Hi...i am a new user here. As per my knowledge if the battery is completely flat it might be a right off. The cells could be damaged, however, if you trickle charge at a milliamp rate you may get some charge back, try 1ooma to start with, this would take about 20 hours to get some charge.

Hi Roy,
Like I already said in similar threads; pay a bit more for major producers, like Panasonic as you mentioned, and you are more likely to get what you paid for.
No names are always a gamble. May be a super start-up, may be a 'let's cash in band wagon jumper! Whoever the 'supplier' is a relatively small number of manufactures are behind them.
Yer pays yer money and yer takes yer choice!
Personally, for the bigger, more expensive anyway, drive batteries I go for a supplier with reputation and known sources and reliable recourse without questions if you are not satisfied. For the little, cheaper, accus. for my Plastic Magic and auxiliary functions I am more inclined to experiment, log the results (manually!) and adjust future purchases accordingly.
Regarding your super duper data logger; how has really it helped you?
I used them often in my professional life, mostly as chief of the Rank Telecomm personal radio test department in the 70s and 80s, but now I don't need to sit in front of the PC analysing such data for hours. Time I could better spend building, sailing or solving other more interesting problems. Without appropriate analysis algorithms data loggers are useless.

I have a thermocouple add-on for my multimeter which tells me if a particular cell is overheating compared to the rest. (The finger tip test usually agrees with the thermocouple!) This gives me a more useful indication than lots of abstruse graphs which you then have to interpret correctly.
The money this cost you might have been better invested in a better charger as Dave M has already advised.
I repeat what I said in the thread Dave mentioned: if I note that an accu. has a weak cell I degrade it to non-essential functions, nothing needed to bring the boat home!
Each to his own; if you want to spend your time pondering over dubious statistics carry on and good luck!
Personally I think Life's too short to waste on such things!
If it woiks, don't fix it. if it don't woik fix or replace it, but don't waste time on bait digging when you could be fishing!
Forgive me if I'm insulting your intelligence, for all I know you may also have 40 years professional experience in electronics. Let go!
Cheers Doug 😎

Gentlemen, many thanks for all the feedback.
To respond to Dave's questions first: the charger I have been using is an MRC Super Brain 989, and the recorder is an Eagletree Micropower E-logger V2; both are off-the-shelf items and not calibrated. I know there can be some discrepancy between the two, but surely not enough to explain the numbers? I cannot check the temperature of the battery. That will be affected by another variable - charge and discharge current, will it not?
As Dave and Doug have pointed out, I made a very, very rough, simple, conservative calculation of mAh looking at the chart. I did not try to be mathematically precise. it was a number for me to explain my question of whether the apparent charge and discharge efficiencies made sense and could they be explained.
I have learnt a lot from this exchange and, since I have already acquired 3 battery chargers over the years, I am reluctant to invest in another one only to measure battery impedance. Question is now, does future potential use of such a charger make better financial sense than simply throwing away the questionable batteries and replacing them. Decisions, decisions.
One final question, what charge and discharge efficiencies would you expect from a new battery past its "running-in" duration?
I am aware that mAh capacities are not based on any recognised standard, but I have been led to believe that the stated capacities of Panasonic and Venom batteries are more reliable than most.
Thanks again,
Roy

Hi Roy,

I like your want for the answers! I have an eagle tree, great bit of kit, but, you are relying on the given info regarding the battery to be correct. As an example, I have an internal resistance meter, I fly planes, so rely on accurate lipo battery info. The "c" rating, and milliamps on most chinese battery stuff is grossly over egged! as are esc ratings and so on. I wouldnt get too hooked on these stats. if the batt is losing a cell, throw it and get another, its not worth the hassle. Overlay that graph with volts, this should correspond to the amp spikes, and if you have a gps then also overlay speed, faster is more amps is quicker volts drop

for enjoyment 😲 check out my amp draw on my Huntsman!



Paul

Hmmm! Looking closer at your chart, and the average current quoted at the bottom, it seems your 'rough calculation' could also account for much of the difference!
You used 2.55 amps, the chart says 2.04A average. delta is 0.51 A or 20%. Looked at that way the situation gets worse!
To get any closer to the real consumption you would need to integrate the whole curve! Rather you than me Gungadin!
Cheers Doug 😎

Absolutely correct Sir Dave!
There are statistics, lies and damned lies!
Test gear calibration was a professional pain in the ... for me for many years.
All comments re ambient temperature, internal resistance absolutely spot on👍 Goes back to my comment in the debate you mentioned, pay a bit more for the batteries and you'll get more out of 'em.
But unfortunately 'what goes in does not (always) come out' 🤔
The capacities printed on the jacket should always be taken with the proverbial pinch! They are only an indication of perfect performance in 'this the most perfect of all possible worlds'!
I remember matched cells too, much more expensive, and not critical for scale ships.
Cheers Doug 😎

Hi Roy
I can understand your conundrum. My work at one time involved gathering and presenting statistics. The difficult part was knowing what answer was expected.
There are statistics and dammed lies and at the end of the day they are at best an indication, but should be taken with a pinch of salt.
To try and shed some light:
What charger were you using and was it calibrated and if so who by?
Same goes for the recorder.
Did you measure the temperature when charging and discharging.
Looking at your chart I not sure I quite agree with your consumption calculation as the 2.55 amps was at best for no more than 40 mins.
Batteries will waste energy in the form of heat and this explains why more energy is required to charge to the retained capacity.
We did discuss the battery internal resistance earlier and this can have a real effect on the capacity and the batteries ability to deliver.
Assuming your measuring equipment is correct then I believe the differences are likely to be due to temperature variations caused by the internal resistance.
If when you charge your pack one cell is noticeably warmer than the rest of the pack it has a higher internal resistance.
At one time you used to be able to buy race packs which had been formed from individual cells that had been matched. Cost a lot more but if you were racing with the pack they gave you the edge.
Cheers
Dave

Here's a conundrum that I'd like your thoughts on.
I recorded some data on one of my boats now running on a couple of the battery packs that were cycled as discussed earlier. They are 6V 5000mAh NiMH packs which, after their final charge, showed they received 5500mAh according to my charger.
For the subsequent run I fitted a recorder which showed the current consumption, as on the attached chart; typically 2.55A, with a maximum of 2.9A. A rough consumption calculation based on the chart, of 2.55A for 70 minutes, is a little less than 3000mAh. When I recharged them after the run, the charger showed they'd taken 3850mAh. Why the difference between the 3000mAh consumption, and the replacement charge of 3850mAh?
The charge and discharge efficiencies are obviously less than 100%, but this data suggests that the two combined are only 78%. So, for example, if the two efficiences are equal (89%), if the charger states a charged value of 5000mAh, the battery has only accepted 4450mAh (=500089/100). When delivering the power, it can only put out 3960mAh.(=445089/100). Or in other words, only 78% of a battery's stated capacity is usable.
Or is there a different explanation?
Roy

Absolutely right Roy 👍
That's why I only use them for non-essential functions - just to squeeze the last drop of use out my hard earned cash! 😎

For someone on a budget, the described methods of recovering a battery may be worth a try. On the other hand, if I have spent a year or two building a scale model, it's not worth risking it for the cost of a battery.
Roy

Hi Dave, Basically I agree with you, I believe I also mentioned that the pack would never reach full rated capacity or be able to deliver high current.
But I still find such 'rescued' packs useful for non essential functions. Would not trust one for high speed motoring.
Regards Doug 😎

Hi Doug
I use a charger that measures the internal resistance of my batteries. Damaged cells exhibit high internal resistance and cannot be recovered to their full capacity. I agree you can flash the cells and they will recover the voltage but will not take or deliver the current. As a result they will quickly discharge under load and prevent the whole pack from delivering its full capacity.
I sail model boats on a large lake and recovering a model due to battery failure is difficult, so I don't risk using sub standard batteries.
I have not used wet Lead acid batteries for many years but I agree the Epsom salts trick did work on such batteries.

I agree flashing is highly dangerous with any battery as there is a high risk of explosion. Great care should be taken and, unless you are competent, should not be attempted.
Take care
Dave

Good one Onetenor 👍
I've also used the 'flash' technique successfully for years, BUT ONLY ON Ni-cad, NiMh or LA. it's the basis of the lead acid battery 'rejuvenators'. in that case it shocks the sulphate off the plates.
DO NOT USE THIS TECHNIQUE FOR LIPOs OR ANY LI*** TYPE ACCU.
The results would be unpredictable and downright bl...y dangerous. 😡
I would not want to be in the same room 😲
Have fun but don't blow yourselves up! 😆
Cheers Doug 😎
PS Roy - there's nowt much wrong with your battery if it charges to 7V.
One cell is a little weaker OK, but if it doesn't show a tendency to collapse under load the accu will be OK if you don't demand too much current from it. Dave M's accus also only charged to 6.8V, an average of 1.36V / cell!
Happy sailing all😊

Good result Just what I was about to suggest as it happened to me.Re testing through insulation Bang Good sell pairs of adaptors to fit over the tips of standard test probes. Sharp as needles but strong enough to pierce most insulation. Nicads NiMhs and others develop needles or dentricles inside which short out the anode to cathode internally These can be dissolved or blasted away by flashing them with a higher voltage .A/C or D/C. I use my welder. Hold one contact to one battery terminal and flash the other with the other welder terminal. Just a glancing blow DO NOT hold it in place.There is no real need to split the pack but can if you would rather treat individual cells.They can be done without unsoldering them just use some wire held in the earth clamp and stickholder or feed gun to reach the ends of each cell.You get the picture .I have done this with battery powered tools for years .I have a 24 volt drill that is still on original battery after more than 10yrs. Another way is connect the supplied charger and plug in and switch on waggle the plug in wall socket in such a way as to make and break contact about 16 or more times. Not as good as the flashing method but works. Wet cells can be improved by putting 2 or three teaspoons of Epson Salts in each cell. Many tops can be chiseled/prized off to access the interior and glued back on with plumbers solvent cement or similar............Gorilla glue would probably do it too. Here 's hoping I've saved a few quid for some of you. BTW I wouldn't know if this would work for Lipos.I wouldn't want to be the first to try 😆🤐Here is the page https://www.banggood.com/search/insulation-piercing-needles.html

Hi both
It would appear that one cell is damaged/faulty. I have had trouble with bad meter connections in the past and it can cause all kinds of trouble & strife.
The pack will be no better than the weakest cell. if you have a good electronic charger it will give you the internal resistance of the whole battery in milliohms. I checked all my NiMh batteries we use for the public sails and found several that had high internal resistance. Replacing the faulty (>1.2v) cell restored the pack to full capacity
Dave

Excellent, more or less what I thought would happen 👍😊
As long as you don't try to take 50A out of it it should be OK for lots more cycles, but the 'weak' cell will remain weak and probably give out before the others.
Whatever - Happy sailing 😎

Gentlemen, after charging the packs at 100mA for 24 hours, I stripped off the heat shrink from one pack for testing. Unloaded, 4 cells in the pack showed 1.40 volts on my multimeter, with one cell showing 1.37V - near as dammit to 7V for the pack. With a load applied, the current started at 0.95A, dropping to 0.9A when the voltage fell to 1.15. The one cell was consistently the lowest, but never more than ~30mV. From this test I calculated that the capacity used was 2250mAh. Not the rated capacity of 2550mAh, but almost there.
During these tests I discovered that the recording meter that I had been relying on for the previous testing was giving inconsistent readings, a fact I only discovered when I started using my multimeter. This possibly resulted in my considering the battery discharged in previous tests, when it was not.
Roy

Hi Roy
Possibly. However as you are charging at 100mA your 1700 mA battery will require charging for at least 17hrs and probably much longer if completely discharged.
It's possible that you have not fully charged the battery but you should get more than 4 mins on 1.2v at 0.95A.
If you have a meter I suggest you connect two sharp pins to the plus and minus leads and try to measure the voltage across each cell through the sleeving- it's tough so some force is needed, but this will indicate each cells voltage. They should all be the same but any less than 1v are toast. Try with and without a load as cells can show a normal voltage but drop under load.
Not the easiest task and you may need to take multiple readings to make sure you have a good connection
Good luck.
Dave

Dave,
I've tested one battery pack as you suggest. With a load of 0.95A the voltage drops immediately from 6.75V to 6.2V, and in just under 4 minutes drops to 6V. I presume this meets your diagnosis that (at least) one cell has failed.
You are correct in presuming that my charger is indicating the 1200mAh capacity. I have been charging at 0.8A, which is the lowest current setting on that charger.
I now have both packs charging on 100mA, which is the closest I can get to the recommendations from Doug and sidley70, but this charger does not indicate mAh.
Roy

Hi
The Formula for charging any battery, Gel cell or NiCad is Divide the amps (milliamps) by 16. This gives the charge rate in Milliamps. The charging time is 14 to 16 hours. With this formula you cannot overcharge the battery. EG (1700 milliamps divided by 16 equals 106.25 milliamps. The approximate charge rate for 14 to 16 hours)😎

Memory effect?
"The packs will charge to full voltage, around 6.8V, and I discharge them down to 6V."
6/5=1.2 which is the nominal voltage of the cells!
Before getting into the 'surgery' I would suggest first discharging to 1.1V / cell, i.e. 5.5V per pack. Not lower or the damage Dave describes may occur.
Then charge at 200mA for 13hours. (12.75 hours to be pedantic for 2550mAh!)
See what happens then😉 it is often effective to repeat this cycle 2 or three times to reduce the memory effect.
If it don't woik then it's back to the surgery, or back to the shop ??
Good luck, Doug 😎

Hi Roy
A NiMh battery when fully charged is 1.4v dropping to 1.2v under load. I use similar batteries for our display boats and they give about 6.8 volts when fully charged.
I am assuming your charger is indicating the 1200mAh capacity?
I suggest you connect the battery to a meter and place a heavy 1 to 2 amp load on the battery and see what the voltage drops to. if it's less than 6v very quickly I would suspect one of the cells has failed. Usually caused by over discharging the pack and one cell becomes damaged by being reversed charged.
If this is a recent purchase I would contact the supplier and ask for help as cells can fail sometimes.
If this is not the case then it is possible to split the pack and check each battery whilst under the load. The faulty one's will have a low voltage. Any less than 1v need replacing as they won't ever take or give the full capacity. I have used pins with the meter to check the cells before splitting the pack but it's difficult.
I solder a new battery of the same capacity in place and heatshrink the pack.
I would check the battery with the replacement cell before sealing as the other cells may be damaged.
I am assuming you are using a NiMh charger set to the correct charge current? AA cells are not as resilient as their larger cousins and do not take well to fast charging with loss of capacity being one of the symptoms.
Hope this helps and you can get a replacement or repair your packs.
Dave

I have a related but different question.

I have 2 battery packs each comprising 5 new AA NiMH batteries of 2550mAh capacity of a well-regarded and long-established manufacturer. They are not charging up to their rated capacity. I have put them through 6 or 7 charge/discharge cycles and the maximum charge they will take is 1200mAh. The packs will charge to full voltage, around 6.8V, and I discharge them down to 6V. More than one cell must be defective for it to be only half capacity, and how likely is that.
Are there any other techniques that might restore the capacity, or are they forever stuck at 1200mAh?
Roy

Hi nutrunner and Mark
If this is a NiMh or NiCad battery I suspect if it is completely flat and has been stored for some years in this condition it will not take a charge.
My experience is that even if you can split the battery and resurrect each cell they will never hold any voltage under load.
If you can find the local club you mentioned in another post there will be an expert available who will advise the best way forward.
I suspect your best option is to buy a new battery. Model cars and planes use the same batteries and there will be a local shop.
Good luck
Dave

HI Nutrunner, if the battery is completely flat it might be a right off. The cells could be damaged, however, if you trickle charge at a milliamp rate you may get some charge back, try 1ooma to start with, this would take about 20 hours to get some charge, enough to test if the battery is ok