A question about computability and Turing machines For any recursively enumerable set theory $T$ (of consistency strength at least superior to KP), if we want to calculate $F(n)=\{F(m)：m∈ω∧mEn\}$ and can determine each $F(n)$ for a Henkin model $(ω,E)$ constructed by $\mathrm{con}(T)$, what is the minimum value of ordinal $α$ so that a Turing machine calculating $α$-recursive function by extension of tape and calculation steps defined in $α$-recursion theory can calculate $F(n)=\{F(m)：m∈ω∧mEn\}$ and can determine each $F(n)$?
Does the $α$ increase with increase of consistency strength of $T$?
 A: There seems to be an assumption underlying your question that from Con(T) we can compute a model $\langle\omega,E\rangle$ for which the function $F(n)=\{m\mid m\mathrel{E} n\}$ makes sense.
But this is not quite right. The function $F$ will be the Mostowski collapse of the structure $\langle \omega,E\rangle$, and in general, this makes sense only when $E$ is a well-founded relation. But in general, from Con(T) alone one does not know there is a well-founded model of T.
Rather, from the theory T we will have to search higher in the constructible hierarchy to find a well-founded model of T.
The height we have to go will be higher than the minimal height of a transitive model of T. For example, in the case of KP itself, this will be height $\omega_1^{CK}$, since $L_{\omega_1^{CK}}$ is the minimal transitive model of KP. But for other c.e. theories such as ZFC, one will have to go much higher, and for ZFC + large cardinals, much higher still. Thus, to answer your final question, yes, in general, the stronger consistency strength theories will have taller models, and so the computations will require bigger ordinals to compute the models.
But the computational difficulty here is not computing $F$ for a given $E$, but rather finding a relation $E$ that is well founded and satisfies the theory in the first place.
