@evoheyax thanks for the write up … yet again
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Good thread, kudos to @evoheyax and @Ackmaniac too bad stuff like this sometimes get lost 
(found it only recently, by a link in another thread 
)
my etwow scooter manufacturer said the battery is “33v, 6.5 ah, generates up to 500 watts.”
my battery is dead now after 320 miles of riding. instead of paying the asking price of $280 for a replacement, i like to buy my own battery packs from HobbyKing. what do i need? from basic math you showed above, i know to get 33v, i would need 8s. and to get 6.5ah, i would need 3p. but 4.2 x 8 x 3 x 2.5 = 252 W, nowhere near the 500 watt that etwow quoted.
what bat packs and configuration should i use to power up my scooter once more? pls advise. thanks!
Seems like your confusing watts and watt hours.
Watts are a a way of measuring how much power there is going through a given system at a given time. It changes depending on the load, what part of acceleration/deceleration/coasting you are in, and many other factors.
Watt hours tells you more or less about range (instead of power output).
4.2 x 8 = 33.6 max voltage.
Since your doing 3 x 2.5 ah batteries, you have 7.5 ah total. 33.6 v x 7.5 ah (amp hours, not amps) means you have 252 Wh. The original battery had 33.6 v x 6.5 ah = 218.4 Wh.
Now, the C rating tells you about max power output. Not sure what battery your planning on using in specific, but say it’s a 20c battery, then you can do 20 c x 7.5 ah = 150 amps max. Now, you want that number into watts, right? So do 33.6 v x 150 amps = 5040 watts max.
20c is pretty normal, 10c is also pretty common. Even at 5c, you will have 1260 W max.
You really don’t need to worry about it. Aim for a 10c or higher to really make sure (they usually exaggerate their ratings), but your not limited with lipos. 18650’s on the other hand though…
ok. i think i got it now. suppose i get two of this 6s1p 40C battery. https://hobbyking.com/en_us/zippy-flightmax-2650mah-6s1p-40c.html i put them in series, so i will have a 12s1p configuration. capacity will then be 12 x 4.2 x 2.5 = 126 Wh, about 1/2 of the 218.4 Wh of the original battery. therefore my travel range will then be half, correct?
as for watt, i will have 40c x 2.5 ah x 50.4 v = 5040 Watt.
should i worry about frying the ESC because the original battery is 33v, while my 12s1p will have 50.4v? maybe i should use 10s1p battery pack?
thanks a lot, this is very enlightening!
i am using 12s1p lipo battery for 1.5 year at this point and im pretty afraid of them (used lipos - fire, you know…)
and i am planning to change it to li-ion 12s2p, but i didnt knew anything about sag… now its clear for me that 12s2p is not enough since i like to push it hard. what should i do? go with 12s3p? will this be enough? i mean compared to 12s1p lipo.
Hey I have a question. Shouldn’t the way to calculate wh be ah times the average voltage of the pack while discharging? Why do some people use nominal voltage and you use max voltage? .5 volts is a lot of difference. Thanks 
You should use nominal, so 3.6 for Li-ion, and 3.7 for LiPos
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This is correct.
Your nominal voltage is the average voltage at a current rate specified by the manufacturer, usually 0.5C (the current it takes to discharge a battery in 2 hours) sometimes even lower, it’s always specified on the datasheets.
It is ok to use this value when the battery isn’t run way higher than the nominal discharge rate. Once you start discharging at 5A+ per cell continuous, you should really take into account the extra voltage sag at 50% discharge but very few people do.
Dead wrong and usually done by people with little understanding on how batteries work.
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Thanks. The voltage sag explained it to me. But @evoheyax confused me
and
Made me rethink my calculations for a second 
Back to trollin, eh?!?! You know that just because your an unhappy and miserable person, that you don’t have to share that with the world, right?
First off, this is literally how 90% of people in the community calculate it, including board manufacturers.
Everything is really an estimate also. The Wh I have on my ride today will be different then tomorrow, based on my riding style and terrain (and for boards ridden by multiple riders, rider weight). This is because higher amp draw results in lower capacity (just look at an 18650 graph). So none of this is really very scientific.
I have actually found that measuring my amp draw per motor and adding it up that based on a max voltage Wh calculation, I get in the real world within 5% of that… It’s close enough…
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I don’t know how to exactly calculate wh, but by charging my lipos with a power supply, I got a wh count similar that of calculated using 3.9v
Again, this would be wrong. Not having much understanding is the culprit.
Lets take a look at some examples from actual test data.
NCR18650GA datasheet: Rated capacity: 3300 mAh Nominal Voltage: 3.6V
Third party test data (not my own):
Energy based on Nominal Voltage and Rated Capacity: 11.88Wh
Energy based on Maximum Voltage and Rated Capacity: 13.86Wh
Looking at the test data you can see that the measured output energy is almost matched with the energy calculated at nominal voltage between the 0.5A and 1A test condition. This is because Panasonic rates their GA cell at 0.67A.
Datasheet:
I can do this for every single cell you specify. Again, the test data isn’t mine. There is no way for me to have manipulated these numbers. This is simple math. These cells are rated to 11.88Wh as the test data above confirms and not 13.86Wh as @evoheyax claims.
I dont need petty insults to prove you wrong. Just look at the data.
I have to agree with @PXSS for this argument. Maybe you’re right, but from all the sources I’ve read, wh is calculated with nominal voltage 3.6. And a battery barely stays above 4v for 20% of its capacity, how can you use 4.2v to calculate energy?
who’s talking
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Lets look at Samsung 30Q datasheet:
This one is even easier, they give us rated capacity and energy on a single table!
Nominal voltage: 3.6 Rated capacity: 3.0Ah Energy calculated based on nominal voltage and capacity: 10.88Wh Energy calculated based on max voltage: 12.6W Energy calculated based on average voltage and capacity discharged @ 0.2C: 10.97Wh
@evoheyax, I’m happy to teach you how to google spec sheets any day.
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So I’m trolling now? PXSS literally comes into every thread and tries to start fights. He literally does this day in and day out. If a bully throws a punch at me, I’m not going to stand there and smile.
@PXSS Your assuming we pull the same amps at all times while we ride. Both you and I know that’s not correct. Wh ratings are estimates, unless it’s at a particular amp draw. Maybe my method is a bit higher than reality, and your method is likely to give a better numbers. But it still is not scientifically accurate. If you just ride and watch your phone with an app, you’ll see how your amps bounce around. So it’s impossible to give an accurate Wh rating for a particular battery.
I had an app that tracked the amps and amp-hours count. Obviously, the current fluctuates, but the watt-hour has to converge to a value which is the average watt-hour in the pack. If you can agree that my way is better then so what if they aren’t scientifically accurate or what’s so important about the scientific way that people still calculate it even if it’s wrong? Sorry if you feel insulted, I didn’t mean to start a quarrel.
The problem is when someone comes in and says people with no understanding of how batteries work calculate it this way. That is insulting to say the least…
I agree that this is a better way. The reason I put the max voltage was because this was how companies like enertion and boosted and evolve calculate it… I guess they all have little understanding of how batteries work…
No I’m not. The examples above were constant current so that is what I used.
Correct
The rating is a guaranteed amount of energy that you will get at an average current.
It is. Energy is a measurement and can be measured. You can measure voltage while you ride, correct? You can measure current while you ride, correct? So at any point in time you can calculate power, correct? Energy is measured by integrating power over time. If you have instantaneous power being transmitted at 5hz then you can calculate the energy used every 0.2 seconds, integrate that over time and you can calculate the energy used throughout your ride very accurately. This is the method used in industry. It is only dependent on the accuracy of your voltage reading, current reading and the rate at which you sample.
If you average your current draw and voltage through that same ride and multiply them together then you get your average power draw throughout the ride, if you multiply this by the duration of your ride then you will have the exact same energy measured as before. The only extra source of error would be any rounding error you introduce when you average your values.
If you calculate your energy based on the average current and maximum voltage, you will have an answer that’s at best 16% higher than actual, at worse, 30+% higher.
My job requies me to write models that predict how much energy a particular battery uses during different missions for UAVs. Trust me, it is very much possible to give an accurate energy measurement and even predict it before your ride.
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