = 10 log [ 3000 / 162 ] = 10 log Noise Reduction = 10 log [ 3000 / 162 ] = 10 log 18.5218.52 = 10 x 1.267 = 12.67 decibels.= 10 x 1.267 = 12.67 decibels.

Reverberation Time = .05 x (9000/3000) = .05 x 3 Reverberation Time = .05 x (9000/3000) = .05 x 3 = .15 sec= .15 sec

Perhaps the only surface that is 100 % absorptive is Perhaps the only surface that is 100 % absorptive is an open window. The sound energy continues an open window. The sound energy continues without interruption until it decays.without interruption until it decays.

The purpose of understanding sound projection The purpose of understanding sound projection and sound absorption is to realize the value of and sound absorption is to realize the value of reinforcing sound where it is needed by utilizing reinforcing sound where it is needed by utilizing reflection, and to diminish sound reverberation reflection, and to diminish sound reverberation where it is not useful by using absorptive materials. where it is not useful by using absorptive materials. Study and application will architecturally reveal Study and application will architecturally reveal where the applications can be made.where the applications can be made.

PRACTICE PROBLEM ONE:PRACTICE PROBLEM ONE:

A 16’ x 20’ x 9’ room has absorptive coefficients A 16’ x 20’ x 9’ room has absorptive coefficients as follows: Ignore doors & windows.as follows: Ignore doors & windows.

WallsWalls .30.30FloorFloor .25.25CeilingCeiling .40.40

A. Find total room absorptionA. Find total room absorptionB. Find reverberation timeB. Find reverberation time

The finishes then change to the following The finishes then change to the following coefficients:coefficients:

WallsWalls .46.46FloorFloor .40.40CeilingCeiling .86.86

C. Find the noise reduction due to the new C. Find the noise reduction due to the new finishesfinishesD. Find the difference in reverberation time.D. Find the difference in reverberation time.

PROBLEM SOLUTIONPROBLEM SOLUTION Wall area = Wall area = 648 x .30648 x .30 == 194.4194.4 Floor area = Floor area = 320 x .25320 x .25 == 80.0 80.0 Ceiling area = Ceiling area = 320 x .40320 x .40 == 128.0128.0 total = 402.4 sabinstotal = 402.4 sabins

T = .05 x (2880 / 402.4) = T = .05 x (2880 / 402.4) = .36 seconds.36 seconds

NEWNEW

Wall area =Wall area = 648 x 648 x .46.46 == 298.08298.08Floor area =Floor area = 320 x 320 x .40.40 == 128.0128.0Ceiling area = Ceiling area = 320 x 320 x .86.86 == 275.2275.2

total = total = 701.28701.28Noise Reduction = 10log (701.28/402.4) = 10 log Noise Reduction = 10log (701.28/402.4) = 10 log

1.7427 =1.7427 =10 x .2412 = 2.41 db. T = .05 x 4.1068 = .205 10 x .2412 = 2.41 db. T = .05 x 4.1068 = .205 secsec

Time difference = .36 - .205 = .1550 sec, = 43%Time difference = .36 - .205 = .1550 sec, = 43%