Positive operators and Hilbert's metric

Question

If an operator L on a finite-dimensional vector space is positive, then it is a contraction in the Hilbert metric $d(x,y)$. The contraction ratio is related to the projective diameter of the operator. Letting $C$ denote the non-negative quadrant, the projective diameter is the diameter of L(C), $\Delta=\sup\{d(x,y): x,y\in L(C)$. The contraction ratio is then $(1/4)\tanh(\Delta)$. Due to Birkhoff. My question is, how does one compute $\Delta$? I would have thought that for a polyhedral cone, it would suffice to compute the maximal distance between the images of the extreme rays of $C$: If $L(x)=Ax$ for an $n\times n$ matrix $A$, this would be the max of the distances of the columns. But it appears to me that open balls in Hilbert's metric are not convex, and if so, the column distance max is not the right answer.