That's right Hans.
By the way, I'm amazed that Joe is allowed to post so much nonsense without getting a reprimand.
Stein
The forum rule #3 in not allowed category, "Intentionally posting incorrect or flagrantly dangerous information". If he believes that it's a correct information, then it's not intentional so he gets away, I suppose.
I have done this and the measurements don’t match the theory. Do you realize how far from optimum a 90 dB per watt driver actually is?
Do you know what SPL at 1M an actual acoustic watt will produce?
Euhhh, no. The power from the amp is still the same, but each driver now gets only 1/4 of that power.
If this would work as you say, you could blast the globe with 1mW if you only took enough drivers.
QED.
Jan
Jan,
The amplifier may be putting out the same voltage, however the load impedance drops and the amplifier produces more current. Thus the power to the loudspeakers is increased.
Just for fun hook up two drivers to an impedance meter and see how it changes if you use different arrangements of the drivers. Try face to face, next to each other line array style and for the interesting bit both face down on a bench top.
If you really want interesting try and measure the impedance of a loudspeakers horn loaded compression driver outdoors.
The amplifier may be putting out the same voltage, however the load impedance drops and the amplifier produces more current. Thus the power to the loudspeakers is increased.
Just for fun hook up two drivers to an impedance meter and see how it changes if you use different arrangements of the drivers. Try face to face, next to each other line array style and for the interesting bit both face down on a bench top.
If you really want interesting try and measure the impedance of a loudspeakers horn loaded compression driver outdoors.
Four identical drivers at distances small compared to a wavelength have four times the motor and four times the surface area, so shouldn't be surprising that they make a composite driver with four times the sensitivity.
To think about the comparative excursion to make, say, 90dB SPL, it's probably easiest to consider how much swept volume is needed to make that SPL at farfield, assuming as given point source simplification. Four times the area means 1/4 the excursion for the same swept volume.
All good fortune,
Chris
To think about the comparative excursion to make, say, 90dB SPL, it's probably easiest to consider how much swept volume is needed to make that SPL at farfield, assuming as given point source simplification. Four times the area means 1/4 the excursion for the same swept volume.
All good fortune,
Chris
When drivers are acoustically coupled to each other, isn't that reflected in their net electrical impedance (given that they are bidirectional transducers)? Seems like that's what Ed was getting at?
Yes, but that's not the simplified point source model we were given. Anyway, regular old speakers are so inefficient that they make poor microphones. Not none, but not much.
All good fortune,
Chris
All good fortune,
Chris
We have an expression down here:
"Hans, you are as mad as a cut snake."
You can't weasel out of this one.
Go back and read:
Did I mention 3.18us and 50,048Hz?
Did I mention Neumann?
Did I not mention Hagerman's RIAA incorporated that 'F4' pole in suibsequent posts?
You know that I did.
You are calling me a liar and yet it is you that is lying.
Instead you should aplogise and you would have walked back.
Try something like this:
"Hi Joe, I think I might have goofed, I did realise what you were saying because I missed out on that time constant you mentioned and that would have meant the response was flat."
I would then have graciously replied:
"Don't worry Hans, we all goof sometimes, but that 17.000V was a good one. Let's just laugh it off as one of those things. Havagoodone!"
That's how real men handle things. Both end up with their self-esteem intact and respect the other person, even they goof.
Sorry Jan, but you have not taken into adjusting for the 'target SPL' of 90dB.
May I please ask that you reread it?
At 96dB it will, as I have stated, be 1 Watt divided by four - that is fully understood. Each driver will see 250mW, absolutely no argument about that. But when we wind it back, to 90dB SPL from 96dB, then it will be 62.5mW per driver. Compare this to a single driver producing 90dB SPL.
So now we get the same 90dB SPL with 250mW with four drivers.
With a single driver, we needed a full 1 Watt to produce 90dB SPL.
So Jan, I would now ask the question of you:
This is about excursions. If the single driver produces an 8mm peak-to-peak excursion with 1 Watt input, how much excursion will there be at 62.5mW - that is essentially the question.
Could you please do the calculation, I would really appreciate it.
Just a few comments re the "what will be the excursion in mm" question.
Wavelength was mentioned. Absolutely.
I could say that if the four drivers occupy the same space, it would be exactly +6dB, no question about it. Can we agree on that?
But in reality the drivers cannot do that, hence wavelength does indeed become an issue. But at LF this is not an issue.
Think of our drivers mounted vertically as a line source and reasonable close to each other. Right in front of those drivers
Tweeters need to be the same way, but it also works when they are in close proximity to each other. It's actually quite straight forward and the +6dB is very real. I have done this more times than I can count, hence I feel confident saying this.
Wavelength was mentioned. Absolutely.
I could say that if the four drivers occupy the same space, it would be exactly +6dB, no question about it. Can we agree on that?
But in reality the drivers cannot do that, hence wavelength does indeed become an issue. But at LF this is not an issue.
Think of our drivers mounted vertically as a line source and reasonable close to each other. Right in front of those drivers
Tweeters need to be the same way, but it also works when they are in close proximity to each other. It's actually quite straight forward and the +6dB is very real. I have done this more times than I can count, hence I feel confident saying this.
I think that you just fabricated a fabrication. 😀😀😀
Bill, are you taking your meds? 😉
I know that I am. 😉
The Neumann pole, also called F4, is a practical realisation of the limitations of the cutter. Take a look below - I didn't write any of it, so if it is a fabrication... not, it wasn't me and I am happy to give the credit to whom it belongs.
I quote above"
"The new 3dB point, according to a Neumann cutting amp manual, is set at 3.18uS - which equates to 50.048Hz..."
Note that is a Lipshitz graph shown and look where is says 'practical' - so he knew about it as well.
I never fabricated it - you fabricated a fabrication. 😀😀😀
PS: Sorry for answering in lumps of posts, it is largely a case of the different time zones, you guys are very active while I am sleeping. 😀
Attachments
Last edited:
This is called "double dipping". Remember that no matter how you slice and dice it, to produce a particular SPL at farfield (meaning that all parts of all radiating surfaces contribute linearly to the farfield SPL) from any driver or composite driver, that SPL requires a certain swept volume displacement.
With four times the radiating surface area, each radiator must move 1/4 the distance to displace the same volume of air. It is independent of sensitivity.
All good fortune,
Chris
Last edited:
I don't think you have done this measurement at all, otherwise, you would not have come to that conclusion. I have done it, many many times.
And please define optimum because I have no idea of what you are talking about?
Who mentioned acoustic watts?
The average sensitivity of modern HiFi speakers, I am told, is circa 87dB or 0.25%, so to produce an acoustic watt, you would need an amp capable of 4000 Watt.
Do you think we have any of those amplifiers lying around? Any of us?
Thank you Chris. I wish you well, as well.
I actually agree with your statement. But this is about a target SPL.
Please re-read it again.
Nobody is arguing that power, or current, or whatever, is not evenly shared between the drivers. It is!
This is about Purifi and their preaching (and rightly so) about the evils (relative sense) of excursions in drivers. The driver goes in, the force factor goes up, the driver goes out, the force factor goes down. This creates very audible IMD distortions, when LF and midrange is mixed - AM distortion. They claim this is worse (and they supply samples) than FM (Doppler) distortion.
So when we take into account that the target SPL is 90dB, then we go from a single driver needed a full 1 Watt to produce the target SPL. But, when four drivers are employed on the same baffle in close proximity, then 1 Watt will produce 96dB SPL. Wind it back to our target 90dB SPL from 96dB, we now only need 1/4 Watt, 250mW.
Now each driver only sees 62.5mW each.
What is the excursion?
Purifi drivers are great, but maybe we can get almost as good a result if we use four drivers to overcome the problem?
That's the nature of the question I have posed here.
Cheers, Joe
This reminds me of a Gilbert and Sullivan, "A paradox, a paradox, a most ingenious paradox".
Farfield SPL of x requires a nearfield volume excursion of y, independent of radiating area and of sensitivity. But I can't see the flaw in your argument. Either you've double dipped or I've passed a brain fart. I'm open to either possibility.
All good fortune,
Chris
Farfield SPL of x requires a nearfield volume excursion of y, independent of radiating area and of sensitivity. But I can't see the flaw in your argument. Either you've double dipped or I've passed a brain fart. I'm open to either possibility.
All good fortune,
Chris
Chris, try think it through.
But apart from anything else, what would be the reduction in an excursion when comparing 1 Watt to 62.5mW - where the excursion is peak-to-peak 8mm, what would 62.5mW reduce it to in mm?
I would be fascinated by your answer.
Joe
But apart from anything else, what would be the reduction in an excursion when comparing 1 Watt to 62.5mW - where the excursion is peak-to-peak 8mm, what would 62.5mW reduce it to in mm?
I would be fascinated by your answer.
Joe
