Sizing the horn — WoodenHorn

This is the part where the thought meets some slightly awkward physics. Before any wood gets cut, we need to work out the shape of the horn: how small the throat should be where the driver sits, how big the mouth has to be, how quickly the walls should flare out, and how long the whole path needs to be. Get those wrong and the horn simply won't do what I'm hoping for.

This is were we need to start reading the amazing book High-Quality Horn Loudspeaker Systems, by Björn Kolbrek and Thomas Dunker. This book has everything from history to theory to design. If you can not find it in this book, it does not exist

The next thing to get your head around is Hornresp, David McBean's horn analysis program. It's not the easiest tool in the world, so the learning curve is steep, but once you get the hang of it you can try out a design in minutes instead of building it and hoping for the best.

The basics

We have decided that the horn should have an exponential growth and is described by a wonderfully compact equation. The cross-sectional area grows exponentially as you move away from the throat:

$$ S(x) = S_0\, e^{m x} $$

$S(x)$ — the cross-sectional area at a distance $x$ along the horn
$S_0$ — the throat area, right at the driver end
$m$ — the flare constant, i.e. how fast the horn opens up
$x$ — the distance measured from the throat

That single constant $m$ is where all the magic (and all the compromise) lives. Make the horn open up slowly — a small $m$ — and it will load down to a low frequency, but it needs to be long. Open it up quickly and it stays compact, but the deep bass slips away.

The flare constant is tied directly to the lowest frequency the horn can support, its cutoff frequency $f_c$:

$$ m = \frac{4 \pi f_c}{c} $$

$f_c$ — the cutoff frequency (below this the horn stops "horn-loading")
$c$ — the speed of sound, about 343 m/s in air

We decided that the target should be 17 Hz and if you plug that in to the formula you get a very gentle flare — which is exactly why the horn ends up so long, and why the space under the floor matters so much.

How big does it have to be?

Here's an awkward detail the theory quietly slips past: the expansion coefficient strictly applies to horns of infinite length. Which is really long — but it does mean I have to land on a practical length and a practical mouth area.

Martin King derives the acoustic impedance for circular and square mouths. He also shows that the impedance is reactive — meaning it produces no sound pressure at all — for small values of the product:

$$ 2 k r_m $$

$r_m$ — the mouth radius
$k$ — the wave number

The wave number itself is just a measure of how much phase the wave picks up per metre travelled. King writes it using the angular frequency $\omega$, but that's the same thing expressed in ordinary hertz — since $\omega = 2\pi f$, the two forms are identical:

$$ k = \frac{\omega}{c} = \frac{2 \pi f}{c} $$

$\omega$ — the angular frequency in radians per second
$f$ — the frequency in hertz
$c$ — the speed of sound, about 343 m/s in air

What happens when $2 k r_m$ is small is that the mass of air sitting just outside the mouth simply refuses to let the sound get out. The low-frequency wavefronts travelling down the horn are mostly reflected straight back into it, and off they go again towards the throat.

As the product grows, though, the resistive part of the impedance grows with it. That's the part that matters: energy starts transferring from the mouth to the surrounding air, and actual sound pressure is produced. The resistive part acts as damping, absorbing the energy that would otherwise have bounced back inside.

The frequency at which sound really starts radiating into the room is defined by

$$ 2 k r_m = 2 $$

At precisely that frequency, the resistive part of the impedance produces half the sound pressure that the horn manages at higher frequencies. It's also the point where the resistive part finally overtakes the reactive part.

King goes on to show that square mouths give essentially the same result as circular ones, provided they have the same area. They also start producing sound pressure at $2 k r_m = 2$. Unfortunately that doesn't hold for rectangular mouths — those only get going at somewhat higher frequencies.

Rearranging $2 k r_m = 2$ gives us the mouth radius:

$$ r_m = \frac{1}{k} $$

And since we already know what $k$ is, the mouth radius — assuming the horn is round — works out as:

$$ r_m = \frac{c}{2 \pi f} $$

which corresponds to a cross-sectional area of:

$$ S_m = \pi r_m^2 = \pi \left( \frac{c}{2 \pi f} \right)^{\!2} = \frac{c^2}{4 \pi f^2} $$

So what does that actually mean in metres? Let's put some numbers in. Taking $f = 17$ Hz and $c = 343$ m/s, the mouth radius comes out as:

$$ r_m = \frac{343}{2 \pi \cdot 17} \approx 3.2 \ \text{m} $$

which gives a mouth area of:

$$ S_m = \pi \cdot 3.2^2 \approx 32 \ \text{m}^2 $$

Let that sink in for a moment. A mouth of 32 square metres — a circle well over six metres across. That is not a loudspeaker, that's a garage door. This is the point where you understand why properly horn-loaded deep bass is so rarely attempted at home.

In practice, of course, no bass horn hangs freely in space. If the horn is instead placed on the ground or on a floor, it can be considered long enough with a mouth area only half the size calculated above. And if it's placed on the floor right next to a wall, the mouth area can be halved once more. Suddenly the numbers look a lot friendlier: those 32 m² drop to 16 m² against the floor, and to around 8 m² in the corner formed by floor and wall. Still enormous — but no longer completely absurd.

What about the throat?

So what area should the horn have at its entrance — the throat area? That one is decided by several factors, all of them to do with how the driver that's going to power the horn gets coupled to the throat. There are various theories about this, and all of them come with significant simplifications baked in. An important question is therefore how large, and how many, drivers I'm going to use. The length of the horn is affected by the throat area too.

The horn's final size comes down to the type of expansion chosen, the cutoff frequency chosen, and how many and how large the drivers are. The reasoning also hints that the bigger the driver used, the shorter the horn can be. That's true in a certain sense, and genuinely interesting — but there are more phenomena to keep in mind.

The horn is also a pipe

Because here's the thing: the horn also behaves like a pipe closed at one end. When the horn's length corresponds to a quarter wavelength of the lowest note it's meant to play, you get a distinct resonance. So the horn's length has to be considerably greater than that.

The wavelength at 17 Hz is:

$$ \lambda = \frac{c}{f} = \frac{343}{17} \approx 20 \ \text{m} $$

So the horn must not be shorter than 5 m — a quarter of 20 m. And there are actually good arguments for making it twice that long. On top of that, the horn's acoustic length is often greater than its physical length. Depending on how the mouth is shaped, the acoustic length can be up to one mouth radius longer. In short horns with a large mouth area, that effect can be substantial.

In practice the size of the horn needs to be adapted to make the best possible use of the space available. The external conditions are, after all, what set the physical limits. But before I get there, there are more investigations to do.

More coming as the design takes shape

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