The original Angel problem, first proposed by Conway, Berlekamp, and Guy has two players: the devil and the angel. The angel is placed at the origin of an infinite $2$-D chessboard. The angel's movement, done on its turn, mimics that of a King in chess that is allowed to move up to $k$ times, where $k$ is called the power of the angel. It is the job of the devil to trap the angel by removing one of the squares on its turn. The angel may leap over any of these barriers (for $k \geq 2$), but cannot end its turn in a removed square. The devil wins if and only if the angel cannot move after a finite series of turns. If play goes on indefinitely, then the angel wins.

I'm considering a variation of the problem on a $3$-D board, where the angel's movement is restricted to changing up to three of its' coordinates, but only by a change of +/- 1. For this case, it is known the angel wins even if its power is one. I noticed here,

http://home.broadpark.no/~oddvark/angel/

that fractional values for $k$ have been considered in $2$-D, where I'm assuming without access to the literature would be defined as the power of the angel divided by the number of squares able to be eliminated by the devil on its turn. It is known for the standard, $3$-D angel problem (the one where the number of squares removed from the board by the devil on its turn exactly one) that the angel wins regardless of the value of its power.

My question is this: have fractional powers, as defined above for the nonstandard problem, been considered for dimensions other than two? It would seem that if we allow the number of squares removed by the devil on each of its turns to be strictly greater than the power of our angel, we have a perhaps interesting, open problem - as there are many variations of this problem.

Note: I posted earlier that if the fractional powers for any two of the same variation of the problem are the equal, then we have equivalent problems. For example, take the power of the angel to be $3$ and the squares removed to be $2$. However, I now don't necessarily think this problem is equivalent to the problem with the power of the angel taken to be $6$ and the squares removed on the devil's turn equal to $4$. I definitely could be wrong though.