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Old 25th April 2004, 09:55 PM   #1
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Default Calculating bent port length

I need some advice in how to calculate a bent port. Basically on my car audio enclosure I posted about previously, I calculated the ports out to be dual 4"ers at 20" long. I'm using the thinner 4" PVC piping for sewer use, as well as 2 90 degree bends and 2 45 degree bends. So far I was very impressed with the fit. Everything fits pretty precisely within the enclosure almost like a puzzle, but it still fits.

My target tuning freq was 33 Hz, however I gave it some tolerance due to my rather inaccurate way to measuring the internal volume of the enclosure. I filled the bottom 1/2 with water and then measured that water then vaccuumed it out with a shop vac then converted the measured water into cubic feet, then did the standard math for the top section. Then I took the driver displacement and tried to calculate the port displacement.

So basically my question is how do I accurately calculate the port length for what I have pictured? I ballparked it by putting a tape measure inside and got it as large and small as possible.

(yes those are popsicle sticks on the right side, the damn rotozip's circle cutter must have moved and the port hole was too large ).

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Old 25th April 2004, 10:11 PM   #2
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Can't really help you as per your questions, but do remember to brace the box well.
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Old 25th April 2004, 10:24 PM   #3
Mikey p is offline Mikey p  Canada
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If you don't know the exact internal volume then it would be a close guess as to how long the ports should be. If you want to be accurate then download test tones off the net (20-50hz, 5 seconds each) and burn to a cd. Install the driver and play the tones. The tone that the driver moves the least is the tuning frequency. If you're lucky it will be right on. If not then add or subtract from the ports depending on if the tuning is to high or to low.
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Old 25th April 2004, 10:55 PM   #4
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The port length is measured down the center end to end. Having the ends of the port that close to the wall will increase it's effective length. From the picture it looks like the interior end is somewhat restricted in that the full 4" cross sectional area isn't available. If so, it will have a huge effect on tuning because you no longer have a 4" port.

How did you come up the ports? They look really big for a small box. What's your interior volume net of the driver and the ports?
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Old 26th April 2004, 12:04 AM   #5
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Thanks all for the replys.

Quote:
Originally posted by johninCR
The port length is measured down the center end to end. Having the ends of the port that close to the wall will increase it's effective length. From the picture it looks like the interior end is somewhat restricted in that the full 4" cross sectional area isn't available. If so, it will have a huge effect on tuning because you no longer have a 4" port.

How did you come up the ports? They look really big for a small box. What's your interior volume net of the driver and the ports?

Came up with the ports using WinISD and Perfect Box, as well as using the manuf. recommendations. The woofer is one of those high excursion woofers so it can move ALOT of air.

I calculated the port to displace approx .3 cubic foot of air, and the driver diplaces .25 according to the manuf. After this, the enclosure is approx 2.5-2.7 cubic foot.

The interior part I think has full 4" cross section available, its hard to see from that 2D pic, but I was careful about this.

Oh well, Ill just plop the driver in and try some test tones. It is for car audio so the application isnt _that_ critical hehe.
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Old 26th April 2004, 11:29 AM   #6
Rocky is offline Rocky  Norway
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With the vent placement you have chosen, you will experience at least some compression.. This makes it very hard to calculate the correct vent size, but you still have the advantage of trial and error. Tune your ports to perfection.

With such large vents, the compression shouldn't be a problem exept in the math..

BTW: the effective length of your 90 degree turn is easily found geometric by (2*pi*r)/4, where r is the centre turn radius, composed of the inner turn radius + ((outter turn darius - inner turn radius) / 2)
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Old 26th April 2004, 12:03 PM   #7
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Just out of curiosity, have you tried dropping the subwoofer into place since installing the ports? From the looks of things, those ports and the speaker's magnet may want to occupy the same real estate.
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Old 26th April 2004, 01:47 PM   #8
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Rocky,
What do you mean by "compression"? That's a new term for me in this context.

Those ports seem very close together which is going to have some kind of effect. Not so much on the push out, but on the in flow they will be fighting each other. Also, that close to the wall, the bottom in this case, the effective length of the ports will be a little longer.

It really does seem like risky port placement, so don't glue the PVC until it sounds perfect. With all that hard work, you deserve perfect sound.

If you have some left over resin and fiberglass, the cab would definitely benefit from bracing. Use part of the circle cutout for a brace in the bottom between the vents, which would prevent interference between the 2. Some braces that attach the baffle to the sides is a definite. Use foam rubber between the ports and bottom, and between the car and the bottom and sides to prevent rattles and give the whole thing more rigidity and resonance dampening.
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Old 26th April 2004, 02:17 PM   #9
Rocky is offline Rocky  Norway
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Quote:
Originally posted by johninCR
Rocky,
What do you mean by "compression"? That's a new term for me in this context.

Compression is what happens when the vents inner mouth is placed to close to an obstacle, like a wall. The air is not properly distributed in the box to create an even pressure, but is "trapped" in a high (or low) pressure "pocket".. It's like if you blow through a tube, keeping your hand or finger in front of the end, you'll notice it takes more effort to blow as closer your hand is to the tube end....
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Old 26th April 2004, 02:28 PM   #10
Rocky is offline Rocky  Norway
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Quote:
Originally posted by johninCR
Those ports seem very close together which is going to have some kind of effect. Not so much on the push out, but on the in flow they will be fighting each other. Also, that close to the wall, the bottom in this case, the effective length of the ports will be a little longer.
Not 100% sure I must say, but I don't think the vents will "fight each other" like you suggests... Assume you are right; wouldn't this phenomenon apply to splitted vents as well? like two intakes joined into one exhaust (or the other way around)? Wouldn't the same laws of physics apply here?
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