On Wed, 20 Sep 2017 14:03:05 +0100, "Dave W"
wrote:

snip

Of course, but 3 x 21A for C/1 is 63A not 93A.

Sorry I'm lost mate. If we were to be judging the worst case capacity
for my real world load, wouldn't that be 30A (electric outboard on
full speed) spread across 3 batteries and so would be 10A each? I'm
not sure where the 21A has come from?

21A is the C/1 rated current of each battery.

Ok ... and was that taken as an example or worse case etc as I'll
never be drawing current at C/1?


From my graph readings of C1 and C20, 21A minus 1.60A gives an extra
voltage
drop of 12.13V minus 11.70V. If we assume that the effective internal
resistance is the same for all currents, it would be V/I = 0.43V/19.4A =
0.022 ohms.

I'm happy to assume that with you but I think it does change in
practice and hence we get Peukert's effect?

As my three values for resistance are so similar, I don't see how
Peukert's
effect comes into it.

Ok but it does ITRW though doesn't it?

I had to google for ITRW.

Sorry. ;-)

I admit I have not investigated Peukert's effect.

Well I think that may be pivotal to the whole issue Dave, both in ITRW
g and mathematically re the calcs etc.

snip

That said, those who say live on boats and rely on a good 12V supply
for a lot of things regularly generator / bank charge their paralleled
batteries (often AGM) at a pretty high rate (as they don't have the
luxury of time etc). ;-)

Ok, you are probably right in keeping them always in parallel.

It wasn't something that sat well with me initially either but if
that's what people do then so be it. ;-)

Martin Brown has raised a very good point that I hadn't considered.
Of course the capacity varies depending on the current taken, and can be
predicted from the current using his formulas.

Ok.

However, the voltage versus percentage discharge graph that you referenced
shows none of that. I wondered how they knew where the 50% discharge
point is, but I now realise it must be 50% of whatever capacity each
current produces.

I believe that is correct yes. As an aside, if this was a conventional
flooded battery with openable cell caps you could simply read the
specific gravity of a cell to glean the current charge condition. No
such possibilities with a sealed gel battery. ;-(

It looks like you will have to keep a running total of how many Ampere-Hours
you have used up by calculating current x time every so often and adding
to the total.

The problem with that Dave is that doesn't automatically compensate
for the change in capacity as the current changes?

Stop when you have used up half the capacity that 10A per
battery gives. Martin's formula times three gives 71Ah capacity at 30A, so
36Ah represents the half discharged point. At full speed you should then
have
36/30 = 1.2 hours before the battery runs out.

Ok.

The question is, how many hours from this point will you get at 15A?

Quite. ;-)

At 15A, Martin's formula predicts 85.5Ah capacity so the half-discharged
point
would be about 43Ah. I can't get my brain around this either, but it seems
to me
that irrespective of how you used up 36Ah, there should be at least 43Ah
left at 15A i.e. 2.9 hours.

Again, (and it still leaves me with a headache) it's all down to the
capacity changing with current drawn. I think thinking of 'remaining
time' is *way* advanced for this project and all I would like to do is
to display the current voltage and the predicted voltage that
represents the 50% DOD at the current current. ;-)

snip

I have skimmed through these links and now my brain hurts.

You and me both Dave!

concludes that it's all too difficult.

I think it concludes it's not realistic, *all_things_considered* to
expect an accurate gauge / outcome. However, I still believe it's not
outside the brains of this group to come up with a formula that would
be better than nothing. ;-)

I may get back to you when I've
done my own analysis.

Ok, and thanks again for your interest in this.

I think may help to picture yourself sitting in a 10' wooden and
folding dinghy, steering via a small electric outboard motor with 3 x
lead acid batteries in parallel in a box in the bow and whilst looking
at a small Arduino driven meter that is displaying:

12.56 V (current battery terminal voltage)
10.8 A (current current drawn from battery)
11.75V (voltage that is calculated from the current voltage and
current to represent the voltage alarm point of 50% DOD).

In theory and if you kept at the same speed especially ...

1) the voltage would initially drop as you turned the outboard on and
then slowly drop over time.

2) The current to mostly stay the same (it would also drop slightly as
the voltage dropped etc).

3) The predicted 50% DOD voltage to remain more or less constant
(allowing for the fluctuations in the above etc).

As I mentioned elsewhere, if you had 10 points taken from a graph that
represented 10 different currents and their resultant voltages (to 50%
DOD) you could simply look for those 10 and display accordingly.

The trick is to not have just 10 but a completely dynamic calculation
that would work with anything from 1A (realistically) and 30A. ;-)

Whilst I'm pretty sure it's not rocket science, the maths might as
well be from my POV. ;-(

Cheers, T i m