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This is amazing work. I don't quite understand how they are detecting the tsunamis though. They mention that this works via significant displacements of air. Is

This is amazing work. I don't quite understand how they are detecting the tsunamis though. They mention that this works via significant displacements of air. Is the amount of air displacement in the open ocean statistically significant for detection or is it being detected via a slightly different source.

hotshot1001 2y ago
i believe its the pressure waves propagating through the air itself is pushing the ionosphere upwards, inducing a detectable signal from fluctuations in the electron density.
magicalhippo 2y ago
The way I understood it is that the wave amplitude of a tsunami isn't typically huge, say only a few feet[1], however it's wavelength is. Thus while normal waves can have a much larger amplitude, they're easily filtered by a low-pass filter. edit: The atmosphere between the surface and the ionosphere forms a natural low-pass filter as well. I imagine typical ocean waves as seen by us are way too high-frequency to make it up to the ionosphere. There are other, natural disturbances in the ionosphere, such as traveling planetary waves[2], but they have a significantly longer wavelength. As such the paper[3] mentions filtering them out using a high-pass filter. In the paper they show some preliminary results trying to invert the parameters in order to estimate the height of the tsunami based on the measured ionosphere disturbance based on synthetic data, and the baseline amplitude is 10cm (4 inches), which the model comes quite close to. [1]: https://earthobservatory.nasa.gov/blogs/fromthefield/2014/04/02/tsunamis-and-the-open-ocean/ https://earthobservatory.nasa.gov/blogs/fromthefield/2014/04... [2]: https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2018JA025720 https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/201... [3]: https://link.springer.com/article/10.1007/s10291-022-01365-6 https://link.springer.com/article/10.1007/s10291-022-01365-6
mgsouth 2y ago
Pretty astounding, isn't it? I don't see a paper, but there was a webinar [1]. There's a technical synopsis at 8:00. The phenomenon they're measuring is actually signficant. It's the total number of (free?) electrons between the satellite and the receiver. Typically its about 10^12 electrons/m^3 (@8:00 in video). The disturbance from the 2011 earthquake and tsunami was, if I'm reading the movie/chart correctly, about +/- 1 TECU, which is 10^16 electrons/m^3 (@10:40). The water elevation may only be a few feet in open ocean, but it's over a vast area. That's a lot of power. They're measuring it by looking for phase differences in the received L-band (~2GHz) signals, rather than amplitude. That eliminates lots of noise. And they're looking for a particular pattern, which lets you get way below the noise floor. For example, the signal strength of the GNSS (GPS) signal itself might be -125 dBm, while the noise level is -110 dBm [2]. That means the signal is 10^-12 _milliwatts_, and the noise is about 30 times larger. But by looking for a pattern the receiver gets a 43 dB processing boost, putting the effective signal well above the noise. [1] https://www.youtube.com/watch?v=BEpZmRPPWFo https://www.youtube.com/watch?v=BEpZmRPPWFo [2] https://www.nxp.com/docs/en/brochure/75016740.pdf https://www.nxp.com/docs/en/brochure/75016740.pdf
- mgsouth 2y ago
  OK, the "typically 10^12 TEC" vs. a +/- 1 TECU (10^16 TEC) disturbance was really bugging me. I think the slide has an error, or there's an apples/oranges issue. The +/- 1 TECU looks to be consistent, but the typical background level is "a few TECU to several hundred" [1]. A Wikipedia page has shows the levels over the US being between 10 - 50 TECU on 2023-11-24, and says that "very small disturbances of 0.1 - 0.5 TEC units" are "primarily generated by gravity waves propagating upward from lower atmosphere." [2]. [1] https://www.swpc.noaa.gov/phenomena/total-electron-content https://www.swpc.noaa.gov/phenomena/total-electron-content [2] https://en.wikipedia.org/wiki/Total_electron_content https://en.wikipedia.org/wiki/Total_electron_content
- Qqqwxs 2y ago
  To expand upon this: >> They're measuring it by looking for phase differences in the received L-band (~2GHz) signals The "L-Band signals" are GNSS signals, for example GPS L1 and L2, which use a carrier wavelength of 1575.42 MHz and 1227.6 MHz, respectively. Both L1 and L2 signals are emitted at the same time, but experience differing levels of delay in the ionosphere during their journey to the receiver. The delay is a function of total electron content (TEC) in the ionosphere and the frequency of the carrier wavelength. Since we already know precisely how carrier frequency affects the ionospheric delay, comparing the delay between L1 and L2 signals allows us to calculate the TEC along the signal path. Another way to think of it is: we have an equation for signal path delay with two unknowns (TEC, freq). Except, it is only one unknown (TEC). Use two signals to solve simultaneously for this unknown. Use additional signals (like L5) to reduce your error and check your variance.
xattt 2y ago
If you can imagine this, can you imagine the forms of sensing-by-proxy that’s possible that’s kept under wraps by various Skunkworks orgs?

hotshot1001 2y ago

i believe its the pressure waves propagating through the air itself is pushing the ionosphere upwards, inducing a detectable signal from fluctuations in the electron density.

magicalhippo 2y ago

The way I understood it is that the wave amplitude of a tsunami isn't typically huge, say only a few feet[1], however it's wavelength is. Thus while normal waves can have a much larger amplitude, they're easily filtered by a low-pass filter. edit: The atmosphere between the surface and the ionosphere forms a natural low-pass filter as well. I imagine typical ocean waves as seen by us are way too high-frequency to make it up to the ionosphere. There are other, natural disturbances in the ionosphere, such as traveling planetary waves[2], but they have a significantly longer wavelength. As such the paper[3] mentions filtering them out using a high-pass filter. In the paper they show some preliminary results trying to invert the parameters in order to estimate the height of the tsunami based on the measured ionosphere disturbance based on synthetic data, and the baseline amplitude is 10cm (4 inches), which the model comes quite close to. [1]: https://earthobservatory.nasa.gov/blogs/fromthefield/2014/04/02/tsunamis-and-the-open-ocean/ https://earthobservatory.nasa.gov/blogs/fromthefield/2014/04... [2]: https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2018JA025720 https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/201... [3]: https://link.springer.com/article/10.1007/s10291-022-01365-6 https://link.springer.com/article/10.1007/s10291-022-01365-6

mgsouth 2y ago

Pretty astounding, isn't it? I don't see a paper, but there was a webinar [1]. There's a technical synopsis at 8:00. The phenomenon they're measuring is actually signficant. It's the total number of (free?) electrons between the satellite and the receiver. Typically its about 10^12 electrons/m^3 (@8:00 in video). The disturbance from the 2011 earthquake and tsunami was, if I'm reading the movie/chart correctly, about +/- 1 TECU, which is 10^16 electrons/m^3 (@10:40). The water elevation may only be a few feet in open ocean, but it's over a vast area. That's a lot of power. They're measuring it by looking for phase differences in the received L-band (~2GHz) signals, rather than amplitude. That eliminates lots of noise. And they're looking for a particular pattern, which lets you get way below the noise floor. For example, the signal strength of the GNSS (GPS) signal itself might be -125 dBm, while the noise level is -110 dBm [2]. That means the signal is 10^-12 _milliwatts_, and the noise is about 30 times larger. But by looking for a pattern the receiver gets a 43 dB processing boost, putting the effective signal well above the noise. [1] https://www.youtube.com/watch?v=BEpZmRPPWFo https://www.youtube.com/watch?v=BEpZmRPPWFo [2] https://www.nxp.com/docs/en/brochure/75016740.pdf https://www.nxp.com/docs/en/brochure/75016740.pdf

mgsouth 2y ago
OK, the "typically 10^12 TEC" vs. a +/- 1 TECU (10^16 TEC) disturbance was really bugging me. I think the slide has an error, or there's an apples/oranges issue. The +/- 1 TECU looks to be consistent, but the typical background level is "a few TECU to several hundred" [1]. A Wikipedia page has shows the levels over the US being between 10 - 50 TECU on 2023-11-24, and says that "very small disturbances of 0.1 - 0.5 TEC units" are "primarily generated by gravity waves propagating upward from lower atmosphere." [2]. [1] https://www.swpc.noaa.gov/phenomena/total-electron-content https://www.swpc.noaa.gov/phenomena/total-electron-content [2] https://en.wikipedia.org/wiki/Total_electron_content https://en.wikipedia.org/wiki/Total_electron_content
Qqqwxs 2y ago
To expand upon this: >> They're measuring it by looking for phase differences in the received L-band (~2GHz) signals The "L-Band signals" are GNSS signals, for example GPS L1 and L2, which use a carrier wavelength of 1575.42 MHz and 1227.6 MHz, respectively. Both L1 and L2 signals are emitted at the same time, but experience differing levels of delay in the ionosphere during their journey to the receiver. The delay is a function of total electron content (TEC) in the ionosphere and the frequency of the carrier wavelength. Since we already know precisely how carrier frequency affects the ionospheric delay, comparing the delay between L1 and L2 signals allows us to calculate the TEC along the signal path. Another way to think of it is: we have an equation for signal path delay with two unknowns (TEC, freq). Except, it is only one unknown (TEC). Use two signals to solve simultaneously for this unknown. Use additional signals (like L5) to reduce your error and check your variance.

xattt 2y ago

If you can imagine this, can you imagine the forms of sensing-by-proxy that’s possible that’s kept under wraps by various Skunkworks orgs?