Monday, January 4, 2016

still quaking!

After the earthquake last year in Katmandu, one of my colleagues, Anupama Singh, mentioned that there fish tank had overflowed.
So, what is the condition for that to happen?
For the water to overflow (dashed line):
                                                                                tan(f) = h/(l/2) = a/g
where a is the horz accn..
Assume h = 5 cms, l = 100 cms, a = 0.1 g.

This corresponds roughly to a Richter 6 quake.

What about the Hyatt swimming pool at Katmandu? The video circulating on WhatsApp showed the water moving in a very complicated way - but it reached the first floor of the adjacent building.

Assume the pool is 25 metres long. The water sloshed up to the 1st floor: i.e. 3.5 metres.

A Richter 8 quake has accelerations of 0.34 to 0.65 g. The quake at Katmandu was less than 7.9, because of the distance of 80 kms to the epicentre, but it was mostly horizontal. For a height of 3.5 metres, a = 0.28 g's. This is reasonable.
I am going to quote verbatim from Phil Plait along with the image he cites and then comment on it after his post:


Take, for example, Tim Ashby-Peckham, who saw that the International Space Station was going to make a nice pass of his location in the early evening of Nov. 1, 2015. He set up his Canon 70D camera on a tripod, aimed it in the right direction, and waited. Once the ISS came into view, he started snapping away.
Take a look at the photo above. You can see the ISS as a long streak, since it moves a lot during the 30-second exposure. But take a closer look. See the wiggle in the streak at the bottom?
Ashby-Peckham took this shot in South Auckland, New Zealand, around 8:57 p.m. local time on Nov. 1, 2015. At 8:58, a magnitude 2.2 earthquake hit a couple of hundred kilometers to the southeast.
My strongest doubt was that a quake that weak could be felt that far away. However, in an email discussion about it, Ashby-Peckham told me that another earthquake about the same distance away was indeed felt in Auckland.
So this is a new one on me. Using the ISS as a virtual seismograph! It’s a pretty funny idea. I suppose if you live in a place with enough earthquakes you could actually calibrate your photography equipment to them, measuring how much the image wiggles versus size/distance of the quake. You need a moving target; stars don’t move quickly enough, and all you get is smeared-out disks for them (note that in the shot, the stars do appear to wiggle a bit). Of course, you could put the camera on a motor, letting it slowly scan the sky over and over again.

My comments:

ISS orbital period = 90 minutes which corresponds to 360°. 

So 30 secs corresponds to an angular displacement of (0.5/90)(360) =  2°
More precisely:
ISS orbital speed is 7.66 km/s. So in 30 secs, ISS moves 229.8 kms.
Assuming Earth radius as 6400 kms and the ISS average altitude as 380 kms, so the total distance from the centre of the Earth is: 6780 kms.
The angle traversed is thus: 229.8/6780 =  0.0339  rads = 1.94°

According to wiki[edia (link above), a Richter 2-3 corresponds to a peak ground acceleration (PGA) of 0.0017 to 0.014 g’s.
Assume a value of 0.0017 for Richter 2. This means a Dq = 0.0017 rads or 0.097°.

 In the image above, if the whole track is FOV is 0.0339 rads, and its corresponding length is 147 mm, while the displacement above the track is about 1 mm. This gives an angle of about 2.31x10-4 rads (i.e. 0.013°).

Richter 2 should give 0.097°. This means that the shaking at Ashby’s location was closer to Richter 1, because the angular displacement of 0.013° is smaller by 7.5X than for Richter 2.

According to the authors, T.Trombetti et al (13th World Conference on Earthquake Engineering, Vancouver, Canada,2004), the Sabetta-Pugliese law (of 1987) states:
Log (A) = - 1.845 + 0.363 (MS) – log (R2+25)1/2+0.195s
Where A is the PGA in g’s, MS is the quake intensity in the Mercalli scale, R is the distance in kms and s is a parameter that is 1.0 for soft soil and 0.0 for bedrock.
According to the S-P law at a low R = 0.001 kms and assuming MS = 2, PGA = 0.0019 for s = 0 and PG = 0.0029 for s = 1. The 1st value is close to that mentioned in wiki.
For s = 0, MS = -2 gives PGA = 6.6x10-5.

Instead we can use the total duration time of earthquake t (in secs) given by Tsimura’s empirical formula:
ML = -2.53 +2.85 log10(t) + 0.0014R
Where ML is the local quake intensity and R is the distance from the epicenter.

The image shows that the quake lasts for about (3/147)(30) = 0.6 secs.

According to Tsimura, for ML = -3, the time t is 0.7 sec at R = 0.001 kms and 0.85 sec at R = 200 kms.
The duration would suggest that the quake was ML= -3 which is very feeble indeed.

On the other hand, as Plait points out the distance is R = 200 kms. With a ML = 2.2, the S-P law gives a PGA = 4.5x10-7 at R = 200 kms. This is much smaller than the actual on-site PGA of 2.31x10-4 from the angular displacement. At that distance, even with ML=6, the PGA is only 1x10-5!

The two arguments from PGA vs R and from duration t are completely inconsistent.

At the end, one would have to invoke – as Phil Plait did – another completely different quake, much closer by to the observation point, to get better consistency. But not a lot, as it happens…

So what values could one try out?

Suppose we go for ML=-3 so that the duration is about right (0.7 sec). The PGA at R = 0.001 kms is 2.9x10-5. So there should have been a mini-quake literally under Ashby’s feet! But this PGA is an order of magnitude too small.
Or a somewhat stronger one a few kilometers away…
If we choose, ML=-0.5, we get PGA=2.3x10-4. Unfortunately, this gives t = 5 secs!
What this means is that we are unable to get consistency between the PGA and the time. What is clear, however, is that it was a mini-quake (weaker than ML = -0.5) and its epicenter was pretty close to where Ashby and his tripod were located.