Recently I was thinking about when a reflex box hits it's tuned frequency, then the cone excursion becomes minimum and at the same time the port output becomes maximum. A very great load is put on the cone at this point to the extent that it can hardly move, while at the same time the port pumps wildly. Also I began to think that the smaller the cone excursion under these conditions the better. Then I woke up from my dream and realised that if the cone excursion is too small then it is not going to be able to impart much acoustic power to the air. This is because to do work you have to have both force *and* displacement, preferably in phase - I think. (I only realised that as I was typing it).
Sooo..... how do we know when the cone is optimally loaded?
Sooo..... how do we know when the cone is optimally loaded?
But if the Q of the box resonance is very high then it means that relatively little energy is escaping to the outside world and most is being used to hold the cone from moving very far.
Just like in an electrical series tuned cct where heaps of oscillating VA's are slopping back and forth and looking like a near short cct to the AC *current* source driving it, while not many AC watts are being diverted to a useful load.
Anybody know anything about radio transmitter final stages and antenna coupling?
Just like in an electrical series tuned cct where heaps of oscillating VA's are slopping back and forth and looking like a near short cct to the AC *current* source driving it, while not many AC watts are being diverted to a useful load.
Anybody know anything about radio transmitter final stages and antenna coupling?
You should have stayed asleep. 
The driver/box is at the alignment's peak efficiency when it stalls out, i.e. applying more power to it doesn't generate any more output. The vent is a separate resonant circuit driven by the driver/box and its theoretical max output occurs when its acoustic impedance = unity (~infinite area), ergo it's ~infinitely long. Fortunately there's a finite point of diminishing returns that in many cases is of reasonable area/length, and can be calculated by finding <5% mach vent 'speed' at the driver's thermal power compression limit. This can be as little as 25% of its rated Pe, and why so often a sub's actual vent output at rated power is nowhere near what it sims.
GM
The driver/box is at the alignment's peak efficiency when it stalls out, i.e. applying more power to it doesn't generate any more output. The vent is a separate resonant circuit driven by the driver/box and its theoretical max output occurs when its acoustic impedance = unity (~infinite area), ergo it's ~infinitely long. Fortunately there's a finite point of diminishing returns that in many cases is of reasonable area/length, and can be calculated by finding <5% mach vent 'speed' at the driver's thermal power compression limit. This can be as little as 25% of its rated Pe, and why so often a sub's actual vent output at rated power is nowhere near what it sims.
GM
Circlotron:
I am not an engineer. The way I visualize the process is that as the tuning frequency is neared, the air in the box "stiffens" and a little movement by the cone sends the box air exploding out the port. Sort of the difference between taking a baseball bat and hitting a nerfball made out of sponge, and at resonance hitting a regular hardball.
The farther you get away from resonance, the more the air in the box "softens".
I am not an engineer. The way I visualize the process is that as the tuning frequency is neared, the air in the box "stiffens" and a little movement by the cone sends the box air exploding out the port. Sort of the difference between taking a baseball bat and hitting a nerfball made out of sponge, and at resonance hitting a regular hardball.
The farther you get away from resonance, the more the air in the box "softens".
Guss:
Well, it depends upon whether the box is the correct volume and the correct tuning for the woofer.
WinISD or any modeling program will give you the frequency response for various volumes and tunings with any given woofer.
Let's take 10" woofers in two different boxes. Both boxes have flat frequency response and are down 3 dB at 40 Hz.
Box A is a closed box. In order to maintain a certain SPL level, it must travel four times as far every octave it drops. So to maintain say, 100 dB SPL, it must travel 4 times as far at 160 Hz as it does at 320 Hz. If it drops from 160 Hz to 80 Hz, it must travel 4 times farther again. And if it goes down to 40 Hz, it must travel four times farther-if it is to maintain 100 dB SPL.
Box B is ported and tuned to 40 Hz. Like Box A, the closed box, it too must travel 4 times farther every time it drops an octave. Except for the lowest octave. When the ported box goes from 80 Hz to 40 Hz at 100 dB SPL, it does NOT travel four times farther. In fact, it only has to travel as far at 40 Hz as it does at 80 Hz.
Put another way, the closed box A has to travel 4 times farther than ported box B to reproduce 40 Hz at 100 dB SPL. Half an octave up, at 56 Hz, ported box B has to travel half as far as closed box A to maintain a 100 dB SPL level. By the time you hit 80 Hz, the two boxes will act identically.
How can the ported box do this? From 80 Hz down to 40 Hz, the port increasingly provides the SPL output. At 40 Hz, the port is putting out most of the output-the woofer very little.
That is the advantage of the ported box. The advantage is confined to the lowest octave only. But then, the lowest octave is where you need it the most.
The excursion advantage of the ported box will vary with differing ported setups. It will be different for setups where the box tuning is well below F3, for instance. But the essential fact remains-the ported box relieves the woofer of having to undergo large excursions to maintain a certain SPL in just about all setups.
Of course, beneath the tuning frequency, there are large excursions, and little output. So a filter is recommended to limit excursions beneath the ported box frequency.
Well, it depends upon whether the box is the correct volume and the correct tuning for the woofer.
WinISD or any modeling program will give you the frequency response for various volumes and tunings with any given woofer.
Let's take 10" woofers in two different boxes. Both boxes have flat frequency response and are down 3 dB at 40 Hz.
Box A is a closed box. In order to maintain a certain SPL level, it must travel four times as far every octave it drops. So to maintain say, 100 dB SPL, it must travel 4 times as far at 160 Hz as it does at 320 Hz. If it drops from 160 Hz to 80 Hz, it must travel 4 times farther again. And if it goes down to 40 Hz, it must travel four times farther-if it is to maintain 100 dB SPL.
Box B is ported and tuned to 40 Hz. Like Box A, the closed box, it too must travel 4 times farther every time it drops an octave. Except for the lowest octave. When the ported box goes from 80 Hz to 40 Hz at 100 dB SPL, it does NOT travel four times farther. In fact, it only has to travel as far at 40 Hz as it does at 80 Hz.
Put another way, the closed box A has to travel 4 times farther than ported box B to reproduce 40 Hz at 100 dB SPL. Half an octave up, at 56 Hz, ported box B has to travel half as far as closed box A to maintain a 100 dB SPL level. By the time you hit 80 Hz, the two boxes will act identically.
How can the ported box do this? From 80 Hz down to 40 Hz, the port increasingly provides the SPL output. At 40 Hz, the port is putting out most of the output-the woofer very little.
That is the advantage of the ported box. The advantage is confined to the lowest octave only. But then, the lowest octave is where you need it the most.
The excursion advantage of the ported box will vary with differing ported setups. It will be different for setups where the box tuning is well below F3, for instance. But the essential fact remains-the ported box relieves the woofer of having to undergo large excursions to maintain a certain SPL in just about all setups.
Of course, beneath the tuning frequency, there are large excursions, and little output. So a filter is recommended to limit excursions beneath the ported box frequency.
>The way I visualize the process is that as the tuning frequency is neared, the air in the box "stiffens" and a little movement by the cone sends the box air exploding out the port.
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Hmm, I don't believe the 'exploding' part is the appropriate analogy since the air mass plug in the vent becomes excited as the cab's Vb 'stiffens up', resonating with increasing frequency with decreasing signal frequency, peaking at Fb.
The animations in this article graphically depicts what's going on quite well:
http://www.hometheaterhifi.com/volume_5_2/cmilleressayporting.html
GM
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Hmm, I don't believe the 'exploding' part is the appropriate analogy since the air mass plug in the vent becomes excited as the cab's Vb 'stiffens up', resonating with increasing frequency with decreasing signal frequency, peaking at Fb.
The animations in this article graphically depicts what's going on quite well:
http://www.hometheaterhifi.com/volume_5_2/cmilleressayporting.html
GM
The way I visualise it is like this - at work I have this nifty 2 metre steel ruler. If I hold it horizontally at the halfway point and move my hand up and down rapidly about 200mm then the ends of the ruler also move up and down at about the same peak to peak distance my hand moves.kelticwizard said:
Now, if I slow the movement down to about 3 Hz it gets to a point where the ruler gets progressively more difficult to move, and the ends of the ruler really start to flap up and down. Maybe 5mm of centre movement (and the ruler is resisting being pushed, believe me...) becomes 300mm movement at the ends of the ruler. We have resonance! Where I hold the ruler in the centre represents the speaker cone, and the ends of the ruler reperesents the port(s).
Anyway... what I meant at the beginning of the thread is that
1/ you have to apply force, (like volts)
2/ you have to push that force a distance (like amps)
3/ peak force corresponds with peak distance (like unity power-factor AC cct)
If distance is too small despite lots of force then you are not feeding very much power into what is by definition a hi Q low loss cct. Big heaps of instantaneous power going back and forth alternately between the box air pressure and the port air-mass velocity, (like inductor current and capacitor voltage respectively) isn't physics cool?
) but not much is being siphoned off - exploding - to the outside world. Hence the huge buildup in resonant energy holding the cone almost stationary.
Rocket science applied to reflex ports.
Anyway, one of the things I am messing with lately is different kinds of ports, and something I mentioned in another post ages ago is that in a box-and-port combination the energy is contained as either air pressure in the box or air-mass movement in the port or a combination of the two as it pumps back and forth.
Now, we know that you can get ports with flared end to reduced the "chuffing" noise produced when the air is really moving quick. but in applications where you are moving really serious amounts of air like for example an automotice turbocharger or perhaps a jet engine or even a rocket exhaust nozzle. you can't really tolerate the least amount of chuffing! The airflow has to be right.
You can see clearly in the turbo compressor pic that the inlet changes cross-section gradually from wide to narrow and for a very good reason -constant *linear* change of velocity of the airstream. Note also that the compressor outlet gradually increases in size as it winds it's way around the housing.
In a speaker box the when the air goes from the box to the plain-ended port it is very suddenly accelerated, and when that air reaches the outside word it is very suddenly decellerated and what's more the kinetic energy that air had as it exits the port is now lost.
If though we had a port with a continuously changing cross section (sort of hourglass shaped) so that when the air enters the port at low velocity high pressure (high *potential* energy) it has lots of space, but when it is halfway along the port the port is narrowest and the velocity and therefore the *kinetic* energy is greatest. Now here is the important part. As the fast moving air eventually approaches the far end of the port, the port grows in cross section again *forcing the air to slow down* and as a result, extracting the kinetic energy from it. This energy is directed to the air further behind in the port airstream as a low pressure, i.e. it yanks the air behind it that is currently in the narrow part of the port.
Summary of rant: I think the port's role in potential-kinetic-potential energy transformation, i.e. pressure-flow-pressure is underdeveloped. It could be a lot better.
Anyway, one of the things I am messing with lately is different kinds of ports, and something I mentioned in another post ages ago is that in a box-and-port combination the energy is contained as either air pressure in the box or air-mass movement in the port or a combination of the two as it pumps back and forth.
Now, we know that you can get ports with flared end to reduced the "chuffing" noise produced when the air is really moving quick. but in applications where you are moving really serious amounts of air like for example an automotice turbocharger or perhaps a jet engine or even a rocket exhaust nozzle. you can't really tolerate the least amount of chuffing! The airflow has to be right.
You can see clearly in the turbo compressor pic that the inlet changes cross-section gradually from wide to narrow and for a very good reason -constant *linear* change of velocity of the airstream. Note also that the compressor outlet gradually increases in size as it winds it's way around the housing.An externally hosted image should be here but it was not working when we last tested it.
In a speaker box the when the air goes from the box to the plain-ended port it is very suddenly accelerated, and when that air reaches the outside word it is very suddenly decellerated and what's more the kinetic energy that air had as it exits the port is now lost.
If though we had a port with a continuously changing cross section (sort of hourglass shaped) so that when the air enters the port at low velocity high pressure (high *potential* energy) it has lots of space, but when it is halfway along the port the port is narrowest and the velocity and therefore the *kinetic* energy is greatest. Now here is the important part. As the fast moving air eventually approaches the far end of the port, the port grows in cross section again *forcing the air to slow down* and as a result, extracting the kinetic energy from it. This energy is directed to the air further behind in the port airstream as a low pressure, i.e. it yanks the air behind it that is currently in the narrow part of the port.
Summary of rant: I think the port's role in potential-kinetic-potential energy transformation, i.e. pressure-flow-pressure is underdeveloped. It could be a lot better.
You'll want to factor the problem of radiation impedance into the idea of a "horn" port. To me port action ceases to be intuitively obvious at this level of thinking and I break out the electrical analogies. Just grab a simple equivalent circuit from an acoustics book and play with SPICE some. That's what I do
When it gets to this point I go by intuition almost exclusively, and if it works then I know the numbers will fall into place later.
I think the curve I need is for a cross sectional area that is inversely proportional the square root of how far you are along toward the halfway point of the port. I have the *feeling* that it is related to the equation for instantaneous displacement of a body undergoing constant acceleration -> S=Ut+0.5at^2.
I think the curve I need is for a cross sectional area that is inversely proportional the square root of how far you are along toward the halfway point of the port. I have the *feeling*
that it is related to the equation for instantaneous displacement of a body undergoing constant acceleration -> S=Ut+0.5at^2.- Status
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