The time between seeing a lightning flash and hearing the resulting thunder gives an accurate estimate of the distance to its location. During one storm a meteorologist observes 29 such flashes, and calculates the distances in miles as follows:
6.02,
3.01,
0.69,
3.38,
2.01,
4.69, 3.54,
3.67,
3.67,
4.55,
5.79,
3.72, 1.05,
3.73,
3.98,
7.28,
2.10,
2.90, 6.95,
6.18,
7.20,
4.89,
6.60,
2.53, 2.09,
1.30,
1.81,
4.75,
6.91
Assume that the cloud is circular, is not moving, and that the lightning strikes are uniformly distributed through the cloud. Determine the diameter of the cloud.
This is a wonderful problem (which means I am struggling with it :-) )
An approximation to the diameter of the cloud is max - min = 7.28 - 0.69 = 6.6 miles and that the near cloud edge is 0.69 miles away. In reality the cloud is both a little bigger and a little closer since the likelihood of a nearest and furthest edge-flash are near 0.
The solution involves 2 steps:
1) for a cloud of radius r and distance D there is a function
prob(d) that a part of the cloud at distance "d" reports, and this is
proportional to the arc length (opening angle theta(r,D,d) ) within the cloud at distance d.
2) A least squares fit of the data (the histogram distribution of 29 "d"s) to the function prob(d) (a function of r and D) gives optimal r and D. I.e., the partial derivative of the sum squared error is set to 0 to minimize the error for the best r and D.
I have decided it is easier to simulate, make a grid of D and r and
find the best match to the histogram in the least square sense.
binning distances 0. - 0.5, 0.5 - 1.0 ... 7.5-8 miles gives
0 1 2 1 3 2 2 6 1 3 0 1 2 3 2 0 which sum to 29 strikes. Note that the data is sparse. A million hits for example would give a smooth function with a well defined maximum, width and shape. We do not have that here, but instead we have small number statistics. So the analytic approach would be somewhat overkill. A simulation seems in order.
Initial results show that having the researcher outside the cloud does not work - the histogram would not be so peaked but rather fairly uniform. So the researcher is likely inside the border of the cloud.
Ah, and the cloud, being a cloud, is up in the air, (duh). This helps make the histogram peak near the cloud altitude.
Edited on February 7, 2019, 3:30 pm
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