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Re: Horn stuff and a post back for Tom

Hi Tom!

You wrote:

>> What your saying is correct then that the speaker, especially a
>> point source direct radiator "moves around in time", it in effect
>> moves front to back depending on frequency.

and then I replied:

And, it is non-linear around system resonance too.

to which you said:

>> Not non linear but its acoustic phase is at its maximum rate of
>> change at resonance.

It's Non-linear. This is where it's shifting the most. Not a slow movement, but a rapid transition. Only after this point does it
become relatively linear.

It is not "non linear", even at resonance, it is acting as its R,L,C components dictate.
Non linear is when the Xmax is reached or if the BL^2 / Rdc changes with respect to radiator position.
Non linear is if the cone bends or deflects such that the volume velocity is different than a perfectly ridged cone depending on position.
Non linear is the portion of its output that is at harmonic frequencies and AT resonance is typically LOW in distortion (non linearity).
Non-linear is when what it does is different than the equivalent circuit predicts.

>> For a woofer, at resonance it's acoustic phase is zero degrees and
>> it goes to near -90 above. At the Rmin point in the impedance,
>> the acoustic phase has risen to zero again and then goes positive
>> above Rmin. By choosing ones crossover, one can intercept the
>> woofer's response with little change from -90 or at zero. In any
>> case one is stuck with what the direct radiating point source does.

One is stuck with it, yes.

Not only the woofers, but the midrange too. If the midrange diaphragm is in resonance in the woofer-midrange overlap
region, it suffers the same fate. And again, this is the case for the compression driver.

Horn loading is a way to make the radiation resistance dominate the moving system and in this case the resonance is damped to invisibility.
Acoustic phase of such a driver DOES NOT show a "box resonance" such as a direct radiator in a box.

You wrote:

>> A highly loaded horn is resistance dominated and so one finds that
>> by the time one has a 50% efficient horn, that the acoustic phase
>> is around zero degrees mid band. Over the region the acoustic
>> phase is zero degrees, the driver does not move in time with a
>> change in frequency.

to which I replied:

But the horn in the Unity is nowhere near as efficient as this,
wouldn't you agree? The midrange drivers aren't even loaded at
the apex; The horn is, as you say, a "lossy" horn or what I have
called a "mal-formed" horn. It does not meet the conditions of
the Webster equation.

and then you said:

>> No, I wouldn't agree, I have measured the acoustic power and
>> impedance, the mid system is about +2DB more efficient than the
>> compression driver. Also my horn simulator program predicts
>> results very close to what is measured with an efficiency around
>> 40%.

But midrange is the most efficient band of the system. For one thing, the horn is tuned for the midrange band and for another,
you have more midrange drivers than any other. If you aren't hitting 50% here, you certainly aren't hitting it anywhere else.
One of the sticking points I've had was that this device was claimed to provide loading from 200Hz to 20Khz, but that's not
possible. It hits a corner just above the midrange band, and must be equalized to compensate just like other comparibly sized
horns.

The efficiency of any horn loaded driver hinges on having the right quantities of motor strength and mass /area.
There is also some confusion here about what the size of the horn does.
One can in fact make the region below "mid" just as efficient as the mid, one can even make a large lf horn and make it as efficient as the mids.
You speak of the conical horn as if it were a "band pass" device which is simply wrong, ALL horns have a "high pass" corner frequency.
The higher "low pass" corner which forms the other half of the peak IS set by the driver and has NOTHING to do with the horn loading.

By the way, what are you using to model this system? What is the "horn simulator" program you are referring to?

I have 2 that work well, one is the NASA funded program written to model any driver on any acoustic passage way.
Each deals with what ever shape "horn" or passage way one described to it.
You can read about it if you go back 10 -12 years to the AES Journal devoted to modeling horns.

>> Loading at the apex is un necessary, one needs to be acoustically
>> close yes, but that is a function of frequency.

That brings us back to the question - How far are the midrange orifices from the compression tweeter diaphragm, at
the apex?

You wrote:

>> Flat amplitude AND zero degrees phase are the conditions needed to
>> preserve the waveshape of the input signal, no wonder so many like
>> horns.

and I replied:

This is true, and it is simultaneously impossible for any filter to accomplish.

then your response was:

>> That is right, only a constant power into resistive radiation load
>> without any significant reactance will do this.

So we are in agreement on this.

Yes and an efficient horn is one of the very few ways to achieve this situation.

You wrote:

>> In the Unity, I can't do anything about the phase response of the
>> woofer, one is stuck with that and in that case the closest one
>> can come is to match the time /phase at and around crossover.

I replied:

I see, yes. You make no attempt to time align the woofer subsystem, finding it impossible.

One is stuck with the inferior phase response of a direct radiating point source if one is using one, all of them are exactly the same in that regard.
The best one can do in that case is to make the transition between it and the higher frequency band seamless.
Once one is in the horns range, the problem is dramatically less as the drivers are resistively loaded not mass loaded.

And then you said:

>> Below crossover it does what all woofers do, at crossover it IS
>> aligned with the mid section of the horn. It is the drivers delay
>> and phase at crossover that one matches.

So what you are saying is that the woofer's time response is skewed, but up near the crossover point, you bring it into
alignment with the mids. Then you further match the mids to the tweeter as it nears the crossover point.

The woofers acoustic phase is what it is, at its low cutoff it is behind in time compared to its high cutoff, ALL direct radiator woofers act this
way.
When one chooses to use a direct radiator, one is stuck with how they work, when one uses a horn to cover that frequency range, one has a
much better acoustic phase but a larger system.
If one could accept a 36 inch square mouth, one could have horn loading down to about 120 Hz.

Putting aside, for a moment, the alignment problems in the overlap region, and putting aside the rapidly changing phase at
diaphragm resonance of the woofers, midranges and tweeters, are you proposing a system that has a slowly changing
apparent location? Or does it rapidly change in the crossover regions? And in either case, how do you propose to keep from
having problems arraying speakers having this kind of "doppler effect?"

Again, there is no problem in the overlap region IF one has the two systems at the correct time and phase.
While horn loaded, the effective position of ALL the drivers changes much less than if direct radiators and to the extent that the acoustic phase
can be made to be at zero degrees, there is NO change in position with frequency.
The rapidly changing phase of the woofer IS AT the BOTTOM end of its response and has nothing to do with the mid range.
So far as array performance, the td-1's re the most arrayble speaker on the market, NO ONE dare show polar plots of even two "arrayable"
boxes because while they sit together nicely, acoustically they clash terribly.
We on the other hand have a polar plot for 3 boxes side by side and it shows little interference and people like Pat Brown (the guy who teaches
Synaudcon since Don Davis retired) says "this is the way arrayable speakers should be made"

>> At 20 kHz, that "impedance matching" is done well before the sound
>> has reached the exit of the driver. 1WL in circumference is .21 inches!!.

No. Impedance matching requires that the flare is of wavelength scale, and in 0.2 inches, you have not even begun a flare. It's
still inside the compression driver, which means you are at the mercy of the motor builder. Whatever flare rate they chose at
20Khz is what you've got.

Wayne think about it.
At 30 Hz, a mouth 1WL in circumference (about 12 feet in diameter) is large enough so that the final radiation resistance is reached.
At 60 Hz, it is 6 feet in diameter.
At 20KHz, it is 1/5 inch in diameter.
In each case, continuing the horn past that point DOES NOT effect the radiation resistance significantly or the load on the radiator.
As the frequency goes up the active part of the horn gets smaller and smaller but the radiation resistance stays the same because it is the acoustic
dimension we are dealing with.

Your horn cannot do anything of scale smaller than one inch if you're using a 1" exit driver device. It looks like free space at
that frequency - Its walls are merely a reflector.

ALL horns of ALL types cannot "load" a radiator once past the point where the critical size is reached, for 20 kHz with the wavelength being
about 5/8 inch, this equals a diameter of about 1/5 inch at 30 HZ, this is about 12 feet in diameter.
Reflections if present would show up in ETC measurements as each represents a different path length and time.
Rather, since the 1 inch exit is too small to form a beam narrower than 90 degrees at 20K, loading into a 60 degree flare DOES NOT cause
reflections.
Once the critical loading size is reached, the horn walls do continue to have an effect, they govern the radiation pattern up to some still larger size
(as defined in the manta ray paper)


>> Even at 5 kHz, the passage way in the compression driver is large
>> enough to fully load the driver. From that view point, one could
>> say that ALL compression drivers do not load any horn above 5K
>> (1 exit.inch driver)

That's right.

>> On the other hand, what the air see's is in the compression driver
>> is a continuation of a conical horn down to a rather small throat
>> area. From that stand point, if you think of the horn as the
>> total air path, it does indeed fully load the driver and does so
>> all the way down to the low cutoff of the horn.

Again, you're at the mercy of the compression driver maker here. What happens in the top octave is not up to you. So you
cannot claim to horn load the device from 200Hz to 20Khz. But that's a dead horse - You've already admitted to using
compensation equalization to address this.

No, this is wrong, the "compensation" has nothing to do with acoustic load and everything to do with the moving mass of the compression driver.
This is why a driver with less moving mass (like a TAD 2001) when mounted to the SAME horn, shows a response easily an octave higher.
That HF roll off HAS NOTHING to do with horn loading, above the low cutoff, the load presented by the horn is essentially constant with
frequency.
That also includes the part of the horn which comes with the compression driver as the real throat is at the diaphragm.

I wrote:

Conical horns become peaky at their low end, this is true. They act
like a series of resonators combined with a high pass filter.

and you replied:

>> No, not peaky, it is the relationship of the low cutoff of the
>> horn and power roll off of the driver which produce that response
>> shape.

The low cutoff of a conical horn system of this scale is relatively peaky, as is shown by the plane wavefront model. It can be
smoothed with a larger mouth or by constrained space, such as using multiple horns in close proximity or installing against a
rigid surface or corner. But the lower cutoff point has a peak, and response after that has a negative slope.

No, this is wrong, peakyness (in a simple straight horn) is an indicator if mis termination at either end exactly the same as with curved wall horns.
Conical horns have a differently shaped radiation curve but all types reach the asymptotic load impedance eventually, conical horns do it more
slowly than curved wall horns and so have to be larger to get to the same low cutoff.
Look at the horn loading curves which show radiation impedance vs size, you will not see anything different at the high end of the response, only
at the low end.

>> If one could say "crank the magnet up to 40,000gauss", one would
>> see the compression drivers high frequency response flat octaves
>> higher.

Not unless the throat were made appropriately small. One cannot have horn loading from a single horn from 20Khz all the
way down to bass frequencies. At this scale, we wouldn't even be talking about throat distortion, because we would have to
solve issues caused by supersonic speeds of the air mass which would be forced to reciprocate in the hyper-constrained
throat.

The horn throat IS appropriately small already, pull off the bug screen and look inside a compression driver.
For a horn driver, velocity = sound level, "X" SPL at 5 KHz requires the same velocity as the same SPL at 20 kHz, there is increase in speed
with frequency. Throat distortion is a function of intensity and expansion rate.

I know that your response is to remove the lower frequency diaphragms from the throat, and to install them along the walls.
But the more we do this, the more we decrease the efficiency of the horn, because the drivers near the mouth are less
horn-loaded. Of course, we can forego the horn altogether but we are talking about horns here, after all.

Efficiency is determined by having the right acoustic and electrical impedance's, no one (except you) is saying that one has to drive the horn from
its end, one can driver it from the half way point (midway between throat and mouth) IF one uses a suitable driver and get the same efficiency
one could get driving it closer to the end.
At some point, the expansion at the apex is so rapid that a driver there is not coupled well to the rest of the horn, in that case moving forward is
the ONLY way to get lower frequencies out of it efficiently.

>> The mid drivers response is flat.

I know that you are basing this statement on your measurements. But before you measured the system, how had you "run the
numbers" on it? What kind of "proof of concept" modeling was done for the midrange and woofer in this configuration?

Like all designs I model it in the computer first, then cut wood and measure.
After I knew what I wanted, I used another program to specify the driver parameters for that size / frequency range horn and asked a speaker
company to make them for me.

There are no suitable horn models for the Unity configuration, so what was the basis?

Again, a commercial program like AKABAK will do this, no problem.
The programs I use are mostly in Mathcad.

>> All horns are a resonator(s), look at the impedance curve of an
>> exponential horn as you make it smaller and smaller.

That's right. That's why there's a peak at cutoff.

Again, a peak is a result of insufficient damping at one end or the other.

>> Flat response happens when the right amount of damping is at each
>> end, electromechanical at the throat and acoustic at the mouth
>> (where its 1WL incircumfrence)

That's true. If the mouth is large or the horn is constrained, then this peak is reduced. I like to build horns having a peak at
cutoff, tuned for just under the motor cabinet's cutoff. In this sense, it is used for extension rather than efficiency.

For normal cone drivers Marshal Leach's paper on horn loading gives superb results

>> Do you use a square horn around this size you could experiment with?

Yes.

I've worked with horns that are very similar to this for over 20 years. The ten π horn is such a device, and I've built them
in various sizes from just a few cubic feet to 60 cubic feet. So I have quite a bit of experience with


OK, if you have one to spare, mount a compression driver on it and measure the response.
Then, go get 4 sealed back drivers (there is a popular 5 1/4 inch which is close enough to work Fs should be ~500Hz).
Drill 8 holes at the corners of the horn where the interior is about 2 3/4 to 3 inches square.
Mount the drivers so that they communicate with the holes and measure the results.
Now, measure the compression driver again and judge the proper crossover by eye and give it a shot.

cheers,

Tom

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