I think that makes sense only for smaller n. At some point your "sufficiently large constant" will need to essentially be computed from the "worse big-O" algorithm to slow the "better big-O" algorithm down enough. E.g. making an O(N) as slow as an O(N^2) algorithm would require a sufficiently large constant roughly equivalent to N^2, at which point you really just have turned the O(N) into O(N^2) in practice.