catapult said:
It seems... last night i dug out my Benson book... it is in the libary now, we;ll see if he corroborates.
dave
Putting the whole thing more simply, half the power gives 0.7 (squareroot 0.5) the cone movement. Two drivers, each receiving 0.5 the power and moving 0.7 as far, move 1.4 times the air.
catapult said:
True, but it can also be looked at as 1/2 the power = 3 dB less output per driver x 2 drivers = the same output.
dave
Another driver with the same SPL adds 6dB as long as they are close enough together to couple acoustically. So it's -3 + 6 = +3dB.
dB get confusing. Just look at the amount of air moved. If you're moving more air it has to be louder.
catapult said:
We know that isn't true... if in parallel, maybe 3 dB from the amp maybe putting out 2 W instead of 1 + the 3 dB we are discussing right now.
The 6 dB thing is just bad "common knowledge"
dave
What do you mean WE? 😉 I've done the measurements.
Simple physics backed up by measurements. No folklore involved and there's nothing "bad" about it. Dave, you really need to do some experiments. Measure SPL and power (volts * amps) with different driver combos.
OK, I'm done. Horses, water.....
catapult said:
We are talking efficiency here not sensitivity... hook 2 drivers in parallel & put 1 W in (not 2.83 V -- ie 2 W, but 1 W)... you don't get 6 dB i guarantee it.
The bad "common knowledge" comes because the 6 dB figure is making the assumption that ALL amps double their power into half the impedance. This again is a blanket assumption that is patently untrue.
dave
Dave, I think you misunderstood what I said.
Forget power for a second and put two speakers with stereo amps next to each other and play a mono signal through them. Then unhook one of the speakers and the SPL will decrease 6dB.
Now drive them at 1 watt with one amp.
Each speaker gets 1/2 watt --> -3dB for each speaker
But you have 2 of them playing that loud --> 6dB louder than one speaker receiving 1/2 watt
Net SPL increase at 1 watt = -3 + 6 = +3dB
That's not really relevant. All it means is you may have to adjust the volume control and/or the amp's performance may suffer with a non-optimum load. But we are talking about speakers here. All we care about is a measured 1 watt going into the speaker system, whatever it takes on the amp end to accomplish that. Measure I and V at the speaker terminals and SPL at 1 meter and the amp becomes irrelevant.
Forget power for a second and put two speakers with stereo amps next to each other and play a mono signal through them. Then unhook one of the speakers and the SPL will decrease 6dB.
Now drive them at 1 watt with one amp.
Each speaker gets 1/2 watt --> -3dB for each speaker
But you have 2 of them playing that loud --> 6dB louder than one speaker receiving 1/2 watt
Net SPL increase at 1 watt = -3 + 6 = +3dB
That's not really relevant. All it means is you may have to adjust the volume control and/or the amp's performance may suffer with a non-optimum load. But we are talking about speakers here. All we care about is a measured 1 watt going into the speaker system, whatever it takes on the amp end to accomplish that. Measure I and V at the speaker terminals and SPL at 1 meter and the amp becomes irrelevant.
catapult said:
half the power & half the drivers
Where does 6 dB come from? you have 2 sources... a doubling of level... that is only 3 dB... so by your math, -3 + 3 dB =0 dB.
you are dragging an extra 3 dB from somewhere... where?
dave
OK, i have to apologize to catapult for bring so harsh... what we all (on this discussion) what to see as an answer that has efficiency increasing with the doubling of drivers. Simple VA math doesn't seem to do it.
Catapult's math is what we all what to see, but where does the extra 3 dB over VA math come from?
I got into the Benson book to-nite -- this book is math HEAVY, it shames many of my math texts -- and found his equation for reference efficiency. It has an Sd^2 term in it. Double the cone area, increase output by 4 (ie 6 dB). So if we assume that 2 cones is the same as 1 cone twice as big (which is probably safe to assume if they are close enuff), then we can say doubling the number of drivers gives us 3 dB more efficiency (only 3 because each driver only sees half the power -- the -3 dB in catapult's equation).
So how close is close. Right next to each other is probably close enough, i would suspect that a line array is spread out sufficiently that we can no longer just add 3 dB for evey doubling of the number of speakers.
Now that i have the background to support my original understanding, time to go hunt down the post that got me thinking and dispute his claim.
dave
Catapult's math is what we all what to see, but where does the extra 3 dB over VA math come from?
I got into the Benson book to-nite -- this book is math HEAVY, it shames many of my math texts -- and found his equation for reference efficiency. It has an Sd^2 term in it. Double the cone area, increase output by 4 (ie 6 dB). So if we assume that 2 cones is the same as 1 cone twice as big (which is probably safe to assume if they are close enuff), then we can say doubling the number of drivers gives us 3 dB more efficiency (only 3 because each driver only sees half the power -- the -3 dB in catapult's equation).
So how close is close. Right next to each other is probably close enough, i would suspect that a line array is spread out sufficiently that we can no longer just add 3 dB for evey doubling of the number of speakers.
Now that i have the background to support my original understanding, time to go hunt down the post that got me thinking and dispute his claim.
dave
No problem, Dave. If the outside edge to edge distance is less than a wavelength, you should get the full 6dB gain for increased area. When experimenting it's best to use a fairly low single tone to avoid lobing and complex impedance issues. As we all know, speakers aren't simple resistors when it comes to their impedance curves and lobing affects the room's power response.
It is an interesting discussion. I would like you to take into consideration what kind of amplifier we are talking about. An OTL-tubeamplifier increase it's output when the impedance is increased! Common tubeamplifiers don't change much output weather it is 4 or 8 Ohm and so forth....
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