Yes, it is normal for the resting voltage to drop below the charging voltage.
When charging a Lithium-Iron-Phosphate cell the voltage will gradually rise as the stored charge increases, then rapidly climb to 3.6V and over as it reaches full charge. This rapid voltage rise is a result of the battery running out of places to store charge, and will not be sustained when it is resting.
In a series pack one cell may reach full charge sooner than the others, and then its voltage could rise to over 4V before the battery reaches peak voltage. To prevent this you should put a 'balancer' circuit across each cell to bypass the current when the cell reaches 3.6V. When all cells reach 3.6V you can stop charging because the pack will be almost full. Alternatively you can gradually reduce charging current to zero while staying at peak voltage, which might get an extra 10% or so into the battery.
In the example plot below the green line is charging current, the red line is voltage across the cell, and the blue line is absorbed charge. You can see that very little charge is accepted after reaching 3.6V, and the resting voltage soon drops to 3.4V after charging current is removed.
Here are some typical discharge curves for AGM lead-acid, in this a case a 12V 60AH battery (40Ah should be similar but with all currents multiplied by 40/60):-
Note that the time scale on this graph is logarithmic so the curves are a bit deceptive. However you can see that as current increases voltage reduces, and capacity is reduced at high current (eg. at 60A it lasts 30 minutes which is only 30Ah). So you can use voltage to estimate remaining capacity, but to get reasonable accuracy the effect of current draw must be factored into your calculations.
how can I continue using the voltage-capacity plot method while taking
into account the changes caused by aging?
Capacity may be reduced due to shedding of active material from the plates, electrolyte loss, sulphation, impurities or contamination etc. Some of these things also increase internal resistance, causing greater voltage drop at higher current.
Capacity loss may occur suddenly and cannot be reliably predicted, so to get the actual capacity you need to measure it. The obvious method is to fully charge the battery and then discharge it until empty. Alternatively you could just discharge until you reach the minimum working voltage in your system at normal current draw, as this is the effective capacity in your application. If only done occasionally this calibration cycle should not significantly affect lifespan.
You can measure internal resistance by applying or waiting for a step change in current, then calculating resistance from the voltage difference. This figure can then be used in the calculations you do to estimate remaining capacity.
Best Answer
Depends on how the manufacturer wants to rate it. Capacity will be reduced at higher current due to internal resistance and chemical effects, but the effect is usually small at rates below 1C (1 hour discharge time).
Here's an example for a 2C rated battery with a nominal capacity of 660mAh. In this case it appears the rated capacity is expected to be achieved at 1C, with about 5% extra at lower discharge rates.