To partially answer this question, in Advanced Macroeconomics by Romer, question 1.5(c) asks
"What saving rate is needed to yield the golden-rule capital stock?"
Answering this tells us what $s_{gold}$ is.
But before we can find this we must find the golden rule level of capital stock $(k_{gold})$. Which we shall derive with the following equations: The intensive form of the production function, the dynamics of capital in the Solow Model, and consumption per unit of effective labour
\begin{cases}
Y_t = (A_t L_t)^{1-\alpha} K_t^{\alpha} = A_t L_t \left( \frac{K_t}{A_t L_t} \right)^{\alpha} = A L k^{\alpha} \\
\Rightarrow Y/AL = y = k^{\alpha} \\
\dot{k} = s y - (n+g+\delta) k \\
c = (1-s)y \\
\end{cases}
Under the balanced growth path, $\dot{k}=0$ so
$$ \displaystyle s k^{* \alpha} = (n+g+\delta)k^* \Rightarrow k^* = \left( \frac{n+g+\delta}{s} \right)^{\frac{1}{\alpha-1}} $$
We maximize $c^*$ with respect to $k^*$ to implicitly define the golden-rule level of capital per unit of effective labour
$$ \displaystyle \frac{\partial }{\partial k^*} c^* = \frac{\partial }{\partial k^*} (1-s)k^{* \alpha} = (1-s)(\alpha k^{* \alpha - 1}) $$
$$ \displaystyle = \alpha (1 - s) \left( \frac{n+g+\delta}{s} \right)^{\frac{\alpha-1}{\alpha-1}} $$
Next replace $s$ with $(n+g+\delta)k^{* 1-\alpha}$
$$ \displaystyle = \alpha (1 - (n+g+\delta)k^{* 1-\alpha}) \left( \frac{n+g+\delta}{(n+g+\delta)k^{* 1-\alpha}} \right) $$
$$ \displaystyle 0 = \alpha (k^{* (\alpha-1)} - (n+g-\delta)) \Rightarrow k^*_{gold} = \left( \frac{\alpha}{n+g+\delta} \right)^{1/(1-\alpha)} $$
We can finally compute, $s_{gold}$
$$ \displaystyle s_{gold} = (n+g+\delta) k_{gold}^{*1-\alpha} = (n+g+\delta) \left( \frac{\alpha}{n+g+\delta}\right)^{(1-\alpha)/(1-\alpha)} = \alpha $$
So for a Cobb-Douglas production function the golden-rule saving rate should be whatever $\alpha$ is.