Now Thats Deep

Assume I design a simple 4th order boom box, single driver, and single port. Assume the driver is 10 inch sub, calculated box volume is .75 ft^3. I am also very meticulous in my measurements. I also take into consideration the port volume and the volume of the back of the woofer in the over all box volume. After construction is complete, I take the time accurately measure the compliance ratio of the box with an impedance curve in its sealed state (port sealed off) and in its vented state. It turns out from (alpha=Vas/Vb) that the box volume is different from what I initially designed and therefore the tuning frequency is different. Now I want to adjust the port length to adjust the tuning frequency of the enclosure to what I thought it was going to be to start with. At the same time by extending the port the box volume increases (or vise versa). My question is. When I calculate the new port length do I add the additional port volume to the port length equation? I don’t know what that is until I calculate the port length from that equation. Any thoughts. Even in the initial design you can’t accommodate for the added box volume from the port because you can’t calculate the port length with out knowing the additional port volume. You can’t get the additional port length because you don’t know the final box volume.
 
When recalculating, the new port volume has to be considered.

Keep at it. Eventually you'll zero in and quit when the difference between the new length and the old is small enough to ignore. This should take only a handful of iterations.

Unless there is a magic formula?
 
Jason, I think see a possible problem. When you sealed off the port, which end did you seal off? If it seems that the sealed box has a larger volume than you expected, and this volume corresponds to the port volume..... the air in the port becomes part of the sealed box volume when you seal it off and allow the air in it to remain connected to the box.

If you're going about this mathematically, it seems to me that the easiest way is to assume (or actually use) a port that's external to the box. Then you figure out the port volume, and make the box exactly that much larger (remember to account for the volume of the tube material).

What I'm trying to say- and i hope this helps- is that additional port volume does NOT count toward box volume for calculating tuning frequency. It's a really strange equation if you try to calculate it with the port inside; it's much easier when the port is outside.

I'd say that the bottom line is- if you're using software, find out what the software's assumptions are (internal vs external port). Also, if you really need to fine-tune your box, consider an adjustable port, or possibly slightly oversizing the box and adding chunks of wood to take up volume.

Joe
 
Additional Port Volume

I sealed the port from the outside but I wasn’t sure if the port inside the box added additional volume. Let's make sure I understand this correctly. You do not assume in the vented state that port volume is added to the total box volume.

I try not to use canned software if possible. I generally write my own. It's kind of like the difference between a frozen biscuit and one made from scratch.
 
I like the way you put that 🙂

Ok, here's how I think of it- are you treating the box as a Helmholtz resonator?

In a helmholtz resonator (as I'm sure you know), you have a spring (the volume of air inside the box) and a mass (the volume of air in the port). When you seal the box and leave the port, that port air is no longer a moving mass- it's just a stationary spring, increasing the box volume.
 
Instead of messing about with all this change in vent length causes change in box volume dilemma (which I must add is a great point that must be overlooked 99% of the time), simply measure the Fb to see if it's what you wanted!

The actual Fb may well be a little out from what equations predict anyway, so going by actual measurements you ensure the tightest result.
 
Rich:

For this post:

Port volume = volume of air in port

Box volume = volume of air in enclosure minus port volume and volume of speaker itself

Enclosure volume = volume of space inside enclosure walls, includes port volume and volume of spekaer itself, bracing, etc.

Your solution-measuring the Fb-is a good one, the one most people use. But it might not work for Jason's box. That is because in most vented boxes, changes in port volume result in negligible changes in box volume, (the port not being considered part of the box volume). In most boxes, the port is no more than 10% of the total volume in the enclosure, so the "box volume", (ie, the volume behind the driver, separate from the port volume) does not change all that significantly.

The problem with Jason is that he has a 10 incher in a box volume-0.75 cubic feet-which is way smaller than usual. And since I am sure he wants to tune it low, the port volume is likely taking up a very significant proportion of the volume of the total enclosure. Hence, his problem.

If Jason lengthens his port to achieve a certain Fb, when he arrives at that Fb it will no longer be the correct Fb because by lengthening the port, he has significantly changed the box volume. Different box volumes require different Fb's.

Example: Suppose Jason's box volume is 0.75 cu ft and his port volume is 0.5 cu ft, for a total enclosure volume of 1.25 cu ft. When he builds his box, he finds that his Fb is different from what he calculated, and that he must lengthen the port. He lengthens the port by 50%, and he now has a port volume of 0.75 cu ft and a box volume of 0.5 cubic feet-a decrease in box volume of 33%. A decrease in box volume of 33% requires different Fb, so he has to change the port volume yet again. And so on.

So measuring the Fb in this case won't get it done, the way it usually gets it done with a vented box where the port takes up considerably less percentage of the total box volume.
 
About the only advice I can give Jason is that small differences in tuning or box volume in the vented box, (+ or -) 10%, really don't result in significant performance changes, so I would just go by iteration. Use a program to calculate something close, and he'll probably end up close to where he wants to go.
 
Are you using a signal generator to sweep and plot the response of the box or are you just calculating the port longth from the math formula?

All factors must be entered into the formula. The volume of the box minus the driver area(back), minus the port area, internal bracing, terminal cup. Anything period.

The way I do it is run the formula and then fine tune with a signal generator and test setup. Once the box is tuned then run a signal sweep and feed the output into your HP plotter and you are done. I like to include a warble test also.
 
Hi,
I used Bullock and put all his equations into a spreadsheet. His explanations are very clear & I really understood what I was trying to achieve. Using someone elses software saves a lot of your time but usually you do not learn to understand the mechanisms that are at work.
How did you do you calcs?
Oh and finally after optimising the box and port I had cut 150mm off the height of the tower and resealed the base. I then found that adding mass to the cone gave a good way to optimise the bass. That must have really upset sensitivity but it worked for me.
regards Andrew T.
 
Back To The Grind

Ok guys I have constructed an enclosure for a PK BL-10 subwoofer. I have a car audio store near buy lets me play with their stuff allot. Anyways I have taken in consideration the volume occupied by the back of woofer and magnet structure. I very carefully added the volume of the duct to the box volume. The port was rectangular not round. After I constructed the enclosure and generated an impedance plot for the enclosure in its sealed state and vented state. I calculated the box compliance and worked backwards using the Vas of the woofer and calculated the box volume. The volume measured 1.53. My target volume was 1.5 five. So this answers my question that the duct volume is included in the overall volume of the enclosure. But the thing that I found that was so interesting was that my calculated tuning frequency was 34Hz but my measured was 26Hz. Why such a huge difference. I was very meticulous with all of the construction and I know the duct cross sectional area and length was exactly correct. From what I have seen on the internet and in literature there is a large amount of error in what is calculated and what is measured in the duct tuning frequency. So my next quest is to find out how to make the duct equation produce better results. I reviewed the derivation of the duct equation and learnt how it was derived. My first thought is that the equation doesn’t take into consideration the air outside of the enclosure directly in front of the port. If the woofer displaces air the amount displaced will exit in or out of the port and then expands. Regardless a certain amount still acts as part of mass of the port. So I figure that the maximum amount of air that can be displaced is the Vd of the driver. Assuming for a peak to peak excursion the volume displacement of the woofer is Vd=2*Xmax*Sd. I re derived the duct equation to consider this. It is given below. Any thoughts on this?

(c^2*R^2)/(4*pi*Fb^2*Vb)-(2*Vd)/(pi*R^2)-1.463*R=Lv

The equation does produce closer results but isn’t entirely correct.
 
you are correct, the effective port length extends beyond the actual duct length. This effect is not entirely consistant however. It varies for flanged and free ends, and also for different flare curvatures and radii. WinISD takes this into consideration somewhat with their standard end correction factors. I have become desensitized to the subtle differences in calcualted tuning and effective tuning. If they're off by less than 1hz you will probably never tell the difference.
 
Did you use this formula to calculate the 34 Hz ?

( ( 8466.4 * R^2 ) / ( Vb * Fb^2 ) ) - (1.463 * R )
So for a 3" port in your 1.5 cu.ft = ( 19049.4 / 1734 ) - 2.1945 = 8.79" long


Then if it's rectangular :

( ( 8466.4 * ( Surface / Pi ) ) / ( Vb * Fb^2 ) ) - (1.463 * sqrt( Surface / Pi ) )
So for a slot port 1" high by 6" long in the same box so Surface = 6 cubic inches...
( 16169.6 / 1734 ) - 2.0218 = 7,30" long

But then my formula might miss a correction factor caused by the way air enter the port in the box by only by one side.

My formula for the round port is the usual dual flared port with correction factors included.
 
I got 26,69" long for 34 Hz.

I guess maybe the box volume is not correct, let me check for the correction factor of a slot port.

Checked WinISD for fun, it gives 26,56" long for a slot port.
I guess the volume is not correct, will continue to search.

If you know the internal box dimensions, we could check that to have a rough estimation.
 
I found out a good website with all the correction factors needed...
http://www.jlaudio.com/tutorials/ports/

I checked and doublechecked...
Are you sure you calculated the port length in the box according to the method listed on the website above?
Could you give us internal box measurements?
Even with the correct correction factor, you're still at 34 Hz with 27 inches. Maybe your measurements about the actual Fb are wrong?

Good luck !!!