Let:

• $V$ be the open-circuit, aka idle, voltage of a batteries bank
• $S(V)$ be the ampere scale function of value from 0% to 100% with respect to the idle voltage, $S(V_{t_1})-S(V_{t_0})=\frac{1}{C} \int_{t_0}^{t_1} I_t d t$ where $C$ is the total ampere capacity of the batteries bank and $V_0$ means having empty voltage. For example of simplicity, it could be $S(V) = \frac{V-20}{7.2}$ then 20 is the empty voltage and 27.2 is the full voltage and $C=600 = \int_{0}^{\infty} I_t d t$ is a batteries bank of capacity 600 ampere-hours.
• $X$ the voltage seen by the charger, for example $X=27.2$
• $r$ be the resistance by the charging circuit and the internal resistance of the batteries bank, so the charging current is $I=\frac{X-V}{r}$, for example $r=0.025$ ohm

What is the needed time $T$ to charge the batteries bank from idle voltage $V_0$ to $V_1$?

Because $C S'(V) d V= I_t d t$, it follows $d t = \frac{C S'(V)}{I_t} d V =\frac{r C S'(V)}{X-V} d V$, therefore $T=r C \int_{V_0}^{V_1} \frac{S'(V)}{X-V} d V$

By the above example, say $V_0=26.6$ and $V_1=27.0$, then

$T=\frac{0.025 \cdot 600}{7.2} \cdot \int_{26.6}^{27.0} \frac{d V}{27.2-V}=\frac{0.025 \cdot 600}{7.2} \cdot \ln(\frac{27.2-26.6}{27.2-27.0})=2.29$ hours

What is the total wh of the batteries bank when it is full?

$\int V_t I_t d t=C \int V S'(V) d V=\frac{600}{7.2} \int_{20}^{27.2} V d V=14160$ wh

The better the battery, the more vertical about the curve $S(V)$ or horizontal if X-axis and Y-axis flipped for the curve. If it is centralized at the voltage $V_{rating}$ then $\int V_t I_t d t=C \int V S'(V) d V \approx C V_{rating}$