Radiation losses are negligible compared to wall losses in narrow pipes

Evidence

According to figure 16.1, the losses due to sound radiation alone would have given a much higher Q, because the reflection amplitude for a 0.95-cm-radius tube at 1180 Hz is 98%. (This is hard to read off figure 16.1 but we recalculated the number using Levine and Schwinger’s theory.) This corresponds to a 4% energy loss for every encounter with the open end, occurring once per round-trip of the pipe. According to the formula in section 10.3, Q = 2π energy remaining/energy lost (per cycle), and we should have had Q = 2π(1/0.04) ~ 157. The measured Q is more than five times smaller, so we can be certain the Q is limited by wall losses and not sound radiation. — Eric J. Heller, Why You Hear What You Hear- An…, 1448

Categorization

• XREF: Connects to Q-factor and resonance advantage curve discussion in chapters 10 and 16 of this same book
• QUESTION: Worth chasing: at what radius/frequency crossover do wall losses stop dominating radiation losses?