T i m wrote:

On Sat, 16 Sep 2017 00:50:48 +0100, (Roger Hayter)
wrote:

snip

And why I'm trying to apply *some* science to whatever I end up with?

If I set the low voltage threshold to a voltage that would represent
50% DOD when discharged at 5A the alarm *will* sound prematurely when
running at a higher current. Now I could take that on the grounds that
at least it wouldn't over-discharge the batteries but I wouldn't get
the full capacity either. ;-(

This is where I think you are slipping into a fallacy. Although the
capacity has been less at the higher current, you have *lost* that
capacity by using a higher current and you can't get it back.

Yep, I understand that.

Even
though 50% DoD has been reached proportionaliy sooner, it *has* been
reached

No, it won't have been reached if the meter is using an arbitrary low
voltage threshold though?

and you can't reclaim the loss of capacity, except by
over-discharging the battery.

Of course (if that was the case).

This is why for battery management, as
opposed to journey management, the reduction of capacity with higher
current is irrelevant to you. You still want to stop at 50% DoD however
soon you reach it.

*Agreed*. Now, what low voltage is *correct* for what current, as one
voltage can't be correct for all currents can it (or they wouldn't
publish a graph showing different voltage cutoff points for different
currents presumably)?

Please don't get embroiled in the thoughts that I want to actually
monitor the actual battery capacity because I don't ... I just want to
do a bit more that use a single 'catch-all' values for what is already
known to be a conditional range. ;-)

Cheers, T i m

p.s. I have various electronic measurement tools including a watt hour
meter but it's no use as a form of protection tool and it doesn't
compensate for the varying capacities at varying loads. It could give
me a 'worse case' value though (30A x the voltage x the period
measured as a percentage (50% say) of the potential at that rate).

The model of the battery as a voltage source in series with a resistance
is quite a good one I think, and the on-load voltage plus predicted
voltage drop is quite a good measure of what the off-load voltage has
fallen to. Which just leaves the temporary loss of capacity at high
load, which would recover if the battery was left off load. The latter
might be a relatively small effect at the moderate current of about 10A
per battery that you are using. But as you might guess I don't know
any better than you.

An interesting experiment would be to run the battery down to the
indicated 50% discharge depth, then, without charging, use it again the
next day and see if you get a significant amount of use before it falls
to the same voltage. If not, then your measurement system is doing
quite an accurate job of protecting the battery without wasting
capacity.


--

Roger Hayter

On Sat, 16 Sep 2017 09:41:25 +0100, (Roger Hayter)
wrote:

snip
p.s. I have various electronic measurement tools including a watt hour
meter but it's no use as a form of protection tool and it doesn't
compensate for the varying capacities at varying loads. It could give
me a 'worse case' value though (30A x the voltage x the period
measured as a percentage (50% say) of the potential at that rate).

The model of the battery as a voltage source in series with a resistance
is quite a good one I think, and the on-load voltage plus predicted
voltage drop is quite a good measure of what the off-load voltage has
fallen to.

Ok.

Which just leaves the temporary loss of capacity at high
load, which would recover if the battery was left off load.

As an aside on that I understand gel batteries aren't quite as linear
as some other LA types. ie, They are quite slow to produce power and
hence why they aren't generally used for engine starting (compared
with flooded or AGM etc).

The latter
might be a relatively small effect at the moderate current of about 10A
per battery that you are using. But as you might guess I don't know
any better than you.

Quite. ;-)

An interesting experiment would be to run the battery down to the
indicated 50% discharge depth, then, without charging, use it again the
next day and see if you get a significant amount of use before it falls
to the same voltage.

Ok, I could try that. My only reservation is that it's not 'good' to
leave LA batteries in a low charge state if possible but I could
certainly discharge it to a higher DOD (say 70%) and try it from
there?

FWIW, when I have also been monitoring the charge voltage (to test the
compatibility of chargers etc, gel once again being a bit 'odd' in
it's charging requirements amongst LA batteries in general), I've
discontinued the charge process say before going to bed and resumed it
the next day and it returns to it's previous 'level' very quickly.

If not, then your measurement system is doing
quite an accurate job of protecting the battery without wasting
capacity.

Ok Roger. I'll give that a go and let you know what I see (so thanks
again for the input). ;-)

Cheers, T i m

On 14/09/2017 21:28, NY wrote:
"newshound" wrote in message
o.uk...
On 14/09/2017 20:14, T i m wrote:
Now, it's obviously not a straight line (Peukert's law), but can you
extrapolate a graph (or create a formula that would be more useful for
my project) from just 3 points please?

p.s. The nearest *I* could get to an answer would be some graph paper
and a Flexicurve. ;-)


Probably no simple answer because it all depends on the shape of the
curve.

I'm not familiar with that particular "law" (I use the term advisedly)
but exponentials are buggers to deal with (especially when, as in this
case, it is obviously only an approximation: exponentials are fine for
radioactive decay, but it will certainly have limits in this case).

I think my approach would be to try to collect some data for your
specific battery, and try to work with that. Quite possibly with a
flexicurve, or with some sort of polynomial fit if there was more data
available.

I'd say the best approach would be to find a transformation (eg y=log(x)
or y=sqrt(x)) which gives a good, well-correlated straight line. Then
extrapolate that and do an inverse transformation (eg antilog or
x-squared) on the predicted value. Obviously the more data points you
have, the better prediction you can make and the better you can
construct a least-squares regression line for extrapolation and then
back-transformation.

I can see that working where your source data are following some kind of
exponential change being driven by a single (or predominate) physical
variable, but it can get rather complicated where you have multiple non
linearities competing in the same data set that have different
weightings at different times in the process. You may find you just end
up setting yourself a task that is of equal or greater difficulty to the
original question.

If you are starting with an empirically collected data set, then you may
find that after manually fitting a line (flexi curve etc) you can
identify sections that behave differently from others, and then attempt
to model them separately. I remember having to do that once years back
with digitised data from a RF power coupler, that was supposed to have a
linear response. Yet clearly the accuracy of the readings varied quite
noticeably over the range of powers it could sense (10s of Watts, to
10+kW). In the end someone had to sit there for a few hours manually
stepping up in 100W increments and recording the readings. Once plotted
it was clear that while there were some nice straight linear sections,
there was also a pronounced curve in the middle. Taking three points on
the curve and treating as a quadratic was accurate enough to get a good
model of the actual transformation for that bit. One then just needed a
little bit pre-scaling decision making in the software before deciding
what conversion to apply based on what range the raw reading was in.




--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On Sat, 16 Sep 2017 08:38:42 +0100
T i m wrote:

No, but if we are stuck without power 2 hours away from the car, in an
open dinghy, in the rain it *would* be a big PITA.

The rowing would keep you warm, and you could sing that song ...

On Sun, 17 Sep 2017 21:04:51 +0100, Rob Morley
wrote:

On Sat, 16 Sep 2017 08:38:42 +0100
T i m wrote:

No, but if we are stuck without power 2 hours away from the car, in an
open dinghy, in the rain it *would* be a big PITA.

The rowing would keep you warm, and you could sing that song ...

Row, row, row the boat ... ? ;-)

As it happens, I do enjoy rowing and because it's difficult to get a
whole days outboarding in a reasonable weight of battery (lead acid
cyclic batteries especially) then rowing was already a part of the
deal (and / or sailing if the weather and space allowed etc).

Like the other day, we outboarded for an hour, I rowed for an hour and
outboarded for an hour again.

The idea for a real trip was to row out and outboard back, hence why I
need a way of monitoring the battery charge status as closely as
possible so I know how far to row out. ;-)

Cheers, T i m

On 17/09/2017 13:26, John Rumm wrote:
On 14/09/2017 21:28, NY wrote:
"newshound" wrote in message
o.uk...
On 14/09/2017 20:14, T i m wrote:
Now, it's obviously not a straight line (Peukert's law), but can you
extrapolate a graph (or create a formula that would be more useful for
my project) from just 3 points please?

p.s. The nearest *I* could get to an answer would be some graph paper
and a Flexicurve. ;-)


Probably no simple answer because it all depends on the shape of the
curve.

I'm not familiar with that particular "law" (I use the term advisedly)
but exponentials are buggers to deal with (especially when, as in this
case, it is obviously only an approximation: exponentials are fine for
radioactive decay, but it will certainly have limits in this case).

I think my approach would be to try to collect some data for your
specific battery, and try to work with that. Quite possibly with a
flexicurve, or with some sort of polynomial fit if there was more data
available.

I'd say the best approach would be to find a transformation (eg y=log(x)
or y=sqrt(x)) which gives a good, well-correlated straight line. Then
extrapolate that and do an inverse transformation (eg antilog or
x-squared) on the predicted value. Obviously the more data points you
have, the better prediction you can make and the better you can
construct a least-squares regression line for extrapolation and then
back-transformation.

I can see that working where your source data are following some kind of
exponential change being driven by a single (or predominate) physical
variable, but it can get rather complicated where you have multiple non
linearities competing in the same data set that have different
weightings at different times in the process. You may find you just end
up setting yourself a task that is of equal or greater difficulty to the
original question.

If you are starting with an empirically collected data set, then you may
find that after manually fitting a line (flexi curve etc) you can
identify sections that behave differently from others, and then attempt
to model them separately. I remember having to do that once years back
with digitised data from a RF power coupler, that was supposed to have a
linear response. Yet clearly the accuracy of the readings varied quite
noticeably over the range of powers it could sense (10s of Watts, to
10+kW). In the end someone had to sit there for a few hours manually
stepping up in 100W increments and recording the readings. Once plotted
it was clear that while there were some nice straight linear sections,
there was also a pronounced curve in the middle. Taking three points on
the curve and treating as a quadratic was accurate enough to get a good
model of the actual transformation for that bit. One then just needed a
little bit pre-scaling decision making in the software before deciding
what conversion to apply based on what range the raw reading was in.


Very clear statement of what you can find in this type of problem. Never
mind what's claimed to be the theoretical equation, make sure you get
the data in the actual region you are interested in, and then fit to
that. If a flexicurve works, use it. Occam's razor.

"T i m" wrote in message
...
On Fri, 15 Sep 2017 14:04:56 +0100, "Dave W"
wrote:

snip

Please state the type no. of the battery;

They are MK 8GU1H

http://www.mkbattery.com/images/8GU1H.pdf
http://www.mkbattery.com/documents/9350MK_GEL_v7_r2.pdf
http://www.mkbattery.com/gel_specs.php?model=8GU1H
http://www.mkbattery.com/documents/1...I&O)_MK_r1.pdf


if more than one how they are
connected;

3, in parallel.

what current range you will be using.

0 to 30A (typically 15A).

Cheers, T i m

There's a great deal to mull over in the references you've given. I will
label them A, B, C, D in order.

Reference 'A' gives:
Current, Capacity
C/100, 36Ah
C/20, 31.6Ah
C/5, 25.6Ah

C is the current that would fully discharge the battery in 1 hour. C/5 is
the current that would fully discharge the battery in 5 hours.

Reference 'B' gives:
Current, Nominal Capacity
5 Hr Rate, 26.8Ah (different from reference 'A')
20 Hr Rate, 31.6Ah

Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)

Reference 'D' gives:
% Charge, Open Circuit Voltage after 24 hours
100%, = 12.85V
75%, 12.65V
50%, 12.35V
25%, 12.00V
0%, 11.80V

On page 6 are graphs of voltage against percentage discharge for currents
C1, C5, C20. (I assume that's another way of saying C/1, C/5, C/20, i.e.
21A, 5.40A, 1.60A respectively from reference 'C').

I magnified the graph, added extra grid lines, and estimated that at 50% the
voltages are 11.70V, 12.05V, 12.13V respectively. These are similar to the
values in your first posting:
11.75V @ 93A (12.4V initial)
12.05V @ 18.2A (12.7V initial)
12.10V @ 4.65A (12.75V initial)
The "initial" voltages correspond to those shown on the graph at zero
current, but I agree with you that they are irrelevant.
The corresponding currents for three batteries in parallel would be (from
reference 'C') 63A, 16.2A, 4.8A, for C/1, C/5, C/20. Similar to your first
posting except your 93A should be 63A.

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.

From my readings of C1 and C5, V/I = 0.35V/15.6A = 0.022 ohms.
From my readings of C5 and C20, V/I = 0.10V/3.8A = 0.026 ohms.

You say you will be drawing between 0A and 10A from each battery, and
typically less than 5A. That's typically near the C/5 rate of 5.40A, so I
would assume that the internal resistance of your arrangement is 0.026/3 =
0.0087 ohms i.e. 8.7 milliohms ignoring lead resistance.

The notional source voltage at the 50% discharge point is, from my C5
reading, 12.05V + 0.026 x 5.40A = 12.19V.

So if you know your current and voltage, you can work out when your notional
source voltage has dropped to 12.19V, which indicates the 50% discharged
point.

I think you should plot a graph of the speed of your normally-laden boat in
still water against current drawn. This will help decide how fast you can go
back before the battery runs out, especially if the relationship is
non-linear. 25.6Ah gives 5 hours at 5A, or 2.5 hours at 10A, your maximum
current.

As an aside, your reference 'B' shows a plot of lifetimes in terms of number
of cycles versus depth of discharge. I multiplied the cycles by the depths,
to give an indication of how the total Ah given by the battery over its life
depends on the depth of discharge:
5700 x 0.1 = 570
2100 x 0.25 = 525
1000 x 0.5 = 500
600 x 0.8 = 480
450 x 1 = 450

This seems to indicate that the life doesn't depend very much on depth of
discharge. The plot says that they are "Based on BSI 2-hour Capacity", and I
found an explanation of the 100% discharge version he
https://www.civicsolar.com/support/i...-gel-batteries

--
Dave W

On Tue, 19 Sep 2017 13:49:30 +0100, "Dave W"
wrote:

"T i m" wrote in message
.. .
On Fri, 15 Sep 2017 14:04:56 +0100, "Dave W"
wrote:

snip

Please state the type no. of the battery;

They are MK 8GU1H

http://www.mkbattery.com/images/8GU1H.pdf
http://www.mkbattery.com/documents/9350MK_GEL_v7_r2.pdf
http://www.mkbattery.com/gel_specs.php?model=8GU1H
http://www.mkbattery.com/documents/1...I&O)_MK_r1.pdf


if more than one how they are
connected;

3, in parallel.

what current range you will be using.

0 to 30A (typically 15A).

Cheers, T i m

There's a great deal to mull over in the references you've given.

Quite ... and I'm not sure all of it is supportive? ;-(

I will
label them A, B, C, D in order.

Reference 'A' gives:
Current, Capacity
C/100, 36Ah
C/20, 31.6Ah
C/5, 25.6Ah

C is the current that would fully discharge the battery in 1 hour. C/5 is
the current that would fully discharge the battery in 5 hours.

Agreed.

Reference 'B' gives:
Current, Nominal Capacity
5 Hr Rate, 26.8Ah (different from reference 'A')
(agreed, but only by a bit). ;-)
20 Hr Rate, 31.6Ah

Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)

Reference 'D' gives:
% Charge, Open Circuit Voltage after 24 hours
100%, = 12.85V
75%, 12.65V
50%, 12.35V
25%, 12.00V
0%, 11.80V

That's the first one in volts so a bit of a loner.

On page 6 are graphs of voltage against percentage discharge for currents
C1, C5, C20. (I assume that's another way of saying C/1, C/5, C/20, i.e.
21A, 5.40A, 1.60A respectively from reference 'C').

Yes, I believe you are correct (they are the sums resolved etc).

I magnified the graph, added extra grid lines, and estimated that at 50% the
voltages are 11.70V, 12.05V, 12.13V respectively.

Ok.

These are similar to the
values in your first posting:
11.75V @ 93A (12.4V initial)
12.05V @ 18.2A (12.7V initial)
12.10V @ 4.65A (12.75V initial)

Ok. (And my way was more 'rough and ready' than yours). ;-)

The "initial" voltages correspond to those shown on the graph at zero
current, but I agree with you that they are irrelevant.

Ok.

The corresponding currents for three batteries in parallel would be (from
reference 'C') 63A, 16.2A, 4.8A, for C/1, C/5, C/20. Similar to your first
posting except your 93A should be 63A.

I was simply using the typical rating which is often at C/20 (x) 3
Dave?

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?

From my readings of C1 and C5, V/I = 0.35V/15.6A = 0.022 ohms.
From my readings of C5 and C20, V/I = 0.10V/3.8A = 0.026 ohms.

You say you will be drawing between 0A and 10A from each battery, and
typically less than 5A. That's typically near the C/5 rate of 5.40A, so I
would assume that the internal resistance of your arrangement is 0.026/3 =
0.0087 ohms i.e. 8.7 milliohms ignoring lead resistance.

Ok (and whilst most of the leads are copper and 20mm^2, they will have
resistance etc).

The notional source voltage at the 50% discharge point is, from my C5
reading, 12.05V + 0.026 x 5.40A = 12.19V.

Sounds like a reasonable number. ;-)

So if you know your current and voltage, you can work out when your notional
source voltage has dropped to 12.19V, which indicates the 50% discharged
point.

Yeahbut, that in itself isn't the solution I was hoping for Dave ...
it would be to be able to calculate that critical voltage at any
current within the likely range?

I think you should plot a graph of the speed of your normally-laden boat in
still water against current drawn. This will help decide how fast you can go
back before the battery runs out, especially if the relationship is
non-linear. 25.6Ah gives 5 hours at 5A, or 2.5 hours at 10A, your maximum
current.

Quite ... and back to the formula that would allow me (or an Arduino)
to calculate that on the fly. ;-)

As an aside, your reference 'B' shows a plot of lifetimes in terms of number
of cycles versus depth of discharge. I multiplied the cycles by the depths,
to give an indication of how the total Ah given by the battery over its life
depends on the depth of discharge:
5700 x 0.1 = 570
2100 x 0.25 = 525
1000 x 0.5 = 500
600 x 0.8 = 480
450 x 1 = 450

This seems to indicate that the life doesn't depend very much on depth of
discharge.

That's interesting Dave, thanks. ;-)

The plot says that they are "Based on BSI 2-hour Capacity", and I
found an explanation of the 100% discharge version he
https://www.civicsolar.com/support/i...-gel-batteries

And the battery they reference there are very close to the spec of
mine as they are the 'Solar use' MK Gel jobbies.

So, Dave, thank you very much for you in depth analysis of the above
and do you think what you have gleaned so far would allow you to
create a formula to calculate the 50% DOD thresholds for the 0-30A
current range?

Funnily enough, whisky-dave posted a link elsewhere to a similar spec
/ capacity lead acid battery where they show the sort of graph I think
we are considering:

https://www.thesafetycentre.co.uk/doc/583.pdf

I'm looking at 'Discharge characteristics at various currents' chart
although it's not particularly clear to what DOD / voltage?

Cheers, T i m

"T i m" wrote in message
...
On Tue, 19 Sep 2017 13:49:30 +0100, "Dave W"
wrote:

I've snipped a lot here for clarity.

The corresponding currents for three batteries in parallel would be (from
reference 'C') 63A, 16.2A, 4.8A, for C/1, C/5, C/20. Similar to your first
posting except your 93A should be 63A.

I was simply using the typical rating which is often at C/20 (x) 3
Dave?

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

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.

From my readings of C1 and C5, V/I = 0.35V/15.6A = 0.022 ohms.
From my readings of C5 and C20, V/I = 0.10V/3.8A = 0.026 ohms.

You say you will be drawing between 0A and 10A from each battery, and
typically less than 5A. That's typically near the C/5 rate of 5.40A, so I
would assume that the internal resistance of your arrangement is 0.026/3 =
0.0087 ohms i.e. 8.7 milliohms ignoring lead resistance.

Ok (and whilst most of the leads are copper and 20mm^2, they will have
resistance etc).

The notional source voltage at the 50% discharge point is, from my C5
reading, 12.05V + 0.026 x 5.40A = 12.19V.

Sounds like a reasonable number. ;-)

So if you know your current and voltage, you can work out when your
notional
source voltage has dropped to 12.19V, which indicates the 50% discharged
point.

Yeahbut, that in itself isn't the solution I was hoping for Dave ...
it would be to be able to calculate that critical voltage at any
current within the likely range?

OK, to make a formula from my paragraph:
Let Vr = Voltage at your measurement point,
Let Vo = 12.19V,
Let Ir = Total Current from the three batteries,
Let Rt = 0.026/3 = 0.0087 ohms, plus whatever your lead resistance is,

Vr = Vo - (Ir x Rt)

As it seems the resistance is roughly independent of current, the formula
will do
for all currents. However if you ever want to find Vr at degrees of
discharge
other than 50%, you will have to work out appropriate values of Vo and Rt
for
each point.


I think you should plot a graph of the speed of your normally-laden boat
in
still water against current drawn. This will help decide how fast you can
go
back before the battery runs out, especially if the relationship is
non-linear. 25.6Ah gives 5 hours at 5A, or 2.5 hours at 10A, your maximum
current.

Quite ... and back to the formula that would allow me (or an Arduino)
to calculate that on the fly. ;-)

As an aside, your reference 'B' shows a plot of lifetimes in terms of
number
of cycles versus depth of discharge. I multiplied the cycles by the
depths,
to give an indication of how the total Ah given by the battery over its
life
depends on the depth of discharge:
5700 x 0.1 = 570
2100 x 0.25 = 525
1000 x 0.5 = 500
600 x 0.8 = 480
450 x 1 = 450

This seems to indicate that the life doesn't depend very much on depth of
discharge.

That's interesting Dave, thanks. ;-)

The plot says that they are "Based on BSI 2-hour Capacity", and I
found an explanation of the 100% discharge version he
https://www.civicsolar.com/support/i...-gel-batteries

And the battery they reference there are very close to the spec of
mine as they are the 'Solar use' MK Gel jobbies.

So, Dave, thank you very much for you in depth analysis of the above
and do you think what you have gleaned so far would allow you to
create a formula to calculate the 50% DOD thresholds for the 0-30A
current range?

Funnily enough, whisky-dave posted a link elsewhere to a similar spec
/ capacity lead acid battery where they show the sort of graph I think
we are considering:

https://www.thesafetycentre.co.uk/doc/583.pdf

I'm looking at 'Discharge characteristics at various currents' chart
although it's not particularly clear to what DOD / voltage?

Yuasa don't say, but I think the 100% discharge must be to that dotted line.
Also the plots are too small to be able to work out source voltage and
resistance.

I would like to add that I think you should charge your batteries
individually
rather than in parallel, otherwise they may not share the current equally.

Regards,
Dave

On Tue, 19 Sep 2017 20:29:26 +0100, "Dave W"
wrote:


"T i m" wrote in message
.. .
On Tue, 19 Sep 2017 13:49:30 +0100, "Dave W"
wrote:

I've snipped a lot here for clarity.

The corresponding currents for three batteries in parallel would be (from
reference 'C') 63A, 16.2A, 4.8A, for C/1, C/5, C/20. Similar to your first
posting except your 93A should be 63A.

I was simply using the typical rating which is often at C/20 (x) 3
Dave?

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?

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?

From my readings of C1 and C5, V/I = 0.35V/15.6A = 0.022 ohms.
From my readings of C5 and C20, V/I = 0.10V/3.8A = 0.026 ohms.

You say you will be drawing between 0A and 10A from each battery, and
typically less than 5A. That's typically near the C/5 rate of 5.40A, so I
would assume that the internal resistance of your arrangement is 0.026/3 =
0.0087 ohms i.e. 8.7 milliohms ignoring lead resistance.

Ok (and whilst most of the leads are copper and 20mm^2, they will have
resistance etc).

The notional source voltage at the 50% discharge point is, from my C5
reading, 12.05V + 0.026 x 5.40A = 12.19V.

Sounds like a reasonable number. ;-)

So if you know your current and voltage, you can work out when your
notional
source voltage has dropped to 12.19V, which indicates the 50% discharged
point.

Yeahbut, that in itself isn't the solution I was hoping for Dave ...
it would be to be able to calculate that critical voltage at any
current within the likely range?

OK, to make a formula from my paragraph:
Let Vr = Voltage at your measurement point,
Let Vo = 12.19V,
Let Ir = Total Current from the three batteries,
Let Rt = 0.026/3 = 0.0087 ohms, plus whatever your lead resistance is,

Vr = Vo - (Ir x Rt)

As it seems the resistance is roughly independent of current, the formula
will do
for all currents. However if you ever want to find Vr at degrees of
discharge
other than 50%, you will have to work out appropriate values of Vo and Rt
for
each point.

Understood.


I think you should plot a graph of the speed of your normally-laden boat
in
still water against current drawn. This will help decide how fast you can
go
back before the battery runs out, especially if the relationship is
non-linear. 25.6Ah gives 5 hours at 5A, or 2.5 hours at 10A, your maximum
current.

Quite ... and back to the formula that would allow me (or an Arduino)
to calculate that on the fly. ;-)

As an aside, your reference 'B' shows a plot of lifetimes in terms of
number
of cycles versus depth of discharge. I multiplied the cycles by the
depths,
to give an indication of how the total Ah given by the battery over its
life
depends on the depth of discharge:
5700 x 0.1 = 570
2100 x 0.25 = 525
1000 x 0.5 = 500
600 x 0.8 = 480
450 x 1 = 450

This seems to indicate that the life doesn't depend very much on depth of
discharge.

That's interesting Dave, thanks. ;-)

The plot says that they are "Based on BSI 2-hour Capacity", and I
found an explanation of the 100% discharge version he
https://www.civicsolar.com/support/i...-gel-batteries

And the battery they reference there are very close to the spec of
mine as they are the 'Solar use' MK Gel jobbies.

So, Dave, thank you very much for you in depth analysis of the above
and do you think what you have gleaned so far would allow you to
create a formula to calculate the 50% DOD thresholds for the 0-30A
current range?

Funnily enough, whisky-dave posted a link elsewhere to a similar spec
/ capacity lead acid battery where they show the sort of graph I think
we are considering:

https://www.thesafetycentre.co.uk/doc/583.pdf

I'm looking at 'Discharge characteristics at various currents' chart
although it's not particularly clear to what DOD / voltage?

Yuasa don't say, but I think the 100% discharge must be to that dotted line.

Ok. Some of the other explanations suggest 10.5 volts (for a nominal
12V LA battery) is what is used to go with any capacity testing.
Therefore, the higher the current the more the voltage will be
depressed and hence the sooner that 10.5V cutoff point is reached.

Also the plots are too small to be able to work out source voltage and
resistance.

Ok.

I would like to add that I think you should charge your batteries
individually
rather than in parallel, otherwise they may not share the current equally.

Again, listening to people who do this all the time it seems like it
can even come down to how you wire all the batteries together. e.g.
Say you have 5 x 12V, 100Ah batteries wired in parallel using busbars,
you take the +Ve off one end and the -Ve off the other as that they
minimises the tiny losses (size being dependent on load and busbar
resistance etc) that would work the batteries on the ends of the bus
more than those in the middle.

And alternative way of wiring them is use a star so that all the wires
between each battery pole and a common connection point are identical
(length and gauge and so hopefully resistance), therefore spreading
the load across all the batteries equally.

I believe the guidance from the battery manufacturers is:

1) All the batteries connected in parallel should *ideally* be of the
same make, model (so capacity), age and cycles.

2) All batteries to be connected together should first be fully
charged so that the float voltages are as similar as possible before
they are joined together.

3) Once so joined, they should then be left joined together for all
subsequent charge and discharge cycles.

I think the logic re parallel charging is that even if the batteries
are slightly different (as they are bound to be), the battery with the
deepest level of discharge (smallest capacity) will depress the charge
voltage more than the others (because it will be taking more of the
current) and as it charges to match the level of the others the charge
current will balance out across all batteries equally.

If one battery reaches it's full charge condition before the others
it's terminal voltage will be higher than the others (would be if
isolated) and so the current would tend to flow though those batteries
that are still a lower terminal voltage (if you were to disconnect
them) and so again, the charge balances out to a reasonable degree.

The third thing is to ensure the maximum charge 'bulk phase' current
is kept to such a level that it can't overdrive any one battery and
similar for the 'absorption phase'.

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). ;-)

Cheers, T i m

p.s. If you like logic and maths, you might find this interesting:

http://www.smartgauge.co.uk/peukert_depth.html

There is even a calculator you can plug values into but whilst it
seems to work for me, I don't understand the outcome: ;-)

http://www.smartgauge.co.uk/calcs/peukert_2.xls


And I can nearly follow this explanation. ;-)

http://preview.tinyurl.com/ycyznwaf


The following seems to reflect my understanding of it all and agrees
that by using Peukerts in any discharge level calculations can lead to
a more realistic display of charge status than when not using it (with
limitations etc).

http://www.bogartengineering.com/wp-...tsComments.pdf

For me the bottom line is still that 'if you discharge a lead acid
battery at an increasing rate the effective capacity *will* decrease.

It is my firm belief that within a pretty strict range of parameters a
formula could be found and used in a realtime display of the predicted
voltage cutoff point. I just don't have the sort of brain that can
handle all the variables, rules and formulae to work out what that is.
;-(

"T i m" wrote in message
...
snip to save scrolling

Are you possibly making too much of this?
If you seem to be losing power, don't you just seek the next charging point
and charge them?
Admittedly with our camper you can just start the engine.
We maybe a tad late with the microwaved gubbins but hey, we have gas.
Worst case, if you can afford a boat you can surely afford new batteries?

On Wed, 20 Sep 2017 00:59:55 +0100, "bm" wrote:


"T i m" wrote in message
.. .
snip to save scrolling

Are you possibly making too much of this?

You are bored aren't you ... but I don't know. Do you have the answer
to my question?

If you seem to be losing power, don't you just seek the next charging point
and charge them?

What, and sit there and wait ~12 hours or so in an open dinghy in the
water?

Admittedly with our camper you can just start the engine.

Well with out rowing dinghy the electric outboard is all we have,
other than the oars.

We maybe a tad late with the microwaved gubbins but hey, we have gas.

That will teach you to eat half cooked food! ;-)

Worst case, if you can afford a boat you can surely afford new batteries?

I'm guessing you either really don't understand much or haven't been
following the thread at all? ;-(

I appreciate, as a Brexiteer you don't need any facts, you just need
an inkling of an idea about something to go jumping off to all sorts
of conclusions. ;-(

Here, try thinking about this (or get 'nurse' to explain it to you)
maybe I don't have a boat that is like your campervan, maybe I don't
have the same needs with my boat as you do with your campervan and
what I'm doing is all about both trying to eeek out the small amount
of energy I have within the largest mass of battery we can manage,
handle and transport and measuring something that is both dynamic and
non-linear at the same time?

Now, if I were a Brexiteer I could just spin the bottle and go along
with whatever comes up and even make that my new crusade in life but I
simply can't do that.

I work on the basis 'You can manage what you can measure' whereas you
work on the 'I'm not sure what it is but I want it right now'.

So, unless you can come up with whatever formula is required to
calculate the real-time voltage on a battery that represents a 50%
discharge level for a given current then it might be best if you went
back to supporting whatever cause that spinning bottle points you at
next. ;-)

Cheers, T i m

"T i m" wrote in message
...
On Wed, 20 Sep 2017 00:59:55 +0100, "bm" wrote:


"T i m" wrote in message
. ..
snip to save scrolling

Are you possibly making too much of this?

You are bored aren't you ... but I don't know. Do you have the answer
to my question?

If you seem to be losing power, don't you just seek the next charging
point
and charge them?

What, and sit there and wait ~12 hours or so in an open dinghy in the
water?
A problem? You chose to be a ****.


Admittedly with our camper you can just start the engine.

Well with out rowing dinghy the electric outboard is all we have,
other than the oars.
Tough titty.

We maybe a tad late with the microwaved gubbins but hey, we have gas.

That will teach you to eat half cooked food! ;-)
****.

Worst case, if you can afford a boat you can surely afford new batteries?

I'm guessing you either really don't understand much or haven't been
following the thread at all? ;-(
I followed it, you're a dick head.

I appreciate, as a Brexiteer you don't need any facts, you just need
an inkling of an idea about something to go jumping off to all sorts
of conclusions. ;-(
Prick.

Here, try thinking about this (or get 'nurse' to explain it to you)
maybe I don't have a boat that is like your campervan, maybe I don't
have the same needs with my boat as you do with your campervan and
what I'm doing is all about both trying to eeek out the small amount
of energy I have within the largest mass of battery we can manage,
handle and transport and measuring something that is both dynamic and
non-linear at the same time?

Now, if I were a Brexiteer I could just spin the bottle and go along
with whatever comes up and even make that my new crusade in life but I
simply can't do that.

I work on the basis 'You can manage what you can measure' whereas you
work on the 'I'm not sure what it is but I want it right now'.

So, unless you can come up with whatever formula is required to
calculate the real-time voltage on a battery that represents a 50%
discharge level for a given current then it might be best if you went
back to supporting whatever cause that spinning bottle points you at
next. ;-)

Cheers, T i m

That settles it old son, you really are a prick of the first order.
Much akin to our Wodney.
Your poor wife.
The formula is D i m = **** and folks like you have a vote (fence sitters)?
LMFAO.
I congratulate you on not giving a toss what others may think.
Well done.
I completely agree.

On 19/09/2017 19:02, T i m wrote:
Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)

The trick is to work out what varies most smoothly with load.

If you work out the battery capacity in Ah and then fit that against
current I you get a reasonable quadratic approximation of capacity

Capacity (Ah) = 35.3 - 1.56*I + 0.04*I^2

It is always going to be a bit rough and ready since individual
batteries will vary from the abuse they have suffered in service.

A linear model would underestimate the low current lifetime

Capacity (Ah) = 33.2 - 0.66I

Datasheet batteries always seem to last longer than real world ones!

You should be able to reproduce these models by putting the numbers into
Excel and plotting I*t against I.

--
Regards,
Martin Brown

On Wed, 20 Sep 2017 02:53:09 +0100, "bm" wrote:

snip BS reply

Ooooh, looks like we struck a nerve there eh! weg

Cheers, T i m

On Sun, 17 Sep 2017 13:26:32 +0100, John Rumm
wrote:

snip

I can see that working where your source data are following some kind of
exponential change being driven by a single (or predominate) physical
variable, but it can get rather complicated where you have multiple non
linearities competing in the same data set that have different
weightings at different times in the process. You may find you just end
up setting yourself a task that is of equal or greater difficulty to the
original question.

You may be right John but I was just hoping that 'maths' would be able
to cope with it, no matter how inaccurate the result might be ITRW.

If you are starting with an empirically collected data set, then you may
find that after manually fitting a line (flexi curve etc) you can
identify sections that behave differently from others, and then attempt
to model them separately.

Looking at the graphs I have seen that represent what I think I'm
looking for, the range of currents I'd be dealing with seem to be in a
reasonably 'linear' (as in say 'exponential') area.

If you look at the graph labeled 'Discharge characteristics at various
currents' he

https://www.thesafetycentre.co.uk/doc/583.pdf

and consider the 2-10A range (= 0.33 to 0.05C when seen across the 3 x
~30Ah batteries in parallel as that battery is of a similar capacity
to one of mine) then it doesn't look like there should be anything
unpredictable.

Even if we only considered it as a straight line, couldn't we
'compute' the (say) 50% DOD cutoff voltage for any current within that
range?

Cheers, T i m

On Sun, 17 Sep 2017 23:19:52 +0100, newshound
wrote:

snip

Never
mind what's claimed to be the theoretical equation, make sure you get
the data in the actual region you are interested in, and then fit to
that. If a flexicurve works, use it. Occam's razor.

The thing is (and assuming we may now have at least a linear slope
rather than just a fixed value (see my reply to John)) even if we had
a graph on paper to work from I still need it in the way of a line of
code (the maths part) to be able to put it to use?

Read CurrentAmps, Read CurrentVolts

If CurrentVolts operator CurrentAmps operatorGraphRule = 50%
DOD voltage, output alarm?

If not, rinse repeat?

Cheers, T i m

On 20/09/17 08:41, T i m wrote:
On Wed, 20 Sep 2017 02:53:09 +0100, "bm" wrote:

snip BS reply

Ooooh, looks like we struck a nerve there eh! weg

Cheers, T i m

No, it looks liek he finally reallised who you are and how unpleasant
and stupid you are.

Typical remoaner, in fact.


--
A lie can travel halfway around the world while the truth is putting on
its shoes.

On Wed, 20 Sep 2017 08:36:50 +0100, Martin Brown
wrote:

On 19/09/2017 19:02, T i m wrote:
Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)

The trick is to work out what varies most smoothly with load.

If you work out the battery capacity in Ah and then fit that against
current I you get a reasonable quadratic approximation of capacity

Capacity (Ah) = 35.3 - 1.56*I + 0.04*I^2

It is always going to be a bit rough and ready since individual
batteries will vary from the abuse they have suffered in service.

A linear model would underestimate the low current lifetime

Capacity (Ah) = 33.2 - 0.66I

Datasheet batteries always seem to last longer than real world ones!

You should be able to reproduce these models by putting the numbers into
Excel and plotting I*t against I.


Hi Martin and thanks for the feedback. I can follow some of it. ;-)

More thinking out loud on it all .... Considering a graph of various
discharge rates shows that the effective capacity does vary as a
function of the discharge rate, I think it would still be worthwhile
trying to accommodate it to some degree if possible (accepting all the
caveats you mention etc). I agree though that whilst a linear function
would be better than nothing at all, it may not reflect the situation
was well as it *might*, if we could find that mathematical solution,
even though it's only ever going to be an approximation at best. I was
never looking for a laboratory solution here. ;-)

On that though, I can (re)evaluate the batteries capacity over time
and modify the numbers accordingly and could even automatically
compensate for temperature (if the typical temperature range
difference we are likely to go boating would be significant etc).

So it's just a matter of (initially especially) seeing if we can make
use / sense of the published data for my specific battery(ies) and if
that isn't sufficient, if we can use the stats from something similar
to at least start the ball rolling?

It's just knowing that there can be such a marked difference in
capacity as a function of the current drawn ... and my potential
maximum current (30A) *is* significant to this, even with 3 batteries
in parallel (~90Ah) there is a very good chance that some massaging of
the readings to predict a particular depth of discharge threshold
would be worthwhile.

If I retained the original resistor based 'speed control', even I,
once I guesstimated what the capacity would be for the currents
typically seen at the 5 set speeds, could possibly write some code
that applied the variable.

Read I read V

If I = 0 and = 7 (where speed 1 was around 5A) then multiply V by .8

Else if I = 10 and = 19 (where speed 2 was around 15A) then multiply
V by .7

(or whatever etc).

It's working out a formula that would give me this variable result
across the entire range (for when using a fully variable and hopefully
more efficient speed controller) that I am unable to deal with (red
mist with such maths etc). ;-(

Cheers, T i m

"T i m" wrote in message
...
On Tue, 19 Sep 2017 20:29:26 +0100, "Dave W"
wrote:


"T i m" wrote in message
. ..
On Tue, 19 Sep 2017 13:49:30 +0100, "Dave W"
wrote:

I've snipped a lot here for clarity.

The corresponding currents for three batteries in parallel would be
(from
reference 'C') 63A, 16.2A, 4.8A, for C/1, C/5, C/20. Similar to your
first
posting except your 93A should be 63A.

I was simply using the typical rating which is often at C/20 (x) 3
Dave?

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.


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. I admit I have not investigated Peukert's effect.

snip


I would like to add that I think you should charge your batteries
individually
rather than in parallel, otherwise they may not share the current equally.

Again, listening to people who do this all the time it seems like it
can even come down to how you wire all the batteries together. e.g.
Say you have 5 x 12V, 100Ah batteries wired in parallel using busbars,
you take the +Ve off one end and the -Ve off the other as that they
minimises the tiny losses (size being dependent on load and busbar
resistance etc) that would work the batteries on the ends of the bus
more than those in the middle.

And alternative way of wiring them is use a star so that all the wires
between each battery pole and a common connection point are identical
(length and gauge and so hopefully resistance), therefore spreading
the load across all the batteries equally.

I believe the guidance from the battery manufacturers is:

1) All the batteries connected in parallel should *ideally* be of the
same make, model (so capacity), age and cycles.

2) All batteries to be connected together should first be fully
charged so that the float voltages are as similar as possible before
they are joined together.

3) Once so joined, they should then be left joined together for all
subsequent charge and discharge cycles.

I think the logic re parallel charging is that even if the batteries
are slightly different (as they are bound to be), the battery with the
deepest level of discharge (smallest capacity) will depress the charge
voltage more than the others (because it will be taking more of the
current) and as it charges to match the level of the others the charge
current will balance out across all batteries equally.

If one battery reaches it's full charge condition before the others
it's terminal voltage will be higher than the others (would be if
isolated) and so the current would tend to flow though those batteries
that are still a lower terminal voltage (if you were to disconnect
them) and so again, the charge balances out to a reasonable degree.

The third thing is to ensure the maximum charge 'bulk phase' current
is kept to such a level that it can't overdrive any one battery and
similar for the 'absorption phase'.

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.

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.

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.

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. 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.

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

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.

Cheers, T i m

p.s. If you like logic and maths, you might find this interesting:

http://www.smartgauge.co.uk/peukert_depth.html

There is even a calculator you can plug values into but whilst it
seems to work for me, I don't understand the outcome: ;-)

http://www.smartgauge.co.uk/calcs/peukert_2.xls


And I can nearly follow this explanation. ;-)

http://preview.tinyurl.com/ycyznwaf


The following seems to reflect my understanding of it all and agrees
that by using Peukerts in any discharge level calculations can lead to
a more realistic display of charge status than when not using it (with
limitations etc).

http://www.bogartengineering.com/wp-...tsComments.pdf

For me the bottom line is still that 'if you discharge a lead acid
battery at an increasing rate the effective capacity *will* decrease.

It is my firm belief that within a pretty strict range of parameters a
formula could be found and used in a realtime display of the predicted
voltage cutoff point. I just don't have the sort of brain that can
handle all the variables, rules and formulae to work out what that is.
;-(

I have skimmed through these links and now my brain hurts. Your last link
concludes that it's all too difficult. I may get back to you when I've
done my own analysis.
Regards,
Dave

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

On Wed, 20 Sep 2017 09:58:11 +0100, The Natural Philosopher
wrote:

On 20/09/17 08:41, T i m wrote:
On Wed, 20 Sep 2017 02:53:09 +0100, "bm" wrote:

snip BS reply

Ooooh, looks like we struck a nerve there eh! weg


No, it looks liek he finally reallised who you are and how unpleasant
and stupid you are.

Typical remoaner, in fact.

If only you could apply the same amount of energy to helping us
resolve this (and what must be to you, 'brain as big as a planet'),
simple *energy* measurement issue as trying to defend the indefensible
(Linux or leaving the EU) we might all get on faster! ;-)


Cheers, T i m

On Wed, 20 Sep 2017 20:24:23 +0100, pamela wrote:

snip bm's drivel

If you find Tim's inquiry to be too boring for you then why don't you
skip to the next thread?

Because he doesn't find it boring, he finds it exciting (because it
confuses him) and he is only too pleased that *someone* will talk to
him (and I do it as part of his virtual care in the community) ;-)

You can't force Tim to adopt your alternative philosophy of "If you
seem to be losing power, don't you just seek the next charging point
and charge them?".

Quite ... he said he had read the entire thread but still had no clue
whatsoever about what was actually going on!

Aww, but bless him for at least having a go pamela. ;-)

Cheers, T i m

"T i m" wrote in message
...
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?

No, I was only correcting the 93A "the manufacturers have provided"
that you gave in your first post.



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. ;-)

Well if 'remaining time' is too advanced, what's wrong with my simple
formula?
It's better than nothing. It might be better to display the source voltage
Vo
rather than the terminal voltage Vr, and when it gets to 12.19V you have
reached the 50% discharged point. Vo = Vr + (Ir x Rt) is the rearranged
formula.


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. ;-)

I can't see the need for that as my three calculations of internal
resistance
were all pretty similar, as would be 10 intermediate curves. You mentioned
that
for a liquid battery you could measure the specific gravity to find the
50% point, so surely that point will also have my predicted source voltage
of
2.19V?

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

Cheers, T i m

Cheers,
Dave

"Dave W" wrote in message
news

snip

Some further thoughts:

The display of Vo should stay fairly constant as the current is ramped up
and down.
If it doesn't, the assumed value for Rt can be adjusted until it does.

I suspect the motor current will be very noisy, making any digital display
very
erratic and difficult to read. Better to use a mechanical meter, with a big
capacitor across it to smooth out wobbles. Or perhaps use software to create
a
running average. A meter would be easy to read in bright sunlight.

The meter could have a big mark at 12.19V, with perhaps everything below
11.5V
marked in red, as that's where your discharge graph starts to drop like a
stone.

If you are using a simple rheostat to adjust the motor current, that's a big
waste of energy in the form of heat. Better to have an efficient
switched-mode
power converter to convert the battery voltage into motor voltage.

Ideally this would have current feedback to compensate for the resistance of
the motor, in effect raising the voltage as more current is taken. The
control
knob could then be calibrated directly in propeller speed.

The feedback could be further modified if you know the relationship between
boat
speed and propeller speed, and the knob could be calibrated in boat speed.
This
would be helpful for estimating journey time.

Regards,
Dave

On Wed, 20 Sep 2017 17:31:23 +0100, T i m wrote:

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

I'll step in here (without, ahem, having read all the thread).

I think looking for a formula won't work here, but as you are using an Arduino:

Take the nominal Ah of the batteries.

Measure the current drawn.

Calculate the capacity of the battery at the current amperage drawn. This will
be a fraction of the nominal capacity. Calculate the inverse ration of this
capacity to the nominal capacity. (100 Ah pack, effectively 75 Ah at current X.
So 100/75= 4/3.)

Multiply current amperage drawn by the time it is drawn (i.e. you measure every
second, 10 Amps, so 10 Amp-seconds). Multiply this by the ratio above, and
subtract from the nominal capacity. So, reduce nominal capacity by 4/3*10 As.

Display remaining capacity as a percentage. Adjust for real-world experience.

Repeat.


If i have thought this through correctly, you reduce the full charge by what you
use, weighted by how much the currently drawn amperage reduces that capacity.

Alternatively, reformulate to use time-to-discharge instead of capacity.

And of course you can just recharge the battery if you run out.

https://youtu.be/38N5OcZx3ko?t=1m44s


Thomas Prufer

On 21/09/2017 00:49, Dave W wrote:
"T i m" wrote in message
...
On Wed, 20 Sep 2017 14:03:05 +0100, "Dave W"
wrote:

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?

OK. I think I understand now what you want to do is make a fuel gauge
for your rather larger than I imagined model boat. In that case you will
need to err on the side of caution (a bit like my car range indicator
which is incredibly pessimistic after a fuel fill showing a maximum
range that is typically 100 miles less than the distance actually
travelled on the last tank). I suspect range or time remaining (at a
chosen speed/drive current) might really be what you should display.

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. ;-)

And that depends how you obtain your 15A. If it consists of equal
amounts of 0A and 30A at a frequency high enough to not be annoying and
low enough not to be too lossy you will go almost twice as far as if you
used a simple series resistor where half the power just ends up as heat.

PWM at a nominally fixed current also simplifies the battery Ah side.

Expect a decent PWM power controller to be expensive and make provision
to switch back to a primitive all or nothing supply in case of failure.

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. ;-)

Well if 'remaining time' is too advanced, what's wrong with my simple
formula?
It's better than nothing. It might be better to display the source voltage
Vo
rather than the terminal voltage Vr, and when it gets to 12.19V you have
reached the 50% discharged point. Vo = Vr + (Ir x Rt) is the rearranged
formula.

I suggest doing some empirical experiments with the boat itself and a
fully charged set of batteries to see how long it lasts run flat out, at
half power, and then half of run flat out followed by half power.

Maybe just use one battery to make the testing a bit quicker.

You need to make sure that if all else fails the panic indicator leaves
you with enough juice in the battery to limp home at a snails pace.

I'd ignore any fancy nonlinear or temperature effects for the first
analysis unless they prove problematic. The change in dynamic impedance
of the electric motor with speed will be more of an issue. The biggest
error by far will be that speed in the water against the current will
not translate to anything like as much forward progress as on dry land.

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. ;-)

I can't see the need for that as my three calculations of internal
resistance
were all pretty similar, as would be 10 intermediate curves. You mentioned
that
for a liquid battery you could measure the specific gravity to find the
50% point, so surely that point will also have my predicted source voltage
of
2.19V?

You really want something that works well enough from an engineering
point of view and fails safe by underestimating remaining capacity.

It is very much an empirical business - and one that most car makers
have not yet mastered in their petrol/diesel range indicators.

Are electric cars any better at this I wonder? I suppose they have to be
or we would see dead and dying electric cars stuck at the roadside.

--
Regards,
Martin Brown

On 20/09/2017 10:09, T i m wrote:
On Wed, 20 Sep 2017 08:36:50 +0100, Martin Brown
wrote:

On 19/09/2017 19:02, T i m wrote:
Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)

The trick is to work out what varies most smoothly with load.

If you work out the battery capacity in Ah and then fit that against
current I you get a reasonable quadratic approximation of capacity

Capacity (Ah) = 35.3 - 1.56*I + 0.04*I^2

It is always going to be a bit rough and ready since individual
batteries will vary from the abuse they have suffered in service.

A linear model would underestimate the low current lifetime

Capacity (Ah) = 33.2 - 0.66I

Datasheet batteries always seem to last longer than real world ones!

You should be able to reproduce these models by putting the numbers into
Excel and plotting I*t against I.


Hi Martin and thanks for the feedback. I can follow some of it. ;-)

More thinking out loud on it all .... Considering a graph of various
discharge rates shows that the effective capacity does vary as a
function of the discharge rate, I think it would still be worthwhile
trying to accommodate it to some degree if possible (accepting all the
caveats you mention etc). I agree though that whilst a linear function
would be better than nothing at all, it may not reflect the situation
was well as it *might*, if we could find that mathematical solution,
even though it's only ever going to be an approximation at best. I was
never looking for a laboratory solution here. ;-)

On that though, I can (re)evaluate the batteries capacity over time
and modify the numbers accordingly and could even automatically
compensate for temperature (if the typical temperature range
difference we are likely to go boating would be significant etc).

So it's just a matter of (initially especially) seeing if we can make
use / sense of the published data for my specific battery(ies) and if
that isn't sufficient, if we can use the stats from something similar
to at least start the ball rolling?

It's just knowing that there can be such a marked difference in
capacity as a function of the current drawn ... and my potential
maximum current (30A) *is* significant to this, even with 3 batteries
in parallel (~90Ah) there is a very good chance that some massaging of
the readings to predict a particular depth of discharge threshold
would be worthwhile.

A quick and dirty value for the discharge threshold is when the terminal
voltage drops below some arbitrary limit ISTR 10.5 or something. You
never want to take an individual cell below some critical value which is
temperature dependent or permanent damage will result. The weakest cell
in a stack of them always gives up the ghost first.

To a good approximation the terminal voltage ends up as

V = V0 - Ir

Where I is the current you are drawing and r is the internal resistance
of the battery (which itself is a weak function of temperature and I).

In practice the difference may help to protect the battery if you just
use a fixed voltage cutoff value since if you are discharging it quickly
it will inevitably be hotter and potentially gassing slightly.

If I retained the original resistor based 'speed control', even I,
once I guesstimated what the capacity would be for the currents
typically seen at the 5 set speeds, could possibly write some code
that applied the variable.

Not sure I understand what you are trying to do but a MOSFET based PWM
speed controller will give you something like 90% efficiency. This will
give you a lot more battery life if you use low power most of the time.

Whereas a resistor will be wasting a lot of power dissipated as heat.
Using a better more efficient speed controller is a far better option
than managing the decline of the poor battery.

--
Regards,
Martin Brown

On Thu, 21 Sep 2017 00:49:18 +0100, "Dave W"
wrote:

snip

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. ;-)

Well if 'remaining time' is too advanced, what's wrong with my simple
formula?

No, nothing Dave ... I'm just still trying to be open to all ideas and
options. ;-)

It's better than nothing.

Quite.

It might be better to display the source voltage
Vo
rather than the terminal voltage Vr, and when it gets to 12.19V you have
reached the 50% discharged point. Vo = Vr + (Ir x Rt) is the rearranged
formula.

I was hoping to display both, the actual voltage and the predicted
cutoff voltage (at that instant current reading) just so that *I* can
get a feel of how things are going (as my feeling on how things are
going may be better in ITRW than any AI). ;-)

snip

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. ;-)

I can't see the need for that as my three calculations of internal
resistance
were all pretty similar, as would be 10 intermediate curves.

Ok

You mentioned
that
for a liquid battery you could measure the specific gravity to find the
50% point, so surely that point will also have my predicted source voltage
of
2.19V?

Well, if all things do tally then yes, it should. However, the
specific gravity of the electrolyte is independent of the charge rate
or terminal voltage, it's just an instantaneous reflection of the
chemical charge state (so not much use ITRW when on the move without
some auto-reading system (and a non gel battery) etc.


Cheers, T i m

On 20/09/17 17:37, T i m wrote:
On Wed, 20 Sep 2017 09:58:11 +0100, The Natural Philosopher
wrote:

On 20/09/17 08:41, T i m wrote:
On Wed, 20 Sep 2017 02:53:09 +0100, "bm" wrote:

snip BS reply

Ooooh, looks like we struck a nerve there eh! weg


No, it looks liek he finally reallised who you are and how unpleasant
and stupid you are.

Typical remoaner, in fact.

If only you could apply the same amount of energy to helping us
resolve this (and what must be to you, 'brain as big as a planet'),
simple *energy* measurement issue as trying to defend the indefensible
(Linux or leaving the EU) we might all get on faster! ;-)



The fact that I have a degree in electrical engineering and a huge
anount of experience in using lithoum ion batteries in model aircraft
completely precludes me from offering what would instantly be shouted
down as 'an opinion'.

But I will give you a hint.

Ohms law.

Cheers, T i m



--
Theres a mighty big difference between good, sound reasons and reasons
that sound good.

Burton Hillis (William Vaughn, American columnist)

On Thu, 21 Sep 2017 08:54:26 +0100, "Dave W"
wrote:


"Dave W" wrote in message
news

snip

Some further thoughts:

Ok. (and thanks for such Dave). ;-)

The display of Vo should stay fairly constant as the current is ramped up
and down.
If it doesn't, the assumed value for Rt can be adjusted until it does.

So Vo is our calculated cutoff voltage Dave?

I suspect the motor current will be very noisy, making any digital display
very
erratic and difficult to read.

All that would be 'damped' by sampling Dave.

Better to use a mechanical meter, with a big
capacitor across it to smooth out wobbles.

I already have such a solution in some monitoring kit I've made up
(moving coil meter and suitable shunt) but that's no good for any
logging of course.

Or perhaps use software to create
a
running average.

That's the idea. See, once I have it 'electronically' I can do loads
of things with it (like logging to an SD car to provide a 'Battery
Life' record.

A meter would be easy to read in bright sunlight.

I only intend using this in England Dave. ;-)

The meter could have a big mark at 12.19V, with perhaps everything below
11.5V
marked in red, as that's where your discharge graph starts to drop like a
stone.

Sure, analogue meters are great for showing stuff (and why I still use
them) and during the initial runs of anything I come up with may well
include some analogue versions of both Volts and Amps, just to provide
a 'simple' comparison. The problem is a moving coil voltmeter is a lot
more expensive than an electronic one (or even an electronic
wattmeter!) and it really only serves as an indication that there is
around 12V there (because you are only interested in the range 10 -
15). I did think of using a 10V zener and a 5V meter as at least then
you would have an expanded range. ;-)

If you are using a simple rheostat to adjust the motor current, that's a big
waste of energy in the form of heat.

Agreed. It's actually what Minn Kota (the people who make the
outboard) call 'Speed coils' that are just resistors in the underwater
unit somewhere and selected via a multi-way switch in the outboard and
controlled by the tiller / twistgrip.

Better to have an efficient
switched-mode
power converter to convert the battery voltage into motor voltage.

Quite, and something (else) I'm working on Dave. ;-)

Ideally this would have current feedback to compensate for the resistance of
the motor, in effect raising the voltage as more current is taken.

Yes, that could be good.

The
control
knob could then be calibrated directly in propeller speed.

Yup. ;-)

The feedback could be further modified if you know the relationship between
boat
speed and propeller speed, and the knob could be calibrated in boat speed.

Whilst that would vary between boats (and there are 5 we may use it
on) it might be of more use than just some arbitrary values.

This
would be helpful for estimating journey time.

As would the current etc.

Cheers, T i m

p.s. Part of the whole 'electric outboard' experiment is to build a
test tank (I now have kindly donated by Mr Lamb of this very group)
and again, use an Arduino micro controller and voltage, current, load
(strain) and water-speed measurement, initially make comparisons
between the efficiency of resistor V PWM speed control and later, prop
/ leg hydrodynamic improvements (or not). ;-)

On Thu, 21 Sep 2017 09:41:53 +0100, pamela wrote:

snip

Aww, but bless him for at least having a go pamela. ;-)


It's one thing for 'bm' to want to do a sloppy job if he was in
the same situation as you but then he went ballistic just because
you want to do it thoroughly.

I really don't think it's a deep as that P, he is really just an
attention starved troll. ;-(

Why does he care so much?

He doesn't of course, he just likes the attention so will say anything
to try to get some. Look how much OT stuff he posts here and how much
weirdness it contains ... ? ;-(


Cheers, T i m

On Thu, 21 Sep 2017 10:45:21 +0100, The Natural Philosopher
wrote:

snip

If only you could apply the same amount of energy to helping us
resolve this (and what must be to you, 'brain as big as a planet'),
simple *energy* measurement issue as trying to defend the indefensible
(Linux or leaving the EU) we might all get on faster! ;-)



The fact that I have a degree in electrical engineering

You say ... ;-)

and a huge
anount of experience in using lithoum ion batteries in model aircraft

;-)

Yeah, I was a 'sponsored' RC car driver so because of that I know all
there is to know about Nicads and NiMh .... (not).

completely precludes me from offering what would instantly be shouted
down as 'an opinion'.

Yes, but if you had an EQ as big as your IQ you would be able to offer
*information* around my question wouldn't you?

But I will give you a hint.

Ohms law.

No use whatsoever mate as it's a linear relationship and this isn't a
linear problem.

As you know all there is to know about LiPo (as the chances are you
*wouldn't* be using Li-Ion in RC aircraft as they are too heavy) you
would know that whilst Peukert's Law also apples to them in theory,
you would also know it is pretty well cancelled out by the increase in
efficiency found as the batteries heat up in use (proportionally to
the current drawn).

So, what with you having a degree in electrical engineering, into
using your (Linux) laptop as a typewriter (writing code for the likes
of Gridwatch?) and like debating things so what's not to like? ;-)

So, cummon, have a go at being the solution, you know you want to. ;-)

Cheers, T i m

On 21/09/17 11:50, T i m wrote:
So, cummon, have a go at being the solution, you know you want to.;-)

Cheers, T i m

Seeing as its you t i m, no, I dont want to.


--
Theres a mighty big difference between good, sound reasons and reasons
that sound good.

Burton Hillis (William Vaughn, American columnist)

On Thu, 21 Sep 2017 09:26:15 +0100, Martin Brown
wrote:

snip

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

OK. I think I understand now what you want to do is make a fuel gauge
for your rather larger than I imagined model boat.

Ah ha. ;-)

In that case you will
need to err on the side of caution

I've already factored that in Martin. The batteries are probably the
best there are out there and so will tolerate (some) abuse with no
real detriment (and partly why I bought them etc). So, if I aim for
say a maximum 50% DOD ... and overshoot a bit, it wouldn't be a big
issue.

(a bit like my car range indicator
which is incredibly pessimistic after a fuel fill showing a maximum
range that is typically 100 miles less than the distance actually
travelled on the last tank).

;-(

I suspect range or time remaining (at a
chosen speed/drive current) might really be what you should display.

True ... but whilst it would be nice, it may well be a step_too_far
when it comes to predicting it accurately and as you say, would only
be able to predict the time at the current speed (current).

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. ;-)

And that depends how you obtain your 15A. If it consists of equal
amounts of 0A and 30A at a frequency high enough to not be annoying and
low enough not to be too lossy you will go almost twice as far as if you
used a simple series resistor where half the power just ends up as heat.

Yup. Or use the 12V for 50% of the time ... etc.

PWM at a nominally fixed current also simplifies the battery Ah side.

Whilst it would ... I do need to use different speeds. ;-)

Expect a decent PWM power controller to be expensive and make provision
to switch back to a primitive all or nothing supply in case of failure.

Sorta ahead of you there Martin. I bought a couple of 30A capable PWM
speed controllers from China ... one is FWD only and the other FWD /
REV (using relays). I have also been looking at commercial speed
controllers or around those powers and they are more flexible and
*could* be controlled by say an Arduino micro controller.

snip

I suggest doing some empirical experiments with the boat itself and a
fully charged set of batteries to see how long it lasts run flat out, at
half power, and then half of run flat out followed by half power.

Again, once we know the currents at each of the 5 existing speed steps
(and we do, 7, 11, 15, 20 and 30A) it wouldn't be any issue to do that
on the bench ... and I have on all 3 batteries individually and at 5A
and got about 4 hours to what I thought was the 50% DOD voltage for
that current and 11.2V

Maybe just use one battery to make the testing a bit quicker.

Quite. ;-)

You need to make sure that if all else fails the panic indicator leaves
you with enough juice in the battery to limp home at a snails pace.

True ... or get the oars out. ;-) Again, once I have a reasonable
idea of the capacity / range / runtime (and I do, worse case at least)
the idea was to row out and outboard back .. or a mixture of them
both.

I'd ignore any fancy nonlinear or temperature effects for the first
analysis unless they prove problematic.

Well, whilst I agree re the temperature (although that would be fairly
easy to implement) I think the non-linearness of the capacity is
mostly what this is all about. eg, I could easy build an Arduino based
LVD (Low Voltage Disconnect (Alarm)) with what I have now (and have
one running in front of me displaying Volts Amps and Watts as we
speak) that triggers on the best case voltage (load at 7A) but then
that would trigger way too soon at 30A. (As the capacities at various
draw rates show).

http://www.mkbattery.com/images/8GU1H.pdf

The change in dynamic impedance
of the electric motor with speed will be more of an issue. The biggest
error by far will be that speed in the water against the current will
not translate to anything like as much forward progress as on dry land.

Quite ... and sorta how some of this started. ;-)

1) We already had a 60Ah cyclic battery that we have used on the
outboard to good effect but with it's very heavy duty copper cables
(the battery needs to go forward to get the balance right) weighs
around 20kg. Whilst I can lift it ok and the Mrs can just on
terra-firma, it's not so easy to lower such 3 ft down off the bank
into a boat bobbing about on the water.

2) I also wanted to extend the range, by a) increasing the battery
capacity and b) the drive efficiency. The battery bit got me with
100Ah worth of Li-Po batteries and whilst they were very light and
very efficient, required more equipment and were a bigger risk ITRW.
The drive efficiency would be improved by (initially at least),
putting the outboard on speed 5 (where the battery is connected
directly to the motor) and using an external PWM controller to do the
rest.

3) Because of the Li-Po risks, I went back to 3 x 31Ah Gel Lead Acid
cyclic batteries that could be transported individually and at only
11.5 kg's or so, much safer for all concerned.

snip

You really want something that works well enough from an engineering
point of view and fails safe by underestimating remaining capacity.

Agreed (with caveats as mentioned above etc).

It is very much an empirical business - and one that most car makers
have not yet mastered in their petrol/diesel range indicators.

Oh, completely understood Martin, it's just that with what I thought
what *might* be sufficient data for someone who could deal with such
things we could do a bit better than a finger in the air. ;-)

Are electric cars any better at this I wonder?

Well, you would like to think so and as few run on Lead Acid batteries
these days (except mine g) and with Li-Ion being pretty popular and
them having a fairy linear capacity across all current ranges ...
*and* some reasonable sensors, cpu and programmers at their disposal,
you would hope so! ;-)

I suppose they have to be
or we would see dead and dying electric cars stuck at the roadside.

And if you do that might be more likely because the nut behind the
wheel overestimated the range of the car or conditions changed (or
weren't predicted, like cold weather or hills) that lost them some
capacity etc.

I knew my EV only had a range of 20 miles and so knew that 10 miles
was my drive out limit and that I also had to consider the terrain and
even wind speed and direction as well. I would happier go 'out' uphill
knowing I could nurse it home easier downhill. ;-)

Cheers, T i m

p.s. When I designed, built and races an electric two wheeler I knew
my batteries had a reserve capacity of 25A for one hour (that is how
reserve capacity is measured, low long at 25A).

So, I kept an eye on my ammeter as I raced and if the out leg of the
course had a reasonably steady average of 25A and the same on the back
section, I could keep that up for the full 1hr 'race'. However, if
there was a hill (like at Cadwell Park), I would have to accept the
say 30A+ load on the uphill sections in the hope that I could save
energy by freewheeling down the downhill bits. In my head I had to
compute the overall drain (considering the reduced capacity at the
higher currents) and hope to cross the line with a *tiny* amount of
charge to spare.

If you ran out of charge before the end it meant you used it up too
quickly (so would have had lower battery capacity and increased
rolling and aerodynamic resistance) and if you had any charge left
over at the end it meant you could have gone faster. ;-)

On Thu, 21 Sep 2017 09:29:18 +0100, Martin Brown
wrote:

On 20/09/2017 10:09, T i m wrote:
On Wed, 20 Sep 2017 08:36:50 +0100, Martin Brown
wrote:

On 19/09/2017 19:02, T i m wrote:
Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)

snip

It's just knowing that there can be such a marked difference in
capacity as a function of the current drawn ... and my potential
maximum current (30A) *is* significant to this, even with 3 batteries
in parallel (~90Ah) there is a very good chance that some massaging of
the readings to predict a particular depth of discharge threshold
would be worthwhile.

A quick and dirty value for the discharge threshold is when the terminal
voltage drops below some arbitrary limit ISTR 10.5 or something.

Agreed, 10.5V is often cited as the on-load 100% discharge point.

e.g.
http://www.chbattery.com.img.800cdn....4153615343.jpg
http://www.coslightindia.in/images/2gel-graph-img1a.jpg

You
never want to take an individual cell below some critical value which is
temperature dependent or permanent damage will result. The weakest cell
in a stack of them always gives up the ghost first.

Agreed.

To a good approximation the terminal voltage ends up as

V = V0 - Ir

Where I is the current you are drawing and r is the internal resistance
of the battery (which itself is a weak function of temperature and I).

Ok.

In practice the difference may help to protect the battery if you just
use a fixed voltage cutoff value since if you are discharging it quickly
it will inevitably be hotter and potentially gassing slightly.

I don't think it would in this case Martin as that's what Lead Acid
and Peukert's law are all about (LA battery temperature doesn't rise
and so you get the non linear characteristics we see).

If I retained the original resistor based 'speed control', even I,
once I guesstimated what the capacity would be for the currents
typically seen at the 5 set speeds, could possibly write some code
that applied the variable.

Not sure I understand what you are trying to do but a MOSFET based PWM
speed controller will give you something like 90% efficiency.

Agreed. I was just saying that *even I* could probably come up with
some code to deal with the std resistor speed controller with it's set
steps.

This will
give you a lot more battery life if you use low power most of the time.

Understood. 15A was about the right speed (for the local rivers and
canals) and so with 3 of the 8GU1H batteries connected in parallel and
a capacity of near 28.5Ah (C/5) at that current, that should give me 3
x 28.5 = 85.5Ah and so a 15A draw to 50% capacity should be 42.75/15 =
2.85 hours.

Whereas a resistor will be wasting a lot of power dissipated as heat.

Agreed, so I'm hoping to be able to do better than say 3 hours. ;-)

Using a better more efficient speed controller is a far better option
than managing the decline of the poor battery.

As you can see I'm actually doing both and more (hydrodynamic fairing
on the tubular prop leg, tail spinner on the prop itself and possibly
some mods to the boat itself).

But, to be able to 'Manage what you can measure', you also need to
know how to make best use of those measurements. ;-)

Cheers, T i m

On Thu, 21 Sep 2017 13:22:57 +0100, The Natural Philosopher
wrote:

On 21/09/17 11:50, T i m wrote:
So, cummon, have a go at being the solution, you know you want to.;-)


Seeing as its you t i m, no, I dont want to.

Yeah, nice cop out mate! Just like the Linux questions I've challenged
you with in the past when you initially portray you know what you are
doing and pretend to offer help but when you are really tested you go
very quiet ... ;-(

What do they say ...'All mouth and no trousers ...'? (Damn, now I have
to get that image out of my head ...). ;-(

Cheers, T i m

On Thu, 21 Sep 2017 14:28:02 +0100, T i m wrote:

On Thu, 21 Sep 2017 13:22:57 +0100, The Natural Philosopher
wrote:

On 21/09/17 11:50, T i m wrote:
So, cummon, have a go at being the solution, you know you want to.;-)


Seeing as its you t i m, no, I dont want to.

Yeah, nice cop out mate! Just like the Linux questions I've challenged
you with in the past when you initially portray you know what you are
doing and pretend to offer help but when you are really tested you go
very quiet ... ;-(

What do they say ...'All mouth and no trousers ...'? (Damn, now I have
to get that image out of my head ...). ;-(

https://i.stack.imgur.com/CdIyt.png

--
My posts are my copyright and if @diy_forums or Home Owners' Hub
wish to copy them they can pay me £1 a message.
Use the BIG mirror service in the UK: http://www.mirrorservice.org
*lightning surge protection* - a w_tom conductor

On 21 Sep 2017 14:52:26 GMT, Bob Eager wrote:

On Thu, 21 Sep 2017 14:28:02 +0100, T i m wrote:

On Thu, 21 Sep 2017 13:22:57 +0100, The Natural Philosopher
wrote:

On 21/09/17 11:50, T i m wrote:
So, cummon, have a go at being the solution, you know you want to.;-)


Seeing as its you t i m, no, I dont want to.

Yeah, nice cop out mate! Just like the Linux questions I've challenged
you with in the past when you initially portray you know what you are
doing and pretend to offer help but when you are really tested you go
very quiet ... ;-(

What do they say ...'All mouth and no trousers ...'? (Damn, now I have
to get that image out of my head ...). ;-(

https://i.stack.imgur.com/CdIyt.png

Thanks for that Bob ... both an appropriate image and a better one to
have stuck in my head. ;-)

Cheers, T i m

T i m wrote:

On Thu, 21 Sep 2017 10:45:21 +0100, The Natural Philosopher
wrote:

snip

If only you could apply the same amount of energy to helping us
resolve this (and what must be to you, 'brain as big as a planet'),
simple *energy* measurement issue as trying to defend the indefensible
(Linux or leaving the EU) we might all get on faster! ;-)



The fact that I have a degree in electrical engineering

You say ... ;-)

and a huge
anount of experience in using lithoum ion batteries in model aircraft

;-)

Yeah, I was a 'sponsored' RC car driver so because of that I know all
there is to know about Nicads and NiMh .... (not).

completely precludes me from offering what would instantly be shouted
down as 'an opinion'.

Yes, but if you had an EQ as big as your IQ you would be able to offer
*information* around my question wouldn't you?

But I will give you a hint.

Ohms law.

No use whatsoever mate as it's a linear relationship and this isn't a
linear problem.

I think we have established that the relation between terminal voltage
and current at a given state of discharge is a linear one.

If you want the *derived* quantity of remaining capacity *at a given
current* for a given state of discharge this can simply be a look up
table and is little or nothing to do with what the battery *has* been
doing, depending only on discharge state, not how you got there. So you
need no other measurements to estimate remaining capacity at a given
current. Specifically it does not depend on how Peukert's law has
applied during the use of the battery up to now, only on how you use it
in future. You could automatically estimate remaining endurance (to a
given future state of discharge) for the prevailing current at the time
of measurement, but you still only need a look up table for capacity
against current and your present state of discharge plus present current
consumption.

You could use an approximation to Peukert's law to calculate capacity at
a given current on the fly, but there is no advantage to doing so as the
future behaviour is apparently not affected by how you have treated the
battery except as represented by current discharge percentage. And, as
the currents you are using are well within the range for which capacity
is published interpolation would be easy. If the 'k' you can calculate
is actually is constant you could program the Peurkert equation, but a
look up table for capacity for each amp of current per battery up to ten
might be even easier.






As you know all there is to know about LiPo (as the chances are you
*wouldn't* be using Li-Ion in RC aircraft as they are too heavy) you
would know that whilst Peukert's Law also apples to them in theory,
you would also know it is pretty well cancelled out by the increase in
efficiency found as the batteries heat up in use (proportionally to
the current drawn).

So, what with you having a degree in electrical engineering, into
using your (Linux) laptop as a typewriter (writing code for the likes
of Gridwatch?) and like debating things so what's not to like? ;-)

So, cummon, have a go at being the solution, you know you want to. ;-)

Cheers, T i m


--

Roger Hayter