gained the runepower. Thor faced him under only his own power and without Mjolnir, and came away with a KO.
Also, what actual showings indicate that Karnilla is above top end herald?
Further, we have enough showings of the Void to indicate that he's WAY above Thor, who is top tier herald...why would he be top end herald when he's probably superior to several guys you agree with in the teambuster rank...
>
> > Thanks to vermin for bringing this up yesterday.
> >
> > Now I want to populate it. Basically, just think of a character who, without too much effort, annihilates a herald character or battles a group of heralds and prime metas.
> >
> > Agree/Disagree and add more!
> >
> > Thanos (for pwning SS among other)
> agree
> > Fernus (for pwning Supes among others)
> Besides (PM), I don't see any (JLA) member winning this guy!
> > Mordru
> To be honest, he's just more powerful b/c of the artifacts he gets/steals, I still believe that (Loki) with prep can probably take hime!
> > Durok
> He's very questionable b/c of (Oeming)'s run, remember that it was (Rune/Odinpower Thor) who took him out! So with that in mind I have to say yes!
> > Desak
> Agree
> > Morg WOL
> Agree
> > Lord Tantalus
> Agree
> > Titannus
> Agree
> > Cyborg Supes w/rings
> Well anybody with (10: QR's) has to be!
> > PE Zuras
> Agree, Zuras IMHO opinion is (Skyfather) level, and is equal in pwr to (Zeus)!
> > Tyrant
> Agree
> > Highfather
> Low End Skyfather
> > Takion
> See Above
> > The Doctor
> See Above
> > Dr. Manhatten
> See Above
> > Solar
> See Above
> > Superboy Prime
> Agree
> > Ion Sodam Yat
> Agree
> > Karnilla
> Low End Skyfather
> > Monarch
> Agree
> > Molecule Man
> Agree
> > HP Doomsday
> Agree
> > Phoenix (GOTG)
> Agree
> > The Void
> High End Herald
> > Super Adaptoid
> Agree
> > Amazo
> Agree
> > Annihilus (Annihilation)
> High End Herald
> > Afro Magus
> High End Herald
> > Runner
> High End Herald
> > Grandmaster
> Agree
> > Champion w/gem
> High End Herald
> > Mangog (recent)
> Agree
> > Destroyer
> Agree
> > Despero (V&V)
> Agree
>
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