Can there be ordinals larger than those contained in Ord, the class of all ordinals,and if so, can these ordinals be used to extend the constructible universe $L$?

In a simplified form, my question is (see last paragraph below): *Given the constructible universe $L$, does there exist a top-extension $N$ of $L$ such that $N$ also satisfies the Axiom of Constructibility?*

To show that the possibility might exist,consider the model $M[H]$ of $ZFC^{-}$ that collapses all sets to be countable. Regarding this model, Andreas Blass writes to Noah S in the comments to mathoverflow question "How can I collapse all cardinals of the ground model except one of them", the following:

"...If I were to think about $M[H]$, I'd pretend that some more powerful set theorists have continued the cumulative hierarchy beyond the hierarchy I see, so that what I think is Ord is actually just some inaccessible cardinal $\kappa$. Then what I call $M[H]$ is, for people in that bigger universe, the result of Levy collapsing below $\kappa$ and and them truncating the universe at $\kappa$. That Levy model is pretty well understood, since it occurs along the way in Solovay's construction of his Lebesgue measure model."

This sounds very similar to some ideas of W.N. Reinhardt found in his paper "Set existence principles of Shoenfield, Ackerman, and Powell" (Fundamenta Mathematicae, vol. 84 (1974), pp. 5-34). Let me quote Shoenfield's principle as given to Reinhardt through private communication (found on page 6 of the paper):

"If $\mathtt C$ is a collection of stages [say $\{V_{\alpha\in Ord}\}$—my comment], and if we can imagine a situation in which all of the stages in $\mathtt C$ have been completed, then there is to be a stage $\mathtt S$ after all the stages in $\mathtt C$." In the comment by Blass to Noah S, the terms "see" and "think" can be fairly deemed to be synonymous with Shoenfield's term "imagine".

Assuming that the stages $L_{\alpha}$ of the 'usual' constructible universe $L$ range over Ord, how does the application of Shoenfield's principle to $L$ (i.e. given that if the union of the elements $L_{\alpha}$ of the collection $\mathtt C$ of stages $L_{\alpha\in Ord}$ form $L$, then there is a stage $L_{\mathtt S}$ after all of the stages in $\mathtt C$, where $\mathtt S$ is a 'new ordinal' such that $\mathtt{S} \notin Ord$ and $L_{\mathtt S}$ is the stage where the constructible subsets of $\mathtt S$ are formed) extend the constructible universe, if at all? (Consider $V$ and $L$ to be classes of sets of ordinals to make better sense of the title.)

Perhaps there is a simpler way to phrase the question. Given two models $M$ and $N$ of ZFC with $Ord(M) \subset Ord(N)$, let $N$ be a Top-extension of $M$ just in case $M$ is a rank initial segment of $N$, that is, $M$ end-extends to $N$ and also all new objects of $N$ have ordinal rank larger than any ordinal in $M$ (this definition comes from Joel David Hamkins, who put it in his answer to mathoverflow question "Elementary end extension of a countable model for ZF"). It should be noted that $M$,$N$ satisfies Shoenfield's principle in that given $M$(which, being a model of ZFC, has its own version of the cumulative hierarchy completed in $M$) there is a stage $\mathtt S$ in $N$ after all the stages in $\mathtt C$, where $\mathtt C$ is the collection of stages in $M$.

What do you mean by "applies Shoenfield's principle to $L$"?Are you asking the following: "Suppose $V\models ZFC+V=L$, $\kappa\in Ord(V)$, and $V_\kappa\models ZFC+V=L$; then what are some ways $V$ and $V_\kappa$ can be different?" Or are you asking, "Given a model $M$ of $ZFC+V=L$, is there another model $N$ such that $M=L_\alpha^N$ for some $\alpha\in N$?" Or what? I'm not being deliberately slow, I honestly don't know precisely what you are asking; and with this sort of question, precision is really important. $\endgroup$ – Noah Schweber Dec 6 '14 at 2:56anymodel of $ZFC$ has an end extension which is a model of $V=L$. So in principal,anymodel arises as part of the constructible universe of another model. Of course, the models Barwise's theorem produces are not well-founded, so their "ordinals" aren't, really; but it's still interesting. You might also find this question of mine relevant: mathoverflow.net/questions/171703/… $\endgroup$ – Noah Schweber Dec 6 '14 at 3:1615more comments