Before and after Room Treatment

[radiomike]

Speed of sound is usually accepted to travel at 343m/s at 20°C and 1Atm Pressure in dry air.

Inverting that by doing 1/343, gives 0.00292 seconds per meter. Multiplied by 1000 to give milliseconds per meter gives 2.92. So sound takes 2.92 milliseconds to travel one meter.

Using 2.6ms, converting that to seconds and multiplying by the speed of sound gives us the actual distance, which is 0.892 metres, probably a more realistic number for the path length difference of the first reflection. Particularly as this seems to be a domestic setting rather than an arena or stadium.

 

[Daverich4]

For the 2.6 ms reflection, I’d keep the test simple: measure L-only ETC and R-only ETC, then temporarily cover one suspected surface at a time with a thick blanket, absorber, or one of the panels. If the 2.6 ms spike drops, that surface is probably the source.

For the panels, I think the idea is to use removal as a diagnostic: measure with everything in place, remove or move one panel, remeasure, then put it back and test the next one. That should show which panels are helping, which are doing little, and whether any might work better somewhere else.

Thanks, I’ll try that out.

 

[Daverich4]

The 2.6ms spike in your ETC represents the path length difference between the direct sound and the reflection. And I calculate a 2.6ms delay to equal 8.92m. It depends on the speed of sound of course, and the speed of sound varies with ambient temperature, elevation, and so on. So it's roughly 8.92m and not exactly 8.92m. I get that all you have are the slides and my book, but:

Speaker to mic: 0ms (i.e. the main impulse)
Any spike after 0ms has travelled an extra distance to get to the mic.

I then gave a few examples. For a front-to-back reflection, you can work it out with simple addition. But for a side wall reflection or ceiling/floor bounce, you have to use Pythagoras' theorem.

One way to work it out is to sit down with a diagram showing measurements where your loudspeakers and mic are, along with the height of the mic and dimensions of the room. Then calculate one by one when each reflection is expected to arrive. This is a bit onerous.

Another way is to take some thick acoustic foam and plonk it on where you think the reflection is coming from. It might be a bit tricky if it's a ceiling reflection! Do a before-after measurement, and compare the spike on the ETC.

View attachment 541945

I don't know what those numbers mean, some of them are clearly room dimensions but I have no idea why your left speaker has "54" and "39" on it. And I am assuming that the listening position labelled "9" is where you are sitting. Assuming that "9" is how far the MLP is from the rear wall, that's about an 18 foot extra distance, 5.5m. That's about 1.6ms. You're the one with the tape measure, so you can work it out

Thanks, I guess I did have the right distance. The numbers are measurements. The speakers are 54” from the side walls, 39” from the front wall. Ten feet between them. The MLP is 12’ from the speakers, 9’ from the back wall. If the problem is the ceiling I can’t do anything about that but the measurements work out almost exactly to a problem on the back wall. I have a suspect there I’m going to start with. And as always, thanks for your help. Your publications and links to other sources have been very helpful. This is the spot I’m going to start with and if it’s not that I’ll locate other things at the same distance. (Those panels just happened to be standing there when I took the picture, I’m going to take the Marlin down and do the measurements again).

 

[Daverich4]

Speed of sound is usually accepted to travel at 343m/s at 20°C and 1Atm Pressure in dry air.

Inverting that by doing 1/343, gives 0.00292 seconds per meter. Multiplied by 1000 to give milliseconds per meter gives 2.92. So sound takes 2.92 milliseconds to travel one meter.

Using 2.6ms, converting that to seconds and multiplying by the speed of sound gives us the actual distance, which is 0.892 metres, probably a more realistic number for the path length difference of the first reflection. Particularly as this seems to be a domestic setting rather than an arena or stadium.

Well, now I’m confused again. Keith_W gave me two formulas to do the calculation. One of them gives a result of just under 9 meters and the other one gives your result. Can the correct result be either of those distances or is one of them wrong? I actually have a suspected reflection point at both distances.

 

[Keith_W]

Well, now I’m confused again. Keith_W gave me two formulas to do the calculation. One of them gives a result of just under 9 meters and the other one gives your result. Can the correct result be either of those distances or is one of them wrong? I actually have a suspected reflection point at both distances.

velocity (m/s) = distance (m) /time (s)
distance (m) = velocity (m/s) * time (s)
distance (m) = 343 * (2.6/1000)
distance = 0.8918m

The slide from my talk that you quoted had the correct formula:

I must have had my brain turned off when I made this post and misplaced the decimal. Sorry about that. And thanks @radiomike.

 

[Daverich4]

velocity (m) = distance (m) /time (s)
distance (m) = velocity (m/s) * time (s)
distance (m) = 343 * (2.6/1000)
distance = 0.8918m

The slide from my talk that you quoted had the correct formula:

View attachment 542012

I must have had my brain turned off when I made this post and misplaced the decimal. Sorry about that. And thanks @radiomike.

That’s where I went wrong. I read the formula as (343*2.6)/1000. Math is not my strong suit. I have two strong candidates for reflections at that distance. Thanks.

 

[Keith_W]

That’s where I went wrong. I read the formula as (343*2.6)/1000. Math is not my strong suit. I have two strong candidates for reflections at that distance. Thanks.

343 * (2.6/1000) = (343*2.6)/1000

 

[Daverich4]

343 * (2.6/1000) = (343*2.6)/1000

As I said, not my strong point.

 

[HiMu]

Well, now I’m confused again. Keith_W gave me two formulas to do the calculation. One of them gives a result of just under 9 meters and the other one gives your result. Can the correct result be either of those distances or is one of them wrong? I actually have a suspected reflection point at both distances.

You don't have to calculate the distances manually. In Impulse response or Filtered IR graph, hold CTRL and drag the distance between the peaks with right mouse button. This is for Windows, could be different button combination in Mac.

 

[Daverich4]

You don't have to calculate the distances manually. In Impulse response or Filtered IR graph, hold CTRL and drag the distance between the peaks with right mouse button. This is for Windows, could be different button combination in Mac.
View attachment 542127

Thanks for the suggestion. I'd used that in other things but didn't know it would provide the distance in that graph. I appreciate your help. (It's the same key combination on a Mac).

 

[Daverich4]

I don't know why I'm having such a hard time with this, but apparently I am. In attempting to correct the 2.6ms reflection I removed all furniture around the MLP. The closest surface to the mic was the floor @ 42". Next closest was the ceiling @ 4'6". Next closest was more than 6' away. The graph looks quite a bit different than it did before I removed the furniture but the reflection @ 0.8918M is still there. I've clearly misunderstood the directions but I have no idea how.

 

[thcdru2k]

I don't know why I'm having such a hard time with this, but apparently I am. In attempting to correct the 2.6ms reflection I removed all furniture around the MLP. The closest surface to the mic was the floor @ 42". Next closest was the ceiling @ 4'6". Next closest was more than 6' away. The graph looks quite a bit different than it did before I removed the furniture but the reflection @ 0.8918M is still there. I've clearly misunderstood the directions but I have no idea how.

I think that means none of the furniture is causing that reflection. I have a feeling it's the ceiling since that is probably the one area you have not tested with treatment

 

[Daverich4]

I think that means none of the furniture is causing that reflection. I have a feeling it's the ceiling since that is probably the one area you have not tested with treatment

According to the graph the ceiling is too far away? Earlier in this thread Dr. Wong posted the original graph and said the distance to the reflection is less than a meter away.

 

[markstein]

I am sorry, but why bother so much about this reflection?

How does it sound to you?

 

[Daverich4]

I am sorry, but why bother so much about this reflection?

How does it sound to you?

It was pointed out to me further upthread with a suggestion for how to eliminate it. The system sounds pretty good to me, I’m mostly trying to learn how to optimize it.

 

[Keith_W]

According to the graph the ceiling is too far away? Earlier in this thread Dr. Wong posted the original graph and said the distance to the reflection is less than a meter away.

It's not less than 1m away. It means the path-length difference is less than 1m. Think of it like this:

- Swimmer A travels from the one side of the swimming pool directly to you. You define the arrival time of Swimmer A as zero.
- Swimmer B, who is swimming at the same speed, leaves the same location at the same time. But he swims to the side of the pool, and then swims to you. He arrives 10 seconds after Swimmer A. Unfortunately you were blindfolded, so you don't know what path Swimmer B took.

--> If you know the speed that both people were swimming at, you can work out how much extra distance Swimmer B has travelled.
--> If you know the shape of the pool, the starting point, and where you are, you can work out the likely path.
--> Maybe he swam to the wall behind you, then directly to you? You measure the distance from yourself to the wall behind you. He had to swim this distance twice (i.e. go past you and then back).
--> Maybe he swam to the side wall, then directly to you. But which side wall? You have to use Pythagoras' theorem to work it out, as shown below.
--> Maybe there is an island (furniture) in the pool, and he swam there first. Again, Pythagoras' theorem.

A path-length difference of 0.892m does not mean that the object is 0.892m away. All it means is that the reflection travelled an extra 0.892m.

 

[markstein]

Here is an easy to use SBIR calculator which lets you understand the rough effects of each reflection. You can also play with REW room simulator.

SBIR calculator

SBIR calculator

tripp.com.au

 

[HiMu]

According to the graph the ceiling is too far away? Earlier in this thread Dr. Wong posted the original graph and said the distance to the reflection is less than a meter away.

Here is another quite handy way of locating reflections and where to place treatments: https://amcoustics.com/tools/amray
Visual path length & time calculator of the concept that Keith_W explained.

Quick side view example of a room. Draw the hard surfaces, select the right side "Raytracing algorithm", place speaker and listener. You can compare those to impulse responses, but do note that the 2d path lengths are different compared to actual 3d distance in a room, so this is not accurate, only approximation.

 

[Bjorn]

What's wrong with using a mirror? Very simple to find reflection points.

 

[Daverich4]

It's not less than 1m away. It means the path-length difference is less than 1m. Think of it like this:

- Swimmer A travels from the one side of the swimming pool directly to you. You define the arrival time of Swimmer A as zero.
- Swimmer B, who is swimming at the same speed, leaves the same location at the same time. But he swims to the side of the pool, and then swims to you. He arrives 10 seconds after Swimmer A. Unfortunately you were blindfolded, so you don't know what path Swimmer B took.

--> If you know the speed that both people were swimming at, you can work out how much extra distance Swimmer B has travelled.
--> If you know the shape of the pool, the starting point, and where you are, you can work out the likely path.
--> Maybe he swam to the wall behind you, then directly to you? You measure the distance from yourself to the wall behind you. He had to swim this distance twice (i.e. go past you and then back).
--> Maybe he swam to the side wall, then directly to you. But which side wall? You have to use Pythagoras' theorem to work it out, as shown below.
--> Maybe there is an island (furniture) in the pool, and he swam there first. Again, Pythagoras' theorem.

View attachment 542608

A path-length difference of 0.892m does not mean that the object is 0.892m away. All it means is that the reflection travelled an extra 0.892m.

Once again, thank you for all the time and effort you put in trying to help me sort this out. However, I think it's time for me to throw in the towel and accept that my room is going to be what it's going to be. I understand the basic problem you're describing to me but haven't been able to gain any insight into what to do about it. You discuss this topic in Pg 52 & 53 in your document REW Book 2. The diagram you included in your latest post is the same one Pg 56 of your Mac Talk Treble document. When I look at it I see squiggly lines, arrows and formulas and can't begin to translate that into a solution for the reflection you pointed out much earlier in this thread. I'm at the point now where I'm so frustrated (and I'm sure you are too) with my lack of ability to optimize my room that I'm not getting any enjoyment out of what I already have. I appreciate your patience with my lack of understanding. Thanks for trying.