Source: https://jmrplens.github.io/phonometry/devices/noise-control/room-to-room/ # Room to room: partition, receiving room, criterion A machine runs in one room, a wall separates that room from an occupied one, and someone wants to know what the occupant hears. It is a short chain, and every link of it is a piece the library already had: the reverberant level the machine builds up on the source side, the transmission loss of the wall, the absorption of the receiving room, and the criterion curve the result is laid against. What was missing was the joint, and the joint is the one place where the arithmetic surprises people: **the noise reduction is not the transmission loss of the wall.** `noise_control.room_to_room` is that joint, and this page walks Norton & Karczub's worked problems through it end to end. The reference throughout is Norton, M. P. & Karczub, D. G., *Fundamentals of Noise and Vibration Analysis for Engineers* (2nd ed., Cambridge University Press 2003), Chapter 4: §4.6 for the sound power models, §4.9 for the room-to-room balance and §4.10 for the enclosure. Its problems 4.16, 4.18 and 4.21 carry printed octave-band answers, and all three are pinned in the [conformance report](https://jmrplens.github.io/phonometry/reference/conformance/). Where the *partition itself* comes from is a different question, answered by [Panel sound insulation](https://jmrplens.github.io/phonometry/buildings/design/panel-sound-insulation/) (mass law, coincidence, double leaves, the plateau method) or by measurement. This page takes the transmission loss as given, exactly as a hand calculation does. ## 1. What the balance says Norton's §4.9 balances the steady-state power in the receiving room: what crosses the partition equals what the room absorbs plus what leaks back through the partition. Replacing intensities with mean-square pressures and taking logarithms gives Equation (4.101), $$ \text{NR} = \text{TL} - 10\log_{10}\!\left[\frac{S_\mathrm{w}}{S_2 \alpha_2 + \tau S_\mathrm{w}}\right], $$ with $\text{NR} = L_{p1} - L_{p2}$ the noise reduction between the two reverberant fields, $\text{TL} = 10\log_{10}(1/\tau)$ the transmission loss of the partition (the quantity UNE calls the sound reduction index $R$), $S_\mathrm{w}$ the area of the partition and $S_2 \alpha_2$ the equivalent absorption area of the receiving room. Read the logarithm and the whole page follows. A **large partition into a hard room** delivers *less* than its transmission loss: a lot of area radiating into a room with nothing to soak it up. A **small partition into a well-absorbing room** delivers *more*. The two rooms and the wall are not separable, which is why quoting a wall's $R_\mathrm{w}$ and stopping is not an answer. The $\tau S_\mathrm{w}$ term is the power the partition itself passes back into the source room. At any realistic transmission loss it is worth a few hundredths of a decibel, so it is off by default and `include_partition_transmission=True` switches it on. ## 2. Three partitions, one receiving room Norton's problem 4.21 (printed pp. 586-587) is the comparison the equation was made for: two 13 mm gypsum wallboards with a 64 mm air gap, a 125 mm plastered brick wall and a double brick wall with a 50 mm cavity, all speaking into the same 8 m x 9 m x 3 m receiving room through the same 8 m x 3 m opening. ```python import numpy as np from phonometry import ( SourceRoom, equivalent_absorption_area, room_to_room_transmission, ) bands = [125.0, 250.0, 500.0, 1000.0, 2000.0, 4000.0] # Receiving room 8 x 9 x 3 m: walls 102 m2, floor and ceiling 72 m2 each. receiving = equivalent_absorption_area([ (102.0, [0.04, 0.04, 0.09, 0.15, 0.17, 0.23]), # walls (72.0, [0.02, 0.06, 0.14, 0.37, 0.60, 0.66]), # floor (72.0, [0.30, 0.20, 0.15, 0.05, 0.05, 0.05]), # ceiling ]) print(np.round(receiving, 1)) # [27.1 22.8 30.1 45.5 64.1 74.6] m2, band by band partitions = { "Two 13 mm wallboards, 64 mm gap": [18, 27, 37, 45, 43, 39], "125 mm plastered brick": [36, 36, 40, 46, 54, 57], "Double brick, 50 mm cavity": [37, 41, 48, 60, 61, 61], } for name, tl in partitions.items(): res = room_to_room_transmission( bands, tl, 8.0 * 3.0, receiving, source=SourceRoom(level=90.0), label=name, ) print(f"{name:32s} {np.round(res.noise_reduction, 1)}") # Two 13 mm wallboards, 64 mm gap [18.5 26.8 38. 47.8 47.3 43.9] # 125 mm plastered brick [36.5 35.8 41. 48.8 58.3 61.9] # Double brick, 50 mm cavity [37.5 40.8 49. 62.8 65.3 65.9] ``` Those are Norton's printed answers to the tenth of a decibel. Look at what the receiving room did to them. At 125 Hz the room absorbs 27.1 m² against a 24 m² partition, so every wall delivers half a decibel *more* than its transmission loss. At 250 Hz the ceiling has stopped absorbing and the floor has not started, the room is down to 22.8 m², and every wall now delivers 0.2 dB *less*. By 4 kHz the carpet and the walls together give 74.6 m² and the same brick wall is worth 4.9 dB more than its transmission loss. The same three walls, ranked the same way, but the numbers a specification would quote move by more than 5 dB depending on a room the wall never touches. ## 3. The chain, end to end Problem 4.18 (printed pp. 585-586) is the whole calculation, and it starts one step earlier: the source-room level is not given, only the blower's free-field sound power level. Two decisions turn that into $L_{p1}$, and both are Norton's. **The reverberant field is what drives the partition.** The receiver of interest is the wall, not a point beside the machine, so the level that matters is $L_{p1} = L_W + 10\log_{10}(4/R_1)$. That is `steady_state_spl` with `distance=None`, the $r \to \infty$ limit of the same Bies Equation (6.43) the library has always used for a receiver at a distance. **The problem asks for a conservative estimate, and conservative has a meaning.** The blower sits on the floor along the middle of a wall, in the intersection of two large flat surfaces, so $Q = 4$. Norton's §4.6 Table 4.5 lists three sound power models: a *constant-power* source radiates the same $\Pi_0$ wherever it stands, a *constant-volume* source is loaded by the nearby boundaries and radiates $\Pi_0 Q$, and a *constant-pressure* source, a theoretical floor, radiates $\Pi_0 / Q$. Real machines sit between the first two whenever they are closer to the boundary than a wavelength, and the constant-volume model is the upper bound. Here it is worth $10\log_{10} 4 = 6.02\ \text{dB}$, and without it the printed answers come out 6 dB low. ```python from phonometry import ( DesignCriterion, SourceRoom, equivalent_absorption_area, mean_absorption, room_constant, room_to_room_transmission, ) bands = [125.0, 250.0, 500.0, 1000.0, 2000.0, 4000.0] ceiling = [0.07, 0.20, 0.40, 0.52, 0.60, 0.67] # absorbent ceiling walls = [0.03, 0.03, 0.03, 0.04, 0.05, 0.07] # both rooms # Plant room 8 x 10 x 3 m: bare floor, absorbent ceiling, 108 m2 of wall. plant = [(80.0, [0.01, 0.01, 0.015, 0.02, 0.02, 0.02]), (80.0, ceiling), (108.0, walls)] # Operator room 5 x 5 x 3 m: carpet, same ceiling and walls. operator = [(25.0, [0.08, 0.24, 0.57, 0.69, 0.71, 0.73]), (25.0, ceiling), (60.0, walls)] chain = room_to_room_transmission( bands, [39.0, 42.0, 50.0, 58.0, 63.0, 67.0], # TL of the separating wall 5.0 * 3.0, # the wall is 5 m x 3 m equivalent_absorption_area(operator), source=SourceRoom( power_level=[105.0, 103.0, 98.0, 108.0, 107.0, 109.0], room_constant=room_constant(268.0, mean_absorption(plant)), directivity=4.0, # floor-wall intersection model="constant_volume", # the conservative bound ), criterion=DesignCriterion(target=45.0), label="Plant room to operator room", ) for row in chain.table(): print(f"{row['label']:<28} " f"{[round(float(v), 1) for v in row['values']]}") ``` ```text Source sound power level [105.0, 103.0, 98.0, 108.0, 107.0, 109.0] Source room level [107.0, 101.7, 93.8, 102.4, 100.6, 101.9] Partition transmission loss [39.0, 42.0, 50.0, 58.0, 63.0, 67.0] Receiving-room absorption [5.5, 12.8, 26.1, 32.6, 35.8, 39.2] Noise reduction [34.7, 41.3, 52.4, 61.4, 66.8, 71.2] Receiving room level [72.4, 60.4, 41.4, 41.0, 33.9, 30.7] NC 45 [60.0, 54.0, 49.0, 46.0, 44.0, 43.0] Required transmission loss [51.4, 48.4, 42.4, 53.0, 52.9, 54.7] ``` Norton's printed answer is 72.3 / 60.4 / 41.4 / 41.0 / 33.8 / 30.7 dB, which this reproduces to a tenth. Two rows are worth staring at. The **noise reduction climbs from 34.7 dB to 71.2 dB** while the transmission loss only climbs from 39 to 67, because the operator room's absorption grows seven fold across the range while the wall is fixed at 15 m²; the little room does half the work at high frequency and none at all at 125 Hz. And the **required transmission loss row is not monotone**: the wall needs 51.4 dB at 125 Hz where it has 39, and 42.4 dB at 500 Hz where it already has 50. A low-frequency problem, in a wall that is over-specified in the middle. *The chain of problem 4.18. The receiving-room spectrum crosses the NC 45 curve between 250 and 500 Hz, so the two low bands are the whole problem. The right axis is the point of Equation (4.101): the noise reduction the wall delivers starts 4 dB below its transmission loss and ends 4 dB above it, and the crossing is where the operator room's absorption grows past the 15 m² of the wall.*

Show the code for this figure

```python import matplotlib.pyplot as plt # `chain` is the RoomToRoomResult built above. # One line: both reverberant spectra, the criterion curve and the noise # reduction against the transmission loss on the twin axis. chain.plot() plt.show() # By hand, from the per-band fields the result carries: fig, ax = plt.subplots() ax.semilogx(chain.frequencies, chain.source_level, "s--", label="Source room") ax.semilogx(chain.frequencies, chain.received_level, "o-", label="Receiving room") ax.semilogx(chain.frequencies, chain.criterion_curve, ":", label="NC 45") twin = ax.twinx() twin.plot(chain.frequencies, chain.transmission_loss, "--", label="TL") twin.plot(chain.frequencies, chain.noise_reduction, "-.", label="NR") plt.show() ```

## 4. The verdict, and the number to specify The result rates itself. `.rating` is the ANSI/ASA S12.2-2019 rating of the received spectrum, `.criterion_curve` is the design curve sampled at the analysis bands, `.exceedance` is the band-by-band excess and `.meets_target` is the plain verdict a design sheet writes down. `.required_transmission_loss` is Equation (4.101) solved for $\text{TL}$, which is the number that goes into a specification. ```python print(chain.meets_target) # False print(round(chain.rating.rating, 1)) # 61.7, governed by 125 Hz print([round(float(v), 1) for v in chain.exceedance]) # [12.4, 6.4, -7.6, -5.0, -10.1, -12.3] deficit = chain.required_transmission_loss - chain.transmission_loss print([round(float(v), 1) for v in deficit]) # [12.4, 6.4, -7.6, -5.0, -10.1, -12.3] ``` The deficit and the exceedance are the same six numbers, which is the whole content of the chain: a decibel too much in the room is a decibel too little in the wall. Whether that low-frequency deficit is worth chasing with mass, a cavity or a second leaf is what [Panel sound insulation](https://jmrplens.github.io/phonometry/buildings/design/panel-sound-insulation/) is for. The verdict is deliberately the band-by-band test, not the rating. A spectrum can rate NC-38 and still poke through the NC 45 curve in one band, because the standard's two-step designation procedure and a design sheet's "no band above the curve" are different questions. ## 5. An enclosure against a criterion Problem 4.16 (printed pp. 584-585) asks the same question about an enclosure instead of a partition: a refrigeration compressor makes a reverberant level in its room, the technician needs that level down to an NC-45 curve, and the enclosure walls have to make up the difference. Norton derives the enclosure equation from the same power balance as §4.9 and gets Equation (4.115), $$ \text{IL} = \text{TL} - 10\log_{10}(S_\mathrm{E} / R_\mathrm{i}), $$ with $S_\mathrm{E}$ the external radiating area and $R_\mathrm{i}$ the room constant of the enclosure interior, machine surface included. Solved for the panels, $\text{TL} = \text{IL} + 10\log_{10}(S_\mathrm{E} / R_\mathrm{i})$, which is `enclosure_required_transmission_loss`. ```python import numpy as np from phonometry import enclosure_required_transmission_loss, mean_absorption bands = [63.0, 125.0, 250.0, 500.0, 1000.0, 2000.0, 4000.0, 8000.0] wool = [0.10, 0.20, 0.45, 0.65, 0.75, 0.80, 0.80, 0.80] # 50 mm blanket concrete = [0.01, 0.01, 0.01, 0.02, 0.02, 0.02, 0.03, 0.03] lp1 = np.array([72.0, 79.0, 81.0, 84.0, 83.0, 81.0, 80.0, 75.0]) # unenclosed nc45 = np.array([67.0, 60.0, 54.0, 49.0, 46.0, 44.0, 43.0, 41.0]) # target # Enclosure 2.5 x 3.5 x 2.5 m on the concrete floor: the walls and roof # radiate (38.75 m2) and their inner faces carry the lining. The rest of the # interior is the uncovered floor plus the five exposed faces of the # 1.5 x 2.5 x 1.5 m machine, both with the absorption of concrete. radiating = 2 * (2.5 * 2.5) + 2 * (3.5 * 2.5) + 2.5 * 3.5 machine = 2 * (1.5 * 1.5) + 2 * (2.5 * 1.5) + 1.5 * 2.5 bare_floor = 2.5 * 3.5 - 1.5 * 2.5 required = enclosure_required_transmission_loss( lp1 - nc45, radiating, radiating + bare_floor + machine, mean_absorption([(radiating, wool), (bare_floor + machine, concrete)]), frequencies=bands, model="norton", ) print(np.round(required.panel_transmission_loss, 1)) # [14.5 25.3 28.9 34.4 35.2 34.7 34.6 31.6] ``` Norton's printed answer is 14.4 / 25.2 / 28.9 / 34.4 / 35.2 / 34.7 / 34.7 / 31.6 dB, matched to 0.11 dB, which is the resolution of his own one-decimal rounding. The shape is the useful part: **the requirement peaks at 1 kHz and falls away at both ends**, because at 63 Hz the mineral wool absorbs almost nothing and the level to lose is only 5 dB, while at 8 kHz the compressor has quietened down. A panel chosen for its 500 Hz to 2 kHz figure is the right panel here. Two footnotes on that snippet, both about being faithful to the source. `model="norton"` matters: the library's default enclosure model is Bies, Hansen & Howard's Equation (7.111), which carries a $0.3$ inside the logarithm and so floors the insertion loss of a fully lined enclosure at $\text{TL} + 5.2\ \text{dB}$. Norton's Equation (4.115) has no such floor. The two agree within a few tenths while the interior is hard and diverge once the lining takes over, and reproducing a published answer means using the model its author used. `ENCLOSURE_MODELS` lists both. And the NC-45 column is Norton's own, taken from his problem statement. It reads 41 dB in the 8 kHz band where ANSI/ASA S12.2-2019 Table 1 (which `nc_curve` implements) reads 42 dB; they agree in every other band. Using the printed column keeps the oracle the published one. ## 6. What the prediction does not know Norton's answer to problem 4.21 ends with the caveat that matters more than any of the arithmetic above: the measured noise reduction runs below the prediction because of **flanking transmission through mechanical connections and air leaks**. A ceiling void carried over the partition, a service penetration, a door undercut, or the wall simply not reaching the structural slab, and the equation's answer becomes an upper bound. The clip below draws what the equation leaves out. The chain of section 3 prices the direct path only, the **Dd** route through the partition; the other three pulses leave the source room over the flanking walls, floor or ceiling — **Ff** flank to flank, **Fd** flank to partition, **Df** partition to flank — and re-radiate on the far side without ever passing through the transmission loss the calculation used. Each path shrinks at every element and junction it crosses, which is why no single one has to be large for the sum of the three to dominate a good partition. [Watch the high-resolution video (WebM)](https://raw.githubusercontent.com/jmrplens/phonometry/main/.github/images/anim_flanking_paths.webm) `DesignCriterion.flanking_penalty` is the explicit debit for that, in decibels off the predicted noise reduction. It is not a model, it is a place to record the allowance and have it show up in `.table()` and in the required transmission loss rather than get lost: ```python # The chain of section 3 again, with 3 dB allowed for flanking and leaks. honest = room_to_room_transmission( bands, [39.0, 42.0, 50.0, 58.0, 63.0, 67.0], 15.0, equivalent_absorption_area(operator), source=SourceRoom(level=chain.source_level), criterion=DesignCriterion(target=45.0, flanking_penalty=3.0), ) print(np.round(honest.received_level, 1)) # [75.4 63.4 44.4 44. 36.9 33.7] every band 3 dB worse ``` Where the flanking paths *are* modelled rather than allowed for, the prediction models of EN 12354 are the right tool, and those live in [Building sound insulation prediction](https://jmrplens.github.io/phonometry/buildings/design/insulation-prediction/) and [Flanking transmission](https://jmrplens.github.io/phonometry/buildings/insulation/flanking-lab/). The other assumption is diffusivity. Equation (4.101) is a diffuse-field balance in both rooms, so it has nothing to say below each room's Schroeder frequency (`schroeder_frequency`), where discrete modes rule. The 5 m x 5 m x 3 m operator room above, 75 m3, has for a half-second reverberation time a Schroeder frequency of 163 Hz, which is exactly why the 125 Hz band of problem 4.18 - the band that governs the whole answer - should be read as an estimate and not as a number. ## What this guide covers **Covered.** The Norton & Karczub §4.9 power balance (Eq. 4.101) as `room_to_room_transmission`, with `SourceRoom`, `equivalent_absorption_area`, `mean_absorption` and `room_constant`; the source-room level from a sound power level through the §4.6 Table 4.5 source models — constant power, constant volume, constant pressure — and the directivity of the machine's position; `DesignCriterion` with its target, its flanking allowance and its ANSI/ASA S12.2-2019 rating, NC by default and RC Mark II with `DesignCriterion(family="RC")`, plus `.required_transmission_loss`, `.exceedance`, `.meets_target` and `.table()`; and the §4.10 enclosure equation (Eq. 4.115) through `enclosure_required_transmission_loss` in both interior models listed by `ENCLOSURE_MODELS`. **Not covered.** The transmission loss of the partition itself, which is measured or predicted elsewhere; flanking as a modelled path rather than the flat allowance used here; and the leaks and openings of an enclosure, which need the composite of [Industrial noise control](https://jmrplens.github.io/phonometry/devices/noise-control/noise-control/). The whole chain is a diffuse-field argument, so it says nothing below either room's Schroeder frequency, or closer to a surface than roughly half a wavelength. ## See also - [Panel sound insulation](https://jmrplens.github.io/phonometry/buildings/design/panel-sound-insulation/): where the transmission loss this page consumes comes from. - [Building sound insulation prediction](https://jmrplens.github.io/phonometry/buildings/design/insulation-prediction/) and [Flanking transmission](https://jmrplens.github.io/phonometry/buildings/insulation/flanking-lab/): the EN 12354 models for the flanking paths this page only debits. - [Field sound insulation](https://jmrplens.github.io/phonometry/buildings/insulation/insulation-field/): how the same room pair is measured, and the $D_\mathrm{nT}$ and $R'$ a test report hands back. - [Room-noise criteria](https://jmrplens.github.io/phonometry/buildings/rooms/room-noise/): the NC and RC Mark II families the verdict is written in. - [Industrial noise control](https://jmrplens.github.io/phonometry/devices/noise-control/noise-control/): the enclosure of section 5 as a construction, with its leaks and its measurement standards. - [Sound power](https://jmrplens.github.io/phonometry/devices/emission/sound-power/): the machine $L_W$ the whole chain starts from. - API reference: [`noise_control.room_to_room`](https://jmrplens.github.io/phonometry/reference/api/noise_control/room-to-room/). ## References - Norton, M. P., & Karczub, D. G. (2003). *Fundamentals of noise and vibration analysis for engineers* (2nd ed.). Cambridge University Press. [doi:10.1017/CBO9781139163927](https://doi.org/10.1017/CBO9781139163927). The sound power models of §4.6 (Table 4.5, Eqs. 4.53-4.56), the room-to-room balance of §4.9 (Eqs. 4.92-4.101), the enclosure design equation of §4.10 (Eqs. 4.102-4.115), and the worked problems 4.16, 4.18 and 4.21 with their printed octave-band answers, which this guide is built around. - Bies, D. A., Hansen, C. H., & Howard, C. Q. (2017). *Engineering noise control* (5th ed.). CRC Press. [doi:10.1201/9781351228152](https://doi.org/10.1201/9781351228152). The steady-state room relations (§6.4, Eqs. 6.43-6.44) behind `steady_state_spl` and `room_constant`, and the enclosure insertion loss of §7.4.2 (Eqs. 7.103, 7.111) that is the library's default enclosure model. - ANSI/ASA S12.2-2019. *Criteria for evaluating room noise*. The NC curves of Table 1 and the RC Mark II family of Annex D that the chain is rated against. - Kuttruff, H. (2016). *Room acoustics* (6th ed.). CRC Press. The Schroeder frequency (Eq. 3.44) that bounds the diffuse-field assumption from below.