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JBL 2404's instead of K77


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It will work in the K- horns, and sound better then the K-77. It will sound even better with a network designed for it.

When I was experimenting with many different midrange horns and drivers I used the ALK network and some Klipsch stock networks. I narrowed it down to the driver, horn and tweeter that I thought sounded the best. Then I sent the drivers and horns off to John Warren ( a Forum member) who has the right testing and measuring equipment to design a network for the components that I chose. I was floored with the difference in the sound that the network made. My midrange driver and horn sounded so much better then the stock Klipschorn stuff, and I thought the sound couldn't get any better. Then when I got the new networks the sound was again, much improved. John found a nasty peak or two in the horn and used notch filters to correct them.

John now sells a tweeter network add-on for the 2404H tweeter.

In my custom networks he provided 3 different settings for the JBL tweeter. I can change the output in 1db increments.

I can't say if it will work in your Cornwall or not.

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Don't make the same mistake everyone else does and

stare at those two peaks until your eyes cross. They are peaks, ignore

them. Look at the response from the 104dB perspective...

-- when you get right down to it we don't really listen to these speakers for their smooth FR.

It really takes a lot to justify that nasty frequency response, doesn't

it? The problem with those peaks is they are probably due to some

resonance in the design. I claim it is very audible though I'm waiting

for some ETF measurements to confirm what I hear. Nevertheless, the

traps in the networks are just going to reduce ("band-aid") the

amplitude of the peaks, but will have no effect on the duration of the

resonation.

Anyways, I would draw the 0 line at 100dB where it is +-5dB from

35-425Hz. But you start seeing cone-breakup beginning right around

300Hz (as indicated by the sharp peaks and dips).

Side note: it makes sense to see cone-breakup earlier on

the khorn than the cornwall because of the increased air load on the

driver.

Anyways, I agree that nobody would ever listen to the khorn (or heck,

any klipsch) for the flat FR response. But at the same time you

absolutely cannot ignore 15dB swings.[+o(]

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Q

Is there any chance you have the schematic that John put together for you. I remember your thread a few years ago when you were putting all this together.

Remind me....what is the setup....Khorn with K33, Altec 311 with 291's (which 291's?), JBL 2404's?

jc

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It really takes a lot to justify that nasty frequency response, doesn't it?

It's the out of network raw response, without smoothing applied. How many raw response curves have you seen -- they all look like crap.:)

The problem with those peaks is they are probably due to some resonance in the design.

Maybe, maybe not -- but there it is.

I claim it is very audible...

Of course it is.

...though I'm waiting for some ETF measurements to confirm what I hear.

You have Klipschorns you can do this with?

Nevertheless, the traps in the networks are just going to reduce ("band-aid") the amplitude of the peaks, but will have no effect on the duration of the resonation.

Take it up with Delgado when you go to Hope.:)

Anyways, I would draw the 0 line at 100dB where it is +-5dB from 35-425Hz.

You can do that if you want, but it doesn't really matter. The response can be adjusted with an EQ or by modding the filter.

But you start seeing cone-breakup beginning right around 300Hz (as indicated by the sharp peaks and dips).

If that's the way you see it. Still, I see it as speculation, and though speculation is always interesting, it doesn't get us anywhere.

"...it makes sense to see cone-breakup earlier on the khorn than the cornwall because of the increased air load on the driver."

PK would have walked you to the guillotine. You need to think that one through.

Anyways, I agree that nobody would ever listen to the khorn (or heck, any klipsch) for the flat FR response. But at the same time you absolutely cannot ignore 15dB swings.

The 15dB swings are a product of your imagination. I told you to ignore the peaks.:) Search the PK patents and study them.

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It really takes a lot to justify that nasty frequency response, doesn't it?

It's the out of network raw response, without smoothing applied. How many raw response curves have you seen -- they all look like crap.:)

A lot more than just this one [:P]

(and they don't all look like crap, but they don't all cost the same either)

...though I'm waiting for some ETF measurements to confirm what I hear.

You have Klipschorns you can do this with?

I don't have the ETF either [:P]

But once I get a setup happening I'm sure I can find someone willing to let me measure

Nevertheless, the traps in the networks are just going to reduce ("band-aid") the amplitude of the peaks, but will have no effect on the duration of the resonation.

Take it up with Delgado when you go to Hope.:)

I'd rather see Heyser take it up with Delgado [;)]

But you start seeing cone-breakup beginning right around 300Hz (as indicated by the sharp peaks and dips).

If that's the way you see it. Still, I see it as speculation, and though speculation is always interesting, it doesn't get us anywhere.

"...it makes sense to see cone-breakup earlier on the khorn than the cornwall because of the increased air load on the driver."

PK would have walked you to the guillotine. You need to think that one through.

We just discussed the issue in acoustics - even got pretty formulas to model it with. Hornloading increases the load on the driver (thus why the impedance increases when a lense is mounted to the driver). Unless I've got my definitions mixed up cone-breakup happens when the VC moves forward but the entire diaphragm doesn't, resulting in a tangential wave down every radius. Since there is more surface area of the driver further away from where the VC connects to the diapghragm there is more 'resistance' on the edges which causes the middle to move forward when the outside doesn't.

Btw, PWK would walk anyone to the guillotine who didn't like his khorn or even considered modding it. I've heard some pretty crazy stories...

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In the Cornwall, the K-33 is at 6 ohms with the network attached. In the Klipschorn it's not much higher.

If I decide to move water using a piston, not much work will get done unless I use a cylinder of some sort. Horn loading a driver works pretty much the same way I think. You put a horn on a driver, and more work can be done with much less movement of the piston, or driver. So, there is higher efficiency, and since there is less movement of the "piston" and attached diaphragm or cone for a given amount of work compared to a direct radiator, distortion is lower. I also think the driver is damped better. Now, I know you know all this, so how did you come to the conclusion that a driver that is moving less and under better control has more cone break-up than the same driver working four times harder to accomplish the same amount of work?

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JC,

I think I have the shematic around here somewhere. I hope so, because I want JW to make me another pair. I'm not sure if it will be of any use to you. The network is the AA, but it's modified to work with the Altec 290-16K driver and 2404H tweeter. The 290 worked with a stock AA, but the impedance was too far off. DJK kept telling me that I was nuts to think that I could just switch drivers & he was right. That's why everything was shipped to JW. Like I said before, the matched network makes a world of difference.

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Looks like John removed all of the data from his website for the 2404. What a shame.

Q-man, in all liklihood you are probably still using an AA Type network, though modified just a bit. I'm sure JW used his 2404 tweeter filter, added a single coil in the midrange circuit to roll off the response of the Altec, and adjusted the primary cap value for a 300Hz crossover. So, probably 13uF to 16uF, .40mH in the midrange, and then his 2404 tweeter filter.

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Dean,

You are right about the network, except the crossover point is still just below 400Hz. I wanted him to make it at 300Hz like the horn and driver, but he insisted that it had to be higher. Something about having to stay a whole octave above Fs. of the horn and driver to keep distortion low. That is why I don't believe in using the Altec 511B horn. It is designed to be crossed over at 500 Hz or above.

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so how did you come to the conclusion that a driver that is moving less and under better control has more cone break-up than the same driver working four times harder to accomplish the same amount of work?

Because it's not really a function of power output, but a function of frequency and acoustical impedances. For example, a driver that starts breaking up at 3kHz is 'always' going to break up at 3kHz. The louder you play the driver, the worse the break up gets, but it still starts at the same frequency.

In your water analogy, it is simply harder for the piston to move in the cylinder. The relative efficiencies and all that are irrelevant for the power independant effects.

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LOL, I never did get the "new math".

So, a driver in a box that's reaching the limit of its exursion has less cone breakup than the same driver loaded by a horn at the same SPL. Am I hearing you right? I thought "Distortion is inversely proportional to efficiency."?

In your water analogy, it is simply harder for the piston to move in the cylinder.

So what? Set up another piston next to that one and see which one has to "work harder" to move the same amount of water.

The relative efficiencies and all that are irrelevant for the power independant effects.

In the context of this discussion, efficiency and power are directly related to distortion levels -- whether it be cone breakup, IMD, THD, or whatever other kind of distortion you want to throw into the mix. The fact that impedance is higher is what is irrelevant, since with climbing impedance the need for current goes down.

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so how did you come to the conclusion that a driver that is moving less and under better control has more cone break-up than the same driver working four times harder to accomplish the same amount of work?

Because it's not really a function of power output, but a function of frequency and acoustical impedances. For example, a driver that starts breaking up at 3kHz is 'always' going to break up at 3kHz. The louder you play the driver, the worse the break up gets, but it still starts at the same frequency.

In your water analogy, it is simply harder for the piston to move in the cylinder. The relative efficiencies and all that are irrelevant for the power independant effects.

I am afraid both generalizations are slightly off the mark, hence neither will prove his point. Dean is on the right track here. Dr. Who, I wonder where you are learning about these issues? I think as you learn more you will see what Dean is getting at. Focus on the function of a horn, why is it used? what does it do? Compare that with the driver used as a direct radiator. The relationship of these two issues is what leads you to understand Dean's point.

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Q

I would still like to have it if you find it. The way Dean mentioned what John might of done would be a simple task.

I just might get a wild hair up my "you know what" and goof around with a 311 horn. I just need to educate myself a little more on all the drivers by Altec or others that will go down to 300Hz.

You don't have a list of possible drivers or know em off the top of your head ...do you?

jc

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It isn't really true that power plays no roll in this. Cone breakup is when a cone isn't moving uniformly as a piston. In a nutshell when does that occur? When the forces acting on the cone are basically greater then the stiffness/strength of the cone itself. All else being equal a cone moving 0.1 inch P-P is going to be stressed less then a cone moving 4" P-P. Even if you forget about air loads it is obvious more force needs to be exerted on the cone to move the cone (and its spider and suspension) 2" from center compared to 0.05 inches from center. Adding in the air load the further the *same* woofer in the *same* cabinet has to move the more power is being transfered from the cone to the air(playing louder). Can't get something for nothing... if in the *same* speaker more power is going into the air that means more power has gone into the woofer and more force has been applied by the woofer. Pushing 4" of air is going to require more force then pushing .1" of air.

Now having said that... the load (impedance) presented by a horn does increase the forces acting on a cone. If the cone was traveling 1" in a Cornwall the same woofer in a K'Horn also traveling 1" would have more forces acting on the cone from the air load in the horn acting upon it. That is the point of a horn... it in effect increases the resistance (impedance) of the air so that the woofer can better couple to the air and therefor more efficiently transfer power to the air. The reality though is a woofer moving 1" in a KHorn would be considerably louder then a woofer moving 1" in a Cornwall. So for the same SPL the K'Horns woofer will be moving less. And therefor would be less stressed from its suspension,spider and such.

As to which would have less cone breakup problems? I don't know. Maybe Klipsch has done some comparisons with this.

I'll add an additional kink to the discussion too. What frequencies are generated by the cone breakup could effect which speaker it would be more audible in. For example if the cone breakup in a K'Horn was generating noise from say 3kHz-6kHz it is possible that would be pretty inaudible, but on a Cornwall it would be more so.

Why? The bass horn is in a very real sense an acoustic low pass filter.

Anyhow, that is my take on it. But I haven't really ever spent a lot of time looking into cone breakup. This is one of those things they test for with "Laser beams" (said in Dr. Evil voice) bouncing off the cone to see when it starts really scattering the beam.

Shawn

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Ya know, this discussion will go nowhere if we aren't all working off the same assumptions...

So, a driver in a box that's reaching the limit of its exursion has less cone breakup than the same driver loaded by a horn at the same SPL. Am I hearing you right?

Why do you keep trying to match the SPL and why the heck are you talking about limits of excursion? Why not compare the drivers for the same excursion? (aka, same drive force on the cone)? And why not discuss the linear behavior of the driver? (especially when the non-linear behavior gets swamped by other factors.)

In the context of this discussion, efficiency and power are directly related to distortion levels -- whether it be cone breakup, IMD, THD, or whatever other kind of distortion you want to throw into the mix. The fact that impedance is higher is what is irrelevant, since with climbing impedance the need for current goes down.

But the frequencies at which those distortions occur is irrelevant of the power (mostly irrelevant anyway). The power/efficiency/watever else mostly just affects the magnitude...the magnitude of the same frequencies. I dunno how to say it more clearly.

Now having said that... the load (impedance) presented by a horn does increase the forces acting on a cone. If the cone was traveling 1" in a Cornwall the same woofer in a K'Horn also traveling 1" would have more forces acting on the cone from the air load in the horn acting upon it. That is the point of a horn... it in effect increases the resistance (impedance) of the air so that the woofer can better couple to the air and therefor more efficiently transfer power to the air. The reality though is a woofer moving 1" in a KHorn would be considerably louder then a woofer moving 1" in a Cornwall. So for the same SPL the K'Horns woofer will be moving less. And therefor would be less stressed from its suspension,spider and such.

This is exactly what I was getting at: comparing the systems at the same excursion / power input...basically the same force on the cone. In such a situation the cone-breakup on the hornloaded driver shifts lower in frequency. I'm kinda busy right now with school and all that, but the derivation isn't 'too hard' (prob three or four pages).

Nevertheless, I wasn't trying to claim that non-hornloaded ever results in less distortion. I was merely commenting on the frequencies at which distortions were taking place (because upon first reading the plot of the khorn bass bin I thought to myself - wow, cone-breakup...and then had to rationalize in my head why that's not occuring in the cornwall).

I was trying to think of an experiment to demonstrate the effects and the best thing I can come up with is to take a cereal box and cut out a circle. Mount a dowel rod to the middle of the circle and push it forward in the air (like a speaker). Notice how the cardboard doesn't flex until you start moving it faster. Now take your contraption and put it in water (where the acoustical impedance is much much larger - just like what the horn does). Notice how you don't have to move it as fast to create the same amount of flex.

In the end it is ultimately a function of the cone velocity relative to the imbalanced forces. Though magnitude marginally affects velocity to an extent, the frequency at which the system vibrates has a much larger effect. Thus why I said "not really" a function of amplitude (because it doesn't change what is effectively the frequency of breakup...especially in the low-signal level world).

And talking about these systems in the large-signal world is a bit silly because the issues get swamped by other non-linear factors. (there's about 3 dozen of them...surely nobody wants to start discussing all of them).

Anyways, I don't claim to be an expert in acoustics (yet), but the source of my information is coming from grad level acoustics courses at UIUC and this topic is covered both in my textbook ("Fundamentals of Acoustics - 4th ed." by Kinsler, Frey, Coppens, and Sanders) and the class I just took taught by William O'Brien (http://www.brl.uiuc.edu/People/william_d_obrien_jr.php).

I thought "Distortion is the inverse of efficiency."?

Depends on the assumptions you start with actually...it's not always true. What's more important is to worry about the audible distortions.

What frequencies are generated by the cone breakup could effect which speaker it would be more audible in. For example if the cone breakup in a K'Horn was generating noise from say 3kHz-6kHz it is possible that would be pretty inaudible, but on a Cornwall it would be more so.

That's my take on it too actually...it's a pretty fundamental rule of speaker building actually - choose a passband for the system where cone-breakup isn't occuring. And to tie all this back into where it started...the khorn bass bin "sucks" above 300Hz...but I suppose you could say it "blows" above 300Hz too...[;)]

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That is why I don't believe in using the Altec 511B horn. It is designed to be crossed over at 500 Hz or above.

Aren't you using a 511B in your custom Khorn center channel?

I use a 311-90 and it's a 300HZ horn. I try to most a picture.

post-2405-13819297896594_thumb.jpg

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Your original comment read: "...it makes sense to see cone-breakup earlier on the khorn than the cornwall because of the increased air load on the driver."

I got stuck on, "...cone-breakup earlier on the khorn than the cornwall..."

At that point my brain locked up. I should have asked what you meant by "earlier". I took it to mean at a lower power level. Turns out you meant "earlier" in the sense of lower in the driver's FR. I probably wouldn't have disputed that, but then would've wanted to know if you've actually seen a response curve for the K-33 loaded in the Cornwall.

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That is why I don't believe in using the Altec 511B horn. It is designed to be crossed over at 500 Hz or above.

Aren't you using a 511B in your custom Khorn center channel?

I use a 311-90 and it's a 300HZ horn. I try to most a picture.

Yeah, that's a pretty substantial chunk of aluminum alright! [;)]

BTW, was that pic taken before or afte you did the custom center channel in your avatar?

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