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Old 30th November 2004, 05:21 PM   #1
jwatts is offline jwatts  United States
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Default 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 donít know what that is until I calculate the port length from that equation. Any thoughts. Even in the initial design you canít accommodate for the added box volume from the port because you canít calculate the port length with out knowing the additional port volume. You canít get the additional port length because you donít know the final box volume.
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Old 30th November 2004, 05:38 PM   #2
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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?
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Old 30th November 2004, 05:46 PM   #3
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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
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Old 30th November 2004, 06:23 PM   #4
jwatts is offline jwatts  United States
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Default Additional Port Volume

I sealed the port from the outside but I wasnít 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.
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Old 30th November 2004, 06:33 PM   #5
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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.
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Old 30th November 2004, 06:44 PM   #6
jwatts is offline jwatts  United States
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Default Additional Moving Mass

Excellent Clarification.

I see said the blind man.
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Old 30th November 2004, 07:21 PM   #7
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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.
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Old 4th December 2004, 10:05 AM   #8
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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.
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Old 4th December 2004, 10:15 AM   #9
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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.
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Old 4th December 2004, 11:21 AM   #10
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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.
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