A "paradox" regarding Voronin's universality theorem The famous and remarkable Voronin's universality theorem states: 
Theorem (Voronin 75): let $0<r<1/4$ and suppose that $g(s)$ is a nonvanishing continuous function on the disk $ \vert s \vert \leq r$ that is analytic in the interior. Then for any $\epsilon>0$, there exists a positive real number $\tau$ such that: $$max_{\vert s \vert \leq r} \vert  \zeta(s+3/4+i \tau)-g(s) \vert <\epsilon. $$
Which practically means that $g(s)$ could be approximated by $\zeta(s)$ for some "high enough" values of $\tau$ - on the right hand side of the critical strip. 
The thing is that there is a certain numerical "mystery" with respect to this theorem - when it comes to the question "how high is high enough"? Of course - there are effective analytic studies in various cases. But its worthwhile to pay attention to direct calculation via computer - it turns out to be really "hard" to verify Voronin numerically (in my eyes, the illustration makes the theorem even more impressive). 
For instance - let us take the constant function $g(z)=e^{3}$. 
Question: Is there any estimate on $\tau$ for which Voronin's approximation works for $g(s)=e^{3}$? 
It is important to point out and compare, for instance, the following graph of $\log \vert \zeta(0.75+e^{0.0001 \tau} i) \vert$ for $\tau = 0,...,250000$ in this case: 

As you can see - for quite big values - the function still doesn't seem to cross the bound $\pm 3$. So when does Voronin's theorem start to kick in this case? 
(It is interesting to note also some implications to zeros of zeta (RH), for instance.)
 A: Consider $e^3$ and $e^{-3}$ separately.
For $e^3$: Think about the order of $\zeta(s)$ in the critical strip: Titchmarsh Chapter V.  Define $\mu(\sigma)$ as the lower bound of the numbers $\xi$ such that
$$
\zeta(\sigma+it)=O(|t|^\xi).
$$
The Lindelof Hypothesis (which is a consequence of the Riemann Hypothesis) is that $\mu(\sigma)=1/2-\sigma$ for $\sigma\le 1/2$ and $\mu(\sigma)=0$ for $\sigma>1/2$.  In this case $\zeta(3/4+it)=O(|t|^\xi)$ is true for every $\xi>0$, so the zeta function grows very slowly on this vertical line.  It's simply going to take a long time to reach $e^3$.
For $e^{-3}$:  $\zeta(s)$ is near $e^{-3}$ when $1/\zeta(s)$ is near $e^3$.  But on the Riemann Hypothesis, $1/\zeta(s)=O(t^\epsilon)$ for $\sigma>1/2$ and every positive $\epsilon$.  (Titchmarsh (14.2.6)).  So again, the function $1/\zeta(s)$ can grow only very slowly on the $3/4$ line.

Update:  "Can you give an estimate...?"  The point of my answer was to show why $\tau$ will likely be quite large for the function you chose, beyond the range where computation is easy. Your comment below indicates you're not really asking about Voronin's Theorem; you're asking about making effective the constants implied by Titchmarsh's use of O.  Many of these can be made effective by paying careful attention to the proofs.  It depends how badly you need them whether it's worth the effort.
