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Acoustic Shadowing: Unpacking Wave Numbers (k) & Introducing The Wave Ratio Thesis

If you’re truly interested in learning about sound propagation and architectural acoustics, it’s something you should be acquainted with, at least casually.

Let’s use an example of a 9” square column that’s 15’ tall. The SigDim of this column will almost always be 9” or 0.75′, not 15′. If someone is sitting behind this column our main concern is the frequencies that are shadowed because of the 9″ width, not the height.

For simplicity, all the following references and formulas are stated as 2D dimensions, but of course 3D dimensions must be considered for anything that isn’t perfectly round. For rectangles and other 3D shapes, using the dimension of the diagonal cross section is recommended.

Now that you know how to identify a SigDim, and a have basic understanding of what ka is, what do we do with the answer? Why do we care?

Let’s plug in some specific values and see what happens. We’ll use 5.57 (k) and 0.75’ (a), which represent 1 kHz and our 9” column. The ka formula and result are: 5.57 * 0.75 = 4.18. This ka tells us that a 9” column will block 4.18 radians, or about 67% of the energy and momentum of a 1 kHz wave.

Unfortunately, the ka number alone isn’t much help without further context. ka is simply another esoteric value uncovered along this journey.

The Importance Of ka = 6.28

6.28 is a useful number to watch for when calculating ka. When ka equals 6.28 the wavelength of a given frequency, and the SigDim of the obstruction being evaluated, are exactly the same.

Example: Using our 9″ SigDim above, the frequency that results in a ka of 6.28 is 1,502 Hz (Figure 4). For a 6″ SigDim obstruction we find that 2,255 Hz results in a ka of 6.28.

Figure 4: A screenshot of my k, ka and kr calculator. kr is explained below.
Wave Ratio Color Code
Green — No Shading. No problem at this and all lower frequencies
Yellow — Partial Shading. Some problems at this and all higher frequencies
Pink — Full Shading. Full blockage at this and all higher frequencies
Here’s a Wave Ratio calculator spreadsheet available for free download

Be patient. Finding the frequency with a ka value of 6.28 isn’t the final answer either. We now have to look well above and below this value to begin seeing the true impact of the obstruction’s acoustic shadowing.

If ka = 12.56, the obstruction in question will effectively block that frequency and all higher frequencies. 12.56 represents 2 * 6.28.

If ka = 1.57, the obstruction will be nearly invisible to that frequency and those lower in pitch. Based in quarter wave theory, 1.57 represents 1/4 of 6.28.

All kas that fall between 1.57 and 12.56 will be partially blocked or diffracted. How much depends on the frequency(s) in question and which end of the SigDim range they’re closest to. Now we’re getting somewhere. Too bad 12.56 and 1.57 are such arcane numbers. Let’s try to make this exercise more user friendly.

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