11 ms·

It was crazy scary, and I was all the way down in Nagoya. There doesn't seem to be too much damage in Miyagi (where the earthquake was) but some fires broke out

It was crazy scary, and I was all the way down in Nagoya. There doesn't seem to be too much damage in Miyagi (where the earthquake was) but some fires broke out apparently. Luckily, it seemed to be a long, hard earthquake rather than a short hard earthquake, which means that buildings are able to withstand the shaking better. The shaking went on for about 3 minutes here, and there were some after shocks that lasted for about a minute. The Great Hanshin Earthquake was only a 7.2, and it was much more destructive.

wisty 16y ago
It's probably also deep. Deep earthquakes do less damage. But still, 8.9, that's almost 100 times bigger than a 7.
- 3pt14159 16y ago
  Closer to 80 times stronger than a hundred, but yeah definitely deep.
  - mrcharles 16y ago
    From Wikipedia: The energy release of an earthquake, which closely correlates to its destructive power, scales with the 3⁄2 power of the shaking amplitude. Thus, a difference in magnitude of 1.0 is equivalent to a factor of 31.6 ( = (101.0)(3 / 2)) in the energy released; a difference in magnitude of 2.0 is equivalent to a factor of 1000 ( = (102.0)(3 / 2) ) in the energy released.[2]
    - borism 16y ago
      so the correct answer is that this one was 1000 times more powerful than Kobe, and also very shallow (depth of 24 km), but fortunately offshore (but unfortunately that means powerful tsunami).
    - jules 16y ago
      How does energy scale with distance from epicenter?
      - ars 16y ago
        Probably goes down by r^2 However earthquakes may be "flat", in which case it's by r (i.e. r^1). It's probably somewhere in between those.
        jules 16y ago
        Wouldn't it go down by r^2 if the earthquake is flat, and by r^3 if it's not?
        wisty 16y ago
        No. A circle with 2X the radius has 2X the diameter, thus the energy will be cut in half. Unless you also expect a linear attenuation, due to energy loss through the bedrock.
      - wisty 16y ago
        Small earthquakes have a lot of energy in the higher end of the frequency scale, which is rapidly dissipated. Larger earthquakes have more low-frequency energy, so they go further. There's no simple equation, they just use regression to fit the attenuation as a factor of 1/r, 1/r^2 (...), and for magnitude, and for frequency. You can start here: http://earthquake.usgs.gov/hazards/about/attenuation.php http://earthquake.usgs.gov/hazards/about/attenuation.php
NZ_Matt 16y ago
The fact the epicenter was 200 km offshore is why the shaking wasn't too destructive. The Great Hanshin earthquake was only 20km from Kobe. The distance from the epicenter is often overlooked by the media. PGV (peak ground velocity) is a more accurate way to estimate the strain put on infrastructure. This shakemap shows the preliminary recordings: http://earthquake.usgs.gov/earthquakes/shakemap/global/shake/c0001xgp/ http://earthquake.usgs.gov/earthquakes/shakemap/global/shake...

wisty 16y ago

It's probably also deep. Deep earthquakes do less damage. But still, 8.9, that's almost 100 times bigger than a 7.

3pt14159 16y ago
Closer to 80 times stronger than a hundred, but yeah definitely deep.
- mrcharles 16y ago
  From Wikipedia: The energy release of an earthquake, which closely correlates to its destructive power, scales with the 3⁄2 power of the shaking amplitude. Thus, a difference in magnitude of 1.0 is equivalent to a factor of 31.6 ( = (101.0)(3 / 2)) in the energy released; a difference in magnitude of 2.0 is equivalent to a factor of 1000 ( = (102.0)(3 / 2) ) in the energy released.[2]
  - borism 16y ago
    so the correct answer is that this one was 1000 times more powerful than Kobe, and also very shallow (depth of 24 km), but fortunately offshore (but unfortunately that means powerful tsunami).
  - jules 16y ago
    How does energy scale with distance from epicenter?
    - ars 16y ago
      Probably goes down by r^2 However earthquakes may be "flat", in which case it's by r (i.e. r^1). It's probably somewhere in between those.
      - jules 16y ago
        Wouldn't it go down by r^2 if the earthquake is flat, and by r^3 if it's not?
        wisty 16y ago
        No. A circle with 2X the radius has 2X the diameter, thus the energy will be cut in half. Unless you also expect a linear attenuation, due to energy loss through the bedrock.
    - wisty 16y ago
      Small earthquakes have a lot of energy in the higher end of the frequency scale, which is rapidly dissipated. Larger earthquakes have more low-frequency energy, so they go further. There's no simple equation, they just use regression to fit the attenuation as a factor of 1/r, 1/r^2 (...), and for magnitude, and for frequency. You can start here: http://earthquake.usgs.gov/hazards/about/attenuation.php http://earthquake.usgs.gov/hazards/about/attenuation.php

NZ_Matt 16y ago

The fact the epicenter was 200 km offshore is why the shaking wasn't too destructive. The Great Hanshin earthquake was only 20km from Kobe. The distance from the epicenter is often overlooked by the media. PGV (peak ground velocity) is a more accurate way to estimate the strain put on infrastructure. This shakemap shows the preliminary recordings: http://earthquake.usgs.gov/earthquakes/shakemap/global/shake/c0001xgp/ http://earthquake.usgs.gov/earthquakes/shakemap/global/shake...