I am interested in showing that certain knots having a surgery description are hyperbolic. Unfortunately I have not had time yet to read Thurston's work, so my understanding of this is vague. But from reading some papers, it seems that the following is correct (so the first question - is it correct?). Given a hyperbolic link, there is a constant, such that filling a subset of components with slopes with denominators bigger than this given constant, one always gets a hyperbolic manifold (complete, finite volume?).


So my real question is: can we explicitly find this constant? We may or may not be given the volume of the initial link.



What constant ensures hyperbolicity of Dehn surgery?
Given a hyperbolic link $L$, [Thurston's Dehn filling theorem](http://en.wikipedia.org/wiki/Hyperbolic_Dehn_surgery) says that there is a finite set of "bad slopes" on each component of $L$ such that every filling avoiding them is hyperbolic. A slope is just a rational number, and since the bad slopes are finite in number, of course if all denominators are sufficiently big to avoid them the result is hyperbolic. So your guess is correct.


A simple way to find these "bad" slopes is to download [SnapPy](http://www.math.uic.edu/t3m/SnapPy/index.html), practice a bit with this user-friendly nice program, then draw your links. If you have some specific Dehn surgery parameters to test, just give them to SnapPy. If you look at these finite sets of slopes to avoid, you must look at the cusp shapes of the boundary tori. The cusp shape of a torus is a complex number $p+qi$. You can then select a cusp section of some area $A$ manually, and from the three numbers $p,q,A$ you can very easily list the finitely many slopes having length at most $6$. These are the "bad slopes" to avoid on that component, by Agol-Lackenby's [6 theorem](http://en.wikipedia.org/wiki/2%CF%80_theorem).


If your link is a knot, then it is conjectured (and almost proved) that if the denominator is >2 then the result is hyperbolic. For links things are however more complicate.


