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Recent AR-2ax Woofer Measurements


Guest kevemaher

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Guest kevemaher

Hi,

I've attached a word file with the results of some recent measurements I've made on an AR-2ax woofer.

I also have a question about the Vas value. I've seen values of 3.5 cu. ft. posted on this forum. The volume of the empty AR-2ax cabinet is about 2.2 cu. ft. How does this affect the woofer's low frequency performance? What is a good reference for me to read up on how this parameter affects the woofer performance? I'm presently planning to install a midrange that needs an enclosure of about 0.18 cu.ft. How will this alter the woofer's performance?

Kevin

AR_Woofer_measurements.doc

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Hi,

I've attached a word file with the results of some recent measurements I've made on an AR-2ax woofer.

I also have a question about the Vas value. I've seen values of 3.5 cu. ft. posted on this forum. The volume of the empty AR-2ax cabinet is about 2.2 cu. ft. How does this affect the woofer's low frequency performance? What is a good reference for me to read up on how this parameter affects the woofer performance? I'm presently planning to install a midrange that needs an enclosure of about 0.18 cu.ft. How will this alter the woofer's performance?

Kevin

AR_Woofer_measurements.doc

A goal in the AR patent for the acoustic suspension loudspeaker was to have the box compliance as seen by the woofer be ten times lower (stiffer) than the speaker suspension. This ratio Vas/Vc(closed box volume) is referred to as alpha in T&S theory and it would be 10 in this case. However, it is difficult to manufacture a driver with a very loose suspension and later models have had an alpha on the order of 3 to 5. There is an interesting characteristic of high alpha systems which is that the closed box resonance Fc, is much more dependent on the box volume and cone mass than with lower alpha systems. The bottom line is that if you measure the driver in system and Fc is correct, and also Qtc then the system will perform as designed.

I suggest you measure the woofer in system with the inductor jumped (shorted) out of the circuit.

2.2 cu ft for an AR-2ax does not sound right at all, check your math, or did you apply a correction for the stuffing?

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Guest kevemaher

Pete,

Thanks for your comments. The AR lit has box dimensions of 13"x24"x11.5". I measured about the same. This is about 2 cu. ft.

I also measured the impedance of the woofer in the box with the series inductor shorted and obtained about the same result; Fs = 31Hz and R=52ohms.

I'm still learning what all these parameters mean and how they interact

.

I am planning to install a 4.5" midrange in the tweeter hole, with an enclosure of about 0.07 cu ft. behind it (this is actually less than the Vas of 0.15 cu. ft qouted by the mfgr, but I figure this is a a good compromise since I will be crossing over at 275Hz or higher with an active 4th order filter and therefore don't need the bass extension). But this will reduce the volume available for the woofer.

I am replacing the fiberglas with some Acoustastuff material. I could pack it in to increase the "stiffness" of the enclosure. I admit this is not a very scientific approach. I am still searching for some analytical guidelines. Perhaps I will learn through experiment.

Further comment and criticism is always appreciated and encouraged.

Kevin

A goal in the AR patent for the acoustic suspension loudspeaker was to have the box compliance as seen by the woofer be ten times lower (stiffer) than the speaker suspension. This ratio Vas/Vc(closed box volume) is referred to as alpha in T&S theory and it would be 10 in this case. However, it is difficult to manufacture a driver with a very loose suspension and later models have had an alpha on the order of 3 to 5. There is an interesting characteristic of high alpha systems which is that the closed box resonance Fc, is much more dependent on the box volume and cone mass than with lower alpha systems. The bottom line is that if you measure the driver in system and Fc is correct, and also Qtc then the system will perform as designed.

I suggest you measure the woofer in system with the inductor jumped (shorted) out of the circuit.

2.2 cu ft for an AR-2ax does not sound right at all, check your math, or did you apply a correction for the stuffing?

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Pete,

Thanks for your comments. The AR lit has box dimensions of 13"x24"x11.5". I measured about the same. This is about 2 cu. ft.

I also measured the impedance of the woofer in the box with the series inductor shorted and obtained about the same result; Fs = 31Hz and R=52ohms.

I'm still learning what all these parameters mean and how they interact

.

I am planning to install a 4.5" midrange in the tweeter hole, with an enclosure of about 0.07 cu ft. behind it (this is actually less than the Vas of 0.15 cu. ft qouted by the mfgr, but I figure this is a a good compromise since I will be crossing over at 275Hz or higher with an active 4th order filter and therefore don't need the bass extension). But this will reduce the volume available for the woofer.

I am replacing the fiberglas with some Acoustastuff material. I could pack it in to increase the "stiffness" of the enclosure. I admit this is not a very scientific approach. I am still searching for some analytical guidelines. Perhaps I will learn through experiment.

Further comment and criticism is always appreciated and encouraged.

Kevin

Hmmm, you really need to read all the literature in the library on the AR-2ax if you want to understand this. You need to subtract the thickness of the wood and account for the front panel set back, you will get a much smaller number than 2 cuft. Obviously, it is the internal volume that matters. Did you seal the tweeter and mid holes when you measured 31 Hz in box? It should be mid 50s if you read the literature which I confirmed here:

http://www.classicspeakerpages.net/IP.Boar...92;pete+b\

All resonable damping materials raise the effective volume to a point and then decrease it as the volume of the fibers begin to significantly detract from the box volume.

The answer is that the small reduction in volume due to the mid chamber is not an issue at all. Fc is inverse square law with regard to volume, you will probably have a hard time measuring the difference in Fc, and Qtc, which are the main parameters that determine the response shape.

I'm somewhat interested in what you're doing because I agree that the AR-2ax would benefit from improved drivers and crossover.

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Guest kevemaher

Pete,

Thanks again for your comments. Of course, the inner box volume will be less than the 2 cu. ft. I will measure it as I replace drivers. I've made all the in-box woofer measurements with a sealed box and the mid and high drivers disconnected, but still in place.

My plans are to put a modern-day mid in the tweeter hole and a modern-day tweeter in the tweeter hole. Both will be bench and in-box characterized. I will gut the insides. I will use a Rane active 4th order LR crossover so I can play with crossover frequency, SPL balance, and phase. My target crossover frequencies are 300 and 3000 Hz. There will be Zobels and DC protection on the drivers as needed. At some point, I will also replace the woofer.

I am still learning about all this. I am particularly weak in the mechanical area. I don't have woodworking skills or tools. I am having fun. I have a very old pair of AR-2ax speakers with new caps and pots that I will use as my sound reference.

I will share my experiences as I proceed.

Kevin

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  • 7 years later...

I just went back into one of my AR2ax speakers in order to further clean up one of the pots.

With regard to the interior volume of the AR2AX speaker cabinet:

The interior volume of the cabinet is 8 3/4 inches deep X 12 inches wide X 22 1/4 inches high.

This works out to be approximately 1.35 cubic feet of volume.

I did not subtract out the volume of the pots, xover or woofer.

I know this is an old but I just had to set the record straight.

AR2AX volume ~ 1.35 cubic feet.

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