Estimating the spl of high frequency early reflections?

Y

yepimonfire

Audioholic Samurai
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423 13 5
#1
I'm mainly talking about frequencies greater than 1khz, where perception of location is mostly controlled by spl and hrtf, and how it might apply to the effectiveness of something like an Atmos module. Let's assume a speaker has uniform 90x90 directivity from 1khz to 14khz (-6dB),such as the klipsch reference premiere speakers I've measured. Each 15 degree increment off axis results in a fairly uniform drop of 2dB above the fc of the speaker. Since an Atmos module is angled at 20 degrees, and is often elevated above ear level, we could assume it would be about 90 degrees off axis, which would translate to the direct sound being - 12dB.

If the speaker is 4' from the ceiling, and the listener is 4.5' from the ceiling, is there a way to mathematically determine how loud the reflection would be? How likely is it that it would be louder than the direct sound?

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highfigh

highfigh

Audioholic Warlord
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2,769 9 5
#2
I'm mainly talking about frequencies greater than 1khz, where perception of location is mostly controlled by spl and hrtf, and how it might apply to the effectiveness of something like an Atmos module. Let's assume a speaker has uniform 90x90 directivity from 1khz to 14khz (-6dB),such as the klipsch reference premiere speakers I've measured. Each 15 degree increment off axis results in a fairly uniform drop of 2dB above the fc of the speaker. Since an Atmos module is angled at 20 degrees, and is often elevated above ear level, we could assume it would be about 90 degrees off axis, which would translate to the direct sound being - 12dB.

If the speaker is 4' from the ceiling, and the listener is 4.5' from the ceiling, is there a way to mathematically determine how loud the reflection would be? How likely is it that it would be louder than the direct sound?

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You can come pretty close if you use the material's NRC (Noise Reduction Coefficient) and calculate the decrease in SPL due to the distance from the speaker to the surface. Honestly, if the sound hits at a low angle, you'll need to do some pretty intense math to figure this out because it's almost impossible to focus sound like a laser and measure the direct reflection.

If the NRC is .5 at a particular frequency, it means that half of the sound will be absorbed, but that's if the sound source is at 90 degrees to the surface.
 
Y

yepimonfire

Audioholic Samurai
Ratings
423 13 5
#3
Any idea what the NRC of drywall is?

You can come pretty close if you use the material's NRC (Noise Reduction Coefficient) and calculate the decrease in SPL due to the distance from the speaker to the surface. Honestly, if the sound hits at a low angle, you'll need to do some pretty intense math to figure this out because it's almost impossible to focus sound like a laser and measure the direct reflection.

If the NRC is .5 at a particular frequency, it means that half of the sound will be absorbed, but that's if the sound source is at 90 degrees to the surface.


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Y

yepimonfire

Audioholic Samurai
Ratings
423 13 5
#5
The charts are available on google- basically, if a material is hard and smooth, it'snot going to reduce the sound much.

https://workingwalls.com/nrc-and-stc-what-are-they-whats-the-difference/
So wouldn't it mainly be dictated by the distance from the speaker to the ceiling (at the angle of incidence) and then the distance from that point to the listening position?

I guess another thing to consider is the fact that the inverse square law assumes a point source radiates spherically at 360 degrees, but if the pattern is restricted to 90 degrees above 1khz because of a waveguide, wouldn't this reduce the sound power lost over distance? I can say from my own experience that according to the popular spl calculator tossed around here that it estimated an spl drop of 9dB but I only experienced a 7dB drop, and this was in a well treated, minimally reverberant room.

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Last edited:
highfigh

highfigh

Audioholic Warlord
Ratings
2,769 9 5
#6
So wouldn't it mainly be dictated by the distance from the speaker to the ceiling (at the angle of incidence) and then the distance from that point to the listening position?

I guess another thing to consider is the fact that the inverse square law assumes a point source radiates spherically at 360 degrees, but if the pattern is restricted to 90 degrees above 1khz because of a waveguide, wouldn't this reduce the sound power lost over distance? I can say from my own experience that according to the popular spl calculator tossed around here that it estimated an spl drop of 9dB but I only experienced a 7dB drop, and this was in a well treated, minimally reverberant room.

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I believe this law is mainly for a free space. The reflective boundaries probably caused the drop off to be less than the formula predicts since this is a geometric formula.

http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html
 

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