I was looking at an unremarkable quake as seen at my station (RF075) and noticed an interesting, relatively pure note on the spectrum display, starting at 30Hz, and drifting down to near 22Hz:
I looked at the same event on my nearest neighboring station (R8005), about four miles away from me, and located in a remote nature preserve, with much less human-caused noise:
The earthquake is much more clearly visible from the quieter location. The same spectral signature is visible here, but it seems to be around two minutes later at the neighboring station than it was at my house. Four miles in two minutes is around 120 mph, which is a reasonable speed for an aircraft, but too slow for seismic waves. 30Hz to 22Hz is low frequency for a small plane propeller, but perhaps a helicopter? We have frequent Coast Guard overflights.
I think I picked up a helicopter, or maybe a different style of aircraft. Does anyone have another plausible explanation?
Note, both units were Raspberry Shakes, not Booms, so they’re not optimized for picking up sounds, but if the sound is loud enough and couples with the ground well enough…
That is almost certainly a helicopter. Small propeller aircraft typically don’t show, because they don’t shake the ground like a helicopter. I have both around here and even a fairly noisy airplane doesn’t show but helicopters do.
The curve shape is the doppler effect. The first time I saw one of these, I spent time carefully estimating the frequencies/shifts and calculating the speed - came out to ~180mph which seemed about right for a helicopter - really needs to go directly overhead for that to work easily.
Thanks for the confirmation. You motivated me to do the math to figure out the speed from doppler shift. If the speed of sound is v, and the speed of the aircraft is x, and assuming the observer is stationary and the air is still, then when the aircraft is approaching, the observed frequency is the frequency as measured on board the aircraft, scaled by v/(v-x), while the frequency gets scaled by v/(v+x) when the aircraft is going away. The speed of sound varies with temperature, but 750 mph is a round number that’s in the right ballpark. The initial sound frequency was 30Hz, while the final frequency was 22Hz, as near as I can tell. So you get two equations:
30 = (750 * f)/(750-x)
22 = (750 * f)/(750+x)
Solving each for f and setting the two fs equal:
30(750-x) = 22(750 +x)
Solving for x gives about 115 mph, which is remarkably consistent with the estimate of 120 mph I got by dividing the distance by the elapsed time between measurements at the two stations. None of my initial measurements are that precise, and I don’t know how well-aligned the course of the helicopter was with the direct path from one station to the other. There was probably wind, which hasn’t been accounted for. So I think the agreement of the speed estimates owes a lot to luck.
Anyway, the purpose of the exercise wasn’t high precision, it was just to do a “sanity check” to see if the idea of a helicopter flying over one station and then the other remains a plausible explanation consistent with the observations. I think it passes the check.
Ha! I just did a reply to another post showing how to do it. The two f’s are actually not equal … but it is probably good enough to assume they are for these purposes.
