Can the lattice stick number of a knot be bounded
by the stick number of the knot? 


The stick number
$S(K)$ of a knot $K$ is the fewest number of segments
needed to realize it by a simple 3D polygon.
The lattice stick number $S\_L(K)$ is the fewest segments in a realization in the cubic
lattice, with all segments parallel to a coordinate axis.
For example, the stick number of the trefoil knot $K=3\_1$
is 6, and its lattice stick number is 12
(the latter a result of Huh and Oh from 2005).


My question is whether it is possible to bound $S\_L(K)$
by $m S(K)$, where $m$ is some multiplier factor. 
Ideally $m$ would be a constant, but perhaps it is more realistic
to expect it to depend
on the complexity of the knot (e.g., on its crossing number
$cr(K)$).
What I have in mind is replacing each stick in a stick realization
by a bounded lattice path.


**Addendum**. Tracy Hall's clever example below indicates that it is unlikely that $m$ could be a constant.



Lattice Stick Number vs. Stick Number of Knot
I wouldn't be surprised by something like a quadratic bound, or possibly something reasonable in terms of another complexity measure for the knot, but I see no hope for making $m$ constant. Consider the following construction: given $m$, choose some large number like $N=(10m)^6$ of points uniformly at random in the unit sphere, and connect them sequentially in a cycle with straight line segments to define a knot $K$. By construction $K$ has stick number no more than $N$, but each stick has a long narrow tunnel that it must traverse in a very precise direction, which is difficult to do with only $m$ lattice sticks. Of course any one tunnel can be made shorter and wider with an affine transformation (or any small collection of tunnels, with a piecewise affine transformation) but I am convinced (without attempting a rigorous proof) that with probability approaching $1$ a knot so constructed has a lattice stick number much higher than $mS\_L(K)$.


