<p>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.</p>
<p>But while the light novels don't give us an accurate size of the crater, they <em>do</em> give us the equivalent energy - 20 megatons (20,000 kilotons) of TNT:</p>
<blockquote>
<p>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.</p>
<p>In accordance with Einstein's Formula, the heat energy equated to
approximately 20,000,000 tons of TNT.</p>
<p><sub>– <em>Volume 7 - Yokohoma Disturbance II</em>, Chapter 13</sub></p>
</blockquote>
<p>Now using this, we can approximately calculate a blast radius.</p>
<p>To get my results, I used this simulation here - <a href="http://nuclearsecrecy.com/nukemap/" rel="nofollow noreferrer">http://nuclearsecrecy.com/nukemap/</a>. First off, I ran a couple of tests by plugging in the data from real-life detonations to see how accurate it was:</p>
<ul>
<li><p>Test 1: <a href="http://en.wikipedia.org/wiki/Ivy_Mike" rel="nofollow noreferrer"><em>Ivy Mike</em> Test</a> - Actual crater diameter: <strong>1900m</strong> Result from simulation: <a href="http://nuclearsecrecy.com/nukemap/?&amp;kt=10400&amp;lat=53.4624326&amp;lng=-1.8402328&amp;airburst=0&amp;hob_ft=0&amp;crater=1&amp;zm=10" rel="nofollow noreferrer"><strong>1680m</strong></a> - 88% accurate</p>
</li>
<li><p>Test 2: <a href="http://en.wikipedia.org/wiki/Castle_Bravo" rel="nofollow noreferrer"><em>Castle Bravo</em> Test</a> - Actual crater diameter: <strong>2000m</strong> Result from simulation: <a href="http://nuclearsecrecy.com/nukemap/?&amp;kt=15000&amp;lat=53.4624326&amp;lng=-1.8402328&amp;airburst=0&amp;hob_ft=0&amp;crater=1&amp;zm=9" rel="nofollow noreferrer"><strong>1900m</strong></a> - 95% accurate</p>
</li>
<li><p>Test 3: <a href="http://en.wikipedia.org/wiki/Sedan_(nuclear_test)" rel="nofollow noreferrer"><em>Sedan</em> Test</a> - Actual crater diameter: <strong>390m</strong> Result from simulation: <a href="http://nuclearsecrecy.com/nukemap/?&amp;kt=104&amp;lat=53.4624326&amp;lng=-1.8402328&amp;airburst=0&amp;hob_ft=0&amp;crater=1&amp;zm=13" rel="nofollow noreferrer"><strong>360m</strong></a> - 92% accurate</p>
</li>
</ul>
<p>Given this information, we can assume that the simulation is roughly 90% accurate, giving an error margin of ±10%.</p>
<p>Setting the yield as 20,000 kilotons, with an airburst of 17m (the beam height of a <a href="http://en.wikipedia.org/wiki/Type_052C_destroyer" rel="nofollow noreferrer">typical modern Chinese ship</a>, or the height at which the flag would be flying) leaves us with a <a href="http://nuclearsecrecy.com/nukemap/?&amp;kt=20000&amp;lat=30.0090526&amp;lng=121.7565488&amp;hob_opt=2&amp;hob_psi=5&amp;hob_ft=56&amp;crater=1&amp;zm=10" rel="nofollow noreferrer">result</a> of a crater diameter of <strong>2100m</strong>, which, accounting for error, could be anywhere between <strong>1890m</strong> and <strong>2310m</strong>.</p>
<p>As for the blast itself, the simulation lists the fireball as having a diameter of <strong>6400m</strong>, which following the previous error margins could be between <strong>5760m</strong> and <strong>7040m</strong>. These and other statistics are all from the simulation and are viewable upon following <a href="http://nuclearsecrecy.com/nukemap/?&amp;kt=20000&amp;lat=30.0090526&amp;lng=121.7565488&amp;hob_opt=2&amp;hob_psi=5&amp;hob_ft=56&amp;crater=1&amp;zm=10" rel="nofollow noreferrer">this link</a>.</p>
<p>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 <a href="http://en.wikipedia.org/wiki/Type_052C_destroyer" rel="nofollow noreferrer">typical modern Chinese ship</a> (~155m), these numbers seem definitely feasible.</p>
<p><img src="https://i.sstatic.net/M3rkn.jpg" alt="Material Burst" /></p>
<p>TL;DR: The <em>crater</em> would be approximately 2.1km across.</p>
<p>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.</p>

CC BY-SA 4.0
2015-05-19T22:32:41.067Z
2023-05-01T20:01:22.120Z
2023-05-01T20:01:22.120Z
Matt
<p>In episode 26 of the anime, Silver/Ooguro/Tatsuya annihilated the Chinese fleet in one shot with <em>Material Burst</em>. No wonder his aunt had his ability locked...</p>

<p>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.</p>

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

<p>So by the images of episode 26, how big was the blast Ooguro caused?</p>

CC BY-SA 3.0
2014-12-05T14:07:01.823Z
It is detonated on the sea, so I don't think there is going to be any crater.
nhahtdh
CC BY-SA 3.0
2014-12-05T23:29:20.957Z
I'me pretty sure any kind of impact like that would leave a crater.
кяαzєя
CC BY-SA 3.0
2014-12-06T17:28:57.507Z
Ok, he created a gulf, bay, you name it. Updated the question.
Mindwin Remember Monica
CC BY-SA 3.0
2014-12-06T18:27:40.383Z
@nhahtdh I rolled back your edit, because the blast of that spell is not a nuclear (since everyone assumes nuclear is also radioactive) explosion. Tatsuya used relativistic/quantic physics to generate energy, ok. But to leave it as nuclear in the title would be misleading. Even in the novels Japan is cleared of the international backlash of doing that stunt exactly because the inspectors did not find evidence of nuclear reactions (i.e. fallout)
Mindwin Remember Monica
<the-irregular-at-magic-high-school>
What is the size of the crater left by the explosion in episode 26?
# What is the size of the crater left by the explosion in episode 26?
Posted by: username_0 on 2014-12-05
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?
**Tags:** the-irregular-at-magic-high-school
> **Comments:**
> **username_1**: It is detonated on the sea, so I don't think there is going to be any crater.  
> **username_2**: I'me pretty sure any kind of impact like that would leave a crater.  
> **username_0**: Ok, he created a gulf, bay, you name it. Updated the question.  
> **username_0**: @nhahtdh I rolled back your edit, because the blast of that spell is not a nuclear (since everyone assumes nuclear is also radioactive) explosion. Tatsuya used relativistic/quantic physics to generate energy, ok. But to leave it as nuclear in the title would be misleading. Even in the novels Japan is cleared of the international backlash of doing that stunt exactly because the inspectors did not find evidence of nuclear reactions (i.e. fallout)  
---
## 1 answers
### Answer by username_3 on 2015-05-19. Score: 5 | Accepted answer
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.sstatic.net/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.
---
