EEVblog #772 - How To Calculate Wasted Battery Capacity

[dk27]

So what exactly are you claiming? Yes, Dave does draw the area under the area of that graph which is incorrect, but he also names the percentage from the graph, which is numerically generated has a Y axis from 0-100%, and is accurate. (To within the bee's dick's margin, as the calculation assumes the battery is dry at 0.8 V, but that was not the issue you complained about to begin with, right?) So if you listen to Dave when he says that 4% of the battery's capacity is left for a 1V cutout voltage, this is accurate no matter what he's filling in under the voltage graph.

Ok, that's a reasonable question, I think I covered it in the following post:  https://www.eevblog.com/forum/blog/eevblog-772-how-to-calculate-wasted-battery-capacity/msg721176/#msg721176, especially possibility 2.

If you accept that the ratio of the areas depicted gives the wrong answer, and you simultaneously use some other means of finding the right answer, then why bother talking about those areas in the first place?  The exercise becomes a meaningless ritual --- we talk about "areas under the curve", but we're not really taking the idea seriously.  I think I'm justified to be concerned about that ritual, because at the same time he's saying that 4% of the battery capacity is left, he's pointing to an area that has a size more like 2% of the total area, so if someone were to mistake the ritual of pointing to areas as something to be taken seriously, although they will hear 4%, they will see about 2%.

I'm sure that for Dave, pointing to areas is not meant to be merely a meaningless ritual, and, to his credit, he later added an annotation to the videos that acknowledged that the areas in question actually need to be integrated down to 0 V to make sense.  However there was also the suggestion that eyeballing the "wrong" areas in order to get a reasonable estimate was a standard practice "in the industry".  This is covered by my possibility 3 in the referenced post.  The point being that this is generally not a very sensible practice, since as an approximation, it is really bad.

[miguelvp]

Dave said in the video that the area under the curve "represents" the energy in the battery. No one claimed that the area "is" the energy in the battery.

Also he mentions that is a "common technique in the industry to compare areas under the curve on 0.8V cutout graphs"

The area under the graph without the integral part of the constant 0.8V added does change the percentage of remaining capacity(which is what we are talking about, hence the title of the thread) but the 0.8V cutout graph is used to "compare" not "compute" the areas, semantics are important here.

What does that mean? well if the battery is considered to be fully discharged at 0.8V using the graph you provided, the ballpark comparison gives you 50% of energy remaining, sure there is an error from the actual 41.(6)% but that is good enough for government work and far away from your "computed" absolute area capacity of 25%.

Edit: the discharge curve on your graph, happens to be the same as the percentage of energy used. So a straight line will do.



If you apply that to your other graph that it doesn't have the cutout, the percentage will be more accurate and closer to the actual 41.(6)%

Engineering even if it's very precise it needs tolerances and constrains to apply the science and math to a "practical" solution, it's not about absolutes because absolutes don't work in engineering, there are way too many external factors.

For example, you can't build a bridge to absolute specifications because external factors out of your control will affect it, so you need to compute what are the best tolerances you are willing to live with to get a practical solution done.

[Fungus]

Dave said in the video that the area under the curve "represents" the energy in the battery. No one claimed that the area "is" the energy in the battery.

This.

[dk27]

Dave said in the video that the area under the curve "represents" the energy in the battery. No one claimed that the area "is" the energy in the battery.

Really?     We're going to get all philosophical about this?  Well, okay...

In your mind, do you have a precise notion of how your area under the curve is meant to "represent" an energy?  For example, is there some quantitative relationship between the area and the energy.  Alternatively do you have a qualitative notion in mind?  Or do you mean that the area bears no particular relationship to the energy at all, and we, by convention, are simply using it as a symbol of the energy?

Also he mentions that is a "common technique in the industry to compare areas under the curve on 0.8V cutout graphs"

Until we know precisely what you mean by an area "representing" an energy, there is no basis to decide whether it is reasonable to compare the areas, or even whether such a comparison is "good enough" for your application.

For example, if you claim that areas are merely symbolic of energies then I can learn nothing by comparing the areas.  In the same way, I learn nothing about the differences between cats and dogs by comparing the symbols "cat" and "dog".

Remove enough meaning from the areas and it's time to ask whether it's worth thinking about them at all.

What does that mean? well if the battery is considered to be fully discharged at 0.8V using the graph you provided, the ballpark comparison gives you 50% of energy remaining, sure there is an error from the actual 41.(6)% but that is good enough for government work and far away from your "computed" absolute area capacity of 25%

Ok, if you are happy with that approximation, that's fine.  However, that approximation has nothing to do with any areas, so why do anything with the areas at all?  Talking about the areas would just be a meaningless ritual.

(Of course, I believe that by treating the areas seriously, and working with the right ones, you can get answers that are even better than those good enough for government work.  You might even be able to get a job in private industry!)

Edit: the discharge curve on your graph, happens to be the same as the percentage of energy used. So a straight line will do.

The percentage of energy dissipated is quadratic in time (recall that my graph was representing a constant current draw), the discharge curve on my graph is not quadratic in time.  The service time remaining is a straight line with respect to time, but that's obvious.

[miguelvp]

Well, the graph clearly shows the 0.8 cutoff so you just have to add the integral of that constant if you really want to calculate the area of that representation. It's pretty trivial, all the information is at hand.

If people don't know how to read graphs, well, that's their problem.

Yeah, 8% error is not ideal by any means at 50% of the time, but a good approximation and not as much as a 25% error as suggested. If some people don't know how to read graphs, then send them back to elementary school.

The thing is that quadratic or not it doesn't matter if you have the discharge curve, and if you don't, I recon a linear approximation will do the job as well for an off the cuff approximation. At the end of the day we are talking about under a minute of battery life after it reaches the 0.8V.

[dk27]

Well, the graph clearly shows the 0.8 cutoff so you just have to add the integral of that constant if you really want to calculate the area of that representation. It's pretty trivial, all the information is at hand.

If people don't know how to read graphs, well, that's their problem.

Yeah, 8% error is not ideal by any means at 50% of the time, but a good approximation and not as much as a 25% error as suggested. If some people don't know how to read graphs, then send them back to elementary school.

All very macho, but perhaps people who don't know how to present graphs so that they are as clear as possible should also be sent back to school.  At the very least, if you're going to present a graph, point to an area you've coloured in, say that it represents the energy remaining, and then also emphasise how small it is, you should remind your audience that they need to imagine adding in the extra rectangles in order to get a true impression of that energy.  Not adding that reminder is a sign of either ignorance, forgetfulness, or dishonesty --- if you're arguing against the claims of people like Batteriser, you don't want to give them the opportunity to accuse you of dishonesty.  People being misled by the graph that you present can very quickly become your problem.  (I realise that you personally might not be trying to refute Batteriser's claims, but that is an activity of a lot of the readers of this forum.)

The thing is that quadratic or not it doesn't matter if you have the discharge curve, and if you don't, I recon a linear approximation will do the job as well for an off the cuff approximation. At the end of the day we are talking about under a minute of battery life after it reaches the 0.8V.

And again the attempt to distract by talking about the battery life after it reaches the 0.8 V --- this is a dishonest tactic, you know full well that my position has nothing to do with this question.

[Fungus]

All very macho, but perhaps people who don't know how to present graphs so that they are as clear as possible should also be sent back to school.

The graph shows voltage against time. There's nothing unclear about it.

Very few people are probably interested in the exact number of joules remaining (and probably even fewer are trying to calculate it using that graph).

Battery lifetime and voltage are FAR more important than the remaining Joules. Reducing the Y axis would be more damaging than chopping it off.

If you really want the rest you can just draw a box underneath. It still won't get you any hard numbers though - you'll need to do the math if you want real numbers.

[EEVblog]

All very macho, but perhaps people who don't know how to present graphs so that they are as clear as possible should also be sent back to school.  At the very least, if you're going to present a graph, point to an area you've coloured in, say that it represents the energy remaining, and then also emphasise how small it is, you should remind your audience that they need to imagine adding in the extra rectangles in order to get a true impression of that energy.  Not adding that reminder is a sign of either ignorance, forgetfulness, or dishonesty

You guys can nerd fight all you want in here over the technical details, just leave me out of it thanks.

[Don Hills]

(Moved to a different thread.)

[EEVblog]

For those playing along at home, dk27 decided to take his bat and ball and go home.

[J4e8a16n]

First, Dave's calculation assumes that there is no energy available once the voltage under load reaches 0.8 volts.  Presumably, if we had a load with a dropout voltage lower than 0.8, we could draw a small amount of extra energy beyond what the calculation considers to be 100%  --- it's pretty clear that the discharge curve becomes close to vertical, however, so this extra energy is pretty much negligable.

This is why the manufactures usually don't provide curves below 0.8, it is the defacto industry standard dropout voltage at which point any extra energy in the battery is consider negligible. As I mentioned in the video, there is some energy left below 0.8V for really small currents, and this is why a joule thief can flash a LED down to bugger all. Very small amount of energy, very niche applications where it can be used.
I didn't want people thinking they should go to the ends of the earth to design products with a 0.5V cutout voltage, that's rarely done in the industry.

Would you develop Integral[V[t]-0.8,{t,t1,t2}] here, for us?



JPD