10-B to 5-A

[Daddybrawl]

This is an incredibly stupid question. Might as well ask though.

A single 10-B’s minimum amount of Input is 40 Joules. A 5-B’s Maximum is 2.7 Yottatons. How many 10-Bs need to combine into a single attack to match the power of 2.7 Yottatons, aka how many 40 Joules does it take to make a 5-B?

Alternate question, same thing but with 2.825 Tenaexafoe, aka 3-A.

Tried looking it up online, the search results literally just sent me to our Attack Potency page, lol. I got curious and decided to ask.

 

[Epsilon_R]

10-B Minimum: 40 Joules

5-B (Maximum): 2.7 Yottatons which is 1.13E+34 Joules

All you need to do is divide 1.13E+34 by 40 and you get 2.825E+32

That means you'll need 2.825E+32 10-Bs combining their strength to be able to generate this much energy

 

[Epsilon_R]

You do the same thing with 2.825 TenaexaFoe which is 2.825E+92 Joules, you get 7.0625E+90 × Baseline 10-B

 

[Daddybrawl]

Oh. That was fast and surprisingly simple. Thanks!

 

[Planck69]

You do the same thing with 2.825 TenaexaFoe which is 2.825E+92 Joules, you get 7.0625E+90 × Baseline 10-B

Jesus, so you need 10 billion times the number of atoms in the universe worth of humans to universe-bust?

 

[DarkDragonMedeus]

Well, combining attacks together like that doesn't quite stack AP in the linear sense.

 

[Greatsage13th]

Well, combining attacks together like that doesn't quite stack AP in the linear sense.

I think fusing attacks makes more sense for AP stacks