2 removed epsilon, as it was not really defined anywhere in the post

It turns out that I was wrong and both methods give correct results. In fact using the two results you can reproduce the scaling parameters $\epsilon$ for the simple Lie algebras given in the Broughton paper linked to by David.

My mistake was in assuming that the length of the roots is always arbitrary. It's well known that the choice of the invariant bilinear on the algebra is unique up to a scale. The restriction of the bilinear onto the Cartan subalgebra is dualized to give the metric on the root space and thus passes on the choice of scale to the roots.

But I had already chosen the Killing form as my bilinear - a necessary ingredient in the first method I presented. Thus the length of my roots was fixed.

Direct construction of a few low rank cases (something that I should have done before posting) confirmed that everything works out ok.

As a quick check and application, we can reproduce the scaling factor given by Broughton for $A_n$. We have $\mathrm{rnk}(A_n)=n$ and $\mathrm{dim}(A_n)=(n+1)^2-1$, so $l=\frac{\mathrm{rnk}(A_n)}{\mathrm{dim}(A_n)-\mathrm{rnk}(A_n)}=\frac{1}{n+1}$.

1

It turns out that I was wrong and both methods give correct results. In fact using the two results you can reproduce the scaling parameters $\epsilon$ for the simple Lie algebras given in the Broughton paper linked to by David.

My mistake was in assuming that the length of the roots is always arbitrary. It's well known that the choice of the invariant bilinear on the algebra is unique up to a scale. The restriction of the bilinear onto the Cartan subalgebra is dualized to give the metric on the root space and thus passes on the choice of scale to the roots.

But I had already chosen the Killing form as my bilinear - a necessary ingredient in the first method I presented. Thus the length of my roots was fixed.

Direct construction of a few low rank cases (something that I should have done before posting) confirmed that everything works out ok.

As a quick check and application, we can reproduce the scaling factor given by Broughton for $A_n$. We have $\mathrm{rnk}(A_n)=n$ and $\mathrm{dim}(A_n)=(n+1)^2-1$, so $l=\frac{\mathrm{rnk}(A_n)}{\mathrm{dim}(A_n)-\mathrm{rnk}(A_n)}=\frac{1}{n+1}$.