In [@donaldson5] Donaldson proves that in the case of the $T_{\nu}$ iteration and starting with a palindromic metric this $\sigma$-value can be computed as $$\sigma_{T_{\nu},k} =
        \frac{(k-1)k}{(k+2)(k+3)} \quad  \mathrm{(if\ }G\mathrm{\ is\ palindromic)}$$ By examining many examples we also observed that if $G$ is not palindromic we get $$\sigma_{T_{\nu},k} =
        \frac{k}{k+2} \quad  \mathrm{(if\ }G\mathrm{\ is\ not\ palindromic)} \\$$ while in the case of the $T$ iteration we have $$\label{sigmaT asymptotic}
    \sigma_{T,k} = \frac{(k-1)(k+6)}{(k+2)(k+3)}$$ and for $T_K$ we get $$\label{sigmaTK asymptotic}
    \sigma_{T_K,k} = \frac{k-1}{k+3} .$$ In neither of these latter two cases does it matter if we start with a palindromic metric or not.

We see that when $k=2$ we have $$\sigma_{T_{\nu},2} \mathrm{(not\ pal.)} > \sigma_{T,2} > \sigma_{T_K,2} > \sigma_{T_{\nu},2} \mathrm{(pal.)}$$ while for $k \geq 3$ we have $$\sigma_{T,k} \geq \sigma_{T_{\nu},k} \mathrm{(not\ pal.)} > \sigma_{T_K,2} > \sigma_{T_{\nu},2} \mathrm{(pal.)}$$ with strict inequalities for every $k>3.$ So in general, if we start with a palindromic metric $G$ we expect that the $T_{\nu}$ iterations will converge the most quickly, followed by $T_K$ and then by $T.$ Starting with a non-palindromic $G$ the $T_{\nu}$ iterations will slow down, and we find that $T_K$ will converge fastest. Here $T$ is still slowest to converge.

The effect on 