Porting

[floppygoat]

I thought I understood porting, but apparently I do not. Can someone explain the idea behind porting. What I really get stuck on is, How can there be a parabolic volume increase at a particular tuning frequency and a inversely related decrease in cone movement? Why does the cone movement decrease at the fb and how is an increase in volume generated?

Thank you,
Tony Williams

[Jeff B.]

I thought I understood porting, but apparently I do not. Can someone explain the idea behind porting. What I really get stuck on is, How can there be a parabolic volume increase at a particular tuning frequency and a inversely related decrease in cone movement? Why does the cone movement decrease at the fb and how is an increase in volume generated?

Thank you,
Tony Williams

The first thing to keep in mind is that you are not tuning the port, rather you are using the port to tune the enclosure volume. The resistance of the air mass in the port acts like a weight attached to the compliant air volume in the enclosure. Picture a weight attached to a spring. As the woofer generates sound to pressurize the enclosure this pressurization starts the mass and spring in motion and together they resonate at a particular Fb, in other words this spring is cycling exactly so many times a second, Fb times per second, actually. Now, what is cycling at Fb is the air inside the enclosure and this air is in direct contact with the cone. At Fb this resonating air mass applies an equal by opposite force against the cone (Remember Newton?) and this force damps the cone's motion at exactly the frequency the air mass is resonating at. It is this back-force from the resonating air mass that puts the brakes on the cone's motion at the resonant frequency. At this frequency the system efficiency is at its maximum and nearly all of the acoustic output is coming from the port. (In a perfect system, anyway. In reality the cone's output is down about 20dB due to losses in the port, ie; Qp, but the primary output is still from the port even with losses due to friction, absorption, etc.).

I hope this explains it clearly enough to picture in your mind.

Jeff B.

[rudyjakubin]

The first thing to keep in mind is that you are not tuning the port, rather you are using the port to tune the enclosure volume. The resistance of the air mass in the port acts like a weight attached to the compliant air volume in the enclosure. Picture a weight attached to a spring. As the woofer generates sound to pressurize the enclosure this pressurization starts the mass and spring in motion and together they resonate at a particular Fb, in other words this spring is cycling exactly so many times a second, Fb times per second, actually. Now, what is cycling at Fb is the air inside the enclosure and this air is in direct contact with the cone. At Fb this resonating air mass applies an equal by opposite force against the cone (Remember Newton?) and this force damps the cone's motion at exactly the frequency the air mass is resonating at. It is this back-force from the resonating air mass that puts the brakes on the cone's motion at the resonant frequency. At this frequency the system efficiency is at its maximum and nearly all of the acoustic output is coming from the port. (In a perfect system, anyway. In reality the cone's output is down about 20dB due to losses in the port, ie; Qp, but the primary output is still from the port even with losses due to friction, absorption, etc.).

I hope this explains it clearly enough to picture in your mind.

Jeff B.

Here's a web site with visuals;
http://www.hometheaterhifi.com/volume_5_2/cmilleressayporting.html

[floppygoat]

Ahhh, the world of audio. The more I learn, the more I don't understand. Over the last year my intense interest has led to growing confusion. It will take a while to digest the responses you have given. Thank you, however, I will get through it. And what exactly does the terms resonant, resonance, etc. mean in relation to driver enclosures? Is it desired or undesired? I have been given varying responses?

[Pete Schumacher ®]

Ahhh, the world of audio. The more I learn, the more I don't understand. Over the last year my intense interest has led to growing confusion. It will take a while to digest the responses you have given. Thank you, however, I will get through it. And what exactly does the terms resonant, resonance, etc. mean in relation to driver enclosures? Is it desired or undesired? I have been given varying responses?

Resonance is a desired characteristic when talking about extending bass response with a vented box.

Here's one way to look at it. A vented box is similar to a bottle. Blow across the top of it and it will "resonate" at it's natural frequency. Now, take a much larger bottle. Usually it will resonate at a lower frequency than the smaller one. However, it can be tuned to the same frequency as the smaller bottle. But exciting its resonance with the same stimulus will produce a "louder" tone than the smaller one.

A woofer with a higher sensitivity will need the larger bottle to augment it's output down low. A lower sensitivity woofer needs the smaller bottle. It's those driver variables (TS Parameters) that determine the size and tuning of the box to compliment the woofer.

[floppygoat]

So a resonant frequency is just the most efficient frequency, db wise, of any type of enclosure, i.e. room, box, bottle, etc.?

Is resonance another term for vibrations creating sound and/or frequencies?

What does the term mean in these senses:

As the woofer generates sound to pressurize the enclosure this pressurization starts the mass and spring in motion and together they resonate at a particular Fb...

At Fb this resonating air mass applies an equal by opposite force against the cone...and this force damps the cone's motion at exactly the frequency the air mass is resonating at.

It is this back-force from the resonating air mass that puts the brakes on the cone's motion at the resonant frequency.

It seems like there are varying principles of the word. Sorry for all the questions, but I think once I grasp this concept, I will grasp the concept of porting and the physics behind it, such as cone movement.

Thank you all,
Tony Williams

[Dirk]

Everything has a resonant frequency..."wub wub wub". Think of a tuning fork. Strike it (ie, create an impulse) and it will ring, resonating at its natural Fs.
You can change the resonant frequency by changing the mass. This is what happens when a guitarist holds down a fret (the string is "shorter", so mass down, tone goes up), or when a trombone player extends his slide (mass of air in the pipes goes up, tone goes down).

Drivers have natural resonant frequencies, as well. At that frequency, you get a maximum of output for a minimum of input. Two different devices can be coupled together, creating a new resonance. This is what happens when you place a woofer in a box. The volume of air in the box has its own resonance, and it combines with the woofer's natural resonance to create the system resonance (Vb).

[Ludo]

So a resonant frequency is just the most efficient frequency, db wise, of any type of enclosure, i.e. room, box, bottle, etc.?

Is resonance another term for vibrations creating sound and/or frequencies?

What does the term mean in these senses:

As the woofer generates sound to pressurize the enclosure this pressurization starts the mass and spring in motion and together they resonate at a particular Fb...

At Fb this resonating air mass applies an equal by opposite force against the cone...and this force damps the cone's motion at exactly the frequency the air mass is resonating at.

It is this back-force from the resonating air mass that puts the brakes on the cone's motion at the resonant frequency.

It seems like there are varying principles of the word. Sorry for all the questions, but I think once I grasp this concept, I will grasp the concept of porting and the physics behind it, such as cone movement.

Thank you all,
Tony Williams

Think of resonance as a gradual release of stored energy which results in some type of periodic phenomena like sound, movement, electrical current flow, etc.

For example; A backyard swing with a child suspended in it has a natural resonance frequency of motion. You push the swing to one side which imparts potential energy to it. When you release it, gravity pulls it back to the middle but because the friction loss from the chain and air is very low the swing does not stop at the bottom but keeps swinging past it to the other side and back and forth at its natural resonance frequency.

A similar thing is happening with the port.

Louis

[whatatrip]

The first thing to keep in mind is that you are not tuning the port, rather you are using the port to tune the enclosure volume. The resistance of the air mass in the port acts like a weight attached to the compliant air volume in the enclosure. Picture a weight attached to a spring. As the woofer generates sound to pressurize the enclosure this pressurization starts the mass and spring in motion and together they resonate at a particular Fb, in other words this spring is cycling exactly so many times a second, Fb times per second, actually. Now, what is cycling at Fb is the air inside the enclosure and this air is in direct contact with the cone. At Fb this resonating air mass applies an equal by opposite force against the cone (Remember Newton?) and this force damps the cone's motion at exactly the frequency the air mass is resonating at. It is this back-force from the resonating air mass that puts the brakes on the cone's motion at the resonant frequency. At this frequency the system efficiency is at its maximum and nearly all of the acoustic output is coming from the port. (In a perfect system, anyway. In reality the cone's output is down about 20dB due to losses in the port, ie; Qp, but the primary output is still from the port even with losses due to friction, absorption, etc.).

I hope this explains it clearly enough to picture in your mind.

Jeff B.

I agree with floppygoat "Ahhh, the world of audio. The more I learn, the more I don't understand."

If the dB is down 20 at Fb then we aren't hearing much at that frequency, right? Is it true that lowering the Fb is a way of extending the audibility of lower frequencies and at the same time we extend the audibilty of the lower frequencies, we are lowering the SPL of of those frequencies close by? I've been using BassBox Pro and can see that a box with Fb of 31 hz has a cut off frequency F3 of 67 Hz and by the time the slope hits 31 Hz, the SPL is down about 11 dB.

This particular speaker, the Dayton DC160S-8 has an Fs of 31 Hz. All things considered can I assume that I won't hear much of this frequency from the speaker at Fs eventhough this is the resonance frequency of the speaker and its output is at its best compared to input? Is it kind of like a car that gets gets 50 mpg at 50 mph but you can't hear as much road noise as you can at 90 mph and 20 mpg?

[Dirk]

If the dB is down 20 at Fb then we aren't hearing much at that frequency, right?


Jeff was saying that the output of the WOOFER was -20dB at Fb. Not the output of the system. That's an important distinction.


Is it true that lowering the Fb is a way of extending the audibility of lower frequencies and at the same time we extend the audibilty of the lower frequencies, we are lowering the SPL of of those frequencies close by?


Yeah, sort of...no. You are not actively lowering the SPL of the frequencies to either side of Fb. Your box is just so big that the driver is rolling off its response before the port can begin to contribute.



I've been using BassBox Pro and can see that a box with Fb of 31 hz has a cut off frequency F3 of 67 Hz and by the time the slope hits 31 Hz, the SPL is down about 11 dB.

This particular speaker, the Dayton DC160S-8 has an Fs of 31 Hz. All things considered can I assume that I won't hear much of this frequency from the speaker at Fs even though this is the resonance frequency of the speaker and its output is at its best compared to input? Is it kind of like a car that gets gets 50 mpg at 50 mph but you can't hear as much road noise as you can at 90 mph and 20 mpg?

First, car analogies are like an Alfa Romeo. Attention-getting, but they rarely work the way you want.

Even though the woofer has an Fs of 31hz, that's the "free air resonance". Once you put it in a box, the woofer's free-air resonance resonance becomes part of a much larger system and the driver's Fs is only one element of many that defines the actual response.

[Jeff B.]

I agree with floppygoat "Ahhh, the world of audio. The more I learn, the more I don't understand."

If the dB is down 20 at Fb then we aren't hearing much at that frequency, right? Is it true that lowering the Fb is a way of extending the audibility of lower frequencies and at the same time we extend the audibilty of the lower frequencies, we are lowering the SPL of of those frequencies close by? I've been using BassBox Pro and can see that a box with Fb of 31 hz has a cut off frequency F3 of 67 Hz and by the time the slope hits 31 Hz, the SPL is down about 11 dB.

This particular speaker, the Dayton DC160S-8 has an Fs of 31 Hz. All things considered can I assume that I won't hear much of this frequency from the speaker at Fs eventhough this is the resonance frequency of the speaker and its output is at its best compared to input? Is it kind of like a car that gets gets 50 mpg at 50 mph but you can't hear as much road noise as you can at 90 mph and 20 mpg?

I didn't say anything like that. Dirk's answers are correct. I will only add a picture to show what we are trying to say. The cone motion, and consequently its output, is damped by the pressure from the port at Fb. At this frequency nearly all of the output is coming from the port. On either side of this frequency the cone takes over again. A vented system's response is the complex summation of these two acoustic outputs.

http://i109.photobucket.com/albums/n70/jeffbagby/VentPhaseGraph2.gif

[whatatrip]

Maybe I should have been more precise in how I phrased what I was asking about. I know the difference between system output and component output. I should have said "If the dB is down 20 at Fb then we aren't hearing much "from the driver"at that frequency, right?" In other words, if the sound coming from the port and the driver could be isolated, only 20 Hz was being fed and a microphone was placed to pick up the relative SPL of each, there would be a 30 dB (based upon your graph) difference between the output of the driver compared to the port? If that is true and most of the sound is coming from the port, how does it compare to what we hear from the driver?

When I was talking about lower SPL in favor of an extended bass, I was comparing SPL levels of a system with extended bass compared to one without the extended bass.

I wonder what I am seeing in the graph posted above. I see a driver SPL of 76 dB at 26 Hz, a vent SPL of 103 dB at 26 Hz and a total SPL of 103 dB at 26 Hz. Does that mean the drive is contributing nothing to the total SPL?

[Jeff B.]

Maybe I should have been more precise in how I phrased what I was asking about. I know the difference between system output and component output. I should have said "If the dB is down 20 at Fb then we aren't hearing much "from the driver"at that frequency, right?" In other words, if the sound coming from the port and the driver could be isolated, only 20 Hz was being fed and a microphone was placed to pick up the relative SPL of each, there would be a 30 dB (based upon your graph) difference between the output of the driver compared to the port? If that is true and most of the sound is coming from the port, how does it compare to what we hear from the driver?

When I was talking about lower SPL in favor of an extended bass, I was comparing SPL levels of a system with extended bass compared to one without the extended bass.

I wonder what I am seeing in the graph posted above. I see a driver SPL of 76 dB at 26 Hz, a vent SPL of 103 dB at 26 Hz and a total SPL of 103 dB at 26 Hz. Does that mean the drive is contributing nothing to the total SPL?

Right - pretty much nothing. I've written it twice already above, but one more time - Regarding the tuning frequency (Fb) - "At this frequency nearly all of the output is coming from the port." Put mics on both and they will measure at a given frequency something pretty close to what I am showing above.

[floppygoat]

The cone motion, and consequently its output, is damped by the pressure from the port at Fb.

How is this pressure created? I always assumed, and was once told, that the port acted as a second "driver" moving a second face of air which created a more efficient system. I have no idea how this pressure can be created in such a way, especially when there is a vent to alleviate pressure. It seems like this phenomena should happen with sealed enclosures, not ported.

[Jeff B.]

How is this pressure created? I always assumed, and was once told, that the port acted as a second "driver" moving a second face of air which created a more efficient system. I have no idea how this pressure can be created in such a way, especially when there is a vent to alleviate pressure. It seems like this phenomena should happen with sealed enclosures, not ported.

The pressure is created by the resonance of the air in enclosure - but it is centered at this frequency (Fb) and is falling off on both sides of Fb, just like its output is.

[Pete Schumacher ®]

How is this pressure created? I always assumed, and was once told, that the port acted as a second "driver" moving a second face of air which created a more efficient system. I have no idea how this pressure can be created in such a way, especially when there is a vent to alleviate pressure. It seems like this phenomena should happen with sealed enclosures, not ported.

Don't forget that air acts like a spring. It compresses behind the woofer as the woofer moves. This pressure is not alleviated instantaneously, but builds up because the port won't let the air out as fast as it's being compressed.

Remember the mass on the spring analogy. The port air slug is a mass attached to the spring of the air in the enclosure. At high frequencies, the spring is compressed without the air in the port moving, much like if you took a weight suspended from a spring and moved the top of the spring rapidly. Very little motion of the mass would occur. As you move down in frequency, the mass will begin moving, and when you hit the resonant point, almost no motion added to the spring will keep the mass oscillating back and forth.

[floppygoat]

Ok, ok, ok...I am about to have an epiphany. A little more time, some more thinking, and maybe some more questions. Thank you so far.

Tony Williams

[rudyjakubin]

Resonance is a desired characteristic when talking about extending bass response with a vented box.

Here's one way to look at it. A vented box is similar to a bottle. Blow across the top of it and it will "resonate" at it's natural frequency. Now, take a much larger bottle. Usually it will resonate at a lower frequency than the smaller one. However, it can be tuned to the same frequency as the smaller bottle. But exciting its resonance with the same stimulus will produce a "louder" tone than the smaller one.

A woofer with a higher sensitivity will need the larger bottle to augment it's output down low. A lower sensitivity woofer needs the smaller bottle. It's those driver variables (TS Parameters) that determine the size and tuning of the box to compliment the woofer.

Hey Pete, You get that bottle analogy from here???
http://www.polkaudio.com/education/tech_article.php?id=53
:)

[Pete Schumacher ®]

Hey Pete, You get that bottle analogy from here???
http://www.polkaudio.com/education/tech_article.php?id=53
:)

Nope. I'm an old resonant device manipulator from way back (played Clarinet and Trumpet in my youth). Smaller resonant devices may be able to play a low note, but a "big" resonant device playing that same note can play it louder. That's where I got my sense of how resonance in Helmholtz systems works.

[rudyjakubin]

Nope. I'm an old resonant device manipulator from way back (played Clarinet and Trumpet in my youth). Smaller resonant devices may be able to play a low note, but a "big" resonant device playing that same note can play it louder. That's where I got my sense of how resonance in Helmholtz systems works.

http://img.photobucket.com/albums/v641/rjakubin/lmao.gif
Hee Hee,
This morning whilst drinking coffee I came across a new techno term ( at least to me) 'Planar Bass Radiator'. While touring Google for an explanation as to it's implementation, I came across the above article by Polk Audio.http://img.photobucket.com/albums/v641/rjakubin/dunno.gif

To me ported bass speakers always have that bass note that quite doesn't sound right and comparing a ported speaker with a sealed one side by side it becomes evident.
Yesterday I was comparing a vented speaker with a passive radiator one.
Together they sounded quite good. The vented one the bass drum sounded solid with a bit of echo or reverb. The PR speakers bass drum note sounded as if the drummer had softened the skin or had a softer head attached.
Together they added a quite good sound to it. I think long ago a speaker outfit had a dual design. The lower chamber had a port and the mid chamber had a passive radiator. Too long ago to put a name on it.
Cheers! http://img.photobucket.com/albums/v641/rjakubin/icon_beer.gif
Didn't mean to derail this thread, but it is somewhat relevant.

[Chris Roemer]

Ok, ok, ok...I am about to have an epiphany. A little more time, some more thinking, and maybe some more questions. Thank you so far.

Tony Williams

Given identical drivers, and port lengths and diameters, but 2 sizes of boxes: the smaller box has less air (less springiness, so it's stiffer, and transfers energy quicker, which means at a higher frequency); the larger box (which ends up being tuned lower) takes longer to transfer the energy.

You can slow a box down (tune it lower) by making it bigger (springier) OR by increasing the mass of air in the port (if you keep the dia. the same) which means making the port longer. Think of the air in the port as a solid "slug" moving back and forth in reaction to the transfer of energy from the air in the box compressing and rarefying.

[Pete Schumacher ®]

I would think a comparo of the exact same driver compliment in the exact same enclosures with one vented and one PR would be the only real way to come to a conclusion about the differences in sound.

Could be fun!!!

[Dirk]

Given identical drivers, and port lengths and diameters, but 2 sizes of boxes: the smaller box has less air (less springiness, so it's stiffer, and transfers energy quicker, which means at a higher frequency); the larger box (which ends up being tuned lower) takes longer to transfer the energy.

You can slow a box down (tune it lower) by making it bigger (springier) OR by increasing the mass of air in the port (if you keep the dia. the same) which means making the port longer. Think of the air in the port as a solid "slug" moving back and forth in reaction to the transfer of energy from the air in the box compressing and rarefying.

That's a brilliant retelling, Chris.

[Mark65]

Resonance is a desired characteristic when talking about extending bass response with a vented box.

Here's one way to look at it. A vented box is similar to a bottle. Blow across the top of it and it will "resonate" at it's natural frequency. Now, take a much larger bottle. Usually it will resonate at a lower frequency than the smaller one. However, it can be tuned to the same frequency as the smaller bottle. But exciting its resonance with the same stimulus will produce a "louder" tone than the smaller one.

A woofer with a higher sensitivity will need the larger bottle to augment it's output down low. A lower sensitivity woofer needs the smaller bottle. It's those driver variables (TS Parameters) that determine the size and tuning of the box to compliment the woofer.


Don't forget that air acts like a spring. It compresses behind the woofer as the woofer moves. This pressure is not alleviated instantaneously, but builds up because the port won't let the air out as fast as it's being compressed.

Remember the mass on the spring analogy. The port air slug is a mass attached to the spring of the air in the enclosure. At high frequencies, the spring is compressed without the air in the port moving, much like if you took a weight suspended from a spring and moved the top of the spring rapidly. Very little motion of the mass would occur. As you move down in frequency, the mass will begin moving, and when you hit the resonant point, almost no motion added to the spring will keep the mass oscillating back and forth.


Those two added together did more to explain resonance than anything else I've read. Although, I believe I heard the bottle analogy from Chris Roemer a while back, also.

Thanks for all of the patient explanation from all of you!

Mark

[billfitzmaurice]

Yesterday I was comparing a vented speaker with a passive radiator one.

The OP might better understand how a port works if he can wrap his head around the fact that it works in almost identically the same way as a passive radiator. It's easy to see how a PR moves back and forth, energized by the motion of the driver cone. It's also easy to understand how the resonant frequency of the PR can be adjusted via making it smaller or larger, and by adding weights to it. A port works in the same way, all that really differs is that with the port the vibrating mass is the air within it, not a cone. You can adjust its resonant frequency by making it larger or smaller, and if you need to add additional mass instead of adding weights to a cone you make the vent air mass heavier by adding a duct to it. There are differences in the final result when using a PR versus a port, but they are subtle. The basic operating principles are the same.

[Pete Schumacher ®]

Those two added together did more to explain resonance than anything else I've read. Although, I believe I heard the bottle analogy from Chris Roemer a while back, also.

Thanks for all of the patient explanation from all of you!

Mark

Aww, Garsh . . . . .

Glad to be of some help . . . ;) :o

[floppygoat]

The only thing that is still confusing me is the cone movement. How at the fb there is so much pressure created, so little movement allowed, yet so much volume?

[rone]

How is this pressure created? I always assumed, and was once told, that the port acted as a second "driver" moving a second face of air which created a more efficient system. I have no idea how this pressure can be created in such a way, especially when there is a vent to alleviate pressure. It seems like this phenomena should happen with sealed enclosures, not ported.

Think phase relationships.
A sealed box acts like a vented box tuned to 0 Hz.

[floppygoat]

Ok, cone movement is minimized at fb. This is do to an increase in internal pressure. However, this pressure decreases at frequencies above and below the fb, right so far? How is this pressure created when the port, in my mind, should alleviates it? For example, when you have a sealed single chamber box with two drivers, pressing on the cone of one will cause the other to move. However, same situation but with a ported enclosure, pressing the cone of one will not cause the other to move, since the pressure is being alleviated by the port. And why does the pressure decrease at levels below fb where excursion is greater, and theoretically, so should pressure? :confused::confused::confused:

[floppygoat]

"As the active driver approaches the tuned frequency of the secondary resonant device, that device begins to actually suck energy so forcefully through the internal air pressure, that it loads the active driver (I don't fully understand what is meant by this). It uses the cone as the fulcrum to bear the force of its own acceleration, thereby limiting the active driver’s excursion, and transferring the greater output to its own opening or surface."

Or is the port alleviating the pressure so efficiently that the driver hardly has to move at all, only to maintain the movement of air through the port? If this is the case, that makes "sense." If not...

[Pete Schumacher ®]

"As the active driver approaches the tuned frequency of the secondary resonant device, that device begins to actually suck energy so forcefully through the internal air pressure, that it loads the active driver (I don't fully understand what is meant by this). It uses the cone as the fulcrum to bear the force of its own acceleration, thereby limiting the active driver’s excursion, and transferring the greater output to its own opening or surface."

Or is the port alleviating the pressure so efficiently that the driver hardly has to move at all, only to maintain the movement of air through the port? If this is the case, that makes "sense." If not...

Now go back to the bottle analogy. It takes very little effort to maintain a good SPL when working at the system resonant frequency.

You can demonstrate the effect yourself, if you have a small, resonant room you can stand or sit in, like a tiled and bare bathroom. Start at a low frequency and just hum, gradually increasing the pitch. What you'll notice is that when you reach a room resonance, the effort required by your voice drops dramatically to keep the same sound level as when you're not at a resonance point.

Do the same thing with a mailing tube a couple feet long. Do the same sweeping hum, and at the standing wave frequencies, you will notice a push back on your vocal cords from the wavefront you set up in the tube.

The pressure system is resonating at the same frequency as the driver, creating a higher pressure than a simple sealed box, when the woofer is pulled into the enclosure, inhibiting its movement, and a lower pressure as the woofer moves out from the enclosure. This resonance makes the enclosure act like a much smaller enclosure than it really is, but only around resonance. This pressure is causing the air to move in and out of the port, which is doing all (or most of) the talking at that frequency.

[floppygoat]

It just isn't going to happen for me I guess.

[floppygoat]

This is what I understand:

A speaker has a resonant frequency
A box has a resonant frequency
A speaker and box create another resonant frequency
"Everything" has a resonant frequency
A sealed box has a permanent resonant frequency
A ported box has an adjustable resonant frequency
Pressure begins to build as this frequency is approached
Pressure is greatest when this frequency is reached

What I don't understand:

What is occurring with the box, port, and speaker to cause this pressure?

Only very simple language and a thorough explanation of everything happening will end in my understanding. If you guys are getting annoyed with this thread, I understand. Sorry this is taking so much explanation, I just can't rap my head around it. Thank you so much for all the help so far.

A simpler and more thorough explanation of these statements may help:

The air mass provided by the box volume and the port can’t react instantly to the pressure induced by the active speaker. Above its own resonant frequency, it’s too slow (what is too slow?) to suck (what is meant by suck?) any energy (what is meant by energy?) off the active driver, and it does almost nothing at all, refusing to move when the direction of force changes too rapidly.

As the active driver approaches the tuned frequency of the secondary resonant device, that device begins to actually suck energy (again suck, energy?) so forcefully through the internal air pressure (what does that mean?), that it loads (what is meant by load?) the active driver. It uses the cone as the fulcrum to bear the force of its own acceleration (what does that mean?), thereby limiting the active driver’s excursion, and transferring the greater output (what is the greater output?) to its own (what is signified by own?) opening or surface (is opening or surface the port?).

Below resonance, the reflex system backs off on loading the active driver, yet still moves quite a bit. But since it is merely venting the back-pressure instead of "cracking the whip" (what is cracking the whip?), it’s no more than a hole in the baffle, which causes cancellation with the active driver’s front output (how is cancellation created?). Because of this, the output rolls off very steeply, at 24 dB per octave. Also, since the port does not limit excursion and the suspension force that would have been provided by the compliance of air in a sealed system doesn’t exist because of the “leak” in the enclosure, the active driver can very easily “bottom out,” or smack itself into itself, below that tuned frequency.

This last sentence is the essence of my confusion. The idea presented in the last sentence makes "sense," that the port would alleviate pressure, not cause it. But, apparently, this only occurs below the tuning frequency, and that is what I don't understand.

Tony Williams

[jclin4]

Well they can't fault you for not making the effort to understand this phenomenon. I admire your tenacity. (I also like your username and the avatar/picture you had before).

I'm a lightweight when it comes to mastering speaker concepts, but this thread has been really interesting reading.


Also, since the port does not limit excursion and the suspension force that would have been provided by the compliance of air in a sealed system doesn’t exist because of the “leak” in the enclosure, the active driver can very easily “bottom out,” or smack itself into itself, below that tuned frequency.

This last sentence is the essence of my confusion. The idea presented in the last sentence makes "sense," that the port would alleviate pressure, not cause it. But, apparently, this only occurs below the tuning frequency, and that is what I don't understand.


From what I read, I think the experts are saying the opposite of your last sentence. Maximum output from the port (and therefore alleviation of pressure) is occuring exactly at the tuning frequency, not below or above it.

The other thing to keep in mind is that when they talk about pressure building up or air compressing and then releasing, they are figuratively slowing time down and examining a tiny slice of it. Sound consists of relatively fast cyclical increases and decreases of air pressure over time. At an Fb of 26Hz for example, this cycle is happening 26 times per second.

[floppygoat]

From what I read, I think the experts are saying the opposite of your last sentence. Maximum output from the port (and therefore alleviation of pressure) is occuring exactly at the tuning frequency, not below or above it.

Yes, maximum output is being derived from the port at fb, but not the alleviation of pressure. Rather, a great amount of internal pressure is being created withing the box at fb, decreasing both above and below fb. AND THIS IS WHAT CONTINUES TO BEFUDDLE ME!!!

[floppygoat]

Can anyone explain the statements above and questions throughout?

[Pete Schumacher ®]

Can anyone explain the statements above and questions throughout?

the system is a spring. It cannot act instantaneously.

Have you ever played with a paddle and a rubber band attached ball? Only when the resonance is reached can you keep that ball bouncing off the paddle.

If the concept is eluding you, then you need to seek out real world examples and play around with them yourself.

Get a spring and attach a mass. Move the spring with your hand. When you reach resonance, your hand will barely be moving to keep the mass oscillating.

[Jeff B.]

Yes, maximum output is being derived from the port at fb, but not the alleviation of pressure. Rather, a great amount of internal pressure is being created withing the box at fb, decreasing both above and below fb. AND THIS IS WHAT CONTINUES TO BEFUDDLE ME!!!

Well, your statement above is correct...probably both of them. I doubt I will explain this fully to your satisfaction, but I will give it one more go.

First, I don't really like the explanation from the text you are quoting from that website. I find it confusing in its analogies. I guess everyone has their own way of explaining things. Theirs just isn't the way I would chose.

Let's go back to the coke bottle analogy that Pete used, as it is the simplest and describes a helmholtz resonator fairly well, which is what we are dealing with. You might look that up if you have time. It might help in your understanding.

OK. When you blow across the top of a coke bottle you can create a resonance which produces a sound. This resonance is created by your blowing action exciting the air inside the bottle and finding the natural resonance frequency of its mass. When it begins to resonate and produce sound it is also creating a pressure that not only results in sound/air coming back out the end of the bottle, but there is actually pressure pushing outward against the inside of the bottle all around inside. You don't feel this, because the bottle is strong enough to resist it, but you experience it in the form of sound and air coming back out the end you are blowing on.

The port in the speaker is no different. The energy that sets the air mass resonating is the cone itself. If the cone wasn't producing any sound then there wouldn't be any energy coming out the port - It is getting it's energy from the speaker. When the natural resonance of the air mass in the BOX is oscillated or excited it does the same thing as the air in the bottle. The oscillation produces a pressure from the air in the box trying to expand and it pushes out in all directions, both forcing sound out of the port and pusing against the cone at the same time. Since this occurs only at the frequency the air is oscillating, that's the only frequency where the pressure is pushing against the cone and resonating out the port, and this is why these affects drop off on both side of this frequency.

You are looking at the fact that there is hole in the box and only thinking that there can't be any pressure because of this hole, but this isn't the case if you start the air inside the box oscillating like we are doing here. Below this oscillation frequency the hole starts to look like a real hole and the cone "unloads". This is what causes the cone to hit high excursions below Fb. The 24dB/oct roll-off is caused by the output of the port going more and more out of phase with the cone as we go down in frequency. You can see this in my pic posted above.

Jeff B.

[floppygoat]

When the natural resonance of the air mass in the cone is oscillated or excited it does the same thing as the air in the bottle.

Did you mean port?

[floppygoat]

the system is a spring. It cannot act instantaneously.

Have you ever played with a paddle and a rubber band attached ball? Only when the resonance is reached can you keep that ball bouncing off the paddle.

Get a spring and attach a mass. Move the spring with your hand. When you reach resonance, your hand will barely be moving to keep the mass oscillating.

I think these are brilliant examples, but how can I use them to explain the pressure. If you could, associate each aspect of and example and its movement with the appropriate part of a reflux system.

[floppygoat]

Well, your statement above is correct...probably both of them. I doubt I will explain this fully to your satisfaction, but I will give it one more go.

First, I don't really like the explanation from the text you are quoting from that website. I find it confusing in its analogies. I guess everyone has their own way of explaining things. Theirs just isn't the way I would chose.

Let's go back to the coke bottle analogy that Pete used, as it is the simplest and describes a helmholtz resonator fairly well, which is what we are dealing with. You might look that up if you have time. It might help in your understanding.

OK. When you blow across the top of a coke bottle you can create a resonance which produces a sound. This resonance is created by your blowing action exciting the air inside the bottle and finding the natural resonance frequency of its mass. When it begins to resonate and produce sound it is also creating a pressure that not only results in sound/air coming back out the end of the bottle, but there is actually pressure pushing outward against the inside of the bottle all around inside. You don't feel this, because the bottle is strong enough to resist it, but you experience it in the form of sound and air coming back out the end you are blowing on.

The port in the speaker is no different. The energy that sets the air mass resonating is the cone itself. If the cone wasn't producing any sound then there wouldn't be any energy coming out the port - It is getting it's energy from the speaker. When the natural resonance of the air mass in the cone is oscillated or excited it does the same thing as the air in the bottle. The oscillation produces a pressure from the air in the box trying to expand and it pushes out in all directions, both forcing sound out of the port and pusing against the cone at the same time. Since this occurs only at the frequency the air is oscillating, that's the only frequency where the pressure is pushing against the cone and resonating out the port, and this is why these affects drop off on both side of this frequency.

You are looking at the fact that there is hole in the box and only thinking that there can't be any pressure because of this hole, but this isn't the case if you start the air inside the box oscillating like we are doing here. Below this oscillation frequency the hole starts to look like a real hole and the cone "unloads". This is what causes the cone to hit high excursions below Fb. The 24dB/oct roll-off is caused by the output of the port going more and more out of phase with the cone as we go down in frequency. You can see this in my pic posted above.

Jeff B.

I would like to say I finally "get it," but would really like to await the further explanations of the examples provided by Pete. :)

[floppygoat]

Thank you everyone for your abunadant and patient help. I, relatively speaking, now understand.

Tony Williams

[Jeff B.]

Did you mean port?

I meant the air in the box. Sorry, I made the change in the text above.

[Jeff B.]

Thank you everyone for your abunadant and patient help. I, relatively speaking, now understand.

Tony Williams

Everything's relative - even in Quantum Physics.

Don't fret it if it takes some time to grasp. I've been working on this a long time now. ;)

[floppygoat]

OHH, so the port is just a device to allow the sound to escape, like the hole in the bottle? It is not what is creating the pressure, the resonance of the box is creating the pressure and suppressing the cone?

[Pete Schumacher ®]

OHH, so the port is just a device to allow the sound to escape, like the hole in the bottle? It is not what is creating the pressure, the resonance of the box is creating the pressure and suppressing the cone?

Yes, the driver is what creates the pressure, not the port. However, the port plays a role, along with the enclosure air spring, in setting the frequency where the driver can create the most pressure, and that is the resonant frequency.

[Jeff B.]

OHH, so the port is just a device to allow the sound to escape, like the hole in the bottle? It is not what is creating the pressure, the resonance of the box is creating the pressure and suppressing the cone?

Not exactly, but I bet you follow this - The air in the box is "compliant" meaning it acts like a spring. You can compress it and it springs back. The port is not a simple hole - it has length, so there's a volume to it. It contains a mass of air that moves like a slug when pushed on by the springy air in the box. This mass slug introduces some resistance to the the springy air, so it acts in the same way as a weight attached to a spring. If you attach a heavier weight (longer port) to a spring (the air in the box) and start it in motion (from the cone's movements) the spring will oscillate slowly, or at a lower frequency. If you do this with a ligher weight it will oscillate faster, or at a higher frequency. This is exactly what is going on in your speaker and how the port dimensions are used to "tune" the enclosure to a specific oscillating or resonance frequency.

Jeff B.

[floppygoat]

The port is not a simple hole - it has length, so there's a volume to it. It contains a mass of air that moves like a slug when pushed on by the springy air in the box. This mass slug introduces some resistance to the the springy air, so it acts in the same way as a weight attached to a spring.

Sorry to disappoint, but I do not understand. I know that when the air is compressed or expanded, by movement of the cone, it wants to return to normal, creating pressure. When the air is compressed or expanded, it begins to naturally move the volume of air found in the port: path of least resistance. Now, this mass of air does or does not respond instantaneously? Why would it not? As the air is compressed, it pushes air out of the port. As it is expanded, it sucks air into the port.

Also, I believe this to be a point you guys may have been trying to address: the full volume of air will not be able escape the length of the port until a particular frequency. To further explain what I mean, at higher frequencies the cone will begin to suck air back into the port before the full volume of air within the port has escaped. At lower frequencies the opposite will happen. An excess volume of air, beyond that which is within the port, will escape the port before the cone begins to suck air back in again. This may or may not have any importance in this argument, but I thought that it might have been a critical point in understanding this concept. If it is, I do not know how you would like me to incorporate it with the explanations given.

If the affect on the port is not instantaneous, will there be a particular frequency were this lag will result in the sucking of air into the port while the cone in compressing the air within the box / the pushing of air out of the port while the cone is expanding the air within the box? This would explain the suppression of cone movement, both during compression and expansion, and a increase in pressure during compression. Is this frequency the Fb?

Tony Williams

P.S: You guys are the best for hanging in there with me!!! :D

[whatatrip]

Floppygoat, if I understand porting correctly, I think you are closer to understanding this than you think you are. I may not be right but I will try to explain my understanding.

You asked "Now, this mass of air does or does not respond instantaneously? Why would it not? As the air is compressed, it pushes air out of the port. As it is expanded, it sucks air into the port."

It does not respond instantaneously --- as a whole --- because it does compress and expand in response to the driver. The air closest to the driver moves the fastest as it compressed against the air next to it. It is kind of like people in a line where one person begins pushing the person in front of them in a line and that person pushing the person in front of them and so on. Each person absorbs a little bit of energy (compression) before exerting force on the person in front of the. The same works in reverse during decompression. It takes time for the "wave" to reach the end of the line. Most of the time the driver is moving air in and out of the port so fast that you mostly have compressed air and decompressed air, the air never really totally leaves the port completely. Well, I don't know for sure about low frequencies but I am pretty sure that is true with higher frequencies. The point is is that by the time the driver moves in the direction of compressing the air in the port and the air begins to compress, the driver is already moving in the other direction and that motion begins decompression of the air.

Others have talked about a spring with a weight on it and how the spring and weight behave when you pull and push on it. If you hang a weight on a spring and pull up one foot very fast on the spring, your hand will move up much faster than the weight (depending upon how much the weight weighs and how stiff the spring is, of course). The spring pulls on the weight but doesn't pull the weight up instantaneouly in direct proportion to the movement of your hand. The spring stores the energy you exerted it and that stored energy is used to move the weight up even after your hand stops moving. The air in the port acts like a spring/weight combination, more like a spring all by itself. Remember the "slinky", that coil of wire used as toy? It has mass and it is a spring. If you move one end fast the other end does not move as fast as your hand does. Same with the mass of air in a port.

You further write "To further explain what I mean, at higher frequencies the cone will begin to suck air back into the port before the full volume of air within the port has escaped. At lower frequencies the opposite will happen. An excess volume of air, beyond that which is within the port, will escape the port before the cone begins to suck air back in again. This may or may not have any importance in this argument, but I thought that it might have been a critical point in understanding this concept."

Yes, I think it is important to understanding this concept. Like I said, I am not sure that all the air is expelled from the port at lower frequencies all the time but you are asking the right questions. If the driver displaces more air than is occupied by the port, then it could expell all the air before it sucks it back in. But this, I think, would be at very low frequencies and the size and length of the port would come into play.

You finally wrote "If the affect on the port is not instantaneous, will there be a particular frequency were this lag will result in the sucking of air into the port while the cone in compressing the air within the box / the pushing of air out of the port while the cone is expanding the air within the box? This would explain the suppression of cone movement, both during compression and expansion, and a increase in pressure during compression. Is this frequency the Fb?"

Yes, I think that is Fb. At high frequencies the the air in the port is just kind of sitting there being compressed and decompressed very little but as you go lower in frequencies, the pushing and pulling by the driver on the air gets slower and slower and the air in the port begins to move with the driver more and more until you reach the point where the mass of air in the port is pushing when the driver is pushing and the mass of air is pulling when the driver is pulling. That is what I understand Fb to be and it looks to me like you got it.

[floppygoat]

Ok then...It seems this can be put to rest for the time being. Thanks again to everyone who helped explain, I appreciate it.


Tony Williams

[cpcarter]

Floppygoat
This is all great info. Just give it time to sink in. I like Jeff's approach showing all the elements in graph form. A picture really is worth a thousand words. Keep looking at the graph and let it soak in. It's all there.

Not to beat a dead horse but....Here's another approach. All of this dicussion shows how to do it right. But what if you don't use the right size box or right size port? Sometimes it helps to understand a phenomenon if you take it to an extreme and observe the possible negative results of doing it wrong.

The resonance of a ported box is a double edge sword. You can use the resonance but you must also control it or damp the oscillations. One analogy that hasn't been touched upon ( if you're still awake) is the suspension in a car. The weight/mass of the car is suspended by springs that allow the suspension to react independently to bumps in the road. The shocks(or dampers as the Brits correctly call them) are tuned to the springs and the mass to control the response to road irregularities. In a perfect response, the suspension would follow the bump pefectly without disturbing the chassis. If you've experianced a car with worn out shocks, that bouncing, floating, out of control, feeling is uncontrolled /underdamped oscillation.

In a perfect speaker response the output would follow the input signal exactly without exciting additional resonances in the box. An underdamped resonance in a speaker will also be out of control. This creates an audible distortion in that you are hearing the box resonate in addition to the speaker motor responding to the electrical input signal. This can result in peaks/nulls in the bass response with the most obvious being the "one note wonder" single frequency boom charachteristic that you hear from the Honda Civic next to you at the stop light.

So I guess what I'm trying to say is that you can characterize the complex elements that make up the compliance of a car suspension or a speaker system mathmatically if that is the way you think or just observe the similarities in any resonating system and the effect that changing any element has on the overall system response.
I know I'll get blasted for being vague and incomplete but for the sake of brevity ...
Just a thought that might help....or maybe not.
Keep trying and Good luck

CC

[rogoll]

[QUOTE=floppygoat;1575259]Ahhh, the world of audio. The more I learn, the more I don't understand. Over the last year my intense interest has led to growing confusion. It will take a while to digest the responses you have given. QUOTE]

"The more one learns, the less one knows. The further one travels, the less one really goes..." -The Beatles.

I found that reading, and re-reading material let's it seep in until finally I come to an understanding. The book, "Speaker Building 201", by Ray Aldon has helped a lot!

[floppygoat]

Thank you everyone, I will continue to read, and read, and re-read. You all have been a lot of help.

Tony Williams