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Speaker driver beaming frequency formula

Music1969

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Hi all

I saw this below formula for calculating speaker driver beaming frequency for driver diameter but when I saw this video by Danny, his numbers for various driver sizes don't match my spreadsheet.

Watch Danny from 7 minutes to 8 minutes. Can anyone help with the correct formula?

Thanks!

1622365745573.png


 
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Jim Matthews

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I thought beaming began when the wavelength produced is less than the diameter of the cone.

Using centimeters:

(Speed of sound) 34400/(Diameter of 8 inch cone)20.32 = 1673 Hz.

Your spreadsheet returns 1074 Hz.

15 inch returns 902 Hz off my Math.
 

Rick Sykora

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I thought beaming began when the wavelength produced is less than the diameter of the cone.

Using centimeters:

(Speed of sound) 34400/(Diameter of 8 inch cone)20.32 = 1673 Hz.

Your spreadsheet returns 1074 Hz.

15 inch returns 902 Hz off my Math.

Will just add that the significant variable here is the cone diameter. The effective cone diameter of a driver is smaller than the driver‘s physical size. For example, have seen others use 17 cm for an 8 inch driver.
 
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Jim Matthews

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Will just add that the significant variable here is the cone diameter. The effective cone diameter of most drivers is smaller than the cutout. For example, have seen other use 17 cm for an 8 inch driver.
That returns a "start of beaming" threshold of 2023 Hz - double the OP spreadsheet!
 

Rick Sykora

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That returns a "start of beaming" threshold of 2023 Hz - double the OP spreadsheet!

Yes, the calc is more a rule of thumb. Tend to use the actual driver off-axis measurements for specific design. :cool:
 

Jim Matthews

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Yes my spreadsheet formula is same as yours x (2 / pi)
Why do you apply those multipliers? They're the source of your discrepancy with DR's video.
 
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Music1969

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Is it anything less than omnidirectional?

Good question, I'll say this quoted, to keep the discussion a bit simpler.

I definitely appreciate things can get very complicated quickly
 

Jim Matthews

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I believe the article is misdirecting your attention. It proports to deal with lobing, where adjacent drivers *might* begin to interfere with each other at a given separation.

Practical crossover design employs the fewest parts to achieve the smoothest transition between drivers. Using your calculator would (effectively) double the number of drivers required.
 
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Music1969

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I believe the article is misdirecting your attention. It proports to deal with lobing, where adjacent drivers *might* begin to interfere with each other at a given separation.

Practical crossover design employs the fewest parts to achieve the smoothest transition between drivers. Using your calculator would (effectively) double the number of drivers required.

Noted. It mentioned beaming specifically and only difference with your formula is (2 / pi) factor:

1622374002750.png
 
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Music1969

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Here they are talking about a simple 2-way design as an example

I guess this formula is more conservative (2 / pi = 0.64 factor) so no harm , if you can find the correct driver?
 

astrex342

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One discrepancy may be due to the "inch" number or the driver not actually being the diameter of the sound recording surface.
The inches in drivers is usually the total diameter of the woofer flange, or the diameter of the screw holes. The actual sound radiating surface is usually an inch or two smaller.
Take the example of the Scanspeak illuminator 6 inch. The actual sound radiating diameter is 108mm or 4.25 inches. The 150mm or 6 inch refers to the woofer flange diameter.

Therefore, I think the formula and the video are both correct. But to use the formula correctly, you should use the actual sound radiating diameter. It can be calculated with 2*(Sd/π)^0.5
 
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Music1969

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One discrepancy may be due to the "inch" number or the driver not actually being the diameter of the sound recording surface.
The inches in drivers is usually the total diameter of the woofer flange, or the diameter of the screw holes. The actual sound radiating surface is usually an inch or two smaller.
Take the example of the Scanspeak illuminator 6 inch. The actual sound radiating diameter is 108mm or 4.25 inches. The 150mm or 6 inch refers to the woofer flange diameter.

Therefore, I think the formula and the video are both correct. But to use the formula correctly, you should use the actual sound radiating diameter. It can be calculated with 2*(Sd/π)^0.5

I think you're right.

That website where I got the formula shows an example with 6.5" Seas: http://www.seas.no/images/stories/prestige/pdfdatasheet/H1215_CA18RNX_Datasheet.pdf

But in their example they state "effective diameter of the driver is 13.16 cm so 0.1316 m" even though 6.5 inch is 0.1651m
 
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Music1969

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One discrepancy may be due to the "inch" number or the driver not actually being the diameter of the sound recording surface.
The inches in drivers is usually the total diameter of the woofer flange, or the diameter of the screw holes. The actual sound radiating surface is usually an inch or two smaller.
Take the example of the Scanspeak illuminator 6 inch. The actual sound radiating diameter is 108mm or 4.25 inches. The 150mm or 6 inch refers to the woofer flange diameter.

Therefore, I think the formula and the video are both correct. But to use the formula correctly, you should use the actual sound radiating diameter. It can be calculated with 2*(Sd/π)^0.5

Ok working backwards to workout their example they've used this from the Seas spec sheet to calculate effective diameter, and then plugged into their formula

1622375414407.png
 
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Music1969

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So I think this is solved

f = (2 * c) / (π * D)

Where :
  • f = frequency where the speaker starts to beam (Hz).
  • c = speed of sound (343 m/s).
  • D = effective diameter of speaker (m) = calculated from effective piston area on driver spec sheet
 

hardisj

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Beaming = speed of sound / driver effective diameter / 2

effective diameter = half surround to half surround

divide by 2 to get half wave

some will divide by 4 but 2 is close enough and makes more sense when talking about CTC spacing.
 
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