Science and Math for Audio Humans – Loudspeaker Curiosities

by Danny Maland

Ladies and gentlemen, please give a warm welcome to The Disclaimer!

Everything that I set before you should be read with the idea that “this is how I've come to understand it.” If somebody catches something that's flat-out wrong, or if you just think that an idea is debatable, please take the time to start a discussion via the comments.

In the last installment of this series, I smacked everyone in the forehead with the pronouncement that getting a 50 Hz tone at 120 dB SPLC slow out of a 1” diameter driver would be rather difficult. At the time, I didn't bother to explain why I would make such a statement. Some of you probably reacted with a mental “But Why?” upon reading that line.

Most pro-audio types instinctively assume that larger loudspeakers “push more air,” and thus can create “loud” sonic events at lower frequencies than small loudspeakers. This is pretty much correct, if lacking in detail. However, a quick thought experiment can lead to a troubling question. Why is it exactly that a 1” diameter driver can't compete with, say, an 18” diameter driver at the game of producing a 50Hz tone?

One of the more compact representations of how this works can be found in a formula that relates sound pressure to the amount of airflow passing by a measurement point.

Please note that I have only ever encountered this formula (or, at least, this statement of it) in a single place: While a certain amount of healthy skepticism is warranted, it would appear that Dr. Svante Granqvist is not just a “random” forum participant. (This is his webpage at The Royal Institute of Technology in Stockholm, Sweden:

Figures 2 - 4 show my “simplified for the sake of easy comparisons” versions of the formula. In my version, the flow rate (U) remains unknown, the value of “rho” (density) is set to Granqvist's suggestion of 1.2 kilograms per cubic meter – the density of air at room temperature – while the frequency in question will be set to 50 Hz. For convenience, the value of “r” will be set to 0.5 meters, as this will let us divide by 1.

  • What we get at the end is that pressure is equal to U60.
  • At this point, you're shouting “So not helpful!” To make this helpful, let's say that “U” refers to the flow rate produced by the 1” diameter driver. The driver is essentially a piston pushing on air molecules. The volume of air that this piston can move every second is directly related to its surface area, as well as how far it can travel. If we're not particularly concerned about absolute numerical correctness, just comparisons, we can look at only the outward excursion of the driver as describing a cylinder of a particular volume. This can give us an idea of how the flow rates of our two drivers might stack up.

If we convert to SI units, assume that the driver can displace outward 0.5mm, and use the formula for the volume of a cylinder, we get this:

Now, what about our 18” diameter driver? Well, if we assume that it can manage a 6mm outward displacement:

A visual comparison of the two air displacements is rather striking.

If we take the 1” driver to have the “unity” value, then dividing the calculated displacement cylinder volume of the 18” driver by that of the 1” driver gives us a ratio of 3887 to 1.

In other words, if the pressure from our 1” driver at 50Hz is U60, the pressure from our 18” driver is U60*3887. The advantage of the 18” driver is enormous. If we decide to use the 1” driver as a decibel reference point, then we can plug that 3887 into our decibel conversion formulas to get a nearly 72 dB SPL difference between the two drivers at maximum excursion. (The appropriate conversion uses 20 as the multiplier of the value from the base-10 log, as these are pressures involved, not power.)

The only way for the 1” driver to keep up with the 18” driver is to have an enormous outward excursion capability – just short of 2 meters! That's just a tiny bit impractical.

The other side of the coin is what happens when we're talking about a frequency of 5000Hz. At such a high frequency, the small driver's lack of excursion and surface area cease to be problematic. In fact, it all makes for a driver much better suited to operating at high speed. The much larger mass of the 18” driver is a challenge to cycle quickly, and that's not to mention all the other potentially contributing factors, such as an impedance curve which may be “through the roof” at 5k, directivity issues, distortion products...

The bottom line is that low frequencies involve moving a lot of air “just quickly enough,” whereas high frequencies involve moving “just enough air” a lot of times every second. One size of driver simply can't do both of those things in an optimum way.

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