## FANDOM

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Mask de Masculine is my favourite sternritter. I like silly characters. I took a look at his profile earlier and saw that his star flash supernova is listed as city block level. I'd like to change that.

This is his ultimate attack that he uses on Renji towards the end of their fight (in chapter 563); he creates a five pointed star in the sky and it shines down a massive blast that destroys a large section of the ground below. This is a simple destruction calculation, the only complexity is in the shape of the crater.

A=(5*A(t))+A(p)

There is an streamlined forumla for working out the area of a five pointed star but I thought that this was easier so I'm using it instead. Basically I'm mapping the points as 5 triangle and the centre as a pentagon.

Green=157.496px
Red=5px
Blue=29.833px
Pink=5px
Orange=95.708px

I'm using 1.5m windows because that's the standard at the OBD and I've seen it used here before. I scaled both the pink and red lines because I wanted to be sure that I was scaling lines to windows that were nearby, but they were both 5 pixels so it was kind of unnecessary.

1.5m=5px
157.496px=47.2488m
29.833px=8.9499m
1.5m=5px
95.708px=28.7124m

First the area of the triangles.

A(t)=(1/2)*b*h

The b above is the length of the base; the b below is the length of the hypotenuse. Don't get confused. They're perfect isosceles triangles so I can work out their heights with this equation here.

h^2=b^2-((a^2)/4)
h^2=47.2488^2-((28.7124^2)/4)
h^2=2232.4491-206.1005
h^2=2026.3486
h=45.0150

Basically the same as the hypotenuse but a little smaller: basically what you'd expect. Let's plug that back into our equation.

A(t)=(1/2)*b*h
A(t)=(1/2)*28.7124*45.0150
A(t)=646.2443m^2

And now we need to work out the area of our pentagram, we can work this out with A=(1/2)*r*p. r being the ampothem (which is basically like radius for polygons), and p is the length of the perimeter.

A(p)=(1/2)*r*p

So we have an equation to work out our ampothem, in it s is the length of one side of the polygon (in this case pentagon). Our s is the same as the base of the triangle above.

r=s/(2tan(180/n))
r=28.7124/(2tan(180/5))
r=28.7124/(2tan36)
r=28.7124/1.4531
r=19.7594m

Working out the perimeter isn't hard, we have a side length and know it's a pentagon.

p=s*5
p=28.7124*5
p=143.562m

We now plug these values into our pentagon equation.

A(p)=(1/2)*r*p
A(p)=(1/2)*19.7594*143.562
A(p)=1418.3494m^2

5 triangles and a single pentagon give us our final area.

A=(5*A(t))+A(p)
A=(5*646.2443)+1418.3494
A=4649.5709m^2

Now to get our volume we're using a depth of 8.9499m; this is the longest motion line I could measure down; this is a major low end because the crater is more than likely a lot deeper, but there is no accurate way to measure its depths as deeper than this.

V=A*h
V=4649.5709*8.9499
V=41613.1946m^3

V=41613194600cc

For our low end we'll use 214J/cc for pulverisation of stone; if you look at the scan above you'll see that the crater is very clean so at the very least this is appropriate.

E=41613194600*214
E=8905223644400J
E=2.128Kt

For our high end we'll use 25700J/cc for vaporisation of stone. I really do think this was vaporised; for evidence look at the following chapter (chapter 564) where every shot of Renji has him surrounded by thick clouds of vapor, or smoke, whatever, it could be dust from pulverisation but I can't imagine pulverisation producing this quantity of dust. It surrounds him for the whole chapter and frames (and to a degree shadows) everything he does. Seriously I cannot express the quantity in words.

E=41613194600*25700
E=1069459101220000J
E=255.607Kt

That gives us our two values.

2.128*10^3 tons of TNT   or   2.45661*10^5 tons of TNT

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