Marc invited me to a party the other day, it’s on Saturday night which is a bit of a downer for me because I’m flying to India the next day but I’ll probably still go over. I hate it when there are too many people at these things though, everyone’s talking over the top of everyone else and you can’t hear what anyone’s saying, I needed some way of making sure the party wasn’t going to be like that.

Fortunately I remembered reading about some research that would be useful, by a chap called William Maclean, he derived a formula to predict the maximum amount of people, or “well mannered guests” as he puts it (this was the 1950′s) are compatible with a quiet party, beyond this number of guests the result is a noisy party, and who wants that?? The whole thing is based around a phenomenon known as positive feedback, basically once a party reaches a critical size people start having to talk louder to be heard in their small groups, then the next group has to increase their volume, etc etc until before you know it you’re bending over shouting words and spit into the ear of the guy next to you who still hasn’t got a clue what you’re going on about.

Anyway, here’s Maclean’s equation, perfect for a lunchtime calculation:

Where (with assumptions):

• N0 is the maximum number of attendees for a “quiet party”
• K is the average number of guests per conversational group – I assumed 5
• a is the sound absorption coefficient of the room – found a handy table here, assumed Marc’s lounge to be a standard room with a value of 0.15
• V is the room volume – see h below, I assumed 180m³.
• h is the properly weighted mean free path of a ray of sound through the room – I found a reference from Sabine (statistical room acoustics) which calculates this with the relationship a=4(V/S), where V and S are the room volume and the room total volume, respectively. Assuming Marc’s lounge to be around 10mx6mx3m this gives V=180m³ and S=216m³ and a value of a = 3.33.
• d0 is the mininum conventional distance between talkers – assumed 0.5m
• Sm is the minimum signal to noise ratio required for intelligible conversation – difficult one, but luckily the US government sets criteria for the minimum signal to noise ratio required for classrooms and we can steal their numbers. They recommend a minimum of 15dB, but at a party I think we can afford to go a lot lower, let’s say 10dB

So, I plugged all the numbers into the equation and the maximum number of people Marc should invite to his party is………………… 5.17 people…. tits….

Something’s gone wrong somewhere, maybe I’m using the wrong units for one of the parameters or something, unfortunately without the original paper I can’t tell exactly what (later I remembered that I was using metric figures and Maclean would have been working in feet and inches, even converting to that only increased to 7 people though).

Anyway, luckily I found another way of calculating it, outlined here, which calculates things a little differently and introduces a reverb time factor, “T”, into the equation (typically between 0.5 and 3.0 seconds, I opted bang in the middle at 1.5 seconds).

The new formula:

N = 2 + 8 x D/d^2

where:

• D = a x S/50
• a = 1 – 10^(-V/(14 x S x T))

I plugged in the numbers and VOILA, we get 14.07 people.

“Great” I thought, that sounds more reasonable, I’ll just give Marc a bell and check how many people he’s invited before deciding whether or not to go. Well, that would WOULD have been a good plan except Marc got wasted the other night and invited half the bar, he can’t remember how many and how many of them will actually turn up. I’m just going to have to chance it and be prepared for a speedy exit.

Actually, in all seriousnous it should be fun, the last party he had ended up with a China Eastern airline pilot (according to him) dropping a full glass of red wine on Marc’s new deep shag rug and then stumbling into his spare room to piss over a pile of stuff, I’d better take my camera!!