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Multiple Faital Pro 3fe25 Back loaded horn
Multiple Faital Pro 3fe25 Back loaded horn
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Old 28th September 2019, 06:17 PM   #1
santitrucco is offline santitrucco  Argentina
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Default Multiple Faital Pro 3fe25 Back loaded horn

Multiple Faital Pro 3fe25 Back loaded horn

Hi at all, i have a dude.

If i use six or eight Faital Pro 3FE25 per Back loaded horn, could i have some problem of dispersion or comb filter ?

Only until 1,8kz frequency..

Santiago
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Old 28th September 2019, 06:49 PM   #2
GM is offline GM  United States
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Find the longest center to center [ctc] spacing to find the 1 WL XO limit with the understanding that it ideally needs to be a 1/4 WL.

For example: it's 20 cm, then 34400/20 = 1720 Hz 1W or 430 Hz 1/4 WL.

Or if 1800 Hz is the limit, then 34400/1800 = 19.11 cm or 4.78 cm.

Note that normally 1/3 WL is acceptable, most commonly used.

GM
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Old 28th September 2019, 09:27 PM   #3
santitrucco is offline santitrucco  Argentina
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Ok , i use 4 drivers , the center of the cones could be at 4.8cms !!

But with 3 colums of two drivers , the distances of cones will be differents between them.
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Old 28th September 2019, 09:29 PM   #4
santitrucco is offline santitrucco  Argentina
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Is possible to use six drivers until 1800hz ?

Is possible to use four drivers until 1800hz ?
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Old 29th September 2019, 08:23 AM   #5
oon_the_kid is offline oon_the_kid  Malaysia
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I believe probably in the original equation they assume a point source. At s distance L apart. However in this case I believe it is a full cone I don't think the equation is valid anymore. All the cones are moving together as if it is one big diaphragm, I don't there will be a combing effect...
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Old 29th September 2019, 10:00 AM   #6
GM is offline GM  United States
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Quote:
Originally Posted by santitrucco View Post
But with 3 colums of two drivers , the distances of cones will be differents between them.
Need to draw a diagonal line.

Again, I posted: "Or if 1800 Hz is the limit, then 34400/1800 = 19.11 cm or 4.78 cm."

So based on a 4.8 cm ctc spacing, you're limited to two rows of three at ~0.794 WL ctc.

GM
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Old 29th September 2019, 10:19 AM   #7
GM is offline GM  United States
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Quote:
Originally Posted by oon_the_kid View Post
I believe probably in the original equation they assume a point source. At s distance L apart. However in this case I believe it is a full cone I don't think the equation is valid anymore. All the cones are moving together as if it is one big diaphragm, I don't there will be a combing effect...
It's valid TTBOMK. They're moving as one down low, but as the frequency increases they become increasingly more independent of each other with obvious combing effect above the ~1 WL point, though this is so dependent on each human's hearing acuity, room, listening distance, etc., like a MTM or line array that it's strictly a guideline.

That said, I've wondered if an average is a better way, i.e. [4.8*9.6]^0.5 = ~6.79" in which case we could just fudge a bit and use 2 rows of four [19.11/4.8 = 3.98].

GM
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Old 29th September 2019, 01:20 PM   #8
xrk971 is offline xrk971  United States
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Multiple Faital Pro 3fe25 Back loaded horn
Try 5 with one in center of square. Like a 5 on a game die.

Series/parallel with 5 is tricky. Or use 5 parallel 16ohms for about 3ohm load assuming amp can handle that.
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Old 30th September 2019, 10:29 PM   #9
turk 182 is offline turk 182  Canada
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maybe i need to stop imbibing scotch today, but with arrays depending on layout wouldn't frequency shading make for a listenable speaker?
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Old 1st October 2019, 05:13 AM   #10
santitrucco is offline santitrucco  Argentina
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Quote:
Originally Posted by turk 182 View Post
maybe i need to stop imbibing scotch today, but with arrays depending on layout wouldn't frequency shading make for a listenable speaker?
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