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In episode 26 of the anime, Silver/Ooguro/Tatsuya annihilated the Chinese
fleet in one shot with _Material Burst_. No wonder his aunt had his ability
locked...

What is the size of that crater? It was shown in satellite images to have
eaten a piece of the coastline! Estimate from a screenshot if no source
exists.

Since it was at the sea, of course the crater is filled with.... duh,
seawater.

So by the images of episode 26, how big was the blast Ooguro caused?


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question
the-irregular-at-magic-high-school
What is the size of the crater left by the explosion in episode 26?


The easiest solution to this would be if the light novels gave us a concise
answer, but of course, nothing in life is that easy.

But while the light novels don't give us an accurate size of the crater, they
_do_ give us the equivalent energy - 20 megatons (20,000 kilotons) of TNT:

> Repeating Kazama's order, Tatsuya pulled the trigger on Third Eye.From
> within Tsushima Base, across the strait, directly into Zhènhai Naval Port.
> Tatsuya's magic transformed 1 kg of matter directly into energy.
>
> In accordance with Einstein's Formula, the heat energy equated to
> approximately 20,000,000 tons of TNT.
>
> – _Volume 7 - Yokohoma Disturbance II_ , Chapter 13

Now using this, we can approximately calculate a blast radius.

To get my results, I used this simulation here -
<http://nuclearsecrecy.com/nukemap/>. First off, I ran a couple of tests by
plugging in the data from real-life detonations to see how accurate it was:

  * Test 1: [_Ivy Mike_ Test](http://en.wikipedia.org/wiki/Ivy_Mike) \- Actual crater diameter: **1900m** Result from simulation: [**1680m**](http://nuclearsecrecy.com/nukemap/?&kt=10400&lat=53.4624326&lng=-1.8402328&airburst=0&hob_ft=0&crater=1&zm=10) \- 88% accurate

  * Test 2: [_Castle Bravo_ Test](http://en.wikipedia.org/wiki/Castle_Bravo) \- Actual crater diameter: **2000m** Result from simulation: [**1900m**](http://nuclearsecrecy.com/nukemap/?&kt=15000&lat=53.4624326&lng=-1.8402328&airburst=0&hob_ft=0&crater=1&zm=9) \- 95% accurate

  * Test 3: [_Sedan_ Test](http://en.wikipedia.org/wiki/Sedan_\(nuclear_test\)) \- Actual crater diameter: **390m** Result from simulation: [**360m**](http://nuclearsecrecy.com/nukemap/?&kt=104&lat=53.4624326&lng=-1.8402328&airburst=0&hob_ft=0&crater=1&zm=13) \- 92% accurate

Given this information, we can assume that the simulation is roughly 90%
accurate, giving an error margin of ±10%.

Setting the yield as 20,000 kilotons, with an airburst of 17m (the beam height
of a [typical modern Chinese
ship](http://en.wikipedia.org/wiki/Type_052C_destroyer), or the height at
which the flag would be flying) leaves us with a
[result](http://nuclearsecrecy.com/nukemap/?&kt=20000&lat=30.0090526&lng=121.7565488&hob_opt=2&hob_psi=5&hob_ft=56&crater=1&zm=10)
of a crater diameter of **2100m** , which, accounting for error, could be
anywhere between **1890m** and **2310m**.

As for the blast itself, the simulation lists the fireball as having a
diameter of **6400m** , which following the previous error margins could be
between **5760m** and **7040m**. These and other statistics are all from the
simulation and are viewable upon following [this
link](http://nuclearsecrecy.com/nukemap/?&kt=20000&lat=30.0090526&lng=121.7565488&hob_opt=2&hob_psi=5&hob_ft=56&crater=1&zm=10).

In the picture below, the blast is still only just beginning, and if we assume
the ships in the shot are the same length as a [typical modern Chinese
ship](http://en.wikipedia.org/wiki/Type_052C_destroyer) (~155m), these numbers
seem definitely feasible.

![Material Burst](https://i.stack.imgur.com/M3rkn.jpg)

TL;DR: The _crater_ would be approximately 2.1km across.

Granted, this simulation is for a nuclear blast and not a mass of pure energy,
but since the energy released is the same, the results should be more or less
equivalent.


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answer