From http://rjlipton.wordpress.com/2009/05/27/arithmetic-hierarchy-and-pnp/,Define, $M_{[x,c]}$ as the deterministic Turing machine that operates  as follows on an input $y$. The machine treats $x$ as a deterministic  program, and simulates $x$ on input $y$. At the same time the machine  runs a counter that stops its execution after steps $|y|^c$. If the  machine accepts before the counter stops, then it accepts; otherwise,  it rejects. Let $f(i,c)$ be the smallest natural number so that $M_{[i,c]}$makes a  mistake on the input $y$. Then, if $P \neq NP$ is true, the function  $f(i,c)$ is always defined.Theorem: Suppose that there are infinite number of $i$ for which there exists a $c$ so that $$f(i,c) > 2^{2^{|i|+c}}$$ Then, for infinitely many $n$, SAT has circuit size $n^{O(\log n)}$. Proof: Let $i>1$ and $c$ be so that $$f(i,c) > 2^{2^{|i|+c}}$$ Define  $n = 2^{|i|+c-1}$. Note, that $c$ is at most $\log n$. Then,  $M_{[i,c]}$ on all $y$ of length $n$ is correct, since $y \leq 2^n = 2^{2^{|i|+c-1}} < f(i,c)$.  The size of the circuit that simulates this Turing machine on inputs  of length $n$ is polynomial in $|i|$, $n$, and the running time of the  machine. The machine, by definition, runs in time $|y|^c \leq n^c \leq n^{\log n}$I am not getting this part. Can anyone explain this (to specify, The size of the circuit that simulates this Turing machine on inputs of length $n$ is polynomial in $|i|$, $n$, and the running time of the machine in the quote)? (So the question is how can we relate the running time of Turing machine to the size of the circuit.)
How to relate circuit size to the running time of Turing machine
complexity theory;circuits