Predicting a graph from 3 (6?) values?

Hi all and knowing there are some scientists / mathematicians walking amongst us ... ;-)
I'm trying to see if there is a (rough even) way of measuring the Depth Of Discharge voltage / cutoff point versus current load value for a specific lead acid battery.
The manufacturers have provided me the following values (@ 50% DOD).
11.75V @ 93A (12.4V initial) 12.05V @ 18.2A (12.7V initial) 12.10V @ 4.65A (12.75V initial)
(The general chart starts at the values marked 'initial' so that would be for those currents but at 0% DOD but I'm not sure if that is relevant).
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?
Maybe Peukert's law itself would give those with a brain wired differently than mine the answer? ;-)
https://en.wikipedia.org/wiki/Peukert%27s_law#Formula
Cheers, T i m
p.s. The nearest *I* could get to an answer would be some graph paper and a Flexicurve. ;-)
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On 14/09/2017 20:14, T i m wrote:

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.
Sometimes you can over-think a problem, you have to work with the data and tools available.
As an example, compare the different "lines" of a Spitfire or a Vanwall with the various front wings of current F1 cars. Modern stuff has the benefit of serious CFD and wing-tunnel testing. And it is justified in that context.
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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.
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On 14/09/2017 21:28, NY wrote:

Sorry, but trying to get the "right" straight line from three points is pretty meaningless. Batteries are intrinsically non-linear. You can (sort of) rely on an exponential for radioactive decay when you are starting with maybe 10^20 atoms. Batteries are more complicated because you have competing processes.
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Yes, I never suggested that the data would be linear. I'd be highly surprised if it was anything like linear. But if you can find a way of transforming the non-linear data so it becomes linear, then you can work out the best-fit line and predict future (transformed) values, from which you can back-transform to get the corresponding real value. Easier to do it that way that to try to fit a flexicurve to a curved line so you can predict an out-of-range value (extrapolating) as opposed to predicting an in-range value (interpolating).
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So, as you sound like you know what you are talking about, can you plug some real values in and see what we get please?
I'm happy to answer any questions re numbers and points etc.
I know the off load starting voltage is 13.10 volts. I know the voltage that represents 50% (and other percentages) for 3 different currents.
We have the formula that should make of it all but I don't have the understanding of how to take what we have and make use of it ITRW (if it's possible)?
Cheers, T i m
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<snip> >I'd say the best approach would be to find a transformation (eg y=log(x) or

Yeah, I was going to say that but thought it was obvious. ;-)
Cheers, T i m
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On 14/09/2017 21:28, NY wrote:

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.
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On 17/09/2017 13:26, John Rumm wrote:

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

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 <operator><GraphRule> =< 50% DOD voltage, output alarm?
If not, rinse repeat?
Cheers, T i m
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On Sun, 17 Sep 2017 13:26:32 +0100, John Rumm

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.

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' here:
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
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On Thu, 14 Sep 2017 21:20:39 +0100, newshound

Catch 22 then. ;-(

Well, I think it translates to this:
t=H(C/IH)k
(k is a power)
where: H is the rated discharge time C is the rated capacity at the discharge rate (Ah) I is the discharge current t is the actual time to discharge the battery in hours K is the Peukert constant.
The value of k is normally between 1.1-1.25 for Gel
(found somewhere on the net).
So, say I this battery: http://www.mkbattery.com/images/8GU1H.pdf
That datasheet may give us some more information if I had the brain to make use of it, eg it says it's:
36Ah at C/100 (so when discharged at 0.36A for 100 hours) 31.6Ah at C/20 (so when discharged at 1.58A for 20 hours) 26.8Ah at C/5 (so when discharged at 5.3A for 5 hours)
However, I would be running 3 of those in parallel so I'm assuming my max 30A load would just pull the discharge time further round the graph (in the more linear section).
And this gives the discharge time versus current (to 10.5V I'm told which is 100% discharged):
http://www.mkbattery.com/gel_specs.php?model=8GU1H
Could we use any of those to give us values the formula for that particular battery?

Understood.

You mean physically? The issue as I see it is that I don't know what the Low Voltage Cutoff voltages should be for any other currents than the ones I listed originally so I'm not sure UI can (and sorta my dilemma). ;-(

Understood.

I'm trying to underthink the problem as anything else makes my brain hurt. ;-(

Understood.
Basically I could put the 3 batteries in my boat, run the electric outboard till it slows to a point that's unusable and charge them back up when I get home. However, while I'm so doing I could damage them and they weren't cheap. ;-(
So, I could assume the worse case discharge (30A) and buy / build a simple voltage alarm that triggers at 'some voltage' (that I think represents 50% depth of discharge) but what if I'm not going a full speed and only drawing 10-15A when the voltage wouldn't be anything like as low as when using 30A?
The idea being, once I know the rules I might be able to include that into the calculations to provide a more dynamic / real-world cutoff / alarm point? (Now I know they might be other variables like temperature and aging but I can also compensate for those).
Cheers, T i m
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T i m wrote:

Three points is a bit tough, 6 points fine. Plot using excel. then use ADD Trendline to try different type fits then on the options tab, you can get it to display the equation of the trend line.
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On Thursday, 14 September 2017 21:40:21 UTC+1, Bob Minchin wrote:

If you do this, be aware that you might need to format the trendline eqn to get a sufficient number of decimal places. Else can get large errors.
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On Thu, 14 Sep 2017 21:41:42 +0100, Bob Minchin

That was my fear Bob. ;-(

Yeah, easy for you to say but because I've never produced a spreadsheet in my life ... ;-(
As posted elsewhere, would this help:
t=H(C/IH)k
(k is a power)
where: H is the rated discharge time C is the rated capacity at the discharge rate (Ah) I is the discharge current t is the actual time to discharge the battery in hours K is the Peukert constant.
The value of k is normally between 1.1-1.25 for Gel
https://en.wikipedia.org/wiki/Peukert%27s_law#Formula
I'm not sure what H would be as it's a function of the other variables? So, we know C at three points. We know I will be between 0 and 30 (we have 3 with other variables). We can take the Peukert constant as 1.2
My brain hurts ... ;-(
Cheers, T i m
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H, C and (I at the standard discharge rate, call it Istd) are constants, measured at the standard conditions for quoting battery capacity (whatever they are). If you know the capacity that the battery is sold at, say 36 Ah, and the discharge current it is standardised at then the capacity C is a constant 36. Istd is the standard discharge current and it is constant, and H is equal to C/Istd and also a constant. This leaves only the particular 'I' you are using as a variable and a fairly simple computation. I am guessing here, but the equation only makes sense if H and C are constants. I am surprised they do no t use 'i' for the independent variable rather than 'I' though.
BICBW
--

Roger Hayter

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On Thu, 14 Sep 2017 23:17:32 +0100, snipped-for-privacy@hayter.org (Roger Hayter) wrote:
<snip> >> Yeah, easy for you to say but because I've never produced a

The thing is I'm not sure there is such a thing Roger, being the battery capacity varies based on the rate of discharge (and what I'm trying to pin down).

Well, I know the capacity at 3 discharge rates because it's on the data sheet:
It's 36Ah at the C/100 rate (0.36A), it's 31.6Ah at the C/20 rate (1.58A) and it's 26.8 at the C/5 rate (5.32A).
I will be discharging it (if using one on it's own or 'them' if using all 3 in parallel) between 0 and 30A, so seeing a capacity between C/0 (= infinity, and better than C/100) and ??/30 (as I don't know the capacity is when the battery(ies) is/are discharged at 30A)?
http://www.mkbattery.com/images/8GU1H.pdf
And we know some other things about the battery from the general data sheet:
http://www.mkbattery.com/documents/19779350MK_GEL_v7_r2.pdf
and I know the voltages that represent 50% DOD for 3 currents because I have taken them from their graph (bottom chart on P6):
http://www.mkbattery.com/documents/13991833%20 (8A-8G%20I&O)_MK_r1.pdf

Ok (I think)? ;-(

But if the capacity is a non linear function of the current drawn ...

Well maybe it isn't as simple as both of us would like / hope Roger. ;-(
Cheers, T i m
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I am assuming the C and the H in your equation are constants but the actual capacity varies according to your equation, which is definitely non-linear.

I'm sure you're right, but if you know C and H at one value of I then your equation should let you work out the time to discharge at another value of I, within the limits of how accurate the equation is. If the equation is meaningful at all the only one item on the RHS is a variable. Admittedly you have to guess a value of k, but you could try two values and see how much difference it makes.
--

Roger Hayter

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On Fri, 15 Sep 2017 10:32:18 +0100, snipped-for-privacy@hayter.org (Roger Hayter) wrote:

Quite.

Ok ...

Well yes, as you say, k for gel batteries lies in between two fairly confined values and reasonably different to other battery chemistries so should be a reasonable starting point.
So, would you mind helping me plug some values in from what we know because I really suffer a form of blindness when it comes to such things (always have).
t=H(C/IH)k
(k is a power)
where: H is the rated discharge time C is the rated capacity at the discharge rate (Ah) I is the discharge current t is the actual time to discharge the battery in hours K is the Peukert constant. The value of k is normally between 1.1-1.25 for Gel
Does this help?
http://batteryuniversity.com/learn/article/bu_503_how_to_calculate_battery_runtime-
Cheers, T i m
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It does occur to me I am solving a problem (how to calculate discharge time) that you are not interested in. You were originally interested in discharge voltage at 50% discharge, as a point to stop using the battery.
Looking at your original figures for voltage at 50% discharge, I think they are adequately modelled within the range 0A to 93A by assuming an open circuit 50% discharge voltage of 12.12 and a series resistance of 4 milliohms. This would give voltages for the cases you mention of 4.65A 12.10 18.2A 12.05 93A 11.75
Which is suspiciously close to their figures. Identical to this degree of precision! I wouldn't be surprised if they used the same method to calculate them. I did this by assuming that the open circuit voltage at 50% discharge is independent of current (not probably true dynamically) and the effective series resistance is the same for all currents, mentally extropolating to 0A and dividing the apparent voltage drop from the 0A level by current to give a notional resistance, slightly different in each case, and choosing the one nearest the high current value. I.e. handwaving plus mental arithmetic.
But this model (12.12 volts minus current times .004) seems to be a very good fit up to 93A. And all linear!
Peurket doesn't help you with this - it tells you how soon you'll get there, but not what criteria to use for 50% discharge. I don't believe it allows for potential battery recovery either, so is probably too conservative. But it does give the same answer as the makers want you to have.
None of the information you have helps with saying whether this relation will still be valid above 93A, but perhaps it should be good up to the rated maximum current of the batteries??
--

Roger Hayter

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