Indeed it looks pretty clean and lots of loading 
I always feel a bit weird about discarding the vertical directivity as if ceiling/floor reflection do not contribute.
However if you want even coverage at the horizontal axis - one can sacrifice the vertical to compensate for the loss of acoustic impedance.
I always feel a bit weird about discarding the vertical directivity as if ceiling/floor reflection do not contribute.
However if you want even coverage at the horizontal axis - one can sacrifice the vertical to compensate for the loss of acoustic impedance.
To me, higher "direct to reverberant ratio" is always desirable, but there are limits to what is achievable in a practical sense.
The highest "direct to reverberant ratio" would be no reverberation at all. I don't think that would sound natural.
Perhaps what Earl means is that the practical limits are still far from that, even with highly directional loudspeakers (if the room is left reflective to a high degree).
Perhaps what Earl means is that the practical limits are still far from that, even with highly directional loudspeakers (if the room is left reflective to a high degree).
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Here's a comparison of three different BIG waveguides (two of which were shown in the above posts), including a simple 2" driver LE model, i.e. showing also actual SPL responses and excursions:
The dashed line in the bottom-right graph shows the required (peak) excursion for 110 dB@1m on axis.
This is the driver model used, I just changed the numbers from a 1" model...
The dashed line in the bottom-right graph shows the required (peak) excursion for 110 dB@1m on axis.
This is the driver model used, I just changed the numbers from a 1" model...
Code:
Def_Driver 'Drv1'
dD=100mm
Mms=8g
Cms=50e-6m/N
Rms=3.0Ns/m
Bl=10Tm
Re=6.3ohm
fre=35kHz ExpoRe=1
Le=0.1mH ExpoLe=0.618
System 'S1'
Driver 'D1' Def='Drv1' Node=1=0=10=20
// Rear volume
Enclosure 'Eb' Node=20
Vb=100cm3 Qb/fo=0.1
// Front volume
Duct 'D1' Node=10=200
dD=100mm Len=0.3mm
// Phase plug (simplification)
Waveguide 'W1' Node=200=300
dTh=18mm dMo=40mm Len=35mm Conical
// Conical section between phase plug and exit
Waveguide 'W2' Node=300=400
dTh=40mm dMo=50mm Len=25mm Conical
RadImp 'Throat' Node=400 DrvGroup=1001
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Maybe I missed it: but is there a way to give an LF source (or both above AND below) also a horn contour? To achieve something like this:
https://www.limmerhorns.de/s/cc_images/teaserbox_1668178.jpg?t=1587496198
https://www.limmerhorns.de/s/cc_images/teaserbox_1668178.jpg?t=1587496198
This explanation makes more sense imo:
"Sound power is the total airborne sound energy radiated by a sound source per unit of time. Sound pressure, on the other hand, is the result of sound sources radiating sound energy that is transferred into a specific acoustical environment and measured at a specific location. Sound power is the cause, and sound pressure is the effect.
Sound power can be determined from sound pressure or sound intensity measurements.
There are two methods for determining sound power from sound pressure measurements, the direct method, and the comparison method. The direct method is most commonly used in essentially free-field environments, but it can also be used in a reverberant sound field. The comparison method, however, is only used in reverberant sound fields."
I really don't understand what can be more simple than this:
The power response curve is a frequency response of an omnidirectional source radiating the same total power as the device under test.
It is calculated as a power average (rms) of the pressures spatially sampled on a imaginary surface enveloping the device.
The power response curve is a frequency response of an omnidirectional source radiating the same total power as the device under test.
It is calculated as a power average (rms) of the pressures spatially sampled on a imaginary surface enveloping the device.
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BTW, here's a "fixed" calculation of the pressure weights, to give the expected values even for an infinite baffle.
So obviously, it's not true what David McBean claimed. Just to be clear on that.
The DI is equal to the difference between the pressure response and the power response in all cases.
So obviously, it's not true what David McBean claimed. Just to be clear on that.
The DI is equal to the difference between the pressure response and the power response in all cases.
Code:
// C++ code for the weights calculation
double cos_a0, cos_a1 = cos(a[0]);
const double cos_a2 = cos(a[n-1]);
//const double St = 1.0 / (cos_a1 - cos_a2);
const double St = 0.5; // <-- FIXED, ALWAYS TAKES 4PI SOLID ANGLE AS A REFERENCE
for(int i = 0; i < n - 1; ++i) {
cos_a0 = cos_a1;
cos_a1 = cos(0.5*(a[i] + a[i+1]));
w[i] = (cos_a0 - cos_a1)*St;
}
w[n - 1] = (cos_a1 - cos_a2)*St;
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An explanation of the fix:
In the previous version I considered the term "omnidirectional" always with respect to the actual radiation conditions, i.e. for half-space radiation I considered "omnidirectional" meaning to radiate equally in "full" half-space, not the real full (4pi) space. This was not in accordance with generally accepted definitions, as it turned out. So now this is fixed and the term omnidirectional in the power response description really means a true omindirectional (4pi) source.
In the previous version I considered the term "omnidirectional" always with respect to the actual radiation conditions, i.e. for half-space radiation I considered "omnidirectional" meaning to radiate equally in "full" half-space, not the real full (4pi) space. This was not in accordance with generally accepted definitions, as it turned out. So now this is fixed and the term omnidirectional in the power response description really means a true omindirectional (4pi) source.
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Physically the Sound Power of a source is the total energy radiated by that source, it does not depend on distance/environment (provided reasonable conditions). If memory serves a grand piano can output close to 1W acoustic, that is seriously loud in a normal size room....
It is traditionally measure in reverberant rooms, Genelec might have published these measurements in their specifications.
For the CEA2034 here is the calculation of the Sound Power taken form what I have done here:
https://www.audiosciencereview.com/...-cea2034-from-the-nfs-data.14141/#post-985452
It is the barycenter (weighted average) of pressure measurements acquired on two orbits (Vertical and Horizontal) around the speaker. It is calculated so that it looks like a FR curve but the true unit is watt.
I think one could see it as the equivalent of the DUT FR if it were converted into a truly omnidirectional source.
It is traditionally measure in reverberant rooms, Genelec might have published these measurements in their specifications.
For the CEA2034 here is the calculation of the Sound Power taken form what I have done here:
https://www.audiosciencereview.com/...-cea2034-from-the-nfs-data.14141/#post-985452
It is the barycenter (weighted average) of pressure measurements acquired on two orbits (Vertical and Horizontal) around the speaker. It is calculated so that it looks like a FR curve but the true unit is watt.
I think one could see it as the equivalent of the DUT FR if it were converted into a truly omnidirectional source.
Code:
% ------------------------------------------------------------------------------
% SP: Sound Power calculation
% weights W are applied to vertical V and Horizontal H
% Energy average of p^2, sum(W*(P/Pref)^2)/sum(w) or sum( W*P^2 )/ sum(w) / Pref^2
% ONp2 = 10 .^ (ON/10)*Pref^2;
% ------------------------------------------------------------------------------
W = [
0.0006044860 % 00 & 180deg - 2
0.0047301890 % +/-10 & +/-170deg V and H - 8
0.0089550270 % +/-20 & +/-160deg V and H - 8
0.0123873540 % +/-30 & +/-150deg V and H - 8
0.0149896110 % +/-40 & +/-140deg V and H - 8
0.0168681540 % +/-50 & +/-130deg V and H - 8
0.0181659620 % +/-60 & +/-120deg V and H - 8
0.0190067440 % +/-70 & +/-110deg V and H - 8
0.0194777870 % +/-80 & +/-100deg V and H - 8
0.0196293730 % +/-90deg V and H - 4
];
SP_calc = 10*log10( (W(1)*ONp2 + ... % Axis (only once!)
W(2)*H10degp2 + W(2)*Hneg10degp2 + W(3)*H20degp2 + W(3)*Hneg20degp2 + W(4)*H30degp2 + W(4)*Hneg30degp2 + ...
W(5)*H40degp2 + W(5)*Hneg40degp2 + W(6)*H50degp2 + W(6)*Hneg50degp2 + W(7)*H60degp2 + W(7)*Hneg60degp2 + ...
W(8)*H70degp2 + W(8)*Hneg70degp2 + W(9)*H80degp2 + W(9)*Hneg80degp2 + W(10)*H90degp2 + W(10)*Hneg90degp2 + ...
W(9)*H100degp2 + W(9)*Hneg100degp2 + W(8)*H110degp2 + W(8)*Hneg110degp2 + W(7)*H120degp2 + W(7)*Hneg120degp2 + ...
W(6)*H130degp2 + W(6)*Hneg130degp2 + W(5)*H140degp2 + W(5)*Hneg140degp2 + W(4)*H150degp2 + W(4)*Hneg150degp2 + ...
W(3)*H160degp2 + W(3)*Hneg160degp2 + W(2)*H170degp2 + W(2)*Hneg170degp2 + ... % Horizontal
W(1)*H180degp2 + ... % Back (only once!)
W(2)*V10degp2 + W(2)*Vneg10degp2 + W(3)*V20degp2 + W(3)*Vneg20degp2 + W(4)*V30degp2 + W(4)*Vneg30degp2 + ...
W(5)*V40degp2 + W(5)*Vneg40degp2 + W(6)*V50degp2 + W(6)*Vneg50degp2 + W(7)*V60degp2 + W(7)*Vneg60degp2 + ...
W(8)*V70degp2 + W(8)*Vneg70degp2 + W(9)*V80degp2 + W(9)*Vneg80degp2 + W(10)*V90degp2 + W(10)*Vneg90degp2 + ...
W(9)*V100degp2 + W(9)*Vneg100degp2 + W(8)*V110degp2 + W(8)*Vneg110degp2 + W(7)*V120degp2 + W(7)*Vneg120degp2 + ...
W(6)*V130degp2 + W(6)*Vneg130degp2 + W(5)*V140degp2 + W(5)*Vneg140degp2 + W(4)*V150degp2 + W(4)*Vneg150degp2 + ...
W(3)*V160degp2 + W(3)*Vneg160degp2 + W(2)*V170degp2 + W(2)*Vneg170degp2) / ... % Vertical
( 2*W(1) + 8*W(2) + 8*W(3) + 8*W(4) + 8*W(5) + 8*W(6) + 8*W(7) + 8*W(8) + 8*W(9) + 4*W(10) )/ Pref^2); % Sum weightsThe unit of power is indeed watt but the unit of a power response as shown e.g. in Ath graphs, spinoramas and elsewhere is certainly dB SPL, i.e. a pressure unit, not power. It is an average pressure after all - square root of sum of squares, that's still the same unit (that's a rms average). It would be watt if you didn't make the square root, which converts it back to (an equivalent) pressure.
- Beware that there's a hidden square root in the formulas: log(sqrt(x)) = log(x^0.5) = 0.5log(x), i.e.:
SPL_power_response = 20*log(sqrt(radiated_power)) = 10*log(radiated_power) [dB SPL]
(The weights are needed only because we don't measure/sample the pressure in a regular grid across the whole sphere, in which case it would be just a mere rms average, without any weights.)
- Beware that there's a hidden square root in the formulas: log(sqrt(x)) = log(x^0.5) = 0.5log(x), i.e.:
SPL_power_response = 20*log(sqrt(radiated_power)) = 10*log(radiated_power) [dB SPL]
For sure! And that's what is really is. I can't see why people still have problems accepting this. It's so intuitive.
(The weights are needed only because we don't measure/sample the pressure in a regular grid across the whole sphere, in which case it would be just a mere rms average, without any weights.)
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So this is your own calculation, and the unit of your "SP_calc" is actually dB SPL (as explained above), not watt.
It's an equivalent pressure, corresponding to that total radiated power, for which there's an intuitive notion - it's a pressure generated by an omnidirectional source radiating that total power.
(For the nitpickers - when I say "pressure" I mean either a pressure (Pa) or SPL of that (dB re 20uPa), that's really a context-dependent detail that doesn't make any difference to what I'm saying.)
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