Converting from free field to reverberant field (Noise break out to outside) If you have a noisy room (reverberant field) and you want to know how loud it sounds to a guy standing in the park outside how do you calculate that. I have a formula in a book about noise break-out that I'm trying to reverse engineer:
L_outside = L_noisyroom - SRI - 20log(r) - 14dB
So from what I can glean they took Q to be 2 (because its a flat surface though it isn't double the noise like a reflected surface) and they subtracted 6 when 'converting from a free field to a reverberant field' (I just want to find out where this idea comes from - saw it in a spreadsheet). Where I'm at atm: 
L_outside = L_wall + 10log(Q/(4.PI.r^2))    (I'm cool with this part)
L_wall = L_noisyroom - SRI + 10logS -6
Anyone know of some literature on this -6 and is it ok to just assume that A goes to 1 by magic and the power of limits at infinity or something?
Because sound reduction index (level difference between rooms caused by partition)is:
SRI = L_source - L_othersideofwall + 10log(S/A)
where A = 0.163*V/RevTime
I imagine it has something to do with this pic but I can't find any more information explaining the graph.  

Muchos gracias :)
 A: Let's divide that into steps. First of all, we know that the field in a room is given by (not citing this but can be found in many textbooks on room acoustics, such as those cited below)
$$L_{p} = L_{w} + 10 \log_{10} \left( \frac{Q}{\Omega r^{2}} + \frac{4}{A} \right)$$
where $L_{p}$ is the sound pressure level at a distance from the source, $L_{w}$ the sound power level emitted by the source (important to note that these are levels, i.e. in dB), $Q$ the directivity factor of the source (important: this is frequency- and angle-dependent), $\Omega$ is the angle of radiation (in steradians) and $A$ is the equivalent absorption area.
Now, regarding the 4 in the numerator of the last term, this is because we assume uniform intensity distribution with respect to angle (diffuse field definition) and this results in the intensity being summed over all incident angles (which is 4$\pi$ steradians and this is where the 4 comes from). For more info, one can have a look at "Room Acoustics" by Heinrich Kuttruff, Chapter 5.
For the SRI one has
$$SRI = L_{p, s} - L_{p, r} + 10 \log_{10} \left( \frac{S}{A_{r}} \right)$$
where $L_{p, s}$ is the pressure in the source room, $L_{p, r}$ the pressure in the receiving room and $S$ the area of radiation (usually this is taken to be the whole area of the partition/façade, although this is not the most correct way to do it). One thing to note here though is that the equivalent absorption area $A_{r}$ is for the receiving room, which I am not sure is something that could be used accurately for outdoor environments. Therefore, I believe that the person who formulated the spreadsheet you are talking about has completely ignored it (equivalent to setting it to 1 as you mention).
If we ignore $A_{r}$ completely and solve the above equation for $L_{p, r}$ we get
$$SRI = L_{p, s} - L_{p, r} + 10 \log_{10} \left( S \right) \implies\\
-L_{p, r} = SRI - L_{p, s} - 10 \log_{10} \left( S \right) \implies\\
L_{p, r} = L_{p, s} + 10 \log_{10} \left( S \right) - SRI$$
Now, combining this last result with the first equation (which represents the $L_{p, s}$ term) we get the sound pressure level on the outdoor side of the wall. This is
$$L_{p, s} = L_{w} + 10 \log_{10} \left( \frac{Q}{\Omega r^{2}} + \frac{4}{A} \right) + 10 \log_{10} \left( S \right) - SRI$$
This whole process is to calculate the sound pressure level on the outside of the partition/façade. Please note that from here on you cannot use the first equation to calculate the sound pressure level at a specific distance from the wall. This is because to do so you would have to know the sound power level radiated by the wall. Instead, what one could do is to use the sound pressure calculated with the above equation and calculate, with the use of the inverse square law, the geometrical attenuation. Then if you would like to assume that you have semi-spherical radiation you could add 3 dB to that. This is because
$$L_{p_{4 \pi}} = L_{w} + 10 \log_{10} \left(\frac{1}{4 \pi r^{2}} \right), \\
L_{p_{2 \pi}} = L_{w} + 10 \log_{10} \left(\frac{1}{2 \pi r^{2}} \right) \implies
L_{p_{2 \pi}} = L_{w} + 10 \log_{10} \left(\frac{1}{4 \pi r^{2}} \cdot 2 \right) \implies \\
L_{p_{2 \pi}} = L_{w} + 10 \log_{10} \left(\frac{1}{4 \pi r^{2}}\right) + 10 \log_{10} \left(2 \right) \implies
L_{p_{2 \pi}} = L_{p_{4 \pi}} + 3$$
So, with the inverse square law, one would get (utilising the third equation)
$$L_{p_{r}} = L_{p} + 10 \log_{10} \left(\frac{1}{r^{2}} \right) \implies
L_{p_{r}} = L_{p, s} + 10 \log_{10} \left( S \right) - SRI - 20 \log_{10} \left(r \right)$$
Finally, one could assume that the $L_{p_{r}}$ in the second equation refers to a diffuse field. So, a possible (not sure how valid this is though!) correction would be to subtract the term that corresponds to the diffuse field. This is to subtract $10 \log_{10} \left( \frac{4}{A} \right)$ from the final result. If we drop $A$ here we are left with what you suggested as $10 \log_{10} \left( 4 \right) \approx -6dB$.
So, the final result would look like
$$L_{p_{r}} = L_{p, s} + 10 \log_{10} \left( S \right) - SRI - 20 \log_{10} \left(r \right) - 6dB$$
To be honest, I am not sure how one would end up with -14dB. You suggested that this number comes from $10 log_{10} \left(\frac{1}{2 \pi} \right) - 6dB$, which seems to be numerically correct. Nevertheless, you should keep in mind that the $10 \log_{10} \left(\frac{Q}{\Omega r^{2}} + \frac{4}{A} \right)$ term is a correction factor on the sound power level of the source (it is derived on an energy base approach) and cannot be arbitrarily/equivalently used on pressure quantities without first calculating the sound power. So, this whole process does make me sceptical as to how valid this process of calculation/estimation is (up to the accuracy of the used formulas as a reference anyway).
I do believe that the most valid result is (replicating from above)
$$L_{p_{r}} = L_{p, s} + 10 \log_{10} \left( S \right) - SRI - 20 \log_{10} \left(r \right)$$
because, as I said, I am not sure that pressure and energy quantities can be easily mixed for "arbitrary" sound field conversions (from direct to diffuse and vice versa).
Finally, for more information on such calculations, one could look in books such as "Engineering Acoustics - An Introduction to Noise Control" by Michael Mösrer, "Engineering Noise Control" by David Bies et al., the corresponding chapter (chapter 10) from "World Health Organisation's publication" and "Building Acoustics" by Tor Erik Vigran.
A: Apparently a sound wave bouncing about a room has four times as much energy as a sound wave in a free field outside. This is where the 4 in the formula for the total sound field comes from
L_outside = L_wall + 10log(Q/(4.PI.r^2 + 4/R)
Considering 10log(4) = 6, this is what you need to subtract when converting from a free field to a reverberant field
Q is 2 because sound goes directly to the receiver as well as bounces off the ground once. 
So the 14 comes from
10log(Q/4PI) - 6 = - 14 
