

Thread Tools 
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
Hey folks,
I wanted to break out a separate thread as not everybody who might benefit from this information is following Marty's and Worm's build threads.
I am just going to copy a post from Marty's thread to start the discussion off here.
Source: http://www.avsforum.com/t/1489541/largeporteddaytonho18s/60#post_23725891
ok, thanks to worm, we really went deep into understanding why winisd miscalculates actual vs. theoretical enclosure tuning.
several people have observed this error between theory and actual results and one great post is here:
http://www.troelsgravesen.dk/vent_tuning.htm
ports behave differently depending on how they are mated to the air at the end of the port. this is called "end correction" and is quantified by a mysterious variable called "k".
for ports that dump cleanly into free space, end correction is low, something like k=0.732 on a baffle and even a little lower if the port has no baffle.
for slot ports that are on the ground, the boundary in front of the port and behind the port have the effect of synthetically extending the length of the port.
as a result, these kinds of ports provide a lower tuning frequency than winisd estimates using k=0.732
this picture shows how end correction (i.e. k values) increase as one moves more toward a slot type port:
source: http://www.diyaudio.com/forums/multiway/97204endcorrectionformularectangularport.html#post1148924
i do not speak german (if that is even the language used here), but i interpret it to be indicating whether there is 1, 2, or 3 boundaries reinforcing the port either in front of or behind the port.
there is also some averaging that occurs, so if it is k=1.5 to the front of the port and k=1.0 to the rear of the port, the average would be k=1.25.
here is the marty sub with the revised 28 inch long slot port.
it can be seen how changing the value of k impacts the tuning frequency.
here is the link to the calculator:
http://www.calculatoredge.com/new/ventlength.htm#factor
for k=0.732, the calculator gives the same results as winisd. however, the version of winisd that i have for windows 7 does not allow changing of the end correction to other value, hence the use of the calculator.
with all this considered, i think that a reasonable best guess with this information and targeting a 1617hz tuning is to use a 28" long slot port with k coming in at somewhere around 1.75.
i will adjust prior posts to reflect this.
So, if folks have more information on this topic, fire away!
I wanted to break out a separate thread as not everybody who might benefit from this information is following Marty's and Worm's build threads.
I am just going to copy a post from Marty's thread to start the discussion off here.
Source: http://www.avsforum.com/t/1489541/largeporteddaytonho18s/60#post_23725891
ok, thanks to worm, we really went deep into understanding why winisd miscalculates actual vs. theoretical enclosure tuning.
several people have observed this error between theory and actual results and one great post is here:
http://www.troelsgravesen.dk/vent_tuning.htm
ports behave differently depending on how they are mated to the air at the end of the port. this is called "end correction" and is quantified by a mysterious variable called "k".
for ports that dump cleanly into free space, end correction is low, something like k=0.732 on a baffle and even a little lower if the port has no baffle.
for slot ports that are on the ground, the boundary in front of the port and behind the port have the effect of synthetically extending the length of the port.
as a result, these kinds of ports provide a lower tuning frequency than winisd estimates using k=0.732
this picture shows how end correction (i.e. k values) increase as one moves more toward a slot type port:
source: http://www.diyaudio.com/forums/multiway/97204endcorrectionformularectangularport.html#post1148924
i do not speak german (if that is even the language used here), but i interpret it to be indicating whether there is 1, 2, or 3 boundaries reinforcing the port either in front of or behind the port.
there is also some averaging that occurs, so if it is k=1.5 to the front of the port and k=1.0 to the rear of the port, the average would be k=1.25.
here is the marty sub with the revised 28 inch long slot port.
it can be seen how changing the value of k impacts the tuning frequency.
here is the link to the calculator:
http://www.calculatoredge.com/new/ventlength.htm#factor
for k=0.732, the calculator gives the same results as winisd. however, the version of winisd that i have for windows 7 does not allow changing of the end correction to other value, hence the use of the calculator.
with all this considered, i think that a reasonable best guess with this information and targeting a 1617hz tuning is to use a 28" long slot port with k coming in at somewhere around 1.75.
i will adjust prior posts to reflect this.
So, if folks have more information on this topic, fire away!
Listen. It's All Good.
Sponsored Links  
Advertisement


AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
Let's look at this a bit more, as I see you have now copied and pasted this in several places.
An overview of the process by which the simulation programs calculate port length is as follows:
L_{ACT} = L_{EFF}  k * D
where k is the end correction factor and D is the diameter of the equivalent cylindrical port having crosssectional area A.
Now, it's been observed that using this equation with k = 0.732 gives incorrect port lengths, and the error results in a port that's usually too long. The website to which you previously linked shows errors between calculated and actual length, but does not establish the why.
For the sake of argument, let's assume your hypothesis is correct, namely that this error is solely due to an error in the end correction factor k. That is, assume there is a value k_{RIGHT} giving the correct length L_{RIGHT} and a value k_{WRONG} giving a wrong length L_{WRONG}. This leads to the following:
L_{RIGHT} = L_{EFF}  k_{RIGHT} * D
L_{WRONG} = L_{EFF}  k_{WRONG} * D
Let's compute the percent error in length as follows:
percent error in length = (L_{RIGHT}  L_{WRONG}) / L_{RIGHT} * 100
In the subtraction of the numerator, L_{EFF} drops out, and the error term in the numerator only depends on the equivalent port diameter D. The result is this:
percent error in length = (k_{WRONG}  k_{RIGHT}) * D / L_{RIGHT} * 100
For a given port area, the numerator does not depend on port length at all. If the port area is constant, the numerator is constant also. That means the percent error in port length would be inversely proportional to the correct port length if this hypothesis were true. Now look at the web site with the data in the table titled "Results, undamped box". The percent error in port length is almost constant (with actual lengtth / calc. length going from 0.81 to 0.83 as the port length varies by a factor of 3). In the case of 10 cm and 20 cm port lengths, the percent error is the same.
So assuming the data of that web site is correct, the "incorrect end correction" hypothesis fails. So yes, there is an error between computed and actual port lengths, but the assumption that this is due entirely to the end correction factor leads to a contradiction.
An overview of the process by which the simulation programs calculate port length is as follows:
 Calculate the effective length L_{EFF} of the port based on box volume, box tuning frequency and port crosssectional area A.
 Calculate the actual length of the port L_{ACT} using an end correction whose value is proportional to the square root of the port crosssectional area A. This formula has the form:
L_{ACT} = L_{EFF}  k * D
where k is the end correction factor and D is the diameter of the equivalent cylindrical port having crosssectional area A.
Now, it's been observed that using this equation with k = 0.732 gives incorrect port lengths, and the error results in a port that's usually too long. The website to which you previously linked shows errors between calculated and actual length, but does not establish the why.
For the sake of argument, let's assume your hypothesis is correct, namely that this error is solely due to an error in the end correction factor k. That is, assume there is a value k_{RIGHT} giving the correct length L_{RIGHT} and a value k_{WRONG} giving a wrong length L_{WRONG}. This leads to the following:
L_{RIGHT} = L_{EFF}  k_{RIGHT} * D
L_{WRONG} = L_{EFF}  k_{WRONG} * D
Let's compute the percent error in length as follows:
percent error in length = (L_{RIGHT}  L_{WRONG}) / L_{RIGHT} * 100
In the subtraction of the numerator, L_{EFF} drops out, and the error term in the numerator only depends on the equivalent port diameter D. The result is this:
percent error in length = (k_{WRONG}  k_{RIGHT}) * D / L_{RIGHT} * 100
For a given port area, the numerator does not depend on port length at all. If the port area is constant, the numerator is constant also. That means the percent error in port length would be inversely proportional to the correct port length if this hypothesis were true. Now look at the web site with the data in the table titled "Results, undamped box". The percent error in port length is almost constant (with actual lengtth / calc. length going from 0.81 to 0.83 as the port length varies by a factor of 3). In the case of 10 cm and 20 cm port lengths, the percent error is the same.
So assuming the data of that web site is correct, the "incorrect end correction" hypothesis fails. So yes, there is an error between computed and actual port lengths, but the assumption that this is due entirely to the end correction factor leads to a contradiction.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
Yes, quite sure.
In this equation:
the expression L + K * D in the denominator is the effective length L_{EFF}
L_{EFF} = L + K * D
Solving for L (the actual length)
L = L_{EFF}  K * D
It's only proportional if the second term on the righthand side involving c is zero. You're using an erroneous definition of proportionality.
In this equation:
the expression L + K * D in the denominator is the effective length L_{EFF}
L_{EFF} = L + K * D
Solving for L (the actual length)
L = L_{EFF}  K * D
Quote:
It's only proportional if the second term on the righthand side involving c is zero. You're using an erroneous definition of proportionality.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
right, but all those terms are the same, kright and kwrong are the only things that change, so when you subtract the second from the first, you end up with length inversely proportional to k.
Listen. It's All Good.
AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
For two different port lengths (the right and the wrong one), Fb is not constant. What's constant is L_{EFF} for a given box tuning frequency, box volume and port area. The only variables under consideration here are how L is calculated from L_{EFF} (using different values of k).
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
yes it is, that is what the whole point of k is for. fb is held constant, so what change in length is required comes through k.
Listen. It's All Good.
AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
But if you use two different values of k to compute the vent length (say, the right one and the wrong one), the actual fb will change as a result of the two different lengths. The desired fb is certainly constant, but you can't just assume the actual fb will not change with a change in vent length.
Edit:
What I'm trying to say is that if you change k in your equation above
Lv = kDv + cD^2/VbFb^2
namely, take two different values of k and subtract the resulting Lv from each other, the second terms on the righthand side of the above will not cancel out, because there will be two different fb values involved. This is the actual fb, not the desired one, and the actual fb is a function of k.
Edit:
What I'm trying to say is that if you change k in your equation above
Lv = kDv + cD^2/VbFb^2
namely, take two different values of k and subtract the resulting Lv from each other, the second terms on the righthand side of the above will not cancel out, because there will be two different fb values involved. This is the actual fb, not the desired one, and the actual fb is a function of k.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
i understand what you are saying.
we just have a different understanding of what that k is doing. i read the equation as k is the link between a varying port length and a constant box tuning. the intuition seems right to me as well. the more boundaries around the exit of the port, the longer it would seem to the exiting airsynthetically extending the port or something like that.
another oddball that pops up is when a port exits fairly close to a perpendicular boundary, that too reduces effective tuning, but somehow that feels like a different effect than k, almost opposite (flow "jam up" vs. continuation of high velocity).
anyways,
we just have a different understanding of what that k is doing. i read the equation as k is the link between a varying port length and a constant box tuning. the intuition seems right to me as well. the more boundaries around the exit of the port, the longer it would seem to the exiting airsynthetically extending the port or something like that.
another oddball that pops up is when a port exits fairly close to a perpendicular boundary, that too reduces effective tuning, but somehow that feels like a different effect than k, almost opposite (flow "jam up" vs. continuation of high velocity).
anyways,
Listen. It's All Good.
AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
I'll try to put some further explanation of end correction on my website, and derive the equation that's shown in the online calculator that's been linked above.
In the meantime, I'll try one more time to expand on this whole end correction idea.
The port is an acoustic mass M_{A}. In the absence of end correction, M_{A} is given by:
M_{A} = rho * L / A (from Beranek)
where
rho = air density
A = crosssectional area
L = length of port
But there is something else going on. There's a radiation impedance (ratio of complex amplitude of acoustic pressure to volume velocity at the boundary) at each end of the tube. Using a lowfrequency approximation, Beranek shows this radiation impedance to be of the mathematical form of an acoustic mass. This is the same exact thing that causes the effective mass of the cone, M_{MS} to be larger than its mechanical mass M_{MD}. This difference between M_{MS} and M_{MD} is the acoustic mass loading of the radiation impedance resulting in an effective mass increase of the cone.
Now, since the two radiation impedance loads on the port are each of the form of an acoustic mass, and the acoustic impedance of the port is also an acoustic mass, the radiation impedances have the effect of making the tube have a larger M_{A} than that given by the formula above, just as the radiation impedance load on the woofer cone makes its mass look bigger. One could think of the larger M_{A} as resulting from an equivalent length L_{EFF} that's longer than its physical length.
In the meantime, I'll try one more time to expand on this whole end correction idea.
The port is an acoustic mass M_{A}. In the absence of end correction, M_{A} is given by:
M_{A} = rho * L / A (from Beranek)
where
rho = air density
A = crosssectional area
L = length of port
But there is something else going on. There's a radiation impedance (ratio of complex amplitude of acoustic pressure to volume velocity at the boundary) at each end of the tube. Using a lowfrequency approximation, Beranek shows this radiation impedance to be of the mathematical form of an acoustic mass. This is the same exact thing that causes the effective mass of the cone, M_{MS} to be larger than its mechanical mass M_{MD}. This difference between M_{MS} and M_{MD} is the acoustic mass loading of the radiation impedance resulting in an effective mass increase of the cone.
Now, since the two radiation impedance loads on the port are each of the form of an acoustic mass, and the acoustic impedance of the port is also an acoustic mass, the radiation impedances have the effect of making the tube have a larger M_{A} than that given by the formula above, just as the radiation impedance load on the woofer cone makes its mass look bigger. One could think of the larger M_{A} as resulting from an equivalent length L_{EFF} that's longer than its physical length.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
is that just semantics or a different idea? it seems that you are saying that a port firing into a smaller space will have greater resistance/mass loading which gives something of a similar effect as a longer port? i'm saying that by extending the boundaries around the exit of the port does the same. what does it matter if we call k end correction factor or mass loading factor?
Listen. It's All Good.
AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
Quote:
The "resistance" part is zero. The tube is an acoustic mass, which is a pure acoustic reactance in the calculation. The radiation impedance at each end is also (given a lowfrequency approximation) a pure acoustic mass (no resistance, just reactance). So you have three acoustic masses in series in the acoustic equivalent circuit: the tube itself without considering the boundary effects, the radiation impedance on the front side, and the radiation impedance on the back side. And yes, it does give the effect of a longer port. This is covered in many places in the literature.
Quote:
Huh? The discussion is about whether the discrepancy in calculated and actual length can be explained by only a different end correction factor k. I just demonstrated that if that were true, the percentage error in port length would be inversely proportional to the correct port length. It isn't. In fact, it's darned near constant, which shows that something else is going on.
Quote:
It doesn't. It was just an attempt to clear up some confusion in communication.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
i wonder what your thoughts might be on roundovers as a method for impedance matching the air in the port to the air in the room.
it came up in another thread and may be a missing variable in the traditional model.
it came up in another thread and may be a missing variable in the traditional model.
Listen. It's All Good.
AVS Addicted Member
Join Date: Apr 1999
Location: Mountain View, CA USA
Posts: 21,395
Mentioned: 2 Post(s)
Tagged: 0 Thread(s)
Quoted: 777 Post(s)
Liked: 266
A 16ft roundover would significantly improve the impedance match
Seriously, if a roundover smooths airflow at the O.D. it should increase the effective diameter of the port, but that may be offset by the reduction in velocity.
I guess the way to tell is to compare the output of two equaldiameter ports with the same core velocity but with different roundover radii.
Seriously, if a roundover smooths airflow at the O.D. it should increase the effective diameter of the port, but that may be offset by the reduction in velocity.
I guess the way to tell is to compare the output of two equaldiameter ports with the same core velocity but with different roundover radii.
Noah
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
I was asking more in the context of the effective tuning frequency. it seems that the impedance mismatch between the air slug in the port and the external air may be creating a condition where the effective tuning frequency of the port is much lower than predicted by the equation above.
Listen. It's All Good.
AVS Special Member
Join Date: Mar 2011
Location: Staten Island NY
Posts: 1,077
Mentioned: 5 Post(s)
Tagged: 0 Thread(s)
Quoted: 377 Post(s)
Liked: 99
Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
"Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?"
yep. that is as I understand it. once you get less than 1 port width to a boundary, turbulence builds up and the flow is restricted. that creates something of the same effect as increasing the length of the port, but it can also cause compression if not done properly.
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?"
yep. that is as I understand it. once you get less than 1 port width to a boundary, turbulence builds up and the flow is restricted. that creates something of the same effect as increasing the length of the port, but it can also cause compression if not done properly.
Listen. It's All Good.
AVS Addicted Member
Join Date: Apr 1999
Location: Mountain View, CA USA
Posts: 21,395
Mentioned: 2 Post(s)
Tagged: 0 Thread(s)
Quoted: 777 Post(s)
Liked: 266
A Helmholz resonator is a mechanical mass/spring system, not an acoustic one.
I don't think impedance matching significantly affects this mechanical behavior, as the acoustic coupling is poor, evidenced by ~1% efficiency.
I don't think impedance matching significantly affects this mechanical behavior, as the acoustic coupling is poor, evidenced by ~1% efficiency.
Noah
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
but what about the impedance mismatch that andy was talking about noah?
Listen. It's All Good.
AVS Special Member
Join Date: Mar 2011
Location: Staten Island NY
Posts: 1,077
Mentioned: 5 Post(s)
Tagged: 0 Thread(s)
Quoted: 377 Post(s)
Liked: 99
Quote:
So then in the annihilator ill only have 4in est. to the back of the enclosure i would need more depth or turn the ports up?
Originally Posted by LTD02
"Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?"
yep. that is as I understand it. once you get less than 1 port width to a boundary, turbulence builds up and the flow is restricted. that creates something of the same effect as increasing the length of the port, but it can also cause compression if not done properly.
"Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?"
yep. that is as I understand it. once you get less than 1 port width to a boundary, turbulence builds up and the flow is restricted. that creates something of the same effect as increasing the length of the port, but it can also cause compression if not done properly.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
that was by design to push tuning down a hair. wouldn't want to go closer than that. if you want to leave more space, that is good too, just gives a higher tune.
no need to futz with angled ports for a simple build.
no need to futz with angled ports for a simple build.
Listen. It's All Good.
AVS Special Member
Join Date: Mar 2011
Location: Staten Island NY
Posts: 1,077
Mentioned: 5 Post(s)
Tagged: 0 Thread(s)
Quoted: 377 Post(s)
Liked: 99
Quote:
I see maybe ill turn them up with a 90deg. Now does this matter as much with flared ports on either end?
AVS Addicted Member
Join Date: Apr 1999
Location: Mountain View, CA USA
Posts: 21,395
Mentioned: 2 Post(s)
Tagged: 0 Thread(s)
Quoted: 777 Post(s)
Liked: 266
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
np.
Listen. It's All Good.
AVS Special Member
Join Date: Feb 2013
Posts: 1,003
Mentioned: 1 Post(s)
Tagged: 0 Thread(s)
Quoted: 17 Post(s)
Liked: 291
Quote:
I didn't say anything at all about impedance matching or mismatch. The purpose of impedance matching is to maximize delivered power, but as Noah correctly points out, that doesn't apply in the case of lowefficiency systems.
AVS Addicted Member
Join Date: Oct 2003
Posts: 20,027
Mentioned: 311 Post(s)
Tagged: 1 Thread(s)
Quoted: 2119 Post(s)
Liked: 2332
summing the radiation impedance load to the acoustic impedance load of the air slug in the port gives the increase in mass that you are talking about.
the question is whether having a large flare would increase or decrease the radiation impedance load and in turn increase or decrease the effective mass.
or put another way, how would adding a large radius flare change the effective mass?
the question is whether having a large flare would increase or decrease the radiation impedance load and in turn increase or decrease the effective mass.
or put another way, how would adding a large radius flare change the effective mass?
Listen. It's All Good.
Sponsored Links  
Advertisement



Posting Rules  