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Math on speaker air movement

Discussion in 'Amps and Cabs [BG]' started by davelowell, Jan 4, 2004.

  1. davelowell


    Jan 18, 2001
    stl, mo
    I am looking for the equation that tells you how much air a particular cabinet moves.

    I am interested in comparing and contrasting some cabs that may be purchasable.

    Thanks in advance,

  2. notanaggie

    notanaggie Guest

    Sep 30, 2003
    That is a bit of variable, since power out into the air is "volume velocity" and "air movement" volume thus changes with frequency.

    But the effective piston area and the max electrically driven excursion will tell you how much air displacement is possible with a particular speaker.

    That said, a tuned box presents a higher impedance to the speaker at the tuning frequency, so the cone movement is LESS at box resonance, but power output can be same or more.
    And a horn speaker's cone hardly moves, but it can be loud as heck.

    So what are you trying to find out specifically?

    Maybe you would be better off comparing efficiency and using the -3 or -6 dB frequency to check low limits.
    Trouble is, those are often fudged a bit to look good......or the cab might be anything but flat, so the effciency is taken at a peak......possibilities are endless.
  3. aladdin


    Mar 7, 2003
    Chiba, Japan
    I think Dave wants to know how many cubic inches of air each driver will move.

    Just square each driver's diameter and add up the products for each cabinet.

    Now, keep in mind, what notanaggie said still holds true. But I think this is the answer you were looking for.


    1x15 = 225 cu. inches
    2x15 = 450 cu. inches
    1x10 = 100 cu. inches
    2x10 = 200 cu. inches
    4x10 = 400 cu. inches
    1x12 = 144 cu. inches

    Etc. You do the math. An SWR Triad (new Model) would move: 325 cu. inches. (1x15 + 1x10)


    {Edit: The above is only meant to give you a ratio regarding the drivers, all other things being equal}
  4. alexclaber

    alexclaber Commercial User

    Jun 19, 2001
    Brighton, UK
    Director - Barefaced Ltd
    Nooooo!!! If you want to find out how much air a given driver can displace you have to figure the travel (Xmax) into the equation.

    However it's not as simple as that. The best bet is to look at sensitivity and frequency response specs but remember that they're often exaggerated.

  5. Air volume = Sd * Xmax

    Sd = piston area
    Xmax = maximum linear cone movement

    These are physical characteristics of the driver, and not tuning dependent.
  6. BruceWane


    Oct 31, 2002
    Houston, TX

    Ummm........the formula to calculate the area of a circle is Pi times (Radius squared).....so for a 15" speaker.......

    3.14159 X 56.25 = 176.714
  7. aladdin


    Mar 7, 2003
    Chiba, Japan
    The guy wants to compare cabinets in a simple fashion. What I wrote is a simple way for him to do so.

    BruceWayne: We are not looking for the surface area of a circle. The speaker is a cone. You would have to use: Acone = pi ~ r ~ (r + (r2 + h2)1/2) and then subtract the area of the missing base. You would also have to take into acount the chunk of the dustcap. Then again, we were never looking for a surface area to begin with. We were looking for a displacement.

    Now Alexclaber, I know that Xmax can be taken into question. I noted that already by stating that what Notanaggie said is still valid. Notanaggie wrote the formula in words, Bgavin wrote the formula in notation above.

    I restate: assume that Xmax is the same for all drivers and the drivers are made of the same material with the same surrounds. Assume you are looking at different cabinets from the same maker and they are the same sensitivity. You are feeding them the same signal with the same wattage. This is not a real world situation, of course, but it gives our friend something to go on. A real world 18" sub might only posess 199 cu inches of effective piston area and move only 200 cu inches of air (Taken from a real 18 inch woofer's specs). A far cry from the 324 that my simplified formula gives.

    Anyway, only Dave knows what he wanted. I gave him the answer that I think he was looking for. I don't think he is looking for a complicated answer that takes into account every factor of speaker building. If he asked for that, I would have given those details.

    I had a feeling that I would take some flack for writing something so simple and not qualifying it. But, good on you guys. We do have to keep things straight here. But do read my post. Notanaggie mentioned those things and I wrote that what he said still mattered. What I meant by that was I was simplifying things. Please read his post also.

  8. bassmantele


    Jul 22, 2003
    Boston MA USA
    Just a guess, but I don't think the original poster asked his question properly. "Moving air" is a common internet buzz term, but unless he's trying to blow out candles it's not particularly useful.

    Let me modify that: if he feels the need to stand in front of his amp and feel his pants flap I stand corrected. I suspect he was speaking metaphorically.
  9. aladdin


    Mar 7, 2003
    Chiba, Japan

    I could blow out candles with some Acoustic Energy Hi-Fi speakers that I had in the early 90's. The cabinet was very small and had 3 ports for a 5.25 inch woofer. Lots of vent wind! No chuffing! Great bass for little guys. I could feel the wind on my face from several meters away when the canons fired in the 1812 Overture!

  10. Jazz Ad

    Jazz Ad Mi la ré sol Supporting Member

    A speaker is a piston with a translation movement.
    The shape of the cap is irrelevant,the air it moves only depends on its flat area, so using the circle area of a speaker is accurate.
           |   identical    |
    Circle |   piston       ( Speaker cone
           |   surface      |
  11. aladdin


    Mar 7, 2003
    Chiba, Japan
    If that is true, then I stand corrected.

  12. Petebass


    Dec 22, 2002
    QLD Australia
    Where's Dave? I'm curious to know exactly what he was trying to find out.
  13. davelowell


    Jan 18, 2001
    stl, mo
    i just wanted a general idea of how "loud" a couple of cabs are, that's all, didn't mean to start the rage! :D

    anyway, i am thinking of trying out an ampeg 4x12 at my dealer, and was curious how it would be different than my mesa RR 2x15. i am running a mid 70's SVT.

    thanks, guys and gals,

  14. BruceWane


    Oct 31, 2002
    Houston, TX
    What you are looking for is called "efficiency". It is expressed as decibels measured at a distance of 1 meter with 1 watt input power. It gives you a pretty good idea of how loud a speaker cabinet is, but it doesn't tell you how much of that sound is lows, mids, or highs.

    Ampeg's current 4x12 (SVT-412HE) is rated at 98Db, which is kinda low for a typical bass cab, but not too bad (compare to an Eden or SWR 4x10 at around 103Db). If your Mesa cab is loaded with EVM's (as a lot of Mesa cabs are), I'd expect the Ampeg to be a lot less loud. EVM 15's are VERY efficient - around 103DB each, so a 2x15 would get you 106DB.
  15. Jerrold Tiers

    Jerrold Tiers

    Nov 14, 2003
    St Louis
    Well, The assumptions above aren't totally correct.

    First, it generally takes about 6 dB to notice a significant difference in "typical" usage. You may be able to "detect" a 3 dB difference, but it may not sound like very much. That's assuming there are no big response differences, which would likely be much more obvious to the ear than 3dB efficiency. Generally a 10 dB increase is accepted as "twice as loud", but that can vary.

    The more dynamic range in the sound, the more difference in effciency it typically takes to notice a difference. With a sine wave, 1 dB or even less may be audible. (1 dB was originally supposed to be the least change that was perceptible to the ear).

    So "a lot less loud" with a 4 dB nominal difference may be stretching a point pretty far.

    Next, two identical speakers are not 2X more efficient than 1, nor will 2 speakers necessarily produce twice the power with the same input (if the impedances are equal).

    Only if the two draw more power than one will they produce more output (if the two are 4 ohms and the one is 8, for instance).
    So two 8 ohm cabinets, one with a single speaker, and one with two, will be the same loudness with the same input if the speakers all have the same efficiency. The power is simply divided between them and each gets less input.
    So two 103 dB speakers don't add up to a 106 dB total unless something else changes also.

    It is possible for multiple speakers to "beam" the sound more, which will put more sound straight out front, instead of spreading it out. That tends to raise the SPL out in front, and lower it elsewhere. A 4 speaker cab has some advantage there over a one or two speaker cab, as it will probably do the "beaming" to a lower frequency.

    If two cabinets are less than about 6 dB different in 1 watt SPL, the response characteristics and any "beaming" effects may make more audible difference than the bare driver efficiency. Your mileage may differ.

    The efficiencies of speaker drivers have been a strange spec. Some companies have rated all speakers with 2.83 volts in , which is 1 watt at 8 ohms, but two watts at 4 ohms. So a 4 ohm speaker would get a higher number than an 8 ohm speaker, since they "called it" 1 watt input regardless.

    Some made 3 ohm or 6 ohm speakers, and called them 4 or 8 ohm speakers, again getting a bigger efficiency number with the "standard" 2.83 volts input.
    Small differences in efficiency may or may not really exist, and may or may not be even audible. I will agree EVM speakers are efficient though.

    It's best to try the systems next to one another and see what you think. Efficiency ain't everything.
  16. The change is amperage.

    Our bass amps are current-limited voltage amplifiers. Connected in parallel, two like drivers always produce +3dB more output, because both are driven at the same voltage in a parallel circuit. The amp provides twice the current, up to the limits of the amp.

    Multiple drivers beam at a lower frequency because they mutually couple. This effectively increases the piston diameter which lowers the onset of beaming where lamba <= diameter.

    Watt-based SPL specs are voltage independent. Liars and Enron executives use 2.83v into 4 ohms so they can raise the SPL claim by +3dB. Worse, they take these measurments at artificial frequencies far higher than the bass region.

    Measurements taken in the bass region are dismal indeed, especially for sealed boxes. The impedance curve peaks at the tuning frequency, and is far higher than the nominal rating. Vented boxes reach the impedance minima at the tuning frequency. IMO, the only meaningful SPL measurement is one displayed against a frequency plot.
  17. Petebass


    Dec 22, 2002
    QLD Australia
    The formula you're looking for the cabinet's maximum SPL:-

    Max SPL = Sensitivity + 10*Log(Watts)

    So to keep the Eden 4x10 example going.

    Max SPL = 103 + 10*Log700 = 131dB.

    If you're powering it with a 350w amp the use 350 instead of 700:-

    Max SPL = 103 + 10*Log350 = 128dB.

    That's the theory! even still one may sound louder than the other even though they've got the same Max SPL. there's a whole bunch of other factors to consider. Ultimately you've gotta let your ears decide.
  18. notanaggie

    notanaggie Guest

    Sep 30, 2003
    That would only be true if the cabinets were different impedances, but that's cheating Enron-style as you said.

    Two 16 ohm drivers give 8 ohms, same as one 8 ohm driver. If you compare a dual 8 ohm (i.e. 4 ohm cab) to a single 8 ohm speaker, of course SPL is different, but the power input is different too.

    One speaker cab isn't twice as efficient as another just because it draws twice the power.

    You could just use a 4 ohm version of the other cab, and get louder that way too.

    I think you kinda argued against yourself there.......
  19. You missed the point entirely.

    Each of the like units produces X acoustic power at a given voltage. When two units producing X acoustic power are connected in parallel, 2x acoustic power is produced.

    Voltage is constant in a parallel circuit, so each driver receives full voltage. The power amp produces additional current as dictated by the reduced load impedance, up to the current maximum of the amp.

    Absolute impedance is not the issue.. it is multiple, like drivers connected in parallel. A doubling of acoustic power is +3dB by definition.
  20. notanaggie

    notanaggie Guest

    Sep 30, 2003
    No, you are IGNORING the point.

    Absolute impedance IS the issue.

    If you connect two 8 ohm speakers producing 103 DB each, you get 106. BUT you "made" a 4 ohm speaker which draws twice the power, and that is where the "extra" 3 dB came from. You are cheating just the way you said was wrong.

    A speaker which has 1 driver at 103 db/1watt will produce the exact same SPL as a speaker with 2 drivers which each produce 103 dB/1watt, as long as the impedances are the same, meaning both systems are 8 ohms.

    Otherwise, you could say a speaker system with 8 speakers was 9dB more efficient than a system with 1. That is nonsense.