At a frequency of 37.5 kHz, a quarter wavelength is about 2 km (1.24 miles) long. How does the black box efficiently radiate anything if it is not attached to 2 km of antenna?


1 Answer 1


The short version of the answer is 37.5KHz is not "radio frequency", it's "audio frequency", so an antenna is not necessary - just a (very loud) speaker, which is basically what the pinger is.

Specifically, the 37.5KHz "pinger" tones used in the Underwater Locator Beacons that are bolted onto Cockpit Voice Recorder / Flight Data Recorder "black boxes" are in the low end of what we call ultrasonic frequencies. Water is generally a good conductor of sound (with at least one notable exception), so it is usually possible for sonar detection equipment to "hear" the pinger tones at reasonable distances (with the interpretation of "reasonable" varying based on the depth of the area being searched - much like in air the sound intensity falls off as distance from the source increases, so if we're listening at the surface of the ocean a pinger in 500 feet of water will have a larger coverage footprint than the same pinger in 5000 feet of water).

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    $\begingroup$ Your question begs the equally-interesting radio-frequency question: "Aviation band radios operate in the frequency range 108MHz-137MHz (about 7ft to 9ft wavelength), so how can our aircraft radios efficiently radiate without being connected to 9 feet of antenna?" - $\endgroup$
    – voretaq7
    Apr 7, 2014 at 5:41
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    $\begingroup$ While the wavelength is 7-9 ft, the optimal antenna is (depending on antenna design) usually half, and sometimes a quarter (as mentioned in the question), so 2-4ft antenna would do the trick, which is, if I'm not mistaken, roughly the size used. $\endgroup$
    – falstro
    Apr 7, 2014 at 7:02
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    $\begingroup$ A quarter-wave antenna is easy to make: you've got the aluminum skin of the airplane to act as your reflecting plane, so it's just a matter of bracing two feet of wire so it's sticking straight out. $\endgroup$
    – Mark
    Apr 7, 2014 at 7:39
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    $\begingroup$ @falstro the point of that comment was to inspire someone to ask the question so we could give a proper answer with the quarter-wave dipole diagram and the RF theory (which will not fit in a comment) :-) $\endgroup$
    – voretaq7
    Apr 7, 2014 at 14:48

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