Fun quiz: what is the size of this room?

[Keith_W]

This freq response graph was posted on another forum:

Two speakers, a Yamaha NS1000 (green) and Monitor Audio Silver 8 (red). The measurements were taken on different days a few months apart (so the mic was moved), but the speakers were in roughly the same position. The question asked was, "how can two different speakers look so similar?"

Anyway, the fun quiz for ASR is: looking at these graphs, can you calculate the dimensions of the room? Assume the room is rectangular, and the standard ceiling height (for Australia) is 2.7m, or 8.85ft.

 

[Deleted member 48726]

Hmm. Should the listening position not be given as well? Otherwise it's a blind shoot.

 

[Sokel]

Treated or not?
If untreated I would say about 25-35m².

 

[Keith_W]

Hmm. Should the listening position not be given as well? Otherwise it's a blind shoot.

Speakers on one end of the room, listening position (and microphone) against the wall on the other end.

Treated or not?
If untreated I would say about 25-35m².

You can see that the T30 is about 400ms.

 

[Sokel]

You can see that the T30 is about 400ms.

You're right,much smaller.

 

[Gruesome]

13' by 6'?

 

[Keith_W]

Sorry gents. There are a few observations you need to make first:

- at what point do the curves diverge, and why?
- there is an obvious bass peak in both speakers. Why?

 

[Gruesome]

Any news on the room size?

 

[Keith_W]

Sorry I forgot about this thread

The key is to look at where the two curves diverge. The reason the two curves below 400Hz look so similar despite being different speakers is because both have approximately the same freq response (except that the Monitor Audio Silver 8 has less bass below 40Hz than the Yammy NS1000). If you place them in the same position, then their behaviour below Schroder will be very close. 400Hz is the end of the transition band, which is 4x the Schroder frequency. From the graphs, transition freq is 400Hz so Schroder freq is 100Hz. From the second graph, T30 is 400ms.

The Schroder freq (Fs) can be predicted by Fs = 2000 * sqrt (T30/V)

Solving for V, we have V = T30 / (Fs/2000)^2 or 160m^3

We also have one room dimension, standard Australian ceiling height of 2.7m. Assuming the bass peak at 45Hz is a room mode, and it is second order, the wavelength is 7.6m. So we have our second room dimension.

So the room size is 7.6m x 7.8m x 2.7m.

In that other forum, my guess was wrong because I had assumed the room was rectangular. It was L-shaped. But I did get the room volume approximately correct and two of the dimensions correct (2.7m and 7.6m).